MODULE 1

DC GENERATOR ldquoEnergy can neither be created nor be destroyed it can only be converted from one form to other rdquo ndashsays the law of conservation of energy Any electric machine does the same A generator converts mechanical energy to electrical energy where as a motor converts electrical energy to mechanical energy These machines generally make use of the principle of electromagnetic induction In case of a dc generator the source of mechanical energy may be a reciprocating or turbine steam engine water falling through a turbine or waterwheel an internal combustion engine a wind turbine a hand crank or any other source of mechanical energy

11 PRINCIPLE

The basic principle behind the working of a generator is the production of dynamically (or motionally) induced emf (Electromotive Force) According to Faradays first Law of Electromagnetic Induction ldquowhenever a conductor cuts magnetic flux dynamically induced emf is produced in itrdquo This emf causes a current to flow if the conductor circuit is closed

Hence the basic essential parts of an electric generator are

A magnetic field and

A conductor or conductors which can so move as to cut the flux

12 AN ELEMENTARY DC GENERATOR

A single-turn rectangular copper coil ABCD as shown in figure 11 moving about its own axis in a magnetic field provided by either permanent magnets or electromagnets The coil is rotated in anticlockwise direction by some source of mechanical energy The instantaneous value of the emf induced according to Faradays second law of electromagnetic induction can be written as e = Em sinɵ where ɵ is the angle that the plane of the coil makes with the magnetic field Thus the induced emf is alternating in natureTo make it unidirectional the two ends of the coil are joined to two split-rings which are insulated from each other and from the central shaft Two collecting brushes (of carbon or copper)

press against the slip rings This split ring - brush arrangement make the emf unidirectional at the load terminals

Fig 11 An elementary dc generator

13 CONSTRUCTION

We have seen the working principle of a dc generator by considering a single loop generator Now coming to the practical construction of DC Generator refer to the cross sectional view of a dc generator as shown in figure 12

A practical DC Generator has the following parts-

1) Yoke2) Pole 3) Field winding4) Armature of dc generator5) Brushes of generator6) Bearing

The functions of the various parts are explained below

Field

Fig 12 Main parts of a dc generator

131 Yoke (Frame)

Yoke of a dc generator serves two purposes

(i) It holds the magnetic pole cores of the generator and acts as cover of the generator(ii) It carries the magnetic field flux

In small generator yoke is made of cast iron Cast iron is cheaper in cost but heavier than steel But for construction of large DC generator where weight of the machine is concerned lighter cast steel or rolled steel is preferable for constructing yoke of dc generator

132 Pole cores and pole shoes

There are mainly two types of construction available-(1) Solid pole core where it made of a solid single piece of cast iron or cast steel (2) Laminated pole core where it made of numbers of thin limitations of annealed steel which are riveted together The pole core is fixed to the inner periphery of the yoke by means of bolts through the yoke and into the pole body The pole shoes are so typically shaped that they spread out the magnetic flux in the air gap and reduce the reluctance of the magnetic path Due to their larger cross ndash section they hold the pole coil at its position

133 Field winding The field coils or pole coils are wound around the pole core These are a simple coil of insulated copper wire or strip which placed on the pole which placed between yoke and pole shoes134 Armature Core The armature core serves the purpose of holding the armature winding and providing a low reluctance path for the flux through the armature from N pole to S pole A cylindrical or drum shaped armature core is build up of circular laminated sheets In every circular lamination slots are either die ndash cut or punched on the outer periphery and the key way is located on the inner periphery Air ducts are also punched of cut on each lamination for circulation of air through the core for providing better cooling Fig 14 shows the armature and field windings

Fig 14 Armature and field windings

135 Armature Winding Armature winding are generally formed wound These are first wound in the form of flat rectangular coils and are then pulled into their proper shape in a coil puller Various conductors of the coils are insulated from each other The conductors are placed in the armature slots which are lined with tough insulating material This slot insulation is folded over above the armature conductors placed in it and secured in place by special hard wooden or fiber wedges There are mainly two types of armature windings namely the lap winding and the wave winding which is shown in figure 15 In lap winding the armature conductors are arranged in P number of parallel paths where P is the number of poles But a wave winding offers only two parallel paths for the flow of armature current Hence a lap winding is used for high current low voltage generator where as a wave wound armature is preferred for a low current high voltage generator

Fig 15 Lap and wave winding

136 Commutator The commutator plays a vital role in dc generator It collects current from armature and sends it to the load as direct current It actually takes alternating current from armature and converts it to direct current and then send it to external load It is cylindrical structured and is build up of wedge ndash shaped segments of high conductivity hard drawn or drop forged copper Each segment is insulated from the shaft by means of insulated commutator segment shown below Each commutator segment is connected with corresponding armature conductor through segment riser or lug Fig 16 shows the arrangement of commutator segments in a dc machine along with its cross sectional view

Fig16 Commutator segments

137 Brushes The brushes are made of carbon These are rectangular block shaped The only function of these carbon brushes of dc generator is to collect current from commutator segments The brushes are housed in the rectangular box shaped brush holder As shown in figure17 the brush face is placed on the commutator segment with attached to the brush holder

Fig 17 Brush in brush holder

138 Bearing For small machine ball bearing is used and for heavy duty dc generator roller bearing is used The bearing must always be lubricated properly for smooth operation and long life of generator

14 EMF EQUATION OF A DC GENERATOR

Let Φ be the flux per pole in WeberZ the total number of armature conductors ndash No of slots x no of armature ConductorsslotP the number of polesA the number of parallel paths in the armature

N the speed of rotation of armature in revolutions per minute (rpm)E the emf induced in any parallel path in armatureEg the emf generated in any parallel path in the armatureThe emf induced per conductor = dΦdt volt (n=1)The flux cut by one conductor in one revolution dΦ = ΦP WbNo of revolutions per second = N60Time for one revolution dt = 60N secondHence EMF generatedconductor is

dφdt = φPN

60

No of conductors in a parallel path = Z A

Hence emf generated per parallal path

Eg = φPN60 Z

A

For a lap wound generator A = P and

For wave wound generator A = 2

15 TYPES OF DC GENERATORS

Fig18 Classification of dc generators

151 Separately Excited DC Generator

This dc generator has a field magnet winding which is excited using a separate voltage source (like battery)The representation in the below figure 19 The output voltage depends on the speed of rotation of armature and field current The higher the speed of rotation and current ndash the higher the output emf

Fig 19 Separately Excited DC Generators

Armature Current Ia = IL

Terminal voltage V = Eg-Ia Ra

Electric Power developed = Eg Ia

Power delivered to load = V IL = (Eg-Ia Ra) Ia = Eg Ia - Ia2

Ra

152 Self Excited DC Generator

These are generators in which the field winding is excited by the output of the generator itself As described before ndash there are three types of self excited dc generators ndash they are 1) Series 2) Shunt and 3) Compound

A series DC generator is shown below in fig110 (a) ndash in which the armature winding is connected in series with the field winding so that the field current flows through the load as well as the field winding Field winding is a low resistance thick wire of few turns Armature current Ia =Ise = IL=I (say)

Terminal voltage V = Eg- I( Ra+ Rse)Power Developed in armature = Eg Ia

Power delivered to load = V IL = Eg Ia- Ia2 (Ra+Rse)

Fig110 (a) Series generator and (b) shunt generator

A shunt DC generator is shown in figure 110 (b) in which the field winding is wired parallel to armature winding so that the voltage across both are same The field winding has high resistance and more number of turns so that only a part of armature current passes through field winding and the rest passes through load Shunt field current Ish = V Rsh

Armature current Ia = IL+ Ish

Terminal voltage V= Eg-Ia Ra

Power developed in armature = EgIa

Power delivered to load = VIL A compound generator is shown in figure 111 below It has two field findings namely Rsh and Rse They are basically shunt winding (Rsh) and series winding (Rse) Compound generator is of two types ndash 1) Short shunt and 2) Long shunt

Short shunt- Here the shunt field winding is wired parallel to armature and series field winding is connected in series to the load It is shown in fig111(a)

In short shunt machines Series field current Ise = IL

Shunt field current Ish iquest V+ I se RseRsh

Terminal voltage V= Eg-Ia Ra-Ise Rse

Power developedin armature = Eg Ia

Power delivered to load = V IL

Fig 111 (a) short shunt generator and (b) Long shunt generator

Long shunt- Here the shunt field winding is parallel to both armature and series field winding (Rse is wired in series to the armature) It is shown in figure 111 (b)

In Long shunt machines Series field current Ise = Ia = IL+ Ish

Shunt field current Ish = V R sh

Terminal voltage V = Eg- Ia (Ra+ Rse) Power developed in armature = Eg Ia

Power delivered to load = VIL

16 GENERATOR CHARACTERISTICS

The speed of a dc machine operated as a generator is fixed by the prime mover For general-purpose operation the prime mover is equipped with a speed governor so that the speed of the generator is practically constant Under such condition the generator performance deals primarily with the relation between excitation terminal voltage and load These relations can be best exhibited graphically by means of curves known as generator characteristics Thesecharacteristics show at a glance the behavior of the generator under differentload conditions

The following are the three most important characteristics of a dc generator161 Open Circuit Characteristic (OCC)

This curve shows the relation between the generated emf at no-load (E0) and the field current (If) at constant speed It is also known as magnetic characteristic or no-load saturation curve Its shape is practically the same for all generators whether separately or self-excited The data for OCC curve are obtained experimentally by operating the generator at no load and constant speed and recording the change in terminal voltage as the field current is varied

162 Internal or Total characteristic (EIa)

This curve shows the relation between the generated emf on load (E) and the armature current (Ia) The emf E is less than E0 due to the demagnetizing effect of armature reaction Therefore this curve will lie below the open circuitcharacteristic (OCC) The internal characteristic is of interest chiefly to thedesigner It cannot be obtained directly by experiment It is because a voltmetercannot read the emf generated on load due to the voltage drop in armatureresistance The internal characteristic can be obtained from externalcharacteristics if winding resistances are known because armature reaction effectis included in both characteristics163 External characteristic (VIL)This curve shows the relation between the terminal voltage (V) and load current(IL) The terminal voltage V will be less than E due to voltage drop in thearmature circuit Therefore this curve will lie below the internal characteristicThis characteristic is very important in determining the suitability of a generatorfor a given purpose It can be obtained by making simultaneous measurements

of terminal voltage and load current (with voltmeter and ammeter) of a loadedgenerator164 Open Circuit Characteristic of a DC Generator

The OCC for a dc generator is determined as follows The field winding of the dc generator (series or shunt) is disconnected from the machine and isseparately excited from an external dc source as shown in Fig 112 Thegenerator is run at fixed speed (ie normal speed) The field current (If) isincreased from zero in steps and the corresponding values of generated emf(E0) read off on a voltmeter connected across the armature terminals On plottingthe relation between E0 and If we get the open circuit characteristic as shown in

Fig113

Fig 112 Experimental setup for plotting OCC

Fig 113 OCC of dc generator

The following points may be noted from OCC When the field current is zero there is some generated emf OA

This is due to the residual magnetism in the field poles Over a fairly wide range of field current (upto point B in the curve)

the curve is linear It is because in this range reluctance of iron is negligible as compared with that of air gap The air gap reluctance is constant and hence linear relationship After point B on the curve the reluctance of iron also comes into picture and reluctance of iron is no longer negligible Consequently the curve deviates from linear relationship

After point C on the curve the magnetic saturation of poles begins and E0 tends to level off

The OCC of even self-excited generator is obtained by running it as a separately excited generator

165 Characteristics of a Separately Excited DCGenerator

The obvious disadvantage of a separately excited dc generator is that we require an external dc source for excitation But since the output voltage may be

controlled more easily and over a wide range (from zero to a maximum) this type of excitation finds many applications

(i) Open circuit characteristic

The OCC of a separately excited generator is determined in a manner described in Sec 164 Fig114 shows the variation of generated em f on noload with field current for various fixed speeds Note that if the value of constant speed is increased the steepness of the curve also increases When the field current is zero the residual magnetism in the poles will give rise to the small initial emf as shown

Fig114 Variation of OCC with speed

Internal and External Characteristics (Load Characteristics)

The external characteristic of a separately excited generator is the curve between the terminal voltage (V) and the load current IL (which is the same as armature current in this case) In order to determine the external characteristic the circuit set up is as shown in Fig 115

Fig 115 Experimental setup to plot load characteristics

As the load current increases the terminal voltage falls due to two reasons(a) The armature reaction weakens the main flux so that actual emf generated E on load is less than that generated (E0) on no load

(b) There is voltage drop across armature resistance (= ILRa = IaRa)Due to these reasons the external characteristic is a drooping curve [curve 3 inFig 116 ] Note that in the absence of armature reaction and armature dropthe generated emf would have been E0 (curve 1)The internal characteristic can be determined from external characteristic by adding IL Ra drop to the external characteristic It is because armature reaction drop is included in the external characteristic Curve 2 is the internal characteristic of the generator and should obviously lie above the external characteristic

Fig 116 Load characteristics of a separately excited dcgenerator

117 Voltage Build-Up in a Self-Excited GeneratorLet us see how voltage builds up in a self-excited generator(i) Shunt generator Consider a shunt generator If the generator is run at a constant speed some emf will be generated due to residual magnetism in the main poles This small emf circulates a field current which in turn produces additional flux toreinforce the original residual flux (provided field winding connections arecorrect) This process continues and the generator builds up the normalgenerated voltage following the OCC shown in Fig 117 (i) The field resistance Rf can be represented by a straight line passing through the origin as shown in Fig 117 (ii) The two curves can be shown on the same diagram as they have the same ordinate as in Fig 117 (iii)

Fig 117

Since the field circuit is inductive there is a delay in the increase in current uponclosing the field circuit switch The rate at which the current increases dependsupon the voltage available for increasing it Suppose at any instant the fieldcurrent is i (= OA) and is increasing at the rate didt Then

E0=i R f +iquest L didt

Where Rf = total field circuit resistance L = inductance of field circuit

At the considered instant the total emf available is AC [See Fig 117 (iii)] Anamount AB of the cmf AC is absorbed by the voltage drop iRf and theremainder part BC is available to overcome L didt Since this surplus voltage isavailable it is possible for the field current to increase above the value OAHowever at point D the available voltage is OM and is all absorbed by i RfD

rop Consequently the field current cannot increase further and the generatorbuild up stops Hence we arrive at a very important conclusion that the voltage build up of the generator is given by the point of intersection of OCC and field resistance line Thus in Fig 117 (iii) D is point of intersection of the two curves Hence the generator will build up a voltage OM(ii) Series generator During initial operation with no current yet flowing a residual voltage will be generated exactly as in the case of a shunt generator The residual voltage will cause a current to flow through the whole series circuit when the circuit isclosed There will then be voltage build up to an equilibrium point exactlyanalogous to the buildup of a shunt generator The voltage build up graph willbe similar to that of shunt generator except that now load current (instead offield current for shunt generator) will be taken along x-axis10487871048787(iii) Compound generator When a compound generator has its series field flux aiding its shunt field flux the machine is said to be cumulative compound When the series field isconnected in reverse so that its field flux opposes the shunt field flux thegenerator is then differential compound The easiest way to build up voltage in a compound generator is to start under no load conditions At no load only the shunt field is effective When no-load voltage build up is achieved the generator is loaded If under load the voltagerises the series field connection is cumulative If the voltage drops significantlythe connection is differential compound118 Critical Field Resistance for a Shunt Generator The voltage build up in a shunt generator depends upon field circuit resistance If the field circuit resistance is R1 (line OA) then generator will build up a voltage OM as shown in Fig 118 If the field circuit resistance is increasedto R2 (tine OB) the generator will build up a voltage OL slightly less than OM As the field circuit resistance is increased the slope of resistance line also increases When the field resistance line becomes tangent (line OC) to OCC the generator would just excite If the field circuit resistance is increased beyond this point (say line OD) the generator will fail to excite The field circuit resistance represented by line OC (tangent to OCC) is called critical field resistance RC for the shunt generator It may be defined as under The maximum field circuit resistance (for a given speed) with which the shunt generator would just excite is known as its critical field resistance It should be noted that shunt generator will build up voltage only if field circuit resistance is less than critical field resistance

Fig 118 Critical field resistance of shunt generator

119 Critical Resistance for a Series Generator

Fig 119 shows the voltage build up in a series generator Here R1 R2 etc represent the total circuit resistance (load resistance and field winding resistance) If the total circuit resistance is R1 then series generator will build up a voltage OL The line OC is tangent to OCC and represents the critical resistance RC for a series generator If the total resistance of the circuit is more than RC (say line OD) the generator will fail to build up voltage Note that Fig 119 is similar to Fig 118 with the following differences

(i) In Fig 118 R1 R2 etc represent the total field circuit resistanceHowever R1 R2 etc in Fig 119 represent the total circuit resistance(load resistance and series field winding resistance etc)

(ii) In Fig 118 field current alone is represented along X-axis However inFig 119 load current IL is represented along Y-axis Note that in aseries generator field current = load current IL

Fig 119 Critical resistance of a series generator

120 Characteristics of Series Generator Fig 120 (i) shows the connections of a series wound generator Since there is only one current (that which flows through the whole machine) the load current is the same as the exciting current(i) OCCCurve 1 shows the open circuit characteristic (OCC) of a series generator Itcan be obtained experimentally by disconnecting the field winding from themachine and exciting it from a separate dc source(ii) Internal characteristic Curve 2 shows the total or internal characteristic of a series generator It gives the relation between the generated emf E on load and armature current Due to armature reaction the flux in the machine will be less than the flux at no load Hence emf E generated under load conditions will be less than the emf E0 generated under no load conditions Consequently internal characteristic curvelies below the OCC curve the difference between them representing the effectof armature reaction [See Fig 120 (ii)]

Fig 120 Characteristics of series generator

(iii) External characteristic

Curve 3 shows the external characteristic of a series generator It gives therelation between terminal voltage and load current ILV = E g- Ia (Ra +Rse )Therefore external characteristic curve will lie below internal characteristiccurve by an amount equal to ohmic drop [ie Ia(Ra + Rse)] in the machine asshown in Fig 120 (ii) The internal and external characteristics ofa dc series generator can be plotted from one another as shown in Fig 121 Supposewe are given the internal characteristic of the generator Let the line OC represent the resistance of the whole machine ie Ra +Rse If the load current is OB drop in the machine is AB ieAB = Ohmic drop in the machine = OB(Ra + Rse)

Fig 121 Load characteristics of series generator

Now raise a perpendicular from point B and mark a point b on this line such that ab = AB Then point b will lie on the external characteristic of the generator

Following similar procedure other points of external characteristic can be located It is easy to see that we can also plot internal characteristic from the external characteristic

121 Characteristics of a Shunt Generator Fig 122 (i) show the connections of a shunt wound generator The armature current Ia splits up into two parts a small fraction Ish flowing through shunt field winding while the major part IL goes to the external load(i) OCC The OCC of a shunt generator is similar in shape to that of a series generator as shown in Fig 122 (ii) The line OA represents the shunt field circuitresistance When the generator is run at normal speed it will build up a voltageOM At no-load the terminal voltage of the generator will be constant (= OM)represented by the horizontal dotted line MC

Fig 122 Characteristics of a Shunt Generator

(ii) Internal characteristic

When the generator is loaded flux per pole is reduced due to armature reaction Therefore emf E generated on load is less than the emf generated at no load As a result the internal characteristic (EIa) drops down slightly as shown in Fig122 (ii)

(iii) External characteristic Curve 2 shows the external characteristic of a shunt generator It gives therelation between terminal voltage V and load current IL

Therefore external characteristic curve will lie below the internal characteristic curve by an amount equal to drop in the armature circuit[ie (IL + Ish)Ra] as shown in Fig 122 (ii)

Note It may be seen from the external characteristic that change in terminalvoltage from no-load to full load is small The terminal voltage can always bemaintained constant by adjusting the field rheostat R automatically122 Critical External Resistance for Shunt Generator

If the load resistance across the terminals of a shunt generator is decreased then load current increase However there is a limit to the increase in

load current with the decrease of load resistance Any decrease of load resistance beyond this point instead of increasing the current ultimately results in reduced current Consequently the external characteristic turns back (dotted curve) as shown in Fig 123 The tangent OA to the curve represents the minimum external resistance required to excite the shunt generator on load and is called critical external resistance If the resistance of the external circuit is less than the critical external resistance (represented by tangent OA in Fig 310) the machine will refuse to excite or will de-excite if already running This means that external resistance is so low as virtually to short circuit the machine and so doing away with its excitation

Note There are two critical resistances for a shunt generator viz (i) criticalfield resistance (ii) critical external resistance For the shunt generator to buildup voltage the former should not be exceeded and the latter must not be gonebelow

123 How to Draw OCC at Different Speeds

If we are given OCC of a generator at a constant speed N1 then we can easily draw the OCC at any other constant speed N2 Fig 123 illustrates theprocedure Here we are given OCC at a constant speed N1 It is desired to findthe OCC at constant speed N2 (it is assumed that n1 lt N2) For constantexcitation E propN

E2

E1 =

N 2

N1 so

E2 iquest E1

N2

N1

As shown in Fig (311) for If = OH E1 = HC Therefore the new value of emf (E2) for the same If but at N2

E2= HCN 2

N 1 = HD

Fig 123 Plotting OCC at another speed

This locates the point D on the new OCC at N2 Similarly other points can belocated taking different values of If The locus of these points will be the OCCat N2

124 Critical Speed (NC)

The critical speed of a shunt generator is the minimum speed below which it fails to excite Clearly it is the speed for which the given shunt field resistance represents the critical resistance In Fig 124 curve 2 corresponds to critical

speed because the shunt field resistance (Rsh) line is tangential to it If the generator runs at full speed N the new OCC moves upward and the Rsh line represents critical resistance for this speed

In order to find critical speed take any convenient point C on excitation axis and erect a perpendicular so as to cut Rsh and Rsh lines at points B and A respectively Then

BCAC

=NcN

Nc=N BCAC

Fig 124 Finding critical speed

125 Conditions for Voltage Build-Up of a Shunt Generator

The necessary conditions for voltage build-up in a shunt generator are

(i) There must be some residual magnetism in generator poles(ii) The connections of the field winding should be such that the field current strengthens the residual magnetism(iii) The resistance of the field circuit should be less than the critical resistance In other words the speed of the generator should be higher

than the critical speed

126 Compound Generator Characteristics

In a compound generator both series and shunt excitation are combined as shown in Fig 123 The shunt winding can be connected either across the armature only (short-shunt connection S) or across armature plus series field (long-shunt connection G) The compound generator can be cumulatively compounded or

differentially compounded generator The latter is rarely used in practice Therefore we shall discuss the characteristics of cumulatively compounded

generator It may be noted that external characteristics of long and short shunt compound generators are almost identical

External characteristic

Fig 124 shows the external characteristics of a cumulatively compoundedgenerator The series excitation aids the shunt excitation The degree of compounding depends upon the increase in series excitation with the increase inload current

(i) If series winding turns are so adjusted that with the increase in load current the terminal voltage increases it is called over-compounded generator In such a case as the load current increases the series field mmf increases and tends to increase the flux and hence the generated voltage The increase in generated voltage is greater than the IaRa drop so that instead of decreasing the terminal voltage increases as shown by curve A in Fig 124

(ii) If series winding turns are so adjusted that with the increase in load current the terminal voltage substantially remains constant it is called Flat-compounded generator The series winding of such a machine has lesser number of turns than the one in over-compounded machine and therefore does not increase the flux as much for a given load current Consequently the full-load voltage is nearly equal to the no-load voltage as indicated by curve B in Fig 124 (iii) If series field winding has lesser number of turns than for a flat compounded machine the terminal voltage falls with increase in load current as indicated by curve C m Fig 124 Such a machine is called under-compounded generator

Fig 123 compound generator

Fig 124 Load characteristics of compound generators

127 LOSSES

The losses in a dc machine (generator or motor) may be divided into three classes viz (i) copper losses or electrical losses (ii) iron or core losses and (iii) mechanical losses All these losses appear as heat and thus raise the temperature of the machine They also lower the efficiency of the machine The loss structure maybe demonstrated as

1Copperlosses

These losses occur due to currents in the various windings of the machine When an electric current of I amps flows in a resistance of R ohms the heat energy is lost at the rate of I2R joules sec and the power loss is I2R watts Generators and motors have one or more field circuits and an armature circuit in which such losses occur All resistance losses of this kind are called copper losses The different copper losses occurring in a dc machine are

(i) Armature copper loss = Ia2Ra

(ii) Shunt field copper loss = Ish2 Rsh

(iii) Series field copper loss = Ise2Rse

Armature copper loss will be around 30 to 40 of full load losses where as total field copper losses will constitute around 30 of full load losses

Note There is also brush contact loss due to brush contact resistance (ie resistance between the surface of brush and surface of commutator) This loss is generally included in armature copper loss The brush contact drop is almost independent of Ia For carbon brushes this drop is around 1 V per brush The power loss due to brush contact resistance is 2ebIa where eb is the voltage drop at one brush

2 IRONORCORELOSSES These losses occur in the armature of a dc machine and are due to the rotation of armature in the magnetic field of the poles They are of two types viz (i) hysteresis loss (ii) eddy current loss

(i) Hysteresis loss

Hysteresis loss occurs in the armature of the dc machine since any given part of the armature is subjected to magnetic field reversals as it passes under successive poles Fig 124 shows an armature rotating in two-pole machine Consider a small piece ab of the armature When the piece ab is under N-pole the magnetic lines pass from a to b Half a revolution later the same piece of iron is under S-pole and magnetic lines pass from b to a so that magnetism in the iron is reversed In order to reverse continuously the molecular magnets in the armature core some amount of power has to be spent which is called hysteresis loss It is given by Steinmetz formula This formula is

Hysteresis Loss Wh = η Bmax 16f V Watts Bmax - Maximum flux density in the core f - Frequency of magnetic reversals V - Volume of armature in m3

η - Steinmetz hysteresis coefficient

In order to reduce this loss in a dc machine armature core is made of such materials which have a low value of Steinmetz hysteresis co-efficient eg silicon steel

Fig 124 Occurrence of hysteresis loss

ii) EDDYCURRENTLOSS

In addition to the voltages induced in the armature conductors there are also voltages induced in the armature core These voltages produce circulating currents in the armature core as shown in Fig 125 These are called eddy currents and power loss due to their flow is called eddy current loss The eddy current loss appears as heat which raises the temperature of the machine and lowers its efficiency Core resistance can be greatly increased by constructing the core of thin round If a continuous solid iron core is used the resistance to eddy current path will be small due to large cross-sectional area of the core Consequently the magnitude of eddy current and hence eddy current loss will be large The magnitude of eddy current can be reduced by making core resistance as high as practical The iron sheets called laminations [See Fig 126] The laminations are insulated from each other with a coating of varnish The insulating coating has a high resistance so very little current flows from one lamination to the other Alsobecause each lamination is very thin the resistance to current flowing through the width of a lamination is also quite large Thus laminating a core increases the core resistance which decreases the eddy current and hence the eddy current loss

Fig 125 Eddy current loss Fig 126 Core laminations

Eddy current loss is biven by the formula Pe= Ke Bmax

2 f2t2 V Watts

Where Ke = Constant depending upon electrical resistace of core Bmax = Maximum flux density in wbm2

f = frequency f magnetic reversals in Hz t = thickness of laminations in m and V = volume of core in m3

It may be noted that eddy current loss depends upon the square of lamination thickness For this reason lamination thickness should be kept as small as possible

3MECHANICALLOSSES

These losses are due to friction and windage(i) Friction losses eg bearing friction brush friction etc(ii) Windage loss ie air friction of rotating armature

These losses depend upon the speed of the machine But for a given speed they are practically constant

Note Iron losses and mechanical losses together are called stray losses

CONSTANT AND VARIABLE LOSSES

The losses in a dc generator (or dc motor) may be sub-divided into

(i) constant losses (ii) variable losses

(i) Constant losses

Those losses in a dc generator which remain constant at all loads are known as constant losses The constant losses in a dc generator are(a) iron losses(b) mechanical losses(c) shunt field losses(ii) Variable lossesThose losses in a dc generator which vary with load are called variable lossesThe variable losses in a dc generator are (a) Copper loss in armature winding (Ia

2 Ra) (b) Copper loss in series field winding (Ise

2Rse)

Total losses = Constant losses + Variable lossesNote Field Cu loss is constant for shunt and compound generators

128 POWER FLOW DIAGRAM

The power flow in a dc generator may be represented as

The different power stages are shown as A B and CA ndash The mechanical power input which is the power output of the prime mover is given as Pin = 2πNT

60 watts

B - The power developed in armature is Pa = Eg Ia watts C- The electrical power output is Pout = VIL watts

129 EFFICIENCY Efficiency of a machine is the ratio of its output power to the input power

Efficiency η = Output powerInput power

= Input powerminusLossesInput power = Output power

Output power+Losses

In case of a dc generator depending upon the power stages three different efficiencies can be defined as-

1 Mechanical Efficiency(ηm)

BA = Electrical power developedisinarmature

Mechanical power input2 Electrical Efficiency(ηe)

CB = Electrical power output

Electrical power developedisinthe armature 3 Overall or commercial efficiency (η)

CA = Electrical power output

Mechanical power input = ηmtimes ηe

130 CONDITION FOR MAXIMUM EFFICIENCY The efficiency of a dc generator is not constant but varies with load Fig 127shows the variation of efficiency with load current Consider a dc shunt generator with terminal voltage V delivering a load current of IL amp Generator output = V IL

Generator input = Output + Losses = V IL + constant losses + Variable losses = V IL + Wc+ Ia

2Ra

= V IL + Wc + (IL+ Ish)2 Rsh

The shunt field current Ish is small compared with IL so may be neglected

Generator input = VIL+ IL2Ra + Wc

Now efficiency η iquest outputinput = VI L

VIL+ Ia 2Ra+Wc

Efficiency is maximum when the denominator of the above equation is minimumie

ddIL ( VI L

VIL+ Ia 2Ra+Wc ) = 0

Which gives IL2 Ra = WC

ie Variable loss = constant lossThe load current for maximum efficiency is given by

IL iquest radicWcRa

Fig 127 Variation of efficiency with load

Hence the efficiency is maximum for that load current at which the variable loss is equal to the constant loss

131 APPLICATIONS

Separately excited generators

These types of generators have special applications such as electroplating electro refining of materials because separate supply is required for field excitation

Shunt generators

These types of generators are used in battery charging and ordinary lighting purposes

Series generators

This type of generators is used commonly as boosters on DC feeders as current generators for welding and arc lamp

Cumulatively compounded generators

This type of generators is used for domestic lighting purposes and to transmit energy over long distances

Differential compound generators

This type of generators is rarely used They are used for special applications such as electric arc welding

132 NUMERICAL EXCERCISES

1 The emf induced in the armature of a 450 kW 250 volt shunt generator is

2588volt when the field current is 200 amp and the generator is supplying power to a load at rated terminal voltage The armature circuit resistance is 0005 ohm Determine (i) load current (ii) power generated (iii) power output and (iv) electrical efficiency Neglect brush contact drop

Solution

Let the load current be I amp Given Ish = 20 A

Armature current = Load current + field current or Ia = I + 20 Induced emf = Terminal voltage + Ia Ra drop

2588 = 250 + (I + 20) times 0005

So (i) Load current I = 1740 amp

(ii) Power generated E Ia = 2588 (1740 + 20) watts = 455488 kW (iii) Power output V I = 250 times 1740 watts

= 435 kW

(iv) Electrical efficiency = 435455488 times 100 = 955

2 A 4 pole wave wound dc shunt generator delivers a load of 45 kW at a terminal voltage of 220 volts Its armature has 150 single turn coils and has a resistance of 001 ohm The air gap flux per pole is 002 weber Shunt field resistance is 50 ohm Calculate the speed at which it is being driven Neglect brush contact drop

Solution Load current I = W

V = 45000220 = 20455 amp

Field Current Ish = V

R s h = 220

50 = 44 amp

Armature current Ia = I +I sh = 20455 +44 = 20895 amp

Armature drop I aRa = 20895 X 001 = 20895 volts Induced emf Eg = V + Ia Ra = 220 + 20895 = 2220895 volts

But induced emf

So N = Eg60 AΦ ZP = 2220895times 60 times2

002 times150 times2 times 4 = 111045 rpm

3 The following data refer to the OCC of a DC separately excited generator at 1000 rpm

Field current in A 00 02 03 04 05 06 07 08 09 10 11 12Armature voltage in V 5 40 75 100 124 145 162 178 188 195 200 205

The machine is now connected as a shunt generator with a total field resistance of 200 Ω and driven at 1000 rpm Estimate graphically (i) the voltage to which the generator will build up at no load (ii) the armature field and load currents when the terminal voltage is found to be 150 V Neglect the effect of armature reaction and brush drop and assume armature resistance Ra to be 08 Ω (iii) Finally estimate the steady state armature current when the machine terminals are shorted

Solution

First of all the OCC of the generator is plotted in a graph paper as shown in figure

(i) The total field circuit resistance is given to be Rf = 200 Ω The Rf line is now drawn on the same graph paper passing through the origin The point of intersection of the Rf line and the OCC decides the final no load voltage and can be read from the graph as 192 V and the corresponding field current is 096 A (ii) Since the terminal voltage V is 150 V (= BM) field current If

is OM = 077 A Generated voltage E is given by AM But we know E = IaRa + V Hence IaRa= E ndash V = AM ndash BM = AB Now from the graph AB = 25 V there4Ia = 2508 = 3125 A So load current IL= Ia ndash If= 3125 ndash 077 = 3048 A

ADDITIONAL PROBLEMS

1 A 4 pole lap wound 115 kW 230 volts dc shunt generator has armatureresistance of 02 ohm and field resistance of 100 ohms Calculate (i) emf

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES

Field

Fig 12 Main parts of a dc generator

131 Yoke (Frame)

Yoke of a dc generator serves two purposes

(i) It holds the magnetic pole cores of the generator and acts as cover of the generator(ii) It carries the magnetic field flux

In small generator yoke is made of cast iron Cast iron is cheaper in cost but heavier than steel But for construction of large DC generator where weight of the machine is concerned lighter cast steel or rolled steel is preferable for constructing yoke of dc generator

132 Pole cores and pole shoes

There are mainly two types of construction available-(1) Solid pole core where it made of a solid single piece of cast iron or cast steel (2) Laminated pole core where it made of numbers of thin limitations of annealed steel which are riveted together The pole core is fixed to the inner periphery of the yoke by means of bolts through the yoke and into the pole body The pole shoes are so typically shaped that they spread out the magnetic flux in the air gap and reduce the reluctance of the magnetic path Due to their larger cross ndash section they hold the pole coil at its position

133 Field winding The field coils or pole coils are wound around the pole core These are a simple coil of insulated copper wire or strip which placed on the pole which placed between yoke and pole shoes134 Armature Core The armature core serves the purpose of holding the armature winding and providing a low reluctance path for the flux through the armature from N pole to S pole A cylindrical or drum shaped armature core is build up of circular laminated sheets In every circular lamination slots are either die ndash cut or punched on the outer periphery and the key way is located on the inner periphery Air ducts are also punched of cut on each lamination for circulation of air through the core for providing better cooling Fig 14 shows the armature and field windings

Fig 14 Armature and field windings

135 Armature Winding Armature winding are generally formed wound These are first wound in the form of flat rectangular coils and are then pulled into their proper shape in a coil puller Various conductors of the coils are insulated from each other The conductors are placed in the armature slots which are lined with tough insulating material This slot insulation is folded over above the armature conductors placed in it and secured in place by special hard wooden or fiber wedges There are mainly two types of armature windings namely the lap winding and the wave winding which is shown in figure 15 In lap winding the armature conductors are arranged in P number of parallel paths where P is the number of poles But a wave winding offers only two parallel paths for the flow of armature current Hence a lap winding is used for high current low voltage generator where as a wave wound armature is preferred for a low current high voltage generator

Fig 15 Lap and wave winding

136 Commutator The commutator plays a vital role in dc generator It collects current from armature and sends it to the load as direct current It actually takes alternating current from armature and converts it to direct current and then send it to external load It is cylindrical structured and is build up of wedge ndash shaped segments of high conductivity hard drawn or drop forged copper Each segment is insulated from the shaft by means of insulated commutator segment shown below Each commutator segment is connected with corresponding armature conductor through segment riser or lug Fig 16 shows the arrangement of commutator segments in a dc machine along with its cross sectional view

Fig16 Commutator segments

137 Brushes The brushes are made of carbon These are rectangular block shaped The only function of these carbon brushes of dc generator is to collect current from commutator segments The brushes are housed in the rectangular box shaped brush holder As shown in figure17 the brush face is placed on the commutator segment with attached to the brush holder

Fig 17 Brush in brush holder

138 Bearing For small machine ball bearing is used and for heavy duty dc generator roller bearing is used The bearing must always be lubricated properly for smooth operation and long life of generator

14 EMF EQUATION OF A DC GENERATOR

Let Φ be the flux per pole in WeberZ the total number of armature conductors ndash No of slots x no of armature ConductorsslotP the number of polesA the number of parallel paths in the armature

N the speed of rotation of armature in revolutions per minute (rpm)E the emf induced in any parallel path in armatureEg the emf generated in any parallel path in the armatureThe emf induced per conductor = dΦdt volt (n=1)The flux cut by one conductor in one revolution dΦ = ΦP WbNo of revolutions per second = N60Time for one revolution dt = 60N secondHence EMF generatedconductor is

dφdt = φPN

60

No of conductors in a parallel path = Z A

Hence emf generated per parallal path

Eg = φPN60 Z

A

For a lap wound generator A = P and

For wave wound generator A = 2

15 TYPES OF DC GENERATORS

Fig18 Classification of dc generators

151 Separately Excited DC Generator

This dc generator has a field magnet winding which is excited using a separate voltage source (like battery)The representation in the below figure 19 The output voltage depends on the speed of rotation of armature and field current The higher the speed of rotation and current ndash the higher the output emf

Fig 19 Separately Excited DC Generators

Armature Current Ia = IL

Terminal voltage V = Eg-Ia Ra

Electric Power developed = Eg Ia

Power delivered to load = V IL = (Eg-Ia Ra) Ia = Eg Ia - Ia2

Ra

152 Self Excited DC Generator

These are generators in which the field winding is excited by the output of the generator itself As described before ndash there are three types of self excited dc generators ndash they are 1) Series 2) Shunt and 3) Compound

A series DC generator is shown below in fig110 (a) ndash in which the armature winding is connected in series with the field winding so that the field current flows through the load as well as the field winding Field winding is a low resistance thick wire of few turns Armature current Ia =Ise = IL=I (say)

Terminal voltage V = Eg- I( Ra+ Rse)Power Developed in armature = Eg Ia

Power delivered to load = V IL = Eg Ia- Ia2 (Ra+Rse)

Fig110 (a) Series generator and (b) shunt generator

A shunt DC generator is shown in figure 110 (b) in which the field winding is wired parallel to armature winding so that the voltage across both are same The field winding has high resistance and more number of turns so that only a part of armature current passes through field winding and the rest passes through load Shunt field current Ish = V Rsh

Armature current Ia = IL+ Ish

Terminal voltage V= Eg-Ia Ra

Power developed in armature = EgIa

Power delivered to load = VIL A compound generator is shown in figure 111 below It has two field findings namely Rsh and Rse They are basically shunt winding (Rsh) and series winding (Rse) Compound generator is of two types ndash 1) Short shunt and 2) Long shunt

Short shunt- Here the shunt field winding is wired parallel to armature and series field winding is connected in series to the load It is shown in fig111(a)

In short shunt machines Series field current Ise = IL

Shunt field current Ish iquest V+ I se RseRsh

Terminal voltage V= Eg-Ia Ra-Ise Rse

Power developedin armature = Eg Ia

Power delivered to load = V IL

Fig 111 (a) short shunt generator and (b) Long shunt generator

Long shunt- Here the shunt field winding is parallel to both armature and series field winding (Rse is wired in series to the armature) It is shown in figure 111 (b)

In Long shunt machines Series field current Ise = Ia = IL+ Ish

Shunt field current Ish = V R sh

Terminal voltage V = Eg- Ia (Ra+ Rse) Power developed in armature = Eg Ia

Power delivered to load = VIL

16 GENERATOR CHARACTERISTICS

The speed of a dc machine operated as a generator is fixed by the prime mover For general-purpose operation the prime mover is equipped with a speed governor so that the speed of the generator is practically constant Under such condition the generator performance deals primarily with the relation between excitation terminal voltage and load These relations can be best exhibited graphically by means of curves known as generator characteristics Thesecharacteristics show at a glance the behavior of the generator under differentload conditions

The following are the three most important characteristics of a dc generator161 Open Circuit Characteristic (OCC)

This curve shows the relation between the generated emf at no-load (E0) and the field current (If) at constant speed It is also known as magnetic characteristic or no-load saturation curve Its shape is practically the same for all generators whether separately or self-excited The data for OCC curve are obtained experimentally by operating the generator at no load and constant speed and recording the change in terminal voltage as the field current is varied

162 Internal or Total characteristic (EIa)

This curve shows the relation between the generated emf on load (E) and the armature current (Ia) The emf E is less than E0 due to the demagnetizing effect of armature reaction Therefore this curve will lie below the open circuitcharacteristic (OCC) The internal characteristic is of interest chiefly to thedesigner It cannot be obtained directly by experiment It is because a voltmetercannot read the emf generated on load due to the voltage drop in armatureresistance The internal characteristic can be obtained from externalcharacteristics if winding resistances are known because armature reaction effectis included in both characteristics163 External characteristic (VIL)This curve shows the relation between the terminal voltage (V) and load current(IL) The terminal voltage V will be less than E due to voltage drop in thearmature circuit Therefore this curve will lie below the internal characteristicThis characteristic is very important in determining the suitability of a generatorfor a given purpose It can be obtained by making simultaneous measurements

of terminal voltage and load current (with voltmeter and ammeter) of a loadedgenerator164 Open Circuit Characteristic of a DC Generator

The OCC for a dc generator is determined as follows The field winding of the dc generator (series or shunt) is disconnected from the machine and isseparately excited from an external dc source as shown in Fig 112 Thegenerator is run at fixed speed (ie normal speed) The field current (If) isincreased from zero in steps and the corresponding values of generated emf(E0) read off on a voltmeter connected across the armature terminals On plottingthe relation between E0 and If we get the open circuit characteristic as shown in

Fig113

Fig 112 Experimental setup for plotting OCC

Fig 113 OCC of dc generator

The following points may be noted from OCC When the field current is zero there is some generated emf OA

This is due to the residual magnetism in the field poles Over a fairly wide range of field current (upto point B in the curve)

the curve is linear It is because in this range reluctance of iron is negligible as compared with that of air gap The air gap reluctance is constant and hence linear relationship After point B on the curve the reluctance of iron also comes into picture and reluctance of iron is no longer negligible Consequently the curve deviates from linear relationship

After point C on the curve the magnetic saturation of poles begins and E0 tends to level off

The OCC of even self-excited generator is obtained by running it as a separately excited generator

165 Characteristics of a Separately Excited DCGenerator

The obvious disadvantage of a separately excited dc generator is that we require an external dc source for excitation But since the output voltage may be

controlled more easily and over a wide range (from zero to a maximum) this type of excitation finds many applications

(i) Open circuit characteristic

The OCC of a separately excited generator is determined in a manner described in Sec 164 Fig114 shows the variation of generated em f on noload with field current for various fixed speeds Note that if the value of constant speed is increased the steepness of the curve also increases When the field current is zero the residual magnetism in the poles will give rise to the small initial emf as shown

Fig114 Variation of OCC with speed

Internal and External Characteristics (Load Characteristics)

The external characteristic of a separately excited generator is the curve between the terminal voltage (V) and the load current IL (which is the same as armature current in this case) In order to determine the external characteristic the circuit set up is as shown in Fig 115

Fig 115 Experimental setup to plot load characteristics

As the load current increases the terminal voltage falls due to two reasons(a) The armature reaction weakens the main flux so that actual emf generated E on load is less than that generated (E0) on no load

(b) There is voltage drop across armature resistance (= ILRa = IaRa)Due to these reasons the external characteristic is a drooping curve [curve 3 inFig 116 ] Note that in the absence of armature reaction and armature dropthe generated emf would have been E0 (curve 1)The internal characteristic can be determined from external characteristic by adding IL Ra drop to the external characteristic It is because armature reaction drop is included in the external characteristic Curve 2 is the internal characteristic of the generator and should obviously lie above the external characteristic

Fig 116 Load characteristics of a separately excited dcgenerator

117 Voltage Build-Up in a Self-Excited GeneratorLet us see how voltage builds up in a self-excited generator(i) Shunt generator Consider a shunt generator If the generator is run at a constant speed some emf will be generated due to residual magnetism in the main poles This small emf circulates a field current which in turn produces additional flux toreinforce the original residual flux (provided field winding connections arecorrect) This process continues and the generator builds up the normalgenerated voltage following the OCC shown in Fig 117 (i) The field resistance Rf can be represented by a straight line passing through the origin as shown in Fig 117 (ii) The two curves can be shown on the same diagram as they have the same ordinate as in Fig 117 (iii)

Fig 117

Since the field circuit is inductive there is a delay in the increase in current uponclosing the field circuit switch The rate at which the current increases dependsupon the voltage available for increasing it Suppose at any instant the fieldcurrent is i (= OA) and is increasing at the rate didt Then

E0=i R f +iquest L didt

Where Rf = total field circuit resistance L = inductance of field circuit

At the considered instant the total emf available is AC [See Fig 117 (iii)] Anamount AB of the cmf AC is absorbed by the voltage drop iRf and theremainder part BC is available to overcome L didt Since this surplus voltage isavailable it is possible for the field current to increase above the value OAHowever at point D the available voltage is OM and is all absorbed by i RfD

rop Consequently the field current cannot increase further and the generatorbuild up stops Hence we arrive at a very important conclusion that the voltage build up of the generator is given by the point of intersection of OCC and field resistance line Thus in Fig 117 (iii) D is point of intersection of the two curves Hence the generator will build up a voltage OM(ii) Series generator During initial operation with no current yet flowing a residual voltage will be generated exactly as in the case of a shunt generator The residual voltage will cause a current to flow through the whole series circuit when the circuit isclosed There will then be voltage build up to an equilibrium point exactlyanalogous to the buildup of a shunt generator The voltage build up graph willbe similar to that of shunt generator except that now load current (instead offield current for shunt generator) will be taken along x-axis10487871048787(iii) Compound generator When a compound generator has its series field flux aiding its shunt field flux the machine is said to be cumulative compound When the series field isconnected in reverse so that its field flux opposes the shunt field flux thegenerator is then differential compound The easiest way to build up voltage in a compound generator is to start under no load conditions At no load only the shunt field is effective When no-load voltage build up is achieved the generator is loaded If under load the voltagerises the series field connection is cumulative If the voltage drops significantlythe connection is differential compound118 Critical Field Resistance for a Shunt Generator The voltage build up in a shunt generator depends upon field circuit resistance If the field circuit resistance is R1 (line OA) then generator will build up a voltage OM as shown in Fig 118 If the field circuit resistance is increasedto R2 (tine OB) the generator will build up a voltage OL slightly less than OM As the field circuit resistance is increased the slope of resistance line also increases When the field resistance line becomes tangent (line OC) to OCC the generator would just excite If the field circuit resistance is increased beyond this point (say line OD) the generator will fail to excite The field circuit resistance represented by line OC (tangent to OCC) is called critical field resistance RC for the shunt generator It may be defined as under The maximum field circuit resistance (for a given speed) with which the shunt generator would just excite is known as its critical field resistance It should be noted that shunt generator will build up voltage only if field circuit resistance is less than critical field resistance

Fig 118 Critical field resistance of shunt generator

119 Critical Resistance for a Series Generator

Fig 119 shows the voltage build up in a series generator Here R1 R2 etc represent the total circuit resistance (load resistance and field winding resistance) If the total circuit resistance is R1 then series generator will build up a voltage OL The line OC is tangent to OCC and represents the critical resistance RC for a series generator If the total resistance of the circuit is more than RC (say line OD) the generator will fail to build up voltage Note that Fig 119 is similar to Fig 118 with the following differences

(i) In Fig 118 R1 R2 etc represent the total field circuit resistanceHowever R1 R2 etc in Fig 119 represent the total circuit resistance(load resistance and series field winding resistance etc)

(ii) In Fig 118 field current alone is represented along X-axis However inFig 119 load current IL is represented along Y-axis Note that in aseries generator field current = load current IL

Fig 119 Critical resistance of a series generator

120 Characteristics of Series Generator Fig 120 (i) shows the connections of a series wound generator Since there is only one current (that which flows through the whole machine) the load current is the same as the exciting current(i) OCCCurve 1 shows the open circuit characteristic (OCC) of a series generator Itcan be obtained experimentally by disconnecting the field winding from themachine and exciting it from a separate dc source(ii) Internal characteristic Curve 2 shows the total or internal characteristic of a series generator It gives the relation between the generated emf E on load and armature current Due to armature reaction the flux in the machine will be less than the flux at no load Hence emf E generated under load conditions will be less than the emf E0 generated under no load conditions Consequently internal characteristic curvelies below the OCC curve the difference between them representing the effectof armature reaction [See Fig 120 (ii)]

Fig 120 Characteristics of series generator

(iii) External characteristic

Curve 3 shows the external characteristic of a series generator It gives therelation between terminal voltage and load current ILV = E g- Ia (Ra +Rse )Therefore external characteristic curve will lie below internal characteristiccurve by an amount equal to ohmic drop [ie Ia(Ra + Rse)] in the machine asshown in Fig 120 (ii) The internal and external characteristics ofa dc series generator can be plotted from one another as shown in Fig 121 Supposewe are given the internal characteristic of the generator Let the line OC represent the resistance of the whole machine ie Ra +Rse If the load current is OB drop in the machine is AB ieAB = Ohmic drop in the machine = OB(Ra + Rse)

Fig 121 Load characteristics of series generator

Now raise a perpendicular from point B and mark a point b on this line such that ab = AB Then point b will lie on the external characteristic of the generator

Following similar procedure other points of external characteristic can be located It is easy to see that we can also plot internal characteristic from the external characteristic

121 Characteristics of a Shunt Generator Fig 122 (i) show the connections of a shunt wound generator The armature current Ia splits up into two parts a small fraction Ish flowing through shunt field winding while the major part IL goes to the external load(i) OCC The OCC of a shunt generator is similar in shape to that of a series generator as shown in Fig 122 (ii) The line OA represents the shunt field circuitresistance When the generator is run at normal speed it will build up a voltageOM At no-load the terminal voltage of the generator will be constant (= OM)represented by the horizontal dotted line MC

Fig 122 Characteristics of a Shunt Generator

(ii) Internal characteristic

When the generator is loaded flux per pole is reduced due to armature reaction Therefore emf E generated on load is less than the emf generated at no load As a result the internal characteristic (EIa) drops down slightly as shown in Fig122 (ii)

(iii) External characteristic Curve 2 shows the external characteristic of a shunt generator It gives therelation between terminal voltage V and load current IL

Therefore external characteristic curve will lie below the internal characteristic curve by an amount equal to drop in the armature circuit[ie (IL + Ish)Ra] as shown in Fig 122 (ii)

Note It may be seen from the external characteristic that change in terminalvoltage from no-load to full load is small The terminal voltage can always bemaintained constant by adjusting the field rheostat R automatically122 Critical External Resistance for Shunt Generator

If the load resistance across the terminals of a shunt generator is decreased then load current increase However there is a limit to the increase in

load current with the decrease of load resistance Any decrease of load resistance beyond this point instead of increasing the current ultimately results in reduced current Consequently the external characteristic turns back (dotted curve) as shown in Fig 123 The tangent OA to the curve represents the minimum external resistance required to excite the shunt generator on load and is called critical external resistance If the resistance of the external circuit is less than the critical external resistance (represented by tangent OA in Fig 310) the machine will refuse to excite or will de-excite if already running This means that external resistance is so low as virtually to short circuit the machine and so doing away with its excitation

Note There are two critical resistances for a shunt generator viz (i) criticalfield resistance (ii) critical external resistance For the shunt generator to buildup voltage the former should not be exceeded and the latter must not be gonebelow

123 How to Draw OCC at Different Speeds

If we are given OCC of a generator at a constant speed N1 then we can easily draw the OCC at any other constant speed N2 Fig 123 illustrates theprocedure Here we are given OCC at a constant speed N1 It is desired to findthe OCC at constant speed N2 (it is assumed that n1 lt N2) For constantexcitation E propN

E2

E1 =

N 2

N1 so

E2 iquest E1

N2

N1

As shown in Fig (311) for If = OH E1 = HC Therefore the new value of emf (E2) for the same If but at N2

E2= HCN 2

N 1 = HD

Fig 123 Plotting OCC at another speed

This locates the point D on the new OCC at N2 Similarly other points can belocated taking different values of If The locus of these points will be the OCCat N2

124 Critical Speed (NC)

The critical speed of a shunt generator is the minimum speed below which it fails to excite Clearly it is the speed for which the given shunt field resistance represents the critical resistance In Fig 124 curve 2 corresponds to critical

speed because the shunt field resistance (Rsh) line is tangential to it If the generator runs at full speed N the new OCC moves upward and the Rsh line represents critical resistance for this speed

In order to find critical speed take any convenient point C on excitation axis and erect a perpendicular so as to cut Rsh and Rsh lines at points B and A respectively Then

BCAC

=NcN

Nc=N BCAC

Fig 124 Finding critical speed

125 Conditions for Voltage Build-Up of a Shunt Generator

The necessary conditions for voltage build-up in a shunt generator are

(i) There must be some residual magnetism in generator poles(ii) The connections of the field winding should be such that the field current strengthens the residual magnetism(iii) The resistance of the field circuit should be less than the critical resistance In other words the speed of the generator should be higher

than the critical speed

126 Compound Generator Characteristics

In a compound generator both series and shunt excitation are combined as shown in Fig 123 The shunt winding can be connected either across the armature only (short-shunt connection S) or across armature plus series field (long-shunt connection G) The compound generator can be cumulatively compounded or

differentially compounded generator The latter is rarely used in practice Therefore we shall discuss the characteristics of cumulatively compounded

generator It may be noted that external characteristics of long and short shunt compound generators are almost identical

External characteristic

Fig 124 shows the external characteristics of a cumulatively compoundedgenerator The series excitation aids the shunt excitation The degree of compounding depends upon the increase in series excitation with the increase inload current

(i) If series winding turns are so adjusted that with the increase in load current the terminal voltage increases it is called over-compounded generator In such a case as the load current increases the series field mmf increases and tends to increase the flux and hence the generated voltage The increase in generated voltage is greater than the IaRa drop so that instead of decreasing the terminal voltage increases as shown by curve A in Fig 124

(ii) If series winding turns are so adjusted that with the increase in load current the terminal voltage substantially remains constant it is called Flat-compounded generator The series winding of such a machine has lesser number of turns than the one in over-compounded machine and therefore does not increase the flux as much for a given load current Consequently the full-load voltage is nearly equal to the no-load voltage as indicated by curve B in Fig 124 (iii) If series field winding has lesser number of turns than for a flat compounded machine the terminal voltage falls with increase in load current as indicated by curve C m Fig 124 Such a machine is called under-compounded generator

Fig 123 compound generator

Fig 124 Load characteristics of compound generators

127 LOSSES

The losses in a dc machine (generator or motor) may be divided into three classes viz (i) copper losses or electrical losses (ii) iron or core losses and (iii) mechanical losses All these losses appear as heat and thus raise the temperature of the machine They also lower the efficiency of the machine The loss structure maybe demonstrated as

1Copperlosses

These losses occur due to currents in the various windings of the machine When an electric current of I amps flows in a resistance of R ohms the heat energy is lost at the rate of I2R joules sec and the power loss is I2R watts Generators and motors have one or more field circuits and an armature circuit in which such losses occur All resistance losses of this kind are called copper losses The different copper losses occurring in a dc machine are

(i) Armature copper loss = Ia2Ra

(ii) Shunt field copper loss = Ish2 Rsh

(iii) Series field copper loss = Ise2Rse

Armature copper loss will be around 30 to 40 of full load losses where as total field copper losses will constitute around 30 of full load losses

Note There is also brush contact loss due to brush contact resistance (ie resistance between the surface of brush and surface of commutator) This loss is generally included in armature copper loss The brush contact drop is almost independent of Ia For carbon brushes this drop is around 1 V per brush The power loss due to brush contact resistance is 2ebIa where eb is the voltage drop at one brush

2 IRONORCORELOSSES These losses occur in the armature of a dc machine and are due to the rotation of armature in the magnetic field of the poles They are of two types viz (i) hysteresis loss (ii) eddy current loss

(i) Hysteresis loss

Hysteresis loss occurs in the armature of the dc machine since any given part of the armature is subjected to magnetic field reversals as it passes under successive poles Fig 124 shows an armature rotating in two-pole machine Consider a small piece ab of the armature When the piece ab is under N-pole the magnetic lines pass from a to b Half a revolution later the same piece of iron is under S-pole and magnetic lines pass from b to a so that magnetism in the iron is reversed In order to reverse continuously the molecular magnets in the armature core some amount of power has to be spent which is called hysteresis loss It is given by Steinmetz formula This formula is

Hysteresis Loss Wh = η Bmax 16f V Watts Bmax - Maximum flux density in the core f - Frequency of magnetic reversals V - Volume of armature in m3

η - Steinmetz hysteresis coefficient

In order to reduce this loss in a dc machine armature core is made of such materials which have a low value of Steinmetz hysteresis co-efficient eg silicon steel

Fig 124 Occurrence of hysteresis loss

ii) EDDYCURRENTLOSS

In addition to the voltages induced in the armature conductors there are also voltages induced in the armature core These voltages produce circulating currents in the armature core as shown in Fig 125 These are called eddy currents and power loss due to their flow is called eddy current loss The eddy current loss appears as heat which raises the temperature of the machine and lowers its efficiency Core resistance can be greatly increased by constructing the core of thin round If a continuous solid iron core is used the resistance to eddy current path will be small due to large cross-sectional area of the core Consequently the magnitude of eddy current and hence eddy current loss will be large The magnitude of eddy current can be reduced by making core resistance as high as practical The iron sheets called laminations [See Fig 126] The laminations are insulated from each other with a coating of varnish The insulating coating has a high resistance so very little current flows from one lamination to the other Alsobecause each lamination is very thin the resistance to current flowing through the width of a lamination is also quite large Thus laminating a core increases the core resistance which decreases the eddy current and hence the eddy current loss

Fig 125 Eddy current loss Fig 126 Core laminations

Eddy current loss is biven by the formula Pe= Ke Bmax

2 f2t2 V Watts

Where Ke = Constant depending upon electrical resistace of core Bmax = Maximum flux density in wbm2

f = frequency f magnetic reversals in Hz t = thickness of laminations in m and V = volume of core in m3

It may be noted that eddy current loss depends upon the square of lamination thickness For this reason lamination thickness should be kept as small as possible

3MECHANICALLOSSES

These losses are due to friction and windage(i) Friction losses eg bearing friction brush friction etc(ii) Windage loss ie air friction of rotating armature

These losses depend upon the speed of the machine But for a given speed they are practically constant

Note Iron losses and mechanical losses together are called stray losses

CONSTANT AND VARIABLE LOSSES

The losses in a dc generator (or dc motor) may be sub-divided into

(i) constant losses (ii) variable losses

(i) Constant losses

Those losses in a dc generator which remain constant at all loads are known as constant losses The constant losses in a dc generator are(a) iron losses(b) mechanical losses(c) shunt field losses(ii) Variable lossesThose losses in a dc generator which vary with load are called variable lossesThe variable losses in a dc generator are (a) Copper loss in armature winding (Ia

2 Ra) (b) Copper loss in series field winding (Ise

2Rse)

Total losses = Constant losses + Variable lossesNote Field Cu loss is constant for shunt and compound generators

128 POWER FLOW DIAGRAM

The power flow in a dc generator may be represented as

The different power stages are shown as A B and CA ndash The mechanical power input which is the power output of the prime mover is given as Pin = 2πNT

60 watts

B - The power developed in armature is Pa = Eg Ia watts C- The electrical power output is Pout = VIL watts

129 EFFICIENCY Efficiency of a machine is the ratio of its output power to the input power

Efficiency η = Output powerInput power

= Input powerminusLossesInput power = Output power

Output power+Losses

In case of a dc generator depending upon the power stages three different efficiencies can be defined as-

1 Mechanical Efficiency(ηm)

BA = Electrical power developedisinarmature

Mechanical power input2 Electrical Efficiency(ηe)

CB = Electrical power output

Electrical power developedisinthe armature 3 Overall or commercial efficiency (η)

CA = Electrical power output

Mechanical power input = ηmtimes ηe

130 CONDITION FOR MAXIMUM EFFICIENCY The efficiency of a dc generator is not constant but varies with load Fig 127shows the variation of efficiency with load current Consider a dc shunt generator with terminal voltage V delivering a load current of IL amp Generator output = V IL

Generator input = Output + Losses = V IL + constant losses + Variable losses = V IL + Wc+ Ia

2Ra

= V IL + Wc + (IL+ Ish)2 Rsh

The shunt field current Ish is small compared with IL so may be neglected

Generator input = VIL+ IL2Ra + Wc

Now efficiency η iquest outputinput = VI L

VIL+ Ia 2Ra+Wc

Efficiency is maximum when the denominator of the above equation is minimumie

ddIL ( VI L

VIL+ Ia 2Ra+Wc ) = 0

Which gives IL2 Ra = WC

ie Variable loss = constant lossThe load current for maximum efficiency is given by

IL iquest radicWcRa

Fig 127 Variation of efficiency with load

Hence the efficiency is maximum for that load current at which the variable loss is equal to the constant loss

131 APPLICATIONS

Separately excited generators

These types of generators have special applications such as electroplating electro refining of materials because separate supply is required for field excitation

Shunt generators

These types of generators are used in battery charging and ordinary lighting purposes

Series generators

This type of generators is used commonly as boosters on DC feeders as current generators for welding and arc lamp

Cumulatively compounded generators

This type of generators is used for domestic lighting purposes and to transmit energy over long distances

Differential compound generators

This type of generators is rarely used They are used for special applications such as electric arc welding

132 NUMERICAL EXCERCISES

1 The emf induced in the armature of a 450 kW 250 volt shunt generator is

2588volt when the field current is 200 amp and the generator is supplying power to a load at rated terminal voltage The armature circuit resistance is 0005 ohm Determine (i) load current (ii) power generated (iii) power output and (iv) electrical efficiency Neglect brush contact drop

Solution

Let the load current be I amp Given Ish = 20 A

Armature current = Load current + field current or Ia = I + 20 Induced emf = Terminal voltage + Ia Ra drop

2588 = 250 + (I + 20) times 0005

So (i) Load current I = 1740 amp

(ii) Power generated E Ia = 2588 (1740 + 20) watts = 455488 kW (iii) Power output V I = 250 times 1740 watts

= 435 kW

(iv) Electrical efficiency = 435455488 times 100 = 955

2 A 4 pole wave wound dc shunt generator delivers a load of 45 kW at a terminal voltage of 220 volts Its armature has 150 single turn coils and has a resistance of 001 ohm The air gap flux per pole is 002 weber Shunt field resistance is 50 ohm Calculate the speed at which it is being driven Neglect brush contact drop

Solution Load current I = W

V = 45000220 = 20455 amp

Field Current Ish = V

R s h = 220

50 = 44 amp

Armature current Ia = I +I sh = 20455 +44 = 20895 amp

Armature drop I aRa = 20895 X 001 = 20895 volts Induced emf Eg = V + Ia Ra = 220 + 20895 = 2220895 volts

But induced emf

So N = Eg60 AΦ ZP = 2220895times 60 times2

002 times150 times2 times 4 = 111045 rpm

3 The following data refer to the OCC of a DC separately excited generator at 1000 rpm

Field current in A 00 02 03 04 05 06 07 08 09 10 11 12Armature voltage in V 5 40 75 100 124 145 162 178 188 195 200 205

The machine is now connected as a shunt generator with a total field resistance of 200 Ω and driven at 1000 rpm Estimate graphically (i) the voltage to which the generator will build up at no load (ii) the armature field and load currents when the terminal voltage is found to be 150 V Neglect the effect of armature reaction and brush drop and assume armature resistance Ra to be 08 Ω (iii) Finally estimate the steady state armature current when the machine terminals are shorted

Solution

First of all the OCC of the generator is plotted in a graph paper as shown in figure

(i) The total field circuit resistance is given to be Rf = 200 Ω The Rf line is now drawn on the same graph paper passing through the origin The point of intersection of the Rf line and the OCC decides the final no load voltage and can be read from the graph as 192 V and the corresponding field current is 096 A (ii) Since the terminal voltage V is 150 V (= BM) field current If

is OM = 077 A Generated voltage E is given by AM But we know E = IaRa + V Hence IaRa= E ndash V = AM ndash BM = AB Now from the graph AB = 25 V there4Ia = 2508 = 3125 A So load current IL= Ia ndash If= 3125 ndash 077 = 3048 A

ADDITIONAL PROBLEMS

1 A 4 pole lap wound 115 kW 230 volts dc shunt generator has armatureresistance of 02 ohm and field resistance of 100 ohms Calculate (i) emf

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES

There are mainly two types of construction available-(1) Solid pole core where it made of a solid single piece of cast iron or cast steel (2) Laminated pole core where it made of numbers of thin limitations of annealed steel which are riveted together The pole core is fixed to the inner periphery of the yoke by means of bolts through the yoke and into the pole body The pole shoes are so typically shaped that they spread out the magnetic flux in the air gap and reduce the reluctance of the magnetic path Due to their larger cross ndash section they hold the pole coil at its position

133 Field winding The field coils or pole coils are wound around the pole core These are a simple coil of insulated copper wire or strip which placed on the pole which placed between yoke and pole shoes134 Armature Core The armature core serves the purpose of holding the armature winding and providing a low reluctance path for the flux through the armature from N pole to S pole A cylindrical or drum shaped armature core is build up of circular laminated sheets In every circular lamination slots are either die ndash cut or punched on the outer periphery and the key way is located on the inner periphery Air ducts are also punched of cut on each lamination for circulation of air through the core for providing better cooling Fig 14 shows the armature and field windings

Fig 14 Armature and field windings

135 Armature Winding Armature winding are generally formed wound These are first wound in the form of flat rectangular coils and are then pulled into their proper shape in a coil puller Various conductors of the coils are insulated from each other The conductors are placed in the armature slots which are lined with tough insulating material This slot insulation is folded over above the armature conductors placed in it and secured in place by special hard wooden or fiber wedges There are mainly two types of armature windings namely the lap winding and the wave winding which is shown in figure 15 In lap winding the armature conductors are arranged in P number of parallel paths where P is the number of poles But a wave winding offers only two parallel paths for the flow of armature current Hence a lap winding is used for high current low voltage generator where as a wave wound armature is preferred for a low current high voltage generator

Fig 15 Lap and wave winding

136 Commutator The commutator plays a vital role in dc generator It collects current from armature and sends it to the load as direct current It actually takes alternating current from armature and converts it to direct current and then send it to external load It is cylindrical structured and is build up of wedge ndash shaped segments of high conductivity hard drawn or drop forged copper Each segment is insulated from the shaft by means of insulated commutator segment shown below Each commutator segment is connected with corresponding armature conductor through segment riser or lug Fig 16 shows the arrangement of commutator segments in a dc machine along with its cross sectional view

Fig16 Commutator segments

137 Brushes The brushes are made of carbon These are rectangular block shaped The only function of these carbon brushes of dc generator is to collect current from commutator segments The brushes are housed in the rectangular box shaped brush holder As shown in figure17 the brush face is placed on the commutator segment with attached to the brush holder

Fig 17 Brush in brush holder

138 Bearing For small machine ball bearing is used and for heavy duty dc generator roller bearing is used The bearing must always be lubricated properly for smooth operation and long life of generator

14 EMF EQUATION OF A DC GENERATOR

Let Φ be the flux per pole in WeberZ the total number of armature conductors ndash No of slots x no of armature ConductorsslotP the number of polesA the number of parallel paths in the armature

N the speed of rotation of armature in revolutions per minute (rpm)E the emf induced in any parallel path in armatureEg the emf generated in any parallel path in the armatureThe emf induced per conductor = dΦdt volt (n=1)The flux cut by one conductor in one revolution dΦ = ΦP WbNo of revolutions per second = N60Time for one revolution dt = 60N secondHence EMF generatedconductor is

dφdt = φPN

60

No of conductors in a parallel path = Z A

Hence emf generated per parallal path

Eg = φPN60 Z

A

For a lap wound generator A = P and

For wave wound generator A = 2

15 TYPES OF DC GENERATORS

Fig18 Classification of dc generators

151 Separately Excited DC Generator

This dc generator has a field magnet winding which is excited using a separate voltage source (like battery)The representation in the below figure 19 The output voltage depends on the speed of rotation of armature and field current The higher the speed of rotation and current ndash the higher the output emf

Fig 19 Separately Excited DC Generators

Armature Current Ia = IL

Terminal voltage V = Eg-Ia Ra

Electric Power developed = Eg Ia

Power delivered to load = V IL = (Eg-Ia Ra) Ia = Eg Ia - Ia2

Ra

152 Self Excited DC Generator

These are generators in which the field winding is excited by the output of the generator itself As described before ndash there are three types of self excited dc generators ndash they are 1) Series 2) Shunt and 3) Compound

A series DC generator is shown below in fig110 (a) ndash in which the armature winding is connected in series with the field winding so that the field current flows through the load as well as the field winding Field winding is a low resistance thick wire of few turns Armature current Ia =Ise = IL=I (say)

Terminal voltage V = Eg- I( Ra+ Rse)Power Developed in armature = Eg Ia

Power delivered to load = V IL = Eg Ia- Ia2 (Ra+Rse)

Fig110 (a) Series generator and (b) shunt generator

A shunt DC generator is shown in figure 110 (b) in which the field winding is wired parallel to armature winding so that the voltage across both are same The field winding has high resistance and more number of turns so that only a part of armature current passes through field winding and the rest passes through load Shunt field current Ish = V Rsh

Armature current Ia = IL+ Ish

Terminal voltage V= Eg-Ia Ra

Power developed in armature = EgIa

Power delivered to load = VIL A compound generator is shown in figure 111 below It has two field findings namely Rsh and Rse They are basically shunt winding (Rsh) and series winding (Rse) Compound generator is of two types ndash 1) Short shunt and 2) Long shunt

Short shunt- Here the shunt field winding is wired parallel to armature and series field winding is connected in series to the load It is shown in fig111(a)

In short shunt machines Series field current Ise = IL

Shunt field current Ish iquest V+ I se RseRsh

Terminal voltage V= Eg-Ia Ra-Ise Rse

Power developedin armature = Eg Ia

Power delivered to load = V IL

Fig 111 (a) short shunt generator and (b) Long shunt generator

Long shunt- Here the shunt field winding is parallel to both armature and series field winding (Rse is wired in series to the armature) It is shown in figure 111 (b)

In Long shunt machines Series field current Ise = Ia = IL+ Ish

Shunt field current Ish = V R sh

Terminal voltage V = Eg- Ia (Ra+ Rse) Power developed in armature = Eg Ia

Power delivered to load = VIL

16 GENERATOR CHARACTERISTICS

The speed of a dc machine operated as a generator is fixed by the prime mover For general-purpose operation the prime mover is equipped with a speed governor so that the speed of the generator is practically constant Under such condition the generator performance deals primarily with the relation between excitation terminal voltage and load These relations can be best exhibited graphically by means of curves known as generator characteristics Thesecharacteristics show at a glance the behavior of the generator under differentload conditions

The following are the three most important characteristics of a dc generator161 Open Circuit Characteristic (OCC)

This curve shows the relation between the generated emf at no-load (E0) and the field current (If) at constant speed It is also known as magnetic characteristic or no-load saturation curve Its shape is practically the same for all generators whether separately or self-excited The data for OCC curve are obtained experimentally by operating the generator at no load and constant speed and recording the change in terminal voltage as the field current is varied

162 Internal or Total characteristic (EIa)

This curve shows the relation between the generated emf on load (E) and the armature current (Ia) The emf E is less than E0 due to the demagnetizing effect of armature reaction Therefore this curve will lie below the open circuitcharacteristic (OCC) The internal characteristic is of interest chiefly to thedesigner It cannot be obtained directly by experiment It is because a voltmetercannot read the emf generated on load due to the voltage drop in armatureresistance The internal characteristic can be obtained from externalcharacteristics if winding resistances are known because armature reaction effectis included in both characteristics163 External characteristic (VIL)This curve shows the relation between the terminal voltage (V) and load current(IL) The terminal voltage V will be less than E due to voltage drop in thearmature circuit Therefore this curve will lie below the internal characteristicThis characteristic is very important in determining the suitability of a generatorfor a given purpose It can be obtained by making simultaneous measurements

of terminal voltage and load current (with voltmeter and ammeter) of a loadedgenerator164 Open Circuit Characteristic of a DC Generator

The OCC for a dc generator is determined as follows The field winding of the dc generator (series or shunt) is disconnected from the machine and isseparately excited from an external dc source as shown in Fig 112 Thegenerator is run at fixed speed (ie normal speed) The field current (If) isincreased from zero in steps and the corresponding values of generated emf(E0) read off on a voltmeter connected across the armature terminals On plottingthe relation between E0 and If we get the open circuit characteristic as shown in

Fig113

Fig 112 Experimental setup for plotting OCC

Fig 113 OCC of dc generator

The following points may be noted from OCC When the field current is zero there is some generated emf OA

This is due to the residual magnetism in the field poles Over a fairly wide range of field current (upto point B in the curve)

the curve is linear It is because in this range reluctance of iron is negligible as compared with that of air gap The air gap reluctance is constant and hence linear relationship After point B on the curve the reluctance of iron also comes into picture and reluctance of iron is no longer negligible Consequently the curve deviates from linear relationship

After point C on the curve the magnetic saturation of poles begins and E0 tends to level off

The OCC of even self-excited generator is obtained by running it as a separately excited generator

165 Characteristics of a Separately Excited DCGenerator

The obvious disadvantage of a separately excited dc generator is that we require an external dc source for excitation But since the output voltage may be

controlled more easily and over a wide range (from zero to a maximum) this type of excitation finds many applications

(i) Open circuit characteristic

The OCC of a separately excited generator is determined in a manner described in Sec 164 Fig114 shows the variation of generated em f on noload with field current for various fixed speeds Note that if the value of constant speed is increased the steepness of the curve also increases When the field current is zero the residual magnetism in the poles will give rise to the small initial emf as shown

Fig114 Variation of OCC with speed

Internal and External Characteristics (Load Characteristics)

The external characteristic of a separately excited generator is the curve between the terminal voltage (V) and the load current IL (which is the same as armature current in this case) In order to determine the external characteristic the circuit set up is as shown in Fig 115

Fig 115 Experimental setup to plot load characteristics

As the load current increases the terminal voltage falls due to two reasons(a) The armature reaction weakens the main flux so that actual emf generated E on load is less than that generated (E0) on no load

(b) There is voltage drop across armature resistance (= ILRa = IaRa)Due to these reasons the external characteristic is a drooping curve [curve 3 inFig 116 ] Note that in the absence of armature reaction and armature dropthe generated emf would have been E0 (curve 1)The internal characteristic can be determined from external characteristic by adding IL Ra drop to the external characteristic It is because armature reaction drop is included in the external characteristic Curve 2 is the internal characteristic of the generator and should obviously lie above the external characteristic

Fig 116 Load characteristics of a separately excited dcgenerator

117 Voltage Build-Up in a Self-Excited GeneratorLet us see how voltage builds up in a self-excited generator(i) Shunt generator Consider a shunt generator If the generator is run at a constant speed some emf will be generated due to residual magnetism in the main poles This small emf circulates a field current which in turn produces additional flux toreinforce the original residual flux (provided field winding connections arecorrect) This process continues and the generator builds up the normalgenerated voltage following the OCC shown in Fig 117 (i) The field resistance Rf can be represented by a straight line passing through the origin as shown in Fig 117 (ii) The two curves can be shown on the same diagram as they have the same ordinate as in Fig 117 (iii)

Fig 117

Since the field circuit is inductive there is a delay in the increase in current uponclosing the field circuit switch The rate at which the current increases dependsupon the voltage available for increasing it Suppose at any instant the fieldcurrent is i (= OA) and is increasing at the rate didt Then

E0=i R f +iquest L didt

Where Rf = total field circuit resistance L = inductance of field circuit

At the considered instant the total emf available is AC [See Fig 117 (iii)] Anamount AB of the cmf AC is absorbed by the voltage drop iRf and theremainder part BC is available to overcome L didt Since this surplus voltage isavailable it is possible for the field current to increase above the value OAHowever at point D the available voltage is OM and is all absorbed by i RfD

rop Consequently the field current cannot increase further and the generatorbuild up stops Hence we arrive at a very important conclusion that the voltage build up of the generator is given by the point of intersection of OCC and field resistance line Thus in Fig 117 (iii) D is point of intersection of the two curves Hence the generator will build up a voltage OM(ii) Series generator During initial operation with no current yet flowing a residual voltage will be generated exactly as in the case of a shunt generator The residual voltage will cause a current to flow through the whole series circuit when the circuit isclosed There will then be voltage build up to an equilibrium point exactlyanalogous to the buildup of a shunt generator The voltage build up graph willbe similar to that of shunt generator except that now load current (instead offield current for shunt generator) will be taken along x-axis10487871048787(iii) Compound generator When a compound generator has its series field flux aiding its shunt field flux the machine is said to be cumulative compound When the series field isconnected in reverse so that its field flux opposes the shunt field flux thegenerator is then differential compound The easiest way to build up voltage in a compound generator is to start under no load conditions At no load only the shunt field is effective When no-load voltage build up is achieved the generator is loaded If under load the voltagerises the series field connection is cumulative If the voltage drops significantlythe connection is differential compound118 Critical Field Resistance for a Shunt Generator The voltage build up in a shunt generator depends upon field circuit resistance If the field circuit resistance is R1 (line OA) then generator will build up a voltage OM as shown in Fig 118 If the field circuit resistance is increasedto R2 (tine OB) the generator will build up a voltage OL slightly less than OM As the field circuit resistance is increased the slope of resistance line also increases When the field resistance line becomes tangent (line OC) to OCC the generator would just excite If the field circuit resistance is increased beyond this point (say line OD) the generator will fail to excite The field circuit resistance represented by line OC (tangent to OCC) is called critical field resistance RC for the shunt generator It may be defined as under The maximum field circuit resistance (for a given speed) with which the shunt generator would just excite is known as its critical field resistance It should be noted that shunt generator will build up voltage only if field circuit resistance is less than critical field resistance

Fig 118 Critical field resistance of shunt generator

119 Critical Resistance for a Series Generator

Fig 119 shows the voltage build up in a series generator Here R1 R2 etc represent the total circuit resistance (load resistance and field winding resistance) If the total circuit resistance is R1 then series generator will build up a voltage OL The line OC is tangent to OCC and represents the critical resistance RC for a series generator If the total resistance of the circuit is more than RC (say line OD) the generator will fail to build up voltage Note that Fig 119 is similar to Fig 118 with the following differences

(i) In Fig 118 R1 R2 etc represent the total field circuit resistanceHowever R1 R2 etc in Fig 119 represent the total circuit resistance(load resistance and series field winding resistance etc)

(ii) In Fig 118 field current alone is represented along X-axis However inFig 119 load current IL is represented along Y-axis Note that in aseries generator field current = load current IL

Fig 119 Critical resistance of a series generator

120 Characteristics of Series Generator Fig 120 (i) shows the connections of a series wound generator Since there is only one current (that which flows through the whole machine) the load current is the same as the exciting current(i) OCCCurve 1 shows the open circuit characteristic (OCC) of a series generator Itcan be obtained experimentally by disconnecting the field winding from themachine and exciting it from a separate dc source(ii) Internal characteristic Curve 2 shows the total or internal characteristic of a series generator It gives the relation between the generated emf E on load and armature current Due to armature reaction the flux in the machine will be less than the flux at no load Hence emf E generated under load conditions will be less than the emf E0 generated under no load conditions Consequently internal characteristic curvelies below the OCC curve the difference between them representing the effectof armature reaction [See Fig 120 (ii)]

Fig 120 Characteristics of series generator

(iii) External characteristic

Curve 3 shows the external characteristic of a series generator It gives therelation between terminal voltage and load current ILV = E g- Ia (Ra +Rse )Therefore external characteristic curve will lie below internal characteristiccurve by an amount equal to ohmic drop [ie Ia(Ra + Rse)] in the machine asshown in Fig 120 (ii) The internal and external characteristics ofa dc series generator can be plotted from one another as shown in Fig 121 Supposewe are given the internal characteristic of the generator Let the line OC represent the resistance of the whole machine ie Ra +Rse If the load current is OB drop in the machine is AB ieAB = Ohmic drop in the machine = OB(Ra + Rse)

Fig 121 Load characteristics of series generator

Now raise a perpendicular from point B and mark a point b on this line such that ab = AB Then point b will lie on the external characteristic of the generator

Following similar procedure other points of external characteristic can be located It is easy to see that we can also plot internal characteristic from the external characteristic

121 Characteristics of a Shunt Generator Fig 122 (i) show the connections of a shunt wound generator The armature current Ia splits up into two parts a small fraction Ish flowing through shunt field winding while the major part IL goes to the external load(i) OCC The OCC of a shunt generator is similar in shape to that of a series generator as shown in Fig 122 (ii) The line OA represents the shunt field circuitresistance When the generator is run at normal speed it will build up a voltageOM At no-load the terminal voltage of the generator will be constant (= OM)represented by the horizontal dotted line MC

Fig 122 Characteristics of a Shunt Generator

(ii) Internal characteristic

When the generator is loaded flux per pole is reduced due to armature reaction Therefore emf E generated on load is less than the emf generated at no load As a result the internal characteristic (EIa) drops down slightly as shown in Fig122 (ii)

(iii) External characteristic Curve 2 shows the external characteristic of a shunt generator It gives therelation between terminal voltage V and load current IL

Therefore external characteristic curve will lie below the internal characteristic curve by an amount equal to drop in the armature circuit[ie (IL + Ish)Ra] as shown in Fig 122 (ii)

Note It may be seen from the external characteristic that change in terminalvoltage from no-load to full load is small The terminal voltage can always bemaintained constant by adjusting the field rheostat R automatically122 Critical External Resistance for Shunt Generator

If the load resistance across the terminals of a shunt generator is decreased then load current increase However there is a limit to the increase in

load current with the decrease of load resistance Any decrease of load resistance beyond this point instead of increasing the current ultimately results in reduced current Consequently the external characteristic turns back (dotted curve) as shown in Fig 123 The tangent OA to the curve represents the minimum external resistance required to excite the shunt generator on load and is called critical external resistance If the resistance of the external circuit is less than the critical external resistance (represented by tangent OA in Fig 310) the machine will refuse to excite or will de-excite if already running This means that external resistance is so low as virtually to short circuit the machine and so doing away with its excitation

Note There are two critical resistances for a shunt generator viz (i) criticalfield resistance (ii) critical external resistance For the shunt generator to buildup voltage the former should not be exceeded and the latter must not be gonebelow

123 How to Draw OCC at Different Speeds

If we are given OCC of a generator at a constant speed N1 then we can easily draw the OCC at any other constant speed N2 Fig 123 illustrates theprocedure Here we are given OCC at a constant speed N1 It is desired to findthe OCC at constant speed N2 (it is assumed that n1 lt N2) For constantexcitation E propN

E2

E1 =

N 2

N1 so

E2 iquest E1

N2

N1

As shown in Fig (311) for If = OH E1 = HC Therefore the new value of emf (E2) for the same If but at N2

E2= HCN 2

N 1 = HD

Fig 123 Plotting OCC at another speed

This locates the point D on the new OCC at N2 Similarly other points can belocated taking different values of If The locus of these points will be the OCCat N2

124 Critical Speed (NC)

The critical speed of a shunt generator is the minimum speed below which it fails to excite Clearly it is the speed for which the given shunt field resistance represents the critical resistance In Fig 124 curve 2 corresponds to critical

speed because the shunt field resistance (Rsh) line is tangential to it If the generator runs at full speed N the new OCC moves upward and the Rsh line represents critical resistance for this speed

In order to find critical speed take any convenient point C on excitation axis and erect a perpendicular so as to cut Rsh and Rsh lines at points B and A respectively Then

BCAC

=NcN

Nc=N BCAC

Fig 124 Finding critical speed

125 Conditions for Voltage Build-Up of a Shunt Generator

The necessary conditions for voltage build-up in a shunt generator are

(i) There must be some residual magnetism in generator poles(ii) The connections of the field winding should be such that the field current strengthens the residual magnetism(iii) The resistance of the field circuit should be less than the critical resistance In other words the speed of the generator should be higher

than the critical speed

126 Compound Generator Characteristics

In a compound generator both series and shunt excitation are combined as shown in Fig 123 The shunt winding can be connected either across the armature only (short-shunt connection S) or across armature plus series field (long-shunt connection G) The compound generator can be cumulatively compounded or

differentially compounded generator The latter is rarely used in practice Therefore we shall discuss the characteristics of cumulatively compounded

generator It may be noted that external characteristics of long and short shunt compound generators are almost identical

External characteristic

Fig 124 shows the external characteristics of a cumulatively compoundedgenerator The series excitation aids the shunt excitation The degree of compounding depends upon the increase in series excitation with the increase inload current

(i) If series winding turns are so adjusted that with the increase in load current the terminal voltage increases it is called over-compounded generator In such a case as the load current increases the series field mmf increases and tends to increase the flux and hence the generated voltage The increase in generated voltage is greater than the IaRa drop so that instead of decreasing the terminal voltage increases as shown by curve A in Fig 124

(ii) If series winding turns are so adjusted that with the increase in load current the terminal voltage substantially remains constant it is called Flat-compounded generator The series winding of such a machine has lesser number of turns than the one in over-compounded machine and therefore does not increase the flux as much for a given load current Consequently the full-load voltage is nearly equal to the no-load voltage as indicated by curve B in Fig 124 (iii) If series field winding has lesser number of turns than for a flat compounded machine the terminal voltage falls with increase in load current as indicated by curve C m Fig 124 Such a machine is called under-compounded generator

Fig 123 compound generator

Fig 124 Load characteristics of compound generators

127 LOSSES

The losses in a dc machine (generator or motor) may be divided into three classes viz (i) copper losses or electrical losses (ii) iron or core losses and (iii) mechanical losses All these losses appear as heat and thus raise the temperature of the machine They also lower the efficiency of the machine The loss structure maybe demonstrated as

1Copperlosses

These losses occur due to currents in the various windings of the machine When an electric current of I amps flows in a resistance of R ohms the heat energy is lost at the rate of I2R joules sec and the power loss is I2R watts Generators and motors have one or more field circuits and an armature circuit in which such losses occur All resistance losses of this kind are called copper losses The different copper losses occurring in a dc machine are

(i) Armature copper loss = Ia2Ra

(ii) Shunt field copper loss = Ish2 Rsh

(iii) Series field copper loss = Ise2Rse

Armature copper loss will be around 30 to 40 of full load losses where as total field copper losses will constitute around 30 of full load losses

Note There is also brush contact loss due to brush contact resistance (ie resistance between the surface of brush and surface of commutator) This loss is generally included in armature copper loss The brush contact drop is almost independent of Ia For carbon brushes this drop is around 1 V per brush The power loss due to brush contact resistance is 2ebIa where eb is the voltage drop at one brush

2 IRONORCORELOSSES These losses occur in the armature of a dc machine and are due to the rotation of armature in the magnetic field of the poles They are of two types viz (i) hysteresis loss (ii) eddy current loss

(i) Hysteresis loss

Hysteresis loss occurs in the armature of the dc machine since any given part of the armature is subjected to magnetic field reversals as it passes under successive poles Fig 124 shows an armature rotating in two-pole machine Consider a small piece ab of the armature When the piece ab is under N-pole the magnetic lines pass from a to b Half a revolution later the same piece of iron is under S-pole and magnetic lines pass from b to a so that magnetism in the iron is reversed In order to reverse continuously the molecular magnets in the armature core some amount of power has to be spent which is called hysteresis loss It is given by Steinmetz formula This formula is

Hysteresis Loss Wh = η Bmax 16f V Watts Bmax - Maximum flux density in the core f - Frequency of magnetic reversals V - Volume of armature in m3

η - Steinmetz hysteresis coefficient

In order to reduce this loss in a dc machine armature core is made of such materials which have a low value of Steinmetz hysteresis co-efficient eg silicon steel

Fig 124 Occurrence of hysteresis loss

ii) EDDYCURRENTLOSS

In addition to the voltages induced in the armature conductors there are also voltages induced in the armature core These voltages produce circulating currents in the armature core as shown in Fig 125 These are called eddy currents and power loss due to their flow is called eddy current loss The eddy current loss appears as heat which raises the temperature of the machine and lowers its efficiency Core resistance can be greatly increased by constructing the core of thin round If a continuous solid iron core is used the resistance to eddy current path will be small due to large cross-sectional area of the core Consequently the magnitude of eddy current and hence eddy current loss will be large The magnitude of eddy current can be reduced by making core resistance as high as practical The iron sheets called laminations [See Fig 126] The laminations are insulated from each other with a coating of varnish The insulating coating has a high resistance so very little current flows from one lamination to the other Alsobecause each lamination is very thin the resistance to current flowing through the width of a lamination is also quite large Thus laminating a core increases the core resistance which decreases the eddy current and hence the eddy current loss

Fig 125 Eddy current loss Fig 126 Core laminations

Eddy current loss is biven by the formula Pe= Ke Bmax

2 f2t2 V Watts

Where Ke = Constant depending upon electrical resistace of core Bmax = Maximum flux density in wbm2

f = frequency f magnetic reversals in Hz t = thickness of laminations in m and V = volume of core in m3

It may be noted that eddy current loss depends upon the square of lamination thickness For this reason lamination thickness should be kept as small as possible

3MECHANICALLOSSES

These losses are due to friction and windage(i) Friction losses eg bearing friction brush friction etc(ii) Windage loss ie air friction of rotating armature

These losses depend upon the speed of the machine But for a given speed they are practically constant

Note Iron losses and mechanical losses together are called stray losses

CONSTANT AND VARIABLE LOSSES

The losses in a dc generator (or dc motor) may be sub-divided into

(i) constant losses (ii) variable losses

(i) Constant losses

Those losses in a dc generator which remain constant at all loads are known as constant losses The constant losses in a dc generator are(a) iron losses(b) mechanical losses(c) shunt field losses(ii) Variable lossesThose losses in a dc generator which vary with load are called variable lossesThe variable losses in a dc generator are (a) Copper loss in armature winding (Ia

2 Ra) (b) Copper loss in series field winding (Ise

2Rse)

Total losses = Constant losses + Variable lossesNote Field Cu loss is constant for shunt and compound generators

128 POWER FLOW DIAGRAM

The power flow in a dc generator may be represented as

The different power stages are shown as A B and CA ndash The mechanical power input which is the power output of the prime mover is given as Pin = 2πNT

60 watts

B - The power developed in armature is Pa = Eg Ia watts C- The electrical power output is Pout = VIL watts

129 EFFICIENCY Efficiency of a machine is the ratio of its output power to the input power

Efficiency η = Output powerInput power

= Input powerminusLossesInput power = Output power

Output power+Losses

In case of a dc generator depending upon the power stages three different efficiencies can be defined as-

1 Mechanical Efficiency(ηm)

BA = Electrical power developedisinarmature

Mechanical power input2 Electrical Efficiency(ηe)

CB = Electrical power output

Electrical power developedisinthe armature 3 Overall or commercial efficiency (η)

CA = Electrical power output

Mechanical power input = ηmtimes ηe

130 CONDITION FOR MAXIMUM EFFICIENCY The efficiency of a dc generator is not constant but varies with load Fig 127shows the variation of efficiency with load current Consider a dc shunt generator with terminal voltage V delivering a load current of IL amp Generator output = V IL

Generator input = Output + Losses = V IL + constant losses + Variable losses = V IL + Wc+ Ia

2Ra

= V IL + Wc + (IL+ Ish)2 Rsh

The shunt field current Ish is small compared with IL so may be neglected

Generator input = VIL+ IL2Ra + Wc

Now efficiency η iquest outputinput = VI L

VIL+ Ia 2Ra+Wc

Efficiency is maximum when the denominator of the above equation is minimumie

ddIL ( VI L

VIL+ Ia 2Ra+Wc ) = 0

Which gives IL2 Ra = WC

ie Variable loss = constant lossThe load current for maximum efficiency is given by

IL iquest radicWcRa

Fig 127 Variation of efficiency with load

Hence the efficiency is maximum for that load current at which the variable loss is equal to the constant loss

131 APPLICATIONS

Separately excited generators

These types of generators have special applications such as electroplating electro refining of materials because separate supply is required for field excitation

Shunt generators

These types of generators are used in battery charging and ordinary lighting purposes

Series generators

This type of generators is used commonly as boosters on DC feeders as current generators for welding and arc lamp

Cumulatively compounded generators

This type of generators is used for domestic lighting purposes and to transmit energy over long distances

Differential compound generators

This type of generators is rarely used They are used for special applications such as electric arc welding

132 NUMERICAL EXCERCISES

1 The emf induced in the armature of a 450 kW 250 volt shunt generator is

2588volt when the field current is 200 amp and the generator is supplying power to a load at rated terminal voltage The armature circuit resistance is 0005 ohm Determine (i) load current (ii) power generated (iii) power output and (iv) electrical efficiency Neglect brush contact drop

Solution

Let the load current be I amp Given Ish = 20 A

Armature current = Load current + field current or Ia = I + 20 Induced emf = Terminal voltage + Ia Ra drop

2588 = 250 + (I + 20) times 0005

So (i) Load current I = 1740 amp

(ii) Power generated E Ia = 2588 (1740 + 20) watts = 455488 kW (iii) Power output V I = 250 times 1740 watts

= 435 kW

(iv) Electrical efficiency = 435455488 times 100 = 955

2 A 4 pole wave wound dc shunt generator delivers a load of 45 kW at a terminal voltage of 220 volts Its armature has 150 single turn coils and has a resistance of 001 ohm The air gap flux per pole is 002 weber Shunt field resistance is 50 ohm Calculate the speed at which it is being driven Neglect brush contact drop

Solution Load current I = W

V = 45000220 = 20455 amp

Field Current Ish = V

R s h = 220

50 = 44 amp

Armature current Ia = I +I sh = 20455 +44 = 20895 amp

Armature drop I aRa = 20895 X 001 = 20895 volts Induced emf Eg = V + Ia Ra = 220 + 20895 = 2220895 volts

But induced emf

So N = Eg60 AΦ ZP = 2220895times 60 times2

002 times150 times2 times 4 = 111045 rpm

3 The following data refer to the OCC of a DC separately excited generator at 1000 rpm

Field current in A 00 02 03 04 05 06 07 08 09 10 11 12Armature voltage in V 5 40 75 100 124 145 162 178 188 195 200 205

The machine is now connected as a shunt generator with a total field resistance of 200 Ω and driven at 1000 rpm Estimate graphically (i) the voltage to which the generator will build up at no load (ii) the armature field and load currents when the terminal voltage is found to be 150 V Neglect the effect of armature reaction and brush drop and assume armature resistance Ra to be 08 Ω (iii) Finally estimate the steady state armature current when the machine terminals are shorted

Solution

First of all the OCC of the generator is plotted in a graph paper as shown in figure

(i) The total field circuit resistance is given to be Rf = 200 Ω The Rf line is now drawn on the same graph paper passing through the origin The point of intersection of the Rf line and the OCC decides the final no load voltage and can be read from the graph as 192 V and the corresponding field current is 096 A (ii) Since the terminal voltage V is 150 V (= BM) field current If

is OM = 077 A Generated voltage E is given by AM But we know E = IaRa + V Hence IaRa= E ndash V = AM ndash BM = AB Now from the graph AB = 25 V there4Ia = 2508 = 3125 A So load current IL= Ia ndash If= 3125 ndash 077 = 3048 A

ADDITIONAL PROBLEMS

1 A 4 pole lap wound 115 kW 230 volts dc shunt generator has armatureresistance of 02 ohm and field resistance of 100 ohms Calculate (i) emf

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES

Fig 14 Armature and field windings

135 Armature Winding Armature winding are generally formed wound These are first wound in the form of flat rectangular coils and are then pulled into their proper shape in a coil puller Various conductors of the coils are insulated from each other The conductors are placed in the armature slots which are lined with tough insulating material This slot insulation is folded over above the armature conductors placed in it and secured in place by special hard wooden or fiber wedges There are mainly two types of armature windings namely the lap winding and the wave winding which is shown in figure 15 In lap winding the armature conductors are arranged in P number of parallel paths where P is the number of poles But a wave winding offers only two parallel paths for the flow of armature current Hence a lap winding is used for high current low voltage generator where as a wave wound armature is preferred for a low current high voltage generator

Fig 15 Lap and wave winding

136 Commutator The commutator plays a vital role in dc generator It collects current from armature and sends it to the load as direct current It actually takes alternating current from armature and converts it to direct current and then send it to external load It is cylindrical structured and is build up of wedge ndash shaped segments of high conductivity hard drawn or drop forged copper Each segment is insulated from the shaft by means of insulated commutator segment shown below Each commutator segment is connected with corresponding armature conductor through segment riser or lug Fig 16 shows the arrangement of commutator segments in a dc machine along with its cross sectional view

Fig16 Commutator segments

137 Brushes The brushes are made of carbon These are rectangular block shaped The only function of these carbon brushes of dc generator is to collect current from commutator segments The brushes are housed in the rectangular box shaped brush holder As shown in figure17 the brush face is placed on the commutator segment with attached to the brush holder

Fig 17 Brush in brush holder

138 Bearing For small machine ball bearing is used and for heavy duty dc generator roller bearing is used The bearing must always be lubricated properly for smooth operation and long life of generator

14 EMF EQUATION OF A DC GENERATOR

Let Φ be the flux per pole in WeberZ the total number of armature conductors ndash No of slots x no of armature ConductorsslotP the number of polesA the number of parallel paths in the armature

N the speed of rotation of armature in revolutions per minute (rpm)E the emf induced in any parallel path in armatureEg the emf generated in any parallel path in the armatureThe emf induced per conductor = dΦdt volt (n=1)The flux cut by one conductor in one revolution dΦ = ΦP WbNo of revolutions per second = N60Time for one revolution dt = 60N secondHence EMF generatedconductor is

dφdt = φPN

60

No of conductors in a parallel path = Z A

Hence emf generated per parallal path

Eg = φPN60 Z

A

For a lap wound generator A = P and

For wave wound generator A = 2

15 TYPES OF DC GENERATORS

Fig18 Classification of dc generators

151 Separately Excited DC Generator

This dc generator has a field magnet winding which is excited using a separate voltage source (like battery)The representation in the below figure 19 The output voltage depends on the speed of rotation of armature and field current The higher the speed of rotation and current ndash the higher the output emf

Fig 19 Separately Excited DC Generators

Armature Current Ia = IL

Terminal voltage V = Eg-Ia Ra

Electric Power developed = Eg Ia

Power delivered to load = V IL = (Eg-Ia Ra) Ia = Eg Ia - Ia2

Ra

152 Self Excited DC Generator

These are generators in which the field winding is excited by the output of the generator itself As described before ndash there are three types of self excited dc generators ndash they are 1) Series 2) Shunt and 3) Compound

A series DC generator is shown below in fig110 (a) ndash in which the armature winding is connected in series with the field winding so that the field current flows through the load as well as the field winding Field winding is a low resistance thick wire of few turns Armature current Ia =Ise = IL=I (say)

Terminal voltage V = Eg- I( Ra+ Rse)Power Developed in armature = Eg Ia

Power delivered to load = V IL = Eg Ia- Ia2 (Ra+Rse)

Fig110 (a) Series generator and (b) shunt generator

A shunt DC generator is shown in figure 110 (b) in which the field winding is wired parallel to armature winding so that the voltage across both are same The field winding has high resistance and more number of turns so that only a part of armature current passes through field winding and the rest passes through load Shunt field current Ish = V Rsh

Armature current Ia = IL+ Ish

Terminal voltage V= Eg-Ia Ra

Power developed in armature = EgIa

Power delivered to load = VIL A compound generator is shown in figure 111 below It has two field findings namely Rsh and Rse They are basically shunt winding (Rsh) and series winding (Rse) Compound generator is of two types ndash 1) Short shunt and 2) Long shunt

Short shunt- Here the shunt field winding is wired parallel to armature and series field winding is connected in series to the load It is shown in fig111(a)

In short shunt machines Series field current Ise = IL

Shunt field current Ish iquest V+ I se RseRsh

Terminal voltage V= Eg-Ia Ra-Ise Rse

Power developedin armature = Eg Ia

Power delivered to load = V IL

Fig 111 (a) short shunt generator and (b) Long shunt generator

Long shunt- Here the shunt field winding is parallel to both armature and series field winding (Rse is wired in series to the armature) It is shown in figure 111 (b)

In Long shunt machines Series field current Ise = Ia = IL+ Ish

Shunt field current Ish = V R sh

Terminal voltage V = Eg- Ia (Ra+ Rse) Power developed in armature = Eg Ia

Power delivered to load = VIL

16 GENERATOR CHARACTERISTICS

The speed of a dc machine operated as a generator is fixed by the prime mover For general-purpose operation the prime mover is equipped with a speed governor so that the speed of the generator is practically constant Under such condition the generator performance deals primarily with the relation between excitation terminal voltage and load These relations can be best exhibited graphically by means of curves known as generator characteristics Thesecharacteristics show at a glance the behavior of the generator under differentload conditions

The following are the three most important characteristics of a dc generator161 Open Circuit Characteristic (OCC)

This curve shows the relation between the generated emf at no-load (E0) and the field current (If) at constant speed It is also known as magnetic characteristic or no-load saturation curve Its shape is practically the same for all generators whether separately or self-excited The data for OCC curve are obtained experimentally by operating the generator at no load and constant speed and recording the change in terminal voltage as the field current is varied

162 Internal or Total characteristic (EIa)

This curve shows the relation between the generated emf on load (E) and the armature current (Ia) The emf E is less than E0 due to the demagnetizing effect of armature reaction Therefore this curve will lie below the open circuitcharacteristic (OCC) The internal characteristic is of interest chiefly to thedesigner It cannot be obtained directly by experiment It is because a voltmetercannot read the emf generated on load due to the voltage drop in armatureresistance The internal characteristic can be obtained from externalcharacteristics if winding resistances are known because armature reaction effectis included in both characteristics163 External characteristic (VIL)This curve shows the relation between the terminal voltage (V) and load current(IL) The terminal voltage V will be less than E due to voltage drop in thearmature circuit Therefore this curve will lie below the internal characteristicThis characteristic is very important in determining the suitability of a generatorfor a given purpose It can be obtained by making simultaneous measurements

of terminal voltage and load current (with voltmeter and ammeter) of a loadedgenerator164 Open Circuit Characteristic of a DC Generator

The OCC for a dc generator is determined as follows The field winding of the dc generator (series or shunt) is disconnected from the machine and isseparately excited from an external dc source as shown in Fig 112 Thegenerator is run at fixed speed (ie normal speed) The field current (If) isincreased from zero in steps and the corresponding values of generated emf(E0) read off on a voltmeter connected across the armature terminals On plottingthe relation between E0 and If we get the open circuit characteristic as shown in

Fig113

Fig 112 Experimental setup for plotting OCC

Fig 113 OCC of dc generator

The following points may be noted from OCC When the field current is zero there is some generated emf OA

This is due to the residual magnetism in the field poles Over a fairly wide range of field current (upto point B in the curve)

the curve is linear It is because in this range reluctance of iron is negligible as compared with that of air gap The air gap reluctance is constant and hence linear relationship After point B on the curve the reluctance of iron also comes into picture and reluctance of iron is no longer negligible Consequently the curve deviates from linear relationship

After point C on the curve the magnetic saturation of poles begins and E0 tends to level off

The OCC of even self-excited generator is obtained by running it as a separately excited generator

165 Characteristics of a Separately Excited DCGenerator

The obvious disadvantage of a separately excited dc generator is that we require an external dc source for excitation But since the output voltage may be

controlled more easily and over a wide range (from zero to a maximum) this type of excitation finds many applications

(i) Open circuit characteristic

The OCC of a separately excited generator is determined in a manner described in Sec 164 Fig114 shows the variation of generated em f on noload with field current for various fixed speeds Note that if the value of constant speed is increased the steepness of the curve also increases When the field current is zero the residual magnetism in the poles will give rise to the small initial emf as shown

Fig114 Variation of OCC with speed

Internal and External Characteristics (Load Characteristics)

The external characteristic of a separately excited generator is the curve between the terminal voltage (V) and the load current IL (which is the same as armature current in this case) In order to determine the external characteristic the circuit set up is as shown in Fig 115

Fig 115 Experimental setup to plot load characteristics

As the load current increases the terminal voltage falls due to two reasons(a) The armature reaction weakens the main flux so that actual emf generated E on load is less than that generated (E0) on no load

(b) There is voltage drop across armature resistance (= ILRa = IaRa)Due to these reasons the external characteristic is a drooping curve [curve 3 inFig 116 ] Note that in the absence of armature reaction and armature dropthe generated emf would have been E0 (curve 1)The internal characteristic can be determined from external characteristic by adding IL Ra drop to the external characteristic It is because armature reaction drop is included in the external characteristic Curve 2 is the internal characteristic of the generator and should obviously lie above the external characteristic

Fig 116 Load characteristics of a separately excited dcgenerator

117 Voltage Build-Up in a Self-Excited GeneratorLet us see how voltage builds up in a self-excited generator(i) Shunt generator Consider a shunt generator If the generator is run at a constant speed some emf will be generated due to residual magnetism in the main poles This small emf circulates a field current which in turn produces additional flux toreinforce the original residual flux (provided field winding connections arecorrect) This process continues and the generator builds up the normalgenerated voltage following the OCC shown in Fig 117 (i) The field resistance Rf can be represented by a straight line passing through the origin as shown in Fig 117 (ii) The two curves can be shown on the same diagram as they have the same ordinate as in Fig 117 (iii)

Fig 117

Since the field circuit is inductive there is a delay in the increase in current uponclosing the field circuit switch The rate at which the current increases dependsupon the voltage available for increasing it Suppose at any instant the fieldcurrent is i (= OA) and is increasing at the rate didt Then

E0=i R f +iquest L didt

Where Rf = total field circuit resistance L = inductance of field circuit

At the considered instant the total emf available is AC [See Fig 117 (iii)] Anamount AB of the cmf AC is absorbed by the voltage drop iRf and theremainder part BC is available to overcome L didt Since this surplus voltage isavailable it is possible for the field current to increase above the value OAHowever at point D the available voltage is OM and is all absorbed by i RfD

rop Consequently the field current cannot increase further and the generatorbuild up stops Hence we arrive at a very important conclusion that the voltage build up of the generator is given by the point of intersection of OCC and field resistance line Thus in Fig 117 (iii) D is point of intersection of the two curves Hence the generator will build up a voltage OM(ii) Series generator During initial operation with no current yet flowing a residual voltage will be generated exactly as in the case of a shunt generator The residual voltage will cause a current to flow through the whole series circuit when the circuit isclosed There will then be voltage build up to an equilibrium point exactlyanalogous to the buildup of a shunt generator The voltage build up graph willbe similar to that of shunt generator except that now load current (instead offield current for shunt generator) will be taken along x-axis10487871048787(iii) Compound generator When a compound generator has its series field flux aiding its shunt field flux the machine is said to be cumulative compound When the series field isconnected in reverse so that its field flux opposes the shunt field flux thegenerator is then differential compound The easiest way to build up voltage in a compound generator is to start under no load conditions At no load only the shunt field is effective When no-load voltage build up is achieved the generator is loaded If under load the voltagerises the series field connection is cumulative If the voltage drops significantlythe connection is differential compound118 Critical Field Resistance for a Shunt Generator The voltage build up in a shunt generator depends upon field circuit resistance If the field circuit resistance is R1 (line OA) then generator will build up a voltage OM as shown in Fig 118 If the field circuit resistance is increasedto R2 (tine OB) the generator will build up a voltage OL slightly less than OM As the field circuit resistance is increased the slope of resistance line also increases When the field resistance line becomes tangent (line OC) to OCC the generator would just excite If the field circuit resistance is increased beyond this point (say line OD) the generator will fail to excite The field circuit resistance represented by line OC (tangent to OCC) is called critical field resistance RC for the shunt generator It may be defined as under The maximum field circuit resistance (for a given speed) with which the shunt generator would just excite is known as its critical field resistance It should be noted that shunt generator will build up voltage only if field circuit resistance is less than critical field resistance

Fig 118 Critical field resistance of shunt generator

119 Critical Resistance for a Series Generator

Fig 119 shows the voltage build up in a series generator Here R1 R2 etc represent the total circuit resistance (load resistance and field winding resistance) If the total circuit resistance is R1 then series generator will build up a voltage OL The line OC is tangent to OCC and represents the critical resistance RC for a series generator If the total resistance of the circuit is more than RC (say line OD) the generator will fail to build up voltage Note that Fig 119 is similar to Fig 118 with the following differences

(i) In Fig 118 R1 R2 etc represent the total field circuit resistanceHowever R1 R2 etc in Fig 119 represent the total circuit resistance(load resistance and series field winding resistance etc)

(ii) In Fig 118 field current alone is represented along X-axis However inFig 119 load current IL is represented along Y-axis Note that in aseries generator field current = load current IL

Fig 119 Critical resistance of a series generator

120 Characteristics of Series Generator Fig 120 (i) shows the connections of a series wound generator Since there is only one current (that which flows through the whole machine) the load current is the same as the exciting current(i) OCCCurve 1 shows the open circuit characteristic (OCC) of a series generator Itcan be obtained experimentally by disconnecting the field winding from themachine and exciting it from a separate dc source(ii) Internal characteristic Curve 2 shows the total or internal characteristic of a series generator It gives the relation between the generated emf E on load and armature current Due to armature reaction the flux in the machine will be less than the flux at no load Hence emf E generated under load conditions will be less than the emf E0 generated under no load conditions Consequently internal characteristic curvelies below the OCC curve the difference between them representing the effectof armature reaction [See Fig 120 (ii)]

Fig 120 Characteristics of series generator

(iii) External characteristic

Curve 3 shows the external characteristic of a series generator It gives therelation between terminal voltage and load current ILV = E g- Ia (Ra +Rse )Therefore external characteristic curve will lie below internal characteristiccurve by an amount equal to ohmic drop [ie Ia(Ra + Rse)] in the machine asshown in Fig 120 (ii) The internal and external characteristics ofa dc series generator can be plotted from one another as shown in Fig 121 Supposewe are given the internal characteristic of the generator Let the line OC represent the resistance of the whole machine ie Ra +Rse If the load current is OB drop in the machine is AB ieAB = Ohmic drop in the machine = OB(Ra + Rse)

Fig 121 Load characteristics of series generator

Now raise a perpendicular from point B and mark a point b on this line such that ab = AB Then point b will lie on the external characteristic of the generator

Following similar procedure other points of external characteristic can be located It is easy to see that we can also plot internal characteristic from the external characteristic

121 Characteristics of a Shunt Generator Fig 122 (i) show the connections of a shunt wound generator The armature current Ia splits up into two parts a small fraction Ish flowing through shunt field winding while the major part IL goes to the external load(i) OCC The OCC of a shunt generator is similar in shape to that of a series generator as shown in Fig 122 (ii) The line OA represents the shunt field circuitresistance When the generator is run at normal speed it will build up a voltageOM At no-load the terminal voltage of the generator will be constant (= OM)represented by the horizontal dotted line MC

Fig 122 Characteristics of a Shunt Generator

(ii) Internal characteristic

When the generator is loaded flux per pole is reduced due to armature reaction Therefore emf E generated on load is less than the emf generated at no load As a result the internal characteristic (EIa) drops down slightly as shown in Fig122 (ii)

(iii) External characteristic Curve 2 shows the external characteristic of a shunt generator It gives therelation between terminal voltage V and load current IL

Therefore external characteristic curve will lie below the internal characteristic curve by an amount equal to drop in the armature circuit[ie (IL + Ish)Ra] as shown in Fig 122 (ii)

Note It may be seen from the external characteristic that change in terminalvoltage from no-load to full load is small The terminal voltage can always bemaintained constant by adjusting the field rheostat R automatically122 Critical External Resistance for Shunt Generator

If the load resistance across the terminals of a shunt generator is decreased then load current increase However there is a limit to the increase in

load current with the decrease of load resistance Any decrease of load resistance beyond this point instead of increasing the current ultimately results in reduced current Consequently the external characteristic turns back (dotted curve) as shown in Fig 123 The tangent OA to the curve represents the minimum external resistance required to excite the shunt generator on load and is called critical external resistance If the resistance of the external circuit is less than the critical external resistance (represented by tangent OA in Fig 310) the machine will refuse to excite or will de-excite if already running This means that external resistance is so low as virtually to short circuit the machine and so doing away with its excitation

Note There are two critical resistances for a shunt generator viz (i) criticalfield resistance (ii) critical external resistance For the shunt generator to buildup voltage the former should not be exceeded and the latter must not be gonebelow

123 How to Draw OCC at Different Speeds

If we are given OCC of a generator at a constant speed N1 then we can easily draw the OCC at any other constant speed N2 Fig 123 illustrates theprocedure Here we are given OCC at a constant speed N1 It is desired to findthe OCC at constant speed N2 (it is assumed that n1 lt N2) For constantexcitation E propN

E2

E1 =

N 2

N1 so

E2 iquest E1

N2

N1

As shown in Fig (311) for If = OH E1 = HC Therefore the new value of emf (E2) for the same If but at N2

E2= HCN 2

N 1 = HD

Fig 123 Plotting OCC at another speed

This locates the point D on the new OCC at N2 Similarly other points can belocated taking different values of If The locus of these points will be the OCCat N2

124 Critical Speed (NC)

The critical speed of a shunt generator is the minimum speed below which it fails to excite Clearly it is the speed for which the given shunt field resistance represents the critical resistance In Fig 124 curve 2 corresponds to critical

speed because the shunt field resistance (Rsh) line is tangential to it If the generator runs at full speed N the new OCC moves upward and the Rsh line represents critical resistance for this speed

In order to find critical speed take any convenient point C on excitation axis and erect a perpendicular so as to cut Rsh and Rsh lines at points B and A respectively Then

BCAC

=NcN

Nc=N BCAC

Fig 124 Finding critical speed

125 Conditions for Voltage Build-Up of a Shunt Generator

The necessary conditions for voltage build-up in a shunt generator are

(i) There must be some residual magnetism in generator poles(ii) The connections of the field winding should be such that the field current strengthens the residual magnetism(iii) The resistance of the field circuit should be less than the critical resistance In other words the speed of the generator should be higher

than the critical speed

126 Compound Generator Characteristics

In a compound generator both series and shunt excitation are combined as shown in Fig 123 The shunt winding can be connected either across the armature only (short-shunt connection S) or across armature plus series field (long-shunt connection G) The compound generator can be cumulatively compounded or

differentially compounded generator The latter is rarely used in practice Therefore we shall discuss the characteristics of cumulatively compounded

generator It may be noted that external characteristics of long and short shunt compound generators are almost identical

External characteristic

Fig 124 shows the external characteristics of a cumulatively compoundedgenerator The series excitation aids the shunt excitation The degree of compounding depends upon the increase in series excitation with the increase inload current

(i) If series winding turns are so adjusted that with the increase in load current the terminal voltage increases it is called over-compounded generator In such a case as the load current increases the series field mmf increases and tends to increase the flux and hence the generated voltage The increase in generated voltage is greater than the IaRa drop so that instead of decreasing the terminal voltage increases as shown by curve A in Fig 124

(ii) If series winding turns are so adjusted that with the increase in load current the terminal voltage substantially remains constant it is called Flat-compounded generator The series winding of such a machine has lesser number of turns than the one in over-compounded machine and therefore does not increase the flux as much for a given load current Consequently the full-load voltage is nearly equal to the no-load voltage as indicated by curve B in Fig 124 (iii) If series field winding has lesser number of turns than for a flat compounded machine the terminal voltage falls with increase in load current as indicated by curve C m Fig 124 Such a machine is called under-compounded generator

Fig 123 compound generator

Fig 124 Load characteristics of compound generators

127 LOSSES

The losses in a dc machine (generator or motor) may be divided into three classes viz (i) copper losses or electrical losses (ii) iron or core losses and (iii) mechanical losses All these losses appear as heat and thus raise the temperature of the machine They also lower the efficiency of the machine The loss structure maybe demonstrated as

1Copperlosses

These losses occur due to currents in the various windings of the machine When an electric current of I amps flows in a resistance of R ohms the heat energy is lost at the rate of I2R joules sec and the power loss is I2R watts Generators and motors have one or more field circuits and an armature circuit in which such losses occur All resistance losses of this kind are called copper losses The different copper losses occurring in a dc machine are

(i) Armature copper loss = Ia2Ra

(ii) Shunt field copper loss = Ish2 Rsh

(iii) Series field copper loss = Ise2Rse

Armature copper loss will be around 30 to 40 of full load losses where as total field copper losses will constitute around 30 of full load losses

Note There is also brush contact loss due to brush contact resistance (ie resistance between the surface of brush and surface of commutator) This loss is generally included in armature copper loss The brush contact drop is almost independent of Ia For carbon brushes this drop is around 1 V per brush The power loss due to brush contact resistance is 2ebIa where eb is the voltage drop at one brush

2 IRONORCORELOSSES These losses occur in the armature of a dc machine and are due to the rotation of armature in the magnetic field of the poles They are of two types viz (i) hysteresis loss (ii) eddy current loss

(i) Hysteresis loss

Hysteresis loss occurs in the armature of the dc machine since any given part of the armature is subjected to magnetic field reversals as it passes under successive poles Fig 124 shows an armature rotating in two-pole machine Consider a small piece ab of the armature When the piece ab is under N-pole the magnetic lines pass from a to b Half a revolution later the same piece of iron is under S-pole and magnetic lines pass from b to a so that magnetism in the iron is reversed In order to reverse continuously the molecular magnets in the armature core some amount of power has to be spent which is called hysteresis loss It is given by Steinmetz formula This formula is

Hysteresis Loss Wh = η Bmax 16f V Watts Bmax - Maximum flux density in the core f - Frequency of magnetic reversals V - Volume of armature in m3

η - Steinmetz hysteresis coefficient

In order to reduce this loss in a dc machine armature core is made of such materials which have a low value of Steinmetz hysteresis co-efficient eg silicon steel

Fig 124 Occurrence of hysteresis loss

ii) EDDYCURRENTLOSS

In addition to the voltages induced in the armature conductors there are also voltages induced in the armature core These voltages produce circulating currents in the armature core as shown in Fig 125 These are called eddy currents and power loss due to their flow is called eddy current loss The eddy current loss appears as heat which raises the temperature of the machine and lowers its efficiency Core resistance can be greatly increased by constructing the core of thin round If a continuous solid iron core is used the resistance to eddy current path will be small due to large cross-sectional area of the core Consequently the magnitude of eddy current and hence eddy current loss will be large The magnitude of eddy current can be reduced by making core resistance as high as practical The iron sheets called laminations [See Fig 126] The laminations are insulated from each other with a coating of varnish The insulating coating has a high resistance so very little current flows from one lamination to the other Alsobecause each lamination is very thin the resistance to current flowing through the width of a lamination is also quite large Thus laminating a core increases the core resistance which decreases the eddy current and hence the eddy current loss

Fig 125 Eddy current loss Fig 126 Core laminations

Eddy current loss is biven by the formula Pe= Ke Bmax

2 f2t2 V Watts

Where Ke = Constant depending upon electrical resistace of core Bmax = Maximum flux density in wbm2

f = frequency f magnetic reversals in Hz t = thickness of laminations in m and V = volume of core in m3

It may be noted that eddy current loss depends upon the square of lamination thickness For this reason lamination thickness should be kept as small as possible

3MECHANICALLOSSES

These losses are due to friction and windage(i) Friction losses eg bearing friction brush friction etc(ii) Windage loss ie air friction of rotating armature

These losses depend upon the speed of the machine But for a given speed they are practically constant

Note Iron losses and mechanical losses together are called stray losses

CONSTANT AND VARIABLE LOSSES

The losses in a dc generator (or dc motor) may be sub-divided into

(i) constant losses (ii) variable losses

(i) Constant losses

Those losses in a dc generator which remain constant at all loads are known as constant losses The constant losses in a dc generator are(a) iron losses(b) mechanical losses(c) shunt field losses(ii) Variable lossesThose losses in a dc generator which vary with load are called variable lossesThe variable losses in a dc generator are (a) Copper loss in armature winding (Ia

2 Ra) (b) Copper loss in series field winding (Ise

2Rse)

Total losses = Constant losses + Variable lossesNote Field Cu loss is constant for shunt and compound generators

128 POWER FLOW DIAGRAM

The power flow in a dc generator may be represented as

The different power stages are shown as A B and CA ndash The mechanical power input which is the power output of the prime mover is given as Pin = 2πNT

60 watts

B - The power developed in armature is Pa = Eg Ia watts C- The electrical power output is Pout = VIL watts

129 EFFICIENCY Efficiency of a machine is the ratio of its output power to the input power

Efficiency η = Output powerInput power

= Input powerminusLossesInput power = Output power

Output power+Losses

In case of a dc generator depending upon the power stages three different efficiencies can be defined as-

1 Mechanical Efficiency(ηm)

BA = Electrical power developedisinarmature

Mechanical power input2 Electrical Efficiency(ηe)

CB = Electrical power output

Electrical power developedisinthe armature 3 Overall or commercial efficiency (η)

CA = Electrical power output

Mechanical power input = ηmtimes ηe

130 CONDITION FOR MAXIMUM EFFICIENCY The efficiency of a dc generator is not constant but varies with load Fig 127shows the variation of efficiency with load current Consider a dc shunt generator with terminal voltage V delivering a load current of IL amp Generator output = V IL

Generator input = Output + Losses = V IL + constant losses + Variable losses = V IL + Wc+ Ia

2Ra

= V IL + Wc + (IL+ Ish)2 Rsh

The shunt field current Ish is small compared with IL so may be neglected

Generator input = VIL+ IL2Ra + Wc

Now efficiency η iquest outputinput = VI L

VIL+ Ia 2Ra+Wc

Efficiency is maximum when the denominator of the above equation is minimumie

ddIL ( VI L

VIL+ Ia 2Ra+Wc ) = 0

Which gives IL2 Ra = WC

ie Variable loss = constant lossThe load current for maximum efficiency is given by

IL iquest radicWcRa

Fig 127 Variation of efficiency with load

Hence the efficiency is maximum for that load current at which the variable loss is equal to the constant loss

131 APPLICATIONS

Separately excited generators

These types of generators have special applications such as electroplating electro refining of materials because separate supply is required for field excitation

Shunt generators

These types of generators are used in battery charging and ordinary lighting purposes

Series generators

This type of generators is used commonly as boosters on DC feeders as current generators for welding and arc lamp

Cumulatively compounded generators

This type of generators is used for domestic lighting purposes and to transmit energy over long distances

Differential compound generators

This type of generators is rarely used They are used for special applications such as electric arc welding

132 NUMERICAL EXCERCISES

1 The emf induced in the armature of a 450 kW 250 volt shunt generator is

2588volt when the field current is 200 amp and the generator is supplying power to a load at rated terminal voltage The armature circuit resistance is 0005 ohm Determine (i) load current (ii) power generated (iii) power output and (iv) electrical efficiency Neglect brush contact drop

Solution

Let the load current be I amp Given Ish = 20 A

Armature current = Load current + field current or Ia = I + 20 Induced emf = Terminal voltage + Ia Ra drop

2588 = 250 + (I + 20) times 0005

So (i) Load current I = 1740 amp

(ii) Power generated E Ia = 2588 (1740 + 20) watts = 455488 kW (iii) Power output V I = 250 times 1740 watts

= 435 kW

(iv) Electrical efficiency = 435455488 times 100 = 955

2 A 4 pole wave wound dc shunt generator delivers a load of 45 kW at a terminal voltage of 220 volts Its armature has 150 single turn coils and has a resistance of 001 ohm The air gap flux per pole is 002 weber Shunt field resistance is 50 ohm Calculate the speed at which it is being driven Neglect brush contact drop

Solution Load current I = W

V = 45000220 = 20455 amp

Field Current Ish = V

R s h = 220

50 = 44 amp

Armature current Ia = I +I sh = 20455 +44 = 20895 amp

Armature drop I aRa = 20895 X 001 = 20895 volts Induced emf Eg = V + Ia Ra = 220 + 20895 = 2220895 volts

But induced emf

So N = Eg60 AΦ ZP = 2220895times 60 times2

002 times150 times2 times 4 = 111045 rpm

3 The following data refer to the OCC of a DC separately excited generator at 1000 rpm

Field current in A 00 02 03 04 05 06 07 08 09 10 11 12Armature voltage in V 5 40 75 100 124 145 162 178 188 195 200 205

The machine is now connected as a shunt generator with a total field resistance of 200 Ω and driven at 1000 rpm Estimate graphically (i) the voltage to which the generator will build up at no load (ii) the armature field and load currents when the terminal voltage is found to be 150 V Neglect the effect of armature reaction and brush drop and assume armature resistance Ra to be 08 Ω (iii) Finally estimate the steady state armature current when the machine terminals are shorted

Solution

First of all the OCC of the generator is plotted in a graph paper as shown in figure

(i) The total field circuit resistance is given to be Rf = 200 Ω The Rf line is now drawn on the same graph paper passing through the origin The point of intersection of the Rf line and the OCC decides the final no load voltage and can be read from the graph as 192 V and the corresponding field current is 096 A (ii) Since the terminal voltage V is 150 V (= BM) field current If

is OM = 077 A Generated voltage E is given by AM But we know E = IaRa + V Hence IaRa= E ndash V = AM ndash BM = AB Now from the graph AB = 25 V there4Ia = 2508 = 3125 A So load current IL= Ia ndash If= 3125 ndash 077 = 3048 A

ADDITIONAL PROBLEMS

1 A 4 pole lap wound 115 kW 230 volts dc shunt generator has armatureresistance of 02 ohm and field resistance of 100 ohms Calculate (i) emf

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES

Fig 15 Lap and wave winding

136 Commutator The commutator plays a vital role in dc generator It collects current from armature and sends it to the load as direct current It actually takes alternating current from armature and converts it to direct current and then send it to external load It is cylindrical structured and is build up of wedge ndash shaped segments of high conductivity hard drawn or drop forged copper Each segment is insulated from the shaft by means of insulated commutator segment shown below Each commutator segment is connected with corresponding armature conductor through segment riser or lug Fig 16 shows the arrangement of commutator segments in a dc machine along with its cross sectional view

Fig16 Commutator segments

137 Brushes The brushes are made of carbon These are rectangular block shaped The only function of these carbon brushes of dc generator is to collect current from commutator segments The brushes are housed in the rectangular box shaped brush holder As shown in figure17 the brush face is placed on the commutator segment with attached to the brush holder

Fig 17 Brush in brush holder

138 Bearing For small machine ball bearing is used and for heavy duty dc generator roller bearing is used The bearing must always be lubricated properly for smooth operation and long life of generator

14 EMF EQUATION OF A DC GENERATOR

Let Φ be the flux per pole in WeberZ the total number of armature conductors ndash No of slots x no of armature ConductorsslotP the number of polesA the number of parallel paths in the armature

N the speed of rotation of armature in revolutions per minute (rpm)E the emf induced in any parallel path in armatureEg the emf generated in any parallel path in the armatureThe emf induced per conductor = dΦdt volt (n=1)The flux cut by one conductor in one revolution dΦ = ΦP WbNo of revolutions per second = N60Time for one revolution dt = 60N secondHence EMF generatedconductor is

dφdt = φPN

60

No of conductors in a parallel path = Z A

Hence emf generated per parallal path

Eg = φPN60 Z

A

For a lap wound generator A = P and

For wave wound generator A = 2

15 TYPES OF DC GENERATORS

Fig18 Classification of dc generators

151 Separately Excited DC Generator

This dc generator has a field magnet winding which is excited using a separate voltage source (like battery)The representation in the below figure 19 The output voltage depends on the speed of rotation of armature and field current The higher the speed of rotation and current ndash the higher the output emf

Fig 19 Separately Excited DC Generators

Armature Current Ia = IL

Terminal voltage V = Eg-Ia Ra

Electric Power developed = Eg Ia

Power delivered to load = V IL = (Eg-Ia Ra) Ia = Eg Ia - Ia2

Ra

152 Self Excited DC Generator

These are generators in which the field winding is excited by the output of the generator itself As described before ndash there are three types of self excited dc generators ndash they are 1) Series 2) Shunt and 3) Compound

A series DC generator is shown below in fig110 (a) ndash in which the armature winding is connected in series with the field winding so that the field current flows through the load as well as the field winding Field winding is a low resistance thick wire of few turns Armature current Ia =Ise = IL=I (say)

Terminal voltage V = Eg- I( Ra+ Rse)Power Developed in armature = Eg Ia

Power delivered to load = V IL = Eg Ia- Ia2 (Ra+Rse)

Fig110 (a) Series generator and (b) shunt generator

A shunt DC generator is shown in figure 110 (b) in which the field winding is wired parallel to armature winding so that the voltage across both are same The field winding has high resistance and more number of turns so that only a part of armature current passes through field winding and the rest passes through load Shunt field current Ish = V Rsh

Armature current Ia = IL+ Ish

Terminal voltage V= Eg-Ia Ra

Power developed in armature = EgIa

Power delivered to load = VIL A compound generator is shown in figure 111 below It has two field findings namely Rsh and Rse They are basically shunt winding (Rsh) and series winding (Rse) Compound generator is of two types ndash 1) Short shunt and 2) Long shunt

Short shunt- Here the shunt field winding is wired parallel to armature and series field winding is connected in series to the load It is shown in fig111(a)

In short shunt machines Series field current Ise = IL

Shunt field current Ish iquest V+ I se RseRsh

Terminal voltage V= Eg-Ia Ra-Ise Rse

Power developedin armature = Eg Ia

Power delivered to load = V IL

Fig 111 (a) short shunt generator and (b) Long shunt generator

Long shunt- Here the shunt field winding is parallel to both armature and series field winding (Rse is wired in series to the armature) It is shown in figure 111 (b)

In Long shunt machines Series field current Ise = Ia = IL+ Ish

Shunt field current Ish = V R sh

Terminal voltage V = Eg- Ia (Ra+ Rse) Power developed in armature = Eg Ia

Power delivered to load = VIL

16 GENERATOR CHARACTERISTICS

The speed of a dc machine operated as a generator is fixed by the prime mover For general-purpose operation the prime mover is equipped with a speed governor so that the speed of the generator is practically constant Under such condition the generator performance deals primarily with the relation between excitation terminal voltage and load These relations can be best exhibited graphically by means of curves known as generator characteristics Thesecharacteristics show at a glance the behavior of the generator under differentload conditions

The following are the three most important characteristics of a dc generator161 Open Circuit Characteristic (OCC)

This curve shows the relation between the generated emf at no-load (E0) and the field current (If) at constant speed It is also known as magnetic characteristic or no-load saturation curve Its shape is practically the same for all generators whether separately or self-excited The data for OCC curve are obtained experimentally by operating the generator at no load and constant speed and recording the change in terminal voltage as the field current is varied

162 Internal or Total characteristic (EIa)

This curve shows the relation between the generated emf on load (E) and the armature current (Ia) The emf E is less than E0 due to the demagnetizing effect of armature reaction Therefore this curve will lie below the open circuitcharacteristic (OCC) The internal characteristic is of interest chiefly to thedesigner It cannot be obtained directly by experiment It is because a voltmetercannot read the emf generated on load due to the voltage drop in armatureresistance The internal characteristic can be obtained from externalcharacteristics if winding resistances are known because armature reaction effectis included in both characteristics163 External characteristic (VIL)This curve shows the relation between the terminal voltage (V) and load current(IL) The terminal voltage V will be less than E due to voltage drop in thearmature circuit Therefore this curve will lie below the internal characteristicThis characteristic is very important in determining the suitability of a generatorfor a given purpose It can be obtained by making simultaneous measurements

of terminal voltage and load current (with voltmeter and ammeter) of a loadedgenerator164 Open Circuit Characteristic of a DC Generator

The OCC for a dc generator is determined as follows The field winding of the dc generator (series or shunt) is disconnected from the machine and isseparately excited from an external dc source as shown in Fig 112 Thegenerator is run at fixed speed (ie normal speed) The field current (If) isincreased from zero in steps and the corresponding values of generated emf(E0) read off on a voltmeter connected across the armature terminals On plottingthe relation between E0 and If we get the open circuit characteristic as shown in

Fig113

Fig 112 Experimental setup for plotting OCC

Fig 113 OCC of dc generator

The following points may be noted from OCC When the field current is zero there is some generated emf OA

This is due to the residual magnetism in the field poles Over a fairly wide range of field current (upto point B in the curve)

the curve is linear It is because in this range reluctance of iron is negligible as compared with that of air gap The air gap reluctance is constant and hence linear relationship After point B on the curve the reluctance of iron also comes into picture and reluctance of iron is no longer negligible Consequently the curve deviates from linear relationship

After point C on the curve the magnetic saturation of poles begins and E0 tends to level off

The OCC of even self-excited generator is obtained by running it as a separately excited generator

165 Characteristics of a Separately Excited DCGenerator

The obvious disadvantage of a separately excited dc generator is that we require an external dc source for excitation But since the output voltage may be

controlled more easily and over a wide range (from zero to a maximum) this type of excitation finds many applications

(i) Open circuit characteristic

The OCC of a separately excited generator is determined in a manner described in Sec 164 Fig114 shows the variation of generated em f on noload with field current for various fixed speeds Note that if the value of constant speed is increased the steepness of the curve also increases When the field current is zero the residual magnetism in the poles will give rise to the small initial emf as shown

Fig114 Variation of OCC with speed

Internal and External Characteristics (Load Characteristics)

The external characteristic of a separately excited generator is the curve between the terminal voltage (V) and the load current IL (which is the same as armature current in this case) In order to determine the external characteristic the circuit set up is as shown in Fig 115

Fig 115 Experimental setup to plot load characteristics

As the load current increases the terminal voltage falls due to two reasons(a) The armature reaction weakens the main flux so that actual emf generated E on load is less than that generated (E0) on no load

(b) There is voltage drop across armature resistance (= ILRa = IaRa)Due to these reasons the external characteristic is a drooping curve [curve 3 inFig 116 ] Note that in the absence of armature reaction and armature dropthe generated emf would have been E0 (curve 1)The internal characteristic can be determined from external characteristic by adding IL Ra drop to the external characteristic It is because armature reaction drop is included in the external characteristic Curve 2 is the internal characteristic of the generator and should obviously lie above the external characteristic

Fig 116 Load characteristics of a separately excited dcgenerator

117 Voltage Build-Up in a Self-Excited GeneratorLet us see how voltage builds up in a self-excited generator(i) Shunt generator Consider a shunt generator If the generator is run at a constant speed some emf will be generated due to residual magnetism in the main poles This small emf circulates a field current which in turn produces additional flux toreinforce the original residual flux (provided field winding connections arecorrect) This process continues and the generator builds up the normalgenerated voltage following the OCC shown in Fig 117 (i) The field resistance Rf can be represented by a straight line passing through the origin as shown in Fig 117 (ii) The two curves can be shown on the same diagram as they have the same ordinate as in Fig 117 (iii)

Fig 117

Since the field circuit is inductive there is a delay in the increase in current uponclosing the field circuit switch The rate at which the current increases dependsupon the voltage available for increasing it Suppose at any instant the fieldcurrent is i (= OA) and is increasing at the rate didt Then

E0=i R f +iquest L didt

Where Rf = total field circuit resistance L = inductance of field circuit

At the considered instant the total emf available is AC [See Fig 117 (iii)] Anamount AB of the cmf AC is absorbed by the voltage drop iRf and theremainder part BC is available to overcome L didt Since this surplus voltage isavailable it is possible for the field current to increase above the value OAHowever at point D the available voltage is OM and is all absorbed by i RfD

rop Consequently the field current cannot increase further and the generatorbuild up stops Hence we arrive at a very important conclusion that the voltage build up of the generator is given by the point of intersection of OCC and field resistance line Thus in Fig 117 (iii) D is point of intersection of the two curves Hence the generator will build up a voltage OM(ii) Series generator During initial operation with no current yet flowing a residual voltage will be generated exactly as in the case of a shunt generator The residual voltage will cause a current to flow through the whole series circuit when the circuit isclosed There will then be voltage build up to an equilibrium point exactlyanalogous to the buildup of a shunt generator The voltage build up graph willbe similar to that of shunt generator except that now load current (instead offield current for shunt generator) will be taken along x-axis10487871048787(iii) Compound generator When a compound generator has its series field flux aiding its shunt field flux the machine is said to be cumulative compound When the series field isconnected in reverse so that its field flux opposes the shunt field flux thegenerator is then differential compound The easiest way to build up voltage in a compound generator is to start under no load conditions At no load only the shunt field is effective When no-load voltage build up is achieved the generator is loaded If under load the voltagerises the series field connection is cumulative If the voltage drops significantlythe connection is differential compound118 Critical Field Resistance for a Shunt Generator The voltage build up in a shunt generator depends upon field circuit resistance If the field circuit resistance is R1 (line OA) then generator will build up a voltage OM as shown in Fig 118 If the field circuit resistance is increasedto R2 (tine OB) the generator will build up a voltage OL slightly less than OM As the field circuit resistance is increased the slope of resistance line also increases When the field resistance line becomes tangent (line OC) to OCC the generator would just excite If the field circuit resistance is increased beyond this point (say line OD) the generator will fail to excite The field circuit resistance represented by line OC (tangent to OCC) is called critical field resistance RC for the shunt generator It may be defined as under The maximum field circuit resistance (for a given speed) with which the shunt generator would just excite is known as its critical field resistance It should be noted that shunt generator will build up voltage only if field circuit resistance is less than critical field resistance

Fig 118 Critical field resistance of shunt generator

119 Critical Resistance for a Series Generator

Fig 119 shows the voltage build up in a series generator Here R1 R2 etc represent the total circuit resistance (load resistance and field winding resistance) If the total circuit resistance is R1 then series generator will build up a voltage OL The line OC is tangent to OCC and represents the critical resistance RC for a series generator If the total resistance of the circuit is more than RC (say line OD) the generator will fail to build up voltage Note that Fig 119 is similar to Fig 118 with the following differences

(i) In Fig 118 R1 R2 etc represent the total field circuit resistanceHowever R1 R2 etc in Fig 119 represent the total circuit resistance(load resistance and series field winding resistance etc)

(ii) In Fig 118 field current alone is represented along X-axis However inFig 119 load current IL is represented along Y-axis Note that in aseries generator field current = load current IL

Fig 119 Critical resistance of a series generator

120 Characteristics of Series Generator Fig 120 (i) shows the connections of a series wound generator Since there is only one current (that which flows through the whole machine) the load current is the same as the exciting current(i) OCCCurve 1 shows the open circuit characteristic (OCC) of a series generator Itcan be obtained experimentally by disconnecting the field winding from themachine and exciting it from a separate dc source(ii) Internal characteristic Curve 2 shows the total or internal characteristic of a series generator It gives the relation between the generated emf E on load and armature current Due to armature reaction the flux in the machine will be less than the flux at no load Hence emf E generated under load conditions will be less than the emf E0 generated under no load conditions Consequently internal characteristic curvelies below the OCC curve the difference between them representing the effectof armature reaction [See Fig 120 (ii)]

Fig 120 Characteristics of series generator

(iii) External characteristic

Curve 3 shows the external characteristic of a series generator It gives therelation between terminal voltage and load current ILV = E g- Ia (Ra +Rse )Therefore external characteristic curve will lie below internal characteristiccurve by an amount equal to ohmic drop [ie Ia(Ra + Rse)] in the machine asshown in Fig 120 (ii) The internal and external characteristics ofa dc series generator can be plotted from one another as shown in Fig 121 Supposewe are given the internal characteristic of the generator Let the line OC represent the resistance of the whole machine ie Ra +Rse If the load current is OB drop in the machine is AB ieAB = Ohmic drop in the machine = OB(Ra + Rse)

Fig 121 Load characteristics of series generator

Now raise a perpendicular from point B and mark a point b on this line such that ab = AB Then point b will lie on the external characteristic of the generator

Following similar procedure other points of external characteristic can be located It is easy to see that we can also plot internal characteristic from the external characteristic

121 Characteristics of a Shunt Generator Fig 122 (i) show the connections of a shunt wound generator The armature current Ia splits up into two parts a small fraction Ish flowing through shunt field winding while the major part IL goes to the external load(i) OCC The OCC of a shunt generator is similar in shape to that of a series generator as shown in Fig 122 (ii) The line OA represents the shunt field circuitresistance When the generator is run at normal speed it will build up a voltageOM At no-load the terminal voltage of the generator will be constant (= OM)represented by the horizontal dotted line MC

Fig 122 Characteristics of a Shunt Generator

(ii) Internal characteristic

When the generator is loaded flux per pole is reduced due to armature reaction Therefore emf E generated on load is less than the emf generated at no load As a result the internal characteristic (EIa) drops down slightly as shown in Fig122 (ii)

(iii) External characteristic Curve 2 shows the external characteristic of a shunt generator It gives therelation between terminal voltage V and load current IL

Therefore external characteristic curve will lie below the internal characteristic curve by an amount equal to drop in the armature circuit[ie (IL + Ish)Ra] as shown in Fig 122 (ii)

Note It may be seen from the external characteristic that change in terminalvoltage from no-load to full load is small The terminal voltage can always bemaintained constant by adjusting the field rheostat R automatically122 Critical External Resistance for Shunt Generator

If the load resistance across the terminals of a shunt generator is decreased then load current increase However there is a limit to the increase in

load current with the decrease of load resistance Any decrease of load resistance beyond this point instead of increasing the current ultimately results in reduced current Consequently the external characteristic turns back (dotted curve) as shown in Fig 123 The tangent OA to the curve represents the minimum external resistance required to excite the shunt generator on load and is called critical external resistance If the resistance of the external circuit is less than the critical external resistance (represented by tangent OA in Fig 310) the machine will refuse to excite or will de-excite if already running This means that external resistance is so low as virtually to short circuit the machine and so doing away with its excitation

Note There are two critical resistances for a shunt generator viz (i) criticalfield resistance (ii) critical external resistance For the shunt generator to buildup voltage the former should not be exceeded and the latter must not be gonebelow

123 How to Draw OCC at Different Speeds

If we are given OCC of a generator at a constant speed N1 then we can easily draw the OCC at any other constant speed N2 Fig 123 illustrates theprocedure Here we are given OCC at a constant speed N1 It is desired to findthe OCC at constant speed N2 (it is assumed that n1 lt N2) For constantexcitation E propN

E2

E1 =

N 2

N1 so

E2 iquest E1

N2

N1

As shown in Fig (311) for If = OH E1 = HC Therefore the new value of emf (E2) for the same If but at N2

E2= HCN 2

N 1 = HD

Fig 123 Plotting OCC at another speed

This locates the point D on the new OCC at N2 Similarly other points can belocated taking different values of If The locus of these points will be the OCCat N2

124 Critical Speed (NC)

The critical speed of a shunt generator is the minimum speed below which it fails to excite Clearly it is the speed for which the given shunt field resistance represents the critical resistance In Fig 124 curve 2 corresponds to critical

speed because the shunt field resistance (Rsh) line is tangential to it If the generator runs at full speed N the new OCC moves upward and the Rsh line represents critical resistance for this speed

In order to find critical speed take any convenient point C on excitation axis and erect a perpendicular so as to cut Rsh and Rsh lines at points B and A respectively Then

BCAC

=NcN

Nc=N BCAC

Fig 124 Finding critical speed

125 Conditions for Voltage Build-Up of a Shunt Generator

The necessary conditions for voltage build-up in a shunt generator are

(i) There must be some residual magnetism in generator poles(ii) The connections of the field winding should be such that the field current strengthens the residual magnetism(iii) The resistance of the field circuit should be less than the critical resistance In other words the speed of the generator should be higher

than the critical speed

126 Compound Generator Characteristics

In a compound generator both series and shunt excitation are combined as shown in Fig 123 The shunt winding can be connected either across the armature only (short-shunt connection S) or across armature plus series field (long-shunt connection G) The compound generator can be cumulatively compounded or

differentially compounded generator The latter is rarely used in practice Therefore we shall discuss the characteristics of cumulatively compounded

generator It may be noted that external characteristics of long and short shunt compound generators are almost identical

External characteristic

Fig 124 shows the external characteristics of a cumulatively compoundedgenerator The series excitation aids the shunt excitation The degree of compounding depends upon the increase in series excitation with the increase inload current

(i) If series winding turns are so adjusted that with the increase in load current the terminal voltage increases it is called over-compounded generator In such a case as the load current increases the series field mmf increases and tends to increase the flux and hence the generated voltage The increase in generated voltage is greater than the IaRa drop so that instead of decreasing the terminal voltage increases as shown by curve A in Fig 124

(ii) If series winding turns are so adjusted that with the increase in load current the terminal voltage substantially remains constant it is called Flat-compounded generator The series winding of such a machine has lesser number of turns than the one in over-compounded machine and therefore does not increase the flux as much for a given load current Consequently the full-load voltage is nearly equal to the no-load voltage as indicated by curve B in Fig 124 (iii) If series field winding has lesser number of turns than for a flat compounded machine the terminal voltage falls with increase in load current as indicated by curve C m Fig 124 Such a machine is called under-compounded generator

Fig 123 compound generator

Fig 124 Load characteristics of compound generators

127 LOSSES

The losses in a dc machine (generator or motor) may be divided into three classes viz (i) copper losses or electrical losses (ii) iron or core losses and (iii) mechanical losses All these losses appear as heat and thus raise the temperature of the machine They also lower the efficiency of the machine The loss structure maybe demonstrated as

1Copperlosses

These losses occur due to currents in the various windings of the machine When an electric current of I amps flows in a resistance of R ohms the heat energy is lost at the rate of I2R joules sec and the power loss is I2R watts Generators and motors have one or more field circuits and an armature circuit in which such losses occur All resistance losses of this kind are called copper losses The different copper losses occurring in a dc machine are

(i) Armature copper loss = Ia2Ra

(ii) Shunt field copper loss = Ish2 Rsh

(iii) Series field copper loss = Ise2Rse

Armature copper loss will be around 30 to 40 of full load losses where as total field copper losses will constitute around 30 of full load losses

Note There is also brush contact loss due to brush contact resistance (ie resistance between the surface of brush and surface of commutator) This loss is generally included in armature copper loss The brush contact drop is almost independent of Ia For carbon brushes this drop is around 1 V per brush The power loss due to brush contact resistance is 2ebIa where eb is the voltage drop at one brush

2 IRONORCORELOSSES These losses occur in the armature of a dc machine and are due to the rotation of armature in the magnetic field of the poles They are of two types viz (i) hysteresis loss (ii) eddy current loss

(i) Hysteresis loss

Hysteresis loss occurs in the armature of the dc machine since any given part of the armature is subjected to magnetic field reversals as it passes under successive poles Fig 124 shows an armature rotating in two-pole machine Consider a small piece ab of the armature When the piece ab is under N-pole the magnetic lines pass from a to b Half a revolution later the same piece of iron is under S-pole and magnetic lines pass from b to a so that magnetism in the iron is reversed In order to reverse continuously the molecular magnets in the armature core some amount of power has to be spent which is called hysteresis loss It is given by Steinmetz formula This formula is

Hysteresis Loss Wh = η Bmax 16f V Watts Bmax - Maximum flux density in the core f - Frequency of magnetic reversals V - Volume of armature in m3

η - Steinmetz hysteresis coefficient

In order to reduce this loss in a dc machine armature core is made of such materials which have a low value of Steinmetz hysteresis co-efficient eg silicon steel

Fig 124 Occurrence of hysteresis loss

ii) EDDYCURRENTLOSS

In addition to the voltages induced in the armature conductors there are also voltages induced in the armature core These voltages produce circulating currents in the armature core as shown in Fig 125 These are called eddy currents and power loss due to their flow is called eddy current loss The eddy current loss appears as heat which raises the temperature of the machine and lowers its efficiency Core resistance can be greatly increased by constructing the core of thin round If a continuous solid iron core is used the resistance to eddy current path will be small due to large cross-sectional area of the core Consequently the magnitude of eddy current and hence eddy current loss will be large The magnitude of eddy current can be reduced by making core resistance as high as practical The iron sheets called laminations [See Fig 126] The laminations are insulated from each other with a coating of varnish The insulating coating has a high resistance so very little current flows from one lamination to the other Alsobecause each lamination is very thin the resistance to current flowing through the width of a lamination is also quite large Thus laminating a core increases the core resistance which decreases the eddy current and hence the eddy current loss

Fig 125 Eddy current loss Fig 126 Core laminations

Eddy current loss is biven by the formula Pe= Ke Bmax

2 f2t2 V Watts

Where Ke = Constant depending upon electrical resistace of core Bmax = Maximum flux density in wbm2

f = frequency f magnetic reversals in Hz t = thickness of laminations in m and V = volume of core in m3

It may be noted that eddy current loss depends upon the square of lamination thickness For this reason lamination thickness should be kept as small as possible

3MECHANICALLOSSES

These losses are due to friction and windage(i) Friction losses eg bearing friction brush friction etc(ii) Windage loss ie air friction of rotating armature

These losses depend upon the speed of the machine But for a given speed they are practically constant

Note Iron losses and mechanical losses together are called stray losses

CONSTANT AND VARIABLE LOSSES

The losses in a dc generator (or dc motor) may be sub-divided into

(i) constant losses (ii) variable losses

(i) Constant losses

Those losses in a dc generator which remain constant at all loads are known as constant losses The constant losses in a dc generator are(a) iron losses(b) mechanical losses(c) shunt field losses(ii) Variable lossesThose losses in a dc generator which vary with load are called variable lossesThe variable losses in a dc generator are (a) Copper loss in armature winding (Ia

2 Ra) (b) Copper loss in series field winding (Ise

2Rse)

Total losses = Constant losses + Variable lossesNote Field Cu loss is constant for shunt and compound generators

128 POWER FLOW DIAGRAM

The power flow in a dc generator may be represented as

The different power stages are shown as A B and CA ndash The mechanical power input which is the power output of the prime mover is given as Pin = 2πNT

60 watts

B - The power developed in armature is Pa = Eg Ia watts C- The electrical power output is Pout = VIL watts

129 EFFICIENCY Efficiency of a machine is the ratio of its output power to the input power

Efficiency η = Output powerInput power

= Input powerminusLossesInput power = Output power

Output power+Losses

In case of a dc generator depending upon the power stages three different efficiencies can be defined as-

1 Mechanical Efficiency(ηm)

BA = Electrical power developedisinarmature

Mechanical power input2 Electrical Efficiency(ηe)

CB = Electrical power output

Electrical power developedisinthe armature 3 Overall or commercial efficiency (η)

CA = Electrical power output

Mechanical power input = ηmtimes ηe

130 CONDITION FOR MAXIMUM EFFICIENCY The efficiency of a dc generator is not constant but varies with load Fig 127shows the variation of efficiency with load current Consider a dc shunt generator with terminal voltage V delivering a load current of IL amp Generator output = V IL

Generator input = Output + Losses = V IL + constant losses + Variable losses = V IL + Wc+ Ia

2Ra

= V IL + Wc + (IL+ Ish)2 Rsh

The shunt field current Ish is small compared with IL so may be neglected

Generator input = VIL+ IL2Ra + Wc

Now efficiency η iquest outputinput = VI L

VIL+ Ia 2Ra+Wc

Efficiency is maximum when the denominator of the above equation is minimumie

ddIL ( VI L

VIL+ Ia 2Ra+Wc ) = 0

Which gives IL2 Ra = WC

ie Variable loss = constant lossThe load current for maximum efficiency is given by

IL iquest radicWcRa

Fig 127 Variation of efficiency with load

Hence the efficiency is maximum for that load current at which the variable loss is equal to the constant loss

131 APPLICATIONS

Separately excited generators

These types of generators have special applications such as electroplating electro refining of materials because separate supply is required for field excitation

Shunt generators

These types of generators are used in battery charging and ordinary lighting purposes

Series generators

This type of generators is used commonly as boosters on DC feeders as current generators for welding and arc lamp

Cumulatively compounded generators

This type of generators is used for domestic lighting purposes and to transmit energy over long distances

Differential compound generators

This type of generators is rarely used They are used for special applications such as electric arc welding

132 NUMERICAL EXCERCISES

1 The emf induced in the armature of a 450 kW 250 volt shunt generator is

2588volt when the field current is 200 amp and the generator is supplying power to a load at rated terminal voltage The armature circuit resistance is 0005 ohm Determine (i) load current (ii) power generated (iii) power output and (iv) electrical efficiency Neglect brush contact drop

Solution

Let the load current be I amp Given Ish = 20 A

Armature current = Load current + field current or Ia = I + 20 Induced emf = Terminal voltage + Ia Ra drop

2588 = 250 + (I + 20) times 0005

So (i) Load current I = 1740 amp

(ii) Power generated E Ia = 2588 (1740 + 20) watts = 455488 kW (iii) Power output V I = 250 times 1740 watts

= 435 kW

(iv) Electrical efficiency = 435455488 times 100 = 955

2 A 4 pole wave wound dc shunt generator delivers a load of 45 kW at a terminal voltage of 220 volts Its armature has 150 single turn coils and has a resistance of 001 ohm The air gap flux per pole is 002 weber Shunt field resistance is 50 ohm Calculate the speed at which it is being driven Neglect brush contact drop

Solution Load current I = W

V = 45000220 = 20455 amp

Field Current Ish = V

R s h = 220

50 = 44 amp

Armature current Ia = I +I sh = 20455 +44 = 20895 amp

Armature drop I aRa = 20895 X 001 = 20895 volts Induced emf Eg = V + Ia Ra = 220 + 20895 = 2220895 volts

But induced emf

So N = Eg60 AΦ ZP = 2220895times 60 times2

002 times150 times2 times 4 = 111045 rpm

3 The following data refer to the OCC of a DC separately excited generator at 1000 rpm

Field current in A 00 02 03 04 05 06 07 08 09 10 11 12Armature voltage in V 5 40 75 100 124 145 162 178 188 195 200 205

The machine is now connected as a shunt generator with a total field resistance of 200 Ω and driven at 1000 rpm Estimate graphically (i) the voltage to which the generator will build up at no load (ii) the armature field and load currents when the terminal voltage is found to be 150 V Neglect the effect of armature reaction and brush drop and assume armature resistance Ra to be 08 Ω (iii) Finally estimate the steady state armature current when the machine terminals are shorted

Solution

First of all the OCC of the generator is plotted in a graph paper as shown in figure

(i) The total field circuit resistance is given to be Rf = 200 Ω The Rf line is now drawn on the same graph paper passing through the origin The point of intersection of the Rf line and the OCC decides the final no load voltage and can be read from the graph as 192 V and the corresponding field current is 096 A (ii) Since the terminal voltage V is 150 V (= BM) field current If

is OM = 077 A Generated voltage E is given by AM But we know E = IaRa + V Hence IaRa= E ndash V = AM ndash BM = AB Now from the graph AB = 25 V there4Ia = 2508 = 3125 A So load current IL= Ia ndash If= 3125 ndash 077 = 3048 A

ADDITIONAL PROBLEMS

1 A 4 pole lap wound 115 kW 230 volts dc shunt generator has armatureresistance of 02 ohm and field resistance of 100 ohms Calculate (i) emf

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES

Fig16 Commutator segments

137 Brushes The brushes are made of carbon These are rectangular block shaped The only function of these carbon brushes of dc generator is to collect current from commutator segments The brushes are housed in the rectangular box shaped brush holder As shown in figure17 the brush face is placed on the commutator segment with attached to the brush holder

Fig 17 Brush in brush holder

138 Bearing For small machine ball bearing is used and for heavy duty dc generator roller bearing is used The bearing must always be lubricated properly for smooth operation and long life of generator

14 EMF EQUATION OF A DC GENERATOR

Let Φ be the flux per pole in WeberZ the total number of armature conductors ndash No of slots x no of armature ConductorsslotP the number of polesA the number of parallel paths in the armature

N the speed of rotation of armature in revolutions per minute (rpm)E the emf induced in any parallel path in armatureEg the emf generated in any parallel path in the armatureThe emf induced per conductor = dΦdt volt (n=1)The flux cut by one conductor in one revolution dΦ = ΦP WbNo of revolutions per second = N60Time for one revolution dt = 60N secondHence EMF generatedconductor is

dφdt = φPN

60

No of conductors in a parallel path = Z A

Hence emf generated per parallal path

Eg = φPN60 Z

A

For a lap wound generator A = P and

For wave wound generator A = 2

15 TYPES OF DC GENERATORS

Fig18 Classification of dc generators

151 Separately Excited DC Generator

This dc generator has a field magnet winding which is excited using a separate voltage source (like battery)The representation in the below figure 19 The output voltage depends on the speed of rotation of armature and field current The higher the speed of rotation and current ndash the higher the output emf

Fig 19 Separately Excited DC Generators

Armature Current Ia = IL

Terminal voltage V = Eg-Ia Ra

Electric Power developed = Eg Ia

Power delivered to load = V IL = (Eg-Ia Ra) Ia = Eg Ia - Ia2

Ra

152 Self Excited DC Generator

These are generators in which the field winding is excited by the output of the generator itself As described before ndash there are three types of self excited dc generators ndash they are 1) Series 2) Shunt and 3) Compound

A series DC generator is shown below in fig110 (a) ndash in which the armature winding is connected in series with the field winding so that the field current flows through the load as well as the field winding Field winding is a low resistance thick wire of few turns Armature current Ia =Ise = IL=I (say)

Terminal voltage V = Eg- I( Ra+ Rse)Power Developed in armature = Eg Ia

Power delivered to load = V IL = Eg Ia- Ia2 (Ra+Rse)

Fig110 (a) Series generator and (b) shunt generator

A shunt DC generator is shown in figure 110 (b) in which the field winding is wired parallel to armature winding so that the voltage across both are same The field winding has high resistance and more number of turns so that only a part of armature current passes through field winding and the rest passes through load Shunt field current Ish = V Rsh

Armature current Ia = IL+ Ish

Terminal voltage V= Eg-Ia Ra

Power developed in armature = EgIa

Power delivered to load = VIL A compound generator is shown in figure 111 below It has two field findings namely Rsh and Rse They are basically shunt winding (Rsh) and series winding (Rse) Compound generator is of two types ndash 1) Short shunt and 2) Long shunt

Short shunt- Here the shunt field winding is wired parallel to armature and series field winding is connected in series to the load It is shown in fig111(a)

In short shunt machines Series field current Ise = IL

Shunt field current Ish iquest V+ I se RseRsh

Terminal voltage V= Eg-Ia Ra-Ise Rse

Power developedin armature = Eg Ia

Power delivered to load = V IL

Fig 111 (a) short shunt generator and (b) Long shunt generator

Long shunt- Here the shunt field winding is parallel to both armature and series field winding (Rse is wired in series to the armature) It is shown in figure 111 (b)

In Long shunt machines Series field current Ise = Ia = IL+ Ish

Shunt field current Ish = V R sh

Terminal voltage V = Eg- Ia (Ra+ Rse) Power developed in armature = Eg Ia

Power delivered to load = VIL

16 GENERATOR CHARACTERISTICS

The speed of a dc machine operated as a generator is fixed by the prime mover For general-purpose operation the prime mover is equipped with a speed governor so that the speed of the generator is practically constant Under such condition the generator performance deals primarily with the relation between excitation terminal voltage and load These relations can be best exhibited graphically by means of curves known as generator characteristics Thesecharacteristics show at a glance the behavior of the generator under differentload conditions

The following are the three most important characteristics of a dc generator161 Open Circuit Characteristic (OCC)

This curve shows the relation between the generated emf at no-load (E0) and the field current (If) at constant speed It is also known as magnetic characteristic or no-load saturation curve Its shape is practically the same for all generators whether separately or self-excited The data for OCC curve are obtained experimentally by operating the generator at no load and constant speed and recording the change in terminal voltage as the field current is varied

162 Internal or Total characteristic (EIa)

This curve shows the relation between the generated emf on load (E) and the armature current (Ia) The emf E is less than E0 due to the demagnetizing effect of armature reaction Therefore this curve will lie below the open circuitcharacteristic (OCC) The internal characteristic is of interest chiefly to thedesigner It cannot be obtained directly by experiment It is because a voltmetercannot read the emf generated on load due to the voltage drop in armatureresistance The internal characteristic can be obtained from externalcharacteristics if winding resistances are known because armature reaction effectis included in both characteristics163 External characteristic (VIL)This curve shows the relation between the terminal voltage (V) and load current(IL) The terminal voltage V will be less than E due to voltage drop in thearmature circuit Therefore this curve will lie below the internal characteristicThis characteristic is very important in determining the suitability of a generatorfor a given purpose It can be obtained by making simultaneous measurements

of terminal voltage and load current (with voltmeter and ammeter) of a loadedgenerator164 Open Circuit Characteristic of a DC Generator

The OCC for a dc generator is determined as follows The field winding of the dc generator (series or shunt) is disconnected from the machine and isseparately excited from an external dc source as shown in Fig 112 Thegenerator is run at fixed speed (ie normal speed) The field current (If) isincreased from zero in steps and the corresponding values of generated emf(E0) read off on a voltmeter connected across the armature terminals On plottingthe relation between E0 and If we get the open circuit characteristic as shown in

Fig113

Fig 112 Experimental setup for plotting OCC

Fig 113 OCC of dc generator

The following points may be noted from OCC When the field current is zero there is some generated emf OA

This is due to the residual magnetism in the field poles Over a fairly wide range of field current (upto point B in the curve)

the curve is linear It is because in this range reluctance of iron is negligible as compared with that of air gap The air gap reluctance is constant and hence linear relationship After point B on the curve the reluctance of iron also comes into picture and reluctance of iron is no longer negligible Consequently the curve deviates from linear relationship

After point C on the curve the magnetic saturation of poles begins and E0 tends to level off

The OCC of even self-excited generator is obtained by running it as a separately excited generator

165 Characteristics of a Separately Excited DCGenerator

The obvious disadvantage of a separately excited dc generator is that we require an external dc source for excitation But since the output voltage may be

controlled more easily and over a wide range (from zero to a maximum) this type of excitation finds many applications

(i) Open circuit characteristic

The OCC of a separately excited generator is determined in a manner described in Sec 164 Fig114 shows the variation of generated em f on noload with field current for various fixed speeds Note that if the value of constant speed is increased the steepness of the curve also increases When the field current is zero the residual magnetism in the poles will give rise to the small initial emf as shown

Fig114 Variation of OCC with speed

Internal and External Characteristics (Load Characteristics)

The external characteristic of a separately excited generator is the curve between the terminal voltage (V) and the load current IL (which is the same as armature current in this case) In order to determine the external characteristic the circuit set up is as shown in Fig 115

Fig 115 Experimental setup to plot load characteristics

As the load current increases the terminal voltage falls due to two reasons(a) The armature reaction weakens the main flux so that actual emf generated E on load is less than that generated (E0) on no load

(b) There is voltage drop across armature resistance (= ILRa = IaRa)Due to these reasons the external characteristic is a drooping curve [curve 3 inFig 116 ] Note that in the absence of armature reaction and armature dropthe generated emf would have been E0 (curve 1)The internal characteristic can be determined from external characteristic by adding IL Ra drop to the external characteristic It is because armature reaction drop is included in the external characteristic Curve 2 is the internal characteristic of the generator and should obviously lie above the external characteristic

Fig 116 Load characteristics of a separately excited dcgenerator

117 Voltage Build-Up in a Self-Excited GeneratorLet us see how voltage builds up in a self-excited generator(i) Shunt generator Consider a shunt generator If the generator is run at a constant speed some emf will be generated due to residual magnetism in the main poles This small emf circulates a field current which in turn produces additional flux toreinforce the original residual flux (provided field winding connections arecorrect) This process continues and the generator builds up the normalgenerated voltage following the OCC shown in Fig 117 (i) The field resistance Rf can be represented by a straight line passing through the origin as shown in Fig 117 (ii) The two curves can be shown on the same diagram as they have the same ordinate as in Fig 117 (iii)

Fig 117

Since the field circuit is inductive there is a delay in the increase in current uponclosing the field circuit switch The rate at which the current increases dependsupon the voltage available for increasing it Suppose at any instant the fieldcurrent is i (= OA) and is increasing at the rate didt Then

E0=i R f +iquest L didt

Where Rf = total field circuit resistance L = inductance of field circuit

At the considered instant the total emf available is AC [See Fig 117 (iii)] Anamount AB of the cmf AC is absorbed by the voltage drop iRf and theremainder part BC is available to overcome L didt Since this surplus voltage isavailable it is possible for the field current to increase above the value OAHowever at point D the available voltage is OM and is all absorbed by i RfD

rop Consequently the field current cannot increase further and the generatorbuild up stops Hence we arrive at a very important conclusion that the voltage build up of the generator is given by the point of intersection of OCC and field resistance line Thus in Fig 117 (iii) D is point of intersection of the two curves Hence the generator will build up a voltage OM(ii) Series generator During initial operation with no current yet flowing a residual voltage will be generated exactly as in the case of a shunt generator The residual voltage will cause a current to flow through the whole series circuit when the circuit isclosed There will then be voltage build up to an equilibrium point exactlyanalogous to the buildup of a shunt generator The voltage build up graph willbe similar to that of shunt generator except that now load current (instead offield current for shunt generator) will be taken along x-axis10487871048787(iii) Compound generator When a compound generator has its series field flux aiding its shunt field flux the machine is said to be cumulative compound When the series field isconnected in reverse so that its field flux opposes the shunt field flux thegenerator is then differential compound The easiest way to build up voltage in a compound generator is to start under no load conditions At no load only the shunt field is effective When no-load voltage build up is achieved the generator is loaded If under load the voltagerises the series field connection is cumulative If the voltage drops significantlythe connection is differential compound118 Critical Field Resistance for a Shunt Generator The voltage build up in a shunt generator depends upon field circuit resistance If the field circuit resistance is R1 (line OA) then generator will build up a voltage OM as shown in Fig 118 If the field circuit resistance is increasedto R2 (tine OB) the generator will build up a voltage OL slightly less than OM As the field circuit resistance is increased the slope of resistance line also increases When the field resistance line becomes tangent (line OC) to OCC the generator would just excite If the field circuit resistance is increased beyond this point (say line OD) the generator will fail to excite The field circuit resistance represented by line OC (tangent to OCC) is called critical field resistance RC for the shunt generator It may be defined as under The maximum field circuit resistance (for a given speed) with which the shunt generator would just excite is known as its critical field resistance It should be noted that shunt generator will build up voltage only if field circuit resistance is less than critical field resistance

Fig 118 Critical field resistance of shunt generator

119 Critical Resistance for a Series Generator

Fig 119 shows the voltage build up in a series generator Here R1 R2 etc represent the total circuit resistance (load resistance and field winding resistance) If the total circuit resistance is R1 then series generator will build up a voltage OL The line OC is tangent to OCC and represents the critical resistance RC for a series generator If the total resistance of the circuit is more than RC (say line OD) the generator will fail to build up voltage Note that Fig 119 is similar to Fig 118 with the following differences

(i) In Fig 118 R1 R2 etc represent the total field circuit resistanceHowever R1 R2 etc in Fig 119 represent the total circuit resistance(load resistance and series field winding resistance etc)

(ii) In Fig 118 field current alone is represented along X-axis However inFig 119 load current IL is represented along Y-axis Note that in aseries generator field current = load current IL

Fig 119 Critical resistance of a series generator

120 Characteristics of Series Generator Fig 120 (i) shows the connections of a series wound generator Since there is only one current (that which flows through the whole machine) the load current is the same as the exciting current(i) OCCCurve 1 shows the open circuit characteristic (OCC) of a series generator Itcan be obtained experimentally by disconnecting the field winding from themachine and exciting it from a separate dc source(ii) Internal characteristic Curve 2 shows the total or internal characteristic of a series generator It gives the relation between the generated emf E on load and armature current Due to armature reaction the flux in the machine will be less than the flux at no load Hence emf E generated under load conditions will be less than the emf E0 generated under no load conditions Consequently internal characteristic curvelies below the OCC curve the difference between them representing the effectof armature reaction [See Fig 120 (ii)]

Fig 120 Characteristics of series generator

(iii) External characteristic

Curve 3 shows the external characteristic of a series generator It gives therelation between terminal voltage and load current ILV = E g- Ia (Ra +Rse )Therefore external characteristic curve will lie below internal characteristiccurve by an amount equal to ohmic drop [ie Ia(Ra + Rse)] in the machine asshown in Fig 120 (ii) The internal and external characteristics ofa dc series generator can be plotted from one another as shown in Fig 121 Supposewe are given the internal characteristic of the generator Let the line OC represent the resistance of the whole machine ie Ra +Rse If the load current is OB drop in the machine is AB ieAB = Ohmic drop in the machine = OB(Ra + Rse)

Fig 121 Load characteristics of series generator

Now raise a perpendicular from point B and mark a point b on this line such that ab = AB Then point b will lie on the external characteristic of the generator

Following similar procedure other points of external characteristic can be located It is easy to see that we can also plot internal characteristic from the external characteristic

121 Characteristics of a Shunt Generator Fig 122 (i) show the connections of a shunt wound generator The armature current Ia splits up into two parts a small fraction Ish flowing through shunt field winding while the major part IL goes to the external load(i) OCC The OCC of a shunt generator is similar in shape to that of a series generator as shown in Fig 122 (ii) The line OA represents the shunt field circuitresistance When the generator is run at normal speed it will build up a voltageOM At no-load the terminal voltage of the generator will be constant (= OM)represented by the horizontal dotted line MC

Fig 122 Characteristics of a Shunt Generator

(ii) Internal characteristic

When the generator is loaded flux per pole is reduced due to armature reaction Therefore emf E generated on load is less than the emf generated at no load As a result the internal characteristic (EIa) drops down slightly as shown in Fig122 (ii)

(iii) External characteristic Curve 2 shows the external characteristic of a shunt generator It gives therelation between terminal voltage V and load current IL

Therefore external characteristic curve will lie below the internal characteristic curve by an amount equal to drop in the armature circuit[ie (IL + Ish)Ra] as shown in Fig 122 (ii)

Note It may be seen from the external characteristic that change in terminalvoltage from no-load to full load is small The terminal voltage can always bemaintained constant by adjusting the field rheostat R automatically122 Critical External Resistance for Shunt Generator

If the load resistance across the terminals of a shunt generator is decreased then load current increase However there is a limit to the increase in

load current with the decrease of load resistance Any decrease of load resistance beyond this point instead of increasing the current ultimately results in reduced current Consequently the external characteristic turns back (dotted curve) as shown in Fig 123 The tangent OA to the curve represents the minimum external resistance required to excite the shunt generator on load and is called critical external resistance If the resistance of the external circuit is less than the critical external resistance (represented by tangent OA in Fig 310) the machine will refuse to excite or will de-excite if already running This means that external resistance is so low as virtually to short circuit the machine and so doing away with its excitation

Note There are two critical resistances for a shunt generator viz (i) criticalfield resistance (ii) critical external resistance For the shunt generator to buildup voltage the former should not be exceeded and the latter must not be gonebelow

123 How to Draw OCC at Different Speeds

If we are given OCC of a generator at a constant speed N1 then we can easily draw the OCC at any other constant speed N2 Fig 123 illustrates theprocedure Here we are given OCC at a constant speed N1 It is desired to findthe OCC at constant speed N2 (it is assumed that n1 lt N2) For constantexcitation E propN

E2

E1 =

N 2

N1 so

E2 iquest E1

N2

N1

As shown in Fig (311) for If = OH E1 = HC Therefore the new value of emf (E2) for the same If but at N2

E2= HCN 2

N 1 = HD

Fig 123 Plotting OCC at another speed

This locates the point D on the new OCC at N2 Similarly other points can belocated taking different values of If The locus of these points will be the OCCat N2

124 Critical Speed (NC)

The critical speed of a shunt generator is the minimum speed below which it fails to excite Clearly it is the speed for which the given shunt field resistance represents the critical resistance In Fig 124 curve 2 corresponds to critical

speed because the shunt field resistance (Rsh) line is tangential to it If the generator runs at full speed N the new OCC moves upward and the Rsh line represents critical resistance for this speed

In order to find critical speed take any convenient point C on excitation axis and erect a perpendicular so as to cut Rsh and Rsh lines at points B and A respectively Then

BCAC

=NcN

Nc=N BCAC

Fig 124 Finding critical speed

125 Conditions for Voltage Build-Up of a Shunt Generator

The necessary conditions for voltage build-up in a shunt generator are

(i) There must be some residual magnetism in generator poles(ii) The connections of the field winding should be such that the field current strengthens the residual magnetism(iii) The resistance of the field circuit should be less than the critical resistance In other words the speed of the generator should be higher

than the critical speed

126 Compound Generator Characteristics

In a compound generator both series and shunt excitation are combined as shown in Fig 123 The shunt winding can be connected either across the armature only (short-shunt connection S) or across armature plus series field (long-shunt connection G) The compound generator can be cumulatively compounded or

differentially compounded generator The latter is rarely used in practice Therefore we shall discuss the characteristics of cumulatively compounded

generator It may be noted that external characteristics of long and short shunt compound generators are almost identical

External characteristic

Fig 124 shows the external characteristics of a cumulatively compoundedgenerator The series excitation aids the shunt excitation The degree of compounding depends upon the increase in series excitation with the increase inload current

(i) If series winding turns are so adjusted that with the increase in load current the terminal voltage increases it is called over-compounded generator In such a case as the load current increases the series field mmf increases and tends to increase the flux and hence the generated voltage The increase in generated voltage is greater than the IaRa drop so that instead of decreasing the terminal voltage increases as shown by curve A in Fig 124

(ii) If series winding turns are so adjusted that with the increase in load current the terminal voltage substantially remains constant it is called Flat-compounded generator The series winding of such a machine has lesser number of turns than the one in over-compounded machine and therefore does not increase the flux as much for a given load current Consequently the full-load voltage is nearly equal to the no-load voltage as indicated by curve B in Fig 124 (iii) If series field winding has lesser number of turns than for a flat compounded machine the terminal voltage falls with increase in load current as indicated by curve C m Fig 124 Such a machine is called under-compounded generator

Fig 123 compound generator

Fig 124 Load characteristics of compound generators

127 LOSSES

The losses in a dc machine (generator or motor) may be divided into three classes viz (i) copper losses or electrical losses (ii) iron or core losses and (iii) mechanical losses All these losses appear as heat and thus raise the temperature of the machine They also lower the efficiency of the machine The loss structure maybe demonstrated as

1Copperlosses

These losses occur due to currents in the various windings of the machine When an electric current of I amps flows in a resistance of R ohms the heat energy is lost at the rate of I2R joules sec and the power loss is I2R watts Generators and motors have one or more field circuits and an armature circuit in which such losses occur All resistance losses of this kind are called copper losses The different copper losses occurring in a dc machine are

(i) Armature copper loss = Ia2Ra

(ii) Shunt field copper loss = Ish2 Rsh

(iii) Series field copper loss = Ise2Rse

Armature copper loss will be around 30 to 40 of full load losses where as total field copper losses will constitute around 30 of full load losses

Note There is also brush contact loss due to brush contact resistance (ie resistance between the surface of brush and surface of commutator) This loss is generally included in armature copper loss The brush contact drop is almost independent of Ia For carbon brushes this drop is around 1 V per brush The power loss due to brush contact resistance is 2ebIa where eb is the voltage drop at one brush

2 IRONORCORELOSSES These losses occur in the armature of a dc machine and are due to the rotation of armature in the magnetic field of the poles They are of two types viz (i) hysteresis loss (ii) eddy current loss

(i) Hysteresis loss

Hysteresis loss occurs in the armature of the dc machine since any given part of the armature is subjected to magnetic field reversals as it passes under successive poles Fig 124 shows an armature rotating in two-pole machine Consider a small piece ab of the armature When the piece ab is under N-pole the magnetic lines pass from a to b Half a revolution later the same piece of iron is under S-pole and magnetic lines pass from b to a so that magnetism in the iron is reversed In order to reverse continuously the molecular magnets in the armature core some amount of power has to be spent which is called hysteresis loss It is given by Steinmetz formula This formula is

Hysteresis Loss Wh = η Bmax 16f V Watts Bmax - Maximum flux density in the core f - Frequency of magnetic reversals V - Volume of armature in m3

η - Steinmetz hysteresis coefficient

In order to reduce this loss in a dc machine armature core is made of such materials which have a low value of Steinmetz hysteresis co-efficient eg silicon steel

Fig 124 Occurrence of hysteresis loss

ii) EDDYCURRENTLOSS

In addition to the voltages induced in the armature conductors there are also voltages induced in the armature core These voltages produce circulating currents in the armature core as shown in Fig 125 These are called eddy currents and power loss due to their flow is called eddy current loss The eddy current loss appears as heat which raises the temperature of the machine and lowers its efficiency Core resistance can be greatly increased by constructing the core of thin round If a continuous solid iron core is used the resistance to eddy current path will be small due to large cross-sectional area of the core Consequently the magnitude of eddy current and hence eddy current loss will be large The magnitude of eddy current can be reduced by making core resistance as high as practical The iron sheets called laminations [See Fig 126] The laminations are insulated from each other with a coating of varnish The insulating coating has a high resistance so very little current flows from one lamination to the other Alsobecause each lamination is very thin the resistance to current flowing through the width of a lamination is also quite large Thus laminating a core increases the core resistance which decreases the eddy current and hence the eddy current loss

Fig 125 Eddy current loss Fig 126 Core laminations

Eddy current loss is biven by the formula Pe= Ke Bmax

2 f2t2 V Watts

Where Ke = Constant depending upon electrical resistace of core Bmax = Maximum flux density in wbm2

f = frequency f magnetic reversals in Hz t = thickness of laminations in m and V = volume of core in m3

It may be noted that eddy current loss depends upon the square of lamination thickness For this reason lamination thickness should be kept as small as possible

3MECHANICALLOSSES

These losses are due to friction and windage(i) Friction losses eg bearing friction brush friction etc(ii) Windage loss ie air friction of rotating armature

These losses depend upon the speed of the machine But for a given speed they are practically constant

Note Iron losses and mechanical losses together are called stray losses

CONSTANT AND VARIABLE LOSSES

The losses in a dc generator (or dc motor) may be sub-divided into

(i) constant losses (ii) variable losses

(i) Constant losses

Those losses in a dc generator which remain constant at all loads are known as constant losses The constant losses in a dc generator are(a) iron losses(b) mechanical losses(c) shunt field losses(ii) Variable lossesThose losses in a dc generator which vary with load are called variable lossesThe variable losses in a dc generator are (a) Copper loss in armature winding (Ia

2 Ra) (b) Copper loss in series field winding (Ise

2Rse)

Total losses = Constant losses + Variable lossesNote Field Cu loss is constant for shunt and compound generators

128 POWER FLOW DIAGRAM

The power flow in a dc generator may be represented as

The different power stages are shown as A B and CA ndash The mechanical power input which is the power output of the prime mover is given as Pin = 2πNT

60 watts

B - The power developed in armature is Pa = Eg Ia watts C- The electrical power output is Pout = VIL watts

129 EFFICIENCY Efficiency of a machine is the ratio of its output power to the input power

Efficiency η = Output powerInput power

= Input powerminusLossesInput power = Output power

Output power+Losses

In case of a dc generator depending upon the power stages three different efficiencies can be defined as-

1 Mechanical Efficiency(ηm)

BA = Electrical power developedisinarmature

Mechanical power input2 Electrical Efficiency(ηe)

CB = Electrical power output

Electrical power developedisinthe armature 3 Overall or commercial efficiency (η)

CA = Electrical power output

Mechanical power input = ηmtimes ηe

130 CONDITION FOR MAXIMUM EFFICIENCY The efficiency of a dc generator is not constant but varies with load Fig 127shows the variation of efficiency with load current Consider a dc shunt generator with terminal voltage V delivering a load current of IL amp Generator output = V IL

Generator input = Output + Losses = V IL + constant losses + Variable losses = V IL + Wc+ Ia

2Ra

= V IL + Wc + (IL+ Ish)2 Rsh

The shunt field current Ish is small compared with IL so may be neglected

Generator input = VIL+ IL2Ra + Wc

Now efficiency η iquest outputinput = VI L

VIL+ Ia 2Ra+Wc

Efficiency is maximum when the denominator of the above equation is minimumie

ddIL ( VI L

VIL+ Ia 2Ra+Wc ) = 0

Which gives IL2 Ra = WC

ie Variable loss = constant lossThe load current for maximum efficiency is given by

IL iquest radicWcRa

Fig 127 Variation of efficiency with load

Hence the efficiency is maximum for that load current at which the variable loss is equal to the constant loss

131 APPLICATIONS

Separately excited generators

These types of generators have special applications such as electroplating electro refining of materials because separate supply is required for field excitation

Shunt generators

These types of generators are used in battery charging and ordinary lighting purposes

Series generators

This type of generators is used commonly as boosters on DC feeders as current generators for welding and arc lamp

Cumulatively compounded generators

This type of generators is used for domestic lighting purposes and to transmit energy over long distances

Differential compound generators

This type of generators is rarely used They are used for special applications such as electric arc welding

132 NUMERICAL EXCERCISES

1 The emf induced in the armature of a 450 kW 250 volt shunt generator is

2588volt when the field current is 200 amp and the generator is supplying power to a load at rated terminal voltage The armature circuit resistance is 0005 ohm Determine (i) load current (ii) power generated (iii) power output and (iv) electrical efficiency Neglect brush contact drop

Solution

Let the load current be I amp Given Ish = 20 A

Armature current = Load current + field current or Ia = I + 20 Induced emf = Terminal voltage + Ia Ra drop

2588 = 250 + (I + 20) times 0005

So (i) Load current I = 1740 amp

(ii) Power generated E Ia = 2588 (1740 + 20) watts = 455488 kW (iii) Power output V I = 250 times 1740 watts

= 435 kW

(iv) Electrical efficiency = 435455488 times 100 = 955

2 A 4 pole wave wound dc shunt generator delivers a load of 45 kW at a terminal voltage of 220 volts Its armature has 150 single turn coils and has a resistance of 001 ohm The air gap flux per pole is 002 weber Shunt field resistance is 50 ohm Calculate the speed at which it is being driven Neglect brush contact drop

Solution Load current I = W

V = 45000220 = 20455 amp

Field Current Ish = V

R s h = 220

50 = 44 amp

Armature current Ia = I +I sh = 20455 +44 = 20895 amp

Armature drop I aRa = 20895 X 001 = 20895 volts Induced emf Eg = V + Ia Ra = 220 + 20895 = 2220895 volts

But induced emf

So N = Eg60 AΦ ZP = 2220895times 60 times2

002 times150 times2 times 4 = 111045 rpm

3 The following data refer to the OCC of a DC separately excited generator at 1000 rpm

Field current in A 00 02 03 04 05 06 07 08 09 10 11 12Armature voltage in V 5 40 75 100 124 145 162 178 188 195 200 205

The machine is now connected as a shunt generator with a total field resistance of 200 Ω and driven at 1000 rpm Estimate graphically (i) the voltage to which the generator will build up at no load (ii) the armature field and load currents when the terminal voltage is found to be 150 V Neglect the effect of armature reaction and brush drop and assume armature resistance Ra to be 08 Ω (iii) Finally estimate the steady state armature current when the machine terminals are shorted

Solution

First of all the OCC of the generator is plotted in a graph paper as shown in figure

(i) The total field circuit resistance is given to be Rf = 200 Ω The Rf line is now drawn on the same graph paper passing through the origin The point of intersection of the Rf line and the OCC decides the final no load voltage and can be read from the graph as 192 V and the corresponding field current is 096 A (ii) Since the terminal voltage V is 150 V (= BM) field current If

is OM = 077 A Generated voltage E is given by AM But we know E = IaRa + V Hence IaRa= E ndash V = AM ndash BM = AB Now from the graph AB = 25 V there4Ia = 2508 = 3125 A So load current IL= Ia ndash If= 3125 ndash 077 = 3048 A

ADDITIONAL PROBLEMS

1 A 4 pole lap wound 115 kW 230 volts dc shunt generator has armatureresistance of 02 ohm and field resistance of 100 ohms Calculate (i) emf

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES

138 Bearing For small machine ball bearing is used and for heavy duty dc generator roller bearing is used The bearing must always be lubricated properly for smooth operation and long life of generator

14 EMF EQUATION OF A DC GENERATOR

Let Φ be the flux per pole in WeberZ the total number of armature conductors ndash No of slots x no of armature ConductorsslotP the number of polesA the number of parallel paths in the armature

N the speed of rotation of armature in revolutions per minute (rpm)E the emf induced in any parallel path in armatureEg the emf generated in any parallel path in the armatureThe emf induced per conductor = dΦdt volt (n=1)The flux cut by one conductor in one revolution dΦ = ΦP WbNo of revolutions per second = N60Time for one revolution dt = 60N secondHence EMF generatedconductor is

dφdt = φPN

60

No of conductors in a parallel path = Z A

Hence emf generated per parallal path

Eg = φPN60 Z

A

For a lap wound generator A = P and

For wave wound generator A = 2

15 TYPES OF DC GENERATORS

Fig18 Classification of dc generators

151 Separately Excited DC Generator

This dc generator has a field magnet winding which is excited using a separate voltage source (like battery)The representation in the below figure 19 The output voltage depends on the speed of rotation of armature and field current The higher the speed of rotation and current ndash the higher the output emf

Fig 19 Separately Excited DC Generators

Armature Current Ia = IL

Terminal voltage V = Eg-Ia Ra

Electric Power developed = Eg Ia

Power delivered to load = V IL = (Eg-Ia Ra) Ia = Eg Ia - Ia2

Ra

152 Self Excited DC Generator

These are generators in which the field winding is excited by the output of the generator itself As described before ndash there are three types of self excited dc generators ndash they are 1) Series 2) Shunt and 3) Compound

A series DC generator is shown below in fig110 (a) ndash in which the armature winding is connected in series with the field winding so that the field current flows through the load as well as the field winding Field winding is a low resistance thick wire of few turns Armature current Ia =Ise = IL=I (say)

Terminal voltage V = Eg- I( Ra+ Rse)Power Developed in armature = Eg Ia

Power delivered to load = V IL = Eg Ia- Ia2 (Ra+Rse)

Fig110 (a) Series generator and (b) shunt generator

A shunt DC generator is shown in figure 110 (b) in which the field winding is wired parallel to armature winding so that the voltage across both are same The field winding has high resistance and more number of turns so that only a part of armature current passes through field winding and the rest passes through load Shunt field current Ish = V Rsh

Armature current Ia = IL+ Ish

Terminal voltage V= Eg-Ia Ra

Power developed in armature = EgIa

Power delivered to load = VIL A compound generator is shown in figure 111 below It has two field findings namely Rsh and Rse They are basically shunt winding (Rsh) and series winding (Rse) Compound generator is of two types ndash 1) Short shunt and 2) Long shunt

Short shunt- Here the shunt field winding is wired parallel to armature and series field winding is connected in series to the load It is shown in fig111(a)

In short shunt machines Series field current Ise = IL

Shunt field current Ish iquest V+ I se RseRsh

Terminal voltage V= Eg-Ia Ra-Ise Rse

Power developedin armature = Eg Ia

Power delivered to load = V IL

Fig 111 (a) short shunt generator and (b) Long shunt generator

Long shunt- Here the shunt field winding is parallel to both armature and series field winding (Rse is wired in series to the armature) It is shown in figure 111 (b)

In Long shunt machines Series field current Ise = Ia = IL+ Ish

Shunt field current Ish = V R sh

Terminal voltage V = Eg- Ia (Ra+ Rse) Power developed in armature = Eg Ia

Power delivered to load = VIL

16 GENERATOR CHARACTERISTICS

The speed of a dc machine operated as a generator is fixed by the prime mover For general-purpose operation the prime mover is equipped with a speed governor so that the speed of the generator is practically constant Under such condition the generator performance deals primarily with the relation between excitation terminal voltage and load These relations can be best exhibited graphically by means of curves known as generator characteristics Thesecharacteristics show at a glance the behavior of the generator under differentload conditions

The following are the three most important characteristics of a dc generator161 Open Circuit Characteristic (OCC)

This curve shows the relation between the generated emf at no-load (E0) and the field current (If) at constant speed It is also known as magnetic characteristic or no-load saturation curve Its shape is practically the same for all generators whether separately or self-excited The data for OCC curve are obtained experimentally by operating the generator at no load and constant speed and recording the change in terminal voltage as the field current is varied

162 Internal or Total characteristic (EIa)

This curve shows the relation between the generated emf on load (E) and the armature current (Ia) The emf E is less than E0 due to the demagnetizing effect of armature reaction Therefore this curve will lie below the open circuitcharacteristic (OCC) The internal characteristic is of interest chiefly to thedesigner It cannot be obtained directly by experiment It is because a voltmetercannot read the emf generated on load due to the voltage drop in armatureresistance The internal characteristic can be obtained from externalcharacteristics if winding resistances are known because armature reaction effectis included in both characteristics163 External characteristic (VIL)This curve shows the relation between the terminal voltage (V) and load current(IL) The terminal voltage V will be less than E due to voltage drop in thearmature circuit Therefore this curve will lie below the internal characteristicThis characteristic is very important in determining the suitability of a generatorfor a given purpose It can be obtained by making simultaneous measurements

of terminal voltage and load current (with voltmeter and ammeter) of a loadedgenerator164 Open Circuit Characteristic of a DC Generator

The OCC for a dc generator is determined as follows The field winding of the dc generator (series or shunt) is disconnected from the machine and isseparately excited from an external dc source as shown in Fig 112 Thegenerator is run at fixed speed (ie normal speed) The field current (If) isincreased from zero in steps and the corresponding values of generated emf(E0) read off on a voltmeter connected across the armature terminals On plottingthe relation between E0 and If we get the open circuit characteristic as shown in

Fig113

Fig 112 Experimental setup for plotting OCC

Fig 113 OCC of dc generator

The following points may be noted from OCC When the field current is zero there is some generated emf OA

This is due to the residual magnetism in the field poles Over a fairly wide range of field current (upto point B in the curve)

the curve is linear It is because in this range reluctance of iron is negligible as compared with that of air gap The air gap reluctance is constant and hence linear relationship After point B on the curve the reluctance of iron also comes into picture and reluctance of iron is no longer negligible Consequently the curve deviates from linear relationship

After point C on the curve the magnetic saturation of poles begins and E0 tends to level off

The OCC of even self-excited generator is obtained by running it as a separately excited generator

165 Characteristics of a Separately Excited DCGenerator

The obvious disadvantage of a separately excited dc generator is that we require an external dc source for excitation But since the output voltage may be

controlled more easily and over a wide range (from zero to a maximum) this type of excitation finds many applications

(i) Open circuit characteristic

The OCC of a separately excited generator is determined in a manner described in Sec 164 Fig114 shows the variation of generated em f on noload with field current for various fixed speeds Note that if the value of constant speed is increased the steepness of the curve also increases When the field current is zero the residual magnetism in the poles will give rise to the small initial emf as shown

Fig114 Variation of OCC with speed

Internal and External Characteristics (Load Characteristics)

The external characteristic of a separately excited generator is the curve between the terminal voltage (V) and the load current IL (which is the same as armature current in this case) In order to determine the external characteristic the circuit set up is as shown in Fig 115

Fig 115 Experimental setup to plot load characteristics

As the load current increases the terminal voltage falls due to two reasons(a) The armature reaction weakens the main flux so that actual emf generated E on load is less than that generated (E0) on no load

(b) There is voltage drop across armature resistance (= ILRa = IaRa)Due to these reasons the external characteristic is a drooping curve [curve 3 inFig 116 ] Note that in the absence of armature reaction and armature dropthe generated emf would have been E0 (curve 1)The internal characteristic can be determined from external characteristic by adding IL Ra drop to the external characteristic It is because armature reaction drop is included in the external characteristic Curve 2 is the internal characteristic of the generator and should obviously lie above the external characteristic

Fig 116 Load characteristics of a separately excited dcgenerator

117 Voltage Build-Up in a Self-Excited GeneratorLet us see how voltage builds up in a self-excited generator(i) Shunt generator Consider a shunt generator If the generator is run at a constant speed some emf will be generated due to residual magnetism in the main poles This small emf circulates a field current which in turn produces additional flux toreinforce the original residual flux (provided field winding connections arecorrect) This process continues and the generator builds up the normalgenerated voltage following the OCC shown in Fig 117 (i) The field resistance Rf can be represented by a straight line passing through the origin as shown in Fig 117 (ii) The two curves can be shown on the same diagram as they have the same ordinate as in Fig 117 (iii)

Fig 117

Since the field circuit is inductive there is a delay in the increase in current uponclosing the field circuit switch The rate at which the current increases dependsupon the voltage available for increasing it Suppose at any instant the fieldcurrent is i (= OA) and is increasing at the rate didt Then

E0=i R f +iquest L didt

Where Rf = total field circuit resistance L = inductance of field circuit

At the considered instant the total emf available is AC [See Fig 117 (iii)] Anamount AB of the cmf AC is absorbed by the voltage drop iRf and theremainder part BC is available to overcome L didt Since this surplus voltage isavailable it is possible for the field current to increase above the value OAHowever at point D the available voltage is OM and is all absorbed by i RfD

rop Consequently the field current cannot increase further and the generatorbuild up stops Hence we arrive at a very important conclusion that the voltage build up of the generator is given by the point of intersection of OCC and field resistance line Thus in Fig 117 (iii) D is point of intersection of the two curves Hence the generator will build up a voltage OM(ii) Series generator During initial operation with no current yet flowing a residual voltage will be generated exactly as in the case of a shunt generator The residual voltage will cause a current to flow through the whole series circuit when the circuit isclosed There will then be voltage build up to an equilibrium point exactlyanalogous to the buildup of a shunt generator The voltage build up graph willbe similar to that of shunt generator except that now load current (instead offield current for shunt generator) will be taken along x-axis10487871048787(iii) Compound generator When a compound generator has its series field flux aiding its shunt field flux the machine is said to be cumulative compound When the series field isconnected in reverse so that its field flux opposes the shunt field flux thegenerator is then differential compound The easiest way to build up voltage in a compound generator is to start under no load conditions At no load only the shunt field is effective When no-load voltage build up is achieved the generator is loaded If under load the voltagerises the series field connection is cumulative If the voltage drops significantlythe connection is differential compound118 Critical Field Resistance for a Shunt Generator The voltage build up in a shunt generator depends upon field circuit resistance If the field circuit resistance is R1 (line OA) then generator will build up a voltage OM as shown in Fig 118 If the field circuit resistance is increasedto R2 (tine OB) the generator will build up a voltage OL slightly less than OM As the field circuit resistance is increased the slope of resistance line also increases When the field resistance line becomes tangent (line OC) to OCC the generator would just excite If the field circuit resistance is increased beyond this point (say line OD) the generator will fail to excite The field circuit resistance represented by line OC (tangent to OCC) is called critical field resistance RC for the shunt generator It may be defined as under The maximum field circuit resistance (for a given speed) with which the shunt generator would just excite is known as its critical field resistance It should be noted that shunt generator will build up voltage only if field circuit resistance is less than critical field resistance

Fig 118 Critical field resistance of shunt generator

119 Critical Resistance for a Series Generator

Fig 119 shows the voltage build up in a series generator Here R1 R2 etc represent the total circuit resistance (load resistance and field winding resistance) If the total circuit resistance is R1 then series generator will build up a voltage OL The line OC is tangent to OCC and represents the critical resistance RC for a series generator If the total resistance of the circuit is more than RC (say line OD) the generator will fail to build up voltage Note that Fig 119 is similar to Fig 118 with the following differences

(i) In Fig 118 R1 R2 etc represent the total field circuit resistanceHowever R1 R2 etc in Fig 119 represent the total circuit resistance(load resistance and series field winding resistance etc)

(ii) In Fig 118 field current alone is represented along X-axis However inFig 119 load current IL is represented along Y-axis Note that in aseries generator field current = load current IL

Fig 119 Critical resistance of a series generator

120 Characteristics of Series Generator Fig 120 (i) shows the connections of a series wound generator Since there is only one current (that which flows through the whole machine) the load current is the same as the exciting current(i) OCCCurve 1 shows the open circuit characteristic (OCC) of a series generator Itcan be obtained experimentally by disconnecting the field winding from themachine and exciting it from a separate dc source(ii) Internal characteristic Curve 2 shows the total or internal characteristic of a series generator It gives the relation between the generated emf E on load and armature current Due to armature reaction the flux in the machine will be less than the flux at no load Hence emf E generated under load conditions will be less than the emf E0 generated under no load conditions Consequently internal characteristic curvelies below the OCC curve the difference between them representing the effectof armature reaction [See Fig 120 (ii)]

Fig 120 Characteristics of series generator

(iii) External characteristic

Curve 3 shows the external characteristic of a series generator It gives therelation between terminal voltage and load current ILV = E g- Ia (Ra +Rse )Therefore external characteristic curve will lie below internal characteristiccurve by an amount equal to ohmic drop [ie Ia(Ra + Rse)] in the machine asshown in Fig 120 (ii) The internal and external characteristics ofa dc series generator can be plotted from one another as shown in Fig 121 Supposewe are given the internal characteristic of the generator Let the line OC represent the resistance of the whole machine ie Ra +Rse If the load current is OB drop in the machine is AB ieAB = Ohmic drop in the machine = OB(Ra + Rse)

Fig 121 Load characteristics of series generator

Now raise a perpendicular from point B and mark a point b on this line such that ab = AB Then point b will lie on the external characteristic of the generator

Following similar procedure other points of external characteristic can be located It is easy to see that we can also plot internal characteristic from the external characteristic

121 Characteristics of a Shunt Generator Fig 122 (i) show the connections of a shunt wound generator The armature current Ia splits up into two parts a small fraction Ish flowing through shunt field winding while the major part IL goes to the external load(i) OCC The OCC of a shunt generator is similar in shape to that of a series generator as shown in Fig 122 (ii) The line OA represents the shunt field circuitresistance When the generator is run at normal speed it will build up a voltageOM At no-load the terminal voltage of the generator will be constant (= OM)represented by the horizontal dotted line MC

Fig 122 Characteristics of a Shunt Generator

(ii) Internal characteristic

When the generator is loaded flux per pole is reduced due to armature reaction Therefore emf E generated on load is less than the emf generated at no load As a result the internal characteristic (EIa) drops down slightly as shown in Fig122 (ii)

(iii) External characteristic Curve 2 shows the external characteristic of a shunt generator It gives therelation between terminal voltage V and load current IL

Therefore external characteristic curve will lie below the internal characteristic curve by an amount equal to drop in the armature circuit[ie (IL + Ish)Ra] as shown in Fig 122 (ii)

Note It may be seen from the external characteristic that change in terminalvoltage from no-load to full load is small The terminal voltage can always bemaintained constant by adjusting the field rheostat R automatically122 Critical External Resistance for Shunt Generator

If the load resistance across the terminals of a shunt generator is decreased then load current increase However there is a limit to the increase in

load current with the decrease of load resistance Any decrease of load resistance beyond this point instead of increasing the current ultimately results in reduced current Consequently the external characteristic turns back (dotted curve) as shown in Fig 123 The tangent OA to the curve represents the minimum external resistance required to excite the shunt generator on load and is called critical external resistance If the resistance of the external circuit is less than the critical external resistance (represented by tangent OA in Fig 310) the machine will refuse to excite or will de-excite if already running This means that external resistance is so low as virtually to short circuit the machine and so doing away with its excitation

Note There are two critical resistances for a shunt generator viz (i) criticalfield resistance (ii) critical external resistance For the shunt generator to buildup voltage the former should not be exceeded and the latter must not be gonebelow

123 How to Draw OCC at Different Speeds

If we are given OCC of a generator at a constant speed N1 then we can easily draw the OCC at any other constant speed N2 Fig 123 illustrates theprocedure Here we are given OCC at a constant speed N1 It is desired to findthe OCC at constant speed N2 (it is assumed that n1 lt N2) For constantexcitation E propN

E2

E1 =

N 2

N1 so

E2 iquest E1

N2

N1

As shown in Fig (311) for If = OH E1 = HC Therefore the new value of emf (E2) for the same If but at N2

E2= HCN 2

N 1 = HD

Fig 123 Plotting OCC at another speed

This locates the point D on the new OCC at N2 Similarly other points can belocated taking different values of If The locus of these points will be the OCCat N2

124 Critical Speed (NC)

The critical speed of a shunt generator is the minimum speed below which it fails to excite Clearly it is the speed for which the given shunt field resistance represents the critical resistance In Fig 124 curve 2 corresponds to critical

speed because the shunt field resistance (Rsh) line is tangential to it If the generator runs at full speed N the new OCC moves upward and the Rsh line represents critical resistance for this speed

In order to find critical speed take any convenient point C on excitation axis and erect a perpendicular so as to cut Rsh and Rsh lines at points B and A respectively Then

BCAC

=NcN

Nc=N BCAC

Fig 124 Finding critical speed

125 Conditions for Voltage Build-Up of a Shunt Generator

The necessary conditions for voltage build-up in a shunt generator are

(i) There must be some residual magnetism in generator poles(ii) The connections of the field winding should be such that the field current strengthens the residual magnetism(iii) The resistance of the field circuit should be less than the critical resistance In other words the speed of the generator should be higher

than the critical speed

126 Compound Generator Characteristics

In a compound generator both series and shunt excitation are combined as shown in Fig 123 The shunt winding can be connected either across the armature only (short-shunt connection S) or across armature plus series field (long-shunt connection G) The compound generator can be cumulatively compounded or

differentially compounded generator The latter is rarely used in practice Therefore we shall discuss the characteristics of cumulatively compounded

generator It may be noted that external characteristics of long and short shunt compound generators are almost identical

External characteristic

Fig 124 shows the external characteristics of a cumulatively compoundedgenerator The series excitation aids the shunt excitation The degree of compounding depends upon the increase in series excitation with the increase inload current

(i) If series winding turns are so adjusted that with the increase in load current the terminal voltage increases it is called over-compounded generator In such a case as the load current increases the series field mmf increases and tends to increase the flux and hence the generated voltage The increase in generated voltage is greater than the IaRa drop so that instead of decreasing the terminal voltage increases as shown by curve A in Fig 124

(ii) If series winding turns are so adjusted that with the increase in load current the terminal voltage substantially remains constant it is called Flat-compounded generator The series winding of such a machine has lesser number of turns than the one in over-compounded machine and therefore does not increase the flux as much for a given load current Consequently the full-load voltage is nearly equal to the no-load voltage as indicated by curve B in Fig 124 (iii) If series field winding has lesser number of turns than for a flat compounded machine the terminal voltage falls with increase in load current as indicated by curve C m Fig 124 Such a machine is called under-compounded generator

Fig 123 compound generator

Fig 124 Load characteristics of compound generators

127 LOSSES

The losses in a dc machine (generator or motor) may be divided into three classes viz (i) copper losses or electrical losses (ii) iron or core losses and (iii) mechanical losses All these losses appear as heat and thus raise the temperature of the machine They also lower the efficiency of the machine The loss structure maybe demonstrated as

1Copperlosses

These losses occur due to currents in the various windings of the machine When an electric current of I amps flows in a resistance of R ohms the heat energy is lost at the rate of I2R joules sec and the power loss is I2R watts Generators and motors have one or more field circuits and an armature circuit in which such losses occur All resistance losses of this kind are called copper losses The different copper losses occurring in a dc machine are

(i) Armature copper loss = Ia2Ra

(ii) Shunt field copper loss = Ish2 Rsh

(iii) Series field copper loss = Ise2Rse

Armature copper loss will be around 30 to 40 of full load losses where as total field copper losses will constitute around 30 of full load losses

Note There is also brush contact loss due to brush contact resistance (ie resistance between the surface of brush and surface of commutator) This loss is generally included in armature copper loss The brush contact drop is almost independent of Ia For carbon brushes this drop is around 1 V per brush The power loss due to brush contact resistance is 2ebIa where eb is the voltage drop at one brush

2 IRONORCORELOSSES These losses occur in the armature of a dc machine and are due to the rotation of armature in the magnetic field of the poles They are of two types viz (i) hysteresis loss (ii) eddy current loss

(i) Hysteresis loss

Hysteresis loss occurs in the armature of the dc machine since any given part of the armature is subjected to magnetic field reversals as it passes under successive poles Fig 124 shows an armature rotating in two-pole machine Consider a small piece ab of the armature When the piece ab is under N-pole the magnetic lines pass from a to b Half a revolution later the same piece of iron is under S-pole and magnetic lines pass from b to a so that magnetism in the iron is reversed In order to reverse continuously the molecular magnets in the armature core some amount of power has to be spent which is called hysteresis loss It is given by Steinmetz formula This formula is

Hysteresis Loss Wh = η Bmax 16f V Watts Bmax - Maximum flux density in the core f - Frequency of magnetic reversals V - Volume of armature in m3

η - Steinmetz hysteresis coefficient

In order to reduce this loss in a dc machine armature core is made of such materials which have a low value of Steinmetz hysteresis co-efficient eg silicon steel

Fig 124 Occurrence of hysteresis loss

ii) EDDYCURRENTLOSS

In addition to the voltages induced in the armature conductors there are also voltages induced in the armature core These voltages produce circulating currents in the armature core as shown in Fig 125 These are called eddy currents and power loss due to their flow is called eddy current loss The eddy current loss appears as heat which raises the temperature of the machine and lowers its efficiency Core resistance can be greatly increased by constructing the core of thin round If a continuous solid iron core is used the resistance to eddy current path will be small due to large cross-sectional area of the core Consequently the magnitude of eddy current and hence eddy current loss will be large The magnitude of eddy current can be reduced by making core resistance as high as practical The iron sheets called laminations [See Fig 126] The laminations are insulated from each other with a coating of varnish The insulating coating has a high resistance so very little current flows from one lamination to the other Alsobecause each lamination is very thin the resistance to current flowing through the width of a lamination is also quite large Thus laminating a core increases the core resistance which decreases the eddy current and hence the eddy current loss

Fig 125 Eddy current loss Fig 126 Core laminations

Eddy current loss is biven by the formula Pe= Ke Bmax

2 f2t2 V Watts

Where Ke = Constant depending upon electrical resistace of core Bmax = Maximum flux density in wbm2

f = frequency f magnetic reversals in Hz t = thickness of laminations in m and V = volume of core in m3

It may be noted that eddy current loss depends upon the square of lamination thickness For this reason lamination thickness should be kept as small as possible

3MECHANICALLOSSES

These losses are due to friction and windage(i) Friction losses eg bearing friction brush friction etc(ii) Windage loss ie air friction of rotating armature

These losses depend upon the speed of the machine But for a given speed they are practically constant

Note Iron losses and mechanical losses together are called stray losses

CONSTANT AND VARIABLE LOSSES

The losses in a dc generator (or dc motor) may be sub-divided into

(i) constant losses (ii) variable losses

(i) Constant losses

Those losses in a dc generator which remain constant at all loads are known as constant losses The constant losses in a dc generator are(a) iron losses(b) mechanical losses(c) shunt field losses(ii) Variable lossesThose losses in a dc generator which vary with load are called variable lossesThe variable losses in a dc generator are (a) Copper loss in armature winding (Ia

2 Ra) (b) Copper loss in series field winding (Ise

2Rse)

Total losses = Constant losses + Variable lossesNote Field Cu loss is constant for shunt and compound generators

128 POWER FLOW DIAGRAM

The power flow in a dc generator may be represented as

The different power stages are shown as A B and CA ndash The mechanical power input which is the power output of the prime mover is given as Pin = 2πNT

60 watts

B - The power developed in armature is Pa = Eg Ia watts C- The electrical power output is Pout = VIL watts

129 EFFICIENCY Efficiency of a machine is the ratio of its output power to the input power

Efficiency η = Output powerInput power

= Input powerminusLossesInput power = Output power

Output power+Losses

In case of a dc generator depending upon the power stages three different efficiencies can be defined as-

1 Mechanical Efficiency(ηm)

BA = Electrical power developedisinarmature

Mechanical power input2 Electrical Efficiency(ηe)

CB = Electrical power output

Electrical power developedisinthe armature 3 Overall or commercial efficiency (η)

CA = Electrical power output

Mechanical power input = ηmtimes ηe

130 CONDITION FOR MAXIMUM EFFICIENCY The efficiency of a dc generator is not constant but varies with load Fig 127shows the variation of efficiency with load current Consider a dc shunt generator with terminal voltage V delivering a load current of IL amp Generator output = V IL

Generator input = Output + Losses = V IL + constant losses + Variable losses = V IL + Wc+ Ia

2Ra

= V IL + Wc + (IL+ Ish)2 Rsh

The shunt field current Ish is small compared with IL so may be neglected

Generator input = VIL+ IL2Ra + Wc

Now efficiency η iquest outputinput = VI L

VIL+ Ia 2Ra+Wc

Efficiency is maximum when the denominator of the above equation is minimumie

ddIL ( VI L

VIL+ Ia 2Ra+Wc ) = 0

Which gives IL2 Ra = WC

ie Variable loss = constant lossThe load current for maximum efficiency is given by

IL iquest radicWcRa

Fig 127 Variation of efficiency with load

Hence the efficiency is maximum for that load current at which the variable loss is equal to the constant loss

131 APPLICATIONS

Separately excited generators

These types of generators have special applications such as electroplating electro refining of materials because separate supply is required for field excitation

Shunt generators

These types of generators are used in battery charging and ordinary lighting purposes

Series generators

This type of generators is used commonly as boosters on DC feeders as current generators for welding and arc lamp

Cumulatively compounded generators

This type of generators is used for domestic lighting purposes and to transmit energy over long distances

Differential compound generators

This type of generators is rarely used They are used for special applications such as electric arc welding

132 NUMERICAL EXCERCISES

1 The emf induced in the armature of a 450 kW 250 volt shunt generator is

2588volt when the field current is 200 amp and the generator is supplying power to a load at rated terminal voltage The armature circuit resistance is 0005 ohm Determine (i) load current (ii) power generated (iii) power output and (iv) electrical efficiency Neglect brush contact drop

Solution

Let the load current be I amp Given Ish = 20 A

Armature current = Load current + field current or Ia = I + 20 Induced emf = Terminal voltage + Ia Ra drop

2588 = 250 + (I + 20) times 0005

So (i) Load current I = 1740 amp

(ii) Power generated E Ia = 2588 (1740 + 20) watts = 455488 kW (iii) Power output V I = 250 times 1740 watts

= 435 kW

(iv) Electrical efficiency = 435455488 times 100 = 955

2 A 4 pole wave wound dc shunt generator delivers a load of 45 kW at a terminal voltage of 220 volts Its armature has 150 single turn coils and has a resistance of 001 ohm The air gap flux per pole is 002 weber Shunt field resistance is 50 ohm Calculate the speed at which it is being driven Neglect brush contact drop

Solution Load current I = W

V = 45000220 = 20455 amp

Field Current Ish = V

R s h = 220

50 = 44 amp

Armature current Ia = I +I sh = 20455 +44 = 20895 amp

Armature drop I aRa = 20895 X 001 = 20895 volts Induced emf Eg = V + Ia Ra = 220 + 20895 = 2220895 volts

But induced emf

So N = Eg60 AΦ ZP = 2220895times 60 times2

002 times150 times2 times 4 = 111045 rpm

3 The following data refer to the OCC of a DC separately excited generator at 1000 rpm

Field current in A 00 02 03 04 05 06 07 08 09 10 11 12Armature voltage in V 5 40 75 100 124 145 162 178 188 195 200 205

The machine is now connected as a shunt generator with a total field resistance of 200 Ω and driven at 1000 rpm Estimate graphically (i) the voltage to which the generator will build up at no load (ii) the armature field and load currents when the terminal voltage is found to be 150 V Neglect the effect of armature reaction and brush drop and assume armature resistance Ra to be 08 Ω (iii) Finally estimate the steady state armature current when the machine terminals are shorted

Solution

First of all the OCC of the generator is plotted in a graph paper as shown in figure

(i) The total field circuit resistance is given to be Rf = 200 Ω The Rf line is now drawn on the same graph paper passing through the origin The point of intersection of the Rf line and the OCC decides the final no load voltage and can be read from the graph as 192 V and the corresponding field current is 096 A (ii) Since the terminal voltage V is 150 V (= BM) field current If

is OM = 077 A Generated voltage E is given by AM But we know E = IaRa + V Hence IaRa= E ndash V = AM ndash BM = AB Now from the graph AB = 25 V there4Ia = 2508 = 3125 A So load current IL= Ia ndash If= 3125 ndash 077 = 3048 A

ADDITIONAL PROBLEMS

1 A 4 pole lap wound 115 kW 230 volts dc shunt generator has armatureresistance of 02 ohm and field resistance of 100 ohms Calculate (i) emf

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES

15 TYPES OF DC GENERATORS

Fig18 Classification of dc generators

151 Separately Excited DC Generator

This dc generator has a field magnet winding which is excited using a separate voltage source (like battery)The representation in the below figure 19 The output voltage depends on the speed of rotation of armature and field current The higher the speed of rotation and current ndash the higher the output emf

Fig 19 Separately Excited DC Generators

Armature Current Ia = IL

Terminal voltage V = Eg-Ia Ra

Electric Power developed = Eg Ia

Power delivered to load = V IL = (Eg-Ia Ra) Ia = Eg Ia - Ia2

Ra

152 Self Excited DC Generator

These are generators in which the field winding is excited by the output of the generator itself As described before ndash there are three types of self excited dc generators ndash they are 1) Series 2) Shunt and 3) Compound

A series DC generator is shown below in fig110 (a) ndash in which the armature winding is connected in series with the field winding so that the field current flows through the load as well as the field winding Field winding is a low resistance thick wire of few turns Armature current Ia =Ise = IL=I (say)

Terminal voltage V = Eg- I( Ra+ Rse)Power Developed in armature = Eg Ia

Power delivered to load = V IL = Eg Ia- Ia2 (Ra+Rse)

Fig110 (a) Series generator and (b) shunt generator

A shunt DC generator is shown in figure 110 (b) in which the field winding is wired parallel to armature winding so that the voltage across both are same The field winding has high resistance and more number of turns so that only a part of armature current passes through field winding and the rest passes through load Shunt field current Ish = V Rsh

Armature current Ia = IL+ Ish

Terminal voltage V= Eg-Ia Ra

Power developed in armature = EgIa

Power delivered to load = VIL A compound generator is shown in figure 111 below It has two field findings namely Rsh and Rse They are basically shunt winding (Rsh) and series winding (Rse) Compound generator is of two types ndash 1) Short shunt and 2) Long shunt

Short shunt- Here the shunt field winding is wired parallel to armature and series field winding is connected in series to the load It is shown in fig111(a)

In short shunt machines Series field current Ise = IL

Shunt field current Ish iquest V+ I se RseRsh

Terminal voltage V= Eg-Ia Ra-Ise Rse

Power developedin armature = Eg Ia

Power delivered to load = V IL

Fig 111 (a) short shunt generator and (b) Long shunt generator

Long shunt- Here the shunt field winding is parallel to both armature and series field winding (Rse is wired in series to the armature) It is shown in figure 111 (b)

In Long shunt machines Series field current Ise = Ia = IL+ Ish

Shunt field current Ish = V R sh

Terminal voltage V = Eg- Ia (Ra+ Rse) Power developed in armature = Eg Ia

Power delivered to load = VIL

16 GENERATOR CHARACTERISTICS

The speed of a dc machine operated as a generator is fixed by the prime mover For general-purpose operation the prime mover is equipped with a speed governor so that the speed of the generator is practically constant Under such condition the generator performance deals primarily with the relation between excitation terminal voltage and load These relations can be best exhibited graphically by means of curves known as generator characteristics Thesecharacteristics show at a glance the behavior of the generator under differentload conditions

The following are the three most important characteristics of a dc generator161 Open Circuit Characteristic (OCC)

This curve shows the relation between the generated emf at no-load (E0) and the field current (If) at constant speed It is also known as magnetic characteristic or no-load saturation curve Its shape is practically the same for all generators whether separately or self-excited The data for OCC curve are obtained experimentally by operating the generator at no load and constant speed and recording the change in terminal voltage as the field current is varied

162 Internal or Total characteristic (EIa)

This curve shows the relation between the generated emf on load (E) and the armature current (Ia) The emf E is less than E0 due to the demagnetizing effect of armature reaction Therefore this curve will lie below the open circuitcharacteristic (OCC) The internal characteristic is of interest chiefly to thedesigner It cannot be obtained directly by experiment It is because a voltmetercannot read the emf generated on load due to the voltage drop in armatureresistance The internal characteristic can be obtained from externalcharacteristics if winding resistances are known because armature reaction effectis included in both characteristics163 External characteristic (VIL)This curve shows the relation between the terminal voltage (V) and load current(IL) The terminal voltage V will be less than E due to voltage drop in thearmature circuit Therefore this curve will lie below the internal characteristicThis characteristic is very important in determining the suitability of a generatorfor a given purpose It can be obtained by making simultaneous measurements

of terminal voltage and load current (with voltmeter and ammeter) of a loadedgenerator164 Open Circuit Characteristic of a DC Generator

The OCC for a dc generator is determined as follows The field winding of the dc generator (series or shunt) is disconnected from the machine and isseparately excited from an external dc source as shown in Fig 112 Thegenerator is run at fixed speed (ie normal speed) The field current (If) isincreased from zero in steps and the corresponding values of generated emf(E0) read off on a voltmeter connected across the armature terminals On plottingthe relation between E0 and If we get the open circuit characteristic as shown in

Fig113

Fig 112 Experimental setup for plotting OCC

Fig 113 OCC of dc generator

The following points may be noted from OCC When the field current is zero there is some generated emf OA

This is due to the residual magnetism in the field poles Over a fairly wide range of field current (upto point B in the curve)

the curve is linear It is because in this range reluctance of iron is negligible as compared with that of air gap The air gap reluctance is constant and hence linear relationship After point B on the curve the reluctance of iron also comes into picture and reluctance of iron is no longer negligible Consequently the curve deviates from linear relationship

After point C on the curve the magnetic saturation of poles begins and E0 tends to level off

The OCC of even self-excited generator is obtained by running it as a separately excited generator

165 Characteristics of a Separately Excited DCGenerator

The obvious disadvantage of a separately excited dc generator is that we require an external dc source for excitation But since the output voltage may be

controlled more easily and over a wide range (from zero to a maximum) this type of excitation finds many applications

(i) Open circuit characteristic

The OCC of a separately excited generator is determined in a manner described in Sec 164 Fig114 shows the variation of generated em f on noload with field current for various fixed speeds Note that if the value of constant speed is increased the steepness of the curve also increases When the field current is zero the residual magnetism in the poles will give rise to the small initial emf as shown

Fig114 Variation of OCC with speed

Internal and External Characteristics (Load Characteristics)

The external characteristic of a separately excited generator is the curve between the terminal voltage (V) and the load current IL (which is the same as armature current in this case) In order to determine the external characteristic the circuit set up is as shown in Fig 115

Fig 115 Experimental setup to plot load characteristics

As the load current increases the terminal voltage falls due to two reasons(a) The armature reaction weakens the main flux so that actual emf generated E on load is less than that generated (E0) on no load

(b) There is voltage drop across armature resistance (= ILRa = IaRa)Due to these reasons the external characteristic is a drooping curve [curve 3 inFig 116 ] Note that in the absence of armature reaction and armature dropthe generated emf would have been E0 (curve 1)The internal characteristic can be determined from external characteristic by adding IL Ra drop to the external characteristic It is because armature reaction drop is included in the external characteristic Curve 2 is the internal characteristic of the generator and should obviously lie above the external characteristic

Fig 116 Load characteristics of a separately excited dcgenerator

117 Voltage Build-Up in a Self-Excited GeneratorLet us see how voltage builds up in a self-excited generator(i) Shunt generator Consider a shunt generator If the generator is run at a constant speed some emf will be generated due to residual magnetism in the main poles This small emf circulates a field current which in turn produces additional flux toreinforce the original residual flux (provided field winding connections arecorrect) This process continues and the generator builds up the normalgenerated voltage following the OCC shown in Fig 117 (i) The field resistance Rf can be represented by a straight line passing through the origin as shown in Fig 117 (ii) The two curves can be shown on the same diagram as they have the same ordinate as in Fig 117 (iii)

Fig 117

Since the field circuit is inductive there is a delay in the increase in current uponclosing the field circuit switch The rate at which the current increases dependsupon the voltage available for increasing it Suppose at any instant the fieldcurrent is i (= OA) and is increasing at the rate didt Then

E0=i R f +iquest L didt

Where Rf = total field circuit resistance L = inductance of field circuit

At the considered instant the total emf available is AC [See Fig 117 (iii)] Anamount AB of the cmf AC is absorbed by the voltage drop iRf and theremainder part BC is available to overcome L didt Since this surplus voltage isavailable it is possible for the field current to increase above the value OAHowever at point D the available voltage is OM and is all absorbed by i RfD

rop Consequently the field current cannot increase further and the generatorbuild up stops Hence we arrive at a very important conclusion that the voltage build up of the generator is given by the point of intersection of OCC and field resistance line Thus in Fig 117 (iii) D is point of intersection of the two curves Hence the generator will build up a voltage OM(ii) Series generator During initial operation with no current yet flowing a residual voltage will be generated exactly as in the case of a shunt generator The residual voltage will cause a current to flow through the whole series circuit when the circuit isclosed There will then be voltage build up to an equilibrium point exactlyanalogous to the buildup of a shunt generator The voltage build up graph willbe similar to that of shunt generator except that now load current (instead offield current for shunt generator) will be taken along x-axis10487871048787(iii) Compound generator When a compound generator has its series field flux aiding its shunt field flux the machine is said to be cumulative compound When the series field isconnected in reverse so that its field flux opposes the shunt field flux thegenerator is then differential compound The easiest way to build up voltage in a compound generator is to start under no load conditions At no load only the shunt field is effective When no-load voltage build up is achieved the generator is loaded If under load the voltagerises the series field connection is cumulative If the voltage drops significantlythe connection is differential compound118 Critical Field Resistance for a Shunt Generator The voltage build up in a shunt generator depends upon field circuit resistance If the field circuit resistance is R1 (line OA) then generator will build up a voltage OM as shown in Fig 118 If the field circuit resistance is increasedto R2 (tine OB) the generator will build up a voltage OL slightly less than OM As the field circuit resistance is increased the slope of resistance line also increases When the field resistance line becomes tangent (line OC) to OCC the generator would just excite If the field circuit resistance is increased beyond this point (say line OD) the generator will fail to excite The field circuit resistance represented by line OC (tangent to OCC) is called critical field resistance RC for the shunt generator It may be defined as under The maximum field circuit resistance (for a given speed) with which the shunt generator would just excite is known as its critical field resistance It should be noted that shunt generator will build up voltage only if field circuit resistance is less than critical field resistance

Fig 118 Critical field resistance of shunt generator

119 Critical Resistance for a Series Generator

Fig 119 shows the voltage build up in a series generator Here R1 R2 etc represent the total circuit resistance (load resistance and field winding resistance) If the total circuit resistance is R1 then series generator will build up a voltage OL The line OC is tangent to OCC and represents the critical resistance RC for a series generator If the total resistance of the circuit is more than RC (say line OD) the generator will fail to build up voltage Note that Fig 119 is similar to Fig 118 with the following differences

(i) In Fig 118 R1 R2 etc represent the total field circuit resistanceHowever R1 R2 etc in Fig 119 represent the total circuit resistance(load resistance and series field winding resistance etc)

(ii) In Fig 118 field current alone is represented along X-axis However inFig 119 load current IL is represented along Y-axis Note that in aseries generator field current = load current IL

Fig 119 Critical resistance of a series generator

120 Characteristics of Series Generator Fig 120 (i) shows the connections of a series wound generator Since there is only one current (that which flows through the whole machine) the load current is the same as the exciting current(i) OCCCurve 1 shows the open circuit characteristic (OCC) of a series generator Itcan be obtained experimentally by disconnecting the field winding from themachine and exciting it from a separate dc source(ii) Internal characteristic Curve 2 shows the total or internal characteristic of a series generator It gives the relation between the generated emf E on load and armature current Due to armature reaction the flux in the machine will be less than the flux at no load Hence emf E generated under load conditions will be less than the emf E0 generated under no load conditions Consequently internal characteristic curvelies below the OCC curve the difference between them representing the effectof armature reaction [See Fig 120 (ii)]

Fig 120 Characteristics of series generator

(iii) External characteristic

Curve 3 shows the external characteristic of a series generator It gives therelation between terminal voltage and load current ILV = E g- Ia (Ra +Rse )Therefore external characteristic curve will lie below internal characteristiccurve by an amount equal to ohmic drop [ie Ia(Ra + Rse)] in the machine asshown in Fig 120 (ii) The internal and external characteristics ofa dc series generator can be plotted from one another as shown in Fig 121 Supposewe are given the internal characteristic of the generator Let the line OC represent the resistance of the whole machine ie Ra +Rse If the load current is OB drop in the machine is AB ieAB = Ohmic drop in the machine = OB(Ra + Rse)

Fig 121 Load characteristics of series generator

Now raise a perpendicular from point B and mark a point b on this line such that ab = AB Then point b will lie on the external characteristic of the generator

Following similar procedure other points of external characteristic can be located It is easy to see that we can also plot internal characteristic from the external characteristic

121 Characteristics of a Shunt Generator Fig 122 (i) show the connections of a shunt wound generator The armature current Ia splits up into two parts a small fraction Ish flowing through shunt field winding while the major part IL goes to the external load(i) OCC The OCC of a shunt generator is similar in shape to that of a series generator as shown in Fig 122 (ii) The line OA represents the shunt field circuitresistance When the generator is run at normal speed it will build up a voltageOM At no-load the terminal voltage of the generator will be constant (= OM)represented by the horizontal dotted line MC

Fig 122 Characteristics of a Shunt Generator

(ii) Internal characteristic

When the generator is loaded flux per pole is reduced due to armature reaction Therefore emf E generated on load is less than the emf generated at no load As a result the internal characteristic (EIa) drops down slightly as shown in Fig122 (ii)

(iii) External characteristic Curve 2 shows the external characteristic of a shunt generator It gives therelation between terminal voltage V and load current IL

Therefore external characteristic curve will lie below the internal characteristic curve by an amount equal to drop in the armature circuit[ie (IL + Ish)Ra] as shown in Fig 122 (ii)

Note It may be seen from the external characteristic that change in terminalvoltage from no-load to full load is small The terminal voltage can always bemaintained constant by adjusting the field rheostat R automatically122 Critical External Resistance for Shunt Generator

If the load resistance across the terminals of a shunt generator is decreased then load current increase However there is a limit to the increase in

load current with the decrease of load resistance Any decrease of load resistance beyond this point instead of increasing the current ultimately results in reduced current Consequently the external characteristic turns back (dotted curve) as shown in Fig 123 The tangent OA to the curve represents the minimum external resistance required to excite the shunt generator on load and is called critical external resistance If the resistance of the external circuit is less than the critical external resistance (represented by tangent OA in Fig 310) the machine will refuse to excite or will de-excite if already running This means that external resistance is so low as virtually to short circuit the machine and so doing away with its excitation

Note There are two critical resistances for a shunt generator viz (i) criticalfield resistance (ii) critical external resistance For the shunt generator to buildup voltage the former should not be exceeded and the latter must not be gonebelow

123 How to Draw OCC at Different Speeds

If we are given OCC of a generator at a constant speed N1 then we can easily draw the OCC at any other constant speed N2 Fig 123 illustrates theprocedure Here we are given OCC at a constant speed N1 It is desired to findthe OCC at constant speed N2 (it is assumed that n1 lt N2) For constantexcitation E propN

E2

E1 =

N 2

N1 so

E2 iquest E1

N2

N1

As shown in Fig (311) for If = OH E1 = HC Therefore the new value of emf (E2) for the same If but at N2

E2= HCN 2

N 1 = HD

Fig 123 Plotting OCC at another speed

This locates the point D on the new OCC at N2 Similarly other points can belocated taking different values of If The locus of these points will be the OCCat N2

124 Critical Speed (NC)

The critical speed of a shunt generator is the minimum speed below which it fails to excite Clearly it is the speed for which the given shunt field resistance represents the critical resistance In Fig 124 curve 2 corresponds to critical

speed because the shunt field resistance (Rsh) line is tangential to it If the generator runs at full speed N the new OCC moves upward and the Rsh line represents critical resistance for this speed

In order to find critical speed take any convenient point C on excitation axis and erect a perpendicular so as to cut Rsh and Rsh lines at points B and A respectively Then

BCAC

=NcN

Nc=N BCAC

Fig 124 Finding critical speed

125 Conditions for Voltage Build-Up of a Shunt Generator

The necessary conditions for voltage build-up in a shunt generator are

(i) There must be some residual magnetism in generator poles(ii) The connections of the field winding should be such that the field current strengthens the residual magnetism(iii) The resistance of the field circuit should be less than the critical resistance In other words the speed of the generator should be higher

than the critical speed

126 Compound Generator Characteristics

In a compound generator both series and shunt excitation are combined as shown in Fig 123 The shunt winding can be connected either across the armature only (short-shunt connection S) or across armature plus series field (long-shunt connection G) The compound generator can be cumulatively compounded or

differentially compounded generator The latter is rarely used in practice Therefore we shall discuss the characteristics of cumulatively compounded

generator It may be noted that external characteristics of long and short shunt compound generators are almost identical

External characteristic

Fig 124 shows the external characteristics of a cumulatively compoundedgenerator The series excitation aids the shunt excitation The degree of compounding depends upon the increase in series excitation with the increase inload current

(i) If series winding turns are so adjusted that with the increase in load current the terminal voltage increases it is called over-compounded generator In such a case as the load current increases the series field mmf increases and tends to increase the flux and hence the generated voltage The increase in generated voltage is greater than the IaRa drop so that instead of decreasing the terminal voltage increases as shown by curve A in Fig 124

(ii) If series winding turns are so adjusted that with the increase in load current the terminal voltage substantially remains constant it is called Flat-compounded generator The series winding of such a machine has lesser number of turns than the one in over-compounded machine and therefore does not increase the flux as much for a given load current Consequently the full-load voltage is nearly equal to the no-load voltage as indicated by curve B in Fig 124 (iii) If series field winding has lesser number of turns than for a flat compounded machine the terminal voltage falls with increase in load current as indicated by curve C m Fig 124 Such a machine is called under-compounded generator

Fig 123 compound generator

Fig 124 Load characteristics of compound generators

127 LOSSES

The losses in a dc machine (generator or motor) may be divided into three classes viz (i) copper losses or electrical losses (ii) iron or core losses and (iii) mechanical losses All these losses appear as heat and thus raise the temperature of the machine They also lower the efficiency of the machine The loss structure maybe demonstrated as

1Copperlosses

These losses occur due to currents in the various windings of the machine When an electric current of I amps flows in a resistance of R ohms the heat energy is lost at the rate of I2R joules sec and the power loss is I2R watts Generators and motors have one or more field circuits and an armature circuit in which such losses occur All resistance losses of this kind are called copper losses The different copper losses occurring in a dc machine are

(i) Armature copper loss = Ia2Ra

(ii) Shunt field copper loss = Ish2 Rsh

(iii) Series field copper loss = Ise2Rse

Armature copper loss will be around 30 to 40 of full load losses where as total field copper losses will constitute around 30 of full load losses

Note There is also brush contact loss due to brush contact resistance (ie resistance between the surface of brush and surface of commutator) This loss is generally included in armature copper loss The brush contact drop is almost independent of Ia For carbon brushes this drop is around 1 V per brush The power loss due to brush contact resistance is 2ebIa where eb is the voltage drop at one brush

2 IRONORCORELOSSES These losses occur in the armature of a dc machine and are due to the rotation of armature in the magnetic field of the poles They are of two types viz (i) hysteresis loss (ii) eddy current loss

(i) Hysteresis loss

Hysteresis loss occurs in the armature of the dc machine since any given part of the armature is subjected to magnetic field reversals as it passes under successive poles Fig 124 shows an armature rotating in two-pole machine Consider a small piece ab of the armature When the piece ab is under N-pole the magnetic lines pass from a to b Half a revolution later the same piece of iron is under S-pole and magnetic lines pass from b to a so that magnetism in the iron is reversed In order to reverse continuously the molecular magnets in the armature core some amount of power has to be spent which is called hysteresis loss It is given by Steinmetz formula This formula is

Hysteresis Loss Wh = η Bmax 16f V Watts Bmax - Maximum flux density in the core f - Frequency of magnetic reversals V - Volume of armature in m3

η - Steinmetz hysteresis coefficient

In order to reduce this loss in a dc machine armature core is made of such materials which have a low value of Steinmetz hysteresis co-efficient eg silicon steel

Fig 124 Occurrence of hysteresis loss

ii) EDDYCURRENTLOSS

In addition to the voltages induced in the armature conductors there are also voltages induced in the armature core These voltages produce circulating currents in the armature core as shown in Fig 125 These are called eddy currents and power loss due to their flow is called eddy current loss The eddy current loss appears as heat which raises the temperature of the machine and lowers its efficiency Core resistance can be greatly increased by constructing the core of thin round If a continuous solid iron core is used the resistance to eddy current path will be small due to large cross-sectional area of the core Consequently the magnitude of eddy current and hence eddy current loss will be large The magnitude of eddy current can be reduced by making core resistance as high as practical The iron sheets called laminations [See Fig 126] The laminations are insulated from each other with a coating of varnish The insulating coating has a high resistance so very little current flows from one lamination to the other Alsobecause each lamination is very thin the resistance to current flowing through the width of a lamination is also quite large Thus laminating a core increases the core resistance which decreases the eddy current and hence the eddy current loss

Fig 125 Eddy current loss Fig 126 Core laminations

Eddy current loss is biven by the formula Pe= Ke Bmax

2 f2t2 V Watts

Where Ke = Constant depending upon electrical resistace of core Bmax = Maximum flux density in wbm2

f = frequency f magnetic reversals in Hz t = thickness of laminations in m and V = volume of core in m3

It may be noted that eddy current loss depends upon the square of lamination thickness For this reason lamination thickness should be kept as small as possible

3MECHANICALLOSSES

These losses are due to friction and windage(i) Friction losses eg bearing friction brush friction etc(ii) Windage loss ie air friction of rotating armature

These losses depend upon the speed of the machine But for a given speed they are practically constant

Note Iron losses and mechanical losses together are called stray losses

CONSTANT AND VARIABLE LOSSES

The losses in a dc generator (or dc motor) may be sub-divided into

(i) constant losses (ii) variable losses

(i) Constant losses

Those losses in a dc generator which remain constant at all loads are known as constant losses The constant losses in a dc generator are(a) iron losses(b) mechanical losses(c) shunt field losses(ii) Variable lossesThose losses in a dc generator which vary with load are called variable lossesThe variable losses in a dc generator are (a) Copper loss in armature winding (Ia

2 Ra) (b) Copper loss in series field winding (Ise

2Rse)

Total losses = Constant losses + Variable lossesNote Field Cu loss is constant for shunt and compound generators

128 POWER FLOW DIAGRAM

The power flow in a dc generator may be represented as

The different power stages are shown as A B and CA ndash The mechanical power input which is the power output of the prime mover is given as Pin = 2πNT

60 watts

B - The power developed in armature is Pa = Eg Ia watts C- The electrical power output is Pout = VIL watts

129 EFFICIENCY Efficiency of a machine is the ratio of its output power to the input power

Efficiency η = Output powerInput power

= Input powerminusLossesInput power = Output power

Output power+Losses

In case of a dc generator depending upon the power stages three different efficiencies can be defined as-

1 Mechanical Efficiency(ηm)

BA = Electrical power developedisinarmature

Mechanical power input2 Electrical Efficiency(ηe)

CB = Electrical power output

Electrical power developedisinthe armature 3 Overall or commercial efficiency (η)

CA = Electrical power output

Mechanical power input = ηmtimes ηe

130 CONDITION FOR MAXIMUM EFFICIENCY The efficiency of a dc generator is not constant but varies with load Fig 127shows the variation of efficiency with load current Consider a dc shunt generator with terminal voltage V delivering a load current of IL amp Generator output = V IL

Generator input = Output + Losses = V IL + constant losses + Variable losses = V IL + Wc+ Ia

2Ra

= V IL + Wc + (IL+ Ish)2 Rsh

The shunt field current Ish is small compared with IL so may be neglected

Generator input = VIL+ IL2Ra + Wc

Now efficiency η iquest outputinput = VI L

VIL+ Ia 2Ra+Wc

Efficiency is maximum when the denominator of the above equation is minimumie

ddIL ( VI L

VIL+ Ia 2Ra+Wc ) = 0

Which gives IL2 Ra = WC

ie Variable loss = constant lossThe load current for maximum efficiency is given by

IL iquest radicWcRa

Fig 127 Variation of efficiency with load

Hence the efficiency is maximum for that load current at which the variable loss is equal to the constant loss

131 APPLICATIONS

Separately excited generators

These types of generators have special applications such as electroplating electro refining of materials because separate supply is required for field excitation

Shunt generators

These types of generators are used in battery charging and ordinary lighting purposes

Series generators

This type of generators is used commonly as boosters on DC feeders as current generators for welding and arc lamp

Cumulatively compounded generators

This type of generators is used for domestic lighting purposes and to transmit energy over long distances

Differential compound generators

This type of generators is rarely used They are used for special applications such as electric arc welding

132 NUMERICAL EXCERCISES

1 The emf induced in the armature of a 450 kW 250 volt shunt generator is

2588volt when the field current is 200 amp and the generator is supplying power to a load at rated terminal voltage The armature circuit resistance is 0005 ohm Determine (i) load current (ii) power generated (iii) power output and (iv) electrical efficiency Neglect brush contact drop

Solution

Let the load current be I amp Given Ish = 20 A

Armature current = Load current + field current or Ia = I + 20 Induced emf = Terminal voltage + Ia Ra drop

2588 = 250 + (I + 20) times 0005

So (i) Load current I = 1740 amp

(ii) Power generated E Ia = 2588 (1740 + 20) watts = 455488 kW (iii) Power output V I = 250 times 1740 watts

= 435 kW

(iv) Electrical efficiency = 435455488 times 100 = 955

2 A 4 pole wave wound dc shunt generator delivers a load of 45 kW at a terminal voltage of 220 volts Its armature has 150 single turn coils and has a resistance of 001 ohm The air gap flux per pole is 002 weber Shunt field resistance is 50 ohm Calculate the speed at which it is being driven Neglect brush contact drop

Solution Load current I = W

V = 45000220 = 20455 amp

Field Current Ish = V

R s h = 220

50 = 44 amp

Armature current Ia = I +I sh = 20455 +44 = 20895 amp

Armature drop I aRa = 20895 X 001 = 20895 volts Induced emf Eg = V + Ia Ra = 220 + 20895 = 2220895 volts

But induced emf

So N = Eg60 AΦ ZP = 2220895times 60 times2

002 times150 times2 times 4 = 111045 rpm

3 The following data refer to the OCC of a DC separately excited generator at 1000 rpm

Field current in A 00 02 03 04 05 06 07 08 09 10 11 12Armature voltage in V 5 40 75 100 124 145 162 178 188 195 200 205

The machine is now connected as a shunt generator with a total field resistance of 200 Ω and driven at 1000 rpm Estimate graphically (i) the voltage to which the generator will build up at no load (ii) the armature field and load currents when the terminal voltage is found to be 150 V Neglect the effect of armature reaction and brush drop and assume armature resistance Ra to be 08 Ω (iii) Finally estimate the steady state armature current when the machine terminals are shorted

Solution

First of all the OCC of the generator is plotted in a graph paper as shown in figure

(i) The total field circuit resistance is given to be Rf = 200 Ω The Rf line is now drawn on the same graph paper passing through the origin The point of intersection of the Rf line and the OCC decides the final no load voltage and can be read from the graph as 192 V and the corresponding field current is 096 A (ii) Since the terminal voltage V is 150 V (= BM) field current If

is OM = 077 A Generated voltage E is given by AM But we know E = IaRa + V Hence IaRa= E ndash V = AM ndash BM = AB Now from the graph AB = 25 V there4Ia = 2508 = 3125 A So load current IL= Ia ndash If= 3125 ndash 077 = 3048 A

ADDITIONAL PROBLEMS

1 A 4 pole lap wound 115 kW 230 volts dc shunt generator has armatureresistance of 02 ohm and field resistance of 100 ohms Calculate (i) emf

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES

Fig 19 Separately Excited DC Generators

Armature Current Ia = IL

Terminal voltage V = Eg-Ia Ra

Electric Power developed = Eg Ia

Power delivered to load = V IL = (Eg-Ia Ra) Ia = Eg Ia - Ia2

Ra

152 Self Excited DC Generator

These are generators in which the field winding is excited by the output of the generator itself As described before ndash there are three types of self excited dc generators ndash they are 1) Series 2) Shunt and 3) Compound

A series DC generator is shown below in fig110 (a) ndash in which the armature winding is connected in series with the field winding so that the field current flows through the load as well as the field winding Field winding is a low resistance thick wire of few turns Armature current Ia =Ise = IL=I (say)

Terminal voltage V = Eg- I( Ra+ Rse)Power Developed in armature = Eg Ia

Power delivered to load = V IL = Eg Ia- Ia2 (Ra+Rse)

Fig110 (a) Series generator and (b) shunt generator

A shunt DC generator is shown in figure 110 (b) in which the field winding is wired parallel to armature winding so that the voltage across both are same The field winding has high resistance and more number of turns so that only a part of armature current passes through field winding and the rest passes through load Shunt field current Ish = V Rsh

Armature current Ia = IL+ Ish

Terminal voltage V= Eg-Ia Ra

Power developed in armature = EgIa

Power delivered to load = VIL A compound generator is shown in figure 111 below It has two field findings namely Rsh and Rse They are basically shunt winding (Rsh) and series winding (Rse) Compound generator is of two types ndash 1) Short shunt and 2) Long shunt

Short shunt- Here the shunt field winding is wired parallel to armature and series field winding is connected in series to the load It is shown in fig111(a)

In short shunt machines Series field current Ise = IL

Shunt field current Ish iquest V+ I se RseRsh

Terminal voltage V= Eg-Ia Ra-Ise Rse

Power developedin armature = Eg Ia

Power delivered to load = V IL

Fig 111 (a) short shunt generator and (b) Long shunt generator

Long shunt- Here the shunt field winding is parallel to both armature and series field winding (Rse is wired in series to the armature) It is shown in figure 111 (b)

In Long shunt machines Series field current Ise = Ia = IL+ Ish

Shunt field current Ish = V R sh

Terminal voltage V = Eg- Ia (Ra+ Rse) Power developed in armature = Eg Ia

Power delivered to load = VIL

16 GENERATOR CHARACTERISTICS

The speed of a dc machine operated as a generator is fixed by the prime mover For general-purpose operation the prime mover is equipped with a speed governor so that the speed of the generator is practically constant Under such condition the generator performance deals primarily with the relation between excitation terminal voltage and load These relations can be best exhibited graphically by means of curves known as generator characteristics Thesecharacteristics show at a glance the behavior of the generator under differentload conditions

The following are the three most important characteristics of a dc generator161 Open Circuit Characteristic (OCC)

This curve shows the relation between the generated emf at no-load (E0) and the field current (If) at constant speed It is also known as magnetic characteristic or no-load saturation curve Its shape is practically the same for all generators whether separately or self-excited The data for OCC curve are obtained experimentally by operating the generator at no load and constant speed and recording the change in terminal voltage as the field current is varied

162 Internal or Total characteristic (EIa)

This curve shows the relation between the generated emf on load (E) and the armature current (Ia) The emf E is less than E0 due to the demagnetizing effect of armature reaction Therefore this curve will lie below the open circuitcharacteristic (OCC) The internal characteristic is of interest chiefly to thedesigner It cannot be obtained directly by experiment It is because a voltmetercannot read the emf generated on load due to the voltage drop in armatureresistance The internal characteristic can be obtained from externalcharacteristics if winding resistances are known because armature reaction effectis included in both characteristics163 External characteristic (VIL)This curve shows the relation between the terminal voltage (V) and load current(IL) The terminal voltage V will be less than E due to voltage drop in thearmature circuit Therefore this curve will lie below the internal characteristicThis characteristic is very important in determining the suitability of a generatorfor a given purpose It can be obtained by making simultaneous measurements

of terminal voltage and load current (with voltmeter and ammeter) of a loadedgenerator164 Open Circuit Characteristic of a DC Generator

The OCC for a dc generator is determined as follows The field winding of the dc generator (series or shunt) is disconnected from the machine and isseparately excited from an external dc source as shown in Fig 112 Thegenerator is run at fixed speed (ie normal speed) The field current (If) isincreased from zero in steps and the corresponding values of generated emf(E0) read off on a voltmeter connected across the armature terminals On plottingthe relation between E0 and If we get the open circuit characteristic as shown in

Fig113

Fig 112 Experimental setup for plotting OCC

Fig 113 OCC of dc generator

The following points may be noted from OCC When the field current is zero there is some generated emf OA

This is due to the residual magnetism in the field poles Over a fairly wide range of field current (upto point B in the curve)

the curve is linear It is because in this range reluctance of iron is negligible as compared with that of air gap The air gap reluctance is constant and hence linear relationship After point B on the curve the reluctance of iron also comes into picture and reluctance of iron is no longer negligible Consequently the curve deviates from linear relationship

After point C on the curve the magnetic saturation of poles begins and E0 tends to level off

The OCC of even self-excited generator is obtained by running it as a separately excited generator

165 Characteristics of a Separately Excited DCGenerator

The obvious disadvantage of a separately excited dc generator is that we require an external dc source for excitation But since the output voltage may be

controlled more easily and over a wide range (from zero to a maximum) this type of excitation finds many applications

(i) Open circuit characteristic

The OCC of a separately excited generator is determined in a manner described in Sec 164 Fig114 shows the variation of generated em f on noload with field current for various fixed speeds Note that if the value of constant speed is increased the steepness of the curve also increases When the field current is zero the residual magnetism in the poles will give rise to the small initial emf as shown

Fig114 Variation of OCC with speed

Internal and External Characteristics (Load Characteristics)

The external characteristic of a separately excited generator is the curve between the terminal voltage (V) and the load current IL (which is the same as armature current in this case) In order to determine the external characteristic the circuit set up is as shown in Fig 115

Fig 115 Experimental setup to plot load characteristics

As the load current increases the terminal voltage falls due to two reasons(a) The armature reaction weakens the main flux so that actual emf generated E on load is less than that generated (E0) on no load

(b) There is voltage drop across armature resistance (= ILRa = IaRa)Due to these reasons the external characteristic is a drooping curve [curve 3 inFig 116 ] Note that in the absence of armature reaction and armature dropthe generated emf would have been E0 (curve 1)The internal characteristic can be determined from external characteristic by adding IL Ra drop to the external characteristic It is because armature reaction drop is included in the external characteristic Curve 2 is the internal characteristic of the generator and should obviously lie above the external characteristic

Fig 116 Load characteristics of a separately excited dcgenerator

117 Voltage Build-Up in a Self-Excited GeneratorLet us see how voltage builds up in a self-excited generator(i) Shunt generator Consider a shunt generator If the generator is run at a constant speed some emf will be generated due to residual magnetism in the main poles This small emf circulates a field current which in turn produces additional flux toreinforce the original residual flux (provided field winding connections arecorrect) This process continues and the generator builds up the normalgenerated voltage following the OCC shown in Fig 117 (i) The field resistance Rf can be represented by a straight line passing through the origin as shown in Fig 117 (ii) The two curves can be shown on the same diagram as they have the same ordinate as in Fig 117 (iii)

Fig 117

Since the field circuit is inductive there is a delay in the increase in current uponclosing the field circuit switch The rate at which the current increases dependsupon the voltage available for increasing it Suppose at any instant the fieldcurrent is i (= OA) and is increasing at the rate didt Then

E0=i R f +iquest L didt

Where Rf = total field circuit resistance L = inductance of field circuit

At the considered instant the total emf available is AC [See Fig 117 (iii)] Anamount AB of the cmf AC is absorbed by the voltage drop iRf and theremainder part BC is available to overcome L didt Since this surplus voltage isavailable it is possible for the field current to increase above the value OAHowever at point D the available voltage is OM and is all absorbed by i RfD

rop Consequently the field current cannot increase further and the generatorbuild up stops Hence we arrive at a very important conclusion that the voltage build up of the generator is given by the point of intersection of OCC and field resistance line Thus in Fig 117 (iii) D is point of intersection of the two curves Hence the generator will build up a voltage OM(ii) Series generator During initial operation with no current yet flowing a residual voltage will be generated exactly as in the case of a shunt generator The residual voltage will cause a current to flow through the whole series circuit when the circuit isclosed There will then be voltage build up to an equilibrium point exactlyanalogous to the buildup of a shunt generator The voltage build up graph willbe similar to that of shunt generator except that now load current (instead offield current for shunt generator) will be taken along x-axis10487871048787(iii) Compound generator When a compound generator has its series field flux aiding its shunt field flux the machine is said to be cumulative compound When the series field isconnected in reverse so that its field flux opposes the shunt field flux thegenerator is then differential compound The easiest way to build up voltage in a compound generator is to start under no load conditions At no load only the shunt field is effective When no-load voltage build up is achieved the generator is loaded If under load the voltagerises the series field connection is cumulative If the voltage drops significantlythe connection is differential compound118 Critical Field Resistance for a Shunt Generator The voltage build up in a shunt generator depends upon field circuit resistance If the field circuit resistance is R1 (line OA) then generator will build up a voltage OM as shown in Fig 118 If the field circuit resistance is increasedto R2 (tine OB) the generator will build up a voltage OL slightly less than OM As the field circuit resistance is increased the slope of resistance line also increases When the field resistance line becomes tangent (line OC) to OCC the generator would just excite If the field circuit resistance is increased beyond this point (say line OD) the generator will fail to excite The field circuit resistance represented by line OC (tangent to OCC) is called critical field resistance RC for the shunt generator It may be defined as under The maximum field circuit resistance (for a given speed) with which the shunt generator would just excite is known as its critical field resistance It should be noted that shunt generator will build up voltage only if field circuit resistance is less than critical field resistance

Fig 118 Critical field resistance of shunt generator

119 Critical Resistance for a Series Generator

Fig 119 shows the voltage build up in a series generator Here R1 R2 etc represent the total circuit resistance (load resistance and field winding resistance) If the total circuit resistance is R1 then series generator will build up a voltage OL The line OC is tangent to OCC and represents the critical resistance RC for a series generator If the total resistance of the circuit is more than RC (say line OD) the generator will fail to build up voltage Note that Fig 119 is similar to Fig 118 with the following differences

(i) In Fig 118 R1 R2 etc represent the total field circuit resistanceHowever R1 R2 etc in Fig 119 represent the total circuit resistance(load resistance and series field winding resistance etc)

(ii) In Fig 118 field current alone is represented along X-axis However inFig 119 load current IL is represented along Y-axis Note that in aseries generator field current = load current IL

Fig 119 Critical resistance of a series generator

120 Characteristics of Series Generator Fig 120 (i) shows the connections of a series wound generator Since there is only one current (that which flows through the whole machine) the load current is the same as the exciting current(i) OCCCurve 1 shows the open circuit characteristic (OCC) of a series generator Itcan be obtained experimentally by disconnecting the field winding from themachine and exciting it from a separate dc source(ii) Internal characteristic Curve 2 shows the total or internal characteristic of a series generator It gives the relation between the generated emf E on load and armature current Due to armature reaction the flux in the machine will be less than the flux at no load Hence emf E generated under load conditions will be less than the emf E0 generated under no load conditions Consequently internal characteristic curvelies below the OCC curve the difference between them representing the effectof armature reaction [See Fig 120 (ii)]

Fig 120 Characteristics of series generator

(iii) External characteristic

Curve 3 shows the external characteristic of a series generator It gives therelation between terminal voltage and load current ILV = E g- Ia (Ra +Rse )Therefore external characteristic curve will lie below internal characteristiccurve by an amount equal to ohmic drop [ie Ia(Ra + Rse)] in the machine asshown in Fig 120 (ii) The internal and external characteristics ofa dc series generator can be plotted from one another as shown in Fig 121 Supposewe are given the internal characteristic of the generator Let the line OC represent the resistance of the whole machine ie Ra +Rse If the load current is OB drop in the machine is AB ieAB = Ohmic drop in the machine = OB(Ra + Rse)

Fig 121 Load characteristics of series generator

Now raise a perpendicular from point B and mark a point b on this line such that ab = AB Then point b will lie on the external characteristic of the generator

Following similar procedure other points of external characteristic can be located It is easy to see that we can also plot internal characteristic from the external characteristic

121 Characteristics of a Shunt Generator Fig 122 (i) show the connections of a shunt wound generator The armature current Ia splits up into two parts a small fraction Ish flowing through shunt field winding while the major part IL goes to the external load(i) OCC The OCC of a shunt generator is similar in shape to that of a series generator as shown in Fig 122 (ii) The line OA represents the shunt field circuitresistance When the generator is run at normal speed it will build up a voltageOM At no-load the terminal voltage of the generator will be constant (= OM)represented by the horizontal dotted line MC

Fig 122 Characteristics of a Shunt Generator

(ii) Internal characteristic

When the generator is loaded flux per pole is reduced due to armature reaction Therefore emf E generated on load is less than the emf generated at no load As a result the internal characteristic (EIa) drops down slightly as shown in Fig122 (ii)

(iii) External characteristic Curve 2 shows the external characteristic of a shunt generator It gives therelation between terminal voltage V and load current IL

Therefore external characteristic curve will lie below the internal characteristic curve by an amount equal to drop in the armature circuit[ie (IL + Ish)Ra] as shown in Fig 122 (ii)

Note It may be seen from the external characteristic that change in terminalvoltage from no-load to full load is small The terminal voltage can always bemaintained constant by adjusting the field rheostat R automatically122 Critical External Resistance for Shunt Generator

If the load resistance across the terminals of a shunt generator is decreased then load current increase However there is a limit to the increase in

load current with the decrease of load resistance Any decrease of load resistance beyond this point instead of increasing the current ultimately results in reduced current Consequently the external characteristic turns back (dotted curve) as shown in Fig 123 The tangent OA to the curve represents the minimum external resistance required to excite the shunt generator on load and is called critical external resistance If the resistance of the external circuit is less than the critical external resistance (represented by tangent OA in Fig 310) the machine will refuse to excite or will de-excite if already running This means that external resistance is so low as virtually to short circuit the machine and so doing away with its excitation

Note There are two critical resistances for a shunt generator viz (i) criticalfield resistance (ii) critical external resistance For the shunt generator to buildup voltage the former should not be exceeded and the latter must not be gonebelow

123 How to Draw OCC at Different Speeds

If we are given OCC of a generator at a constant speed N1 then we can easily draw the OCC at any other constant speed N2 Fig 123 illustrates theprocedure Here we are given OCC at a constant speed N1 It is desired to findthe OCC at constant speed N2 (it is assumed that n1 lt N2) For constantexcitation E propN

E2

E1 =

N 2

N1 so

E2 iquest E1

N2

N1

As shown in Fig (311) for If = OH E1 = HC Therefore the new value of emf (E2) for the same If but at N2

E2= HCN 2

N 1 = HD

Fig 123 Plotting OCC at another speed

This locates the point D on the new OCC at N2 Similarly other points can belocated taking different values of If The locus of these points will be the OCCat N2

124 Critical Speed (NC)

The critical speed of a shunt generator is the minimum speed below which it fails to excite Clearly it is the speed for which the given shunt field resistance represents the critical resistance In Fig 124 curve 2 corresponds to critical

speed because the shunt field resistance (Rsh) line is tangential to it If the generator runs at full speed N the new OCC moves upward and the Rsh line represents critical resistance for this speed

In order to find critical speed take any convenient point C on excitation axis and erect a perpendicular so as to cut Rsh and Rsh lines at points B and A respectively Then

BCAC

=NcN

Nc=N BCAC

Fig 124 Finding critical speed

125 Conditions for Voltage Build-Up of a Shunt Generator

The necessary conditions for voltage build-up in a shunt generator are

(i) There must be some residual magnetism in generator poles(ii) The connections of the field winding should be such that the field current strengthens the residual magnetism(iii) The resistance of the field circuit should be less than the critical resistance In other words the speed of the generator should be higher

than the critical speed

126 Compound Generator Characteristics

In a compound generator both series and shunt excitation are combined as shown in Fig 123 The shunt winding can be connected either across the armature only (short-shunt connection S) or across armature plus series field (long-shunt connection G) The compound generator can be cumulatively compounded or

differentially compounded generator The latter is rarely used in practice Therefore we shall discuss the characteristics of cumulatively compounded

generator It may be noted that external characteristics of long and short shunt compound generators are almost identical

External characteristic

Fig 124 shows the external characteristics of a cumulatively compoundedgenerator The series excitation aids the shunt excitation The degree of compounding depends upon the increase in series excitation with the increase inload current

(i) If series winding turns are so adjusted that with the increase in load current the terminal voltage increases it is called over-compounded generator In such a case as the load current increases the series field mmf increases and tends to increase the flux and hence the generated voltage The increase in generated voltage is greater than the IaRa drop so that instead of decreasing the terminal voltage increases as shown by curve A in Fig 124

(ii) If series winding turns are so adjusted that with the increase in load current the terminal voltage substantially remains constant it is called Flat-compounded generator The series winding of such a machine has lesser number of turns than the one in over-compounded machine and therefore does not increase the flux as much for a given load current Consequently the full-load voltage is nearly equal to the no-load voltage as indicated by curve B in Fig 124 (iii) If series field winding has lesser number of turns than for a flat compounded machine the terminal voltage falls with increase in load current as indicated by curve C m Fig 124 Such a machine is called under-compounded generator

Fig 123 compound generator

Fig 124 Load characteristics of compound generators

127 LOSSES

The losses in a dc machine (generator or motor) may be divided into three classes viz (i) copper losses or electrical losses (ii) iron or core losses and (iii) mechanical losses All these losses appear as heat and thus raise the temperature of the machine They also lower the efficiency of the machine The loss structure maybe demonstrated as

1Copperlosses

These losses occur due to currents in the various windings of the machine When an electric current of I amps flows in a resistance of R ohms the heat energy is lost at the rate of I2R joules sec and the power loss is I2R watts Generators and motors have one or more field circuits and an armature circuit in which such losses occur All resistance losses of this kind are called copper losses The different copper losses occurring in a dc machine are

(i) Armature copper loss = Ia2Ra

(ii) Shunt field copper loss = Ish2 Rsh

(iii) Series field copper loss = Ise2Rse

Armature copper loss will be around 30 to 40 of full load losses where as total field copper losses will constitute around 30 of full load losses

Note There is also brush contact loss due to brush contact resistance (ie resistance between the surface of brush and surface of commutator) This loss is generally included in armature copper loss The brush contact drop is almost independent of Ia For carbon brushes this drop is around 1 V per brush The power loss due to brush contact resistance is 2ebIa where eb is the voltage drop at one brush

2 IRONORCORELOSSES These losses occur in the armature of a dc machine and are due to the rotation of armature in the magnetic field of the poles They are of two types viz (i) hysteresis loss (ii) eddy current loss

(i) Hysteresis loss

Hysteresis loss occurs in the armature of the dc machine since any given part of the armature is subjected to magnetic field reversals as it passes under successive poles Fig 124 shows an armature rotating in two-pole machine Consider a small piece ab of the armature When the piece ab is under N-pole the magnetic lines pass from a to b Half a revolution later the same piece of iron is under S-pole and magnetic lines pass from b to a so that magnetism in the iron is reversed In order to reverse continuously the molecular magnets in the armature core some amount of power has to be spent which is called hysteresis loss It is given by Steinmetz formula This formula is

Hysteresis Loss Wh = η Bmax 16f V Watts Bmax - Maximum flux density in the core f - Frequency of magnetic reversals V - Volume of armature in m3

η - Steinmetz hysteresis coefficient

In order to reduce this loss in a dc machine armature core is made of such materials which have a low value of Steinmetz hysteresis co-efficient eg silicon steel

Fig 124 Occurrence of hysteresis loss

ii) EDDYCURRENTLOSS

In addition to the voltages induced in the armature conductors there are also voltages induced in the armature core These voltages produce circulating currents in the armature core as shown in Fig 125 These are called eddy currents and power loss due to their flow is called eddy current loss The eddy current loss appears as heat which raises the temperature of the machine and lowers its efficiency Core resistance can be greatly increased by constructing the core of thin round If a continuous solid iron core is used the resistance to eddy current path will be small due to large cross-sectional area of the core Consequently the magnitude of eddy current and hence eddy current loss will be large The magnitude of eddy current can be reduced by making core resistance as high as practical The iron sheets called laminations [See Fig 126] The laminations are insulated from each other with a coating of varnish The insulating coating has a high resistance so very little current flows from one lamination to the other Alsobecause each lamination is very thin the resistance to current flowing through the width of a lamination is also quite large Thus laminating a core increases the core resistance which decreases the eddy current and hence the eddy current loss

Fig 125 Eddy current loss Fig 126 Core laminations

Eddy current loss is biven by the formula Pe= Ke Bmax

2 f2t2 V Watts

Where Ke = Constant depending upon electrical resistace of core Bmax = Maximum flux density in wbm2

f = frequency f magnetic reversals in Hz t = thickness of laminations in m and V = volume of core in m3

It may be noted that eddy current loss depends upon the square of lamination thickness For this reason lamination thickness should be kept as small as possible

3MECHANICALLOSSES

These losses are due to friction and windage(i) Friction losses eg bearing friction brush friction etc(ii) Windage loss ie air friction of rotating armature

These losses depend upon the speed of the machine But for a given speed they are practically constant

Note Iron losses and mechanical losses together are called stray losses

CONSTANT AND VARIABLE LOSSES

The losses in a dc generator (or dc motor) may be sub-divided into

(i) constant losses (ii) variable losses

(i) Constant losses

Those losses in a dc generator which remain constant at all loads are known as constant losses The constant losses in a dc generator are(a) iron losses(b) mechanical losses(c) shunt field losses(ii) Variable lossesThose losses in a dc generator which vary with load are called variable lossesThe variable losses in a dc generator are (a) Copper loss in armature winding (Ia

2 Ra) (b) Copper loss in series field winding (Ise

2Rse)

Total losses = Constant losses + Variable lossesNote Field Cu loss is constant for shunt and compound generators

128 POWER FLOW DIAGRAM

The power flow in a dc generator may be represented as

The different power stages are shown as A B and CA ndash The mechanical power input which is the power output of the prime mover is given as Pin = 2πNT

60 watts

B - The power developed in armature is Pa = Eg Ia watts C- The electrical power output is Pout = VIL watts

129 EFFICIENCY Efficiency of a machine is the ratio of its output power to the input power

Efficiency η = Output powerInput power

= Input powerminusLossesInput power = Output power

Output power+Losses

In case of a dc generator depending upon the power stages three different efficiencies can be defined as-

1 Mechanical Efficiency(ηm)

BA = Electrical power developedisinarmature

Mechanical power input2 Electrical Efficiency(ηe)

CB = Electrical power output

Electrical power developedisinthe armature 3 Overall or commercial efficiency (η)

CA = Electrical power output

Mechanical power input = ηmtimes ηe

130 CONDITION FOR MAXIMUM EFFICIENCY The efficiency of a dc generator is not constant but varies with load Fig 127shows the variation of efficiency with load current Consider a dc shunt generator with terminal voltage V delivering a load current of IL amp Generator output = V IL

Generator input = Output + Losses = V IL + constant losses + Variable losses = V IL + Wc+ Ia

2Ra

= V IL + Wc + (IL+ Ish)2 Rsh

The shunt field current Ish is small compared with IL so may be neglected

Generator input = VIL+ IL2Ra + Wc

Now efficiency η iquest outputinput = VI L

VIL+ Ia 2Ra+Wc

Efficiency is maximum when the denominator of the above equation is minimumie

ddIL ( VI L

VIL+ Ia 2Ra+Wc ) = 0

Which gives IL2 Ra = WC

ie Variable loss = constant lossThe load current for maximum efficiency is given by

IL iquest radicWcRa

Fig 127 Variation of efficiency with load

Hence the efficiency is maximum for that load current at which the variable loss is equal to the constant loss

131 APPLICATIONS

Separately excited generators

These types of generators have special applications such as electroplating electro refining of materials because separate supply is required for field excitation

Shunt generators

These types of generators are used in battery charging and ordinary lighting purposes

Series generators

This type of generators is used commonly as boosters on DC feeders as current generators for welding and arc lamp

Cumulatively compounded generators

This type of generators is used for domestic lighting purposes and to transmit energy over long distances

Differential compound generators

This type of generators is rarely used They are used for special applications such as electric arc welding

132 NUMERICAL EXCERCISES

1 The emf induced in the armature of a 450 kW 250 volt shunt generator is

2588volt when the field current is 200 amp and the generator is supplying power to a load at rated terminal voltage The armature circuit resistance is 0005 ohm Determine (i) load current (ii) power generated (iii) power output and (iv) electrical efficiency Neglect brush contact drop

Solution

Let the load current be I amp Given Ish = 20 A

Armature current = Load current + field current or Ia = I + 20 Induced emf = Terminal voltage + Ia Ra drop

2588 = 250 + (I + 20) times 0005

So (i) Load current I = 1740 amp

(ii) Power generated E Ia = 2588 (1740 + 20) watts = 455488 kW (iii) Power output V I = 250 times 1740 watts

= 435 kW

(iv) Electrical efficiency = 435455488 times 100 = 955

2 A 4 pole wave wound dc shunt generator delivers a load of 45 kW at a terminal voltage of 220 volts Its armature has 150 single turn coils and has a resistance of 001 ohm The air gap flux per pole is 002 weber Shunt field resistance is 50 ohm Calculate the speed at which it is being driven Neglect brush contact drop

Solution Load current I = W

V = 45000220 = 20455 amp

Field Current Ish = V

R s h = 220

50 = 44 amp

Armature current Ia = I +I sh = 20455 +44 = 20895 amp

Armature drop I aRa = 20895 X 001 = 20895 volts Induced emf Eg = V + Ia Ra = 220 + 20895 = 2220895 volts

But induced emf

So N = Eg60 AΦ ZP = 2220895times 60 times2

002 times150 times2 times 4 = 111045 rpm

3 The following data refer to the OCC of a DC separately excited generator at 1000 rpm

Field current in A 00 02 03 04 05 06 07 08 09 10 11 12Armature voltage in V 5 40 75 100 124 145 162 178 188 195 200 205

The machine is now connected as a shunt generator with a total field resistance of 200 Ω and driven at 1000 rpm Estimate graphically (i) the voltage to which the generator will build up at no load (ii) the armature field and load currents when the terminal voltage is found to be 150 V Neglect the effect of armature reaction and brush drop and assume armature resistance Ra to be 08 Ω (iii) Finally estimate the steady state armature current when the machine terminals are shorted

Solution

First of all the OCC of the generator is plotted in a graph paper as shown in figure

(i) The total field circuit resistance is given to be Rf = 200 Ω The Rf line is now drawn on the same graph paper passing through the origin The point of intersection of the Rf line and the OCC decides the final no load voltage and can be read from the graph as 192 V and the corresponding field current is 096 A (ii) Since the terminal voltage V is 150 V (= BM) field current If

is OM = 077 A Generated voltage E is given by AM But we know E = IaRa + V Hence IaRa= E ndash V = AM ndash BM = AB Now from the graph AB = 25 V there4Ia = 2508 = 3125 A So load current IL= Ia ndash If= 3125 ndash 077 = 3048 A

ADDITIONAL PROBLEMS

1 A 4 pole lap wound 115 kW 230 volts dc shunt generator has armatureresistance of 02 ohm and field resistance of 100 ohms Calculate (i) emf

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES

Fig110 (a) Series generator and (b) shunt generator

A shunt DC generator is shown in figure 110 (b) in which the field winding is wired parallel to armature winding so that the voltage across both are same The field winding has high resistance and more number of turns so that only a part of armature current passes through field winding and the rest passes through load Shunt field current Ish = V Rsh

Armature current Ia = IL+ Ish

Terminal voltage V= Eg-Ia Ra

Power developed in armature = EgIa

Power delivered to load = VIL A compound generator is shown in figure 111 below It has two field findings namely Rsh and Rse They are basically shunt winding (Rsh) and series winding (Rse) Compound generator is of two types ndash 1) Short shunt and 2) Long shunt

Short shunt- Here the shunt field winding is wired parallel to armature and series field winding is connected in series to the load It is shown in fig111(a)

In short shunt machines Series field current Ise = IL

Shunt field current Ish iquest V+ I se RseRsh

Terminal voltage V= Eg-Ia Ra-Ise Rse

Power developedin armature = Eg Ia

Power delivered to load = V IL

Fig 111 (a) short shunt generator and (b) Long shunt generator

Long shunt- Here the shunt field winding is parallel to both armature and series field winding (Rse is wired in series to the armature) It is shown in figure 111 (b)

In Long shunt machines Series field current Ise = Ia = IL+ Ish

Shunt field current Ish = V R sh

Terminal voltage V = Eg- Ia (Ra+ Rse) Power developed in armature = Eg Ia

Power delivered to load = VIL

16 GENERATOR CHARACTERISTICS

The speed of a dc machine operated as a generator is fixed by the prime mover For general-purpose operation the prime mover is equipped with a speed governor so that the speed of the generator is practically constant Under such condition the generator performance deals primarily with the relation between excitation terminal voltage and load These relations can be best exhibited graphically by means of curves known as generator characteristics Thesecharacteristics show at a glance the behavior of the generator under differentload conditions

The following are the three most important characteristics of a dc generator161 Open Circuit Characteristic (OCC)

This curve shows the relation between the generated emf at no-load (E0) and the field current (If) at constant speed It is also known as magnetic characteristic or no-load saturation curve Its shape is practically the same for all generators whether separately or self-excited The data for OCC curve are obtained experimentally by operating the generator at no load and constant speed and recording the change in terminal voltage as the field current is varied

162 Internal or Total characteristic (EIa)

This curve shows the relation between the generated emf on load (E) and the armature current (Ia) The emf E is less than E0 due to the demagnetizing effect of armature reaction Therefore this curve will lie below the open circuitcharacteristic (OCC) The internal characteristic is of interest chiefly to thedesigner It cannot be obtained directly by experiment It is because a voltmetercannot read the emf generated on load due to the voltage drop in armatureresistance The internal characteristic can be obtained from externalcharacteristics if winding resistances are known because armature reaction effectis included in both characteristics163 External characteristic (VIL)This curve shows the relation between the terminal voltage (V) and load current(IL) The terminal voltage V will be less than E due to voltage drop in thearmature circuit Therefore this curve will lie below the internal characteristicThis characteristic is very important in determining the suitability of a generatorfor a given purpose It can be obtained by making simultaneous measurements

of terminal voltage and load current (with voltmeter and ammeter) of a loadedgenerator164 Open Circuit Characteristic of a DC Generator

The OCC for a dc generator is determined as follows The field winding of the dc generator (series or shunt) is disconnected from the machine and isseparately excited from an external dc source as shown in Fig 112 Thegenerator is run at fixed speed (ie normal speed) The field current (If) isincreased from zero in steps and the corresponding values of generated emf(E0) read off on a voltmeter connected across the armature terminals On plottingthe relation between E0 and If we get the open circuit characteristic as shown in

Fig113

Fig 112 Experimental setup for plotting OCC

Fig 113 OCC of dc generator

The following points may be noted from OCC When the field current is zero there is some generated emf OA

This is due to the residual magnetism in the field poles Over a fairly wide range of field current (upto point B in the curve)

the curve is linear It is because in this range reluctance of iron is negligible as compared with that of air gap The air gap reluctance is constant and hence linear relationship After point B on the curve the reluctance of iron also comes into picture and reluctance of iron is no longer negligible Consequently the curve deviates from linear relationship

After point C on the curve the magnetic saturation of poles begins and E0 tends to level off

The OCC of even self-excited generator is obtained by running it as a separately excited generator

165 Characteristics of a Separately Excited DCGenerator

The obvious disadvantage of a separately excited dc generator is that we require an external dc source for excitation But since the output voltage may be

controlled more easily and over a wide range (from zero to a maximum) this type of excitation finds many applications

(i) Open circuit characteristic

The OCC of a separately excited generator is determined in a manner described in Sec 164 Fig114 shows the variation of generated em f on noload with field current for various fixed speeds Note that if the value of constant speed is increased the steepness of the curve also increases When the field current is zero the residual magnetism in the poles will give rise to the small initial emf as shown

Fig114 Variation of OCC with speed

Internal and External Characteristics (Load Characteristics)

The external characteristic of a separately excited generator is the curve between the terminal voltage (V) and the load current IL (which is the same as armature current in this case) In order to determine the external characteristic the circuit set up is as shown in Fig 115

Fig 115 Experimental setup to plot load characteristics

As the load current increases the terminal voltage falls due to two reasons(a) The armature reaction weakens the main flux so that actual emf generated E on load is less than that generated (E0) on no load

(b) There is voltage drop across armature resistance (= ILRa = IaRa)Due to these reasons the external characteristic is a drooping curve [curve 3 inFig 116 ] Note that in the absence of armature reaction and armature dropthe generated emf would have been E0 (curve 1)The internal characteristic can be determined from external characteristic by adding IL Ra drop to the external characteristic It is because armature reaction drop is included in the external characteristic Curve 2 is the internal characteristic of the generator and should obviously lie above the external characteristic

Fig 116 Load characteristics of a separately excited dcgenerator

117 Voltage Build-Up in a Self-Excited GeneratorLet us see how voltage builds up in a self-excited generator(i) Shunt generator Consider a shunt generator If the generator is run at a constant speed some emf will be generated due to residual magnetism in the main poles This small emf circulates a field current which in turn produces additional flux toreinforce the original residual flux (provided field winding connections arecorrect) This process continues and the generator builds up the normalgenerated voltage following the OCC shown in Fig 117 (i) The field resistance Rf can be represented by a straight line passing through the origin as shown in Fig 117 (ii) The two curves can be shown on the same diagram as they have the same ordinate as in Fig 117 (iii)

Fig 117

Since the field circuit is inductive there is a delay in the increase in current uponclosing the field circuit switch The rate at which the current increases dependsupon the voltage available for increasing it Suppose at any instant the fieldcurrent is i (= OA) and is increasing at the rate didt Then

E0=i R f +iquest L didt

Where Rf = total field circuit resistance L = inductance of field circuit

At the considered instant the total emf available is AC [See Fig 117 (iii)] Anamount AB of the cmf AC is absorbed by the voltage drop iRf and theremainder part BC is available to overcome L didt Since this surplus voltage isavailable it is possible for the field current to increase above the value OAHowever at point D the available voltage is OM and is all absorbed by i RfD

rop Consequently the field current cannot increase further and the generatorbuild up stops Hence we arrive at a very important conclusion that the voltage build up of the generator is given by the point of intersection of OCC and field resistance line Thus in Fig 117 (iii) D is point of intersection of the two curves Hence the generator will build up a voltage OM(ii) Series generator During initial operation with no current yet flowing a residual voltage will be generated exactly as in the case of a shunt generator The residual voltage will cause a current to flow through the whole series circuit when the circuit isclosed There will then be voltage build up to an equilibrium point exactlyanalogous to the buildup of a shunt generator The voltage build up graph willbe similar to that of shunt generator except that now load current (instead offield current for shunt generator) will be taken along x-axis10487871048787(iii) Compound generator When a compound generator has its series field flux aiding its shunt field flux the machine is said to be cumulative compound When the series field isconnected in reverse so that its field flux opposes the shunt field flux thegenerator is then differential compound The easiest way to build up voltage in a compound generator is to start under no load conditions At no load only the shunt field is effective When no-load voltage build up is achieved the generator is loaded If under load the voltagerises the series field connection is cumulative If the voltage drops significantlythe connection is differential compound118 Critical Field Resistance for a Shunt Generator The voltage build up in a shunt generator depends upon field circuit resistance If the field circuit resistance is R1 (line OA) then generator will build up a voltage OM as shown in Fig 118 If the field circuit resistance is increasedto R2 (tine OB) the generator will build up a voltage OL slightly less than OM As the field circuit resistance is increased the slope of resistance line also increases When the field resistance line becomes tangent (line OC) to OCC the generator would just excite If the field circuit resistance is increased beyond this point (say line OD) the generator will fail to excite The field circuit resistance represented by line OC (tangent to OCC) is called critical field resistance RC for the shunt generator It may be defined as under The maximum field circuit resistance (for a given speed) with which the shunt generator would just excite is known as its critical field resistance It should be noted that shunt generator will build up voltage only if field circuit resistance is less than critical field resistance

Fig 118 Critical field resistance of shunt generator

119 Critical Resistance for a Series Generator

Fig 119 shows the voltage build up in a series generator Here R1 R2 etc represent the total circuit resistance (load resistance and field winding resistance) If the total circuit resistance is R1 then series generator will build up a voltage OL The line OC is tangent to OCC and represents the critical resistance RC for a series generator If the total resistance of the circuit is more than RC (say line OD) the generator will fail to build up voltage Note that Fig 119 is similar to Fig 118 with the following differences

(i) In Fig 118 R1 R2 etc represent the total field circuit resistanceHowever R1 R2 etc in Fig 119 represent the total circuit resistance(load resistance and series field winding resistance etc)

(ii) In Fig 118 field current alone is represented along X-axis However inFig 119 load current IL is represented along Y-axis Note that in aseries generator field current = load current IL

Fig 119 Critical resistance of a series generator

120 Characteristics of Series Generator Fig 120 (i) shows the connections of a series wound generator Since there is only one current (that which flows through the whole machine) the load current is the same as the exciting current(i) OCCCurve 1 shows the open circuit characteristic (OCC) of a series generator Itcan be obtained experimentally by disconnecting the field winding from themachine and exciting it from a separate dc source(ii) Internal characteristic Curve 2 shows the total or internal characteristic of a series generator It gives the relation between the generated emf E on load and armature current Due to armature reaction the flux in the machine will be less than the flux at no load Hence emf E generated under load conditions will be less than the emf E0 generated under no load conditions Consequently internal characteristic curvelies below the OCC curve the difference between them representing the effectof armature reaction [See Fig 120 (ii)]

Fig 120 Characteristics of series generator

(iii) External characteristic

Curve 3 shows the external characteristic of a series generator It gives therelation between terminal voltage and load current ILV = E g- Ia (Ra +Rse )Therefore external characteristic curve will lie below internal characteristiccurve by an amount equal to ohmic drop [ie Ia(Ra + Rse)] in the machine asshown in Fig 120 (ii) The internal and external characteristics ofa dc series generator can be plotted from one another as shown in Fig 121 Supposewe are given the internal characteristic of the generator Let the line OC represent the resistance of the whole machine ie Ra +Rse If the load current is OB drop in the machine is AB ieAB = Ohmic drop in the machine = OB(Ra + Rse)

Fig 121 Load characteristics of series generator

Now raise a perpendicular from point B and mark a point b on this line such that ab = AB Then point b will lie on the external characteristic of the generator

Following similar procedure other points of external characteristic can be located It is easy to see that we can also plot internal characteristic from the external characteristic

121 Characteristics of a Shunt Generator Fig 122 (i) show the connections of a shunt wound generator The armature current Ia splits up into two parts a small fraction Ish flowing through shunt field winding while the major part IL goes to the external load(i) OCC The OCC of a shunt generator is similar in shape to that of a series generator as shown in Fig 122 (ii) The line OA represents the shunt field circuitresistance When the generator is run at normal speed it will build up a voltageOM At no-load the terminal voltage of the generator will be constant (= OM)represented by the horizontal dotted line MC

Fig 122 Characteristics of a Shunt Generator

(ii) Internal characteristic

When the generator is loaded flux per pole is reduced due to armature reaction Therefore emf E generated on load is less than the emf generated at no load As a result the internal characteristic (EIa) drops down slightly as shown in Fig122 (ii)

(iii) External characteristic Curve 2 shows the external characteristic of a shunt generator It gives therelation between terminal voltage V and load current IL

Therefore external characteristic curve will lie below the internal characteristic curve by an amount equal to drop in the armature circuit[ie (IL + Ish)Ra] as shown in Fig 122 (ii)

Note It may be seen from the external characteristic that change in terminalvoltage from no-load to full load is small The terminal voltage can always bemaintained constant by adjusting the field rheostat R automatically122 Critical External Resistance for Shunt Generator

If the load resistance across the terminals of a shunt generator is decreased then load current increase However there is a limit to the increase in

load current with the decrease of load resistance Any decrease of load resistance beyond this point instead of increasing the current ultimately results in reduced current Consequently the external characteristic turns back (dotted curve) as shown in Fig 123 The tangent OA to the curve represents the minimum external resistance required to excite the shunt generator on load and is called critical external resistance If the resistance of the external circuit is less than the critical external resistance (represented by tangent OA in Fig 310) the machine will refuse to excite or will de-excite if already running This means that external resistance is so low as virtually to short circuit the machine and so doing away with its excitation

Note There are two critical resistances for a shunt generator viz (i) criticalfield resistance (ii) critical external resistance For the shunt generator to buildup voltage the former should not be exceeded and the latter must not be gonebelow

123 How to Draw OCC at Different Speeds

If we are given OCC of a generator at a constant speed N1 then we can easily draw the OCC at any other constant speed N2 Fig 123 illustrates theprocedure Here we are given OCC at a constant speed N1 It is desired to findthe OCC at constant speed N2 (it is assumed that n1 lt N2) For constantexcitation E propN

E2

E1 =

N 2

N1 so

E2 iquest E1

N2

N1

As shown in Fig (311) for If = OH E1 = HC Therefore the new value of emf (E2) for the same If but at N2

E2= HCN 2

N 1 = HD

Fig 123 Plotting OCC at another speed

This locates the point D on the new OCC at N2 Similarly other points can belocated taking different values of If The locus of these points will be the OCCat N2

124 Critical Speed (NC)

The critical speed of a shunt generator is the minimum speed below which it fails to excite Clearly it is the speed for which the given shunt field resistance represents the critical resistance In Fig 124 curve 2 corresponds to critical

speed because the shunt field resistance (Rsh) line is tangential to it If the generator runs at full speed N the new OCC moves upward and the Rsh line represents critical resistance for this speed

In order to find critical speed take any convenient point C on excitation axis and erect a perpendicular so as to cut Rsh and Rsh lines at points B and A respectively Then

BCAC

=NcN

Nc=N BCAC

Fig 124 Finding critical speed

125 Conditions for Voltage Build-Up of a Shunt Generator

The necessary conditions for voltage build-up in a shunt generator are

(i) There must be some residual magnetism in generator poles(ii) The connections of the field winding should be such that the field current strengthens the residual magnetism(iii) The resistance of the field circuit should be less than the critical resistance In other words the speed of the generator should be higher

than the critical speed

126 Compound Generator Characteristics

In a compound generator both series and shunt excitation are combined as shown in Fig 123 The shunt winding can be connected either across the armature only (short-shunt connection S) or across armature plus series field (long-shunt connection G) The compound generator can be cumulatively compounded or

differentially compounded generator The latter is rarely used in practice Therefore we shall discuss the characteristics of cumulatively compounded

generator It may be noted that external characteristics of long and short shunt compound generators are almost identical

External characteristic

Fig 124 shows the external characteristics of a cumulatively compoundedgenerator The series excitation aids the shunt excitation The degree of compounding depends upon the increase in series excitation with the increase inload current

(i) If series winding turns are so adjusted that with the increase in load current the terminal voltage increases it is called over-compounded generator In such a case as the load current increases the series field mmf increases and tends to increase the flux and hence the generated voltage The increase in generated voltage is greater than the IaRa drop so that instead of decreasing the terminal voltage increases as shown by curve A in Fig 124

(ii) If series winding turns are so adjusted that with the increase in load current the terminal voltage substantially remains constant it is called Flat-compounded generator The series winding of such a machine has lesser number of turns than the one in over-compounded machine and therefore does not increase the flux as much for a given load current Consequently the full-load voltage is nearly equal to the no-load voltage as indicated by curve B in Fig 124 (iii) If series field winding has lesser number of turns than for a flat compounded machine the terminal voltage falls with increase in load current as indicated by curve C m Fig 124 Such a machine is called under-compounded generator

Fig 123 compound generator

Fig 124 Load characteristics of compound generators

127 LOSSES

The losses in a dc machine (generator or motor) may be divided into three classes viz (i) copper losses or electrical losses (ii) iron or core losses and (iii) mechanical losses All these losses appear as heat and thus raise the temperature of the machine They also lower the efficiency of the machine The loss structure maybe demonstrated as

1Copperlosses

These losses occur due to currents in the various windings of the machine When an electric current of I amps flows in a resistance of R ohms the heat energy is lost at the rate of I2R joules sec and the power loss is I2R watts Generators and motors have one or more field circuits and an armature circuit in which such losses occur All resistance losses of this kind are called copper losses The different copper losses occurring in a dc machine are

(i) Armature copper loss = Ia2Ra

(ii) Shunt field copper loss = Ish2 Rsh

(iii) Series field copper loss = Ise2Rse

Armature copper loss will be around 30 to 40 of full load losses where as total field copper losses will constitute around 30 of full load losses

Note There is also brush contact loss due to brush contact resistance (ie resistance between the surface of brush and surface of commutator) This loss is generally included in armature copper loss The brush contact drop is almost independent of Ia For carbon brushes this drop is around 1 V per brush The power loss due to brush contact resistance is 2ebIa where eb is the voltage drop at one brush

2 IRONORCORELOSSES These losses occur in the armature of a dc machine and are due to the rotation of armature in the magnetic field of the poles They are of two types viz (i) hysteresis loss (ii) eddy current loss

(i) Hysteresis loss

Hysteresis loss occurs in the armature of the dc machine since any given part of the armature is subjected to magnetic field reversals as it passes under successive poles Fig 124 shows an armature rotating in two-pole machine Consider a small piece ab of the armature When the piece ab is under N-pole the magnetic lines pass from a to b Half a revolution later the same piece of iron is under S-pole and magnetic lines pass from b to a so that magnetism in the iron is reversed In order to reverse continuously the molecular magnets in the armature core some amount of power has to be spent which is called hysteresis loss It is given by Steinmetz formula This formula is

Hysteresis Loss Wh = η Bmax 16f V Watts Bmax - Maximum flux density in the core f - Frequency of magnetic reversals V - Volume of armature in m3

η - Steinmetz hysteresis coefficient

In order to reduce this loss in a dc machine armature core is made of such materials which have a low value of Steinmetz hysteresis co-efficient eg silicon steel

Fig 124 Occurrence of hysteresis loss

ii) EDDYCURRENTLOSS

In addition to the voltages induced in the armature conductors there are also voltages induced in the armature core These voltages produce circulating currents in the armature core as shown in Fig 125 These are called eddy currents and power loss due to their flow is called eddy current loss The eddy current loss appears as heat which raises the temperature of the machine and lowers its efficiency Core resistance can be greatly increased by constructing the core of thin round If a continuous solid iron core is used the resistance to eddy current path will be small due to large cross-sectional area of the core Consequently the magnitude of eddy current and hence eddy current loss will be large The magnitude of eddy current can be reduced by making core resistance as high as practical The iron sheets called laminations [See Fig 126] The laminations are insulated from each other with a coating of varnish The insulating coating has a high resistance so very little current flows from one lamination to the other Alsobecause each lamination is very thin the resistance to current flowing through the width of a lamination is also quite large Thus laminating a core increases the core resistance which decreases the eddy current and hence the eddy current loss

Fig 125 Eddy current loss Fig 126 Core laminations

Eddy current loss is biven by the formula Pe= Ke Bmax

2 f2t2 V Watts

Where Ke = Constant depending upon electrical resistace of core Bmax = Maximum flux density in wbm2

f = frequency f magnetic reversals in Hz t = thickness of laminations in m and V = volume of core in m3

It may be noted that eddy current loss depends upon the square of lamination thickness For this reason lamination thickness should be kept as small as possible

3MECHANICALLOSSES

These losses are due to friction and windage(i) Friction losses eg bearing friction brush friction etc(ii) Windage loss ie air friction of rotating armature

These losses depend upon the speed of the machine But for a given speed they are practically constant

Note Iron losses and mechanical losses together are called stray losses

CONSTANT AND VARIABLE LOSSES

The losses in a dc generator (or dc motor) may be sub-divided into

(i) constant losses (ii) variable losses

(i) Constant losses

Those losses in a dc generator which remain constant at all loads are known as constant losses The constant losses in a dc generator are(a) iron losses(b) mechanical losses(c) shunt field losses(ii) Variable lossesThose losses in a dc generator which vary with load are called variable lossesThe variable losses in a dc generator are (a) Copper loss in armature winding (Ia

2 Ra) (b) Copper loss in series field winding (Ise

2Rse)

Total losses = Constant losses + Variable lossesNote Field Cu loss is constant for shunt and compound generators

128 POWER FLOW DIAGRAM

The power flow in a dc generator may be represented as

The different power stages are shown as A B and CA ndash The mechanical power input which is the power output of the prime mover is given as Pin = 2πNT

60 watts

B - The power developed in armature is Pa = Eg Ia watts C- The electrical power output is Pout = VIL watts

129 EFFICIENCY Efficiency of a machine is the ratio of its output power to the input power

Efficiency η = Output powerInput power

= Input powerminusLossesInput power = Output power

Output power+Losses

In case of a dc generator depending upon the power stages three different efficiencies can be defined as-

1 Mechanical Efficiency(ηm)

BA = Electrical power developedisinarmature

Mechanical power input2 Electrical Efficiency(ηe)

CB = Electrical power output

Electrical power developedisinthe armature 3 Overall or commercial efficiency (η)

CA = Electrical power output

Mechanical power input = ηmtimes ηe

130 CONDITION FOR MAXIMUM EFFICIENCY The efficiency of a dc generator is not constant but varies with load Fig 127shows the variation of efficiency with load current Consider a dc shunt generator with terminal voltage V delivering a load current of IL amp Generator output = V IL

Generator input = Output + Losses = V IL + constant losses + Variable losses = V IL + Wc+ Ia

2Ra

= V IL + Wc + (IL+ Ish)2 Rsh

The shunt field current Ish is small compared with IL so may be neglected

Generator input = VIL+ IL2Ra + Wc

Now efficiency η iquest outputinput = VI L

VIL+ Ia 2Ra+Wc

Efficiency is maximum when the denominator of the above equation is minimumie

ddIL ( VI L

VIL+ Ia 2Ra+Wc ) = 0

Which gives IL2 Ra = WC

ie Variable loss = constant lossThe load current for maximum efficiency is given by

IL iquest radicWcRa

Fig 127 Variation of efficiency with load

Hence the efficiency is maximum for that load current at which the variable loss is equal to the constant loss

131 APPLICATIONS

Separately excited generators

These types of generators have special applications such as electroplating electro refining of materials because separate supply is required for field excitation

Shunt generators

These types of generators are used in battery charging and ordinary lighting purposes

Series generators

This type of generators is used commonly as boosters on DC feeders as current generators for welding and arc lamp

Cumulatively compounded generators

This type of generators is used for domestic lighting purposes and to transmit energy over long distances

Differential compound generators

This type of generators is rarely used They are used for special applications such as electric arc welding

132 NUMERICAL EXCERCISES

1 The emf induced in the armature of a 450 kW 250 volt shunt generator is

2588volt when the field current is 200 amp and the generator is supplying power to a load at rated terminal voltage The armature circuit resistance is 0005 ohm Determine (i) load current (ii) power generated (iii) power output and (iv) electrical efficiency Neglect brush contact drop

Solution

Let the load current be I amp Given Ish = 20 A

Armature current = Load current + field current or Ia = I + 20 Induced emf = Terminal voltage + Ia Ra drop

2588 = 250 + (I + 20) times 0005

So (i) Load current I = 1740 amp

(ii) Power generated E Ia = 2588 (1740 + 20) watts = 455488 kW (iii) Power output V I = 250 times 1740 watts

= 435 kW

(iv) Electrical efficiency = 435455488 times 100 = 955

2 A 4 pole wave wound dc shunt generator delivers a load of 45 kW at a terminal voltage of 220 volts Its armature has 150 single turn coils and has a resistance of 001 ohm The air gap flux per pole is 002 weber Shunt field resistance is 50 ohm Calculate the speed at which it is being driven Neglect brush contact drop

Solution Load current I = W

V = 45000220 = 20455 amp

Field Current Ish = V

R s h = 220

50 = 44 amp

Armature current Ia = I +I sh = 20455 +44 = 20895 amp

Armature drop I aRa = 20895 X 001 = 20895 volts Induced emf Eg = V + Ia Ra = 220 + 20895 = 2220895 volts

But induced emf

So N = Eg60 AΦ ZP = 2220895times 60 times2

002 times150 times2 times 4 = 111045 rpm

3 The following data refer to the OCC of a DC separately excited generator at 1000 rpm

Field current in A 00 02 03 04 05 06 07 08 09 10 11 12Armature voltage in V 5 40 75 100 124 145 162 178 188 195 200 205

The machine is now connected as a shunt generator with a total field resistance of 200 Ω and driven at 1000 rpm Estimate graphically (i) the voltage to which the generator will build up at no load (ii) the armature field and load currents when the terminal voltage is found to be 150 V Neglect the effect of armature reaction and brush drop and assume armature resistance Ra to be 08 Ω (iii) Finally estimate the steady state armature current when the machine terminals are shorted

Solution

First of all the OCC of the generator is plotted in a graph paper as shown in figure

(i) The total field circuit resistance is given to be Rf = 200 Ω The Rf line is now drawn on the same graph paper passing through the origin The point of intersection of the Rf line and the OCC decides the final no load voltage and can be read from the graph as 192 V and the corresponding field current is 096 A (ii) Since the terminal voltage V is 150 V (= BM) field current If

is OM = 077 A Generated voltage E is given by AM But we know E = IaRa + V Hence IaRa= E ndash V = AM ndash BM = AB Now from the graph AB = 25 V there4Ia = 2508 = 3125 A So load current IL= Ia ndash If= 3125 ndash 077 = 3048 A

ADDITIONAL PROBLEMS

1 A 4 pole lap wound 115 kW 230 volts dc shunt generator has armatureresistance of 02 ohm and field resistance of 100 ohms Calculate (i) emf

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES

Power developedin armature = Eg Ia

Power delivered to load = V IL

Fig 111 (a) short shunt generator and (b) Long shunt generator

Long shunt- Here the shunt field winding is parallel to both armature and series field winding (Rse is wired in series to the armature) It is shown in figure 111 (b)

In Long shunt machines Series field current Ise = Ia = IL+ Ish

Shunt field current Ish = V R sh

Terminal voltage V = Eg- Ia (Ra+ Rse) Power developed in armature = Eg Ia

Power delivered to load = VIL

16 GENERATOR CHARACTERISTICS

The speed of a dc machine operated as a generator is fixed by the prime mover For general-purpose operation the prime mover is equipped with a speed governor so that the speed of the generator is practically constant Under such condition the generator performance deals primarily with the relation between excitation terminal voltage and load These relations can be best exhibited graphically by means of curves known as generator characteristics Thesecharacteristics show at a glance the behavior of the generator under differentload conditions

The following are the three most important characteristics of a dc generator161 Open Circuit Characteristic (OCC)

This curve shows the relation between the generated emf at no-load (E0) and the field current (If) at constant speed It is also known as magnetic characteristic or no-load saturation curve Its shape is practically the same for all generators whether separately or self-excited The data for OCC curve are obtained experimentally by operating the generator at no load and constant speed and recording the change in terminal voltage as the field current is varied

162 Internal or Total characteristic (EIa)

This curve shows the relation between the generated emf on load (E) and the armature current (Ia) The emf E is less than E0 due to the demagnetizing effect of armature reaction Therefore this curve will lie below the open circuitcharacteristic (OCC) The internal characteristic is of interest chiefly to thedesigner It cannot be obtained directly by experiment It is because a voltmetercannot read the emf generated on load due to the voltage drop in armatureresistance The internal characteristic can be obtained from externalcharacteristics if winding resistances are known because armature reaction effectis included in both characteristics163 External characteristic (VIL)This curve shows the relation between the terminal voltage (V) and load current(IL) The terminal voltage V will be less than E due to voltage drop in thearmature circuit Therefore this curve will lie below the internal characteristicThis characteristic is very important in determining the suitability of a generatorfor a given purpose It can be obtained by making simultaneous measurements

of terminal voltage and load current (with voltmeter and ammeter) of a loadedgenerator164 Open Circuit Characteristic of a DC Generator

The OCC for a dc generator is determined as follows The field winding of the dc generator (series or shunt) is disconnected from the machine and isseparately excited from an external dc source as shown in Fig 112 Thegenerator is run at fixed speed (ie normal speed) The field current (If) isincreased from zero in steps and the corresponding values of generated emf(E0) read off on a voltmeter connected across the armature terminals On plottingthe relation between E0 and If we get the open circuit characteristic as shown in

Fig113

Fig 112 Experimental setup for plotting OCC

Fig 113 OCC of dc generator

The following points may be noted from OCC When the field current is zero there is some generated emf OA

This is due to the residual magnetism in the field poles Over a fairly wide range of field current (upto point B in the curve)

the curve is linear It is because in this range reluctance of iron is negligible as compared with that of air gap The air gap reluctance is constant and hence linear relationship After point B on the curve the reluctance of iron also comes into picture and reluctance of iron is no longer negligible Consequently the curve deviates from linear relationship

After point C on the curve the magnetic saturation of poles begins and E0 tends to level off

The OCC of even self-excited generator is obtained by running it as a separately excited generator

165 Characteristics of a Separately Excited DCGenerator

The obvious disadvantage of a separately excited dc generator is that we require an external dc source for excitation But since the output voltage may be

controlled more easily and over a wide range (from zero to a maximum) this type of excitation finds many applications

(i) Open circuit characteristic

The OCC of a separately excited generator is determined in a manner described in Sec 164 Fig114 shows the variation of generated em f on noload with field current for various fixed speeds Note that if the value of constant speed is increased the steepness of the curve also increases When the field current is zero the residual magnetism in the poles will give rise to the small initial emf as shown

Fig114 Variation of OCC with speed

Internal and External Characteristics (Load Characteristics)

The external characteristic of a separately excited generator is the curve between the terminal voltage (V) and the load current IL (which is the same as armature current in this case) In order to determine the external characteristic the circuit set up is as shown in Fig 115

Fig 115 Experimental setup to plot load characteristics

As the load current increases the terminal voltage falls due to two reasons(a) The armature reaction weakens the main flux so that actual emf generated E on load is less than that generated (E0) on no load

(b) There is voltage drop across armature resistance (= ILRa = IaRa)Due to these reasons the external characteristic is a drooping curve [curve 3 inFig 116 ] Note that in the absence of armature reaction and armature dropthe generated emf would have been E0 (curve 1)The internal characteristic can be determined from external characteristic by adding IL Ra drop to the external characteristic It is because armature reaction drop is included in the external characteristic Curve 2 is the internal characteristic of the generator and should obviously lie above the external characteristic

Fig 116 Load characteristics of a separately excited dcgenerator

117 Voltage Build-Up in a Self-Excited GeneratorLet us see how voltage builds up in a self-excited generator(i) Shunt generator Consider a shunt generator If the generator is run at a constant speed some emf will be generated due to residual magnetism in the main poles This small emf circulates a field current which in turn produces additional flux toreinforce the original residual flux (provided field winding connections arecorrect) This process continues and the generator builds up the normalgenerated voltage following the OCC shown in Fig 117 (i) The field resistance Rf can be represented by a straight line passing through the origin as shown in Fig 117 (ii) The two curves can be shown on the same diagram as they have the same ordinate as in Fig 117 (iii)

Fig 117

Since the field circuit is inductive there is a delay in the increase in current uponclosing the field circuit switch The rate at which the current increases dependsupon the voltage available for increasing it Suppose at any instant the fieldcurrent is i (= OA) and is increasing at the rate didt Then

E0=i R f +iquest L didt

Where Rf = total field circuit resistance L = inductance of field circuit

At the considered instant the total emf available is AC [See Fig 117 (iii)] Anamount AB of the cmf AC is absorbed by the voltage drop iRf and theremainder part BC is available to overcome L didt Since this surplus voltage isavailable it is possible for the field current to increase above the value OAHowever at point D the available voltage is OM and is all absorbed by i RfD

rop Consequently the field current cannot increase further and the generatorbuild up stops Hence we arrive at a very important conclusion that the voltage build up of the generator is given by the point of intersection of OCC and field resistance line Thus in Fig 117 (iii) D is point of intersection of the two curves Hence the generator will build up a voltage OM(ii) Series generator During initial operation with no current yet flowing a residual voltage will be generated exactly as in the case of a shunt generator The residual voltage will cause a current to flow through the whole series circuit when the circuit isclosed There will then be voltage build up to an equilibrium point exactlyanalogous to the buildup of a shunt generator The voltage build up graph willbe similar to that of shunt generator except that now load current (instead offield current for shunt generator) will be taken along x-axis10487871048787(iii) Compound generator When a compound generator has its series field flux aiding its shunt field flux the machine is said to be cumulative compound When the series field isconnected in reverse so that its field flux opposes the shunt field flux thegenerator is then differential compound The easiest way to build up voltage in a compound generator is to start under no load conditions At no load only the shunt field is effective When no-load voltage build up is achieved the generator is loaded If under load the voltagerises the series field connection is cumulative If the voltage drops significantlythe connection is differential compound118 Critical Field Resistance for a Shunt Generator The voltage build up in a shunt generator depends upon field circuit resistance If the field circuit resistance is R1 (line OA) then generator will build up a voltage OM as shown in Fig 118 If the field circuit resistance is increasedto R2 (tine OB) the generator will build up a voltage OL slightly less than OM As the field circuit resistance is increased the slope of resistance line also increases When the field resistance line becomes tangent (line OC) to OCC the generator would just excite If the field circuit resistance is increased beyond this point (say line OD) the generator will fail to excite The field circuit resistance represented by line OC (tangent to OCC) is called critical field resistance RC for the shunt generator It may be defined as under The maximum field circuit resistance (for a given speed) with which the shunt generator would just excite is known as its critical field resistance It should be noted that shunt generator will build up voltage only if field circuit resistance is less than critical field resistance

Fig 118 Critical field resistance of shunt generator

119 Critical Resistance for a Series Generator

Fig 119 shows the voltage build up in a series generator Here R1 R2 etc represent the total circuit resistance (load resistance and field winding resistance) If the total circuit resistance is R1 then series generator will build up a voltage OL The line OC is tangent to OCC and represents the critical resistance RC for a series generator If the total resistance of the circuit is more than RC (say line OD) the generator will fail to build up voltage Note that Fig 119 is similar to Fig 118 with the following differences

(i) In Fig 118 R1 R2 etc represent the total field circuit resistanceHowever R1 R2 etc in Fig 119 represent the total circuit resistance(load resistance and series field winding resistance etc)

(ii) In Fig 118 field current alone is represented along X-axis However inFig 119 load current IL is represented along Y-axis Note that in aseries generator field current = load current IL

Fig 119 Critical resistance of a series generator

120 Characteristics of Series Generator Fig 120 (i) shows the connections of a series wound generator Since there is only one current (that which flows through the whole machine) the load current is the same as the exciting current(i) OCCCurve 1 shows the open circuit characteristic (OCC) of a series generator Itcan be obtained experimentally by disconnecting the field winding from themachine and exciting it from a separate dc source(ii) Internal characteristic Curve 2 shows the total or internal characteristic of a series generator It gives the relation between the generated emf E on load and armature current Due to armature reaction the flux in the machine will be less than the flux at no load Hence emf E generated under load conditions will be less than the emf E0 generated under no load conditions Consequently internal characteristic curvelies below the OCC curve the difference between them representing the effectof armature reaction [See Fig 120 (ii)]

Fig 120 Characteristics of series generator

(iii) External characteristic

Curve 3 shows the external characteristic of a series generator It gives therelation between terminal voltage and load current ILV = E g- Ia (Ra +Rse )Therefore external characteristic curve will lie below internal characteristiccurve by an amount equal to ohmic drop [ie Ia(Ra + Rse)] in the machine asshown in Fig 120 (ii) The internal and external characteristics ofa dc series generator can be plotted from one another as shown in Fig 121 Supposewe are given the internal characteristic of the generator Let the line OC represent the resistance of the whole machine ie Ra +Rse If the load current is OB drop in the machine is AB ieAB = Ohmic drop in the machine = OB(Ra + Rse)

Fig 121 Load characteristics of series generator

Now raise a perpendicular from point B and mark a point b on this line such that ab = AB Then point b will lie on the external characteristic of the generator

Following similar procedure other points of external characteristic can be located It is easy to see that we can also plot internal characteristic from the external characteristic

121 Characteristics of a Shunt Generator Fig 122 (i) show the connections of a shunt wound generator The armature current Ia splits up into two parts a small fraction Ish flowing through shunt field winding while the major part IL goes to the external load(i) OCC The OCC of a shunt generator is similar in shape to that of a series generator as shown in Fig 122 (ii) The line OA represents the shunt field circuitresistance When the generator is run at normal speed it will build up a voltageOM At no-load the terminal voltage of the generator will be constant (= OM)represented by the horizontal dotted line MC

Fig 122 Characteristics of a Shunt Generator

(ii) Internal characteristic

When the generator is loaded flux per pole is reduced due to armature reaction Therefore emf E generated on load is less than the emf generated at no load As a result the internal characteristic (EIa) drops down slightly as shown in Fig122 (ii)

(iii) External characteristic Curve 2 shows the external characteristic of a shunt generator It gives therelation between terminal voltage V and load current IL

Therefore external characteristic curve will lie below the internal characteristic curve by an amount equal to drop in the armature circuit[ie (IL + Ish)Ra] as shown in Fig 122 (ii)

Note It may be seen from the external characteristic that change in terminalvoltage from no-load to full load is small The terminal voltage can always bemaintained constant by adjusting the field rheostat R automatically122 Critical External Resistance for Shunt Generator

If the load resistance across the terminals of a shunt generator is decreased then load current increase However there is a limit to the increase in

load current with the decrease of load resistance Any decrease of load resistance beyond this point instead of increasing the current ultimately results in reduced current Consequently the external characteristic turns back (dotted curve) as shown in Fig 123 The tangent OA to the curve represents the minimum external resistance required to excite the shunt generator on load and is called critical external resistance If the resistance of the external circuit is less than the critical external resistance (represented by tangent OA in Fig 310) the machine will refuse to excite or will de-excite if already running This means that external resistance is so low as virtually to short circuit the machine and so doing away with its excitation

Note There are two critical resistances for a shunt generator viz (i) criticalfield resistance (ii) critical external resistance For the shunt generator to buildup voltage the former should not be exceeded and the latter must not be gonebelow

123 How to Draw OCC at Different Speeds

If we are given OCC of a generator at a constant speed N1 then we can easily draw the OCC at any other constant speed N2 Fig 123 illustrates theprocedure Here we are given OCC at a constant speed N1 It is desired to findthe OCC at constant speed N2 (it is assumed that n1 lt N2) For constantexcitation E propN

E2

E1 =

N 2

N1 so

E2 iquest E1

N2

N1

As shown in Fig (311) for If = OH E1 = HC Therefore the new value of emf (E2) for the same If but at N2

E2= HCN 2

N 1 = HD

Fig 123 Plotting OCC at another speed

This locates the point D on the new OCC at N2 Similarly other points can belocated taking different values of If The locus of these points will be the OCCat N2

124 Critical Speed (NC)

The critical speed of a shunt generator is the minimum speed below which it fails to excite Clearly it is the speed for which the given shunt field resistance represents the critical resistance In Fig 124 curve 2 corresponds to critical

speed because the shunt field resistance (Rsh) line is tangential to it If the generator runs at full speed N the new OCC moves upward and the Rsh line represents critical resistance for this speed

In order to find critical speed take any convenient point C on excitation axis and erect a perpendicular so as to cut Rsh and Rsh lines at points B and A respectively Then

BCAC

=NcN

Nc=N BCAC

Fig 124 Finding critical speed

125 Conditions for Voltage Build-Up of a Shunt Generator

The necessary conditions for voltage build-up in a shunt generator are

(i) There must be some residual magnetism in generator poles(ii) The connections of the field winding should be such that the field current strengthens the residual magnetism(iii) The resistance of the field circuit should be less than the critical resistance In other words the speed of the generator should be higher

than the critical speed

126 Compound Generator Characteristics

In a compound generator both series and shunt excitation are combined as shown in Fig 123 The shunt winding can be connected either across the armature only (short-shunt connection S) or across armature plus series field (long-shunt connection G) The compound generator can be cumulatively compounded or

differentially compounded generator The latter is rarely used in practice Therefore we shall discuss the characteristics of cumulatively compounded

generator It may be noted that external characteristics of long and short shunt compound generators are almost identical

External characteristic

Fig 124 shows the external characteristics of a cumulatively compoundedgenerator The series excitation aids the shunt excitation The degree of compounding depends upon the increase in series excitation with the increase inload current

(i) If series winding turns are so adjusted that with the increase in load current the terminal voltage increases it is called over-compounded generator In such a case as the load current increases the series field mmf increases and tends to increase the flux and hence the generated voltage The increase in generated voltage is greater than the IaRa drop so that instead of decreasing the terminal voltage increases as shown by curve A in Fig 124

(ii) If series winding turns are so adjusted that with the increase in load current the terminal voltage substantially remains constant it is called Flat-compounded generator The series winding of such a machine has lesser number of turns than the one in over-compounded machine and therefore does not increase the flux as much for a given load current Consequently the full-load voltage is nearly equal to the no-load voltage as indicated by curve B in Fig 124 (iii) If series field winding has lesser number of turns than for a flat compounded machine the terminal voltage falls with increase in load current as indicated by curve C m Fig 124 Such a machine is called under-compounded generator

Fig 123 compound generator

Fig 124 Load characteristics of compound generators

127 LOSSES

The losses in a dc machine (generator or motor) may be divided into three classes viz (i) copper losses or electrical losses (ii) iron or core losses and (iii) mechanical losses All these losses appear as heat and thus raise the temperature of the machine They also lower the efficiency of the machine The loss structure maybe demonstrated as

1Copperlosses

These losses occur due to currents in the various windings of the machine When an electric current of I amps flows in a resistance of R ohms the heat energy is lost at the rate of I2R joules sec and the power loss is I2R watts Generators and motors have one or more field circuits and an armature circuit in which such losses occur All resistance losses of this kind are called copper losses The different copper losses occurring in a dc machine are

(i) Armature copper loss = Ia2Ra

(ii) Shunt field copper loss = Ish2 Rsh

(iii) Series field copper loss = Ise2Rse

Armature copper loss will be around 30 to 40 of full load losses where as total field copper losses will constitute around 30 of full load losses

Note There is also brush contact loss due to brush contact resistance (ie resistance between the surface of brush and surface of commutator) This loss is generally included in armature copper loss The brush contact drop is almost independent of Ia For carbon brushes this drop is around 1 V per brush The power loss due to brush contact resistance is 2ebIa where eb is the voltage drop at one brush

2 IRONORCORELOSSES These losses occur in the armature of a dc machine and are due to the rotation of armature in the magnetic field of the poles They are of two types viz (i) hysteresis loss (ii) eddy current loss

(i) Hysteresis loss

Hysteresis loss occurs in the armature of the dc machine since any given part of the armature is subjected to magnetic field reversals as it passes under successive poles Fig 124 shows an armature rotating in two-pole machine Consider a small piece ab of the armature When the piece ab is under N-pole the magnetic lines pass from a to b Half a revolution later the same piece of iron is under S-pole and magnetic lines pass from b to a so that magnetism in the iron is reversed In order to reverse continuously the molecular magnets in the armature core some amount of power has to be spent which is called hysteresis loss It is given by Steinmetz formula This formula is

Hysteresis Loss Wh = η Bmax 16f V Watts Bmax - Maximum flux density in the core f - Frequency of magnetic reversals V - Volume of armature in m3

η - Steinmetz hysteresis coefficient

In order to reduce this loss in a dc machine armature core is made of such materials which have a low value of Steinmetz hysteresis co-efficient eg silicon steel

Fig 124 Occurrence of hysteresis loss

ii) EDDYCURRENTLOSS

In addition to the voltages induced in the armature conductors there are also voltages induced in the armature core These voltages produce circulating currents in the armature core as shown in Fig 125 These are called eddy currents and power loss due to their flow is called eddy current loss The eddy current loss appears as heat which raises the temperature of the machine and lowers its efficiency Core resistance can be greatly increased by constructing the core of thin round If a continuous solid iron core is used the resistance to eddy current path will be small due to large cross-sectional area of the core Consequently the magnitude of eddy current and hence eddy current loss will be large The magnitude of eddy current can be reduced by making core resistance as high as practical The iron sheets called laminations [See Fig 126] The laminations are insulated from each other with a coating of varnish The insulating coating has a high resistance so very little current flows from one lamination to the other Alsobecause each lamination is very thin the resistance to current flowing through the width of a lamination is also quite large Thus laminating a core increases the core resistance which decreases the eddy current and hence the eddy current loss

Fig 125 Eddy current loss Fig 126 Core laminations

Eddy current loss is biven by the formula Pe= Ke Bmax

2 f2t2 V Watts

Where Ke = Constant depending upon electrical resistace of core Bmax = Maximum flux density in wbm2

f = frequency f magnetic reversals in Hz t = thickness of laminations in m and V = volume of core in m3

It may be noted that eddy current loss depends upon the square of lamination thickness For this reason lamination thickness should be kept as small as possible

3MECHANICALLOSSES

These losses are due to friction and windage(i) Friction losses eg bearing friction brush friction etc(ii) Windage loss ie air friction of rotating armature

These losses depend upon the speed of the machine But for a given speed they are practically constant

Note Iron losses and mechanical losses together are called stray losses

CONSTANT AND VARIABLE LOSSES

The losses in a dc generator (or dc motor) may be sub-divided into

(i) constant losses (ii) variable losses

(i) Constant losses

Those losses in a dc generator which remain constant at all loads are known as constant losses The constant losses in a dc generator are(a) iron losses(b) mechanical losses(c) shunt field losses(ii) Variable lossesThose losses in a dc generator which vary with load are called variable lossesThe variable losses in a dc generator are (a) Copper loss in armature winding (Ia

2 Ra) (b) Copper loss in series field winding (Ise

2Rse)

Total losses = Constant losses + Variable lossesNote Field Cu loss is constant for shunt and compound generators

128 POWER FLOW DIAGRAM

The power flow in a dc generator may be represented as

The different power stages are shown as A B and CA ndash The mechanical power input which is the power output of the prime mover is given as Pin = 2πNT

60 watts

B - The power developed in armature is Pa = Eg Ia watts C- The electrical power output is Pout = VIL watts

129 EFFICIENCY Efficiency of a machine is the ratio of its output power to the input power

Efficiency η = Output powerInput power

= Input powerminusLossesInput power = Output power

Output power+Losses

In case of a dc generator depending upon the power stages three different efficiencies can be defined as-

1 Mechanical Efficiency(ηm)

BA = Electrical power developedisinarmature

Mechanical power input2 Electrical Efficiency(ηe)

CB = Electrical power output

Electrical power developedisinthe armature 3 Overall or commercial efficiency (η)

CA = Electrical power output

Mechanical power input = ηmtimes ηe

130 CONDITION FOR MAXIMUM EFFICIENCY The efficiency of a dc generator is not constant but varies with load Fig 127shows the variation of efficiency with load current Consider a dc shunt generator with terminal voltage V delivering a load current of IL amp Generator output = V IL

Generator input = Output + Losses = V IL + constant losses + Variable losses = V IL + Wc+ Ia

2Ra

= V IL + Wc + (IL+ Ish)2 Rsh

The shunt field current Ish is small compared with IL so may be neglected

Generator input = VIL+ IL2Ra + Wc

Now efficiency η iquest outputinput = VI L

VIL+ Ia 2Ra+Wc

Efficiency is maximum when the denominator of the above equation is minimumie

ddIL ( VI L

VIL+ Ia 2Ra+Wc ) = 0

Which gives IL2 Ra = WC

ie Variable loss = constant lossThe load current for maximum efficiency is given by

IL iquest radicWcRa

Fig 127 Variation of efficiency with load

Hence the efficiency is maximum for that load current at which the variable loss is equal to the constant loss

131 APPLICATIONS

Separately excited generators

These types of generators have special applications such as electroplating electro refining of materials because separate supply is required for field excitation

Shunt generators

These types of generators are used in battery charging and ordinary lighting purposes

Series generators

This type of generators is used commonly as boosters on DC feeders as current generators for welding and arc lamp

Cumulatively compounded generators

This type of generators is used for domestic lighting purposes and to transmit energy over long distances

Differential compound generators

This type of generators is rarely used They are used for special applications such as electric arc welding

132 NUMERICAL EXCERCISES

1 The emf induced in the armature of a 450 kW 250 volt shunt generator is

2588volt when the field current is 200 amp and the generator is supplying power to a load at rated terminal voltage The armature circuit resistance is 0005 ohm Determine (i) load current (ii) power generated (iii) power output and (iv) electrical efficiency Neglect brush contact drop

Solution

Let the load current be I amp Given Ish = 20 A

Armature current = Load current + field current or Ia = I + 20 Induced emf = Terminal voltage + Ia Ra drop

2588 = 250 + (I + 20) times 0005

So (i) Load current I = 1740 amp

(ii) Power generated E Ia = 2588 (1740 + 20) watts = 455488 kW (iii) Power output V I = 250 times 1740 watts

= 435 kW

(iv) Electrical efficiency = 435455488 times 100 = 955

2 A 4 pole wave wound dc shunt generator delivers a load of 45 kW at a terminal voltage of 220 volts Its armature has 150 single turn coils and has a resistance of 001 ohm The air gap flux per pole is 002 weber Shunt field resistance is 50 ohm Calculate the speed at which it is being driven Neglect brush contact drop

Solution Load current I = W

V = 45000220 = 20455 amp

Field Current Ish = V

R s h = 220

50 = 44 amp

Armature current Ia = I +I sh = 20455 +44 = 20895 amp

Armature drop I aRa = 20895 X 001 = 20895 volts Induced emf Eg = V + Ia Ra = 220 + 20895 = 2220895 volts

But induced emf

So N = Eg60 AΦ ZP = 2220895times 60 times2

002 times150 times2 times 4 = 111045 rpm

3 The following data refer to the OCC of a DC separately excited generator at 1000 rpm

Field current in A 00 02 03 04 05 06 07 08 09 10 11 12Armature voltage in V 5 40 75 100 124 145 162 178 188 195 200 205

The machine is now connected as a shunt generator with a total field resistance of 200 Ω and driven at 1000 rpm Estimate graphically (i) the voltage to which the generator will build up at no load (ii) the armature field and load currents when the terminal voltage is found to be 150 V Neglect the effect of armature reaction and brush drop and assume armature resistance Ra to be 08 Ω (iii) Finally estimate the steady state armature current when the machine terminals are shorted

Solution

First of all the OCC of the generator is plotted in a graph paper as shown in figure

(i) The total field circuit resistance is given to be Rf = 200 Ω The Rf line is now drawn on the same graph paper passing through the origin The point of intersection of the Rf line and the OCC decides the final no load voltage and can be read from the graph as 192 V and the corresponding field current is 096 A (ii) Since the terminal voltage V is 150 V (= BM) field current If

is OM = 077 A Generated voltage E is given by AM But we know E = IaRa + V Hence IaRa= E ndash V = AM ndash BM = AB Now from the graph AB = 25 V there4Ia = 2508 = 3125 A So load current IL= Ia ndash If= 3125 ndash 077 = 3048 A

ADDITIONAL PROBLEMS

1 A 4 pole lap wound 115 kW 230 volts dc shunt generator has armatureresistance of 02 ohm and field resistance of 100 ohms Calculate (i) emf

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES

16 GENERATOR CHARACTERISTICS

The speed of a dc machine operated as a generator is fixed by the prime mover For general-purpose operation the prime mover is equipped with a speed governor so that the speed of the generator is practically constant Under such condition the generator performance deals primarily with the relation between excitation terminal voltage and load These relations can be best exhibited graphically by means of curves known as generator characteristics Thesecharacteristics show at a glance the behavior of the generator under differentload conditions

The following are the three most important characteristics of a dc generator161 Open Circuit Characteristic (OCC)

This curve shows the relation between the generated emf at no-load (E0) and the field current (If) at constant speed It is also known as magnetic characteristic or no-load saturation curve Its shape is practically the same for all generators whether separately or self-excited The data for OCC curve are obtained experimentally by operating the generator at no load and constant speed and recording the change in terminal voltage as the field current is varied

162 Internal or Total characteristic (EIa)

This curve shows the relation between the generated emf on load (E) and the armature current (Ia) The emf E is less than E0 due to the demagnetizing effect of armature reaction Therefore this curve will lie below the open circuitcharacteristic (OCC) The internal characteristic is of interest chiefly to thedesigner It cannot be obtained directly by experiment It is because a voltmetercannot read the emf generated on load due to the voltage drop in armatureresistance The internal characteristic can be obtained from externalcharacteristics if winding resistances are known because armature reaction effectis included in both characteristics163 External characteristic (VIL)This curve shows the relation between the terminal voltage (V) and load current(IL) The terminal voltage V will be less than E due to voltage drop in thearmature circuit Therefore this curve will lie below the internal characteristicThis characteristic is very important in determining the suitability of a generatorfor a given purpose It can be obtained by making simultaneous measurements

of terminal voltage and load current (with voltmeter and ammeter) of a loadedgenerator164 Open Circuit Characteristic of a DC Generator

The OCC for a dc generator is determined as follows The field winding of the dc generator (series or shunt) is disconnected from the machine and isseparately excited from an external dc source as shown in Fig 112 Thegenerator is run at fixed speed (ie normal speed) The field current (If) isincreased from zero in steps and the corresponding values of generated emf(E0) read off on a voltmeter connected across the armature terminals On plottingthe relation between E0 and If we get the open circuit characteristic as shown in

Fig113

Fig 112 Experimental setup for plotting OCC

Fig 113 OCC of dc generator

The following points may be noted from OCC When the field current is zero there is some generated emf OA

This is due to the residual magnetism in the field poles Over a fairly wide range of field current (upto point B in the curve)

the curve is linear It is because in this range reluctance of iron is negligible as compared with that of air gap The air gap reluctance is constant and hence linear relationship After point B on the curve the reluctance of iron also comes into picture and reluctance of iron is no longer negligible Consequently the curve deviates from linear relationship

After point C on the curve the magnetic saturation of poles begins and E0 tends to level off

The OCC of even self-excited generator is obtained by running it as a separately excited generator

165 Characteristics of a Separately Excited DCGenerator

The obvious disadvantage of a separately excited dc generator is that we require an external dc source for excitation But since the output voltage may be

controlled more easily and over a wide range (from zero to a maximum) this type of excitation finds many applications

(i) Open circuit characteristic

The OCC of a separately excited generator is determined in a manner described in Sec 164 Fig114 shows the variation of generated em f on noload with field current for various fixed speeds Note that if the value of constant speed is increased the steepness of the curve also increases When the field current is zero the residual magnetism in the poles will give rise to the small initial emf as shown

Fig114 Variation of OCC with speed

Internal and External Characteristics (Load Characteristics)

The external characteristic of a separately excited generator is the curve between the terminal voltage (V) and the load current IL (which is the same as armature current in this case) In order to determine the external characteristic the circuit set up is as shown in Fig 115

Fig 115 Experimental setup to plot load characteristics

As the load current increases the terminal voltage falls due to two reasons(a) The armature reaction weakens the main flux so that actual emf generated E on load is less than that generated (E0) on no load

(b) There is voltage drop across armature resistance (= ILRa = IaRa)Due to these reasons the external characteristic is a drooping curve [curve 3 inFig 116 ] Note that in the absence of armature reaction and armature dropthe generated emf would have been E0 (curve 1)The internal characteristic can be determined from external characteristic by adding IL Ra drop to the external characteristic It is because armature reaction drop is included in the external characteristic Curve 2 is the internal characteristic of the generator and should obviously lie above the external characteristic

Fig 116 Load characteristics of a separately excited dcgenerator

117 Voltage Build-Up in a Self-Excited GeneratorLet us see how voltage builds up in a self-excited generator(i) Shunt generator Consider a shunt generator If the generator is run at a constant speed some emf will be generated due to residual magnetism in the main poles This small emf circulates a field current which in turn produces additional flux toreinforce the original residual flux (provided field winding connections arecorrect) This process continues and the generator builds up the normalgenerated voltage following the OCC shown in Fig 117 (i) The field resistance Rf can be represented by a straight line passing through the origin as shown in Fig 117 (ii) The two curves can be shown on the same diagram as they have the same ordinate as in Fig 117 (iii)

Fig 117

Since the field circuit is inductive there is a delay in the increase in current uponclosing the field circuit switch The rate at which the current increases dependsupon the voltage available for increasing it Suppose at any instant the fieldcurrent is i (= OA) and is increasing at the rate didt Then

E0=i R f +iquest L didt

Where Rf = total field circuit resistance L = inductance of field circuit

At the considered instant the total emf available is AC [See Fig 117 (iii)] Anamount AB of the cmf AC is absorbed by the voltage drop iRf and theremainder part BC is available to overcome L didt Since this surplus voltage isavailable it is possible for the field current to increase above the value OAHowever at point D the available voltage is OM and is all absorbed by i RfD

rop Consequently the field current cannot increase further and the generatorbuild up stops Hence we arrive at a very important conclusion that the voltage build up of the generator is given by the point of intersection of OCC and field resistance line Thus in Fig 117 (iii) D is point of intersection of the two curves Hence the generator will build up a voltage OM(ii) Series generator During initial operation with no current yet flowing a residual voltage will be generated exactly as in the case of a shunt generator The residual voltage will cause a current to flow through the whole series circuit when the circuit isclosed There will then be voltage build up to an equilibrium point exactlyanalogous to the buildup of a shunt generator The voltage build up graph willbe similar to that of shunt generator except that now load current (instead offield current for shunt generator) will be taken along x-axis10487871048787(iii) Compound generator When a compound generator has its series field flux aiding its shunt field flux the machine is said to be cumulative compound When the series field isconnected in reverse so that its field flux opposes the shunt field flux thegenerator is then differential compound The easiest way to build up voltage in a compound generator is to start under no load conditions At no load only the shunt field is effective When no-load voltage build up is achieved the generator is loaded If under load the voltagerises the series field connection is cumulative If the voltage drops significantlythe connection is differential compound118 Critical Field Resistance for a Shunt Generator The voltage build up in a shunt generator depends upon field circuit resistance If the field circuit resistance is R1 (line OA) then generator will build up a voltage OM as shown in Fig 118 If the field circuit resistance is increasedto R2 (tine OB) the generator will build up a voltage OL slightly less than OM As the field circuit resistance is increased the slope of resistance line also increases When the field resistance line becomes tangent (line OC) to OCC the generator would just excite If the field circuit resistance is increased beyond this point (say line OD) the generator will fail to excite The field circuit resistance represented by line OC (tangent to OCC) is called critical field resistance RC for the shunt generator It may be defined as under The maximum field circuit resistance (for a given speed) with which the shunt generator would just excite is known as its critical field resistance It should be noted that shunt generator will build up voltage only if field circuit resistance is less than critical field resistance

Fig 118 Critical field resistance of shunt generator

119 Critical Resistance for a Series Generator

Fig 119 shows the voltage build up in a series generator Here R1 R2 etc represent the total circuit resistance (load resistance and field winding resistance) If the total circuit resistance is R1 then series generator will build up a voltage OL The line OC is tangent to OCC and represents the critical resistance RC for a series generator If the total resistance of the circuit is more than RC (say line OD) the generator will fail to build up voltage Note that Fig 119 is similar to Fig 118 with the following differences

(i) In Fig 118 R1 R2 etc represent the total field circuit resistanceHowever R1 R2 etc in Fig 119 represent the total circuit resistance(load resistance and series field winding resistance etc)

(ii) In Fig 118 field current alone is represented along X-axis However inFig 119 load current IL is represented along Y-axis Note that in aseries generator field current = load current IL

Fig 119 Critical resistance of a series generator

120 Characteristics of Series Generator Fig 120 (i) shows the connections of a series wound generator Since there is only one current (that which flows through the whole machine) the load current is the same as the exciting current(i) OCCCurve 1 shows the open circuit characteristic (OCC) of a series generator Itcan be obtained experimentally by disconnecting the field winding from themachine and exciting it from a separate dc source(ii) Internal characteristic Curve 2 shows the total or internal characteristic of a series generator It gives the relation between the generated emf E on load and armature current Due to armature reaction the flux in the machine will be less than the flux at no load Hence emf E generated under load conditions will be less than the emf E0 generated under no load conditions Consequently internal characteristic curvelies below the OCC curve the difference between them representing the effectof armature reaction [See Fig 120 (ii)]

Fig 120 Characteristics of series generator

(iii) External characteristic

Curve 3 shows the external characteristic of a series generator It gives therelation between terminal voltage and load current ILV = E g- Ia (Ra +Rse )Therefore external characteristic curve will lie below internal characteristiccurve by an amount equal to ohmic drop [ie Ia(Ra + Rse)] in the machine asshown in Fig 120 (ii) The internal and external characteristics ofa dc series generator can be plotted from one another as shown in Fig 121 Supposewe are given the internal characteristic of the generator Let the line OC represent the resistance of the whole machine ie Ra +Rse If the load current is OB drop in the machine is AB ieAB = Ohmic drop in the machine = OB(Ra + Rse)

Fig 121 Load characteristics of series generator

Now raise a perpendicular from point B and mark a point b on this line such that ab = AB Then point b will lie on the external characteristic of the generator

Following similar procedure other points of external characteristic can be located It is easy to see that we can also plot internal characteristic from the external characteristic

121 Characteristics of a Shunt Generator Fig 122 (i) show the connections of a shunt wound generator The armature current Ia splits up into two parts a small fraction Ish flowing through shunt field winding while the major part IL goes to the external load(i) OCC The OCC of a shunt generator is similar in shape to that of a series generator as shown in Fig 122 (ii) The line OA represents the shunt field circuitresistance When the generator is run at normal speed it will build up a voltageOM At no-load the terminal voltage of the generator will be constant (= OM)represented by the horizontal dotted line MC

Fig 122 Characteristics of a Shunt Generator

(ii) Internal characteristic

When the generator is loaded flux per pole is reduced due to armature reaction Therefore emf E generated on load is less than the emf generated at no load As a result the internal characteristic (EIa) drops down slightly as shown in Fig122 (ii)

(iii) External characteristic Curve 2 shows the external characteristic of a shunt generator It gives therelation between terminal voltage V and load current IL

Therefore external characteristic curve will lie below the internal characteristic curve by an amount equal to drop in the armature circuit[ie (IL + Ish)Ra] as shown in Fig 122 (ii)

Note It may be seen from the external characteristic that change in terminalvoltage from no-load to full load is small The terminal voltage can always bemaintained constant by adjusting the field rheostat R automatically122 Critical External Resistance for Shunt Generator

If the load resistance across the terminals of a shunt generator is decreased then load current increase However there is a limit to the increase in

load current with the decrease of load resistance Any decrease of load resistance beyond this point instead of increasing the current ultimately results in reduced current Consequently the external characteristic turns back (dotted curve) as shown in Fig 123 The tangent OA to the curve represents the minimum external resistance required to excite the shunt generator on load and is called critical external resistance If the resistance of the external circuit is less than the critical external resistance (represented by tangent OA in Fig 310) the machine will refuse to excite or will de-excite if already running This means that external resistance is so low as virtually to short circuit the machine and so doing away with its excitation

Note There are two critical resistances for a shunt generator viz (i) criticalfield resistance (ii) critical external resistance For the shunt generator to buildup voltage the former should not be exceeded and the latter must not be gonebelow

123 How to Draw OCC at Different Speeds

If we are given OCC of a generator at a constant speed N1 then we can easily draw the OCC at any other constant speed N2 Fig 123 illustrates theprocedure Here we are given OCC at a constant speed N1 It is desired to findthe OCC at constant speed N2 (it is assumed that n1 lt N2) For constantexcitation E propN

E2

E1 =

N 2

N1 so

E2 iquest E1

N2

N1

As shown in Fig (311) for If = OH E1 = HC Therefore the new value of emf (E2) for the same If but at N2

E2= HCN 2

N 1 = HD

Fig 123 Plotting OCC at another speed

This locates the point D on the new OCC at N2 Similarly other points can belocated taking different values of If The locus of these points will be the OCCat N2

124 Critical Speed (NC)

The critical speed of a shunt generator is the minimum speed below which it fails to excite Clearly it is the speed for which the given shunt field resistance represents the critical resistance In Fig 124 curve 2 corresponds to critical

speed because the shunt field resistance (Rsh) line is tangential to it If the generator runs at full speed N the new OCC moves upward and the Rsh line represents critical resistance for this speed

In order to find critical speed take any convenient point C on excitation axis and erect a perpendicular so as to cut Rsh and Rsh lines at points B and A respectively Then

BCAC

=NcN

Nc=N BCAC

Fig 124 Finding critical speed

125 Conditions for Voltage Build-Up of a Shunt Generator

The necessary conditions for voltage build-up in a shunt generator are

(i) There must be some residual magnetism in generator poles(ii) The connections of the field winding should be such that the field current strengthens the residual magnetism(iii) The resistance of the field circuit should be less than the critical resistance In other words the speed of the generator should be higher

than the critical speed

126 Compound Generator Characteristics

In a compound generator both series and shunt excitation are combined as shown in Fig 123 The shunt winding can be connected either across the armature only (short-shunt connection S) or across armature plus series field (long-shunt connection G) The compound generator can be cumulatively compounded or

differentially compounded generator The latter is rarely used in practice Therefore we shall discuss the characteristics of cumulatively compounded

generator It may be noted that external characteristics of long and short shunt compound generators are almost identical

External characteristic

Fig 124 shows the external characteristics of a cumulatively compoundedgenerator The series excitation aids the shunt excitation The degree of compounding depends upon the increase in series excitation with the increase inload current

(i) If series winding turns are so adjusted that with the increase in load current the terminal voltage increases it is called over-compounded generator In such a case as the load current increases the series field mmf increases and tends to increase the flux and hence the generated voltage The increase in generated voltage is greater than the IaRa drop so that instead of decreasing the terminal voltage increases as shown by curve A in Fig 124

(ii) If series winding turns are so adjusted that with the increase in load current the terminal voltage substantially remains constant it is called Flat-compounded generator The series winding of such a machine has lesser number of turns than the one in over-compounded machine and therefore does not increase the flux as much for a given load current Consequently the full-load voltage is nearly equal to the no-load voltage as indicated by curve B in Fig 124 (iii) If series field winding has lesser number of turns than for a flat compounded machine the terminal voltage falls with increase in load current as indicated by curve C m Fig 124 Such a machine is called under-compounded generator

Fig 123 compound generator

Fig 124 Load characteristics of compound generators

127 LOSSES

The losses in a dc machine (generator or motor) may be divided into three classes viz (i) copper losses or electrical losses (ii) iron or core losses and (iii) mechanical losses All these losses appear as heat and thus raise the temperature of the machine They also lower the efficiency of the machine The loss structure maybe demonstrated as

1Copperlosses

These losses occur due to currents in the various windings of the machine When an electric current of I amps flows in a resistance of R ohms the heat energy is lost at the rate of I2R joules sec and the power loss is I2R watts Generators and motors have one or more field circuits and an armature circuit in which such losses occur All resistance losses of this kind are called copper losses The different copper losses occurring in a dc machine are

(i) Armature copper loss = Ia2Ra

(ii) Shunt field copper loss = Ish2 Rsh

(iii) Series field copper loss = Ise2Rse

Armature copper loss will be around 30 to 40 of full load losses where as total field copper losses will constitute around 30 of full load losses

Note There is also brush contact loss due to brush contact resistance (ie resistance between the surface of brush and surface of commutator) This loss is generally included in armature copper loss The brush contact drop is almost independent of Ia For carbon brushes this drop is around 1 V per brush The power loss due to brush contact resistance is 2ebIa where eb is the voltage drop at one brush

2 IRONORCORELOSSES These losses occur in the armature of a dc machine and are due to the rotation of armature in the magnetic field of the poles They are of two types viz (i) hysteresis loss (ii) eddy current loss

(i) Hysteresis loss

Hysteresis loss occurs in the armature of the dc machine since any given part of the armature is subjected to magnetic field reversals as it passes under successive poles Fig 124 shows an armature rotating in two-pole machine Consider a small piece ab of the armature When the piece ab is under N-pole the magnetic lines pass from a to b Half a revolution later the same piece of iron is under S-pole and magnetic lines pass from b to a so that magnetism in the iron is reversed In order to reverse continuously the molecular magnets in the armature core some amount of power has to be spent which is called hysteresis loss It is given by Steinmetz formula This formula is

Hysteresis Loss Wh = η Bmax 16f V Watts Bmax - Maximum flux density in the core f - Frequency of magnetic reversals V - Volume of armature in m3

η - Steinmetz hysteresis coefficient

In order to reduce this loss in a dc machine armature core is made of such materials which have a low value of Steinmetz hysteresis co-efficient eg silicon steel

Fig 124 Occurrence of hysteresis loss

ii) EDDYCURRENTLOSS

In addition to the voltages induced in the armature conductors there are also voltages induced in the armature core These voltages produce circulating currents in the armature core as shown in Fig 125 These are called eddy currents and power loss due to their flow is called eddy current loss The eddy current loss appears as heat which raises the temperature of the machine and lowers its efficiency Core resistance can be greatly increased by constructing the core of thin round If a continuous solid iron core is used the resistance to eddy current path will be small due to large cross-sectional area of the core Consequently the magnitude of eddy current and hence eddy current loss will be large The magnitude of eddy current can be reduced by making core resistance as high as practical The iron sheets called laminations [See Fig 126] The laminations are insulated from each other with a coating of varnish The insulating coating has a high resistance so very little current flows from one lamination to the other Alsobecause each lamination is very thin the resistance to current flowing through the width of a lamination is also quite large Thus laminating a core increases the core resistance which decreases the eddy current and hence the eddy current loss

Fig 125 Eddy current loss Fig 126 Core laminations

Eddy current loss is biven by the formula Pe= Ke Bmax

2 f2t2 V Watts

Where Ke = Constant depending upon electrical resistace of core Bmax = Maximum flux density in wbm2

f = frequency f magnetic reversals in Hz t = thickness of laminations in m and V = volume of core in m3

It may be noted that eddy current loss depends upon the square of lamination thickness For this reason lamination thickness should be kept as small as possible

3MECHANICALLOSSES

These losses are due to friction and windage(i) Friction losses eg bearing friction brush friction etc(ii) Windage loss ie air friction of rotating armature

These losses depend upon the speed of the machine But for a given speed they are practically constant

Note Iron losses and mechanical losses together are called stray losses

CONSTANT AND VARIABLE LOSSES

The losses in a dc generator (or dc motor) may be sub-divided into

(i) constant losses (ii) variable losses

(i) Constant losses

Those losses in a dc generator which remain constant at all loads are known as constant losses The constant losses in a dc generator are(a) iron losses(b) mechanical losses(c) shunt field losses(ii) Variable lossesThose losses in a dc generator which vary with load are called variable lossesThe variable losses in a dc generator are (a) Copper loss in armature winding (Ia

2 Ra) (b) Copper loss in series field winding (Ise

2Rse)

Total losses = Constant losses + Variable lossesNote Field Cu loss is constant for shunt and compound generators

128 POWER FLOW DIAGRAM

The power flow in a dc generator may be represented as

The different power stages are shown as A B and CA ndash The mechanical power input which is the power output of the prime mover is given as Pin = 2πNT

60 watts

B - The power developed in armature is Pa = Eg Ia watts C- The electrical power output is Pout = VIL watts

129 EFFICIENCY Efficiency of a machine is the ratio of its output power to the input power

Efficiency η = Output powerInput power

= Input powerminusLossesInput power = Output power

Output power+Losses

In case of a dc generator depending upon the power stages three different efficiencies can be defined as-

1 Mechanical Efficiency(ηm)

BA = Electrical power developedisinarmature

Mechanical power input2 Electrical Efficiency(ηe)

CB = Electrical power output

Electrical power developedisinthe armature 3 Overall or commercial efficiency (η)

CA = Electrical power output

Mechanical power input = ηmtimes ηe

130 CONDITION FOR MAXIMUM EFFICIENCY The efficiency of a dc generator is not constant but varies with load Fig 127shows the variation of efficiency with load current Consider a dc shunt generator with terminal voltage V delivering a load current of IL amp Generator output = V IL

Generator input = Output + Losses = V IL + constant losses + Variable losses = V IL + Wc+ Ia

2Ra

= V IL + Wc + (IL+ Ish)2 Rsh

The shunt field current Ish is small compared with IL so may be neglected

Generator input = VIL+ IL2Ra + Wc

Now efficiency η iquest outputinput = VI L

VIL+ Ia 2Ra+Wc

Efficiency is maximum when the denominator of the above equation is minimumie

ddIL ( VI L

VIL+ Ia 2Ra+Wc ) = 0

Which gives IL2 Ra = WC

ie Variable loss = constant lossThe load current for maximum efficiency is given by

IL iquest radicWcRa

Fig 127 Variation of efficiency with load

Hence the efficiency is maximum for that load current at which the variable loss is equal to the constant loss

131 APPLICATIONS

Separately excited generators

These types of generators have special applications such as electroplating electro refining of materials because separate supply is required for field excitation

Shunt generators

These types of generators are used in battery charging and ordinary lighting purposes

Series generators

This type of generators is used commonly as boosters on DC feeders as current generators for welding and arc lamp

Cumulatively compounded generators

This type of generators is used for domestic lighting purposes and to transmit energy over long distances

Differential compound generators

This type of generators is rarely used They are used for special applications such as electric arc welding

132 NUMERICAL EXCERCISES

1 The emf induced in the armature of a 450 kW 250 volt shunt generator is

2588volt when the field current is 200 amp and the generator is supplying power to a load at rated terminal voltage The armature circuit resistance is 0005 ohm Determine (i) load current (ii) power generated (iii) power output and (iv) electrical efficiency Neglect brush contact drop

Solution

Let the load current be I amp Given Ish = 20 A

Armature current = Load current + field current or Ia = I + 20 Induced emf = Terminal voltage + Ia Ra drop

2588 = 250 + (I + 20) times 0005

So (i) Load current I = 1740 amp

(ii) Power generated E Ia = 2588 (1740 + 20) watts = 455488 kW (iii) Power output V I = 250 times 1740 watts

= 435 kW

(iv) Electrical efficiency = 435455488 times 100 = 955

2 A 4 pole wave wound dc shunt generator delivers a load of 45 kW at a terminal voltage of 220 volts Its armature has 150 single turn coils and has a resistance of 001 ohm The air gap flux per pole is 002 weber Shunt field resistance is 50 ohm Calculate the speed at which it is being driven Neglect brush contact drop

Solution Load current I = W

V = 45000220 = 20455 amp

Field Current Ish = V

R s h = 220

50 = 44 amp

Armature current Ia = I +I sh = 20455 +44 = 20895 amp

Armature drop I aRa = 20895 X 001 = 20895 volts Induced emf Eg = V + Ia Ra = 220 + 20895 = 2220895 volts

But induced emf

So N = Eg60 AΦ ZP = 2220895times 60 times2

002 times150 times2 times 4 = 111045 rpm

3 The following data refer to the OCC of a DC separately excited generator at 1000 rpm

Field current in A 00 02 03 04 05 06 07 08 09 10 11 12Armature voltage in V 5 40 75 100 124 145 162 178 188 195 200 205

The machine is now connected as a shunt generator with a total field resistance of 200 Ω and driven at 1000 rpm Estimate graphically (i) the voltage to which the generator will build up at no load (ii) the armature field and load currents when the terminal voltage is found to be 150 V Neglect the effect of armature reaction and brush drop and assume armature resistance Ra to be 08 Ω (iii) Finally estimate the steady state armature current when the machine terminals are shorted

Solution

First of all the OCC of the generator is plotted in a graph paper as shown in figure

(i) The total field circuit resistance is given to be Rf = 200 Ω The Rf line is now drawn on the same graph paper passing through the origin The point of intersection of the Rf line and the OCC decides the final no load voltage and can be read from the graph as 192 V and the corresponding field current is 096 A (ii) Since the terminal voltage V is 150 V (= BM) field current If

is OM = 077 A Generated voltage E is given by AM But we know E = IaRa + V Hence IaRa= E ndash V = AM ndash BM = AB Now from the graph AB = 25 V there4Ia = 2508 = 3125 A So load current IL= Ia ndash If= 3125 ndash 077 = 3048 A

ADDITIONAL PROBLEMS

1 A 4 pole lap wound 115 kW 230 volts dc shunt generator has armatureresistance of 02 ohm and field resistance of 100 ohms Calculate (i) emf

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES

of terminal voltage and load current (with voltmeter and ammeter) of a loadedgenerator164 Open Circuit Characteristic of a DC Generator

The OCC for a dc generator is determined as follows The field winding of the dc generator (series or shunt) is disconnected from the machine and isseparately excited from an external dc source as shown in Fig 112 Thegenerator is run at fixed speed (ie normal speed) The field current (If) isincreased from zero in steps and the corresponding values of generated emf(E0) read off on a voltmeter connected across the armature terminals On plottingthe relation between E0 and If we get the open circuit characteristic as shown in

Fig113

Fig 112 Experimental setup for plotting OCC

Fig 113 OCC of dc generator

The following points may be noted from OCC When the field current is zero there is some generated emf OA

This is due to the residual magnetism in the field poles Over a fairly wide range of field current (upto point B in the curve)

the curve is linear It is because in this range reluctance of iron is negligible as compared with that of air gap The air gap reluctance is constant and hence linear relationship After point B on the curve the reluctance of iron also comes into picture and reluctance of iron is no longer negligible Consequently the curve deviates from linear relationship

After point C on the curve the magnetic saturation of poles begins and E0 tends to level off

The OCC of even self-excited generator is obtained by running it as a separately excited generator

165 Characteristics of a Separately Excited DCGenerator

The obvious disadvantage of a separately excited dc generator is that we require an external dc source for excitation But since the output voltage may be

controlled more easily and over a wide range (from zero to a maximum) this type of excitation finds many applications

(i) Open circuit characteristic

The OCC of a separately excited generator is determined in a manner described in Sec 164 Fig114 shows the variation of generated em f on noload with field current for various fixed speeds Note that if the value of constant speed is increased the steepness of the curve also increases When the field current is zero the residual magnetism in the poles will give rise to the small initial emf as shown

Fig114 Variation of OCC with speed

Internal and External Characteristics (Load Characteristics)

The external characteristic of a separately excited generator is the curve between the terminal voltage (V) and the load current IL (which is the same as armature current in this case) In order to determine the external characteristic the circuit set up is as shown in Fig 115

Fig 115 Experimental setup to plot load characteristics

As the load current increases the terminal voltage falls due to two reasons(a) The armature reaction weakens the main flux so that actual emf generated E on load is less than that generated (E0) on no load

(b) There is voltage drop across armature resistance (= ILRa = IaRa)Due to these reasons the external characteristic is a drooping curve [curve 3 inFig 116 ] Note that in the absence of armature reaction and armature dropthe generated emf would have been E0 (curve 1)The internal characteristic can be determined from external characteristic by adding IL Ra drop to the external characteristic It is because armature reaction drop is included in the external characteristic Curve 2 is the internal characteristic of the generator and should obviously lie above the external characteristic

Fig 116 Load characteristics of a separately excited dcgenerator

117 Voltage Build-Up in a Self-Excited GeneratorLet us see how voltage builds up in a self-excited generator(i) Shunt generator Consider a shunt generator If the generator is run at a constant speed some emf will be generated due to residual magnetism in the main poles This small emf circulates a field current which in turn produces additional flux toreinforce the original residual flux (provided field winding connections arecorrect) This process continues and the generator builds up the normalgenerated voltage following the OCC shown in Fig 117 (i) The field resistance Rf can be represented by a straight line passing through the origin as shown in Fig 117 (ii) The two curves can be shown on the same diagram as they have the same ordinate as in Fig 117 (iii)

Fig 117

Since the field circuit is inductive there is a delay in the increase in current uponclosing the field circuit switch The rate at which the current increases dependsupon the voltage available for increasing it Suppose at any instant the fieldcurrent is i (= OA) and is increasing at the rate didt Then

E0=i R f +iquest L didt

Where Rf = total field circuit resistance L = inductance of field circuit

At the considered instant the total emf available is AC [See Fig 117 (iii)] Anamount AB of the cmf AC is absorbed by the voltage drop iRf and theremainder part BC is available to overcome L didt Since this surplus voltage isavailable it is possible for the field current to increase above the value OAHowever at point D the available voltage is OM and is all absorbed by i RfD

rop Consequently the field current cannot increase further and the generatorbuild up stops Hence we arrive at a very important conclusion that the voltage build up of the generator is given by the point of intersection of OCC and field resistance line Thus in Fig 117 (iii) D is point of intersection of the two curves Hence the generator will build up a voltage OM(ii) Series generator During initial operation with no current yet flowing a residual voltage will be generated exactly as in the case of a shunt generator The residual voltage will cause a current to flow through the whole series circuit when the circuit isclosed There will then be voltage build up to an equilibrium point exactlyanalogous to the buildup of a shunt generator The voltage build up graph willbe similar to that of shunt generator except that now load current (instead offield current for shunt generator) will be taken along x-axis10487871048787(iii) Compound generator When a compound generator has its series field flux aiding its shunt field flux the machine is said to be cumulative compound When the series field isconnected in reverse so that its field flux opposes the shunt field flux thegenerator is then differential compound The easiest way to build up voltage in a compound generator is to start under no load conditions At no load only the shunt field is effective When no-load voltage build up is achieved the generator is loaded If under load the voltagerises the series field connection is cumulative If the voltage drops significantlythe connection is differential compound118 Critical Field Resistance for a Shunt Generator The voltage build up in a shunt generator depends upon field circuit resistance If the field circuit resistance is R1 (line OA) then generator will build up a voltage OM as shown in Fig 118 If the field circuit resistance is increasedto R2 (tine OB) the generator will build up a voltage OL slightly less than OM As the field circuit resistance is increased the slope of resistance line also increases When the field resistance line becomes tangent (line OC) to OCC the generator would just excite If the field circuit resistance is increased beyond this point (say line OD) the generator will fail to excite The field circuit resistance represented by line OC (tangent to OCC) is called critical field resistance RC for the shunt generator It may be defined as under The maximum field circuit resistance (for a given speed) with which the shunt generator would just excite is known as its critical field resistance It should be noted that shunt generator will build up voltage only if field circuit resistance is less than critical field resistance

Fig 118 Critical field resistance of shunt generator

119 Critical Resistance for a Series Generator

Fig 119 shows the voltage build up in a series generator Here R1 R2 etc represent the total circuit resistance (load resistance and field winding resistance) If the total circuit resistance is R1 then series generator will build up a voltage OL The line OC is tangent to OCC and represents the critical resistance RC for a series generator If the total resistance of the circuit is more than RC (say line OD) the generator will fail to build up voltage Note that Fig 119 is similar to Fig 118 with the following differences

(i) In Fig 118 R1 R2 etc represent the total field circuit resistanceHowever R1 R2 etc in Fig 119 represent the total circuit resistance(load resistance and series field winding resistance etc)

(ii) In Fig 118 field current alone is represented along X-axis However inFig 119 load current IL is represented along Y-axis Note that in aseries generator field current = load current IL

Fig 119 Critical resistance of a series generator

120 Characteristics of Series Generator Fig 120 (i) shows the connections of a series wound generator Since there is only one current (that which flows through the whole machine) the load current is the same as the exciting current(i) OCCCurve 1 shows the open circuit characteristic (OCC) of a series generator Itcan be obtained experimentally by disconnecting the field winding from themachine and exciting it from a separate dc source(ii) Internal characteristic Curve 2 shows the total or internal characteristic of a series generator It gives the relation between the generated emf E on load and armature current Due to armature reaction the flux in the machine will be less than the flux at no load Hence emf E generated under load conditions will be less than the emf E0 generated under no load conditions Consequently internal characteristic curvelies below the OCC curve the difference between them representing the effectof armature reaction [See Fig 120 (ii)]

Fig 120 Characteristics of series generator

(iii) External characteristic

Curve 3 shows the external characteristic of a series generator It gives therelation between terminal voltage and load current ILV = E g- Ia (Ra +Rse )Therefore external characteristic curve will lie below internal characteristiccurve by an amount equal to ohmic drop [ie Ia(Ra + Rse)] in the machine asshown in Fig 120 (ii) The internal and external characteristics ofa dc series generator can be plotted from one another as shown in Fig 121 Supposewe are given the internal characteristic of the generator Let the line OC represent the resistance of the whole machine ie Ra +Rse If the load current is OB drop in the machine is AB ieAB = Ohmic drop in the machine = OB(Ra + Rse)

Fig 121 Load characteristics of series generator

Now raise a perpendicular from point B and mark a point b on this line such that ab = AB Then point b will lie on the external characteristic of the generator

Following similar procedure other points of external characteristic can be located It is easy to see that we can also plot internal characteristic from the external characteristic

121 Characteristics of a Shunt Generator Fig 122 (i) show the connections of a shunt wound generator The armature current Ia splits up into two parts a small fraction Ish flowing through shunt field winding while the major part IL goes to the external load(i) OCC The OCC of a shunt generator is similar in shape to that of a series generator as shown in Fig 122 (ii) The line OA represents the shunt field circuitresistance When the generator is run at normal speed it will build up a voltageOM At no-load the terminal voltage of the generator will be constant (= OM)represented by the horizontal dotted line MC

Fig 122 Characteristics of a Shunt Generator

(ii) Internal characteristic

When the generator is loaded flux per pole is reduced due to armature reaction Therefore emf E generated on load is less than the emf generated at no load As a result the internal characteristic (EIa) drops down slightly as shown in Fig122 (ii)

(iii) External characteristic Curve 2 shows the external characteristic of a shunt generator It gives therelation between terminal voltage V and load current IL

Therefore external characteristic curve will lie below the internal characteristic curve by an amount equal to drop in the armature circuit[ie (IL + Ish)Ra] as shown in Fig 122 (ii)

Note It may be seen from the external characteristic that change in terminalvoltage from no-load to full load is small The terminal voltage can always bemaintained constant by adjusting the field rheostat R automatically122 Critical External Resistance for Shunt Generator

If the load resistance across the terminals of a shunt generator is decreased then load current increase However there is a limit to the increase in

load current with the decrease of load resistance Any decrease of load resistance beyond this point instead of increasing the current ultimately results in reduced current Consequently the external characteristic turns back (dotted curve) as shown in Fig 123 The tangent OA to the curve represents the minimum external resistance required to excite the shunt generator on load and is called critical external resistance If the resistance of the external circuit is less than the critical external resistance (represented by tangent OA in Fig 310) the machine will refuse to excite or will de-excite if already running This means that external resistance is so low as virtually to short circuit the machine and so doing away with its excitation

Note There are two critical resistances for a shunt generator viz (i) criticalfield resistance (ii) critical external resistance For the shunt generator to buildup voltage the former should not be exceeded and the latter must not be gonebelow

123 How to Draw OCC at Different Speeds

If we are given OCC of a generator at a constant speed N1 then we can easily draw the OCC at any other constant speed N2 Fig 123 illustrates theprocedure Here we are given OCC at a constant speed N1 It is desired to findthe OCC at constant speed N2 (it is assumed that n1 lt N2) For constantexcitation E propN

E2

E1 =

N 2

N1 so

E2 iquest E1

N2

N1

As shown in Fig (311) for If = OH E1 = HC Therefore the new value of emf (E2) for the same If but at N2

E2= HCN 2

N 1 = HD

Fig 123 Plotting OCC at another speed

This locates the point D on the new OCC at N2 Similarly other points can belocated taking different values of If The locus of these points will be the OCCat N2

124 Critical Speed (NC)

The critical speed of a shunt generator is the minimum speed below which it fails to excite Clearly it is the speed for which the given shunt field resistance represents the critical resistance In Fig 124 curve 2 corresponds to critical

speed because the shunt field resistance (Rsh) line is tangential to it If the generator runs at full speed N the new OCC moves upward and the Rsh line represents critical resistance for this speed

In order to find critical speed take any convenient point C on excitation axis and erect a perpendicular so as to cut Rsh and Rsh lines at points B and A respectively Then

BCAC

=NcN

Nc=N BCAC

Fig 124 Finding critical speed

125 Conditions for Voltage Build-Up of a Shunt Generator

The necessary conditions for voltage build-up in a shunt generator are

(i) There must be some residual magnetism in generator poles(ii) The connections of the field winding should be such that the field current strengthens the residual magnetism(iii) The resistance of the field circuit should be less than the critical resistance In other words the speed of the generator should be higher

than the critical speed

126 Compound Generator Characteristics

In a compound generator both series and shunt excitation are combined as shown in Fig 123 The shunt winding can be connected either across the armature only (short-shunt connection S) or across armature plus series field (long-shunt connection G) The compound generator can be cumulatively compounded or

differentially compounded generator The latter is rarely used in practice Therefore we shall discuss the characteristics of cumulatively compounded

generator It may be noted that external characteristics of long and short shunt compound generators are almost identical

External characteristic

Fig 124 shows the external characteristics of a cumulatively compoundedgenerator The series excitation aids the shunt excitation The degree of compounding depends upon the increase in series excitation with the increase inload current

(i) If series winding turns are so adjusted that with the increase in load current the terminal voltage increases it is called over-compounded generator In such a case as the load current increases the series field mmf increases and tends to increase the flux and hence the generated voltage The increase in generated voltage is greater than the IaRa drop so that instead of decreasing the terminal voltage increases as shown by curve A in Fig 124

(ii) If series winding turns are so adjusted that with the increase in load current the terminal voltage substantially remains constant it is called Flat-compounded generator The series winding of such a machine has lesser number of turns than the one in over-compounded machine and therefore does not increase the flux as much for a given load current Consequently the full-load voltage is nearly equal to the no-load voltage as indicated by curve B in Fig 124 (iii) If series field winding has lesser number of turns than for a flat compounded machine the terminal voltage falls with increase in load current as indicated by curve C m Fig 124 Such a machine is called under-compounded generator

Fig 123 compound generator

Fig 124 Load characteristics of compound generators

127 LOSSES

The losses in a dc machine (generator or motor) may be divided into three classes viz (i) copper losses or electrical losses (ii) iron or core losses and (iii) mechanical losses All these losses appear as heat and thus raise the temperature of the machine They also lower the efficiency of the machine The loss structure maybe demonstrated as

1Copperlosses

These losses occur due to currents in the various windings of the machine When an electric current of I amps flows in a resistance of R ohms the heat energy is lost at the rate of I2R joules sec and the power loss is I2R watts Generators and motors have one or more field circuits and an armature circuit in which such losses occur All resistance losses of this kind are called copper losses The different copper losses occurring in a dc machine are

(i) Armature copper loss = Ia2Ra

(ii) Shunt field copper loss = Ish2 Rsh

(iii) Series field copper loss = Ise2Rse

Armature copper loss will be around 30 to 40 of full load losses where as total field copper losses will constitute around 30 of full load losses

Note There is also brush contact loss due to brush contact resistance (ie resistance between the surface of brush and surface of commutator) This loss is generally included in armature copper loss The brush contact drop is almost independent of Ia For carbon brushes this drop is around 1 V per brush The power loss due to brush contact resistance is 2ebIa where eb is the voltage drop at one brush

2 IRONORCORELOSSES These losses occur in the armature of a dc machine and are due to the rotation of armature in the magnetic field of the poles They are of two types viz (i) hysteresis loss (ii) eddy current loss

(i) Hysteresis loss

Hysteresis loss occurs in the armature of the dc machine since any given part of the armature is subjected to magnetic field reversals as it passes under successive poles Fig 124 shows an armature rotating in two-pole machine Consider a small piece ab of the armature When the piece ab is under N-pole the magnetic lines pass from a to b Half a revolution later the same piece of iron is under S-pole and magnetic lines pass from b to a so that magnetism in the iron is reversed In order to reverse continuously the molecular magnets in the armature core some amount of power has to be spent which is called hysteresis loss It is given by Steinmetz formula This formula is

Hysteresis Loss Wh = η Bmax 16f V Watts Bmax - Maximum flux density in the core f - Frequency of magnetic reversals V - Volume of armature in m3

η - Steinmetz hysteresis coefficient

In order to reduce this loss in a dc machine armature core is made of such materials which have a low value of Steinmetz hysteresis co-efficient eg silicon steel

Fig 124 Occurrence of hysteresis loss

ii) EDDYCURRENTLOSS

In addition to the voltages induced in the armature conductors there are also voltages induced in the armature core These voltages produce circulating currents in the armature core as shown in Fig 125 These are called eddy currents and power loss due to their flow is called eddy current loss The eddy current loss appears as heat which raises the temperature of the machine and lowers its efficiency Core resistance can be greatly increased by constructing the core of thin round If a continuous solid iron core is used the resistance to eddy current path will be small due to large cross-sectional area of the core Consequently the magnitude of eddy current and hence eddy current loss will be large The magnitude of eddy current can be reduced by making core resistance as high as practical The iron sheets called laminations [See Fig 126] The laminations are insulated from each other with a coating of varnish The insulating coating has a high resistance so very little current flows from one lamination to the other Alsobecause each lamination is very thin the resistance to current flowing through the width of a lamination is also quite large Thus laminating a core increases the core resistance which decreases the eddy current and hence the eddy current loss

Fig 125 Eddy current loss Fig 126 Core laminations

Eddy current loss is biven by the formula Pe= Ke Bmax

2 f2t2 V Watts

Where Ke = Constant depending upon electrical resistace of core Bmax = Maximum flux density in wbm2

f = frequency f magnetic reversals in Hz t = thickness of laminations in m and V = volume of core in m3

It may be noted that eddy current loss depends upon the square of lamination thickness For this reason lamination thickness should be kept as small as possible

3MECHANICALLOSSES

These losses are due to friction and windage(i) Friction losses eg bearing friction brush friction etc(ii) Windage loss ie air friction of rotating armature

These losses depend upon the speed of the machine But for a given speed they are practically constant

Note Iron losses and mechanical losses together are called stray losses

CONSTANT AND VARIABLE LOSSES

The losses in a dc generator (or dc motor) may be sub-divided into

(i) constant losses (ii) variable losses

(i) Constant losses

Those losses in a dc generator which remain constant at all loads are known as constant losses The constant losses in a dc generator are(a) iron losses(b) mechanical losses(c) shunt field losses(ii) Variable lossesThose losses in a dc generator which vary with load are called variable lossesThe variable losses in a dc generator are (a) Copper loss in armature winding (Ia

2 Ra) (b) Copper loss in series field winding (Ise

2Rse)

Total losses = Constant losses + Variable lossesNote Field Cu loss is constant for shunt and compound generators

128 POWER FLOW DIAGRAM

The power flow in a dc generator may be represented as

The different power stages are shown as A B and CA ndash The mechanical power input which is the power output of the prime mover is given as Pin = 2πNT

60 watts

B - The power developed in armature is Pa = Eg Ia watts C- The electrical power output is Pout = VIL watts

129 EFFICIENCY Efficiency of a machine is the ratio of its output power to the input power

Efficiency η = Output powerInput power

= Input powerminusLossesInput power = Output power

Output power+Losses

In case of a dc generator depending upon the power stages three different efficiencies can be defined as-

1 Mechanical Efficiency(ηm)

BA = Electrical power developedisinarmature

Mechanical power input2 Electrical Efficiency(ηe)

CB = Electrical power output

Electrical power developedisinthe armature 3 Overall or commercial efficiency (η)

CA = Electrical power output

Mechanical power input = ηmtimes ηe

130 CONDITION FOR MAXIMUM EFFICIENCY The efficiency of a dc generator is not constant but varies with load Fig 127shows the variation of efficiency with load current Consider a dc shunt generator with terminal voltage V delivering a load current of IL amp Generator output = V IL

Generator input = Output + Losses = V IL + constant losses + Variable losses = V IL + Wc+ Ia

2Ra

= V IL + Wc + (IL+ Ish)2 Rsh

The shunt field current Ish is small compared with IL so may be neglected

Generator input = VIL+ IL2Ra + Wc

Now efficiency η iquest outputinput = VI L

VIL+ Ia 2Ra+Wc

Efficiency is maximum when the denominator of the above equation is minimumie

ddIL ( VI L

VIL+ Ia 2Ra+Wc ) = 0

Which gives IL2 Ra = WC

ie Variable loss = constant lossThe load current for maximum efficiency is given by

IL iquest radicWcRa

Fig 127 Variation of efficiency with load

Hence the efficiency is maximum for that load current at which the variable loss is equal to the constant loss

131 APPLICATIONS

Separately excited generators

These types of generators have special applications such as electroplating electro refining of materials because separate supply is required for field excitation

Shunt generators

These types of generators are used in battery charging and ordinary lighting purposes

Series generators

This type of generators is used commonly as boosters on DC feeders as current generators for welding and arc lamp

Cumulatively compounded generators

This type of generators is used for domestic lighting purposes and to transmit energy over long distances

Differential compound generators

This type of generators is rarely used They are used for special applications such as electric arc welding

132 NUMERICAL EXCERCISES

1 The emf induced in the armature of a 450 kW 250 volt shunt generator is

2588volt when the field current is 200 amp and the generator is supplying power to a load at rated terminal voltage The armature circuit resistance is 0005 ohm Determine (i) load current (ii) power generated (iii) power output and (iv) electrical efficiency Neglect brush contact drop

Solution

Let the load current be I amp Given Ish = 20 A

Armature current = Load current + field current or Ia = I + 20 Induced emf = Terminal voltage + Ia Ra drop

2588 = 250 + (I + 20) times 0005

So (i) Load current I = 1740 amp

(ii) Power generated E Ia = 2588 (1740 + 20) watts = 455488 kW (iii) Power output V I = 250 times 1740 watts

= 435 kW

(iv) Electrical efficiency = 435455488 times 100 = 955

2 A 4 pole wave wound dc shunt generator delivers a load of 45 kW at a terminal voltage of 220 volts Its armature has 150 single turn coils and has a resistance of 001 ohm The air gap flux per pole is 002 weber Shunt field resistance is 50 ohm Calculate the speed at which it is being driven Neglect brush contact drop

Solution Load current I = W

V = 45000220 = 20455 amp

Field Current Ish = V

R s h = 220

50 = 44 amp

Armature current Ia = I +I sh = 20455 +44 = 20895 amp

Armature drop I aRa = 20895 X 001 = 20895 volts Induced emf Eg = V + Ia Ra = 220 + 20895 = 2220895 volts

But induced emf

So N = Eg60 AΦ ZP = 2220895times 60 times2

002 times150 times2 times 4 = 111045 rpm

3 The following data refer to the OCC of a DC separately excited generator at 1000 rpm

Field current in A 00 02 03 04 05 06 07 08 09 10 11 12Armature voltage in V 5 40 75 100 124 145 162 178 188 195 200 205

The machine is now connected as a shunt generator with a total field resistance of 200 Ω and driven at 1000 rpm Estimate graphically (i) the voltage to which the generator will build up at no load (ii) the armature field and load currents when the terminal voltage is found to be 150 V Neglect the effect of armature reaction and brush drop and assume armature resistance Ra to be 08 Ω (iii) Finally estimate the steady state armature current when the machine terminals are shorted

Solution

First of all the OCC of the generator is plotted in a graph paper as shown in figure

(i) The total field circuit resistance is given to be Rf = 200 Ω The Rf line is now drawn on the same graph paper passing through the origin The point of intersection of the Rf line and the OCC decides the final no load voltage and can be read from the graph as 192 V and the corresponding field current is 096 A (ii) Since the terminal voltage V is 150 V (= BM) field current If

is OM = 077 A Generated voltage E is given by AM But we know E = IaRa + V Hence IaRa= E ndash V = AM ndash BM = AB Now from the graph AB = 25 V there4Ia = 2508 = 3125 A So load current IL= Ia ndash If= 3125 ndash 077 = 3048 A

ADDITIONAL PROBLEMS

1 A 4 pole lap wound 115 kW 230 volts dc shunt generator has armatureresistance of 02 ohm and field resistance of 100 ohms Calculate (i) emf

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES

Fig 113 OCC of dc generator

The following points may be noted from OCC When the field current is zero there is some generated emf OA

This is due to the residual magnetism in the field poles Over a fairly wide range of field current (upto point B in the curve)

the curve is linear It is because in this range reluctance of iron is negligible as compared with that of air gap The air gap reluctance is constant and hence linear relationship After point B on the curve the reluctance of iron also comes into picture and reluctance of iron is no longer negligible Consequently the curve deviates from linear relationship

After point C on the curve the magnetic saturation of poles begins and E0 tends to level off

The OCC of even self-excited generator is obtained by running it as a separately excited generator

165 Characteristics of a Separately Excited DCGenerator

The obvious disadvantage of a separately excited dc generator is that we require an external dc source for excitation But since the output voltage may be

controlled more easily and over a wide range (from zero to a maximum) this type of excitation finds many applications

(i) Open circuit characteristic

The OCC of a separately excited generator is determined in a manner described in Sec 164 Fig114 shows the variation of generated em f on noload with field current for various fixed speeds Note that if the value of constant speed is increased the steepness of the curve also increases When the field current is zero the residual magnetism in the poles will give rise to the small initial emf as shown

Fig114 Variation of OCC with speed

Internal and External Characteristics (Load Characteristics)

The external characteristic of a separately excited generator is the curve between the terminal voltage (V) and the load current IL (which is the same as armature current in this case) In order to determine the external characteristic the circuit set up is as shown in Fig 115

Fig 115 Experimental setup to plot load characteristics

As the load current increases the terminal voltage falls due to two reasons(a) The armature reaction weakens the main flux so that actual emf generated E on load is less than that generated (E0) on no load

(b) There is voltage drop across armature resistance (= ILRa = IaRa)Due to these reasons the external characteristic is a drooping curve [curve 3 inFig 116 ] Note that in the absence of armature reaction and armature dropthe generated emf would have been E0 (curve 1)The internal characteristic can be determined from external characteristic by adding IL Ra drop to the external characteristic It is because armature reaction drop is included in the external characteristic Curve 2 is the internal characteristic of the generator and should obviously lie above the external characteristic

Fig 116 Load characteristics of a separately excited dcgenerator

117 Voltage Build-Up in a Self-Excited GeneratorLet us see how voltage builds up in a self-excited generator(i) Shunt generator Consider a shunt generator If the generator is run at a constant speed some emf will be generated due to residual magnetism in the main poles This small emf circulates a field current which in turn produces additional flux toreinforce the original residual flux (provided field winding connections arecorrect) This process continues and the generator builds up the normalgenerated voltage following the OCC shown in Fig 117 (i) The field resistance Rf can be represented by a straight line passing through the origin as shown in Fig 117 (ii) The two curves can be shown on the same diagram as they have the same ordinate as in Fig 117 (iii)

Fig 117

Since the field circuit is inductive there is a delay in the increase in current uponclosing the field circuit switch The rate at which the current increases dependsupon the voltage available for increasing it Suppose at any instant the fieldcurrent is i (= OA) and is increasing at the rate didt Then

E0=i R f +iquest L didt

Where Rf = total field circuit resistance L = inductance of field circuit

At the considered instant the total emf available is AC [See Fig 117 (iii)] Anamount AB of the cmf AC is absorbed by the voltage drop iRf and theremainder part BC is available to overcome L didt Since this surplus voltage isavailable it is possible for the field current to increase above the value OAHowever at point D the available voltage is OM and is all absorbed by i RfD

rop Consequently the field current cannot increase further and the generatorbuild up stops Hence we arrive at a very important conclusion that the voltage build up of the generator is given by the point of intersection of OCC and field resistance line Thus in Fig 117 (iii) D is point of intersection of the two curves Hence the generator will build up a voltage OM(ii) Series generator During initial operation with no current yet flowing a residual voltage will be generated exactly as in the case of a shunt generator The residual voltage will cause a current to flow through the whole series circuit when the circuit isclosed There will then be voltage build up to an equilibrium point exactlyanalogous to the buildup of a shunt generator The voltage build up graph willbe similar to that of shunt generator except that now load current (instead offield current for shunt generator) will be taken along x-axis10487871048787(iii) Compound generator When a compound generator has its series field flux aiding its shunt field flux the machine is said to be cumulative compound When the series field isconnected in reverse so that its field flux opposes the shunt field flux thegenerator is then differential compound The easiest way to build up voltage in a compound generator is to start under no load conditions At no load only the shunt field is effective When no-load voltage build up is achieved the generator is loaded If under load the voltagerises the series field connection is cumulative If the voltage drops significantlythe connection is differential compound118 Critical Field Resistance for a Shunt Generator The voltage build up in a shunt generator depends upon field circuit resistance If the field circuit resistance is R1 (line OA) then generator will build up a voltage OM as shown in Fig 118 If the field circuit resistance is increasedto R2 (tine OB) the generator will build up a voltage OL slightly less than OM As the field circuit resistance is increased the slope of resistance line also increases When the field resistance line becomes tangent (line OC) to OCC the generator would just excite If the field circuit resistance is increased beyond this point (say line OD) the generator will fail to excite The field circuit resistance represented by line OC (tangent to OCC) is called critical field resistance RC for the shunt generator It may be defined as under The maximum field circuit resistance (for a given speed) with which the shunt generator would just excite is known as its critical field resistance It should be noted that shunt generator will build up voltage only if field circuit resistance is less than critical field resistance

Fig 118 Critical field resistance of shunt generator

119 Critical Resistance for a Series Generator

Fig 119 shows the voltage build up in a series generator Here R1 R2 etc represent the total circuit resistance (load resistance and field winding resistance) If the total circuit resistance is R1 then series generator will build up a voltage OL The line OC is tangent to OCC and represents the critical resistance RC for a series generator If the total resistance of the circuit is more than RC (say line OD) the generator will fail to build up voltage Note that Fig 119 is similar to Fig 118 with the following differences

(i) In Fig 118 R1 R2 etc represent the total field circuit resistanceHowever R1 R2 etc in Fig 119 represent the total circuit resistance(load resistance and series field winding resistance etc)

(ii) In Fig 118 field current alone is represented along X-axis However inFig 119 load current IL is represented along Y-axis Note that in aseries generator field current = load current IL

Fig 119 Critical resistance of a series generator

120 Characteristics of Series Generator Fig 120 (i) shows the connections of a series wound generator Since there is only one current (that which flows through the whole machine) the load current is the same as the exciting current(i) OCCCurve 1 shows the open circuit characteristic (OCC) of a series generator Itcan be obtained experimentally by disconnecting the field winding from themachine and exciting it from a separate dc source(ii) Internal characteristic Curve 2 shows the total or internal characteristic of a series generator It gives the relation between the generated emf E on load and armature current Due to armature reaction the flux in the machine will be less than the flux at no load Hence emf E generated under load conditions will be less than the emf E0 generated under no load conditions Consequently internal characteristic curvelies below the OCC curve the difference between them representing the effectof armature reaction [See Fig 120 (ii)]

Fig 120 Characteristics of series generator

(iii) External characteristic

Curve 3 shows the external characteristic of a series generator It gives therelation between terminal voltage and load current ILV = E g- Ia (Ra +Rse )Therefore external characteristic curve will lie below internal characteristiccurve by an amount equal to ohmic drop [ie Ia(Ra + Rse)] in the machine asshown in Fig 120 (ii) The internal and external characteristics ofa dc series generator can be plotted from one another as shown in Fig 121 Supposewe are given the internal characteristic of the generator Let the line OC represent the resistance of the whole machine ie Ra +Rse If the load current is OB drop in the machine is AB ieAB = Ohmic drop in the machine = OB(Ra + Rse)

Fig 121 Load characteristics of series generator

Now raise a perpendicular from point B and mark a point b on this line such that ab = AB Then point b will lie on the external characteristic of the generator

Following similar procedure other points of external characteristic can be located It is easy to see that we can also plot internal characteristic from the external characteristic

121 Characteristics of a Shunt Generator Fig 122 (i) show the connections of a shunt wound generator The armature current Ia splits up into two parts a small fraction Ish flowing through shunt field winding while the major part IL goes to the external load(i) OCC The OCC of a shunt generator is similar in shape to that of a series generator as shown in Fig 122 (ii) The line OA represents the shunt field circuitresistance When the generator is run at normal speed it will build up a voltageOM At no-load the terminal voltage of the generator will be constant (= OM)represented by the horizontal dotted line MC

Fig 122 Characteristics of a Shunt Generator

(ii) Internal characteristic

When the generator is loaded flux per pole is reduced due to armature reaction Therefore emf E generated on load is less than the emf generated at no load As a result the internal characteristic (EIa) drops down slightly as shown in Fig122 (ii)

(iii) External characteristic Curve 2 shows the external characteristic of a shunt generator It gives therelation between terminal voltage V and load current IL

Therefore external characteristic curve will lie below the internal characteristic curve by an amount equal to drop in the armature circuit[ie (IL + Ish)Ra] as shown in Fig 122 (ii)

Note It may be seen from the external characteristic that change in terminalvoltage from no-load to full load is small The terminal voltage can always bemaintained constant by adjusting the field rheostat R automatically122 Critical External Resistance for Shunt Generator

If the load resistance across the terminals of a shunt generator is decreased then load current increase However there is a limit to the increase in

load current with the decrease of load resistance Any decrease of load resistance beyond this point instead of increasing the current ultimately results in reduced current Consequently the external characteristic turns back (dotted curve) as shown in Fig 123 The tangent OA to the curve represents the minimum external resistance required to excite the shunt generator on load and is called critical external resistance If the resistance of the external circuit is less than the critical external resistance (represented by tangent OA in Fig 310) the machine will refuse to excite or will de-excite if already running This means that external resistance is so low as virtually to short circuit the machine and so doing away with its excitation

Note There are two critical resistances for a shunt generator viz (i) criticalfield resistance (ii) critical external resistance For the shunt generator to buildup voltage the former should not be exceeded and the latter must not be gonebelow

123 How to Draw OCC at Different Speeds

If we are given OCC of a generator at a constant speed N1 then we can easily draw the OCC at any other constant speed N2 Fig 123 illustrates theprocedure Here we are given OCC at a constant speed N1 It is desired to findthe OCC at constant speed N2 (it is assumed that n1 lt N2) For constantexcitation E propN

E2

E1 =

N 2

N1 so

E2 iquest E1

N2

N1

As shown in Fig (311) for If = OH E1 = HC Therefore the new value of emf (E2) for the same If but at N2

E2= HCN 2

N 1 = HD

Fig 123 Plotting OCC at another speed

This locates the point D on the new OCC at N2 Similarly other points can belocated taking different values of If The locus of these points will be the OCCat N2

124 Critical Speed (NC)

The critical speed of a shunt generator is the minimum speed below which it fails to excite Clearly it is the speed for which the given shunt field resistance represents the critical resistance In Fig 124 curve 2 corresponds to critical

speed because the shunt field resistance (Rsh) line is tangential to it If the generator runs at full speed N the new OCC moves upward and the Rsh line represents critical resistance for this speed

In order to find critical speed take any convenient point C on excitation axis and erect a perpendicular so as to cut Rsh and Rsh lines at points B and A respectively Then

BCAC

=NcN

Nc=N BCAC

Fig 124 Finding critical speed

125 Conditions for Voltage Build-Up of a Shunt Generator

The necessary conditions for voltage build-up in a shunt generator are

(i) There must be some residual magnetism in generator poles(ii) The connections of the field winding should be such that the field current strengthens the residual magnetism(iii) The resistance of the field circuit should be less than the critical resistance In other words the speed of the generator should be higher

than the critical speed

126 Compound Generator Characteristics

In a compound generator both series and shunt excitation are combined as shown in Fig 123 The shunt winding can be connected either across the armature only (short-shunt connection S) or across armature plus series field (long-shunt connection G) The compound generator can be cumulatively compounded or

differentially compounded generator The latter is rarely used in practice Therefore we shall discuss the characteristics of cumulatively compounded

generator It may be noted that external characteristics of long and short shunt compound generators are almost identical

External characteristic

Fig 124 shows the external characteristics of a cumulatively compoundedgenerator The series excitation aids the shunt excitation The degree of compounding depends upon the increase in series excitation with the increase inload current

(i) If series winding turns are so adjusted that with the increase in load current the terminal voltage increases it is called over-compounded generator In such a case as the load current increases the series field mmf increases and tends to increase the flux and hence the generated voltage The increase in generated voltage is greater than the IaRa drop so that instead of decreasing the terminal voltage increases as shown by curve A in Fig 124

(ii) If series winding turns are so adjusted that with the increase in load current the terminal voltage substantially remains constant it is called Flat-compounded generator The series winding of such a machine has lesser number of turns than the one in over-compounded machine and therefore does not increase the flux as much for a given load current Consequently the full-load voltage is nearly equal to the no-load voltage as indicated by curve B in Fig 124 (iii) If series field winding has lesser number of turns than for a flat compounded machine the terminal voltage falls with increase in load current as indicated by curve C m Fig 124 Such a machine is called under-compounded generator

Fig 123 compound generator

Fig 124 Load characteristics of compound generators

127 LOSSES

The losses in a dc machine (generator or motor) may be divided into three classes viz (i) copper losses or electrical losses (ii) iron or core losses and (iii) mechanical losses All these losses appear as heat and thus raise the temperature of the machine They also lower the efficiency of the machine The loss structure maybe demonstrated as

1Copperlosses

These losses occur due to currents in the various windings of the machine When an electric current of I amps flows in a resistance of R ohms the heat energy is lost at the rate of I2R joules sec and the power loss is I2R watts Generators and motors have one or more field circuits and an armature circuit in which such losses occur All resistance losses of this kind are called copper losses The different copper losses occurring in a dc machine are

(i) Armature copper loss = Ia2Ra

(ii) Shunt field copper loss = Ish2 Rsh

(iii) Series field copper loss = Ise2Rse

Armature copper loss will be around 30 to 40 of full load losses where as total field copper losses will constitute around 30 of full load losses

Note There is also brush contact loss due to brush contact resistance (ie resistance between the surface of brush and surface of commutator) This loss is generally included in armature copper loss The brush contact drop is almost independent of Ia For carbon brushes this drop is around 1 V per brush The power loss due to brush contact resistance is 2ebIa where eb is the voltage drop at one brush

2 IRONORCORELOSSES These losses occur in the armature of a dc machine and are due to the rotation of armature in the magnetic field of the poles They are of two types viz (i) hysteresis loss (ii) eddy current loss

(i) Hysteresis loss

Hysteresis loss occurs in the armature of the dc machine since any given part of the armature is subjected to magnetic field reversals as it passes under successive poles Fig 124 shows an armature rotating in two-pole machine Consider a small piece ab of the armature When the piece ab is under N-pole the magnetic lines pass from a to b Half a revolution later the same piece of iron is under S-pole and magnetic lines pass from b to a so that magnetism in the iron is reversed In order to reverse continuously the molecular magnets in the armature core some amount of power has to be spent which is called hysteresis loss It is given by Steinmetz formula This formula is

Hysteresis Loss Wh = η Bmax 16f V Watts Bmax - Maximum flux density in the core f - Frequency of magnetic reversals V - Volume of armature in m3

η - Steinmetz hysteresis coefficient

In order to reduce this loss in a dc machine armature core is made of such materials which have a low value of Steinmetz hysteresis co-efficient eg silicon steel

Fig 124 Occurrence of hysteresis loss

ii) EDDYCURRENTLOSS

In addition to the voltages induced in the armature conductors there are also voltages induced in the armature core These voltages produce circulating currents in the armature core as shown in Fig 125 These are called eddy currents and power loss due to their flow is called eddy current loss The eddy current loss appears as heat which raises the temperature of the machine and lowers its efficiency Core resistance can be greatly increased by constructing the core of thin round If a continuous solid iron core is used the resistance to eddy current path will be small due to large cross-sectional area of the core Consequently the magnitude of eddy current and hence eddy current loss will be large The magnitude of eddy current can be reduced by making core resistance as high as practical The iron sheets called laminations [See Fig 126] The laminations are insulated from each other with a coating of varnish The insulating coating has a high resistance so very little current flows from one lamination to the other Alsobecause each lamination is very thin the resistance to current flowing through the width of a lamination is also quite large Thus laminating a core increases the core resistance which decreases the eddy current and hence the eddy current loss

Fig 125 Eddy current loss Fig 126 Core laminations

Eddy current loss is biven by the formula Pe= Ke Bmax

2 f2t2 V Watts

Where Ke = Constant depending upon electrical resistace of core Bmax = Maximum flux density in wbm2

f = frequency f magnetic reversals in Hz t = thickness of laminations in m and V = volume of core in m3

It may be noted that eddy current loss depends upon the square of lamination thickness For this reason lamination thickness should be kept as small as possible

3MECHANICALLOSSES

These losses are due to friction and windage(i) Friction losses eg bearing friction brush friction etc(ii) Windage loss ie air friction of rotating armature

These losses depend upon the speed of the machine But for a given speed they are practically constant

Note Iron losses and mechanical losses together are called stray losses

CONSTANT AND VARIABLE LOSSES

The losses in a dc generator (or dc motor) may be sub-divided into

(i) constant losses (ii) variable losses

(i) Constant losses

Those losses in a dc generator which remain constant at all loads are known as constant losses The constant losses in a dc generator are(a) iron losses(b) mechanical losses(c) shunt field losses(ii) Variable lossesThose losses in a dc generator which vary with load are called variable lossesThe variable losses in a dc generator are (a) Copper loss in armature winding (Ia

2 Ra) (b) Copper loss in series field winding (Ise

2Rse)

Total losses = Constant losses + Variable lossesNote Field Cu loss is constant for shunt and compound generators

128 POWER FLOW DIAGRAM

The power flow in a dc generator may be represented as

The different power stages are shown as A B and CA ndash The mechanical power input which is the power output of the prime mover is given as Pin = 2πNT

60 watts

B - The power developed in armature is Pa = Eg Ia watts C- The electrical power output is Pout = VIL watts

129 EFFICIENCY Efficiency of a machine is the ratio of its output power to the input power

Efficiency η = Output powerInput power

= Input powerminusLossesInput power = Output power

Output power+Losses

In case of a dc generator depending upon the power stages three different efficiencies can be defined as-

1 Mechanical Efficiency(ηm)

BA = Electrical power developedisinarmature

Mechanical power input2 Electrical Efficiency(ηe)

CB = Electrical power output

Electrical power developedisinthe armature 3 Overall or commercial efficiency (η)

CA = Electrical power output

Mechanical power input = ηmtimes ηe

130 CONDITION FOR MAXIMUM EFFICIENCY The efficiency of a dc generator is not constant but varies with load Fig 127shows the variation of efficiency with load current Consider a dc shunt generator with terminal voltage V delivering a load current of IL amp Generator output = V IL

Generator input = Output + Losses = V IL + constant losses + Variable losses = V IL + Wc+ Ia

2Ra

= V IL + Wc + (IL+ Ish)2 Rsh

The shunt field current Ish is small compared with IL so may be neglected

Generator input = VIL+ IL2Ra + Wc

Now efficiency η iquest outputinput = VI L

VIL+ Ia 2Ra+Wc

Efficiency is maximum when the denominator of the above equation is minimumie

ddIL ( VI L

VIL+ Ia 2Ra+Wc ) = 0

Which gives IL2 Ra = WC

ie Variable loss = constant lossThe load current for maximum efficiency is given by

IL iquest radicWcRa

Fig 127 Variation of efficiency with load

Hence the efficiency is maximum for that load current at which the variable loss is equal to the constant loss

131 APPLICATIONS

Separately excited generators

These types of generators have special applications such as electroplating electro refining of materials because separate supply is required for field excitation

Shunt generators

These types of generators are used in battery charging and ordinary lighting purposes

Series generators

This type of generators is used commonly as boosters on DC feeders as current generators for welding and arc lamp

Cumulatively compounded generators

This type of generators is used for domestic lighting purposes and to transmit energy over long distances

Differential compound generators

This type of generators is rarely used They are used for special applications such as electric arc welding

132 NUMERICAL EXCERCISES

1 The emf induced in the armature of a 450 kW 250 volt shunt generator is

2588volt when the field current is 200 amp and the generator is supplying power to a load at rated terminal voltage The armature circuit resistance is 0005 ohm Determine (i) load current (ii) power generated (iii) power output and (iv) electrical efficiency Neglect brush contact drop

Solution

Let the load current be I amp Given Ish = 20 A

Armature current = Load current + field current or Ia = I + 20 Induced emf = Terminal voltage + Ia Ra drop

2588 = 250 + (I + 20) times 0005

So (i) Load current I = 1740 amp

(ii) Power generated E Ia = 2588 (1740 + 20) watts = 455488 kW (iii) Power output V I = 250 times 1740 watts

= 435 kW

(iv) Electrical efficiency = 435455488 times 100 = 955

2 A 4 pole wave wound dc shunt generator delivers a load of 45 kW at a terminal voltage of 220 volts Its armature has 150 single turn coils and has a resistance of 001 ohm The air gap flux per pole is 002 weber Shunt field resistance is 50 ohm Calculate the speed at which it is being driven Neglect brush contact drop

Solution Load current I = W

V = 45000220 = 20455 amp

Field Current Ish = V

R s h = 220

50 = 44 amp

Armature current Ia = I +I sh = 20455 +44 = 20895 amp

Armature drop I aRa = 20895 X 001 = 20895 volts Induced emf Eg = V + Ia Ra = 220 + 20895 = 2220895 volts

But induced emf

So N = Eg60 AΦ ZP = 2220895times 60 times2

002 times150 times2 times 4 = 111045 rpm

3 The following data refer to the OCC of a DC separately excited generator at 1000 rpm

Field current in A 00 02 03 04 05 06 07 08 09 10 11 12Armature voltage in V 5 40 75 100 124 145 162 178 188 195 200 205

The machine is now connected as a shunt generator with a total field resistance of 200 Ω and driven at 1000 rpm Estimate graphically (i) the voltage to which the generator will build up at no load (ii) the armature field and load currents when the terminal voltage is found to be 150 V Neglect the effect of armature reaction and brush drop and assume armature resistance Ra to be 08 Ω (iii) Finally estimate the steady state armature current when the machine terminals are shorted

Solution

First of all the OCC of the generator is plotted in a graph paper as shown in figure

(i) The total field circuit resistance is given to be Rf = 200 Ω The Rf line is now drawn on the same graph paper passing through the origin The point of intersection of the Rf line and the OCC decides the final no load voltage and can be read from the graph as 192 V and the corresponding field current is 096 A (ii) Since the terminal voltage V is 150 V (= BM) field current If

is OM = 077 A Generated voltage E is given by AM But we know E = IaRa + V Hence IaRa= E ndash V = AM ndash BM = AB Now from the graph AB = 25 V there4Ia = 2508 = 3125 A So load current IL= Ia ndash If= 3125 ndash 077 = 3048 A

ADDITIONAL PROBLEMS

1 A 4 pole lap wound 115 kW 230 volts dc shunt generator has armatureresistance of 02 ohm and field resistance of 100 ohms Calculate (i) emf

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES

controlled more easily and over a wide range (from zero to a maximum) this type of excitation finds many applications

(i) Open circuit characteristic

The OCC of a separately excited generator is determined in a manner described in Sec 164 Fig114 shows the variation of generated em f on noload with field current for various fixed speeds Note that if the value of constant speed is increased the steepness of the curve also increases When the field current is zero the residual magnetism in the poles will give rise to the small initial emf as shown

Fig114 Variation of OCC with speed

Internal and External Characteristics (Load Characteristics)

The external characteristic of a separately excited generator is the curve between the terminal voltage (V) and the load current IL (which is the same as armature current in this case) In order to determine the external characteristic the circuit set up is as shown in Fig 115

Fig 115 Experimental setup to plot load characteristics

As the load current increases the terminal voltage falls due to two reasons(a) The armature reaction weakens the main flux so that actual emf generated E on load is less than that generated (E0) on no load

(b) There is voltage drop across armature resistance (= ILRa = IaRa)Due to these reasons the external characteristic is a drooping curve [curve 3 inFig 116 ] Note that in the absence of armature reaction and armature dropthe generated emf would have been E0 (curve 1)The internal characteristic can be determined from external characteristic by adding IL Ra drop to the external characteristic It is because armature reaction drop is included in the external characteristic Curve 2 is the internal characteristic of the generator and should obviously lie above the external characteristic

Fig 116 Load characteristics of a separately excited dcgenerator

117 Voltage Build-Up in a Self-Excited GeneratorLet us see how voltage builds up in a self-excited generator(i) Shunt generator Consider a shunt generator If the generator is run at a constant speed some emf will be generated due to residual magnetism in the main poles This small emf circulates a field current which in turn produces additional flux toreinforce the original residual flux (provided field winding connections arecorrect) This process continues and the generator builds up the normalgenerated voltage following the OCC shown in Fig 117 (i) The field resistance Rf can be represented by a straight line passing through the origin as shown in Fig 117 (ii) The two curves can be shown on the same diagram as they have the same ordinate as in Fig 117 (iii)

Fig 117

Since the field circuit is inductive there is a delay in the increase in current uponclosing the field circuit switch The rate at which the current increases dependsupon the voltage available for increasing it Suppose at any instant the fieldcurrent is i (= OA) and is increasing at the rate didt Then

E0=i R f +iquest L didt

Where Rf = total field circuit resistance L = inductance of field circuit

At the considered instant the total emf available is AC [See Fig 117 (iii)] Anamount AB of the cmf AC is absorbed by the voltage drop iRf and theremainder part BC is available to overcome L didt Since this surplus voltage isavailable it is possible for the field current to increase above the value OAHowever at point D the available voltage is OM and is all absorbed by i RfD

rop Consequently the field current cannot increase further and the generatorbuild up stops Hence we arrive at a very important conclusion that the voltage build up of the generator is given by the point of intersection of OCC and field resistance line Thus in Fig 117 (iii) D is point of intersection of the two curves Hence the generator will build up a voltage OM(ii) Series generator During initial operation with no current yet flowing a residual voltage will be generated exactly as in the case of a shunt generator The residual voltage will cause a current to flow through the whole series circuit when the circuit isclosed There will then be voltage build up to an equilibrium point exactlyanalogous to the buildup of a shunt generator The voltage build up graph willbe similar to that of shunt generator except that now load current (instead offield current for shunt generator) will be taken along x-axis10487871048787(iii) Compound generator When a compound generator has its series field flux aiding its shunt field flux the machine is said to be cumulative compound When the series field isconnected in reverse so that its field flux opposes the shunt field flux thegenerator is then differential compound The easiest way to build up voltage in a compound generator is to start under no load conditions At no load only the shunt field is effective When no-load voltage build up is achieved the generator is loaded If under load the voltagerises the series field connection is cumulative If the voltage drops significantlythe connection is differential compound118 Critical Field Resistance for a Shunt Generator The voltage build up in a shunt generator depends upon field circuit resistance If the field circuit resistance is R1 (line OA) then generator will build up a voltage OM as shown in Fig 118 If the field circuit resistance is increasedto R2 (tine OB) the generator will build up a voltage OL slightly less than OM As the field circuit resistance is increased the slope of resistance line also increases When the field resistance line becomes tangent (line OC) to OCC the generator would just excite If the field circuit resistance is increased beyond this point (say line OD) the generator will fail to excite The field circuit resistance represented by line OC (tangent to OCC) is called critical field resistance RC for the shunt generator It may be defined as under The maximum field circuit resistance (for a given speed) with which the shunt generator would just excite is known as its critical field resistance It should be noted that shunt generator will build up voltage only if field circuit resistance is less than critical field resistance

Fig 118 Critical field resistance of shunt generator

119 Critical Resistance for a Series Generator

Fig 119 shows the voltage build up in a series generator Here R1 R2 etc represent the total circuit resistance (load resistance and field winding resistance) If the total circuit resistance is R1 then series generator will build up a voltage OL The line OC is tangent to OCC and represents the critical resistance RC for a series generator If the total resistance of the circuit is more than RC (say line OD) the generator will fail to build up voltage Note that Fig 119 is similar to Fig 118 with the following differences

(i) In Fig 118 R1 R2 etc represent the total field circuit resistanceHowever R1 R2 etc in Fig 119 represent the total circuit resistance(load resistance and series field winding resistance etc)

(ii) In Fig 118 field current alone is represented along X-axis However inFig 119 load current IL is represented along Y-axis Note that in aseries generator field current = load current IL

Fig 119 Critical resistance of a series generator

120 Characteristics of Series Generator Fig 120 (i) shows the connections of a series wound generator Since there is only one current (that which flows through the whole machine) the load current is the same as the exciting current(i) OCCCurve 1 shows the open circuit characteristic (OCC) of a series generator Itcan be obtained experimentally by disconnecting the field winding from themachine and exciting it from a separate dc source(ii) Internal characteristic Curve 2 shows the total or internal characteristic of a series generator It gives the relation between the generated emf E on load and armature current Due to armature reaction the flux in the machine will be less than the flux at no load Hence emf E generated under load conditions will be less than the emf E0 generated under no load conditions Consequently internal characteristic curvelies below the OCC curve the difference between them representing the effectof armature reaction [See Fig 120 (ii)]

Fig 120 Characteristics of series generator

(iii) External characteristic

Curve 3 shows the external characteristic of a series generator It gives therelation between terminal voltage and load current ILV = E g- Ia (Ra +Rse )Therefore external characteristic curve will lie below internal characteristiccurve by an amount equal to ohmic drop [ie Ia(Ra + Rse)] in the machine asshown in Fig 120 (ii) The internal and external characteristics ofa dc series generator can be plotted from one another as shown in Fig 121 Supposewe are given the internal characteristic of the generator Let the line OC represent the resistance of the whole machine ie Ra +Rse If the load current is OB drop in the machine is AB ieAB = Ohmic drop in the machine = OB(Ra + Rse)

Fig 121 Load characteristics of series generator

Now raise a perpendicular from point B and mark a point b on this line such that ab = AB Then point b will lie on the external characteristic of the generator

Following similar procedure other points of external characteristic can be located It is easy to see that we can also plot internal characteristic from the external characteristic

121 Characteristics of a Shunt Generator Fig 122 (i) show the connections of a shunt wound generator The armature current Ia splits up into two parts a small fraction Ish flowing through shunt field winding while the major part IL goes to the external load(i) OCC The OCC of a shunt generator is similar in shape to that of a series generator as shown in Fig 122 (ii) The line OA represents the shunt field circuitresistance When the generator is run at normal speed it will build up a voltageOM At no-load the terminal voltage of the generator will be constant (= OM)represented by the horizontal dotted line MC

Fig 122 Characteristics of a Shunt Generator

(ii) Internal characteristic

When the generator is loaded flux per pole is reduced due to armature reaction Therefore emf E generated on load is less than the emf generated at no load As a result the internal characteristic (EIa) drops down slightly as shown in Fig122 (ii)

(iii) External characteristic Curve 2 shows the external characteristic of a shunt generator It gives therelation between terminal voltage V and load current IL

Therefore external characteristic curve will lie below the internal characteristic curve by an amount equal to drop in the armature circuit[ie (IL + Ish)Ra] as shown in Fig 122 (ii)

Note It may be seen from the external characteristic that change in terminalvoltage from no-load to full load is small The terminal voltage can always bemaintained constant by adjusting the field rheostat R automatically122 Critical External Resistance for Shunt Generator

If the load resistance across the terminals of a shunt generator is decreased then load current increase However there is a limit to the increase in

load current with the decrease of load resistance Any decrease of load resistance beyond this point instead of increasing the current ultimately results in reduced current Consequently the external characteristic turns back (dotted curve) as shown in Fig 123 The tangent OA to the curve represents the minimum external resistance required to excite the shunt generator on load and is called critical external resistance If the resistance of the external circuit is less than the critical external resistance (represented by tangent OA in Fig 310) the machine will refuse to excite or will de-excite if already running This means that external resistance is so low as virtually to short circuit the machine and so doing away with its excitation

Note There are two critical resistances for a shunt generator viz (i) criticalfield resistance (ii) critical external resistance For the shunt generator to buildup voltage the former should not be exceeded and the latter must not be gonebelow

123 How to Draw OCC at Different Speeds

If we are given OCC of a generator at a constant speed N1 then we can easily draw the OCC at any other constant speed N2 Fig 123 illustrates theprocedure Here we are given OCC at a constant speed N1 It is desired to findthe OCC at constant speed N2 (it is assumed that n1 lt N2) For constantexcitation E propN

E2

E1 =

N 2

N1 so

E2 iquest E1

N2

N1

As shown in Fig (311) for If = OH E1 = HC Therefore the new value of emf (E2) for the same If but at N2

E2= HCN 2

N 1 = HD

Fig 123 Plotting OCC at another speed

This locates the point D on the new OCC at N2 Similarly other points can belocated taking different values of If The locus of these points will be the OCCat N2

124 Critical Speed (NC)

The critical speed of a shunt generator is the minimum speed below which it fails to excite Clearly it is the speed for which the given shunt field resistance represents the critical resistance In Fig 124 curve 2 corresponds to critical

speed because the shunt field resistance (Rsh) line is tangential to it If the generator runs at full speed N the new OCC moves upward and the Rsh line represents critical resistance for this speed

In order to find critical speed take any convenient point C on excitation axis and erect a perpendicular so as to cut Rsh and Rsh lines at points B and A respectively Then

BCAC

=NcN

Nc=N BCAC

Fig 124 Finding critical speed

125 Conditions for Voltage Build-Up of a Shunt Generator

The necessary conditions for voltage build-up in a shunt generator are

(i) There must be some residual magnetism in generator poles(ii) The connections of the field winding should be such that the field current strengthens the residual magnetism(iii) The resistance of the field circuit should be less than the critical resistance In other words the speed of the generator should be higher

than the critical speed

126 Compound Generator Characteristics

In a compound generator both series and shunt excitation are combined as shown in Fig 123 The shunt winding can be connected either across the armature only (short-shunt connection S) or across armature plus series field (long-shunt connection G) The compound generator can be cumulatively compounded or

differentially compounded generator The latter is rarely used in practice Therefore we shall discuss the characteristics of cumulatively compounded

generator It may be noted that external characteristics of long and short shunt compound generators are almost identical

External characteristic

Fig 124 shows the external characteristics of a cumulatively compoundedgenerator The series excitation aids the shunt excitation The degree of compounding depends upon the increase in series excitation with the increase inload current

(i) If series winding turns are so adjusted that with the increase in load current the terminal voltage increases it is called over-compounded generator In such a case as the load current increases the series field mmf increases and tends to increase the flux and hence the generated voltage The increase in generated voltage is greater than the IaRa drop so that instead of decreasing the terminal voltage increases as shown by curve A in Fig 124

(ii) If series winding turns are so adjusted that with the increase in load current the terminal voltage substantially remains constant it is called Flat-compounded generator The series winding of such a machine has lesser number of turns than the one in over-compounded machine and therefore does not increase the flux as much for a given load current Consequently the full-load voltage is nearly equal to the no-load voltage as indicated by curve B in Fig 124 (iii) If series field winding has lesser number of turns than for a flat compounded machine the terminal voltage falls with increase in load current as indicated by curve C m Fig 124 Such a machine is called under-compounded generator

Fig 123 compound generator

Fig 124 Load characteristics of compound generators

127 LOSSES

The losses in a dc machine (generator or motor) may be divided into three classes viz (i) copper losses or electrical losses (ii) iron or core losses and (iii) mechanical losses All these losses appear as heat and thus raise the temperature of the machine They also lower the efficiency of the machine The loss structure maybe demonstrated as

1Copperlosses

These losses occur due to currents in the various windings of the machine When an electric current of I amps flows in a resistance of R ohms the heat energy is lost at the rate of I2R joules sec and the power loss is I2R watts Generators and motors have one or more field circuits and an armature circuit in which such losses occur All resistance losses of this kind are called copper losses The different copper losses occurring in a dc machine are

(i) Armature copper loss = Ia2Ra

(ii) Shunt field copper loss = Ish2 Rsh

(iii) Series field copper loss = Ise2Rse

Armature copper loss will be around 30 to 40 of full load losses where as total field copper losses will constitute around 30 of full load losses

Note There is also brush contact loss due to brush contact resistance (ie resistance between the surface of brush and surface of commutator) This loss is generally included in armature copper loss The brush contact drop is almost independent of Ia For carbon brushes this drop is around 1 V per brush The power loss due to brush contact resistance is 2ebIa where eb is the voltage drop at one brush

2 IRONORCORELOSSES These losses occur in the armature of a dc machine and are due to the rotation of armature in the magnetic field of the poles They are of two types viz (i) hysteresis loss (ii) eddy current loss

(i) Hysteresis loss

Hysteresis loss occurs in the armature of the dc machine since any given part of the armature is subjected to magnetic field reversals as it passes under successive poles Fig 124 shows an armature rotating in two-pole machine Consider a small piece ab of the armature When the piece ab is under N-pole the magnetic lines pass from a to b Half a revolution later the same piece of iron is under S-pole and magnetic lines pass from b to a so that magnetism in the iron is reversed In order to reverse continuously the molecular magnets in the armature core some amount of power has to be spent which is called hysteresis loss It is given by Steinmetz formula This formula is

Hysteresis Loss Wh = η Bmax 16f V Watts Bmax - Maximum flux density in the core f - Frequency of magnetic reversals V - Volume of armature in m3

η - Steinmetz hysteresis coefficient

In order to reduce this loss in a dc machine armature core is made of such materials which have a low value of Steinmetz hysteresis co-efficient eg silicon steel

Fig 124 Occurrence of hysteresis loss

ii) EDDYCURRENTLOSS

In addition to the voltages induced in the armature conductors there are also voltages induced in the armature core These voltages produce circulating currents in the armature core as shown in Fig 125 These are called eddy currents and power loss due to their flow is called eddy current loss The eddy current loss appears as heat which raises the temperature of the machine and lowers its efficiency Core resistance can be greatly increased by constructing the core of thin round If a continuous solid iron core is used the resistance to eddy current path will be small due to large cross-sectional area of the core Consequently the magnitude of eddy current and hence eddy current loss will be large The magnitude of eddy current can be reduced by making core resistance as high as practical The iron sheets called laminations [See Fig 126] The laminations are insulated from each other with a coating of varnish The insulating coating has a high resistance so very little current flows from one lamination to the other Alsobecause each lamination is very thin the resistance to current flowing through the width of a lamination is also quite large Thus laminating a core increases the core resistance which decreases the eddy current and hence the eddy current loss

Fig 125 Eddy current loss Fig 126 Core laminations

Eddy current loss is biven by the formula Pe= Ke Bmax

2 f2t2 V Watts

Where Ke = Constant depending upon electrical resistace of core Bmax = Maximum flux density in wbm2

f = frequency f magnetic reversals in Hz t = thickness of laminations in m and V = volume of core in m3

It may be noted that eddy current loss depends upon the square of lamination thickness For this reason lamination thickness should be kept as small as possible

3MECHANICALLOSSES

These losses are due to friction and windage(i) Friction losses eg bearing friction brush friction etc(ii) Windage loss ie air friction of rotating armature

These losses depend upon the speed of the machine But for a given speed they are practically constant

Note Iron losses and mechanical losses together are called stray losses

CONSTANT AND VARIABLE LOSSES

The losses in a dc generator (or dc motor) may be sub-divided into

(i) constant losses (ii) variable losses

(i) Constant losses

Those losses in a dc generator which remain constant at all loads are known as constant losses The constant losses in a dc generator are(a) iron losses(b) mechanical losses(c) shunt field losses(ii) Variable lossesThose losses in a dc generator which vary with load are called variable lossesThe variable losses in a dc generator are (a) Copper loss in armature winding (Ia

2 Ra) (b) Copper loss in series field winding (Ise

2Rse)

Total losses = Constant losses + Variable lossesNote Field Cu loss is constant for shunt and compound generators

128 POWER FLOW DIAGRAM

The power flow in a dc generator may be represented as

The different power stages are shown as A B and CA ndash The mechanical power input which is the power output of the prime mover is given as Pin = 2πNT

60 watts

B - The power developed in armature is Pa = Eg Ia watts C- The electrical power output is Pout = VIL watts

129 EFFICIENCY Efficiency of a machine is the ratio of its output power to the input power

Efficiency η = Output powerInput power

= Input powerminusLossesInput power = Output power

Output power+Losses

In case of a dc generator depending upon the power stages three different efficiencies can be defined as-

1 Mechanical Efficiency(ηm)

BA = Electrical power developedisinarmature

Mechanical power input2 Electrical Efficiency(ηe)

CB = Electrical power output

Electrical power developedisinthe armature 3 Overall or commercial efficiency (η)

CA = Electrical power output

Mechanical power input = ηmtimes ηe

130 CONDITION FOR MAXIMUM EFFICIENCY The efficiency of a dc generator is not constant but varies with load Fig 127shows the variation of efficiency with load current Consider a dc shunt generator with terminal voltage V delivering a load current of IL amp Generator output = V IL

Generator input = Output + Losses = V IL + constant losses + Variable losses = V IL + Wc+ Ia

2Ra

= V IL + Wc + (IL+ Ish)2 Rsh

The shunt field current Ish is small compared with IL so may be neglected

Generator input = VIL+ IL2Ra + Wc

Now efficiency η iquest outputinput = VI L

VIL+ Ia 2Ra+Wc

Efficiency is maximum when the denominator of the above equation is minimumie

ddIL ( VI L

VIL+ Ia 2Ra+Wc ) = 0

Which gives IL2 Ra = WC

ie Variable loss = constant lossThe load current for maximum efficiency is given by

IL iquest radicWcRa

Fig 127 Variation of efficiency with load

Hence the efficiency is maximum for that load current at which the variable loss is equal to the constant loss

131 APPLICATIONS

Separately excited generators

These types of generators have special applications such as electroplating electro refining of materials because separate supply is required for field excitation

Shunt generators

These types of generators are used in battery charging and ordinary lighting purposes

Series generators

This type of generators is used commonly as boosters on DC feeders as current generators for welding and arc lamp

Cumulatively compounded generators

This type of generators is used for domestic lighting purposes and to transmit energy over long distances

Differential compound generators

This type of generators is rarely used They are used for special applications such as electric arc welding

132 NUMERICAL EXCERCISES

1 The emf induced in the armature of a 450 kW 250 volt shunt generator is

2588volt when the field current is 200 amp and the generator is supplying power to a load at rated terminal voltage The armature circuit resistance is 0005 ohm Determine (i) load current (ii) power generated (iii) power output and (iv) electrical efficiency Neglect brush contact drop

Solution

Let the load current be I amp Given Ish = 20 A

Armature current = Load current + field current or Ia = I + 20 Induced emf = Terminal voltage + Ia Ra drop

2588 = 250 + (I + 20) times 0005

So (i) Load current I = 1740 amp

(ii) Power generated E Ia = 2588 (1740 + 20) watts = 455488 kW (iii) Power output V I = 250 times 1740 watts

= 435 kW

(iv) Electrical efficiency = 435455488 times 100 = 955

2 A 4 pole wave wound dc shunt generator delivers a load of 45 kW at a terminal voltage of 220 volts Its armature has 150 single turn coils and has a resistance of 001 ohm The air gap flux per pole is 002 weber Shunt field resistance is 50 ohm Calculate the speed at which it is being driven Neglect brush contact drop

Solution Load current I = W

V = 45000220 = 20455 amp

Field Current Ish = V

R s h = 220

50 = 44 amp

Armature current Ia = I +I sh = 20455 +44 = 20895 amp

Armature drop I aRa = 20895 X 001 = 20895 volts Induced emf Eg = V + Ia Ra = 220 + 20895 = 2220895 volts

But induced emf

So N = Eg60 AΦ ZP = 2220895times 60 times2

002 times150 times2 times 4 = 111045 rpm

3 The following data refer to the OCC of a DC separately excited generator at 1000 rpm

Field current in A 00 02 03 04 05 06 07 08 09 10 11 12Armature voltage in V 5 40 75 100 124 145 162 178 188 195 200 205

The machine is now connected as a shunt generator with a total field resistance of 200 Ω and driven at 1000 rpm Estimate graphically (i) the voltage to which the generator will build up at no load (ii) the armature field and load currents when the terminal voltage is found to be 150 V Neglect the effect of armature reaction and brush drop and assume armature resistance Ra to be 08 Ω (iii) Finally estimate the steady state armature current when the machine terminals are shorted

Solution

First of all the OCC of the generator is plotted in a graph paper as shown in figure

(i) The total field circuit resistance is given to be Rf = 200 Ω The Rf line is now drawn on the same graph paper passing through the origin The point of intersection of the Rf line and the OCC decides the final no load voltage and can be read from the graph as 192 V and the corresponding field current is 096 A (ii) Since the terminal voltage V is 150 V (= BM) field current If

is OM = 077 A Generated voltage E is given by AM But we know E = IaRa + V Hence IaRa= E ndash V = AM ndash BM = AB Now from the graph AB = 25 V there4Ia = 2508 = 3125 A So load current IL= Ia ndash If= 3125 ndash 077 = 3048 A

ADDITIONAL PROBLEMS

1 A 4 pole lap wound 115 kW 230 volts dc shunt generator has armatureresistance of 02 ohm and field resistance of 100 ohms Calculate (i) emf

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES

Fig 115 Experimental setup to plot load characteristics

As the load current increases the terminal voltage falls due to two reasons(a) The armature reaction weakens the main flux so that actual emf generated E on load is less than that generated (E0) on no load

(b) There is voltage drop across armature resistance (= ILRa = IaRa)Due to these reasons the external characteristic is a drooping curve [curve 3 inFig 116 ] Note that in the absence of armature reaction and armature dropthe generated emf would have been E0 (curve 1)The internal characteristic can be determined from external characteristic by adding IL Ra drop to the external characteristic It is because armature reaction drop is included in the external characteristic Curve 2 is the internal characteristic of the generator and should obviously lie above the external characteristic

Fig 116 Load characteristics of a separately excited dcgenerator

117 Voltage Build-Up in a Self-Excited GeneratorLet us see how voltage builds up in a self-excited generator(i) Shunt generator Consider a shunt generator If the generator is run at a constant speed some emf will be generated due to residual magnetism in the main poles This small emf circulates a field current which in turn produces additional flux toreinforce the original residual flux (provided field winding connections arecorrect) This process continues and the generator builds up the normalgenerated voltage following the OCC shown in Fig 117 (i) The field resistance Rf can be represented by a straight line passing through the origin as shown in Fig 117 (ii) The two curves can be shown on the same diagram as they have the same ordinate as in Fig 117 (iii)

Fig 117

Since the field circuit is inductive there is a delay in the increase in current uponclosing the field circuit switch The rate at which the current increases dependsupon the voltage available for increasing it Suppose at any instant the fieldcurrent is i (= OA) and is increasing at the rate didt Then

E0=i R f +iquest L didt

Where Rf = total field circuit resistance L = inductance of field circuit

At the considered instant the total emf available is AC [See Fig 117 (iii)] Anamount AB of the cmf AC is absorbed by the voltage drop iRf and theremainder part BC is available to overcome L didt Since this surplus voltage isavailable it is possible for the field current to increase above the value OAHowever at point D the available voltage is OM and is all absorbed by i RfD

rop Consequently the field current cannot increase further and the generatorbuild up stops Hence we arrive at a very important conclusion that the voltage build up of the generator is given by the point of intersection of OCC and field resistance line Thus in Fig 117 (iii) D is point of intersection of the two curves Hence the generator will build up a voltage OM(ii) Series generator During initial operation with no current yet flowing a residual voltage will be generated exactly as in the case of a shunt generator The residual voltage will cause a current to flow through the whole series circuit when the circuit isclosed There will then be voltage build up to an equilibrium point exactlyanalogous to the buildup of a shunt generator The voltage build up graph willbe similar to that of shunt generator except that now load current (instead offield current for shunt generator) will be taken along x-axis10487871048787(iii) Compound generator When a compound generator has its series field flux aiding its shunt field flux the machine is said to be cumulative compound When the series field isconnected in reverse so that its field flux opposes the shunt field flux thegenerator is then differential compound The easiest way to build up voltage in a compound generator is to start under no load conditions At no load only the shunt field is effective When no-load voltage build up is achieved the generator is loaded If under load the voltagerises the series field connection is cumulative If the voltage drops significantlythe connection is differential compound118 Critical Field Resistance for a Shunt Generator The voltage build up in a shunt generator depends upon field circuit resistance If the field circuit resistance is R1 (line OA) then generator will build up a voltage OM as shown in Fig 118 If the field circuit resistance is increasedto R2 (tine OB) the generator will build up a voltage OL slightly less than OM As the field circuit resistance is increased the slope of resistance line also increases When the field resistance line becomes tangent (line OC) to OCC the generator would just excite If the field circuit resistance is increased beyond this point (say line OD) the generator will fail to excite The field circuit resistance represented by line OC (tangent to OCC) is called critical field resistance RC for the shunt generator It may be defined as under The maximum field circuit resistance (for a given speed) with which the shunt generator would just excite is known as its critical field resistance It should be noted that shunt generator will build up voltage only if field circuit resistance is less than critical field resistance

Fig 118 Critical field resistance of shunt generator

119 Critical Resistance for a Series Generator

Fig 119 shows the voltage build up in a series generator Here R1 R2 etc represent the total circuit resistance (load resistance and field winding resistance) If the total circuit resistance is R1 then series generator will build up a voltage OL The line OC is tangent to OCC and represents the critical resistance RC for a series generator If the total resistance of the circuit is more than RC (say line OD) the generator will fail to build up voltage Note that Fig 119 is similar to Fig 118 with the following differences

(i) In Fig 118 R1 R2 etc represent the total field circuit resistanceHowever R1 R2 etc in Fig 119 represent the total circuit resistance(load resistance and series field winding resistance etc)

(ii) In Fig 118 field current alone is represented along X-axis However inFig 119 load current IL is represented along Y-axis Note that in aseries generator field current = load current IL

Fig 119 Critical resistance of a series generator

120 Characteristics of Series Generator Fig 120 (i) shows the connections of a series wound generator Since there is only one current (that which flows through the whole machine) the load current is the same as the exciting current(i) OCCCurve 1 shows the open circuit characteristic (OCC) of a series generator Itcan be obtained experimentally by disconnecting the field winding from themachine and exciting it from a separate dc source(ii) Internal characteristic Curve 2 shows the total or internal characteristic of a series generator It gives the relation between the generated emf E on load and armature current Due to armature reaction the flux in the machine will be less than the flux at no load Hence emf E generated under load conditions will be less than the emf E0 generated under no load conditions Consequently internal characteristic curvelies below the OCC curve the difference between them representing the effectof armature reaction [See Fig 120 (ii)]

Fig 120 Characteristics of series generator

(iii) External characteristic

Curve 3 shows the external characteristic of a series generator It gives therelation between terminal voltage and load current ILV = E g- Ia (Ra +Rse )Therefore external characteristic curve will lie below internal characteristiccurve by an amount equal to ohmic drop [ie Ia(Ra + Rse)] in the machine asshown in Fig 120 (ii) The internal and external characteristics ofa dc series generator can be plotted from one another as shown in Fig 121 Supposewe are given the internal characteristic of the generator Let the line OC represent the resistance of the whole machine ie Ra +Rse If the load current is OB drop in the machine is AB ieAB = Ohmic drop in the machine = OB(Ra + Rse)

Fig 121 Load characteristics of series generator

Now raise a perpendicular from point B and mark a point b on this line such that ab = AB Then point b will lie on the external characteristic of the generator

Following similar procedure other points of external characteristic can be located It is easy to see that we can also plot internal characteristic from the external characteristic

121 Characteristics of a Shunt Generator Fig 122 (i) show the connections of a shunt wound generator The armature current Ia splits up into two parts a small fraction Ish flowing through shunt field winding while the major part IL goes to the external load(i) OCC The OCC of a shunt generator is similar in shape to that of a series generator as shown in Fig 122 (ii) The line OA represents the shunt field circuitresistance When the generator is run at normal speed it will build up a voltageOM At no-load the terminal voltage of the generator will be constant (= OM)represented by the horizontal dotted line MC

Fig 122 Characteristics of a Shunt Generator

(ii) Internal characteristic

When the generator is loaded flux per pole is reduced due to armature reaction Therefore emf E generated on load is less than the emf generated at no load As a result the internal characteristic (EIa) drops down slightly as shown in Fig122 (ii)

(iii) External characteristic Curve 2 shows the external characteristic of a shunt generator It gives therelation between terminal voltage V and load current IL

Therefore external characteristic curve will lie below the internal characteristic curve by an amount equal to drop in the armature circuit[ie (IL + Ish)Ra] as shown in Fig 122 (ii)

Note It may be seen from the external characteristic that change in terminalvoltage from no-load to full load is small The terminal voltage can always bemaintained constant by adjusting the field rheostat R automatically122 Critical External Resistance for Shunt Generator

If the load resistance across the terminals of a shunt generator is decreased then load current increase However there is a limit to the increase in

load current with the decrease of load resistance Any decrease of load resistance beyond this point instead of increasing the current ultimately results in reduced current Consequently the external characteristic turns back (dotted curve) as shown in Fig 123 The tangent OA to the curve represents the minimum external resistance required to excite the shunt generator on load and is called critical external resistance If the resistance of the external circuit is less than the critical external resistance (represented by tangent OA in Fig 310) the machine will refuse to excite or will de-excite if already running This means that external resistance is so low as virtually to short circuit the machine and so doing away with its excitation

Note There are two critical resistances for a shunt generator viz (i) criticalfield resistance (ii) critical external resistance For the shunt generator to buildup voltage the former should not be exceeded and the latter must not be gonebelow

123 How to Draw OCC at Different Speeds

If we are given OCC of a generator at a constant speed N1 then we can easily draw the OCC at any other constant speed N2 Fig 123 illustrates theprocedure Here we are given OCC at a constant speed N1 It is desired to findthe OCC at constant speed N2 (it is assumed that n1 lt N2) For constantexcitation E propN

E2

E1 =

N 2

N1 so

E2 iquest E1

N2

N1

As shown in Fig (311) for If = OH E1 = HC Therefore the new value of emf (E2) for the same If but at N2

E2= HCN 2

N 1 = HD

Fig 123 Plotting OCC at another speed

This locates the point D on the new OCC at N2 Similarly other points can belocated taking different values of If The locus of these points will be the OCCat N2

124 Critical Speed (NC)

The critical speed of a shunt generator is the minimum speed below which it fails to excite Clearly it is the speed for which the given shunt field resistance represents the critical resistance In Fig 124 curve 2 corresponds to critical

speed because the shunt field resistance (Rsh) line is tangential to it If the generator runs at full speed N the new OCC moves upward and the Rsh line represents critical resistance for this speed

In order to find critical speed take any convenient point C on excitation axis and erect a perpendicular so as to cut Rsh and Rsh lines at points B and A respectively Then

BCAC

=NcN

Nc=N BCAC

Fig 124 Finding critical speed

125 Conditions for Voltage Build-Up of a Shunt Generator

The necessary conditions for voltage build-up in a shunt generator are

(i) There must be some residual magnetism in generator poles(ii) The connections of the field winding should be such that the field current strengthens the residual magnetism(iii) The resistance of the field circuit should be less than the critical resistance In other words the speed of the generator should be higher

than the critical speed

126 Compound Generator Characteristics

In a compound generator both series and shunt excitation are combined as shown in Fig 123 The shunt winding can be connected either across the armature only (short-shunt connection S) or across armature plus series field (long-shunt connection G) The compound generator can be cumulatively compounded or

differentially compounded generator The latter is rarely used in practice Therefore we shall discuss the characteristics of cumulatively compounded

generator It may be noted that external characteristics of long and short shunt compound generators are almost identical

External characteristic

Fig 124 shows the external characteristics of a cumulatively compoundedgenerator The series excitation aids the shunt excitation The degree of compounding depends upon the increase in series excitation with the increase inload current

(i) If series winding turns are so adjusted that with the increase in load current the terminal voltage increases it is called over-compounded generator In such a case as the load current increases the series field mmf increases and tends to increase the flux and hence the generated voltage The increase in generated voltage is greater than the IaRa drop so that instead of decreasing the terminal voltage increases as shown by curve A in Fig 124

(ii) If series winding turns are so adjusted that with the increase in load current the terminal voltage substantially remains constant it is called Flat-compounded generator The series winding of such a machine has lesser number of turns than the one in over-compounded machine and therefore does not increase the flux as much for a given load current Consequently the full-load voltage is nearly equal to the no-load voltage as indicated by curve B in Fig 124 (iii) If series field winding has lesser number of turns than for a flat compounded machine the terminal voltage falls with increase in load current as indicated by curve C m Fig 124 Such a machine is called under-compounded generator

Fig 123 compound generator

Fig 124 Load characteristics of compound generators

127 LOSSES

The losses in a dc machine (generator or motor) may be divided into three classes viz (i) copper losses or electrical losses (ii) iron or core losses and (iii) mechanical losses All these losses appear as heat and thus raise the temperature of the machine They also lower the efficiency of the machine The loss structure maybe demonstrated as

1Copperlosses

These losses occur due to currents in the various windings of the machine When an electric current of I amps flows in a resistance of R ohms the heat energy is lost at the rate of I2R joules sec and the power loss is I2R watts Generators and motors have one or more field circuits and an armature circuit in which such losses occur All resistance losses of this kind are called copper losses The different copper losses occurring in a dc machine are

(i) Armature copper loss = Ia2Ra

(ii) Shunt field copper loss = Ish2 Rsh

(iii) Series field copper loss = Ise2Rse

Armature copper loss will be around 30 to 40 of full load losses where as total field copper losses will constitute around 30 of full load losses

Note There is also brush contact loss due to brush contact resistance (ie resistance between the surface of brush and surface of commutator) This loss is generally included in armature copper loss The brush contact drop is almost independent of Ia For carbon brushes this drop is around 1 V per brush The power loss due to brush contact resistance is 2ebIa where eb is the voltage drop at one brush

2 IRONORCORELOSSES These losses occur in the armature of a dc machine and are due to the rotation of armature in the magnetic field of the poles They are of two types viz (i) hysteresis loss (ii) eddy current loss

(i) Hysteresis loss

Hysteresis loss occurs in the armature of the dc machine since any given part of the armature is subjected to magnetic field reversals as it passes under successive poles Fig 124 shows an armature rotating in two-pole machine Consider a small piece ab of the armature When the piece ab is under N-pole the magnetic lines pass from a to b Half a revolution later the same piece of iron is under S-pole and magnetic lines pass from b to a so that magnetism in the iron is reversed In order to reverse continuously the molecular magnets in the armature core some amount of power has to be spent which is called hysteresis loss It is given by Steinmetz formula This formula is

Hysteresis Loss Wh = η Bmax 16f V Watts Bmax - Maximum flux density in the core f - Frequency of magnetic reversals V - Volume of armature in m3

η - Steinmetz hysteresis coefficient

In order to reduce this loss in a dc machine armature core is made of such materials which have a low value of Steinmetz hysteresis co-efficient eg silicon steel

Fig 124 Occurrence of hysteresis loss

ii) EDDYCURRENTLOSS

In addition to the voltages induced in the armature conductors there are also voltages induced in the armature core These voltages produce circulating currents in the armature core as shown in Fig 125 These are called eddy currents and power loss due to their flow is called eddy current loss The eddy current loss appears as heat which raises the temperature of the machine and lowers its efficiency Core resistance can be greatly increased by constructing the core of thin round If a continuous solid iron core is used the resistance to eddy current path will be small due to large cross-sectional area of the core Consequently the magnitude of eddy current and hence eddy current loss will be large The magnitude of eddy current can be reduced by making core resistance as high as practical The iron sheets called laminations [See Fig 126] The laminations are insulated from each other with a coating of varnish The insulating coating has a high resistance so very little current flows from one lamination to the other Alsobecause each lamination is very thin the resistance to current flowing through the width of a lamination is also quite large Thus laminating a core increases the core resistance which decreases the eddy current and hence the eddy current loss

Fig 125 Eddy current loss Fig 126 Core laminations

Eddy current loss is biven by the formula Pe= Ke Bmax

2 f2t2 V Watts

Where Ke = Constant depending upon electrical resistace of core Bmax = Maximum flux density in wbm2

f = frequency f magnetic reversals in Hz t = thickness of laminations in m and V = volume of core in m3

It may be noted that eddy current loss depends upon the square of lamination thickness For this reason lamination thickness should be kept as small as possible

3MECHANICALLOSSES

These losses are due to friction and windage(i) Friction losses eg bearing friction brush friction etc(ii) Windage loss ie air friction of rotating armature

These losses depend upon the speed of the machine But for a given speed they are practically constant

Note Iron losses and mechanical losses together are called stray losses

CONSTANT AND VARIABLE LOSSES

The losses in a dc generator (or dc motor) may be sub-divided into

(i) constant losses (ii) variable losses

(i) Constant losses

Those losses in a dc generator which remain constant at all loads are known as constant losses The constant losses in a dc generator are(a) iron losses(b) mechanical losses(c) shunt field losses(ii) Variable lossesThose losses in a dc generator which vary with load are called variable lossesThe variable losses in a dc generator are (a) Copper loss in armature winding (Ia

2 Ra) (b) Copper loss in series field winding (Ise

2Rse)

Total losses = Constant losses + Variable lossesNote Field Cu loss is constant for shunt and compound generators

128 POWER FLOW DIAGRAM

The power flow in a dc generator may be represented as

The different power stages are shown as A B and CA ndash The mechanical power input which is the power output of the prime mover is given as Pin = 2πNT

60 watts

B - The power developed in armature is Pa = Eg Ia watts C- The electrical power output is Pout = VIL watts

129 EFFICIENCY Efficiency of a machine is the ratio of its output power to the input power

Efficiency η = Output powerInput power

= Input powerminusLossesInput power = Output power

Output power+Losses

In case of a dc generator depending upon the power stages three different efficiencies can be defined as-

1 Mechanical Efficiency(ηm)

BA = Electrical power developedisinarmature

Mechanical power input2 Electrical Efficiency(ηe)

CB = Electrical power output

Electrical power developedisinthe armature 3 Overall or commercial efficiency (η)

CA = Electrical power output

Mechanical power input = ηmtimes ηe

130 CONDITION FOR MAXIMUM EFFICIENCY The efficiency of a dc generator is not constant but varies with load Fig 127shows the variation of efficiency with load current Consider a dc shunt generator with terminal voltage V delivering a load current of IL amp Generator output = V IL

Generator input = Output + Losses = V IL + constant losses + Variable losses = V IL + Wc+ Ia

2Ra

= V IL + Wc + (IL+ Ish)2 Rsh

The shunt field current Ish is small compared with IL so may be neglected

Generator input = VIL+ IL2Ra + Wc

Now efficiency η iquest outputinput = VI L

VIL+ Ia 2Ra+Wc

Efficiency is maximum when the denominator of the above equation is minimumie

ddIL ( VI L

VIL+ Ia 2Ra+Wc ) = 0

Which gives IL2 Ra = WC

ie Variable loss = constant lossThe load current for maximum efficiency is given by

IL iquest radicWcRa

Fig 127 Variation of efficiency with load

Hence the efficiency is maximum for that load current at which the variable loss is equal to the constant loss

131 APPLICATIONS

Separately excited generators

These types of generators have special applications such as electroplating electro refining of materials because separate supply is required for field excitation

Shunt generators

These types of generators are used in battery charging and ordinary lighting purposes

Series generators

This type of generators is used commonly as boosters on DC feeders as current generators for welding and arc lamp

Cumulatively compounded generators

This type of generators is used for domestic lighting purposes and to transmit energy over long distances

Differential compound generators

This type of generators is rarely used They are used for special applications such as electric arc welding

132 NUMERICAL EXCERCISES

1 The emf induced in the armature of a 450 kW 250 volt shunt generator is

2588volt when the field current is 200 amp and the generator is supplying power to a load at rated terminal voltage The armature circuit resistance is 0005 ohm Determine (i) load current (ii) power generated (iii) power output and (iv) electrical efficiency Neglect brush contact drop

Solution

Let the load current be I amp Given Ish = 20 A

Armature current = Load current + field current or Ia = I + 20 Induced emf = Terminal voltage + Ia Ra drop

2588 = 250 + (I + 20) times 0005

So (i) Load current I = 1740 amp

(ii) Power generated E Ia = 2588 (1740 + 20) watts = 455488 kW (iii) Power output V I = 250 times 1740 watts

= 435 kW

(iv) Electrical efficiency = 435455488 times 100 = 955

2 A 4 pole wave wound dc shunt generator delivers a load of 45 kW at a terminal voltage of 220 volts Its armature has 150 single turn coils and has a resistance of 001 ohm The air gap flux per pole is 002 weber Shunt field resistance is 50 ohm Calculate the speed at which it is being driven Neglect brush contact drop

Solution Load current I = W

V = 45000220 = 20455 amp

Field Current Ish = V

R s h = 220

50 = 44 amp

Armature current Ia = I +I sh = 20455 +44 = 20895 amp

Armature drop I aRa = 20895 X 001 = 20895 volts Induced emf Eg = V + Ia Ra = 220 + 20895 = 2220895 volts

But induced emf

So N = Eg60 AΦ ZP = 2220895times 60 times2

002 times150 times2 times 4 = 111045 rpm

3 The following data refer to the OCC of a DC separately excited generator at 1000 rpm

Field current in A 00 02 03 04 05 06 07 08 09 10 11 12Armature voltage in V 5 40 75 100 124 145 162 178 188 195 200 205

The machine is now connected as a shunt generator with a total field resistance of 200 Ω and driven at 1000 rpm Estimate graphically (i) the voltage to which the generator will build up at no load (ii) the armature field and load currents when the terminal voltage is found to be 150 V Neglect the effect of armature reaction and brush drop and assume armature resistance Ra to be 08 Ω (iii) Finally estimate the steady state armature current when the machine terminals are shorted

Solution

First of all the OCC of the generator is plotted in a graph paper as shown in figure

(i) The total field circuit resistance is given to be Rf = 200 Ω The Rf line is now drawn on the same graph paper passing through the origin The point of intersection of the Rf line and the OCC decides the final no load voltage and can be read from the graph as 192 V and the corresponding field current is 096 A (ii) Since the terminal voltage V is 150 V (= BM) field current If

is OM = 077 A Generated voltage E is given by AM But we know E = IaRa + V Hence IaRa= E ndash V = AM ndash BM = AB Now from the graph AB = 25 V there4Ia = 2508 = 3125 A So load current IL= Ia ndash If= 3125 ndash 077 = 3048 A

ADDITIONAL PROBLEMS

1 A 4 pole lap wound 115 kW 230 volts dc shunt generator has armatureresistance of 02 ohm and field resistance of 100 ohms Calculate (i) emf

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES

117 Voltage Build-Up in a Self-Excited GeneratorLet us see how voltage builds up in a self-excited generator(i) Shunt generator Consider a shunt generator If the generator is run at a constant speed some emf will be generated due to residual magnetism in the main poles This small emf circulates a field current which in turn produces additional flux toreinforce the original residual flux (provided field winding connections arecorrect) This process continues and the generator builds up the normalgenerated voltage following the OCC shown in Fig 117 (i) The field resistance Rf can be represented by a straight line passing through the origin as shown in Fig 117 (ii) The two curves can be shown on the same diagram as they have the same ordinate as in Fig 117 (iii)

Fig 117

Since the field circuit is inductive there is a delay in the increase in current uponclosing the field circuit switch The rate at which the current increases dependsupon the voltage available for increasing it Suppose at any instant the fieldcurrent is i (= OA) and is increasing at the rate didt Then

E0=i R f +iquest L didt

Where Rf = total field circuit resistance L = inductance of field circuit

At the considered instant the total emf available is AC [See Fig 117 (iii)] Anamount AB of the cmf AC is absorbed by the voltage drop iRf and theremainder part BC is available to overcome L didt Since this surplus voltage isavailable it is possible for the field current to increase above the value OAHowever at point D the available voltage is OM and is all absorbed by i RfD

rop Consequently the field current cannot increase further and the generatorbuild up stops Hence we arrive at a very important conclusion that the voltage build up of the generator is given by the point of intersection of OCC and field resistance line Thus in Fig 117 (iii) D is point of intersection of the two curves Hence the generator will build up a voltage OM(ii) Series generator During initial operation with no current yet flowing a residual voltage will be generated exactly as in the case of a shunt generator The residual voltage will cause a current to flow through the whole series circuit when the circuit isclosed There will then be voltage build up to an equilibrium point exactlyanalogous to the buildup of a shunt generator The voltage build up graph willbe similar to that of shunt generator except that now load current (instead offield current for shunt generator) will be taken along x-axis10487871048787(iii) Compound generator When a compound generator has its series field flux aiding its shunt field flux the machine is said to be cumulative compound When the series field isconnected in reverse so that its field flux opposes the shunt field flux thegenerator is then differential compound The easiest way to build up voltage in a compound generator is to start under no load conditions At no load only the shunt field is effective When no-load voltage build up is achieved the generator is loaded If under load the voltagerises the series field connection is cumulative If the voltage drops significantlythe connection is differential compound118 Critical Field Resistance for a Shunt Generator The voltage build up in a shunt generator depends upon field circuit resistance If the field circuit resistance is R1 (line OA) then generator will build up a voltage OM as shown in Fig 118 If the field circuit resistance is increasedto R2 (tine OB) the generator will build up a voltage OL slightly less than OM As the field circuit resistance is increased the slope of resistance line also increases When the field resistance line becomes tangent (line OC) to OCC the generator would just excite If the field circuit resistance is increased beyond this point (say line OD) the generator will fail to excite The field circuit resistance represented by line OC (tangent to OCC) is called critical field resistance RC for the shunt generator It may be defined as under The maximum field circuit resistance (for a given speed) with which the shunt generator would just excite is known as its critical field resistance It should be noted that shunt generator will build up voltage only if field circuit resistance is less than critical field resistance

Fig 118 Critical field resistance of shunt generator

119 Critical Resistance for a Series Generator

Fig 119 shows the voltage build up in a series generator Here R1 R2 etc represent the total circuit resistance (load resistance and field winding resistance) If the total circuit resistance is R1 then series generator will build up a voltage OL The line OC is tangent to OCC and represents the critical resistance RC for a series generator If the total resistance of the circuit is more than RC (say line OD) the generator will fail to build up voltage Note that Fig 119 is similar to Fig 118 with the following differences

(i) In Fig 118 R1 R2 etc represent the total field circuit resistanceHowever R1 R2 etc in Fig 119 represent the total circuit resistance(load resistance and series field winding resistance etc)

(ii) In Fig 118 field current alone is represented along X-axis However inFig 119 load current IL is represented along Y-axis Note that in aseries generator field current = load current IL

Fig 119 Critical resistance of a series generator

120 Characteristics of Series Generator Fig 120 (i) shows the connections of a series wound generator Since there is only one current (that which flows through the whole machine) the load current is the same as the exciting current(i) OCCCurve 1 shows the open circuit characteristic (OCC) of a series generator Itcan be obtained experimentally by disconnecting the field winding from themachine and exciting it from a separate dc source(ii) Internal characteristic Curve 2 shows the total or internal characteristic of a series generator It gives the relation between the generated emf E on load and armature current Due to armature reaction the flux in the machine will be less than the flux at no load Hence emf E generated under load conditions will be less than the emf E0 generated under no load conditions Consequently internal characteristic curvelies below the OCC curve the difference between them representing the effectof armature reaction [See Fig 120 (ii)]

Fig 120 Characteristics of series generator

(iii) External characteristic

Curve 3 shows the external characteristic of a series generator It gives therelation between terminal voltage and load current ILV = E g- Ia (Ra +Rse )Therefore external characteristic curve will lie below internal characteristiccurve by an amount equal to ohmic drop [ie Ia(Ra + Rse)] in the machine asshown in Fig 120 (ii) The internal and external characteristics ofa dc series generator can be plotted from one another as shown in Fig 121 Supposewe are given the internal characteristic of the generator Let the line OC represent the resistance of the whole machine ie Ra +Rse If the load current is OB drop in the machine is AB ieAB = Ohmic drop in the machine = OB(Ra + Rse)

Fig 121 Load characteristics of series generator

Now raise a perpendicular from point B and mark a point b on this line such that ab = AB Then point b will lie on the external characteristic of the generator

Following similar procedure other points of external characteristic can be located It is easy to see that we can also plot internal characteristic from the external characteristic

121 Characteristics of a Shunt Generator Fig 122 (i) show the connections of a shunt wound generator The armature current Ia splits up into two parts a small fraction Ish flowing through shunt field winding while the major part IL goes to the external load(i) OCC The OCC of a shunt generator is similar in shape to that of a series generator as shown in Fig 122 (ii) The line OA represents the shunt field circuitresistance When the generator is run at normal speed it will build up a voltageOM At no-load the terminal voltage of the generator will be constant (= OM)represented by the horizontal dotted line MC

Fig 122 Characteristics of a Shunt Generator

(ii) Internal characteristic

When the generator is loaded flux per pole is reduced due to armature reaction Therefore emf E generated on load is less than the emf generated at no load As a result the internal characteristic (EIa) drops down slightly as shown in Fig122 (ii)

(iii) External characteristic Curve 2 shows the external characteristic of a shunt generator It gives therelation between terminal voltage V and load current IL

Therefore external characteristic curve will lie below the internal characteristic curve by an amount equal to drop in the armature circuit[ie (IL + Ish)Ra] as shown in Fig 122 (ii)

Note It may be seen from the external characteristic that change in terminalvoltage from no-load to full load is small The terminal voltage can always bemaintained constant by adjusting the field rheostat R automatically122 Critical External Resistance for Shunt Generator

If the load resistance across the terminals of a shunt generator is decreased then load current increase However there is a limit to the increase in

load current with the decrease of load resistance Any decrease of load resistance beyond this point instead of increasing the current ultimately results in reduced current Consequently the external characteristic turns back (dotted curve) as shown in Fig 123 The tangent OA to the curve represents the minimum external resistance required to excite the shunt generator on load and is called critical external resistance If the resistance of the external circuit is less than the critical external resistance (represented by tangent OA in Fig 310) the machine will refuse to excite or will de-excite if already running This means that external resistance is so low as virtually to short circuit the machine and so doing away with its excitation

Note There are two critical resistances for a shunt generator viz (i) criticalfield resistance (ii) critical external resistance For the shunt generator to buildup voltage the former should not be exceeded and the latter must not be gonebelow

123 How to Draw OCC at Different Speeds

If we are given OCC of a generator at a constant speed N1 then we can easily draw the OCC at any other constant speed N2 Fig 123 illustrates theprocedure Here we are given OCC at a constant speed N1 It is desired to findthe OCC at constant speed N2 (it is assumed that n1 lt N2) For constantexcitation E propN

E2

E1 =

N 2

N1 so

E2 iquest E1

N2

N1

As shown in Fig (311) for If = OH E1 = HC Therefore the new value of emf (E2) for the same If but at N2

E2= HCN 2

N 1 = HD

Fig 123 Plotting OCC at another speed

This locates the point D on the new OCC at N2 Similarly other points can belocated taking different values of If The locus of these points will be the OCCat N2

124 Critical Speed (NC)

The critical speed of a shunt generator is the minimum speed below which it fails to excite Clearly it is the speed for which the given shunt field resistance represents the critical resistance In Fig 124 curve 2 corresponds to critical

speed because the shunt field resistance (Rsh) line is tangential to it If the generator runs at full speed N the new OCC moves upward and the Rsh line represents critical resistance for this speed

In order to find critical speed take any convenient point C on excitation axis and erect a perpendicular so as to cut Rsh and Rsh lines at points B and A respectively Then

BCAC

=NcN

Nc=N BCAC

Fig 124 Finding critical speed

125 Conditions for Voltage Build-Up of a Shunt Generator

The necessary conditions for voltage build-up in a shunt generator are

(i) There must be some residual magnetism in generator poles(ii) The connections of the field winding should be such that the field current strengthens the residual magnetism(iii) The resistance of the field circuit should be less than the critical resistance In other words the speed of the generator should be higher

than the critical speed

126 Compound Generator Characteristics

In a compound generator both series and shunt excitation are combined as shown in Fig 123 The shunt winding can be connected either across the armature only (short-shunt connection S) or across armature plus series field (long-shunt connection G) The compound generator can be cumulatively compounded or

differentially compounded generator The latter is rarely used in practice Therefore we shall discuss the characteristics of cumulatively compounded

generator It may be noted that external characteristics of long and short shunt compound generators are almost identical

External characteristic

Fig 124 shows the external characteristics of a cumulatively compoundedgenerator The series excitation aids the shunt excitation The degree of compounding depends upon the increase in series excitation with the increase inload current

(i) If series winding turns are so adjusted that with the increase in load current the terminal voltage increases it is called over-compounded generator In such a case as the load current increases the series field mmf increases and tends to increase the flux and hence the generated voltage The increase in generated voltage is greater than the IaRa drop so that instead of decreasing the terminal voltage increases as shown by curve A in Fig 124

(ii) If series winding turns are so adjusted that with the increase in load current the terminal voltage substantially remains constant it is called Flat-compounded generator The series winding of such a machine has lesser number of turns than the one in over-compounded machine and therefore does not increase the flux as much for a given load current Consequently the full-load voltage is nearly equal to the no-load voltage as indicated by curve B in Fig 124 (iii) If series field winding has lesser number of turns than for a flat compounded machine the terminal voltage falls with increase in load current as indicated by curve C m Fig 124 Such a machine is called under-compounded generator

Fig 123 compound generator

Fig 124 Load characteristics of compound generators

127 LOSSES

The losses in a dc machine (generator or motor) may be divided into three classes viz (i) copper losses or electrical losses (ii) iron or core losses and (iii) mechanical losses All these losses appear as heat and thus raise the temperature of the machine They also lower the efficiency of the machine The loss structure maybe demonstrated as

1Copperlosses

These losses occur due to currents in the various windings of the machine When an electric current of I amps flows in a resistance of R ohms the heat energy is lost at the rate of I2R joules sec and the power loss is I2R watts Generators and motors have one or more field circuits and an armature circuit in which such losses occur All resistance losses of this kind are called copper losses The different copper losses occurring in a dc machine are

(i) Armature copper loss = Ia2Ra

(ii) Shunt field copper loss = Ish2 Rsh

(iii) Series field copper loss = Ise2Rse

Armature copper loss will be around 30 to 40 of full load losses where as total field copper losses will constitute around 30 of full load losses

Note There is also brush contact loss due to brush contact resistance (ie resistance between the surface of brush and surface of commutator) This loss is generally included in armature copper loss The brush contact drop is almost independent of Ia For carbon brushes this drop is around 1 V per brush The power loss due to brush contact resistance is 2ebIa where eb is the voltage drop at one brush

2 IRONORCORELOSSES These losses occur in the armature of a dc machine and are due to the rotation of armature in the magnetic field of the poles They are of two types viz (i) hysteresis loss (ii) eddy current loss

(i) Hysteresis loss

Hysteresis loss occurs in the armature of the dc machine since any given part of the armature is subjected to magnetic field reversals as it passes under successive poles Fig 124 shows an armature rotating in two-pole machine Consider a small piece ab of the armature When the piece ab is under N-pole the magnetic lines pass from a to b Half a revolution later the same piece of iron is under S-pole and magnetic lines pass from b to a so that magnetism in the iron is reversed In order to reverse continuously the molecular magnets in the armature core some amount of power has to be spent which is called hysteresis loss It is given by Steinmetz formula This formula is

Hysteresis Loss Wh = η Bmax 16f V Watts Bmax - Maximum flux density in the core f - Frequency of magnetic reversals V - Volume of armature in m3

η - Steinmetz hysteresis coefficient

In order to reduce this loss in a dc machine armature core is made of such materials which have a low value of Steinmetz hysteresis co-efficient eg silicon steel

Fig 124 Occurrence of hysteresis loss

ii) EDDYCURRENTLOSS

In addition to the voltages induced in the armature conductors there are also voltages induced in the armature core These voltages produce circulating currents in the armature core as shown in Fig 125 These are called eddy currents and power loss due to their flow is called eddy current loss The eddy current loss appears as heat which raises the temperature of the machine and lowers its efficiency Core resistance can be greatly increased by constructing the core of thin round If a continuous solid iron core is used the resistance to eddy current path will be small due to large cross-sectional area of the core Consequently the magnitude of eddy current and hence eddy current loss will be large The magnitude of eddy current can be reduced by making core resistance as high as practical The iron sheets called laminations [See Fig 126] The laminations are insulated from each other with a coating of varnish The insulating coating has a high resistance so very little current flows from one lamination to the other Alsobecause each lamination is very thin the resistance to current flowing through the width of a lamination is also quite large Thus laminating a core increases the core resistance which decreases the eddy current and hence the eddy current loss

Fig 125 Eddy current loss Fig 126 Core laminations

Eddy current loss is biven by the formula Pe= Ke Bmax

2 f2t2 V Watts

Where Ke = Constant depending upon electrical resistace of core Bmax = Maximum flux density in wbm2

f = frequency f magnetic reversals in Hz t = thickness of laminations in m and V = volume of core in m3

It may be noted that eddy current loss depends upon the square of lamination thickness For this reason lamination thickness should be kept as small as possible

3MECHANICALLOSSES

These losses are due to friction and windage(i) Friction losses eg bearing friction brush friction etc(ii) Windage loss ie air friction of rotating armature

These losses depend upon the speed of the machine But for a given speed they are practically constant

Note Iron losses and mechanical losses together are called stray losses

CONSTANT AND VARIABLE LOSSES

The losses in a dc generator (or dc motor) may be sub-divided into

(i) constant losses (ii) variable losses

(i) Constant losses

Those losses in a dc generator which remain constant at all loads are known as constant losses The constant losses in a dc generator are(a) iron losses(b) mechanical losses(c) shunt field losses(ii) Variable lossesThose losses in a dc generator which vary with load are called variable lossesThe variable losses in a dc generator are (a) Copper loss in armature winding (Ia

2 Ra) (b) Copper loss in series field winding (Ise

2Rse)

Total losses = Constant losses + Variable lossesNote Field Cu loss is constant for shunt and compound generators

128 POWER FLOW DIAGRAM

The power flow in a dc generator may be represented as

The different power stages are shown as A B and CA ndash The mechanical power input which is the power output of the prime mover is given as Pin = 2πNT

60 watts

B - The power developed in armature is Pa = Eg Ia watts C- The electrical power output is Pout = VIL watts

129 EFFICIENCY Efficiency of a machine is the ratio of its output power to the input power

Efficiency η = Output powerInput power

= Input powerminusLossesInput power = Output power

Output power+Losses

In case of a dc generator depending upon the power stages three different efficiencies can be defined as-

1 Mechanical Efficiency(ηm)

BA = Electrical power developedisinarmature

Mechanical power input2 Electrical Efficiency(ηe)

CB = Electrical power output

Electrical power developedisinthe armature 3 Overall or commercial efficiency (η)

CA = Electrical power output

Mechanical power input = ηmtimes ηe

130 CONDITION FOR MAXIMUM EFFICIENCY The efficiency of a dc generator is not constant but varies with load Fig 127shows the variation of efficiency with load current Consider a dc shunt generator with terminal voltage V delivering a load current of IL amp Generator output = V IL

Generator input = Output + Losses = V IL + constant losses + Variable losses = V IL + Wc+ Ia

2Ra

= V IL + Wc + (IL+ Ish)2 Rsh

The shunt field current Ish is small compared with IL so may be neglected

Generator input = VIL+ IL2Ra + Wc

Now efficiency η iquest outputinput = VI L

VIL+ Ia 2Ra+Wc

Efficiency is maximum when the denominator of the above equation is minimumie

ddIL ( VI L

VIL+ Ia 2Ra+Wc ) = 0

Which gives IL2 Ra = WC

ie Variable loss = constant lossThe load current for maximum efficiency is given by

IL iquest radicWcRa

Fig 127 Variation of efficiency with load

Hence the efficiency is maximum for that load current at which the variable loss is equal to the constant loss

131 APPLICATIONS

Separately excited generators

These types of generators have special applications such as electroplating electro refining of materials because separate supply is required for field excitation

Shunt generators

These types of generators are used in battery charging and ordinary lighting purposes

Series generators

This type of generators is used commonly as boosters on DC feeders as current generators for welding and arc lamp

Cumulatively compounded generators

This type of generators is used for domestic lighting purposes and to transmit energy over long distances

Differential compound generators

This type of generators is rarely used They are used for special applications such as electric arc welding

132 NUMERICAL EXCERCISES

1 The emf induced in the armature of a 450 kW 250 volt shunt generator is

2588volt when the field current is 200 amp and the generator is supplying power to a load at rated terminal voltage The armature circuit resistance is 0005 ohm Determine (i) load current (ii) power generated (iii) power output and (iv) electrical efficiency Neglect brush contact drop

Solution

Let the load current be I amp Given Ish = 20 A

Armature current = Load current + field current or Ia = I + 20 Induced emf = Terminal voltage + Ia Ra drop

2588 = 250 + (I + 20) times 0005

So (i) Load current I = 1740 amp

(ii) Power generated E Ia = 2588 (1740 + 20) watts = 455488 kW (iii) Power output V I = 250 times 1740 watts

= 435 kW

(iv) Electrical efficiency = 435455488 times 100 = 955

2 A 4 pole wave wound dc shunt generator delivers a load of 45 kW at a terminal voltage of 220 volts Its armature has 150 single turn coils and has a resistance of 001 ohm The air gap flux per pole is 002 weber Shunt field resistance is 50 ohm Calculate the speed at which it is being driven Neglect brush contact drop

Solution Load current I = W

V = 45000220 = 20455 amp

Field Current Ish = V

R s h = 220

50 = 44 amp

Armature current Ia = I +I sh = 20455 +44 = 20895 amp

Armature drop I aRa = 20895 X 001 = 20895 volts Induced emf Eg = V + Ia Ra = 220 + 20895 = 2220895 volts

But induced emf

So N = Eg60 AΦ ZP = 2220895times 60 times2

002 times150 times2 times 4 = 111045 rpm

3 The following data refer to the OCC of a DC separately excited generator at 1000 rpm

Field current in A 00 02 03 04 05 06 07 08 09 10 11 12Armature voltage in V 5 40 75 100 124 145 162 178 188 195 200 205

The machine is now connected as a shunt generator with a total field resistance of 200 Ω and driven at 1000 rpm Estimate graphically (i) the voltage to which the generator will build up at no load (ii) the armature field and load currents when the terminal voltage is found to be 150 V Neglect the effect of armature reaction and brush drop and assume armature resistance Ra to be 08 Ω (iii) Finally estimate the steady state armature current when the machine terminals are shorted

Solution

First of all the OCC of the generator is plotted in a graph paper as shown in figure

(i) The total field circuit resistance is given to be Rf = 200 Ω The Rf line is now drawn on the same graph paper passing through the origin The point of intersection of the Rf line and the OCC decides the final no load voltage and can be read from the graph as 192 V and the corresponding field current is 096 A (ii) Since the terminal voltage V is 150 V (= BM) field current If

is OM = 077 A Generated voltage E is given by AM But we know E = IaRa + V Hence IaRa= E ndash V = AM ndash BM = AB Now from the graph AB = 25 V there4Ia = 2508 = 3125 A So load current IL= Ia ndash If= 3125 ndash 077 = 3048 A

ADDITIONAL PROBLEMS

1 A 4 pole lap wound 115 kW 230 volts dc shunt generator has armatureresistance of 02 ohm and field resistance of 100 ohms Calculate (i) emf

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES

Where Rf = total field circuit resistance L = inductance of field circuit

At the considered instant the total emf available is AC [See Fig 117 (iii)] Anamount AB of the cmf AC is absorbed by the voltage drop iRf and theremainder part BC is available to overcome L didt Since this surplus voltage isavailable it is possible for the field current to increase above the value OAHowever at point D the available voltage is OM and is all absorbed by i RfD

rop Consequently the field current cannot increase further and the generatorbuild up stops Hence we arrive at a very important conclusion that the voltage build up of the generator is given by the point of intersection of OCC and field resistance line Thus in Fig 117 (iii) D is point of intersection of the two curves Hence the generator will build up a voltage OM(ii) Series generator During initial operation with no current yet flowing a residual voltage will be generated exactly as in the case of a shunt generator The residual voltage will cause a current to flow through the whole series circuit when the circuit isclosed There will then be voltage build up to an equilibrium point exactlyanalogous to the buildup of a shunt generator The voltage build up graph willbe similar to that of shunt generator except that now load current (instead offield current for shunt generator) will be taken along x-axis10487871048787(iii) Compound generator When a compound generator has its series field flux aiding its shunt field flux the machine is said to be cumulative compound When the series field isconnected in reverse so that its field flux opposes the shunt field flux thegenerator is then differential compound The easiest way to build up voltage in a compound generator is to start under no load conditions At no load only the shunt field is effective When no-load voltage build up is achieved the generator is loaded If under load the voltagerises the series field connection is cumulative If the voltage drops significantlythe connection is differential compound118 Critical Field Resistance for a Shunt Generator The voltage build up in a shunt generator depends upon field circuit resistance If the field circuit resistance is R1 (line OA) then generator will build up a voltage OM as shown in Fig 118 If the field circuit resistance is increasedto R2 (tine OB) the generator will build up a voltage OL slightly less than OM As the field circuit resistance is increased the slope of resistance line also increases When the field resistance line becomes tangent (line OC) to OCC the generator would just excite If the field circuit resistance is increased beyond this point (say line OD) the generator will fail to excite The field circuit resistance represented by line OC (tangent to OCC) is called critical field resistance RC for the shunt generator It may be defined as under The maximum field circuit resistance (for a given speed) with which the shunt generator would just excite is known as its critical field resistance It should be noted that shunt generator will build up voltage only if field circuit resistance is less than critical field resistance

Fig 118 Critical field resistance of shunt generator

119 Critical Resistance for a Series Generator

Fig 119 shows the voltage build up in a series generator Here R1 R2 etc represent the total circuit resistance (load resistance and field winding resistance) If the total circuit resistance is R1 then series generator will build up a voltage OL The line OC is tangent to OCC and represents the critical resistance RC for a series generator If the total resistance of the circuit is more than RC (say line OD) the generator will fail to build up voltage Note that Fig 119 is similar to Fig 118 with the following differences

(i) In Fig 118 R1 R2 etc represent the total field circuit resistanceHowever R1 R2 etc in Fig 119 represent the total circuit resistance(load resistance and series field winding resistance etc)

(ii) In Fig 118 field current alone is represented along X-axis However inFig 119 load current IL is represented along Y-axis Note that in aseries generator field current = load current IL

Fig 119 Critical resistance of a series generator

120 Characteristics of Series Generator Fig 120 (i) shows the connections of a series wound generator Since there is only one current (that which flows through the whole machine) the load current is the same as the exciting current(i) OCCCurve 1 shows the open circuit characteristic (OCC) of a series generator Itcan be obtained experimentally by disconnecting the field winding from themachine and exciting it from a separate dc source(ii) Internal characteristic Curve 2 shows the total or internal characteristic of a series generator It gives the relation between the generated emf E on load and armature current Due to armature reaction the flux in the machine will be less than the flux at no load Hence emf E generated under load conditions will be less than the emf E0 generated under no load conditions Consequently internal characteristic curvelies below the OCC curve the difference between them representing the effectof armature reaction [See Fig 120 (ii)]

Fig 120 Characteristics of series generator

(iii) External characteristic

Curve 3 shows the external characteristic of a series generator It gives therelation between terminal voltage and load current ILV = E g- Ia (Ra +Rse )Therefore external characteristic curve will lie below internal characteristiccurve by an amount equal to ohmic drop [ie Ia(Ra + Rse)] in the machine asshown in Fig 120 (ii) The internal and external characteristics ofa dc series generator can be plotted from one another as shown in Fig 121 Supposewe are given the internal characteristic of the generator Let the line OC represent the resistance of the whole machine ie Ra +Rse If the load current is OB drop in the machine is AB ieAB = Ohmic drop in the machine = OB(Ra + Rse)

Fig 121 Load characteristics of series generator

Now raise a perpendicular from point B and mark a point b on this line such that ab = AB Then point b will lie on the external characteristic of the generator

Following similar procedure other points of external characteristic can be located It is easy to see that we can also plot internal characteristic from the external characteristic

121 Characteristics of a Shunt Generator Fig 122 (i) show the connections of a shunt wound generator The armature current Ia splits up into two parts a small fraction Ish flowing through shunt field winding while the major part IL goes to the external load(i) OCC The OCC of a shunt generator is similar in shape to that of a series generator as shown in Fig 122 (ii) The line OA represents the shunt field circuitresistance When the generator is run at normal speed it will build up a voltageOM At no-load the terminal voltage of the generator will be constant (= OM)represented by the horizontal dotted line MC

Fig 122 Characteristics of a Shunt Generator

(ii) Internal characteristic

When the generator is loaded flux per pole is reduced due to armature reaction Therefore emf E generated on load is less than the emf generated at no load As a result the internal characteristic (EIa) drops down slightly as shown in Fig122 (ii)

(iii) External characteristic Curve 2 shows the external characteristic of a shunt generator It gives therelation between terminal voltage V and load current IL

Therefore external characteristic curve will lie below the internal characteristic curve by an amount equal to drop in the armature circuit[ie (IL + Ish)Ra] as shown in Fig 122 (ii)

Note It may be seen from the external characteristic that change in terminalvoltage from no-load to full load is small The terminal voltage can always bemaintained constant by adjusting the field rheostat R automatically122 Critical External Resistance for Shunt Generator

If the load resistance across the terminals of a shunt generator is decreased then load current increase However there is a limit to the increase in

load current with the decrease of load resistance Any decrease of load resistance beyond this point instead of increasing the current ultimately results in reduced current Consequently the external characteristic turns back (dotted curve) as shown in Fig 123 The tangent OA to the curve represents the minimum external resistance required to excite the shunt generator on load and is called critical external resistance If the resistance of the external circuit is less than the critical external resistance (represented by tangent OA in Fig 310) the machine will refuse to excite or will de-excite if already running This means that external resistance is so low as virtually to short circuit the machine and so doing away with its excitation

Note There are two critical resistances for a shunt generator viz (i) criticalfield resistance (ii) critical external resistance For the shunt generator to buildup voltage the former should not be exceeded and the latter must not be gonebelow

123 How to Draw OCC at Different Speeds

If we are given OCC of a generator at a constant speed N1 then we can easily draw the OCC at any other constant speed N2 Fig 123 illustrates theprocedure Here we are given OCC at a constant speed N1 It is desired to findthe OCC at constant speed N2 (it is assumed that n1 lt N2) For constantexcitation E propN

E2

E1 =

N 2

N1 so

E2 iquest E1

N2

N1

As shown in Fig (311) for If = OH E1 = HC Therefore the new value of emf (E2) for the same If but at N2

E2= HCN 2

N 1 = HD

Fig 123 Plotting OCC at another speed

This locates the point D on the new OCC at N2 Similarly other points can belocated taking different values of If The locus of these points will be the OCCat N2

124 Critical Speed (NC)

The critical speed of a shunt generator is the minimum speed below which it fails to excite Clearly it is the speed for which the given shunt field resistance represents the critical resistance In Fig 124 curve 2 corresponds to critical

speed because the shunt field resistance (Rsh) line is tangential to it If the generator runs at full speed N the new OCC moves upward and the Rsh line represents critical resistance for this speed

In order to find critical speed take any convenient point C on excitation axis and erect a perpendicular so as to cut Rsh and Rsh lines at points B and A respectively Then

BCAC

=NcN

Nc=N BCAC

Fig 124 Finding critical speed

125 Conditions for Voltage Build-Up of a Shunt Generator

The necessary conditions for voltage build-up in a shunt generator are

(i) There must be some residual magnetism in generator poles(ii) The connections of the field winding should be such that the field current strengthens the residual magnetism(iii) The resistance of the field circuit should be less than the critical resistance In other words the speed of the generator should be higher

than the critical speed

126 Compound Generator Characteristics

In a compound generator both series and shunt excitation are combined as shown in Fig 123 The shunt winding can be connected either across the armature only (short-shunt connection S) or across armature plus series field (long-shunt connection G) The compound generator can be cumulatively compounded or

differentially compounded generator The latter is rarely used in practice Therefore we shall discuss the characteristics of cumulatively compounded

generator It may be noted that external characteristics of long and short shunt compound generators are almost identical

External characteristic

Fig 124 shows the external characteristics of a cumulatively compoundedgenerator The series excitation aids the shunt excitation The degree of compounding depends upon the increase in series excitation with the increase inload current

(i) If series winding turns are so adjusted that with the increase in load current the terminal voltage increases it is called over-compounded generator In such a case as the load current increases the series field mmf increases and tends to increase the flux and hence the generated voltage The increase in generated voltage is greater than the IaRa drop so that instead of decreasing the terminal voltage increases as shown by curve A in Fig 124

(ii) If series winding turns are so adjusted that with the increase in load current the terminal voltage substantially remains constant it is called Flat-compounded generator The series winding of such a machine has lesser number of turns than the one in over-compounded machine and therefore does not increase the flux as much for a given load current Consequently the full-load voltage is nearly equal to the no-load voltage as indicated by curve B in Fig 124 (iii) If series field winding has lesser number of turns than for a flat compounded machine the terminal voltage falls with increase in load current as indicated by curve C m Fig 124 Such a machine is called under-compounded generator

Fig 123 compound generator

Fig 124 Load characteristics of compound generators

127 LOSSES

The losses in a dc machine (generator or motor) may be divided into three classes viz (i) copper losses or electrical losses (ii) iron or core losses and (iii) mechanical losses All these losses appear as heat and thus raise the temperature of the machine They also lower the efficiency of the machine The loss structure maybe demonstrated as

1Copperlosses

These losses occur due to currents in the various windings of the machine When an electric current of I amps flows in a resistance of R ohms the heat energy is lost at the rate of I2R joules sec and the power loss is I2R watts Generators and motors have one or more field circuits and an armature circuit in which such losses occur All resistance losses of this kind are called copper losses The different copper losses occurring in a dc machine are

(i) Armature copper loss = Ia2Ra

(ii) Shunt field copper loss = Ish2 Rsh

(iii) Series field copper loss = Ise2Rse

Armature copper loss will be around 30 to 40 of full load losses where as total field copper losses will constitute around 30 of full load losses

Note There is also brush contact loss due to brush contact resistance (ie resistance between the surface of brush and surface of commutator) This loss is generally included in armature copper loss The brush contact drop is almost independent of Ia For carbon brushes this drop is around 1 V per brush The power loss due to brush contact resistance is 2ebIa where eb is the voltage drop at one brush

2 IRONORCORELOSSES These losses occur in the armature of a dc machine and are due to the rotation of armature in the magnetic field of the poles They are of two types viz (i) hysteresis loss (ii) eddy current loss

(i) Hysteresis loss

Hysteresis loss occurs in the armature of the dc machine since any given part of the armature is subjected to magnetic field reversals as it passes under successive poles Fig 124 shows an armature rotating in two-pole machine Consider a small piece ab of the armature When the piece ab is under N-pole the magnetic lines pass from a to b Half a revolution later the same piece of iron is under S-pole and magnetic lines pass from b to a so that magnetism in the iron is reversed In order to reverse continuously the molecular magnets in the armature core some amount of power has to be spent which is called hysteresis loss It is given by Steinmetz formula This formula is

Hysteresis Loss Wh = η Bmax 16f V Watts Bmax - Maximum flux density in the core f - Frequency of magnetic reversals V - Volume of armature in m3

η - Steinmetz hysteresis coefficient

In order to reduce this loss in a dc machine armature core is made of such materials which have a low value of Steinmetz hysteresis co-efficient eg silicon steel

Fig 124 Occurrence of hysteresis loss

ii) EDDYCURRENTLOSS

In addition to the voltages induced in the armature conductors there are also voltages induced in the armature core These voltages produce circulating currents in the armature core as shown in Fig 125 These are called eddy currents and power loss due to their flow is called eddy current loss The eddy current loss appears as heat which raises the temperature of the machine and lowers its efficiency Core resistance can be greatly increased by constructing the core of thin round If a continuous solid iron core is used the resistance to eddy current path will be small due to large cross-sectional area of the core Consequently the magnitude of eddy current and hence eddy current loss will be large The magnitude of eddy current can be reduced by making core resistance as high as practical The iron sheets called laminations [See Fig 126] The laminations are insulated from each other with a coating of varnish The insulating coating has a high resistance so very little current flows from one lamination to the other Alsobecause each lamination is very thin the resistance to current flowing through the width of a lamination is also quite large Thus laminating a core increases the core resistance which decreases the eddy current and hence the eddy current loss

Fig 125 Eddy current loss Fig 126 Core laminations

Eddy current loss is biven by the formula Pe= Ke Bmax

2 f2t2 V Watts

Where Ke = Constant depending upon electrical resistace of core Bmax = Maximum flux density in wbm2

f = frequency f magnetic reversals in Hz t = thickness of laminations in m and V = volume of core in m3

It may be noted that eddy current loss depends upon the square of lamination thickness For this reason lamination thickness should be kept as small as possible

3MECHANICALLOSSES

These losses are due to friction and windage(i) Friction losses eg bearing friction brush friction etc(ii) Windage loss ie air friction of rotating armature

These losses depend upon the speed of the machine But for a given speed they are practically constant

Note Iron losses and mechanical losses together are called stray losses

CONSTANT AND VARIABLE LOSSES

The losses in a dc generator (or dc motor) may be sub-divided into

(i) constant losses (ii) variable losses

(i) Constant losses

Those losses in a dc generator which remain constant at all loads are known as constant losses The constant losses in a dc generator are(a) iron losses(b) mechanical losses(c) shunt field losses(ii) Variable lossesThose losses in a dc generator which vary with load are called variable lossesThe variable losses in a dc generator are (a) Copper loss in armature winding (Ia

2 Ra) (b) Copper loss in series field winding (Ise

2Rse)

Total losses = Constant losses + Variable lossesNote Field Cu loss is constant for shunt and compound generators

128 POWER FLOW DIAGRAM

The power flow in a dc generator may be represented as

The different power stages are shown as A B and CA ndash The mechanical power input which is the power output of the prime mover is given as Pin = 2πNT

60 watts

B - The power developed in armature is Pa = Eg Ia watts C- The electrical power output is Pout = VIL watts

129 EFFICIENCY Efficiency of a machine is the ratio of its output power to the input power

Efficiency η = Output powerInput power

= Input powerminusLossesInput power = Output power

Output power+Losses

In case of a dc generator depending upon the power stages three different efficiencies can be defined as-

1 Mechanical Efficiency(ηm)

BA = Electrical power developedisinarmature

Mechanical power input2 Electrical Efficiency(ηe)

CB = Electrical power output

Electrical power developedisinthe armature 3 Overall or commercial efficiency (η)

CA = Electrical power output

Mechanical power input = ηmtimes ηe

130 CONDITION FOR MAXIMUM EFFICIENCY The efficiency of a dc generator is not constant but varies with load Fig 127shows the variation of efficiency with load current Consider a dc shunt generator with terminal voltage V delivering a load current of IL amp Generator output = V IL

Generator input = Output + Losses = V IL + constant losses + Variable losses = V IL + Wc+ Ia

2Ra

= V IL + Wc + (IL+ Ish)2 Rsh

The shunt field current Ish is small compared with IL so may be neglected

Generator input = VIL+ IL2Ra + Wc

Now efficiency η iquest outputinput = VI L

VIL+ Ia 2Ra+Wc

Efficiency is maximum when the denominator of the above equation is minimumie

ddIL ( VI L

VIL+ Ia 2Ra+Wc ) = 0

Which gives IL2 Ra = WC

ie Variable loss = constant lossThe load current for maximum efficiency is given by

IL iquest radicWcRa

Fig 127 Variation of efficiency with load

Hence the efficiency is maximum for that load current at which the variable loss is equal to the constant loss

131 APPLICATIONS

Separately excited generators

These types of generators have special applications such as electroplating electro refining of materials because separate supply is required for field excitation

Shunt generators

These types of generators are used in battery charging and ordinary lighting purposes

Series generators

This type of generators is used commonly as boosters on DC feeders as current generators for welding and arc lamp

Cumulatively compounded generators

This type of generators is used for domestic lighting purposes and to transmit energy over long distances

Differential compound generators

This type of generators is rarely used They are used for special applications such as electric arc welding

132 NUMERICAL EXCERCISES

1 The emf induced in the armature of a 450 kW 250 volt shunt generator is

2588volt when the field current is 200 amp and the generator is supplying power to a load at rated terminal voltage The armature circuit resistance is 0005 ohm Determine (i) load current (ii) power generated (iii) power output and (iv) electrical efficiency Neglect brush contact drop

Solution

Let the load current be I amp Given Ish = 20 A

Armature current = Load current + field current or Ia = I + 20 Induced emf = Terminal voltage + Ia Ra drop

2588 = 250 + (I + 20) times 0005

So (i) Load current I = 1740 amp

(ii) Power generated E Ia = 2588 (1740 + 20) watts = 455488 kW (iii) Power output V I = 250 times 1740 watts

= 435 kW

(iv) Electrical efficiency = 435455488 times 100 = 955

2 A 4 pole wave wound dc shunt generator delivers a load of 45 kW at a terminal voltage of 220 volts Its armature has 150 single turn coils and has a resistance of 001 ohm The air gap flux per pole is 002 weber Shunt field resistance is 50 ohm Calculate the speed at which it is being driven Neglect brush contact drop

Solution Load current I = W

V = 45000220 = 20455 amp

Field Current Ish = V

R s h = 220

50 = 44 amp

Armature current Ia = I +I sh = 20455 +44 = 20895 amp

Armature drop I aRa = 20895 X 001 = 20895 volts Induced emf Eg = V + Ia Ra = 220 + 20895 = 2220895 volts

But induced emf

So N = Eg60 AΦ ZP = 2220895times 60 times2

002 times150 times2 times 4 = 111045 rpm

3 The following data refer to the OCC of a DC separately excited generator at 1000 rpm

Field current in A 00 02 03 04 05 06 07 08 09 10 11 12Armature voltage in V 5 40 75 100 124 145 162 178 188 195 200 205

The machine is now connected as a shunt generator with a total field resistance of 200 Ω and driven at 1000 rpm Estimate graphically (i) the voltage to which the generator will build up at no load (ii) the armature field and load currents when the terminal voltage is found to be 150 V Neglect the effect of armature reaction and brush drop and assume armature resistance Ra to be 08 Ω (iii) Finally estimate the steady state armature current when the machine terminals are shorted

Solution

First of all the OCC of the generator is plotted in a graph paper as shown in figure

(i) The total field circuit resistance is given to be Rf = 200 Ω The Rf line is now drawn on the same graph paper passing through the origin The point of intersection of the Rf line and the OCC decides the final no load voltage and can be read from the graph as 192 V and the corresponding field current is 096 A (ii) Since the terminal voltage V is 150 V (= BM) field current If

is OM = 077 A Generated voltage E is given by AM But we know E = IaRa + V Hence IaRa= E ndash V = AM ndash BM = AB Now from the graph AB = 25 V there4Ia = 2508 = 3125 A So load current IL= Ia ndash If= 3125 ndash 077 = 3048 A

ADDITIONAL PROBLEMS

1 A 4 pole lap wound 115 kW 230 volts dc shunt generator has armatureresistance of 02 ohm and field resistance of 100 ohms Calculate (i) emf

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES

rop Consequently the field current cannot increase further and the generatorbuild up stops Hence we arrive at a very important conclusion that the voltage build up of the generator is given by the point of intersection of OCC and field resistance line Thus in Fig 117 (iii) D is point of intersection of the two curves Hence the generator will build up a voltage OM(ii) Series generator During initial operation with no current yet flowing a residual voltage will be generated exactly as in the case of a shunt generator The residual voltage will cause a current to flow through the whole series circuit when the circuit isclosed There will then be voltage build up to an equilibrium point exactlyanalogous to the buildup of a shunt generator The voltage build up graph willbe similar to that of shunt generator except that now load current (instead offield current for shunt generator) will be taken along x-axis10487871048787(iii) Compound generator When a compound generator has its series field flux aiding its shunt field flux the machine is said to be cumulative compound When the series field isconnected in reverse so that its field flux opposes the shunt field flux thegenerator is then differential compound The easiest way to build up voltage in a compound generator is to start under no load conditions At no load only the shunt field is effective When no-load voltage build up is achieved the generator is loaded If under load the voltagerises the series field connection is cumulative If the voltage drops significantlythe connection is differential compound118 Critical Field Resistance for a Shunt Generator The voltage build up in a shunt generator depends upon field circuit resistance If the field circuit resistance is R1 (line OA) then generator will build up a voltage OM as shown in Fig 118 If the field circuit resistance is increasedto R2 (tine OB) the generator will build up a voltage OL slightly less than OM As the field circuit resistance is increased the slope of resistance line also increases When the field resistance line becomes tangent (line OC) to OCC the generator would just excite If the field circuit resistance is increased beyond this point (say line OD) the generator will fail to excite The field circuit resistance represented by line OC (tangent to OCC) is called critical field resistance RC for the shunt generator It may be defined as under The maximum field circuit resistance (for a given speed) with which the shunt generator would just excite is known as its critical field resistance It should be noted that shunt generator will build up voltage only if field circuit resistance is less than critical field resistance

Fig 118 Critical field resistance of shunt generator

119 Critical Resistance for a Series Generator

Fig 119 shows the voltage build up in a series generator Here R1 R2 etc represent the total circuit resistance (load resistance and field winding resistance) If the total circuit resistance is R1 then series generator will build up a voltage OL The line OC is tangent to OCC and represents the critical resistance RC for a series generator If the total resistance of the circuit is more than RC (say line OD) the generator will fail to build up voltage Note that Fig 119 is similar to Fig 118 with the following differences

(i) In Fig 118 R1 R2 etc represent the total field circuit resistanceHowever R1 R2 etc in Fig 119 represent the total circuit resistance(load resistance and series field winding resistance etc)

(ii) In Fig 118 field current alone is represented along X-axis However inFig 119 load current IL is represented along Y-axis Note that in aseries generator field current = load current IL

Fig 119 Critical resistance of a series generator

120 Characteristics of Series Generator Fig 120 (i) shows the connections of a series wound generator Since there is only one current (that which flows through the whole machine) the load current is the same as the exciting current(i) OCCCurve 1 shows the open circuit characteristic (OCC) of a series generator Itcan be obtained experimentally by disconnecting the field winding from themachine and exciting it from a separate dc source(ii) Internal characteristic Curve 2 shows the total or internal characteristic of a series generator It gives the relation between the generated emf E on load and armature current Due to armature reaction the flux in the machine will be less than the flux at no load Hence emf E generated under load conditions will be less than the emf E0 generated under no load conditions Consequently internal characteristic curvelies below the OCC curve the difference between them representing the effectof armature reaction [See Fig 120 (ii)]

Fig 120 Characteristics of series generator

(iii) External characteristic

Curve 3 shows the external characteristic of a series generator It gives therelation between terminal voltage and load current ILV = E g- Ia (Ra +Rse )Therefore external characteristic curve will lie below internal characteristiccurve by an amount equal to ohmic drop [ie Ia(Ra + Rse)] in the machine asshown in Fig 120 (ii) The internal and external characteristics ofa dc series generator can be plotted from one another as shown in Fig 121 Supposewe are given the internal characteristic of the generator Let the line OC represent the resistance of the whole machine ie Ra +Rse If the load current is OB drop in the machine is AB ieAB = Ohmic drop in the machine = OB(Ra + Rse)

Fig 121 Load characteristics of series generator

Now raise a perpendicular from point B and mark a point b on this line such that ab = AB Then point b will lie on the external characteristic of the generator

Following similar procedure other points of external characteristic can be located It is easy to see that we can also plot internal characteristic from the external characteristic

121 Characteristics of a Shunt Generator Fig 122 (i) show the connections of a shunt wound generator The armature current Ia splits up into two parts a small fraction Ish flowing through shunt field winding while the major part IL goes to the external load(i) OCC The OCC of a shunt generator is similar in shape to that of a series generator as shown in Fig 122 (ii) The line OA represents the shunt field circuitresistance When the generator is run at normal speed it will build up a voltageOM At no-load the terminal voltage of the generator will be constant (= OM)represented by the horizontal dotted line MC

Fig 122 Characteristics of a Shunt Generator

(ii) Internal characteristic

When the generator is loaded flux per pole is reduced due to armature reaction Therefore emf E generated on load is less than the emf generated at no load As a result the internal characteristic (EIa) drops down slightly as shown in Fig122 (ii)

(iii) External characteristic Curve 2 shows the external characteristic of a shunt generator It gives therelation between terminal voltage V and load current IL

Therefore external characteristic curve will lie below the internal characteristic curve by an amount equal to drop in the armature circuit[ie (IL + Ish)Ra] as shown in Fig 122 (ii)

Note It may be seen from the external characteristic that change in terminalvoltage from no-load to full load is small The terminal voltage can always bemaintained constant by adjusting the field rheostat R automatically122 Critical External Resistance for Shunt Generator

If the load resistance across the terminals of a shunt generator is decreased then load current increase However there is a limit to the increase in

load current with the decrease of load resistance Any decrease of load resistance beyond this point instead of increasing the current ultimately results in reduced current Consequently the external characteristic turns back (dotted curve) as shown in Fig 123 The tangent OA to the curve represents the minimum external resistance required to excite the shunt generator on load and is called critical external resistance If the resistance of the external circuit is less than the critical external resistance (represented by tangent OA in Fig 310) the machine will refuse to excite or will de-excite if already running This means that external resistance is so low as virtually to short circuit the machine and so doing away with its excitation

Note There are two critical resistances for a shunt generator viz (i) criticalfield resistance (ii) critical external resistance For the shunt generator to buildup voltage the former should not be exceeded and the latter must not be gonebelow

123 How to Draw OCC at Different Speeds

If we are given OCC of a generator at a constant speed N1 then we can easily draw the OCC at any other constant speed N2 Fig 123 illustrates theprocedure Here we are given OCC at a constant speed N1 It is desired to findthe OCC at constant speed N2 (it is assumed that n1 lt N2) For constantexcitation E propN

E2

E1 =

N 2

N1 so

E2 iquest E1

N2

N1

As shown in Fig (311) for If = OH E1 = HC Therefore the new value of emf (E2) for the same If but at N2

E2= HCN 2

N 1 = HD

Fig 123 Plotting OCC at another speed

This locates the point D on the new OCC at N2 Similarly other points can belocated taking different values of If The locus of these points will be the OCCat N2

124 Critical Speed (NC)

The critical speed of a shunt generator is the minimum speed below which it fails to excite Clearly it is the speed for which the given shunt field resistance represents the critical resistance In Fig 124 curve 2 corresponds to critical

speed because the shunt field resistance (Rsh) line is tangential to it If the generator runs at full speed N the new OCC moves upward and the Rsh line represents critical resistance for this speed

In order to find critical speed take any convenient point C on excitation axis and erect a perpendicular so as to cut Rsh and Rsh lines at points B and A respectively Then

BCAC

=NcN

Nc=N BCAC

Fig 124 Finding critical speed

125 Conditions for Voltage Build-Up of a Shunt Generator

The necessary conditions for voltage build-up in a shunt generator are

(i) There must be some residual magnetism in generator poles(ii) The connections of the field winding should be such that the field current strengthens the residual magnetism(iii) The resistance of the field circuit should be less than the critical resistance In other words the speed of the generator should be higher

than the critical speed

126 Compound Generator Characteristics

In a compound generator both series and shunt excitation are combined as shown in Fig 123 The shunt winding can be connected either across the armature only (short-shunt connection S) or across armature plus series field (long-shunt connection G) The compound generator can be cumulatively compounded or

differentially compounded generator The latter is rarely used in practice Therefore we shall discuss the characteristics of cumulatively compounded

generator It may be noted that external characteristics of long and short shunt compound generators are almost identical

External characteristic

Fig 124 shows the external characteristics of a cumulatively compoundedgenerator The series excitation aids the shunt excitation The degree of compounding depends upon the increase in series excitation with the increase inload current

(i) If series winding turns are so adjusted that with the increase in load current the terminal voltage increases it is called over-compounded generator In such a case as the load current increases the series field mmf increases and tends to increase the flux and hence the generated voltage The increase in generated voltage is greater than the IaRa drop so that instead of decreasing the terminal voltage increases as shown by curve A in Fig 124

(ii) If series winding turns are so adjusted that with the increase in load current the terminal voltage substantially remains constant it is called Flat-compounded generator The series winding of such a machine has lesser number of turns than the one in over-compounded machine and therefore does not increase the flux as much for a given load current Consequently the full-load voltage is nearly equal to the no-load voltage as indicated by curve B in Fig 124 (iii) If series field winding has lesser number of turns than for a flat compounded machine the terminal voltage falls with increase in load current as indicated by curve C m Fig 124 Such a machine is called under-compounded generator

Fig 123 compound generator

Fig 124 Load characteristics of compound generators

127 LOSSES

The losses in a dc machine (generator or motor) may be divided into three classes viz (i) copper losses or electrical losses (ii) iron or core losses and (iii) mechanical losses All these losses appear as heat and thus raise the temperature of the machine They also lower the efficiency of the machine The loss structure maybe demonstrated as

1Copperlosses

These losses occur due to currents in the various windings of the machine When an electric current of I amps flows in a resistance of R ohms the heat energy is lost at the rate of I2R joules sec and the power loss is I2R watts Generators and motors have one or more field circuits and an armature circuit in which such losses occur All resistance losses of this kind are called copper losses The different copper losses occurring in a dc machine are

(i) Armature copper loss = Ia2Ra

(ii) Shunt field copper loss = Ish2 Rsh

(iii) Series field copper loss = Ise2Rse

Armature copper loss will be around 30 to 40 of full load losses where as total field copper losses will constitute around 30 of full load losses

Note There is also brush contact loss due to brush contact resistance (ie resistance between the surface of brush and surface of commutator) This loss is generally included in armature copper loss The brush contact drop is almost independent of Ia For carbon brushes this drop is around 1 V per brush The power loss due to brush contact resistance is 2ebIa where eb is the voltage drop at one brush

2 IRONORCORELOSSES These losses occur in the armature of a dc machine and are due to the rotation of armature in the magnetic field of the poles They are of two types viz (i) hysteresis loss (ii) eddy current loss

(i) Hysteresis loss

Hysteresis loss occurs in the armature of the dc machine since any given part of the armature is subjected to magnetic field reversals as it passes under successive poles Fig 124 shows an armature rotating in two-pole machine Consider a small piece ab of the armature When the piece ab is under N-pole the magnetic lines pass from a to b Half a revolution later the same piece of iron is under S-pole and magnetic lines pass from b to a so that magnetism in the iron is reversed In order to reverse continuously the molecular magnets in the armature core some amount of power has to be spent which is called hysteresis loss It is given by Steinmetz formula This formula is

Hysteresis Loss Wh = η Bmax 16f V Watts Bmax - Maximum flux density in the core f - Frequency of magnetic reversals V - Volume of armature in m3

η - Steinmetz hysteresis coefficient

In order to reduce this loss in a dc machine armature core is made of such materials which have a low value of Steinmetz hysteresis co-efficient eg silicon steel

Fig 124 Occurrence of hysteresis loss

ii) EDDYCURRENTLOSS

In addition to the voltages induced in the armature conductors there are also voltages induced in the armature core These voltages produce circulating currents in the armature core as shown in Fig 125 These are called eddy currents and power loss due to their flow is called eddy current loss The eddy current loss appears as heat which raises the temperature of the machine and lowers its efficiency Core resistance can be greatly increased by constructing the core of thin round If a continuous solid iron core is used the resistance to eddy current path will be small due to large cross-sectional area of the core Consequently the magnitude of eddy current and hence eddy current loss will be large The magnitude of eddy current can be reduced by making core resistance as high as practical The iron sheets called laminations [See Fig 126] The laminations are insulated from each other with a coating of varnish The insulating coating has a high resistance so very little current flows from one lamination to the other Alsobecause each lamination is very thin the resistance to current flowing through the width of a lamination is also quite large Thus laminating a core increases the core resistance which decreases the eddy current and hence the eddy current loss

Fig 125 Eddy current loss Fig 126 Core laminations

Eddy current loss is biven by the formula Pe= Ke Bmax

2 f2t2 V Watts

Where Ke = Constant depending upon electrical resistace of core Bmax = Maximum flux density in wbm2

f = frequency f magnetic reversals in Hz t = thickness of laminations in m and V = volume of core in m3

It may be noted that eddy current loss depends upon the square of lamination thickness For this reason lamination thickness should be kept as small as possible

3MECHANICALLOSSES

These losses are due to friction and windage(i) Friction losses eg bearing friction brush friction etc(ii) Windage loss ie air friction of rotating armature

These losses depend upon the speed of the machine But for a given speed they are practically constant

Note Iron losses and mechanical losses together are called stray losses

CONSTANT AND VARIABLE LOSSES

The losses in a dc generator (or dc motor) may be sub-divided into

(i) constant losses (ii) variable losses

(i) Constant losses

Those losses in a dc generator which remain constant at all loads are known as constant losses The constant losses in a dc generator are(a) iron losses(b) mechanical losses(c) shunt field losses(ii) Variable lossesThose losses in a dc generator which vary with load are called variable lossesThe variable losses in a dc generator are (a) Copper loss in armature winding (Ia

2 Ra) (b) Copper loss in series field winding (Ise

2Rse)

Total losses = Constant losses + Variable lossesNote Field Cu loss is constant for shunt and compound generators

128 POWER FLOW DIAGRAM

The power flow in a dc generator may be represented as

The different power stages are shown as A B and CA ndash The mechanical power input which is the power output of the prime mover is given as Pin = 2πNT

60 watts

B - The power developed in armature is Pa = Eg Ia watts C- The electrical power output is Pout = VIL watts

129 EFFICIENCY Efficiency of a machine is the ratio of its output power to the input power

Efficiency η = Output powerInput power

= Input powerminusLossesInput power = Output power

Output power+Losses

In case of a dc generator depending upon the power stages three different efficiencies can be defined as-

1 Mechanical Efficiency(ηm)

BA = Electrical power developedisinarmature

Mechanical power input2 Electrical Efficiency(ηe)

CB = Electrical power output

Electrical power developedisinthe armature 3 Overall or commercial efficiency (η)

CA = Electrical power output

Mechanical power input = ηmtimes ηe

130 CONDITION FOR MAXIMUM EFFICIENCY The efficiency of a dc generator is not constant but varies with load Fig 127shows the variation of efficiency with load current Consider a dc shunt generator with terminal voltage V delivering a load current of IL amp Generator output = V IL

Generator input = Output + Losses = V IL + constant losses + Variable losses = V IL + Wc+ Ia

2Ra

= V IL + Wc + (IL+ Ish)2 Rsh

The shunt field current Ish is small compared with IL so may be neglected

Generator input = VIL+ IL2Ra + Wc

Now efficiency η iquest outputinput = VI L

VIL+ Ia 2Ra+Wc

Efficiency is maximum when the denominator of the above equation is minimumie

ddIL ( VI L

VIL+ Ia 2Ra+Wc ) = 0

Which gives IL2 Ra = WC

ie Variable loss = constant lossThe load current for maximum efficiency is given by

IL iquest radicWcRa

Fig 127 Variation of efficiency with load

Hence the efficiency is maximum for that load current at which the variable loss is equal to the constant loss

131 APPLICATIONS

Separately excited generators

These types of generators have special applications such as electroplating electro refining of materials because separate supply is required for field excitation

Shunt generators

These types of generators are used in battery charging and ordinary lighting purposes

Series generators

This type of generators is used commonly as boosters on DC feeders as current generators for welding and arc lamp

Cumulatively compounded generators

This type of generators is used for domestic lighting purposes and to transmit energy over long distances

Differential compound generators

This type of generators is rarely used They are used for special applications such as electric arc welding

132 NUMERICAL EXCERCISES

1 The emf induced in the armature of a 450 kW 250 volt shunt generator is

2588volt when the field current is 200 amp and the generator is supplying power to a load at rated terminal voltage The armature circuit resistance is 0005 ohm Determine (i) load current (ii) power generated (iii) power output and (iv) electrical efficiency Neglect brush contact drop

Solution

Let the load current be I amp Given Ish = 20 A

Armature current = Load current + field current or Ia = I + 20 Induced emf = Terminal voltage + Ia Ra drop

2588 = 250 + (I + 20) times 0005

So (i) Load current I = 1740 amp

(ii) Power generated E Ia = 2588 (1740 + 20) watts = 455488 kW (iii) Power output V I = 250 times 1740 watts

= 435 kW

(iv) Electrical efficiency = 435455488 times 100 = 955

2 A 4 pole wave wound dc shunt generator delivers a load of 45 kW at a terminal voltage of 220 volts Its armature has 150 single turn coils and has a resistance of 001 ohm The air gap flux per pole is 002 weber Shunt field resistance is 50 ohm Calculate the speed at which it is being driven Neglect brush contact drop

Solution Load current I = W

V = 45000220 = 20455 amp

Field Current Ish = V

R s h = 220

50 = 44 amp

Armature current Ia = I +I sh = 20455 +44 = 20895 amp

Armature drop I aRa = 20895 X 001 = 20895 volts Induced emf Eg = V + Ia Ra = 220 + 20895 = 2220895 volts

But induced emf

So N = Eg60 AΦ ZP = 2220895times 60 times2

002 times150 times2 times 4 = 111045 rpm

3 The following data refer to the OCC of a DC separately excited generator at 1000 rpm

Field current in A 00 02 03 04 05 06 07 08 09 10 11 12Armature voltage in V 5 40 75 100 124 145 162 178 188 195 200 205

The machine is now connected as a shunt generator with a total field resistance of 200 Ω and driven at 1000 rpm Estimate graphically (i) the voltage to which the generator will build up at no load (ii) the armature field and load currents when the terminal voltage is found to be 150 V Neglect the effect of armature reaction and brush drop and assume armature resistance Ra to be 08 Ω (iii) Finally estimate the steady state armature current when the machine terminals are shorted

Solution

First of all the OCC of the generator is plotted in a graph paper as shown in figure

(i) The total field circuit resistance is given to be Rf = 200 Ω The Rf line is now drawn on the same graph paper passing through the origin The point of intersection of the Rf line and the OCC decides the final no load voltage and can be read from the graph as 192 V and the corresponding field current is 096 A (ii) Since the terminal voltage V is 150 V (= BM) field current If

is OM = 077 A Generated voltage E is given by AM But we know E = IaRa + V Hence IaRa= E ndash V = AM ndash BM = AB Now from the graph AB = 25 V there4Ia = 2508 = 3125 A So load current IL= Ia ndash If= 3125 ndash 077 = 3048 A

ADDITIONAL PROBLEMS

1 A 4 pole lap wound 115 kW 230 volts dc shunt generator has armatureresistance of 02 ohm and field resistance of 100 ohms Calculate (i) emf

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES

Fig 118 Critical field resistance of shunt generator

119 Critical Resistance for a Series Generator

Fig 119 shows the voltage build up in a series generator Here R1 R2 etc represent the total circuit resistance (load resistance and field winding resistance) If the total circuit resistance is R1 then series generator will build up a voltage OL The line OC is tangent to OCC and represents the critical resistance RC for a series generator If the total resistance of the circuit is more than RC (say line OD) the generator will fail to build up voltage Note that Fig 119 is similar to Fig 118 with the following differences

(i) In Fig 118 R1 R2 etc represent the total field circuit resistanceHowever R1 R2 etc in Fig 119 represent the total circuit resistance(load resistance and series field winding resistance etc)

(ii) In Fig 118 field current alone is represented along X-axis However inFig 119 load current IL is represented along Y-axis Note that in aseries generator field current = load current IL

Fig 119 Critical resistance of a series generator

120 Characteristics of Series Generator Fig 120 (i) shows the connections of a series wound generator Since there is only one current (that which flows through the whole machine) the load current is the same as the exciting current(i) OCCCurve 1 shows the open circuit characteristic (OCC) of a series generator Itcan be obtained experimentally by disconnecting the field winding from themachine and exciting it from a separate dc source(ii) Internal characteristic Curve 2 shows the total or internal characteristic of a series generator It gives the relation between the generated emf E on load and armature current Due to armature reaction the flux in the machine will be less than the flux at no load Hence emf E generated under load conditions will be less than the emf E0 generated under no load conditions Consequently internal characteristic curvelies below the OCC curve the difference between them representing the effectof armature reaction [See Fig 120 (ii)]

Fig 120 Characteristics of series generator

(iii) External characteristic

Curve 3 shows the external characteristic of a series generator It gives therelation between terminal voltage and load current ILV = E g- Ia (Ra +Rse )Therefore external characteristic curve will lie below internal characteristiccurve by an amount equal to ohmic drop [ie Ia(Ra + Rse)] in the machine asshown in Fig 120 (ii) The internal and external characteristics ofa dc series generator can be plotted from one another as shown in Fig 121 Supposewe are given the internal characteristic of the generator Let the line OC represent the resistance of the whole machine ie Ra +Rse If the load current is OB drop in the machine is AB ieAB = Ohmic drop in the machine = OB(Ra + Rse)

Fig 121 Load characteristics of series generator

Now raise a perpendicular from point B and mark a point b on this line such that ab = AB Then point b will lie on the external characteristic of the generator

Following similar procedure other points of external characteristic can be located It is easy to see that we can also plot internal characteristic from the external characteristic

121 Characteristics of a Shunt Generator Fig 122 (i) show the connections of a shunt wound generator The armature current Ia splits up into two parts a small fraction Ish flowing through shunt field winding while the major part IL goes to the external load(i) OCC The OCC of a shunt generator is similar in shape to that of a series generator as shown in Fig 122 (ii) The line OA represents the shunt field circuitresistance When the generator is run at normal speed it will build up a voltageOM At no-load the terminal voltage of the generator will be constant (= OM)represented by the horizontal dotted line MC

Fig 122 Characteristics of a Shunt Generator

(ii) Internal characteristic

When the generator is loaded flux per pole is reduced due to armature reaction Therefore emf E generated on load is less than the emf generated at no load As a result the internal characteristic (EIa) drops down slightly as shown in Fig122 (ii)

(iii) External characteristic Curve 2 shows the external characteristic of a shunt generator It gives therelation between terminal voltage V and load current IL

Therefore external characteristic curve will lie below the internal characteristic curve by an amount equal to drop in the armature circuit[ie (IL + Ish)Ra] as shown in Fig 122 (ii)

Note It may be seen from the external characteristic that change in terminalvoltage from no-load to full load is small The terminal voltage can always bemaintained constant by adjusting the field rheostat R automatically122 Critical External Resistance for Shunt Generator

If the load resistance across the terminals of a shunt generator is decreased then load current increase However there is a limit to the increase in

load current with the decrease of load resistance Any decrease of load resistance beyond this point instead of increasing the current ultimately results in reduced current Consequently the external characteristic turns back (dotted curve) as shown in Fig 123 The tangent OA to the curve represents the minimum external resistance required to excite the shunt generator on load and is called critical external resistance If the resistance of the external circuit is less than the critical external resistance (represented by tangent OA in Fig 310) the machine will refuse to excite or will de-excite if already running This means that external resistance is so low as virtually to short circuit the machine and so doing away with its excitation

Note There are two critical resistances for a shunt generator viz (i) criticalfield resistance (ii) critical external resistance For the shunt generator to buildup voltage the former should not be exceeded and the latter must not be gonebelow

123 How to Draw OCC at Different Speeds

If we are given OCC of a generator at a constant speed N1 then we can easily draw the OCC at any other constant speed N2 Fig 123 illustrates theprocedure Here we are given OCC at a constant speed N1 It is desired to findthe OCC at constant speed N2 (it is assumed that n1 lt N2) For constantexcitation E propN

E2

E1 =

N 2

N1 so

E2 iquest E1

N2

N1

As shown in Fig (311) for If = OH E1 = HC Therefore the new value of emf (E2) for the same If but at N2

E2= HCN 2

N 1 = HD

Fig 123 Plotting OCC at another speed

This locates the point D on the new OCC at N2 Similarly other points can belocated taking different values of If The locus of these points will be the OCCat N2

124 Critical Speed (NC)

The critical speed of a shunt generator is the minimum speed below which it fails to excite Clearly it is the speed for which the given shunt field resistance represents the critical resistance In Fig 124 curve 2 corresponds to critical

speed because the shunt field resistance (Rsh) line is tangential to it If the generator runs at full speed N the new OCC moves upward and the Rsh line represents critical resistance for this speed

In order to find critical speed take any convenient point C on excitation axis and erect a perpendicular so as to cut Rsh and Rsh lines at points B and A respectively Then

BCAC

=NcN

Nc=N BCAC

Fig 124 Finding critical speed

125 Conditions for Voltage Build-Up of a Shunt Generator

The necessary conditions for voltage build-up in a shunt generator are

(i) There must be some residual magnetism in generator poles(ii) The connections of the field winding should be such that the field current strengthens the residual magnetism(iii) The resistance of the field circuit should be less than the critical resistance In other words the speed of the generator should be higher

than the critical speed

126 Compound Generator Characteristics

In a compound generator both series and shunt excitation are combined as shown in Fig 123 The shunt winding can be connected either across the armature only (short-shunt connection S) or across armature plus series field (long-shunt connection G) The compound generator can be cumulatively compounded or

differentially compounded generator The latter is rarely used in practice Therefore we shall discuss the characteristics of cumulatively compounded

generator It may be noted that external characteristics of long and short shunt compound generators are almost identical

External characteristic

Fig 124 shows the external characteristics of a cumulatively compoundedgenerator The series excitation aids the shunt excitation The degree of compounding depends upon the increase in series excitation with the increase inload current

(i) If series winding turns are so adjusted that with the increase in load current the terminal voltage increases it is called over-compounded generator In such a case as the load current increases the series field mmf increases and tends to increase the flux and hence the generated voltage The increase in generated voltage is greater than the IaRa drop so that instead of decreasing the terminal voltage increases as shown by curve A in Fig 124

(ii) If series winding turns are so adjusted that with the increase in load current the terminal voltage substantially remains constant it is called Flat-compounded generator The series winding of such a machine has lesser number of turns than the one in over-compounded machine and therefore does not increase the flux as much for a given load current Consequently the full-load voltage is nearly equal to the no-load voltage as indicated by curve B in Fig 124 (iii) If series field winding has lesser number of turns than for a flat compounded machine the terminal voltage falls with increase in load current as indicated by curve C m Fig 124 Such a machine is called under-compounded generator

Fig 123 compound generator

Fig 124 Load characteristics of compound generators

127 LOSSES

The losses in a dc machine (generator or motor) may be divided into three classes viz (i) copper losses or electrical losses (ii) iron or core losses and (iii) mechanical losses All these losses appear as heat and thus raise the temperature of the machine They also lower the efficiency of the machine The loss structure maybe demonstrated as

1Copperlosses

These losses occur due to currents in the various windings of the machine When an electric current of I amps flows in a resistance of R ohms the heat energy is lost at the rate of I2R joules sec and the power loss is I2R watts Generators and motors have one or more field circuits and an armature circuit in which such losses occur All resistance losses of this kind are called copper losses The different copper losses occurring in a dc machine are

(i) Armature copper loss = Ia2Ra

(ii) Shunt field copper loss = Ish2 Rsh

(iii) Series field copper loss = Ise2Rse

Armature copper loss will be around 30 to 40 of full load losses where as total field copper losses will constitute around 30 of full load losses

Note There is also brush contact loss due to brush contact resistance (ie resistance between the surface of brush and surface of commutator) This loss is generally included in armature copper loss The brush contact drop is almost independent of Ia For carbon brushes this drop is around 1 V per brush The power loss due to brush contact resistance is 2ebIa where eb is the voltage drop at one brush

2 IRONORCORELOSSES These losses occur in the armature of a dc machine and are due to the rotation of armature in the magnetic field of the poles They are of two types viz (i) hysteresis loss (ii) eddy current loss

(i) Hysteresis loss

Hysteresis loss occurs in the armature of the dc machine since any given part of the armature is subjected to magnetic field reversals as it passes under successive poles Fig 124 shows an armature rotating in two-pole machine Consider a small piece ab of the armature When the piece ab is under N-pole the magnetic lines pass from a to b Half a revolution later the same piece of iron is under S-pole and magnetic lines pass from b to a so that magnetism in the iron is reversed In order to reverse continuously the molecular magnets in the armature core some amount of power has to be spent which is called hysteresis loss It is given by Steinmetz formula This formula is

Hysteresis Loss Wh = η Bmax 16f V Watts Bmax - Maximum flux density in the core f - Frequency of magnetic reversals V - Volume of armature in m3

η - Steinmetz hysteresis coefficient

In order to reduce this loss in a dc machine armature core is made of such materials which have a low value of Steinmetz hysteresis co-efficient eg silicon steel

Fig 124 Occurrence of hysteresis loss

ii) EDDYCURRENTLOSS

In addition to the voltages induced in the armature conductors there are also voltages induced in the armature core These voltages produce circulating currents in the armature core as shown in Fig 125 These are called eddy currents and power loss due to their flow is called eddy current loss The eddy current loss appears as heat which raises the temperature of the machine and lowers its efficiency Core resistance can be greatly increased by constructing the core of thin round If a continuous solid iron core is used the resistance to eddy current path will be small due to large cross-sectional area of the core Consequently the magnitude of eddy current and hence eddy current loss will be large The magnitude of eddy current can be reduced by making core resistance as high as practical The iron sheets called laminations [See Fig 126] The laminations are insulated from each other with a coating of varnish The insulating coating has a high resistance so very little current flows from one lamination to the other Alsobecause each lamination is very thin the resistance to current flowing through the width of a lamination is also quite large Thus laminating a core increases the core resistance which decreases the eddy current and hence the eddy current loss

Fig 125 Eddy current loss Fig 126 Core laminations

Eddy current loss is biven by the formula Pe= Ke Bmax

2 f2t2 V Watts

Where Ke = Constant depending upon electrical resistace of core Bmax = Maximum flux density in wbm2

f = frequency f magnetic reversals in Hz t = thickness of laminations in m and V = volume of core in m3

It may be noted that eddy current loss depends upon the square of lamination thickness For this reason lamination thickness should be kept as small as possible

3MECHANICALLOSSES

These losses are due to friction and windage(i) Friction losses eg bearing friction brush friction etc(ii) Windage loss ie air friction of rotating armature

These losses depend upon the speed of the machine But for a given speed they are practically constant

Note Iron losses and mechanical losses together are called stray losses

CONSTANT AND VARIABLE LOSSES

The losses in a dc generator (or dc motor) may be sub-divided into

(i) constant losses (ii) variable losses

(i) Constant losses

Those losses in a dc generator which remain constant at all loads are known as constant losses The constant losses in a dc generator are(a) iron losses(b) mechanical losses(c) shunt field losses(ii) Variable lossesThose losses in a dc generator which vary with load are called variable lossesThe variable losses in a dc generator are (a) Copper loss in armature winding (Ia

2 Ra) (b) Copper loss in series field winding (Ise

2Rse)

Total losses = Constant losses + Variable lossesNote Field Cu loss is constant for shunt and compound generators

128 POWER FLOW DIAGRAM

The power flow in a dc generator may be represented as

The different power stages are shown as A B and CA ndash The mechanical power input which is the power output of the prime mover is given as Pin = 2πNT

60 watts

B - The power developed in armature is Pa = Eg Ia watts C- The electrical power output is Pout = VIL watts

129 EFFICIENCY Efficiency of a machine is the ratio of its output power to the input power

Efficiency η = Output powerInput power

= Input powerminusLossesInput power = Output power

Output power+Losses

In case of a dc generator depending upon the power stages three different efficiencies can be defined as-

1 Mechanical Efficiency(ηm)

BA = Electrical power developedisinarmature

Mechanical power input2 Electrical Efficiency(ηe)

CB = Electrical power output

Electrical power developedisinthe armature 3 Overall or commercial efficiency (η)

CA = Electrical power output

Mechanical power input = ηmtimes ηe

130 CONDITION FOR MAXIMUM EFFICIENCY The efficiency of a dc generator is not constant but varies with load Fig 127shows the variation of efficiency with load current Consider a dc shunt generator with terminal voltage V delivering a load current of IL amp Generator output = V IL

Generator input = Output + Losses = V IL + constant losses + Variable losses = V IL + Wc+ Ia

2Ra

= V IL + Wc + (IL+ Ish)2 Rsh

The shunt field current Ish is small compared with IL so may be neglected

Generator input = VIL+ IL2Ra + Wc

Now efficiency η iquest outputinput = VI L

VIL+ Ia 2Ra+Wc

Efficiency is maximum when the denominator of the above equation is minimumie

ddIL ( VI L

VIL+ Ia 2Ra+Wc ) = 0

Which gives IL2 Ra = WC

ie Variable loss = constant lossThe load current for maximum efficiency is given by

IL iquest radicWcRa

Fig 127 Variation of efficiency with load

Hence the efficiency is maximum for that load current at which the variable loss is equal to the constant loss

131 APPLICATIONS

Separately excited generators

These types of generators have special applications such as electroplating electro refining of materials because separate supply is required for field excitation

Shunt generators

These types of generators are used in battery charging and ordinary lighting purposes

Series generators

This type of generators is used commonly as boosters on DC feeders as current generators for welding and arc lamp

Cumulatively compounded generators

This type of generators is used for domestic lighting purposes and to transmit energy over long distances

Differential compound generators

This type of generators is rarely used They are used for special applications such as electric arc welding

132 NUMERICAL EXCERCISES

1 The emf induced in the armature of a 450 kW 250 volt shunt generator is

2588volt when the field current is 200 amp and the generator is supplying power to a load at rated terminal voltage The armature circuit resistance is 0005 ohm Determine (i) load current (ii) power generated (iii) power output and (iv) electrical efficiency Neglect brush contact drop

Solution

Let the load current be I amp Given Ish = 20 A

Armature current = Load current + field current or Ia = I + 20 Induced emf = Terminal voltage + Ia Ra drop

2588 = 250 + (I + 20) times 0005

So (i) Load current I = 1740 amp

(ii) Power generated E Ia = 2588 (1740 + 20) watts = 455488 kW (iii) Power output V I = 250 times 1740 watts

= 435 kW

(iv) Electrical efficiency = 435455488 times 100 = 955

2 A 4 pole wave wound dc shunt generator delivers a load of 45 kW at a terminal voltage of 220 volts Its armature has 150 single turn coils and has a resistance of 001 ohm The air gap flux per pole is 002 weber Shunt field resistance is 50 ohm Calculate the speed at which it is being driven Neglect brush contact drop

Solution Load current I = W

V = 45000220 = 20455 amp

Field Current Ish = V

R s h = 220

50 = 44 amp

Armature current Ia = I +I sh = 20455 +44 = 20895 amp

Armature drop I aRa = 20895 X 001 = 20895 volts Induced emf Eg = V + Ia Ra = 220 + 20895 = 2220895 volts

But induced emf

So N = Eg60 AΦ ZP = 2220895times 60 times2

002 times150 times2 times 4 = 111045 rpm

3 The following data refer to the OCC of a DC separately excited generator at 1000 rpm

Field current in A 00 02 03 04 05 06 07 08 09 10 11 12Armature voltage in V 5 40 75 100 124 145 162 178 188 195 200 205

The machine is now connected as a shunt generator with a total field resistance of 200 Ω and driven at 1000 rpm Estimate graphically (i) the voltage to which the generator will build up at no load (ii) the armature field and load currents when the terminal voltage is found to be 150 V Neglect the effect of armature reaction and brush drop and assume armature resistance Ra to be 08 Ω (iii) Finally estimate the steady state armature current when the machine terminals are shorted

Solution

First of all the OCC of the generator is plotted in a graph paper as shown in figure

(i) The total field circuit resistance is given to be Rf = 200 Ω The Rf line is now drawn on the same graph paper passing through the origin The point of intersection of the Rf line and the OCC decides the final no load voltage and can be read from the graph as 192 V and the corresponding field current is 096 A (ii) Since the terminal voltage V is 150 V (= BM) field current If

is OM = 077 A Generated voltage E is given by AM But we know E = IaRa + V Hence IaRa= E ndash V = AM ndash BM = AB Now from the graph AB = 25 V there4Ia = 2508 = 3125 A So load current IL= Ia ndash If= 3125 ndash 077 = 3048 A

ADDITIONAL PROBLEMS

1 A 4 pole lap wound 115 kW 230 volts dc shunt generator has armatureresistance of 02 ohm and field resistance of 100 ohms Calculate (i) emf

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES

(ii) In Fig 118 field current alone is represented along X-axis However inFig 119 load current IL is represented along Y-axis Note that in aseries generator field current = load current IL

Fig 119 Critical resistance of a series generator

120 Characteristics of Series Generator Fig 120 (i) shows the connections of a series wound generator Since there is only one current (that which flows through the whole machine) the load current is the same as the exciting current(i) OCCCurve 1 shows the open circuit characteristic (OCC) of a series generator Itcan be obtained experimentally by disconnecting the field winding from themachine and exciting it from a separate dc source(ii) Internal characteristic Curve 2 shows the total or internal characteristic of a series generator It gives the relation between the generated emf E on load and armature current Due to armature reaction the flux in the machine will be less than the flux at no load Hence emf E generated under load conditions will be less than the emf E0 generated under no load conditions Consequently internal characteristic curvelies below the OCC curve the difference between them representing the effectof armature reaction [See Fig 120 (ii)]

Fig 120 Characteristics of series generator

(iii) External characteristic

Curve 3 shows the external characteristic of a series generator It gives therelation between terminal voltage and load current ILV = E g- Ia (Ra +Rse )Therefore external characteristic curve will lie below internal characteristiccurve by an amount equal to ohmic drop [ie Ia(Ra + Rse)] in the machine asshown in Fig 120 (ii) The internal and external characteristics ofa dc series generator can be plotted from one another as shown in Fig 121 Supposewe are given the internal characteristic of the generator Let the line OC represent the resistance of the whole machine ie Ra +Rse If the load current is OB drop in the machine is AB ieAB = Ohmic drop in the machine = OB(Ra + Rse)

Fig 121 Load characteristics of series generator

Now raise a perpendicular from point B and mark a point b on this line such that ab = AB Then point b will lie on the external characteristic of the generator

Following similar procedure other points of external characteristic can be located It is easy to see that we can also plot internal characteristic from the external characteristic

121 Characteristics of a Shunt Generator Fig 122 (i) show the connections of a shunt wound generator The armature current Ia splits up into two parts a small fraction Ish flowing through shunt field winding while the major part IL goes to the external load(i) OCC The OCC of a shunt generator is similar in shape to that of a series generator as shown in Fig 122 (ii) The line OA represents the shunt field circuitresistance When the generator is run at normal speed it will build up a voltageOM At no-load the terminal voltage of the generator will be constant (= OM)represented by the horizontal dotted line MC

Fig 122 Characteristics of a Shunt Generator

(ii) Internal characteristic

When the generator is loaded flux per pole is reduced due to armature reaction Therefore emf E generated on load is less than the emf generated at no load As a result the internal characteristic (EIa) drops down slightly as shown in Fig122 (ii)

(iii) External characteristic Curve 2 shows the external characteristic of a shunt generator It gives therelation between terminal voltage V and load current IL

Therefore external characteristic curve will lie below the internal characteristic curve by an amount equal to drop in the armature circuit[ie (IL + Ish)Ra] as shown in Fig 122 (ii)

Note It may be seen from the external characteristic that change in terminalvoltage from no-load to full load is small The terminal voltage can always bemaintained constant by adjusting the field rheostat R automatically122 Critical External Resistance for Shunt Generator

If the load resistance across the terminals of a shunt generator is decreased then load current increase However there is a limit to the increase in

load current with the decrease of load resistance Any decrease of load resistance beyond this point instead of increasing the current ultimately results in reduced current Consequently the external characteristic turns back (dotted curve) as shown in Fig 123 The tangent OA to the curve represents the minimum external resistance required to excite the shunt generator on load and is called critical external resistance If the resistance of the external circuit is less than the critical external resistance (represented by tangent OA in Fig 310) the machine will refuse to excite or will de-excite if already running This means that external resistance is so low as virtually to short circuit the machine and so doing away with its excitation

Note There are two critical resistances for a shunt generator viz (i) criticalfield resistance (ii) critical external resistance For the shunt generator to buildup voltage the former should not be exceeded and the latter must not be gonebelow

123 How to Draw OCC at Different Speeds

If we are given OCC of a generator at a constant speed N1 then we can easily draw the OCC at any other constant speed N2 Fig 123 illustrates theprocedure Here we are given OCC at a constant speed N1 It is desired to findthe OCC at constant speed N2 (it is assumed that n1 lt N2) For constantexcitation E propN

E2

E1 =

N 2

N1 so

E2 iquest E1

N2

N1

As shown in Fig (311) for If = OH E1 = HC Therefore the new value of emf (E2) for the same If but at N2

E2= HCN 2

N 1 = HD

Fig 123 Plotting OCC at another speed

This locates the point D on the new OCC at N2 Similarly other points can belocated taking different values of If The locus of these points will be the OCCat N2

124 Critical Speed (NC)

The critical speed of a shunt generator is the minimum speed below which it fails to excite Clearly it is the speed for which the given shunt field resistance represents the critical resistance In Fig 124 curve 2 corresponds to critical

speed because the shunt field resistance (Rsh) line is tangential to it If the generator runs at full speed N the new OCC moves upward and the Rsh line represents critical resistance for this speed

In order to find critical speed take any convenient point C on excitation axis and erect a perpendicular so as to cut Rsh and Rsh lines at points B and A respectively Then

BCAC

=NcN

Nc=N BCAC

Fig 124 Finding critical speed

125 Conditions for Voltage Build-Up of a Shunt Generator

The necessary conditions for voltage build-up in a shunt generator are

(i) There must be some residual magnetism in generator poles(ii) The connections of the field winding should be such that the field current strengthens the residual magnetism(iii) The resistance of the field circuit should be less than the critical resistance In other words the speed of the generator should be higher

than the critical speed

126 Compound Generator Characteristics

In a compound generator both series and shunt excitation are combined as shown in Fig 123 The shunt winding can be connected either across the armature only (short-shunt connection S) or across armature plus series field (long-shunt connection G) The compound generator can be cumulatively compounded or

differentially compounded generator The latter is rarely used in practice Therefore we shall discuss the characteristics of cumulatively compounded

generator It may be noted that external characteristics of long and short shunt compound generators are almost identical

External characteristic

Fig 124 shows the external characteristics of a cumulatively compoundedgenerator The series excitation aids the shunt excitation The degree of compounding depends upon the increase in series excitation with the increase inload current

(i) If series winding turns are so adjusted that with the increase in load current the terminal voltage increases it is called over-compounded generator In such a case as the load current increases the series field mmf increases and tends to increase the flux and hence the generated voltage The increase in generated voltage is greater than the IaRa drop so that instead of decreasing the terminal voltage increases as shown by curve A in Fig 124

(ii) If series winding turns are so adjusted that with the increase in load current the terminal voltage substantially remains constant it is called Flat-compounded generator The series winding of such a machine has lesser number of turns than the one in over-compounded machine and therefore does not increase the flux as much for a given load current Consequently the full-load voltage is nearly equal to the no-load voltage as indicated by curve B in Fig 124 (iii) If series field winding has lesser number of turns than for a flat compounded machine the terminal voltage falls with increase in load current as indicated by curve C m Fig 124 Such a machine is called under-compounded generator

Fig 123 compound generator

Fig 124 Load characteristics of compound generators

127 LOSSES

The losses in a dc machine (generator or motor) may be divided into three classes viz (i) copper losses or electrical losses (ii) iron or core losses and (iii) mechanical losses All these losses appear as heat and thus raise the temperature of the machine They also lower the efficiency of the machine The loss structure maybe demonstrated as

1Copperlosses

These losses occur due to currents in the various windings of the machine When an electric current of I amps flows in a resistance of R ohms the heat energy is lost at the rate of I2R joules sec and the power loss is I2R watts Generators and motors have one or more field circuits and an armature circuit in which such losses occur All resistance losses of this kind are called copper losses The different copper losses occurring in a dc machine are

(i) Armature copper loss = Ia2Ra

(ii) Shunt field copper loss = Ish2 Rsh

(iii) Series field copper loss = Ise2Rse

Armature copper loss will be around 30 to 40 of full load losses where as total field copper losses will constitute around 30 of full load losses

Note There is also brush contact loss due to brush contact resistance (ie resistance between the surface of brush and surface of commutator) This loss is generally included in armature copper loss The brush contact drop is almost independent of Ia For carbon brushes this drop is around 1 V per brush The power loss due to brush contact resistance is 2ebIa where eb is the voltage drop at one brush

2 IRONORCORELOSSES These losses occur in the armature of a dc machine and are due to the rotation of armature in the magnetic field of the poles They are of two types viz (i) hysteresis loss (ii) eddy current loss

(i) Hysteresis loss

Hysteresis loss occurs in the armature of the dc machine since any given part of the armature is subjected to magnetic field reversals as it passes under successive poles Fig 124 shows an armature rotating in two-pole machine Consider a small piece ab of the armature When the piece ab is under N-pole the magnetic lines pass from a to b Half a revolution later the same piece of iron is under S-pole and magnetic lines pass from b to a so that magnetism in the iron is reversed In order to reverse continuously the molecular magnets in the armature core some amount of power has to be spent which is called hysteresis loss It is given by Steinmetz formula This formula is

Hysteresis Loss Wh = η Bmax 16f V Watts Bmax - Maximum flux density in the core f - Frequency of magnetic reversals V - Volume of armature in m3

η - Steinmetz hysteresis coefficient

In order to reduce this loss in a dc machine armature core is made of such materials which have a low value of Steinmetz hysteresis co-efficient eg silicon steel

Fig 124 Occurrence of hysteresis loss

ii) EDDYCURRENTLOSS

In addition to the voltages induced in the armature conductors there are also voltages induced in the armature core These voltages produce circulating currents in the armature core as shown in Fig 125 These are called eddy currents and power loss due to their flow is called eddy current loss The eddy current loss appears as heat which raises the temperature of the machine and lowers its efficiency Core resistance can be greatly increased by constructing the core of thin round If a continuous solid iron core is used the resistance to eddy current path will be small due to large cross-sectional area of the core Consequently the magnitude of eddy current and hence eddy current loss will be large The magnitude of eddy current can be reduced by making core resistance as high as practical The iron sheets called laminations [See Fig 126] The laminations are insulated from each other with a coating of varnish The insulating coating has a high resistance so very little current flows from one lamination to the other Alsobecause each lamination is very thin the resistance to current flowing through the width of a lamination is also quite large Thus laminating a core increases the core resistance which decreases the eddy current and hence the eddy current loss

Fig 125 Eddy current loss Fig 126 Core laminations

Eddy current loss is biven by the formula Pe= Ke Bmax

2 f2t2 V Watts

Where Ke = Constant depending upon electrical resistace of core Bmax = Maximum flux density in wbm2

f = frequency f magnetic reversals in Hz t = thickness of laminations in m and V = volume of core in m3

It may be noted that eddy current loss depends upon the square of lamination thickness For this reason lamination thickness should be kept as small as possible

3MECHANICALLOSSES

These losses are due to friction and windage(i) Friction losses eg bearing friction brush friction etc(ii) Windage loss ie air friction of rotating armature

These losses depend upon the speed of the machine But for a given speed they are practically constant

Note Iron losses and mechanical losses together are called stray losses

CONSTANT AND VARIABLE LOSSES

The losses in a dc generator (or dc motor) may be sub-divided into

(i) constant losses (ii) variable losses

(i) Constant losses

Those losses in a dc generator which remain constant at all loads are known as constant losses The constant losses in a dc generator are(a) iron losses(b) mechanical losses(c) shunt field losses(ii) Variable lossesThose losses in a dc generator which vary with load are called variable lossesThe variable losses in a dc generator are (a) Copper loss in armature winding (Ia

2 Ra) (b) Copper loss in series field winding (Ise

2Rse)

Total losses = Constant losses + Variable lossesNote Field Cu loss is constant for shunt and compound generators

128 POWER FLOW DIAGRAM

The power flow in a dc generator may be represented as

The different power stages are shown as A B and CA ndash The mechanical power input which is the power output of the prime mover is given as Pin = 2πNT

60 watts

B - The power developed in armature is Pa = Eg Ia watts C- The electrical power output is Pout = VIL watts

129 EFFICIENCY Efficiency of a machine is the ratio of its output power to the input power

Efficiency η = Output powerInput power

= Input powerminusLossesInput power = Output power

Output power+Losses

In case of a dc generator depending upon the power stages three different efficiencies can be defined as-

1 Mechanical Efficiency(ηm)

BA = Electrical power developedisinarmature

Mechanical power input2 Electrical Efficiency(ηe)

CB = Electrical power output

Electrical power developedisinthe armature 3 Overall or commercial efficiency (η)

CA = Electrical power output

Mechanical power input = ηmtimes ηe

130 CONDITION FOR MAXIMUM EFFICIENCY The efficiency of a dc generator is not constant but varies with load Fig 127shows the variation of efficiency with load current Consider a dc shunt generator with terminal voltage V delivering a load current of IL amp Generator output = V IL

Generator input = Output + Losses = V IL + constant losses + Variable losses = V IL + Wc+ Ia

2Ra

= V IL + Wc + (IL+ Ish)2 Rsh

The shunt field current Ish is small compared with IL so may be neglected

Generator input = VIL+ IL2Ra + Wc

Now efficiency η iquest outputinput = VI L

VIL+ Ia 2Ra+Wc

Efficiency is maximum when the denominator of the above equation is minimumie

ddIL ( VI L

VIL+ Ia 2Ra+Wc ) = 0

Which gives IL2 Ra = WC

ie Variable loss = constant lossThe load current for maximum efficiency is given by

IL iquest radicWcRa

Fig 127 Variation of efficiency with load

Hence the efficiency is maximum for that load current at which the variable loss is equal to the constant loss

131 APPLICATIONS

Separately excited generators

These types of generators have special applications such as electroplating electro refining of materials because separate supply is required for field excitation

Shunt generators

These types of generators are used in battery charging and ordinary lighting purposes

Series generators

This type of generators is used commonly as boosters on DC feeders as current generators for welding and arc lamp

Cumulatively compounded generators

This type of generators is used for domestic lighting purposes and to transmit energy over long distances

Differential compound generators

This type of generators is rarely used They are used for special applications such as electric arc welding

132 NUMERICAL EXCERCISES

1 The emf induced in the armature of a 450 kW 250 volt shunt generator is

2588volt when the field current is 200 amp and the generator is supplying power to a load at rated terminal voltage The armature circuit resistance is 0005 ohm Determine (i) load current (ii) power generated (iii) power output and (iv) electrical efficiency Neglect brush contact drop

Solution

Let the load current be I amp Given Ish = 20 A

Armature current = Load current + field current or Ia = I + 20 Induced emf = Terminal voltage + Ia Ra drop

2588 = 250 + (I + 20) times 0005

So (i) Load current I = 1740 amp

(ii) Power generated E Ia = 2588 (1740 + 20) watts = 455488 kW (iii) Power output V I = 250 times 1740 watts

= 435 kW

(iv) Electrical efficiency = 435455488 times 100 = 955

2 A 4 pole wave wound dc shunt generator delivers a load of 45 kW at a terminal voltage of 220 volts Its armature has 150 single turn coils and has a resistance of 001 ohm The air gap flux per pole is 002 weber Shunt field resistance is 50 ohm Calculate the speed at which it is being driven Neglect brush contact drop

Solution Load current I = W

V = 45000220 = 20455 amp

Field Current Ish = V

R s h = 220

50 = 44 amp

Armature current Ia = I +I sh = 20455 +44 = 20895 amp

Armature drop I aRa = 20895 X 001 = 20895 volts Induced emf Eg = V + Ia Ra = 220 + 20895 = 2220895 volts

But induced emf

So N = Eg60 AΦ ZP = 2220895times 60 times2

002 times150 times2 times 4 = 111045 rpm

3 The following data refer to the OCC of a DC separately excited generator at 1000 rpm

Field current in A 00 02 03 04 05 06 07 08 09 10 11 12Armature voltage in V 5 40 75 100 124 145 162 178 188 195 200 205

The machine is now connected as a shunt generator with a total field resistance of 200 Ω and driven at 1000 rpm Estimate graphically (i) the voltage to which the generator will build up at no load (ii) the armature field and load currents when the terminal voltage is found to be 150 V Neglect the effect of armature reaction and brush drop and assume armature resistance Ra to be 08 Ω (iii) Finally estimate the steady state armature current when the machine terminals are shorted

Solution

First of all the OCC of the generator is plotted in a graph paper as shown in figure

(i) The total field circuit resistance is given to be Rf = 200 Ω The Rf line is now drawn on the same graph paper passing through the origin The point of intersection of the Rf line and the OCC decides the final no load voltage and can be read from the graph as 192 V and the corresponding field current is 096 A (ii) Since the terminal voltage V is 150 V (= BM) field current If

is OM = 077 A Generated voltage E is given by AM But we know E = IaRa + V Hence IaRa= E ndash V = AM ndash BM = AB Now from the graph AB = 25 V there4Ia = 2508 = 3125 A So load current IL= Ia ndash If= 3125 ndash 077 = 3048 A

ADDITIONAL PROBLEMS

1 A 4 pole lap wound 115 kW 230 volts dc shunt generator has armatureresistance of 02 ohm and field resistance of 100 ohms Calculate (i) emf

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES

Fig 120 Characteristics of series generator

(iii) External characteristic

Curve 3 shows the external characteristic of a series generator It gives therelation between terminal voltage and load current ILV = E g- Ia (Ra +Rse )Therefore external characteristic curve will lie below internal characteristiccurve by an amount equal to ohmic drop [ie Ia(Ra + Rse)] in the machine asshown in Fig 120 (ii) The internal and external characteristics ofa dc series generator can be plotted from one another as shown in Fig 121 Supposewe are given the internal characteristic of the generator Let the line OC represent the resistance of the whole machine ie Ra +Rse If the load current is OB drop in the machine is AB ieAB = Ohmic drop in the machine = OB(Ra + Rse)

Fig 121 Load characteristics of series generator

Now raise a perpendicular from point B and mark a point b on this line such that ab = AB Then point b will lie on the external characteristic of the generator

Following similar procedure other points of external characteristic can be located It is easy to see that we can also plot internal characteristic from the external characteristic

121 Characteristics of a Shunt Generator Fig 122 (i) show the connections of a shunt wound generator The armature current Ia splits up into two parts a small fraction Ish flowing through shunt field winding while the major part IL goes to the external load(i) OCC The OCC of a shunt generator is similar in shape to that of a series generator as shown in Fig 122 (ii) The line OA represents the shunt field circuitresistance When the generator is run at normal speed it will build up a voltageOM At no-load the terminal voltage of the generator will be constant (= OM)represented by the horizontal dotted line MC

Fig 122 Characteristics of a Shunt Generator

(ii) Internal characteristic

When the generator is loaded flux per pole is reduced due to armature reaction Therefore emf E generated on load is less than the emf generated at no load As a result the internal characteristic (EIa) drops down slightly as shown in Fig122 (ii)

(iii) External characteristic Curve 2 shows the external characteristic of a shunt generator It gives therelation between terminal voltage V and load current IL

Therefore external characteristic curve will lie below the internal characteristic curve by an amount equal to drop in the armature circuit[ie (IL + Ish)Ra] as shown in Fig 122 (ii)

Note It may be seen from the external characteristic that change in terminalvoltage from no-load to full load is small The terminal voltage can always bemaintained constant by adjusting the field rheostat R automatically122 Critical External Resistance for Shunt Generator

If the load resistance across the terminals of a shunt generator is decreased then load current increase However there is a limit to the increase in

load current with the decrease of load resistance Any decrease of load resistance beyond this point instead of increasing the current ultimately results in reduced current Consequently the external characteristic turns back (dotted curve) as shown in Fig 123 The tangent OA to the curve represents the minimum external resistance required to excite the shunt generator on load and is called critical external resistance If the resistance of the external circuit is less than the critical external resistance (represented by tangent OA in Fig 310) the machine will refuse to excite or will de-excite if already running This means that external resistance is so low as virtually to short circuit the machine and so doing away with its excitation

Note There are two critical resistances for a shunt generator viz (i) criticalfield resistance (ii) critical external resistance For the shunt generator to buildup voltage the former should not be exceeded and the latter must not be gonebelow

123 How to Draw OCC at Different Speeds

If we are given OCC of a generator at a constant speed N1 then we can easily draw the OCC at any other constant speed N2 Fig 123 illustrates theprocedure Here we are given OCC at a constant speed N1 It is desired to findthe OCC at constant speed N2 (it is assumed that n1 lt N2) For constantexcitation E propN

E2

E1 =

N 2

N1 so

E2 iquest E1

N2

N1

As shown in Fig (311) for If = OH E1 = HC Therefore the new value of emf (E2) for the same If but at N2

E2= HCN 2

N 1 = HD

Fig 123 Plotting OCC at another speed

This locates the point D on the new OCC at N2 Similarly other points can belocated taking different values of If The locus of these points will be the OCCat N2

124 Critical Speed (NC)

The critical speed of a shunt generator is the minimum speed below which it fails to excite Clearly it is the speed for which the given shunt field resistance represents the critical resistance In Fig 124 curve 2 corresponds to critical

speed because the shunt field resistance (Rsh) line is tangential to it If the generator runs at full speed N the new OCC moves upward and the Rsh line represents critical resistance for this speed

In order to find critical speed take any convenient point C on excitation axis and erect a perpendicular so as to cut Rsh and Rsh lines at points B and A respectively Then

BCAC

=NcN

Nc=N BCAC

Fig 124 Finding critical speed

125 Conditions for Voltage Build-Up of a Shunt Generator

The necessary conditions for voltage build-up in a shunt generator are

(i) There must be some residual magnetism in generator poles(ii) The connections of the field winding should be such that the field current strengthens the residual magnetism(iii) The resistance of the field circuit should be less than the critical resistance In other words the speed of the generator should be higher

than the critical speed

126 Compound Generator Characteristics

In a compound generator both series and shunt excitation are combined as shown in Fig 123 The shunt winding can be connected either across the armature only (short-shunt connection S) or across armature plus series field (long-shunt connection G) The compound generator can be cumulatively compounded or

differentially compounded generator The latter is rarely used in practice Therefore we shall discuss the characteristics of cumulatively compounded

generator It may be noted that external characteristics of long and short shunt compound generators are almost identical

External characteristic

Fig 124 shows the external characteristics of a cumulatively compoundedgenerator The series excitation aids the shunt excitation The degree of compounding depends upon the increase in series excitation with the increase inload current

(i) If series winding turns are so adjusted that with the increase in load current the terminal voltage increases it is called over-compounded generator In such a case as the load current increases the series field mmf increases and tends to increase the flux and hence the generated voltage The increase in generated voltage is greater than the IaRa drop so that instead of decreasing the terminal voltage increases as shown by curve A in Fig 124

(ii) If series winding turns are so adjusted that with the increase in load current the terminal voltage substantially remains constant it is called Flat-compounded generator The series winding of such a machine has lesser number of turns than the one in over-compounded machine and therefore does not increase the flux as much for a given load current Consequently the full-load voltage is nearly equal to the no-load voltage as indicated by curve B in Fig 124 (iii) If series field winding has lesser number of turns than for a flat compounded machine the terminal voltage falls with increase in load current as indicated by curve C m Fig 124 Such a machine is called under-compounded generator

Fig 123 compound generator

Fig 124 Load characteristics of compound generators

127 LOSSES

The losses in a dc machine (generator or motor) may be divided into three classes viz (i) copper losses or electrical losses (ii) iron or core losses and (iii) mechanical losses All these losses appear as heat and thus raise the temperature of the machine They also lower the efficiency of the machine The loss structure maybe demonstrated as

1Copperlosses

These losses occur due to currents in the various windings of the machine When an electric current of I amps flows in a resistance of R ohms the heat energy is lost at the rate of I2R joules sec and the power loss is I2R watts Generators and motors have one or more field circuits and an armature circuit in which such losses occur All resistance losses of this kind are called copper losses The different copper losses occurring in a dc machine are

(i) Armature copper loss = Ia2Ra

(ii) Shunt field copper loss = Ish2 Rsh

(iii) Series field copper loss = Ise2Rse

Armature copper loss will be around 30 to 40 of full load losses where as total field copper losses will constitute around 30 of full load losses

Note There is also brush contact loss due to brush contact resistance (ie resistance between the surface of brush and surface of commutator) This loss is generally included in armature copper loss The brush contact drop is almost independent of Ia For carbon brushes this drop is around 1 V per brush The power loss due to brush contact resistance is 2ebIa where eb is the voltage drop at one brush

2 IRONORCORELOSSES These losses occur in the armature of a dc machine and are due to the rotation of armature in the magnetic field of the poles They are of two types viz (i) hysteresis loss (ii) eddy current loss

(i) Hysteresis loss

Hysteresis loss occurs in the armature of the dc machine since any given part of the armature is subjected to magnetic field reversals as it passes under successive poles Fig 124 shows an armature rotating in two-pole machine Consider a small piece ab of the armature When the piece ab is under N-pole the magnetic lines pass from a to b Half a revolution later the same piece of iron is under S-pole and magnetic lines pass from b to a so that magnetism in the iron is reversed In order to reverse continuously the molecular magnets in the armature core some amount of power has to be spent which is called hysteresis loss It is given by Steinmetz formula This formula is

Hysteresis Loss Wh = η Bmax 16f V Watts Bmax - Maximum flux density in the core f - Frequency of magnetic reversals V - Volume of armature in m3

η - Steinmetz hysteresis coefficient

In order to reduce this loss in a dc machine armature core is made of such materials which have a low value of Steinmetz hysteresis co-efficient eg silicon steel

Fig 124 Occurrence of hysteresis loss

ii) EDDYCURRENTLOSS

In addition to the voltages induced in the armature conductors there are also voltages induced in the armature core These voltages produce circulating currents in the armature core as shown in Fig 125 These are called eddy currents and power loss due to their flow is called eddy current loss The eddy current loss appears as heat which raises the temperature of the machine and lowers its efficiency Core resistance can be greatly increased by constructing the core of thin round If a continuous solid iron core is used the resistance to eddy current path will be small due to large cross-sectional area of the core Consequently the magnitude of eddy current and hence eddy current loss will be large The magnitude of eddy current can be reduced by making core resistance as high as practical The iron sheets called laminations [See Fig 126] The laminations are insulated from each other with a coating of varnish The insulating coating has a high resistance so very little current flows from one lamination to the other Alsobecause each lamination is very thin the resistance to current flowing through the width of a lamination is also quite large Thus laminating a core increases the core resistance which decreases the eddy current and hence the eddy current loss

Fig 125 Eddy current loss Fig 126 Core laminations

Eddy current loss is biven by the formula Pe= Ke Bmax

2 f2t2 V Watts

Where Ke = Constant depending upon electrical resistace of core Bmax = Maximum flux density in wbm2

f = frequency f magnetic reversals in Hz t = thickness of laminations in m and V = volume of core in m3

It may be noted that eddy current loss depends upon the square of lamination thickness For this reason lamination thickness should be kept as small as possible

3MECHANICALLOSSES

These losses are due to friction and windage(i) Friction losses eg bearing friction brush friction etc(ii) Windage loss ie air friction of rotating armature

These losses depend upon the speed of the machine But for a given speed they are practically constant

Note Iron losses and mechanical losses together are called stray losses

CONSTANT AND VARIABLE LOSSES

The losses in a dc generator (or dc motor) may be sub-divided into

(i) constant losses (ii) variable losses

(i) Constant losses

Those losses in a dc generator which remain constant at all loads are known as constant losses The constant losses in a dc generator are(a) iron losses(b) mechanical losses(c) shunt field losses(ii) Variable lossesThose losses in a dc generator which vary with load are called variable lossesThe variable losses in a dc generator are (a) Copper loss in armature winding (Ia

2 Ra) (b) Copper loss in series field winding (Ise

2Rse)

Total losses = Constant losses + Variable lossesNote Field Cu loss is constant for shunt and compound generators

128 POWER FLOW DIAGRAM

The power flow in a dc generator may be represented as

The different power stages are shown as A B and CA ndash The mechanical power input which is the power output of the prime mover is given as Pin = 2πNT

60 watts

B - The power developed in armature is Pa = Eg Ia watts C- The electrical power output is Pout = VIL watts

129 EFFICIENCY Efficiency of a machine is the ratio of its output power to the input power

Efficiency η = Output powerInput power

= Input powerminusLossesInput power = Output power

Output power+Losses

In case of a dc generator depending upon the power stages three different efficiencies can be defined as-

1 Mechanical Efficiency(ηm)

BA = Electrical power developedisinarmature

Mechanical power input2 Electrical Efficiency(ηe)

CB = Electrical power output

Electrical power developedisinthe armature 3 Overall or commercial efficiency (η)

CA = Electrical power output

Mechanical power input = ηmtimes ηe

130 CONDITION FOR MAXIMUM EFFICIENCY The efficiency of a dc generator is not constant but varies with load Fig 127shows the variation of efficiency with load current Consider a dc shunt generator with terminal voltage V delivering a load current of IL amp Generator output = V IL

Generator input = Output + Losses = V IL + constant losses + Variable losses = V IL + Wc+ Ia

2Ra

= V IL + Wc + (IL+ Ish)2 Rsh

The shunt field current Ish is small compared with IL so may be neglected

Generator input = VIL+ IL2Ra + Wc

Now efficiency η iquest outputinput = VI L

VIL+ Ia 2Ra+Wc

Efficiency is maximum when the denominator of the above equation is minimumie

ddIL ( VI L

VIL+ Ia 2Ra+Wc ) = 0

Which gives IL2 Ra = WC

ie Variable loss = constant lossThe load current for maximum efficiency is given by

IL iquest radicWcRa

Fig 127 Variation of efficiency with load

Hence the efficiency is maximum for that load current at which the variable loss is equal to the constant loss

131 APPLICATIONS

Separately excited generators

These types of generators have special applications such as electroplating electro refining of materials because separate supply is required for field excitation

Shunt generators

These types of generators are used in battery charging and ordinary lighting purposes

Series generators

This type of generators is used commonly as boosters on DC feeders as current generators for welding and arc lamp

Cumulatively compounded generators

This type of generators is used for domestic lighting purposes and to transmit energy over long distances

Differential compound generators

This type of generators is rarely used They are used for special applications such as electric arc welding

132 NUMERICAL EXCERCISES

1 The emf induced in the armature of a 450 kW 250 volt shunt generator is

2588volt when the field current is 200 amp and the generator is supplying power to a load at rated terminal voltage The armature circuit resistance is 0005 ohm Determine (i) load current (ii) power generated (iii) power output and (iv) electrical efficiency Neglect brush contact drop

Solution

Let the load current be I amp Given Ish = 20 A

Armature current = Load current + field current or Ia = I + 20 Induced emf = Terminal voltage + Ia Ra drop

2588 = 250 + (I + 20) times 0005

So (i) Load current I = 1740 amp

(ii) Power generated E Ia = 2588 (1740 + 20) watts = 455488 kW (iii) Power output V I = 250 times 1740 watts

= 435 kW

(iv) Electrical efficiency = 435455488 times 100 = 955

2 A 4 pole wave wound dc shunt generator delivers a load of 45 kW at a terminal voltage of 220 volts Its armature has 150 single turn coils and has a resistance of 001 ohm The air gap flux per pole is 002 weber Shunt field resistance is 50 ohm Calculate the speed at which it is being driven Neglect brush contact drop

Solution Load current I = W

V = 45000220 = 20455 amp

Field Current Ish = V

R s h = 220

50 = 44 amp

Armature current Ia = I +I sh = 20455 +44 = 20895 amp

Armature drop I aRa = 20895 X 001 = 20895 volts Induced emf Eg = V + Ia Ra = 220 + 20895 = 2220895 volts

But induced emf

So N = Eg60 AΦ ZP = 2220895times 60 times2

002 times150 times2 times 4 = 111045 rpm

3 The following data refer to the OCC of a DC separately excited generator at 1000 rpm

Field current in A 00 02 03 04 05 06 07 08 09 10 11 12Armature voltage in V 5 40 75 100 124 145 162 178 188 195 200 205

The machine is now connected as a shunt generator with a total field resistance of 200 Ω and driven at 1000 rpm Estimate graphically (i) the voltage to which the generator will build up at no load (ii) the armature field and load currents when the terminal voltage is found to be 150 V Neglect the effect of armature reaction and brush drop and assume armature resistance Ra to be 08 Ω (iii) Finally estimate the steady state armature current when the machine terminals are shorted

Solution

First of all the OCC of the generator is plotted in a graph paper as shown in figure

(i) The total field circuit resistance is given to be Rf = 200 Ω The Rf line is now drawn on the same graph paper passing through the origin The point of intersection of the Rf line and the OCC decides the final no load voltage and can be read from the graph as 192 V and the corresponding field current is 096 A (ii) Since the terminal voltage V is 150 V (= BM) field current If

is OM = 077 A Generated voltage E is given by AM But we know E = IaRa + V Hence IaRa= E ndash V = AM ndash BM = AB Now from the graph AB = 25 V there4Ia = 2508 = 3125 A So load current IL= Ia ndash If= 3125 ndash 077 = 3048 A

ADDITIONAL PROBLEMS

1 A 4 pole lap wound 115 kW 230 volts dc shunt generator has armatureresistance of 02 ohm and field resistance of 100 ohms Calculate (i) emf

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES

Fig 121 Load characteristics of series generator

Now raise a perpendicular from point B and mark a point b on this line such that ab = AB Then point b will lie on the external characteristic of the generator

Following similar procedure other points of external characteristic can be located It is easy to see that we can also plot internal characteristic from the external characteristic

121 Characteristics of a Shunt Generator Fig 122 (i) show the connections of a shunt wound generator The armature current Ia splits up into two parts a small fraction Ish flowing through shunt field winding while the major part IL goes to the external load(i) OCC The OCC of a shunt generator is similar in shape to that of a series generator as shown in Fig 122 (ii) The line OA represents the shunt field circuitresistance When the generator is run at normal speed it will build up a voltageOM At no-load the terminal voltage of the generator will be constant (= OM)represented by the horizontal dotted line MC

Fig 122 Characteristics of a Shunt Generator

(ii) Internal characteristic

When the generator is loaded flux per pole is reduced due to armature reaction Therefore emf E generated on load is less than the emf generated at no load As a result the internal characteristic (EIa) drops down slightly as shown in Fig122 (ii)

(iii) External characteristic Curve 2 shows the external characteristic of a shunt generator It gives therelation between terminal voltage V and load current IL

Therefore external characteristic curve will lie below the internal characteristic curve by an amount equal to drop in the armature circuit[ie (IL + Ish)Ra] as shown in Fig 122 (ii)

Note It may be seen from the external characteristic that change in terminalvoltage from no-load to full load is small The terminal voltage can always bemaintained constant by adjusting the field rheostat R automatically122 Critical External Resistance for Shunt Generator

If the load resistance across the terminals of a shunt generator is decreased then load current increase However there is a limit to the increase in

load current with the decrease of load resistance Any decrease of load resistance beyond this point instead of increasing the current ultimately results in reduced current Consequently the external characteristic turns back (dotted curve) as shown in Fig 123 The tangent OA to the curve represents the minimum external resistance required to excite the shunt generator on load and is called critical external resistance If the resistance of the external circuit is less than the critical external resistance (represented by tangent OA in Fig 310) the machine will refuse to excite or will de-excite if already running This means that external resistance is so low as virtually to short circuit the machine and so doing away with its excitation

Note There are two critical resistances for a shunt generator viz (i) criticalfield resistance (ii) critical external resistance For the shunt generator to buildup voltage the former should not be exceeded and the latter must not be gonebelow

123 How to Draw OCC at Different Speeds

If we are given OCC of a generator at a constant speed N1 then we can easily draw the OCC at any other constant speed N2 Fig 123 illustrates theprocedure Here we are given OCC at a constant speed N1 It is desired to findthe OCC at constant speed N2 (it is assumed that n1 lt N2) For constantexcitation E propN

E2

E1 =

N 2

N1 so

E2 iquest E1

N2

N1

As shown in Fig (311) for If = OH E1 = HC Therefore the new value of emf (E2) for the same If but at N2

E2= HCN 2

N 1 = HD

Fig 123 Plotting OCC at another speed

This locates the point D on the new OCC at N2 Similarly other points can belocated taking different values of If The locus of these points will be the OCCat N2

124 Critical Speed (NC)

The critical speed of a shunt generator is the minimum speed below which it fails to excite Clearly it is the speed for which the given shunt field resistance represents the critical resistance In Fig 124 curve 2 corresponds to critical

speed because the shunt field resistance (Rsh) line is tangential to it If the generator runs at full speed N the new OCC moves upward and the Rsh line represents critical resistance for this speed

In order to find critical speed take any convenient point C on excitation axis and erect a perpendicular so as to cut Rsh and Rsh lines at points B and A respectively Then

BCAC

=NcN

Nc=N BCAC

Fig 124 Finding critical speed

125 Conditions for Voltage Build-Up of a Shunt Generator

The necessary conditions for voltage build-up in a shunt generator are

(i) There must be some residual magnetism in generator poles(ii) The connections of the field winding should be such that the field current strengthens the residual magnetism(iii) The resistance of the field circuit should be less than the critical resistance In other words the speed of the generator should be higher

than the critical speed

126 Compound Generator Characteristics

In a compound generator both series and shunt excitation are combined as shown in Fig 123 The shunt winding can be connected either across the armature only (short-shunt connection S) or across armature plus series field (long-shunt connection G) The compound generator can be cumulatively compounded or

differentially compounded generator The latter is rarely used in practice Therefore we shall discuss the characteristics of cumulatively compounded

generator It may be noted that external characteristics of long and short shunt compound generators are almost identical

External characteristic

Fig 124 shows the external characteristics of a cumulatively compoundedgenerator The series excitation aids the shunt excitation The degree of compounding depends upon the increase in series excitation with the increase inload current

(i) If series winding turns are so adjusted that with the increase in load current the terminal voltage increases it is called over-compounded generator In such a case as the load current increases the series field mmf increases and tends to increase the flux and hence the generated voltage The increase in generated voltage is greater than the IaRa drop so that instead of decreasing the terminal voltage increases as shown by curve A in Fig 124

(ii) If series winding turns are so adjusted that with the increase in load current the terminal voltage substantially remains constant it is called Flat-compounded generator The series winding of such a machine has lesser number of turns than the one in over-compounded machine and therefore does not increase the flux as much for a given load current Consequently the full-load voltage is nearly equal to the no-load voltage as indicated by curve B in Fig 124 (iii) If series field winding has lesser number of turns than for a flat compounded machine the terminal voltage falls with increase in load current as indicated by curve C m Fig 124 Such a machine is called under-compounded generator

Fig 123 compound generator

Fig 124 Load characteristics of compound generators

127 LOSSES

The losses in a dc machine (generator or motor) may be divided into three classes viz (i) copper losses or electrical losses (ii) iron or core losses and (iii) mechanical losses All these losses appear as heat and thus raise the temperature of the machine They also lower the efficiency of the machine The loss structure maybe demonstrated as

1Copperlosses

These losses occur due to currents in the various windings of the machine When an electric current of I amps flows in a resistance of R ohms the heat energy is lost at the rate of I2R joules sec and the power loss is I2R watts Generators and motors have one or more field circuits and an armature circuit in which such losses occur All resistance losses of this kind are called copper losses The different copper losses occurring in a dc machine are

(i) Armature copper loss = Ia2Ra

(ii) Shunt field copper loss = Ish2 Rsh

(iii) Series field copper loss = Ise2Rse

Armature copper loss will be around 30 to 40 of full load losses where as total field copper losses will constitute around 30 of full load losses

Note There is also brush contact loss due to brush contact resistance (ie resistance between the surface of brush and surface of commutator) This loss is generally included in armature copper loss The brush contact drop is almost independent of Ia For carbon brushes this drop is around 1 V per brush The power loss due to brush contact resistance is 2ebIa where eb is the voltage drop at one brush

2 IRONORCORELOSSES These losses occur in the armature of a dc machine and are due to the rotation of armature in the magnetic field of the poles They are of two types viz (i) hysteresis loss (ii) eddy current loss

(i) Hysteresis loss

Hysteresis loss occurs in the armature of the dc machine since any given part of the armature is subjected to magnetic field reversals as it passes under successive poles Fig 124 shows an armature rotating in two-pole machine Consider a small piece ab of the armature When the piece ab is under N-pole the magnetic lines pass from a to b Half a revolution later the same piece of iron is under S-pole and magnetic lines pass from b to a so that magnetism in the iron is reversed In order to reverse continuously the molecular magnets in the armature core some amount of power has to be spent which is called hysteresis loss It is given by Steinmetz formula This formula is

Hysteresis Loss Wh = η Bmax 16f V Watts Bmax - Maximum flux density in the core f - Frequency of magnetic reversals V - Volume of armature in m3

η - Steinmetz hysteresis coefficient

In order to reduce this loss in a dc machine armature core is made of such materials which have a low value of Steinmetz hysteresis co-efficient eg silicon steel

Fig 124 Occurrence of hysteresis loss

ii) EDDYCURRENTLOSS

In addition to the voltages induced in the armature conductors there are also voltages induced in the armature core These voltages produce circulating currents in the armature core as shown in Fig 125 These are called eddy currents and power loss due to their flow is called eddy current loss The eddy current loss appears as heat which raises the temperature of the machine and lowers its efficiency Core resistance can be greatly increased by constructing the core of thin round If a continuous solid iron core is used the resistance to eddy current path will be small due to large cross-sectional area of the core Consequently the magnitude of eddy current and hence eddy current loss will be large The magnitude of eddy current can be reduced by making core resistance as high as practical The iron sheets called laminations [See Fig 126] The laminations are insulated from each other with a coating of varnish The insulating coating has a high resistance so very little current flows from one lamination to the other Alsobecause each lamination is very thin the resistance to current flowing through the width of a lamination is also quite large Thus laminating a core increases the core resistance which decreases the eddy current and hence the eddy current loss

Fig 125 Eddy current loss Fig 126 Core laminations

Eddy current loss is biven by the formula Pe= Ke Bmax

2 f2t2 V Watts

Where Ke = Constant depending upon electrical resistace of core Bmax = Maximum flux density in wbm2

f = frequency f magnetic reversals in Hz t = thickness of laminations in m and V = volume of core in m3

It may be noted that eddy current loss depends upon the square of lamination thickness For this reason lamination thickness should be kept as small as possible

3MECHANICALLOSSES

These losses are due to friction and windage(i) Friction losses eg bearing friction brush friction etc(ii) Windage loss ie air friction of rotating armature

These losses depend upon the speed of the machine But for a given speed they are practically constant

Note Iron losses and mechanical losses together are called stray losses

CONSTANT AND VARIABLE LOSSES

The losses in a dc generator (or dc motor) may be sub-divided into

(i) constant losses (ii) variable losses

(i) Constant losses

Those losses in a dc generator which remain constant at all loads are known as constant losses The constant losses in a dc generator are(a) iron losses(b) mechanical losses(c) shunt field losses(ii) Variable lossesThose losses in a dc generator which vary with load are called variable lossesThe variable losses in a dc generator are (a) Copper loss in armature winding (Ia

2 Ra) (b) Copper loss in series field winding (Ise

2Rse)

Total losses = Constant losses + Variable lossesNote Field Cu loss is constant for shunt and compound generators

128 POWER FLOW DIAGRAM

The power flow in a dc generator may be represented as

The different power stages are shown as A B and CA ndash The mechanical power input which is the power output of the prime mover is given as Pin = 2πNT

60 watts

B - The power developed in armature is Pa = Eg Ia watts C- The electrical power output is Pout = VIL watts

129 EFFICIENCY Efficiency of a machine is the ratio of its output power to the input power

Efficiency η = Output powerInput power

= Input powerminusLossesInput power = Output power

Output power+Losses

In case of a dc generator depending upon the power stages three different efficiencies can be defined as-

1 Mechanical Efficiency(ηm)

BA = Electrical power developedisinarmature

Mechanical power input2 Electrical Efficiency(ηe)

CB = Electrical power output

Electrical power developedisinthe armature 3 Overall or commercial efficiency (η)

CA = Electrical power output

Mechanical power input = ηmtimes ηe

130 CONDITION FOR MAXIMUM EFFICIENCY The efficiency of a dc generator is not constant but varies with load Fig 127shows the variation of efficiency with load current Consider a dc shunt generator with terminal voltage V delivering a load current of IL amp Generator output = V IL

Generator input = Output + Losses = V IL + constant losses + Variable losses = V IL + Wc+ Ia

2Ra

= V IL + Wc + (IL+ Ish)2 Rsh

The shunt field current Ish is small compared with IL so may be neglected

Generator input = VIL+ IL2Ra + Wc

Now efficiency η iquest outputinput = VI L

VIL+ Ia 2Ra+Wc

Efficiency is maximum when the denominator of the above equation is minimumie

ddIL ( VI L

VIL+ Ia 2Ra+Wc ) = 0

Which gives IL2 Ra = WC

ie Variable loss = constant lossThe load current for maximum efficiency is given by

IL iquest radicWcRa

Fig 127 Variation of efficiency with load

Hence the efficiency is maximum for that load current at which the variable loss is equal to the constant loss

131 APPLICATIONS

Separately excited generators

These types of generators have special applications such as electroplating electro refining of materials because separate supply is required for field excitation

Shunt generators

These types of generators are used in battery charging and ordinary lighting purposes

Series generators

This type of generators is used commonly as boosters on DC feeders as current generators for welding and arc lamp

Cumulatively compounded generators

This type of generators is used for domestic lighting purposes and to transmit energy over long distances

Differential compound generators

This type of generators is rarely used They are used for special applications such as electric arc welding

132 NUMERICAL EXCERCISES

1 The emf induced in the armature of a 450 kW 250 volt shunt generator is

2588volt when the field current is 200 amp and the generator is supplying power to a load at rated terminal voltage The armature circuit resistance is 0005 ohm Determine (i) load current (ii) power generated (iii) power output and (iv) electrical efficiency Neglect brush contact drop

Solution

Let the load current be I amp Given Ish = 20 A

Armature current = Load current + field current or Ia = I + 20 Induced emf = Terminal voltage + Ia Ra drop

2588 = 250 + (I + 20) times 0005

So (i) Load current I = 1740 amp

(ii) Power generated E Ia = 2588 (1740 + 20) watts = 455488 kW (iii) Power output V I = 250 times 1740 watts

= 435 kW

(iv) Electrical efficiency = 435455488 times 100 = 955

2 A 4 pole wave wound dc shunt generator delivers a load of 45 kW at a terminal voltage of 220 volts Its armature has 150 single turn coils and has a resistance of 001 ohm The air gap flux per pole is 002 weber Shunt field resistance is 50 ohm Calculate the speed at which it is being driven Neglect brush contact drop

Solution Load current I = W

V = 45000220 = 20455 amp

Field Current Ish = V

R s h = 220

50 = 44 amp

Armature current Ia = I +I sh = 20455 +44 = 20895 amp

Armature drop I aRa = 20895 X 001 = 20895 volts Induced emf Eg = V + Ia Ra = 220 + 20895 = 2220895 volts

But induced emf

So N = Eg60 AΦ ZP = 2220895times 60 times2

002 times150 times2 times 4 = 111045 rpm

3 The following data refer to the OCC of a DC separately excited generator at 1000 rpm

Field current in A 00 02 03 04 05 06 07 08 09 10 11 12Armature voltage in V 5 40 75 100 124 145 162 178 188 195 200 205

The machine is now connected as a shunt generator with a total field resistance of 200 Ω and driven at 1000 rpm Estimate graphically (i) the voltage to which the generator will build up at no load (ii) the armature field and load currents when the terminal voltage is found to be 150 V Neglect the effect of armature reaction and brush drop and assume armature resistance Ra to be 08 Ω (iii) Finally estimate the steady state armature current when the machine terminals are shorted

Solution

First of all the OCC of the generator is plotted in a graph paper as shown in figure

(i) The total field circuit resistance is given to be Rf = 200 Ω The Rf line is now drawn on the same graph paper passing through the origin The point of intersection of the Rf line and the OCC decides the final no load voltage and can be read from the graph as 192 V and the corresponding field current is 096 A (ii) Since the terminal voltage V is 150 V (= BM) field current If

is OM = 077 A Generated voltage E is given by AM But we know E = IaRa + V Hence IaRa= E ndash V = AM ndash BM = AB Now from the graph AB = 25 V there4Ia = 2508 = 3125 A So load current IL= Ia ndash If= 3125 ndash 077 = 3048 A

ADDITIONAL PROBLEMS

1 A 4 pole lap wound 115 kW 230 volts dc shunt generator has armatureresistance of 02 ohm and field resistance of 100 ohms Calculate (i) emf

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES

Fig 122 Characteristics of a Shunt Generator

(ii) Internal characteristic

When the generator is loaded flux per pole is reduced due to armature reaction Therefore emf E generated on load is less than the emf generated at no load As a result the internal characteristic (EIa) drops down slightly as shown in Fig122 (ii)

(iii) External characteristic Curve 2 shows the external characteristic of a shunt generator It gives therelation between terminal voltage V and load current IL

Therefore external characteristic curve will lie below the internal characteristic curve by an amount equal to drop in the armature circuit[ie (IL + Ish)Ra] as shown in Fig 122 (ii)

Note It may be seen from the external characteristic that change in terminalvoltage from no-load to full load is small The terminal voltage can always bemaintained constant by adjusting the field rheostat R automatically122 Critical External Resistance for Shunt Generator

If the load resistance across the terminals of a shunt generator is decreased then load current increase However there is a limit to the increase in

load current with the decrease of load resistance Any decrease of load resistance beyond this point instead of increasing the current ultimately results in reduced current Consequently the external characteristic turns back (dotted curve) as shown in Fig 123 The tangent OA to the curve represents the minimum external resistance required to excite the shunt generator on load and is called critical external resistance If the resistance of the external circuit is less than the critical external resistance (represented by tangent OA in Fig 310) the machine will refuse to excite or will de-excite if already running This means that external resistance is so low as virtually to short circuit the machine and so doing away with its excitation

Note There are two critical resistances for a shunt generator viz (i) criticalfield resistance (ii) critical external resistance For the shunt generator to buildup voltage the former should not be exceeded and the latter must not be gonebelow

123 How to Draw OCC at Different Speeds

If we are given OCC of a generator at a constant speed N1 then we can easily draw the OCC at any other constant speed N2 Fig 123 illustrates theprocedure Here we are given OCC at a constant speed N1 It is desired to findthe OCC at constant speed N2 (it is assumed that n1 lt N2) For constantexcitation E propN

E2

E1 =

N 2

N1 so

E2 iquest E1

N2

N1

As shown in Fig (311) for If = OH E1 = HC Therefore the new value of emf (E2) for the same If but at N2

E2= HCN 2

N 1 = HD

Fig 123 Plotting OCC at another speed

This locates the point D on the new OCC at N2 Similarly other points can belocated taking different values of If The locus of these points will be the OCCat N2

124 Critical Speed (NC)

The critical speed of a shunt generator is the minimum speed below which it fails to excite Clearly it is the speed for which the given shunt field resistance represents the critical resistance In Fig 124 curve 2 corresponds to critical

speed because the shunt field resistance (Rsh) line is tangential to it If the generator runs at full speed N the new OCC moves upward and the Rsh line represents critical resistance for this speed

In order to find critical speed take any convenient point C on excitation axis and erect a perpendicular so as to cut Rsh and Rsh lines at points B and A respectively Then

BCAC

=NcN

Nc=N BCAC

Fig 124 Finding critical speed

125 Conditions for Voltage Build-Up of a Shunt Generator

The necessary conditions for voltage build-up in a shunt generator are

(i) There must be some residual magnetism in generator poles(ii) The connections of the field winding should be such that the field current strengthens the residual magnetism(iii) The resistance of the field circuit should be less than the critical resistance In other words the speed of the generator should be higher

than the critical speed

126 Compound Generator Characteristics

In a compound generator both series and shunt excitation are combined as shown in Fig 123 The shunt winding can be connected either across the armature only (short-shunt connection S) or across armature plus series field (long-shunt connection G) The compound generator can be cumulatively compounded or

differentially compounded generator The latter is rarely used in practice Therefore we shall discuss the characteristics of cumulatively compounded

generator It may be noted that external characteristics of long and short shunt compound generators are almost identical

External characteristic

Fig 124 shows the external characteristics of a cumulatively compoundedgenerator The series excitation aids the shunt excitation The degree of compounding depends upon the increase in series excitation with the increase inload current

(i) If series winding turns are so adjusted that with the increase in load current the terminal voltage increases it is called over-compounded generator In such a case as the load current increases the series field mmf increases and tends to increase the flux and hence the generated voltage The increase in generated voltage is greater than the IaRa drop so that instead of decreasing the terminal voltage increases as shown by curve A in Fig 124

(ii) If series winding turns are so adjusted that with the increase in load current the terminal voltage substantially remains constant it is called Flat-compounded generator The series winding of such a machine has lesser number of turns than the one in over-compounded machine and therefore does not increase the flux as much for a given load current Consequently the full-load voltage is nearly equal to the no-load voltage as indicated by curve B in Fig 124 (iii) If series field winding has lesser number of turns than for a flat compounded machine the terminal voltage falls with increase in load current as indicated by curve C m Fig 124 Such a machine is called under-compounded generator

Fig 123 compound generator

Fig 124 Load characteristics of compound generators

127 LOSSES

The losses in a dc machine (generator or motor) may be divided into three classes viz (i) copper losses or electrical losses (ii) iron or core losses and (iii) mechanical losses All these losses appear as heat and thus raise the temperature of the machine They also lower the efficiency of the machine The loss structure maybe demonstrated as

1Copperlosses

These losses occur due to currents in the various windings of the machine When an electric current of I amps flows in a resistance of R ohms the heat energy is lost at the rate of I2R joules sec and the power loss is I2R watts Generators and motors have one or more field circuits and an armature circuit in which such losses occur All resistance losses of this kind are called copper losses The different copper losses occurring in a dc machine are

(i) Armature copper loss = Ia2Ra

(ii) Shunt field copper loss = Ish2 Rsh

(iii) Series field copper loss = Ise2Rse

Armature copper loss will be around 30 to 40 of full load losses where as total field copper losses will constitute around 30 of full load losses

Note There is also brush contact loss due to brush contact resistance (ie resistance between the surface of brush and surface of commutator) This loss is generally included in armature copper loss The brush contact drop is almost independent of Ia For carbon brushes this drop is around 1 V per brush The power loss due to brush contact resistance is 2ebIa where eb is the voltage drop at one brush

2 IRONORCORELOSSES These losses occur in the armature of a dc machine and are due to the rotation of armature in the magnetic field of the poles They are of two types viz (i) hysteresis loss (ii) eddy current loss

(i) Hysteresis loss

Hysteresis loss occurs in the armature of the dc machine since any given part of the armature is subjected to magnetic field reversals as it passes under successive poles Fig 124 shows an armature rotating in two-pole machine Consider a small piece ab of the armature When the piece ab is under N-pole the magnetic lines pass from a to b Half a revolution later the same piece of iron is under S-pole and magnetic lines pass from b to a so that magnetism in the iron is reversed In order to reverse continuously the molecular magnets in the armature core some amount of power has to be spent which is called hysteresis loss It is given by Steinmetz formula This formula is

Hysteresis Loss Wh = η Bmax 16f V Watts Bmax - Maximum flux density in the core f - Frequency of magnetic reversals V - Volume of armature in m3

η - Steinmetz hysteresis coefficient

In order to reduce this loss in a dc machine armature core is made of such materials which have a low value of Steinmetz hysteresis co-efficient eg silicon steel

Fig 124 Occurrence of hysteresis loss

ii) EDDYCURRENTLOSS

In addition to the voltages induced in the armature conductors there are also voltages induced in the armature core These voltages produce circulating currents in the armature core as shown in Fig 125 These are called eddy currents and power loss due to their flow is called eddy current loss The eddy current loss appears as heat which raises the temperature of the machine and lowers its efficiency Core resistance can be greatly increased by constructing the core of thin round If a continuous solid iron core is used the resistance to eddy current path will be small due to large cross-sectional area of the core Consequently the magnitude of eddy current and hence eddy current loss will be large The magnitude of eddy current can be reduced by making core resistance as high as practical The iron sheets called laminations [See Fig 126] The laminations are insulated from each other with a coating of varnish The insulating coating has a high resistance so very little current flows from one lamination to the other Alsobecause each lamination is very thin the resistance to current flowing through the width of a lamination is also quite large Thus laminating a core increases the core resistance which decreases the eddy current and hence the eddy current loss

Fig 125 Eddy current loss Fig 126 Core laminations

Eddy current loss is biven by the formula Pe= Ke Bmax

2 f2t2 V Watts

Where Ke = Constant depending upon electrical resistace of core Bmax = Maximum flux density in wbm2

f = frequency f magnetic reversals in Hz t = thickness of laminations in m and V = volume of core in m3

It may be noted that eddy current loss depends upon the square of lamination thickness For this reason lamination thickness should be kept as small as possible

3MECHANICALLOSSES

These losses are due to friction and windage(i) Friction losses eg bearing friction brush friction etc(ii) Windage loss ie air friction of rotating armature

These losses depend upon the speed of the machine But for a given speed they are practically constant

Note Iron losses and mechanical losses together are called stray losses

CONSTANT AND VARIABLE LOSSES

The losses in a dc generator (or dc motor) may be sub-divided into

(i) constant losses (ii) variable losses

(i) Constant losses

Those losses in a dc generator which remain constant at all loads are known as constant losses The constant losses in a dc generator are(a) iron losses(b) mechanical losses(c) shunt field losses(ii) Variable lossesThose losses in a dc generator which vary with load are called variable lossesThe variable losses in a dc generator are (a) Copper loss in armature winding (Ia

2 Ra) (b) Copper loss in series field winding (Ise

2Rse)

Total losses = Constant losses + Variable lossesNote Field Cu loss is constant for shunt and compound generators

128 POWER FLOW DIAGRAM

The power flow in a dc generator may be represented as

The different power stages are shown as A B and CA ndash The mechanical power input which is the power output of the prime mover is given as Pin = 2πNT

60 watts

B - The power developed in armature is Pa = Eg Ia watts C- The electrical power output is Pout = VIL watts

129 EFFICIENCY Efficiency of a machine is the ratio of its output power to the input power

Efficiency η = Output powerInput power

= Input powerminusLossesInput power = Output power

Output power+Losses

In case of a dc generator depending upon the power stages three different efficiencies can be defined as-

1 Mechanical Efficiency(ηm)

BA = Electrical power developedisinarmature

Mechanical power input2 Electrical Efficiency(ηe)

CB = Electrical power output

Electrical power developedisinthe armature 3 Overall or commercial efficiency (η)

CA = Electrical power output

Mechanical power input = ηmtimes ηe

130 CONDITION FOR MAXIMUM EFFICIENCY The efficiency of a dc generator is not constant but varies with load Fig 127shows the variation of efficiency with load current Consider a dc shunt generator with terminal voltage V delivering a load current of IL amp Generator output = V IL

Generator input = Output + Losses = V IL + constant losses + Variable losses = V IL + Wc+ Ia

2Ra

= V IL + Wc + (IL+ Ish)2 Rsh

The shunt field current Ish is small compared with IL so may be neglected

Generator input = VIL+ IL2Ra + Wc

Now efficiency η iquest outputinput = VI L

VIL+ Ia 2Ra+Wc

Efficiency is maximum when the denominator of the above equation is minimumie

ddIL ( VI L

VIL+ Ia 2Ra+Wc ) = 0

Which gives IL2 Ra = WC

ie Variable loss = constant lossThe load current for maximum efficiency is given by

IL iquest radicWcRa

Fig 127 Variation of efficiency with load

Hence the efficiency is maximum for that load current at which the variable loss is equal to the constant loss

131 APPLICATIONS

Separately excited generators

These types of generators have special applications such as electroplating electro refining of materials because separate supply is required for field excitation

Shunt generators

These types of generators are used in battery charging and ordinary lighting purposes

Series generators

This type of generators is used commonly as boosters on DC feeders as current generators for welding and arc lamp

Cumulatively compounded generators

This type of generators is used for domestic lighting purposes and to transmit energy over long distances

Differential compound generators

This type of generators is rarely used They are used for special applications such as electric arc welding

132 NUMERICAL EXCERCISES

1 The emf induced in the armature of a 450 kW 250 volt shunt generator is

2588volt when the field current is 200 amp and the generator is supplying power to a load at rated terminal voltage The armature circuit resistance is 0005 ohm Determine (i) load current (ii) power generated (iii) power output and (iv) electrical efficiency Neglect brush contact drop

Solution

Let the load current be I amp Given Ish = 20 A

Armature current = Load current + field current or Ia = I + 20 Induced emf = Terminal voltage + Ia Ra drop

2588 = 250 + (I + 20) times 0005

So (i) Load current I = 1740 amp

(ii) Power generated E Ia = 2588 (1740 + 20) watts = 455488 kW (iii) Power output V I = 250 times 1740 watts

= 435 kW

(iv) Electrical efficiency = 435455488 times 100 = 955

2 A 4 pole wave wound dc shunt generator delivers a load of 45 kW at a terminal voltage of 220 volts Its armature has 150 single turn coils and has a resistance of 001 ohm The air gap flux per pole is 002 weber Shunt field resistance is 50 ohm Calculate the speed at which it is being driven Neglect brush contact drop

Solution Load current I = W

V = 45000220 = 20455 amp

Field Current Ish = V

R s h = 220

50 = 44 amp

Armature current Ia = I +I sh = 20455 +44 = 20895 amp

Armature drop I aRa = 20895 X 001 = 20895 volts Induced emf Eg = V + Ia Ra = 220 + 20895 = 2220895 volts

But induced emf

So N = Eg60 AΦ ZP = 2220895times 60 times2

002 times150 times2 times 4 = 111045 rpm

3 The following data refer to the OCC of a DC separately excited generator at 1000 rpm

Field current in A 00 02 03 04 05 06 07 08 09 10 11 12Armature voltage in V 5 40 75 100 124 145 162 178 188 195 200 205

The machine is now connected as a shunt generator with a total field resistance of 200 Ω and driven at 1000 rpm Estimate graphically (i) the voltage to which the generator will build up at no load (ii) the armature field and load currents when the terminal voltage is found to be 150 V Neglect the effect of armature reaction and brush drop and assume armature resistance Ra to be 08 Ω (iii) Finally estimate the steady state armature current when the machine terminals are shorted

Solution

First of all the OCC of the generator is plotted in a graph paper as shown in figure

(i) The total field circuit resistance is given to be Rf = 200 Ω The Rf line is now drawn on the same graph paper passing through the origin The point of intersection of the Rf line and the OCC decides the final no load voltage and can be read from the graph as 192 V and the corresponding field current is 096 A (ii) Since the terminal voltage V is 150 V (= BM) field current If

is OM = 077 A Generated voltage E is given by AM But we know E = IaRa + V Hence IaRa= E ndash V = AM ndash BM = AB Now from the graph AB = 25 V there4Ia = 2508 = 3125 A So load current IL= Ia ndash If= 3125 ndash 077 = 3048 A

ADDITIONAL PROBLEMS

1 A 4 pole lap wound 115 kW 230 volts dc shunt generator has armatureresistance of 02 ohm and field resistance of 100 ohms Calculate (i) emf

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES

load current with the decrease of load resistance Any decrease of load resistance beyond this point instead of increasing the current ultimately results in reduced current Consequently the external characteristic turns back (dotted curve) as shown in Fig 123 The tangent OA to the curve represents the minimum external resistance required to excite the shunt generator on load and is called critical external resistance If the resistance of the external circuit is less than the critical external resistance (represented by tangent OA in Fig 310) the machine will refuse to excite or will de-excite if already running This means that external resistance is so low as virtually to short circuit the machine and so doing away with its excitation

Note There are two critical resistances for a shunt generator viz (i) criticalfield resistance (ii) critical external resistance For the shunt generator to buildup voltage the former should not be exceeded and the latter must not be gonebelow

123 How to Draw OCC at Different Speeds

If we are given OCC of a generator at a constant speed N1 then we can easily draw the OCC at any other constant speed N2 Fig 123 illustrates theprocedure Here we are given OCC at a constant speed N1 It is desired to findthe OCC at constant speed N2 (it is assumed that n1 lt N2) For constantexcitation E propN

E2

E1 =

N 2

N1 so

E2 iquest E1

N2

N1

As shown in Fig (311) for If = OH E1 = HC Therefore the new value of emf (E2) for the same If but at N2

E2= HCN 2

N 1 = HD

Fig 123 Plotting OCC at another speed

This locates the point D on the new OCC at N2 Similarly other points can belocated taking different values of If The locus of these points will be the OCCat N2

124 Critical Speed (NC)

The critical speed of a shunt generator is the minimum speed below which it fails to excite Clearly it is the speed for which the given shunt field resistance represents the critical resistance In Fig 124 curve 2 corresponds to critical

speed because the shunt field resistance (Rsh) line is tangential to it If the generator runs at full speed N the new OCC moves upward and the Rsh line represents critical resistance for this speed

In order to find critical speed take any convenient point C on excitation axis and erect a perpendicular so as to cut Rsh and Rsh lines at points B and A respectively Then

BCAC

=NcN

Nc=N BCAC

Fig 124 Finding critical speed

125 Conditions for Voltage Build-Up of a Shunt Generator

The necessary conditions for voltage build-up in a shunt generator are

(i) There must be some residual magnetism in generator poles(ii) The connections of the field winding should be such that the field current strengthens the residual magnetism(iii) The resistance of the field circuit should be less than the critical resistance In other words the speed of the generator should be higher

than the critical speed

126 Compound Generator Characteristics

In a compound generator both series and shunt excitation are combined as shown in Fig 123 The shunt winding can be connected either across the armature only (short-shunt connection S) or across armature plus series field (long-shunt connection G) The compound generator can be cumulatively compounded or

differentially compounded generator The latter is rarely used in practice Therefore we shall discuss the characteristics of cumulatively compounded

generator It may be noted that external characteristics of long and short shunt compound generators are almost identical

External characteristic

Fig 124 shows the external characteristics of a cumulatively compoundedgenerator The series excitation aids the shunt excitation The degree of compounding depends upon the increase in series excitation with the increase inload current

(i) If series winding turns are so adjusted that with the increase in load current the terminal voltage increases it is called over-compounded generator In such a case as the load current increases the series field mmf increases and tends to increase the flux and hence the generated voltage The increase in generated voltage is greater than the IaRa drop so that instead of decreasing the terminal voltage increases as shown by curve A in Fig 124

(ii) If series winding turns are so adjusted that with the increase in load current the terminal voltage substantially remains constant it is called Flat-compounded generator The series winding of such a machine has lesser number of turns than the one in over-compounded machine and therefore does not increase the flux as much for a given load current Consequently the full-load voltage is nearly equal to the no-load voltage as indicated by curve B in Fig 124 (iii) If series field winding has lesser number of turns than for a flat compounded machine the terminal voltage falls with increase in load current as indicated by curve C m Fig 124 Such a machine is called under-compounded generator

Fig 123 compound generator

Fig 124 Load characteristics of compound generators

127 LOSSES

The losses in a dc machine (generator or motor) may be divided into three classes viz (i) copper losses or electrical losses (ii) iron or core losses and (iii) mechanical losses All these losses appear as heat and thus raise the temperature of the machine They also lower the efficiency of the machine The loss structure maybe demonstrated as

1Copperlosses

These losses occur due to currents in the various windings of the machine When an electric current of I amps flows in a resistance of R ohms the heat energy is lost at the rate of I2R joules sec and the power loss is I2R watts Generators and motors have one or more field circuits and an armature circuit in which such losses occur All resistance losses of this kind are called copper losses The different copper losses occurring in a dc machine are

(i) Armature copper loss = Ia2Ra

(ii) Shunt field copper loss = Ish2 Rsh

(iii) Series field copper loss = Ise2Rse

Armature copper loss will be around 30 to 40 of full load losses where as total field copper losses will constitute around 30 of full load losses

Note There is also brush contact loss due to brush contact resistance (ie resistance between the surface of brush and surface of commutator) This loss is generally included in armature copper loss The brush contact drop is almost independent of Ia For carbon brushes this drop is around 1 V per brush The power loss due to brush contact resistance is 2ebIa where eb is the voltage drop at one brush

2 IRONORCORELOSSES These losses occur in the armature of a dc machine and are due to the rotation of armature in the magnetic field of the poles They are of two types viz (i) hysteresis loss (ii) eddy current loss

(i) Hysteresis loss

Hysteresis loss occurs in the armature of the dc machine since any given part of the armature is subjected to magnetic field reversals as it passes under successive poles Fig 124 shows an armature rotating in two-pole machine Consider a small piece ab of the armature When the piece ab is under N-pole the magnetic lines pass from a to b Half a revolution later the same piece of iron is under S-pole and magnetic lines pass from b to a so that magnetism in the iron is reversed In order to reverse continuously the molecular magnets in the armature core some amount of power has to be spent which is called hysteresis loss It is given by Steinmetz formula This formula is

Hysteresis Loss Wh = η Bmax 16f V Watts Bmax - Maximum flux density in the core f - Frequency of magnetic reversals V - Volume of armature in m3

η - Steinmetz hysteresis coefficient

In order to reduce this loss in a dc machine armature core is made of such materials which have a low value of Steinmetz hysteresis co-efficient eg silicon steel

Fig 124 Occurrence of hysteresis loss

ii) EDDYCURRENTLOSS

In addition to the voltages induced in the armature conductors there are also voltages induced in the armature core These voltages produce circulating currents in the armature core as shown in Fig 125 These are called eddy currents and power loss due to their flow is called eddy current loss The eddy current loss appears as heat which raises the temperature of the machine and lowers its efficiency Core resistance can be greatly increased by constructing the core of thin round If a continuous solid iron core is used the resistance to eddy current path will be small due to large cross-sectional area of the core Consequently the magnitude of eddy current and hence eddy current loss will be large The magnitude of eddy current can be reduced by making core resistance as high as practical The iron sheets called laminations [See Fig 126] The laminations are insulated from each other with a coating of varnish The insulating coating has a high resistance so very little current flows from one lamination to the other Alsobecause each lamination is very thin the resistance to current flowing through the width of a lamination is also quite large Thus laminating a core increases the core resistance which decreases the eddy current and hence the eddy current loss

Fig 125 Eddy current loss Fig 126 Core laminations

Eddy current loss is biven by the formula Pe= Ke Bmax

2 f2t2 V Watts

Where Ke = Constant depending upon electrical resistace of core Bmax = Maximum flux density in wbm2

f = frequency f magnetic reversals in Hz t = thickness of laminations in m and V = volume of core in m3

It may be noted that eddy current loss depends upon the square of lamination thickness For this reason lamination thickness should be kept as small as possible

3MECHANICALLOSSES

These losses are due to friction and windage(i) Friction losses eg bearing friction brush friction etc(ii) Windage loss ie air friction of rotating armature

These losses depend upon the speed of the machine But for a given speed they are practically constant

Note Iron losses and mechanical losses together are called stray losses

CONSTANT AND VARIABLE LOSSES

The losses in a dc generator (or dc motor) may be sub-divided into

(i) constant losses (ii) variable losses

(i) Constant losses

Those losses in a dc generator which remain constant at all loads are known as constant losses The constant losses in a dc generator are(a) iron losses(b) mechanical losses(c) shunt field losses(ii) Variable lossesThose losses in a dc generator which vary with load are called variable lossesThe variable losses in a dc generator are (a) Copper loss in armature winding (Ia

2 Ra) (b) Copper loss in series field winding (Ise

2Rse)

Total losses = Constant losses + Variable lossesNote Field Cu loss is constant for shunt and compound generators

128 POWER FLOW DIAGRAM

The power flow in a dc generator may be represented as

The different power stages are shown as A B and CA ndash The mechanical power input which is the power output of the prime mover is given as Pin = 2πNT

60 watts

B - The power developed in armature is Pa = Eg Ia watts C- The electrical power output is Pout = VIL watts

129 EFFICIENCY Efficiency of a machine is the ratio of its output power to the input power

Efficiency η = Output powerInput power

= Input powerminusLossesInput power = Output power

Output power+Losses

In case of a dc generator depending upon the power stages three different efficiencies can be defined as-

1 Mechanical Efficiency(ηm)

BA = Electrical power developedisinarmature

Mechanical power input2 Electrical Efficiency(ηe)

CB = Electrical power output

Electrical power developedisinthe armature 3 Overall or commercial efficiency (η)

CA = Electrical power output

Mechanical power input = ηmtimes ηe

130 CONDITION FOR MAXIMUM EFFICIENCY The efficiency of a dc generator is not constant but varies with load Fig 127shows the variation of efficiency with load current Consider a dc shunt generator with terminal voltage V delivering a load current of IL amp Generator output = V IL

Generator input = Output + Losses = V IL + constant losses + Variable losses = V IL + Wc+ Ia

2Ra

= V IL + Wc + (IL+ Ish)2 Rsh

The shunt field current Ish is small compared with IL so may be neglected

Generator input = VIL+ IL2Ra + Wc

Now efficiency η iquest outputinput = VI L

VIL+ Ia 2Ra+Wc

Efficiency is maximum when the denominator of the above equation is minimumie

ddIL ( VI L

VIL+ Ia 2Ra+Wc ) = 0

Which gives IL2 Ra = WC

ie Variable loss = constant lossThe load current for maximum efficiency is given by

IL iquest radicWcRa

Fig 127 Variation of efficiency with load

Hence the efficiency is maximum for that load current at which the variable loss is equal to the constant loss

131 APPLICATIONS

Separately excited generators

These types of generators have special applications such as electroplating electro refining of materials because separate supply is required for field excitation

Shunt generators

These types of generators are used in battery charging and ordinary lighting purposes

Series generators

This type of generators is used commonly as boosters on DC feeders as current generators for welding and arc lamp

Cumulatively compounded generators

This type of generators is used for domestic lighting purposes and to transmit energy over long distances

Differential compound generators

This type of generators is rarely used They are used for special applications such as electric arc welding

132 NUMERICAL EXCERCISES

1 The emf induced in the armature of a 450 kW 250 volt shunt generator is

2588volt when the field current is 200 amp and the generator is supplying power to a load at rated terminal voltage The armature circuit resistance is 0005 ohm Determine (i) load current (ii) power generated (iii) power output and (iv) electrical efficiency Neglect brush contact drop

Solution

Let the load current be I amp Given Ish = 20 A

Armature current = Load current + field current or Ia = I + 20 Induced emf = Terminal voltage + Ia Ra drop

2588 = 250 + (I + 20) times 0005

So (i) Load current I = 1740 amp

(ii) Power generated E Ia = 2588 (1740 + 20) watts = 455488 kW (iii) Power output V I = 250 times 1740 watts

= 435 kW

(iv) Electrical efficiency = 435455488 times 100 = 955

2 A 4 pole wave wound dc shunt generator delivers a load of 45 kW at a terminal voltage of 220 volts Its armature has 150 single turn coils and has a resistance of 001 ohm The air gap flux per pole is 002 weber Shunt field resistance is 50 ohm Calculate the speed at which it is being driven Neglect brush contact drop

Solution Load current I = W

V = 45000220 = 20455 amp

Field Current Ish = V

R s h = 220

50 = 44 amp

Armature current Ia = I +I sh = 20455 +44 = 20895 amp

Armature drop I aRa = 20895 X 001 = 20895 volts Induced emf Eg = V + Ia Ra = 220 + 20895 = 2220895 volts

But induced emf

So N = Eg60 AΦ ZP = 2220895times 60 times2

002 times150 times2 times 4 = 111045 rpm

3 The following data refer to the OCC of a DC separately excited generator at 1000 rpm

Field current in A 00 02 03 04 05 06 07 08 09 10 11 12Armature voltage in V 5 40 75 100 124 145 162 178 188 195 200 205

The machine is now connected as a shunt generator with a total field resistance of 200 Ω and driven at 1000 rpm Estimate graphically (i) the voltage to which the generator will build up at no load (ii) the armature field and load currents when the terminal voltage is found to be 150 V Neglect the effect of armature reaction and brush drop and assume armature resistance Ra to be 08 Ω (iii) Finally estimate the steady state armature current when the machine terminals are shorted

Solution

First of all the OCC of the generator is plotted in a graph paper as shown in figure

(i) The total field circuit resistance is given to be Rf = 200 Ω The Rf line is now drawn on the same graph paper passing through the origin The point of intersection of the Rf line and the OCC decides the final no load voltage and can be read from the graph as 192 V and the corresponding field current is 096 A (ii) Since the terminal voltage V is 150 V (= BM) field current If

is OM = 077 A Generated voltage E is given by AM But we know E = IaRa + V Hence IaRa= E ndash V = AM ndash BM = AB Now from the graph AB = 25 V there4Ia = 2508 = 3125 A So load current IL= Ia ndash If= 3125 ndash 077 = 3048 A

ADDITIONAL PROBLEMS

1 A 4 pole lap wound 115 kW 230 volts dc shunt generator has armatureresistance of 02 ohm and field resistance of 100 ohms Calculate (i) emf

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES

Fig 123 Plotting OCC at another speed

This locates the point D on the new OCC at N2 Similarly other points can belocated taking different values of If The locus of these points will be the OCCat N2

124 Critical Speed (NC)

The critical speed of a shunt generator is the minimum speed below which it fails to excite Clearly it is the speed for which the given shunt field resistance represents the critical resistance In Fig 124 curve 2 corresponds to critical

speed because the shunt field resistance (Rsh) line is tangential to it If the generator runs at full speed N the new OCC moves upward and the Rsh line represents critical resistance for this speed

In order to find critical speed take any convenient point C on excitation axis and erect a perpendicular so as to cut Rsh and Rsh lines at points B and A respectively Then

BCAC

=NcN

Nc=N BCAC

Fig 124 Finding critical speed

125 Conditions for Voltage Build-Up of a Shunt Generator

The necessary conditions for voltage build-up in a shunt generator are

(i) There must be some residual magnetism in generator poles(ii) The connections of the field winding should be such that the field current strengthens the residual magnetism(iii) The resistance of the field circuit should be less than the critical resistance In other words the speed of the generator should be higher

than the critical speed

126 Compound Generator Characteristics

In a compound generator both series and shunt excitation are combined as shown in Fig 123 The shunt winding can be connected either across the armature only (short-shunt connection S) or across armature plus series field (long-shunt connection G) The compound generator can be cumulatively compounded or

differentially compounded generator The latter is rarely used in practice Therefore we shall discuss the characteristics of cumulatively compounded

generator It may be noted that external characteristics of long and short shunt compound generators are almost identical

External characteristic

Fig 124 shows the external characteristics of a cumulatively compoundedgenerator The series excitation aids the shunt excitation The degree of compounding depends upon the increase in series excitation with the increase inload current

(i) If series winding turns are so adjusted that with the increase in load current the terminal voltage increases it is called over-compounded generator In such a case as the load current increases the series field mmf increases and tends to increase the flux and hence the generated voltage The increase in generated voltage is greater than the IaRa drop so that instead of decreasing the terminal voltage increases as shown by curve A in Fig 124

(ii) If series winding turns are so adjusted that with the increase in load current the terminal voltage substantially remains constant it is called Flat-compounded generator The series winding of such a machine has lesser number of turns than the one in over-compounded machine and therefore does not increase the flux as much for a given load current Consequently the full-load voltage is nearly equal to the no-load voltage as indicated by curve B in Fig 124 (iii) If series field winding has lesser number of turns than for a flat compounded machine the terminal voltage falls with increase in load current as indicated by curve C m Fig 124 Such a machine is called under-compounded generator

Fig 123 compound generator

Fig 124 Load characteristics of compound generators

127 LOSSES

The losses in a dc machine (generator or motor) may be divided into three classes viz (i) copper losses or electrical losses (ii) iron or core losses and (iii) mechanical losses All these losses appear as heat and thus raise the temperature of the machine They also lower the efficiency of the machine The loss structure maybe demonstrated as

1Copperlosses

These losses occur due to currents in the various windings of the machine When an electric current of I amps flows in a resistance of R ohms the heat energy is lost at the rate of I2R joules sec and the power loss is I2R watts Generators and motors have one or more field circuits and an armature circuit in which such losses occur All resistance losses of this kind are called copper losses The different copper losses occurring in a dc machine are

(i) Armature copper loss = Ia2Ra

(ii) Shunt field copper loss = Ish2 Rsh

(iii) Series field copper loss = Ise2Rse

Armature copper loss will be around 30 to 40 of full load losses where as total field copper losses will constitute around 30 of full load losses

Note There is also brush contact loss due to brush contact resistance (ie resistance between the surface of brush and surface of commutator) This loss is generally included in armature copper loss The brush contact drop is almost independent of Ia For carbon brushes this drop is around 1 V per brush The power loss due to brush contact resistance is 2ebIa where eb is the voltage drop at one brush

2 IRONORCORELOSSES These losses occur in the armature of a dc machine and are due to the rotation of armature in the magnetic field of the poles They are of two types viz (i) hysteresis loss (ii) eddy current loss

(i) Hysteresis loss

Hysteresis loss occurs in the armature of the dc machine since any given part of the armature is subjected to magnetic field reversals as it passes under successive poles Fig 124 shows an armature rotating in two-pole machine Consider a small piece ab of the armature When the piece ab is under N-pole the magnetic lines pass from a to b Half a revolution later the same piece of iron is under S-pole and magnetic lines pass from b to a so that magnetism in the iron is reversed In order to reverse continuously the molecular magnets in the armature core some amount of power has to be spent which is called hysteresis loss It is given by Steinmetz formula This formula is

Hysteresis Loss Wh = η Bmax 16f V Watts Bmax - Maximum flux density in the core f - Frequency of magnetic reversals V - Volume of armature in m3

η - Steinmetz hysteresis coefficient

In order to reduce this loss in a dc machine armature core is made of such materials which have a low value of Steinmetz hysteresis co-efficient eg silicon steel

Fig 124 Occurrence of hysteresis loss

ii) EDDYCURRENTLOSS

In addition to the voltages induced in the armature conductors there are also voltages induced in the armature core These voltages produce circulating currents in the armature core as shown in Fig 125 These are called eddy currents and power loss due to their flow is called eddy current loss The eddy current loss appears as heat which raises the temperature of the machine and lowers its efficiency Core resistance can be greatly increased by constructing the core of thin round If a continuous solid iron core is used the resistance to eddy current path will be small due to large cross-sectional area of the core Consequently the magnitude of eddy current and hence eddy current loss will be large The magnitude of eddy current can be reduced by making core resistance as high as practical The iron sheets called laminations [See Fig 126] The laminations are insulated from each other with a coating of varnish The insulating coating has a high resistance so very little current flows from one lamination to the other Alsobecause each lamination is very thin the resistance to current flowing through the width of a lamination is also quite large Thus laminating a core increases the core resistance which decreases the eddy current and hence the eddy current loss

Fig 125 Eddy current loss Fig 126 Core laminations

Eddy current loss is biven by the formula Pe= Ke Bmax

2 f2t2 V Watts

Where Ke = Constant depending upon electrical resistace of core Bmax = Maximum flux density in wbm2

f = frequency f magnetic reversals in Hz t = thickness of laminations in m and V = volume of core in m3

It may be noted that eddy current loss depends upon the square of lamination thickness For this reason lamination thickness should be kept as small as possible

3MECHANICALLOSSES

These losses are due to friction and windage(i) Friction losses eg bearing friction brush friction etc(ii) Windage loss ie air friction of rotating armature

These losses depend upon the speed of the machine But for a given speed they are practically constant

Note Iron losses and mechanical losses together are called stray losses

CONSTANT AND VARIABLE LOSSES

The losses in a dc generator (or dc motor) may be sub-divided into

(i) constant losses (ii) variable losses

(i) Constant losses

Those losses in a dc generator which remain constant at all loads are known as constant losses The constant losses in a dc generator are(a) iron losses(b) mechanical losses(c) shunt field losses(ii) Variable lossesThose losses in a dc generator which vary with load are called variable lossesThe variable losses in a dc generator are (a) Copper loss in armature winding (Ia

2 Ra) (b) Copper loss in series field winding (Ise

2Rse)

Total losses = Constant losses + Variable lossesNote Field Cu loss is constant for shunt and compound generators

128 POWER FLOW DIAGRAM

The power flow in a dc generator may be represented as

The different power stages are shown as A B and CA ndash The mechanical power input which is the power output of the prime mover is given as Pin = 2πNT

60 watts

B - The power developed in armature is Pa = Eg Ia watts C- The electrical power output is Pout = VIL watts

129 EFFICIENCY Efficiency of a machine is the ratio of its output power to the input power

Efficiency η = Output powerInput power

= Input powerminusLossesInput power = Output power

Output power+Losses

In case of a dc generator depending upon the power stages three different efficiencies can be defined as-

1 Mechanical Efficiency(ηm)

BA = Electrical power developedisinarmature

Mechanical power input2 Electrical Efficiency(ηe)

CB = Electrical power output

Electrical power developedisinthe armature 3 Overall or commercial efficiency (η)

CA = Electrical power output

Mechanical power input = ηmtimes ηe

130 CONDITION FOR MAXIMUM EFFICIENCY The efficiency of a dc generator is not constant but varies with load Fig 127shows the variation of efficiency with load current Consider a dc shunt generator with terminal voltage V delivering a load current of IL amp Generator output = V IL

Generator input = Output + Losses = V IL + constant losses + Variable losses = V IL + Wc+ Ia

2Ra

= V IL + Wc + (IL+ Ish)2 Rsh

The shunt field current Ish is small compared with IL so may be neglected

Generator input = VIL+ IL2Ra + Wc

Now efficiency η iquest outputinput = VI L

VIL+ Ia 2Ra+Wc

Efficiency is maximum when the denominator of the above equation is minimumie

ddIL ( VI L

VIL+ Ia 2Ra+Wc ) = 0

Which gives IL2 Ra = WC

ie Variable loss = constant lossThe load current for maximum efficiency is given by

IL iquest radicWcRa

Fig 127 Variation of efficiency with load

Hence the efficiency is maximum for that load current at which the variable loss is equal to the constant loss

131 APPLICATIONS

Separately excited generators

These types of generators have special applications such as electroplating electro refining of materials because separate supply is required for field excitation

Shunt generators

These types of generators are used in battery charging and ordinary lighting purposes

Series generators

This type of generators is used commonly as boosters on DC feeders as current generators for welding and arc lamp

Cumulatively compounded generators

This type of generators is used for domestic lighting purposes and to transmit energy over long distances

Differential compound generators

This type of generators is rarely used They are used for special applications such as electric arc welding

132 NUMERICAL EXCERCISES

1 The emf induced in the armature of a 450 kW 250 volt shunt generator is

2588volt when the field current is 200 amp and the generator is supplying power to a load at rated terminal voltage The armature circuit resistance is 0005 ohm Determine (i) load current (ii) power generated (iii) power output and (iv) electrical efficiency Neglect brush contact drop

Solution

Let the load current be I amp Given Ish = 20 A

Armature current = Load current + field current or Ia = I + 20 Induced emf = Terminal voltage + Ia Ra drop

2588 = 250 + (I + 20) times 0005

So (i) Load current I = 1740 amp

(ii) Power generated E Ia = 2588 (1740 + 20) watts = 455488 kW (iii) Power output V I = 250 times 1740 watts

= 435 kW

(iv) Electrical efficiency = 435455488 times 100 = 955

2 A 4 pole wave wound dc shunt generator delivers a load of 45 kW at a terminal voltage of 220 volts Its armature has 150 single turn coils and has a resistance of 001 ohm The air gap flux per pole is 002 weber Shunt field resistance is 50 ohm Calculate the speed at which it is being driven Neglect brush contact drop

Solution Load current I = W

V = 45000220 = 20455 amp

Field Current Ish = V

R s h = 220

50 = 44 amp

Armature current Ia = I +I sh = 20455 +44 = 20895 amp

Armature drop I aRa = 20895 X 001 = 20895 volts Induced emf Eg = V + Ia Ra = 220 + 20895 = 2220895 volts

But induced emf

So N = Eg60 AΦ ZP = 2220895times 60 times2

002 times150 times2 times 4 = 111045 rpm

3 The following data refer to the OCC of a DC separately excited generator at 1000 rpm

Field current in A 00 02 03 04 05 06 07 08 09 10 11 12Armature voltage in V 5 40 75 100 124 145 162 178 188 195 200 205

The machine is now connected as a shunt generator with a total field resistance of 200 Ω and driven at 1000 rpm Estimate graphically (i) the voltage to which the generator will build up at no load (ii) the armature field and load currents when the terminal voltage is found to be 150 V Neglect the effect of armature reaction and brush drop and assume armature resistance Ra to be 08 Ω (iii) Finally estimate the steady state armature current when the machine terminals are shorted

Solution

First of all the OCC of the generator is plotted in a graph paper as shown in figure

(i) The total field circuit resistance is given to be Rf = 200 Ω The Rf line is now drawn on the same graph paper passing through the origin The point of intersection of the Rf line and the OCC decides the final no load voltage and can be read from the graph as 192 V and the corresponding field current is 096 A (ii) Since the terminal voltage V is 150 V (= BM) field current If

is OM = 077 A Generated voltage E is given by AM But we know E = IaRa + V Hence IaRa= E ndash V = AM ndash BM = AB Now from the graph AB = 25 V there4Ia = 2508 = 3125 A So load current IL= Ia ndash If= 3125 ndash 077 = 3048 A

ADDITIONAL PROBLEMS

1 A 4 pole lap wound 115 kW 230 volts dc shunt generator has armatureresistance of 02 ohm and field resistance of 100 ohms Calculate (i) emf

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES

BCAC

=NcN

Nc=N BCAC

Fig 124 Finding critical speed

125 Conditions for Voltage Build-Up of a Shunt Generator

The necessary conditions for voltage build-up in a shunt generator are

(i) There must be some residual magnetism in generator poles(ii) The connections of the field winding should be such that the field current strengthens the residual magnetism(iii) The resistance of the field circuit should be less than the critical resistance In other words the speed of the generator should be higher

than the critical speed

126 Compound Generator Characteristics

In a compound generator both series and shunt excitation are combined as shown in Fig 123 The shunt winding can be connected either across the armature only (short-shunt connection S) or across armature plus series field (long-shunt connection G) The compound generator can be cumulatively compounded or

differentially compounded generator The latter is rarely used in practice Therefore we shall discuss the characteristics of cumulatively compounded

generator It may be noted that external characteristics of long and short shunt compound generators are almost identical

External characteristic

Fig 124 shows the external characteristics of a cumulatively compoundedgenerator The series excitation aids the shunt excitation The degree of compounding depends upon the increase in series excitation with the increase inload current

(i) If series winding turns are so adjusted that with the increase in load current the terminal voltage increases it is called over-compounded generator In such a case as the load current increases the series field mmf increases and tends to increase the flux and hence the generated voltage The increase in generated voltage is greater than the IaRa drop so that instead of decreasing the terminal voltage increases as shown by curve A in Fig 124

(ii) If series winding turns are so adjusted that with the increase in load current the terminal voltage substantially remains constant it is called Flat-compounded generator The series winding of such a machine has lesser number of turns than the one in over-compounded machine and therefore does not increase the flux as much for a given load current Consequently the full-load voltage is nearly equal to the no-load voltage as indicated by curve B in Fig 124 (iii) If series field winding has lesser number of turns than for a flat compounded machine the terminal voltage falls with increase in load current as indicated by curve C m Fig 124 Such a machine is called under-compounded generator

Fig 123 compound generator

Fig 124 Load characteristics of compound generators

127 LOSSES

The losses in a dc machine (generator or motor) may be divided into three classes viz (i) copper losses or electrical losses (ii) iron or core losses and (iii) mechanical losses All these losses appear as heat and thus raise the temperature of the machine They also lower the efficiency of the machine The loss structure maybe demonstrated as

1Copperlosses

These losses occur due to currents in the various windings of the machine When an electric current of I amps flows in a resistance of R ohms the heat energy is lost at the rate of I2R joules sec and the power loss is I2R watts Generators and motors have one or more field circuits and an armature circuit in which such losses occur All resistance losses of this kind are called copper losses The different copper losses occurring in a dc machine are

(i) Armature copper loss = Ia2Ra

(ii) Shunt field copper loss = Ish2 Rsh

(iii) Series field copper loss = Ise2Rse

Armature copper loss will be around 30 to 40 of full load losses where as total field copper losses will constitute around 30 of full load losses

Note There is also brush contact loss due to brush contact resistance (ie resistance between the surface of brush and surface of commutator) This loss is generally included in armature copper loss The brush contact drop is almost independent of Ia For carbon brushes this drop is around 1 V per brush The power loss due to brush contact resistance is 2ebIa where eb is the voltage drop at one brush

2 IRONORCORELOSSES These losses occur in the armature of a dc machine and are due to the rotation of armature in the magnetic field of the poles They are of two types viz (i) hysteresis loss (ii) eddy current loss

(i) Hysteresis loss

Hysteresis loss occurs in the armature of the dc machine since any given part of the armature is subjected to magnetic field reversals as it passes under successive poles Fig 124 shows an armature rotating in two-pole machine Consider a small piece ab of the armature When the piece ab is under N-pole the magnetic lines pass from a to b Half a revolution later the same piece of iron is under S-pole and magnetic lines pass from b to a so that magnetism in the iron is reversed In order to reverse continuously the molecular magnets in the armature core some amount of power has to be spent which is called hysteresis loss It is given by Steinmetz formula This formula is

Hysteresis Loss Wh = η Bmax 16f V Watts Bmax - Maximum flux density in the core f - Frequency of magnetic reversals V - Volume of armature in m3

η - Steinmetz hysteresis coefficient

In order to reduce this loss in a dc machine armature core is made of such materials which have a low value of Steinmetz hysteresis co-efficient eg silicon steel

Fig 124 Occurrence of hysteresis loss

ii) EDDYCURRENTLOSS

In addition to the voltages induced in the armature conductors there are also voltages induced in the armature core These voltages produce circulating currents in the armature core as shown in Fig 125 These are called eddy currents and power loss due to their flow is called eddy current loss The eddy current loss appears as heat which raises the temperature of the machine and lowers its efficiency Core resistance can be greatly increased by constructing the core of thin round If a continuous solid iron core is used the resistance to eddy current path will be small due to large cross-sectional area of the core Consequently the magnitude of eddy current and hence eddy current loss will be large The magnitude of eddy current can be reduced by making core resistance as high as practical The iron sheets called laminations [See Fig 126] The laminations are insulated from each other with a coating of varnish The insulating coating has a high resistance so very little current flows from one lamination to the other Alsobecause each lamination is very thin the resistance to current flowing through the width of a lamination is also quite large Thus laminating a core increases the core resistance which decreases the eddy current and hence the eddy current loss

Fig 125 Eddy current loss Fig 126 Core laminations

Eddy current loss is biven by the formula Pe= Ke Bmax

2 f2t2 V Watts

Where Ke = Constant depending upon electrical resistace of core Bmax = Maximum flux density in wbm2

f = frequency f magnetic reversals in Hz t = thickness of laminations in m and V = volume of core in m3

It may be noted that eddy current loss depends upon the square of lamination thickness For this reason lamination thickness should be kept as small as possible

3MECHANICALLOSSES

These losses are due to friction and windage(i) Friction losses eg bearing friction brush friction etc(ii) Windage loss ie air friction of rotating armature

These losses depend upon the speed of the machine But for a given speed they are practically constant

Note Iron losses and mechanical losses together are called stray losses

CONSTANT AND VARIABLE LOSSES

The losses in a dc generator (or dc motor) may be sub-divided into

(i) constant losses (ii) variable losses

(i) Constant losses

Those losses in a dc generator which remain constant at all loads are known as constant losses The constant losses in a dc generator are(a) iron losses(b) mechanical losses(c) shunt field losses(ii) Variable lossesThose losses in a dc generator which vary with load are called variable lossesThe variable losses in a dc generator are (a) Copper loss in armature winding (Ia

2 Ra) (b) Copper loss in series field winding (Ise

2Rse)

Total losses = Constant losses + Variable lossesNote Field Cu loss is constant for shunt and compound generators

128 POWER FLOW DIAGRAM

The power flow in a dc generator may be represented as

The different power stages are shown as A B and CA ndash The mechanical power input which is the power output of the prime mover is given as Pin = 2πNT

60 watts

B - The power developed in armature is Pa = Eg Ia watts C- The electrical power output is Pout = VIL watts

129 EFFICIENCY Efficiency of a machine is the ratio of its output power to the input power

Efficiency η = Output powerInput power

= Input powerminusLossesInput power = Output power

Output power+Losses

In case of a dc generator depending upon the power stages three different efficiencies can be defined as-

1 Mechanical Efficiency(ηm)

BA = Electrical power developedisinarmature

Mechanical power input2 Electrical Efficiency(ηe)

CB = Electrical power output

Electrical power developedisinthe armature 3 Overall or commercial efficiency (η)

CA = Electrical power output

Mechanical power input = ηmtimes ηe

130 CONDITION FOR MAXIMUM EFFICIENCY The efficiency of a dc generator is not constant but varies with load Fig 127shows the variation of efficiency with load current Consider a dc shunt generator with terminal voltage V delivering a load current of IL amp Generator output = V IL

Generator input = Output + Losses = V IL + constant losses + Variable losses = V IL + Wc+ Ia

2Ra

= V IL + Wc + (IL+ Ish)2 Rsh

The shunt field current Ish is small compared with IL so may be neglected

Generator input = VIL+ IL2Ra + Wc

Now efficiency η iquest outputinput = VI L

VIL+ Ia 2Ra+Wc

Efficiency is maximum when the denominator of the above equation is minimumie

ddIL ( VI L

VIL+ Ia 2Ra+Wc ) = 0

Which gives IL2 Ra = WC

ie Variable loss = constant lossThe load current for maximum efficiency is given by

IL iquest radicWcRa

Fig 127 Variation of efficiency with load

Hence the efficiency is maximum for that load current at which the variable loss is equal to the constant loss

131 APPLICATIONS

Separately excited generators

These types of generators have special applications such as electroplating electro refining of materials because separate supply is required for field excitation

Shunt generators

These types of generators are used in battery charging and ordinary lighting purposes

Series generators

This type of generators is used commonly as boosters on DC feeders as current generators for welding and arc lamp

Cumulatively compounded generators

This type of generators is used for domestic lighting purposes and to transmit energy over long distances

Differential compound generators

This type of generators is rarely used They are used for special applications such as electric arc welding

132 NUMERICAL EXCERCISES

1 The emf induced in the armature of a 450 kW 250 volt shunt generator is

2588volt when the field current is 200 amp and the generator is supplying power to a load at rated terminal voltage The armature circuit resistance is 0005 ohm Determine (i) load current (ii) power generated (iii) power output and (iv) electrical efficiency Neglect brush contact drop

Solution

Let the load current be I amp Given Ish = 20 A

Armature current = Load current + field current or Ia = I + 20 Induced emf = Terminal voltage + Ia Ra drop

2588 = 250 + (I + 20) times 0005

So (i) Load current I = 1740 amp

(ii) Power generated E Ia = 2588 (1740 + 20) watts = 455488 kW (iii) Power output V I = 250 times 1740 watts

= 435 kW

(iv) Electrical efficiency = 435455488 times 100 = 955

2 A 4 pole wave wound dc shunt generator delivers a load of 45 kW at a terminal voltage of 220 volts Its armature has 150 single turn coils and has a resistance of 001 ohm The air gap flux per pole is 002 weber Shunt field resistance is 50 ohm Calculate the speed at which it is being driven Neglect brush contact drop

Solution Load current I = W

V = 45000220 = 20455 amp

Field Current Ish = V

R s h = 220

50 = 44 amp

Armature current Ia = I +I sh = 20455 +44 = 20895 amp

Armature drop I aRa = 20895 X 001 = 20895 volts Induced emf Eg = V + Ia Ra = 220 + 20895 = 2220895 volts

But induced emf

So N = Eg60 AΦ ZP = 2220895times 60 times2

002 times150 times2 times 4 = 111045 rpm

3 The following data refer to the OCC of a DC separately excited generator at 1000 rpm

Field current in A 00 02 03 04 05 06 07 08 09 10 11 12Armature voltage in V 5 40 75 100 124 145 162 178 188 195 200 205

The machine is now connected as a shunt generator with a total field resistance of 200 Ω and driven at 1000 rpm Estimate graphically (i) the voltage to which the generator will build up at no load (ii) the armature field and load currents when the terminal voltage is found to be 150 V Neglect the effect of armature reaction and brush drop and assume armature resistance Ra to be 08 Ω (iii) Finally estimate the steady state armature current when the machine terminals are shorted

Solution

First of all the OCC of the generator is plotted in a graph paper as shown in figure

(i) The total field circuit resistance is given to be Rf = 200 Ω The Rf line is now drawn on the same graph paper passing through the origin The point of intersection of the Rf line and the OCC decides the final no load voltage and can be read from the graph as 192 V and the corresponding field current is 096 A (ii) Since the terminal voltage V is 150 V (= BM) field current If

is OM = 077 A Generated voltage E is given by AM But we know E = IaRa + V Hence IaRa= E ndash V = AM ndash BM = AB Now from the graph AB = 25 V there4Ia = 2508 = 3125 A So load current IL= Ia ndash If= 3125 ndash 077 = 3048 A

ADDITIONAL PROBLEMS

1 A 4 pole lap wound 115 kW 230 volts dc shunt generator has armatureresistance of 02 ohm and field resistance of 100 ohms Calculate (i) emf

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES

differentially compounded generator The latter is rarely used in practice Therefore we shall discuss the characteristics of cumulatively compounded

generator It may be noted that external characteristics of long and short shunt compound generators are almost identical

External characteristic

Fig 124 shows the external characteristics of a cumulatively compoundedgenerator The series excitation aids the shunt excitation The degree of compounding depends upon the increase in series excitation with the increase inload current

(i) If series winding turns are so adjusted that with the increase in load current the terminal voltage increases it is called over-compounded generator In such a case as the load current increases the series field mmf increases and tends to increase the flux and hence the generated voltage The increase in generated voltage is greater than the IaRa drop so that instead of decreasing the terminal voltage increases as shown by curve A in Fig 124

(ii) If series winding turns are so adjusted that with the increase in load current the terminal voltage substantially remains constant it is called Flat-compounded generator The series winding of such a machine has lesser number of turns than the one in over-compounded machine and therefore does not increase the flux as much for a given load current Consequently the full-load voltage is nearly equal to the no-load voltage as indicated by curve B in Fig 124 (iii) If series field winding has lesser number of turns than for a flat compounded machine the terminal voltage falls with increase in load current as indicated by curve C m Fig 124 Such a machine is called under-compounded generator

Fig 123 compound generator

Fig 124 Load characteristics of compound generators

127 LOSSES

The losses in a dc machine (generator or motor) may be divided into three classes viz (i) copper losses or electrical losses (ii) iron or core losses and (iii) mechanical losses All these losses appear as heat and thus raise the temperature of the machine They also lower the efficiency of the machine The loss structure maybe demonstrated as

1Copperlosses

These losses occur due to currents in the various windings of the machine When an electric current of I amps flows in a resistance of R ohms the heat energy is lost at the rate of I2R joules sec and the power loss is I2R watts Generators and motors have one or more field circuits and an armature circuit in which such losses occur All resistance losses of this kind are called copper losses The different copper losses occurring in a dc machine are

(i) Armature copper loss = Ia2Ra

(ii) Shunt field copper loss = Ish2 Rsh

(iii) Series field copper loss = Ise2Rse

Armature copper loss will be around 30 to 40 of full load losses where as total field copper losses will constitute around 30 of full load losses

Note There is also brush contact loss due to brush contact resistance (ie resistance between the surface of brush and surface of commutator) This loss is generally included in armature copper loss The brush contact drop is almost independent of Ia For carbon brushes this drop is around 1 V per brush The power loss due to brush contact resistance is 2ebIa where eb is the voltage drop at one brush

2 IRONORCORELOSSES These losses occur in the armature of a dc machine and are due to the rotation of armature in the magnetic field of the poles They are of two types viz (i) hysteresis loss (ii) eddy current loss

(i) Hysteresis loss

Hysteresis loss occurs in the armature of the dc machine since any given part of the armature is subjected to magnetic field reversals as it passes under successive poles Fig 124 shows an armature rotating in two-pole machine Consider a small piece ab of the armature When the piece ab is under N-pole the magnetic lines pass from a to b Half a revolution later the same piece of iron is under S-pole and magnetic lines pass from b to a so that magnetism in the iron is reversed In order to reverse continuously the molecular magnets in the armature core some amount of power has to be spent which is called hysteresis loss It is given by Steinmetz formula This formula is

Hysteresis Loss Wh = η Bmax 16f V Watts Bmax - Maximum flux density in the core f - Frequency of magnetic reversals V - Volume of armature in m3

η - Steinmetz hysteresis coefficient

In order to reduce this loss in a dc machine armature core is made of such materials which have a low value of Steinmetz hysteresis co-efficient eg silicon steel

Fig 124 Occurrence of hysteresis loss

ii) EDDYCURRENTLOSS

In addition to the voltages induced in the armature conductors there are also voltages induced in the armature core These voltages produce circulating currents in the armature core as shown in Fig 125 These are called eddy currents and power loss due to their flow is called eddy current loss The eddy current loss appears as heat which raises the temperature of the machine and lowers its efficiency Core resistance can be greatly increased by constructing the core of thin round If a continuous solid iron core is used the resistance to eddy current path will be small due to large cross-sectional area of the core Consequently the magnitude of eddy current and hence eddy current loss will be large The magnitude of eddy current can be reduced by making core resistance as high as practical The iron sheets called laminations [See Fig 126] The laminations are insulated from each other with a coating of varnish The insulating coating has a high resistance so very little current flows from one lamination to the other Alsobecause each lamination is very thin the resistance to current flowing through the width of a lamination is also quite large Thus laminating a core increases the core resistance which decreases the eddy current and hence the eddy current loss

Fig 125 Eddy current loss Fig 126 Core laminations

Eddy current loss is biven by the formula Pe= Ke Bmax

2 f2t2 V Watts

Where Ke = Constant depending upon electrical resistace of core Bmax = Maximum flux density in wbm2

f = frequency f magnetic reversals in Hz t = thickness of laminations in m and V = volume of core in m3

It may be noted that eddy current loss depends upon the square of lamination thickness For this reason lamination thickness should be kept as small as possible

3MECHANICALLOSSES

These losses are due to friction and windage(i) Friction losses eg bearing friction brush friction etc(ii) Windage loss ie air friction of rotating armature

These losses depend upon the speed of the machine But for a given speed they are practically constant

Note Iron losses and mechanical losses together are called stray losses

CONSTANT AND VARIABLE LOSSES

The losses in a dc generator (or dc motor) may be sub-divided into

(i) constant losses (ii) variable losses

(i) Constant losses

Those losses in a dc generator which remain constant at all loads are known as constant losses The constant losses in a dc generator are(a) iron losses(b) mechanical losses(c) shunt field losses(ii) Variable lossesThose losses in a dc generator which vary with load are called variable lossesThe variable losses in a dc generator are (a) Copper loss in armature winding (Ia

2 Ra) (b) Copper loss in series field winding (Ise

2Rse)

Total losses = Constant losses + Variable lossesNote Field Cu loss is constant for shunt and compound generators

128 POWER FLOW DIAGRAM

The power flow in a dc generator may be represented as

The different power stages are shown as A B and CA ndash The mechanical power input which is the power output of the prime mover is given as Pin = 2πNT

60 watts

B - The power developed in armature is Pa = Eg Ia watts C- The electrical power output is Pout = VIL watts

129 EFFICIENCY Efficiency of a machine is the ratio of its output power to the input power

Efficiency η = Output powerInput power

= Input powerminusLossesInput power = Output power

Output power+Losses

In case of a dc generator depending upon the power stages three different efficiencies can be defined as-

1 Mechanical Efficiency(ηm)

BA = Electrical power developedisinarmature

Mechanical power input2 Electrical Efficiency(ηe)

CB = Electrical power output

Electrical power developedisinthe armature 3 Overall or commercial efficiency (η)

CA = Electrical power output

Mechanical power input = ηmtimes ηe

130 CONDITION FOR MAXIMUM EFFICIENCY The efficiency of a dc generator is not constant but varies with load Fig 127shows the variation of efficiency with load current Consider a dc shunt generator with terminal voltage V delivering a load current of IL amp Generator output = V IL

Generator input = Output + Losses = V IL + constant losses + Variable losses = V IL + Wc+ Ia

2Ra

= V IL + Wc + (IL+ Ish)2 Rsh

The shunt field current Ish is small compared with IL so may be neglected

Generator input = VIL+ IL2Ra + Wc

Now efficiency η iquest outputinput = VI L

VIL+ Ia 2Ra+Wc

Efficiency is maximum when the denominator of the above equation is minimumie

ddIL ( VI L

VIL+ Ia 2Ra+Wc ) = 0

Which gives IL2 Ra = WC

ie Variable loss = constant lossThe load current for maximum efficiency is given by

IL iquest radicWcRa

Fig 127 Variation of efficiency with load

Hence the efficiency is maximum for that load current at which the variable loss is equal to the constant loss

131 APPLICATIONS

Separately excited generators

These types of generators have special applications such as electroplating electro refining of materials because separate supply is required for field excitation

Shunt generators

These types of generators are used in battery charging and ordinary lighting purposes

Series generators

This type of generators is used commonly as boosters on DC feeders as current generators for welding and arc lamp

Cumulatively compounded generators

This type of generators is used for domestic lighting purposes and to transmit energy over long distances

Differential compound generators

This type of generators is rarely used They are used for special applications such as electric arc welding

132 NUMERICAL EXCERCISES

1 The emf induced in the armature of a 450 kW 250 volt shunt generator is

2588volt when the field current is 200 amp and the generator is supplying power to a load at rated terminal voltage The armature circuit resistance is 0005 ohm Determine (i) load current (ii) power generated (iii) power output and (iv) electrical efficiency Neglect brush contact drop

Solution

Let the load current be I amp Given Ish = 20 A

Armature current = Load current + field current or Ia = I + 20 Induced emf = Terminal voltage + Ia Ra drop

2588 = 250 + (I + 20) times 0005

So (i) Load current I = 1740 amp

(ii) Power generated E Ia = 2588 (1740 + 20) watts = 455488 kW (iii) Power output V I = 250 times 1740 watts

= 435 kW

(iv) Electrical efficiency = 435455488 times 100 = 955

2 A 4 pole wave wound dc shunt generator delivers a load of 45 kW at a terminal voltage of 220 volts Its armature has 150 single turn coils and has a resistance of 001 ohm The air gap flux per pole is 002 weber Shunt field resistance is 50 ohm Calculate the speed at which it is being driven Neglect brush contact drop

Solution Load current I = W

V = 45000220 = 20455 amp

Field Current Ish = V

R s h = 220

50 = 44 amp

Armature current Ia = I +I sh = 20455 +44 = 20895 amp

Armature drop I aRa = 20895 X 001 = 20895 volts Induced emf Eg = V + Ia Ra = 220 + 20895 = 2220895 volts

But induced emf

So N = Eg60 AΦ ZP = 2220895times 60 times2

002 times150 times2 times 4 = 111045 rpm

3 The following data refer to the OCC of a DC separately excited generator at 1000 rpm

Field current in A 00 02 03 04 05 06 07 08 09 10 11 12Armature voltage in V 5 40 75 100 124 145 162 178 188 195 200 205

The machine is now connected as a shunt generator with a total field resistance of 200 Ω and driven at 1000 rpm Estimate graphically (i) the voltage to which the generator will build up at no load (ii) the armature field and load currents when the terminal voltage is found to be 150 V Neglect the effect of armature reaction and brush drop and assume armature resistance Ra to be 08 Ω (iii) Finally estimate the steady state armature current when the machine terminals are shorted

Solution

First of all the OCC of the generator is plotted in a graph paper as shown in figure

(i) The total field circuit resistance is given to be Rf = 200 Ω The Rf line is now drawn on the same graph paper passing through the origin The point of intersection of the Rf line and the OCC decides the final no load voltage and can be read from the graph as 192 V and the corresponding field current is 096 A (ii) Since the terminal voltage V is 150 V (= BM) field current If

is OM = 077 A Generated voltage E is given by AM But we know E = IaRa + V Hence IaRa= E ndash V = AM ndash BM = AB Now from the graph AB = 25 V there4Ia = 2508 = 3125 A So load current IL= Ia ndash If= 3125 ndash 077 = 3048 A

ADDITIONAL PROBLEMS

1 A 4 pole lap wound 115 kW 230 volts dc shunt generator has armatureresistance of 02 ohm and field resistance of 100 ohms Calculate (i) emf

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES

Fig 123 compound generator

Fig 124 Load characteristics of compound generators

127 LOSSES

The losses in a dc machine (generator or motor) may be divided into three classes viz (i) copper losses or electrical losses (ii) iron or core losses and (iii) mechanical losses All these losses appear as heat and thus raise the temperature of the machine They also lower the efficiency of the machine The loss structure maybe demonstrated as

1Copperlosses

These losses occur due to currents in the various windings of the machine When an electric current of I amps flows in a resistance of R ohms the heat energy is lost at the rate of I2R joules sec and the power loss is I2R watts Generators and motors have one or more field circuits and an armature circuit in which such losses occur All resistance losses of this kind are called copper losses The different copper losses occurring in a dc machine are

(i) Armature copper loss = Ia2Ra

(ii) Shunt field copper loss = Ish2 Rsh

(iii) Series field copper loss = Ise2Rse

Armature copper loss will be around 30 to 40 of full load losses where as total field copper losses will constitute around 30 of full load losses

Note There is also brush contact loss due to brush contact resistance (ie resistance between the surface of brush and surface of commutator) This loss is generally included in armature copper loss The brush contact drop is almost independent of Ia For carbon brushes this drop is around 1 V per brush The power loss due to brush contact resistance is 2ebIa where eb is the voltage drop at one brush

2 IRONORCORELOSSES These losses occur in the armature of a dc machine and are due to the rotation of armature in the magnetic field of the poles They are of two types viz (i) hysteresis loss (ii) eddy current loss

(i) Hysteresis loss

Hysteresis loss occurs in the armature of the dc machine since any given part of the armature is subjected to magnetic field reversals as it passes under successive poles Fig 124 shows an armature rotating in two-pole machine Consider a small piece ab of the armature When the piece ab is under N-pole the magnetic lines pass from a to b Half a revolution later the same piece of iron is under S-pole and magnetic lines pass from b to a so that magnetism in the iron is reversed In order to reverse continuously the molecular magnets in the armature core some amount of power has to be spent which is called hysteresis loss It is given by Steinmetz formula This formula is

Hysteresis Loss Wh = η Bmax 16f V Watts Bmax - Maximum flux density in the core f - Frequency of magnetic reversals V - Volume of armature in m3

η - Steinmetz hysteresis coefficient

In order to reduce this loss in a dc machine armature core is made of such materials which have a low value of Steinmetz hysteresis co-efficient eg silicon steel

Fig 124 Occurrence of hysteresis loss

ii) EDDYCURRENTLOSS

In addition to the voltages induced in the armature conductors there are also voltages induced in the armature core These voltages produce circulating currents in the armature core as shown in Fig 125 These are called eddy currents and power loss due to their flow is called eddy current loss The eddy current loss appears as heat which raises the temperature of the machine and lowers its efficiency Core resistance can be greatly increased by constructing the core of thin round If a continuous solid iron core is used the resistance to eddy current path will be small due to large cross-sectional area of the core Consequently the magnitude of eddy current and hence eddy current loss will be large The magnitude of eddy current can be reduced by making core resistance as high as practical The iron sheets called laminations [See Fig 126] The laminations are insulated from each other with a coating of varnish The insulating coating has a high resistance so very little current flows from one lamination to the other Alsobecause each lamination is very thin the resistance to current flowing through the width of a lamination is also quite large Thus laminating a core increases the core resistance which decreases the eddy current and hence the eddy current loss

Fig 125 Eddy current loss Fig 126 Core laminations

Eddy current loss is biven by the formula Pe= Ke Bmax

2 f2t2 V Watts

Where Ke = Constant depending upon electrical resistace of core Bmax = Maximum flux density in wbm2

f = frequency f magnetic reversals in Hz t = thickness of laminations in m and V = volume of core in m3

It may be noted that eddy current loss depends upon the square of lamination thickness For this reason lamination thickness should be kept as small as possible

3MECHANICALLOSSES

These losses are due to friction and windage(i) Friction losses eg bearing friction brush friction etc(ii) Windage loss ie air friction of rotating armature

These losses depend upon the speed of the machine But for a given speed they are practically constant

Note Iron losses and mechanical losses together are called stray losses

CONSTANT AND VARIABLE LOSSES

The losses in a dc generator (or dc motor) may be sub-divided into

(i) constant losses (ii) variable losses

(i) Constant losses

Those losses in a dc generator which remain constant at all loads are known as constant losses The constant losses in a dc generator are(a) iron losses(b) mechanical losses(c) shunt field losses(ii) Variable lossesThose losses in a dc generator which vary with load are called variable lossesThe variable losses in a dc generator are (a) Copper loss in armature winding (Ia

2 Ra) (b) Copper loss in series field winding (Ise

2Rse)

Total losses = Constant losses + Variable lossesNote Field Cu loss is constant for shunt and compound generators

128 POWER FLOW DIAGRAM

The power flow in a dc generator may be represented as

The different power stages are shown as A B and CA ndash The mechanical power input which is the power output of the prime mover is given as Pin = 2πNT

60 watts

B - The power developed in armature is Pa = Eg Ia watts C- The electrical power output is Pout = VIL watts

129 EFFICIENCY Efficiency of a machine is the ratio of its output power to the input power

Efficiency η = Output powerInput power

= Input powerminusLossesInput power = Output power

Output power+Losses

In case of a dc generator depending upon the power stages three different efficiencies can be defined as-

1 Mechanical Efficiency(ηm)

BA = Electrical power developedisinarmature

Mechanical power input2 Electrical Efficiency(ηe)

CB = Electrical power output

Electrical power developedisinthe armature 3 Overall or commercial efficiency (η)

CA = Electrical power output

Mechanical power input = ηmtimes ηe

130 CONDITION FOR MAXIMUM EFFICIENCY The efficiency of a dc generator is not constant but varies with load Fig 127shows the variation of efficiency with load current Consider a dc shunt generator with terminal voltage V delivering a load current of IL amp Generator output = V IL

Generator input = Output + Losses = V IL + constant losses + Variable losses = V IL + Wc+ Ia

2Ra

= V IL + Wc + (IL+ Ish)2 Rsh

The shunt field current Ish is small compared with IL so may be neglected

Generator input = VIL+ IL2Ra + Wc

Now efficiency η iquest outputinput = VI L

VIL+ Ia 2Ra+Wc

Efficiency is maximum when the denominator of the above equation is minimumie

ddIL ( VI L

VIL+ Ia 2Ra+Wc ) = 0

Which gives IL2 Ra = WC

ie Variable loss = constant lossThe load current for maximum efficiency is given by

IL iquest radicWcRa

Fig 127 Variation of efficiency with load

Hence the efficiency is maximum for that load current at which the variable loss is equal to the constant loss

131 APPLICATIONS

Separately excited generators

These types of generators have special applications such as electroplating electro refining of materials because separate supply is required for field excitation

Shunt generators

These types of generators are used in battery charging and ordinary lighting purposes

Series generators

This type of generators is used commonly as boosters on DC feeders as current generators for welding and arc lamp

Cumulatively compounded generators

This type of generators is used for domestic lighting purposes and to transmit energy over long distances

Differential compound generators

This type of generators is rarely used They are used for special applications such as electric arc welding

132 NUMERICAL EXCERCISES

1 The emf induced in the armature of a 450 kW 250 volt shunt generator is

2588volt when the field current is 200 amp and the generator is supplying power to a load at rated terminal voltage The armature circuit resistance is 0005 ohm Determine (i) load current (ii) power generated (iii) power output and (iv) electrical efficiency Neglect brush contact drop

Solution

Let the load current be I amp Given Ish = 20 A

Armature current = Load current + field current or Ia = I + 20 Induced emf = Terminal voltage + Ia Ra drop

2588 = 250 + (I + 20) times 0005

So (i) Load current I = 1740 amp

(ii) Power generated E Ia = 2588 (1740 + 20) watts = 455488 kW (iii) Power output V I = 250 times 1740 watts

= 435 kW

(iv) Electrical efficiency = 435455488 times 100 = 955

2 A 4 pole wave wound dc shunt generator delivers a load of 45 kW at a terminal voltage of 220 volts Its armature has 150 single turn coils and has a resistance of 001 ohm The air gap flux per pole is 002 weber Shunt field resistance is 50 ohm Calculate the speed at which it is being driven Neglect brush contact drop

Solution Load current I = W

V = 45000220 = 20455 amp

Field Current Ish = V

R s h = 220

50 = 44 amp

Armature current Ia = I +I sh = 20455 +44 = 20895 amp

Armature drop I aRa = 20895 X 001 = 20895 volts Induced emf Eg = V + Ia Ra = 220 + 20895 = 2220895 volts

But induced emf

So N = Eg60 AΦ ZP = 2220895times 60 times2

002 times150 times2 times 4 = 111045 rpm

3 The following data refer to the OCC of a DC separately excited generator at 1000 rpm

Field current in A 00 02 03 04 05 06 07 08 09 10 11 12Armature voltage in V 5 40 75 100 124 145 162 178 188 195 200 205

The machine is now connected as a shunt generator with a total field resistance of 200 Ω and driven at 1000 rpm Estimate graphically (i) the voltage to which the generator will build up at no load (ii) the armature field and load currents when the terminal voltage is found to be 150 V Neglect the effect of armature reaction and brush drop and assume armature resistance Ra to be 08 Ω (iii) Finally estimate the steady state armature current when the machine terminals are shorted

Solution

First of all the OCC of the generator is plotted in a graph paper as shown in figure

(i) The total field circuit resistance is given to be Rf = 200 Ω The Rf line is now drawn on the same graph paper passing through the origin The point of intersection of the Rf line and the OCC decides the final no load voltage and can be read from the graph as 192 V and the corresponding field current is 096 A (ii) Since the terminal voltage V is 150 V (= BM) field current If

is OM = 077 A Generated voltage E is given by AM But we know E = IaRa + V Hence IaRa= E ndash V = AM ndash BM = AB Now from the graph AB = 25 V there4Ia = 2508 = 3125 A So load current IL= Ia ndash If= 3125 ndash 077 = 3048 A

ADDITIONAL PROBLEMS

1 A 4 pole lap wound 115 kW 230 volts dc shunt generator has armatureresistance of 02 ohm and field resistance of 100 ohms Calculate (i) emf

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES

The losses in a dc machine (generator or motor) may be divided into three classes viz (i) copper losses or electrical losses (ii) iron or core losses and (iii) mechanical losses All these losses appear as heat and thus raise the temperature of the machine They also lower the efficiency of the machine The loss structure maybe demonstrated as

1Copperlosses

These losses occur due to currents in the various windings of the machine When an electric current of I amps flows in a resistance of R ohms the heat energy is lost at the rate of I2R joules sec and the power loss is I2R watts Generators and motors have one or more field circuits and an armature circuit in which such losses occur All resistance losses of this kind are called copper losses The different copper losses occurring in a dc machine are

(i) Armature copper loss = Ia2Ra

(ii) Shunt field copper loss = Ish2 Rsh

(iii) Series field copper loss = Ise2Rse

Armature copper loss will be around 30 to 40 of full load losses where as total field copper losses will constitute around 30 of full load losses

Note There is also brush contact loss due to brush contact resistance (ie resistance between the surface of brush and surface of commutator) This loss is generally included in armature copper loss The brush contact drop is almost independent of Ia For carbon brushes this drop is around 1 V per brush The power loss due to brush contact resistance is 2ebIa where eb is the voltage drop at one brush

2 IRONORCORELOSSES These losses occur in the armature of a dc machine and are due to the rotation of armature in the magnetic field of the poles They are of two types viz (i) hysteresis loss (ii) eddy current loss

(i) Hysteresis loss

Hysteresis loss occurs in the armature of the dc machine since any given part of the armature is subjected to magnetic field reversals as it passes under successive poles Fig 124 shows an armature rotating in two-pole machine Consider a small piece ab of the armature When the piece ab is under N-pole the magnetic lines pass from a to b Half a revolution later the same piece of iron is under S-pole and magnetic lines pass from b to a so that magnetism in the iron is reversed In order to reverse continuously the molecular magnets in the armature core some amount of power has to be spent which is called hysteresis loss It is given by Steinmetz formula This formula is

Hysteresis Loss Wh = η Bmax 16f V Watts Bmax - Maximum flux density in the core f - Frequency of magnetic reversals V - Volume of armature in m3

η - Steinmetz hysteresis coefficient

In order to reduce this loss in a dc machine armature core is made of such materials which have a low value of Steinmetz hysteresis co-efficient eg silicon steel

Fig 124 Occurrence of hysteresis loss

ii) EDDYCURRENTLOSS

In addition to the voltages induced in the armature conductors there are also voltages induced in the armature core These voltages produce circulating currents in the armature core as shown in Fig 125 These are called eddy currents and power loss due to their flow is called eddy current loss The eddy current loss appears as heat which raises the temperature of the machine and lowers its efficiency Core resistance can be greatly increased by constructing the core of thin round If a continuous solid iron core is used the resistance to eddy current path will be small due to large cross-sectional area of the core Consequently the magnitude of eddy current and hence eddy current loss will be large The magnitude of eddy current can be reduced by making core resistance as high as practical The iron sheets called laminations [See Fig 126] The laminations are insulated from each other with a coating of varnish The insulating coating has a high resistance so very little current flows from one lamination to the other Alsobecause each lamination is very thin the resistance to current flowing through the width of a lamination is also quite large Thus laminating a core increases the core resistance which decreases the eddy current and hence the eddy current loss

Fig 125 Eddy current loss Fig 126 Core laminations

Eddy current loss is biven by the formula Pe= Ke Bmax

2 f2t2 V Watts

Where Ke = Constant depending upon electrical resistace of core Bmax = Maximum flux density in wbm2

f = frequency f magnetic reversals in Hz t = thickness of laminations in m and V = volume of core in m3

It may be noted that eddy current loss depends upon the square of lamination thickness For this reason lamination thickness should be kept as small as possible

3MECHANICALLOSSES

These losses are due to friction and windage(i) Friction losses eg bearing friction brush friction etc(ii) Windage loss ie air friction of rotating armature

These losses depend upon the speed of the machine But for a given speed they are practically constant

Note Iron losses and mechanical losses together are called stray losses

CONSTANT AND VARIABLE LOSSES

The losses in a dc generator (or dc motor) may be sub-divided into

(i) constant losses (ii) variable losses

(i) Constant losses

Those losses in a dc generator which remain constant at all loads are known as constant losses The constant losses in a dc generator are(a) iron losses(b) mechanical losses(c) shunt field losses(ii) Variable lossesThose losses in a dc generator which vary with load are called variable lossesThe variable losses in a dc generator are (a) Copper loss in armature winding (Ia

2 Ra) (b) Copper loss in series field winding (Ise

2Rse)

Total losses = Constant losses + Variable lossesNote Field Cu loss is constant for shunt and compound generators

128 POWER FLOW DIAGRAM

The power flow in a dc generator may be represented as

The different power stages are shown as A B and CA ndash The mechanical power input which is the power output of the prime mover is given as Pin = 2πNT

60 watts

B - The power developed in armature is Pa = Eg Ia watts C- The electrical power output is Pout = VIL watts

129 EFFICIENCY Efficiency of a machine is the ratio of its output power to the input power

Efficiency η = Output powerInput power

= Input powerminusLossesInput power = Output power

Output power+Losses

In case of a dc generator depending upon the power stages three different efficiencies can be defined as-

1 Mechanical Efficiency(ηm)

BA = Electrical power developedisinarmature

Mechanical power input2 Electrical Efficiency(ηe)

CB = Electrical power output

Electrical power developedisinthe armature 3 Overall or commercial efficiency (η)

CA = Electrical power output

Mechanical power input = ηmtimes ηe

130 CONDITION FOR MAXIMUM EFFICIENCY The efficiency of a dc generator is not constant but varies with load Fig 127shows the variation of efficiency with load current Consider a dc shunt generator with terminal voltage V delivering a load current of IL amp Generator output = V IL

Generator input = Output + Losses = V IL + constant losses + Variable losses = V IL + Wc+ Ia

2Ra

= V IL + Wc + (IL+ Ish)2 Rsh

The shunt field current Ish is small compared with IL so may be neglected

Generator input = VIL+ IL2Ra + Wc

Now efficiency η iquest outputinput = VI L

VIL+ Ia 2Ra+Wc

Efficiency is maximum when the denominator of the above equation is minimumie

ddIL ( VI L

VIL+ Ia 2Ra+Wc ) = 0

Which gives IL2 Ra = WC

ie Variable loss = constant lossThe load current for maximum efficiency is given by

IL iquest radicWcRa

Fig 127 Variation of efficiency with load

Hence the efficiency is maximum for that load current at which the variable loss is equal to the constant loss

131 APPLICATIONS

Separately excited generators

These types of generators have special applications such as electroplating electro refining of materials because separate supply is required for field excitation

Shunt generators

These types of generators are used in battery charging and ordinary lighting purposes

Series generators

This type of generators is used commonly as boosters on DC feeders as current generators for welding and arc lamp

Cumulatively compounded generators

This type of generators is used for domestic lighting purposes and to transmit energy over long distances

Differential compound generators

This type of generators is rarely used They are used for special applications such as electric arc welding

132 NUMERICAL EXCERCISES

1 The emf induced in the armature of a 450 kW 250 volt shunt generator is

2588volt when the field current is 200 amp and the generator is supplying power to a load at rated terminal voltage The armature circuit resistance is 0005 ohm Determine (i) load current (ii) power generated (iii) power output and (iv) electrical efficiency Neglect brush contact drop

Solution

Let the load current be I amp Given Ish = 20 A

Armature current = Load current + field current or Ia = I + 20 Induced emf = Terminal voltage + Ia Ra drop

2588 = 250 + (I + 20) times 0005

So (i) Load current I = 1740 amp

(ii) Power generated E Ia = 2588 (1740 + 20) watts = 455488 kW (iii) Power output V I = 250 times 1740 watts

= 435 kW

(iv) Electrical efficiency = 435455488 times 100 = 955

2 A 4 pole wave wound dc shunt generator delivers a load of 45 kW at a terminal voltage of 220 volts Its armature has 150 single turn coils and has a resistance of 001 ohm The air gap flux per pole is 002 weber Shunt field resistance is 50 ohm Calculate the speed at which it is being driven Neglect brush contact drop

Solution Load current I = W

V = 45000220 = 20455 amp

Field Current Ish = V

R s h = 220

50 = 44 amp

Armature current Ia = I +I sh = 20455 +44 = 20895 amp

Armature drop I aRa = 20895 X 001 = 20895 volts Induced emf Eg = V + Ia Ra = 220 + 20895 = 2220895 volts

But induced emf

So N = Eg60 AΦ ZP = 2220895times 60 times2

002 times150 times2 times 4 = 111045 rpm

3 The following data refer to the OCC of a DC separately excited generator at 1000 rpm

Field current in A 00 02 03 04 05 06 07 08 09 10 11 12Armature voltage in V 5 40 75 100 124 145 162 178 188 195 200 205

The machine is now connected as a shunt generator with a total field resistance of 200 Ω and driven at 1000 rpm Estimate graphically (i) the voltage to which the generator will build up at no load (ii) the armature field and load currents when the terminal voltage is found to be 150 V Neglect the effect of armature reaction and brush drop and assume armature resistance Ra to be 08 Ω (iii) Finally estimate the steady state armature current when the machine terminals are shorted

Solution

First of all the OCC of the generator is plotted in a graph paper as shown in figure

(i) The total field circuit resistance is given to be Rf = 200 Ω The Rf line is now drawn on the same graph paper passing through the origin The point of intersection of the Rf line and the OCC decides the final no load voltage and can be read from the graph as 192 V and the corresponding field current is 096 A (ii) Since the terminal voltage V is 150 V (= BM) field current If

is OM = 077 A Generated voltage E is given by AM But we know E = IaRa + V Hence IaRa= E ndash V = AM ndash BM = AB Now from the graph AB = 25 V there4Ia = 2508 = 3125 A So load current IL= Ia ndash If= 3125 ndash 077 = 3048 A

ADDITIONAL PROBLEMS

1 A 4 pole lap wound 115 kW 230 volts dc shunt generator has armatureresistance of 02 ohm and field resistance of 100 ohms Calculate (i) emf

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES

Note There is also brush contact loss due to brush contact resistance (ie resistance between the surface of brush and surface of commutator) This loss is generally included in armature copper loss The brush contact drop is almost independent of Ia For carbon brushes this drop is around 1 V per brush The power loss due to brush contact resistance is 2ebIa where eb is the voltage drop at one brush

2 IRONORCORELOSSES These losses occur in the armature of a dc machine and are due to the rotation of armature in the magnetic field of the poles They are of two types viz (i) hysteresis loss (ii) eddy current loss

(i) Hysteresis loss

Hysteresis loss occurs in the armature of the dc machine since any given part of the armature is subjected to magnetic field reversals as it passes under successive poles Fig 124 shows an armature rotating in two-pole machine Consider a small piece ab of the armature When the piece ab is under N-pole the magnetic lines pass from a to b Half a revolution later the same piece of iron is under S-pole and magnetic lines pass from b to a so that magnetism in the iron is reversed In order to reverse continuously the molecular magnets in the armature core some amount of power has to be spent which is called hysteresis loss It is given by Steinmetz formula This formula is

Hysteresis Loss Wh = η Bmax 16f V Watts Bmax - Maximum flux density in the core f - Frequency of magnetic reversals V - Volume of armature in m3

η - Steinmetz hysteresis coefficient

In order to reduce this loss in a dc machine armature core is made of such materials which have a low value of Steinmetz hysteresis co-efficient eg silicon steel

Fig 124 Occurrence of hysteresis loss

ii) EDDYCURRENTLOSS

In addition to the voltages induced in the armature conductors there are also voltages induced in the armature core These voltages produce circulating currents in the armature core as shown in Fig 125 These are called eddy currents and power loss due to their flow is called eddy current loss The eddy current loss appears as heat which raises the temperature of the machine and lowers its efficiency Core resistance can be greatly increased by constructing the core of thin round If a continuous solid iron core is used the resistance to eddy current path will be small due to large cross-sectional area of the core Consequently the magnitude of eddy current and hence eddy current loss will be large The magnitude of eddy current can be reduced by making core resistance as high as practical The iron sheets called laminations [See Fig 126] The laminations are insulated from each other with a coating of varnish The insulating coating has a high resistance so very little current flows from one lamination to the other Alsobecause each lamination is very thin the resistance to current flowing through the width of a lamination is also quite large Thus laminating a core increases the core resistance which decreases the eddy current and hence the eddy current loss

Fig 125 Eddy current loss Fig 126 Core laminations

Eddy current loss is biven by the formula Pe= Ke Bmax

2 f2t2 V Watts

Where Ke = Constant depending upon electrical resistace of core Bmax = Maximum flux density in wbm2

f = frequency f magnetic reversals in Hz t = thickness of laminations in m and V = volume of core in m3

It may be noted that eddy current loss depends upon the square of lamination thickness For this reason lamination thickness should be kept as small as possible

3MECHANICALLOSSES

These losses are due to friction and windage(i) Friction losses eg bearing friction brush friction etc(ii) Windage loss ie air friction of rotating armature

These losses depend upon the speed of the machine But for a given speed they are practically constant

Note Iron losses and mechanical losses together are called stray losses

CONSTANT AND VARIABLE LOSSES

The losses in a dc generator (or dc motor) may be sub-divided into

(i) constant losses (ii) variable losses

(i) Constant losses

Those losses in a dc generator which remain constant at all loads are known as constant losses The constant losses in a dc generator are(a) iron losses(b) mechanical losses(c) shunt field losses(ii) Variable lossesThose losses in a dc generator which vary with load are called variable lossesThe variable losses in a dc generator are (a) Copper loss in armature winding (Ia

2 Ra) (b) Copper loss in series field winding (Ise

2Rse)

Total losses = Constant losses + Variable lossesNote Field Cu loss is constant for shunt and compound generators

128 POWER FLOW DIAGRAM

The power flow in a dc generator may be represented as

The different power stages are shown as A B and CA ndash The mechanical power input which is the power output of the prime mover is given as Pin = 2πNT

60 watts

B - The power developed in armature is Pa = Eg Ia watts C- The electrical power output is Pout = VIL watts

129 EFFICIENCY Efficiency of a machine is the ratio of its output power to the input power

Efficiency η = Output powerInput power

= Input powerminusLossesInput power = Output power

Output power+Losses

In case of a dc generator depending upon the power stages three different efficiencies can be defined as-

1 Mechanical Efficiency(ηm)

BA = Electrical power developedisinarmature

Mechanical power input2 Electrical Efficiency(ηe)

CB = Electrical power output

Electrical power developedisinthe armature 3 Overall or commercial efficiency (η)

CA = Electrical power output

Mechanical power input = ηmtimes ηe

130 CONDITION FOR MAXIMUM EFFICIENCY The efficiency of a dc generator is not constant but varies with load Fig 127shows the variation of efficiency with load current Consider a dc shunt generator with terminal voltage V delivering a load current of IL amp Generator output = V IL

Generator input = Output + Losses = V IL + constant losses + Variable losses = V IL + Wc+ Ia

2Ra

= V IL + Wc + (IL+ Ish)2 Rsh

The shunt field current Ish is small compared with IL so may be neglected

Generator input = VIL+ IL2Ra + Wc

Now efficiency η iquest outputinput = VI L

VIL+ Ia 2Ra+Wc

Efficiency is maximum when the denominator of the above equation is minimumie

ddIL ( VI L

VIL+ Ia 2Ra+Wc ) = 0

Which gives IL2 Ra = WC

ie Variable loss = constant lossThe load current for maximum efficiency is given by

IL iquest radicWcRa

Fig 127 Variation of efficiency with load

Hence the efficiency is maximum for that load current at which the variable loss is equal to the constant loss

131 APPLICATIONS

Separately excited generators

These types of generators have special applications such as electroplating electro refining of materials because separate supply is required for field excitation

Shunt generators

These types of generators are used in battery charging and ordinary lighting purposes

Series generators

This type of generators is used commonly as boosters on DC feeders as current generators for welding and arc lamp

Cumulatively compounded generators

This type of generators is used for domestic lighting purposes and to transmit energy over long distances

Differential compound generators

This type of generators is rarely used They are used for special applications such as electric arc welding

132 NUMERICAL EXCERCISES

1 The emf induced in the armature of a 450 kW 250 volt shunt generator is

2588volt when the field current is 200 amp and the generator is supplying power to a load at rated terminal voltage The armature circuit resistance is 0005 ohm Determine (i) load current (ii) power generated (iii) power output and (iv) electrical efficiency Neglect brush contact drop

Solution

Let the load current be I amp Given Ish = 20 A

Armature current = Load current + field current or Ia = I + 20 Induced emf = Terminal voltage + Ia Ra drop

2588 = 250 + (I + 20) times 0005

So (i) Load current I = 1740 amp

(ii) Power generated E Ia = 2588 (1740 + 20) watts = 455488 kW (iii) Power output V I = 250 times 1740 watts

= 435 kW

(iv) Electrical efficiency = 435455488 times 100 = 955

2 A 4 pole wave wound dc shunt generator delivers a load of 45 kW at a terminal voltage of 220 volts Its armature has 150 single turn coils and has a resistance of 001 ohm The air gap flux per pole is 002 weber Shunt field resistance is 50 ohm Calculate the speed at which it is being driven Neglect brush contact drop

Solution Load current I = W

V = 45000220 = 20455 amp

Field Current Ish = V

R s h = 220

50 = 44 amp

Armature current Ia = I +I sh = 20455 +44 = 20895 amp

Armature drop I aRa = 20895 X 001 = 20895 volts Induced emf Eg = V + Ia Ra = 220 + 20895 = 2220895 volts

But induced emf

So N = Eg60 AΦ ZP = 2220895times 60 times2

002 times150 times2 times 4 = 111045 rpm

3 The following data refer to the OCC of a DC separately excited generator at 1000 rpm

Field current in A 00 02 03 04 05 06 07 08 09 10 11 12Armature voltage in V 5 40 75 100 124 145 162 178 188 195 200 205

The machine is now connected as a shunt generator with a total field resistance of 200 Ω and driven at 1000 rpm Estimate graphically (i) the voltage to which the generator will build up at no load (ii) the armature field and load currents when the terminal voltage is found to be 150 V Neglect the effect of armature reaction and brush drop and assume armature resistance Ra to be 08 Ω (iii) Finally estimate the steady state armature current when the machine terminals are shorted

Solution

First of all the OCC of the generator is plotted in a graph paper as shown in figure

(i) The total field circuit resistance is given to be Rf = 200 Ω The Rf line is now drawn on the same graph paper passing through the origin The point of intersection of the Rf line and the OCC decides the final no load voltage and can be read from the graph as 192 V and the corresponding field current is 096 A (ii) Since the terminal voltage V is 150 V (= BM) field current If

is OM = 077 A Generated voltage E is given by AM But we know E = IaRa + V Hence IaRa= E ndash V = AM ndash BM = AB Now from the graph AB = 25 V there4Ia = 2508 = 3125 A So load current IL= Ia ndash If= 3125 ndash 077 = 3048 A

ADDITIONAL PROBLEMS

1 A 4 pole lap wound 115 kW 230 volts dc shunt generator has armatureresistance of 02 ohm and field resistance of 100 ohms Calculate (i) emf

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES

Fig 124 Occurrence of hysteresis loss

ii) EDDYCURRENTLOSS

In addition to the voltages induced in the armature conductors there are also voltages induced in the armature core These voltages produce circulating currents in the armature core as shown in Fig 125 These are called eddy currents and power loss due to their flow is called eddy current loss The eddy current loss appears as heat which raises the temperature of the machine and lowers its efficiency Core resistance can be greatly increased by constructing the core of thin round If a continuous solid iron core is used the resistance to eddy current path will be small due to large cross-sectional area of the core Consequently the magnitude of eddy current and hence eddy current loss will be large The magnitude of eddy current can be reduced by making core resistance as high as practical The iron sheets called laminations [See Fig 126] The laminations are insulated from each other with a coating of varnish The insulating coating has a high resistance so very little current flows from one lamination to the other Alsobecause each lamination is very thin the resistance to current flowing through the width of a lamination is also quite large Thus laminating a core increases the core resistance which decreases the eddy current and hence the eddy current loss

Fig 125 Eddy current loss Fig 126 Core laminations

Eddy current loss is biven by the formula Pe= Ke Bmax

2 f2t2 V Watts

Where Ke = Constant depending upon electrical resistace of core Bmax = Maximum flux density in wbm2

f = frequency f magnetic reversals in Hz t = thickness of laminations in m and V = volume of core in m3

It may be noted that eddy current loss depends upon the square of lamination thickness For this reason lamination thickness should be kept as small as possible

3MECHANICALLOSSES

These losses are due to friction and windage(i) Friction losses eg bearing friction brush friction etc(ii) Windage loss ie air friction of rotating armature

These losses depend upon the speed of the machine But for a given speed they are practically constant

Note Iron losses and mechanical losses together are called stray losses

CONSTANT AND VARIABLE LOSSES

The losses in a dc generator (or dc motor) may be sub-divided into

(i) constant losses (ii) variable losses

(i) Constant losses

Those losses in a dc generator which remain constant at all loads are known as constant losses The constant losses in a dc generator are(a) iron losses(b) mechanical losses(c) shunt field losses(ii) Variable lossesThose losses in a dc generator which vary with load are called variable lossesThe variable losses in a dc generator are (a) Copper loss in armature winding (Ia

2 Ra) (b) Copper loss in series field winding (Ise

2Rse)

Total losses = Constant losses + Variable lossesNote Field Cu loss is constant for shunt and compound generators

128 POWER FLOW DIAGRAM

The power flow in a dc generator may be represented as

The different power stages are shown as A B and CA ndash The mechanical power input which is the power output of the prime mover is given as Pin = 2πNT

60 watts

B - The power developed in armature is Pa = Eg Ia watts C- The electrical power output is Pout = VIL watts

129 EFFICIENCY Efficiency of a machine is the ratio of its output power to the input power

Efficiency η = Output powerInput power

= Input powerminusLossesInput power = Output power

Output power+Losses

In case of a dc generator depending upon the power stages three different efficiencies can be defined as-

1 Mechanical Efficiency(ηm)

BA = Electrical power developedisinarmature

Mechanical power input2 Electrical Efficiency(ηe)

CB = Electrical power output

Electrical power developedisinthe armature 3 Overall or commercial efficiency (η)

CA = Electrical power output

Mechanical power input = ηmtimes ηe

130 CONDITION FOR MAXIMUM EFFICIENCY The efficiency of a dc generator is not constant but varies with load Fig 127shows the variation of efficiency with load current Consider a dc shunt generator with terminal voltage V delivering a load current of IL amp Generator output = V IL

Generator input = Output + Losses = V IL + constant losses + Variable losses = V IL + Wc+ Ia

2Ra

= V IL + Wc + (IL+ Ish)2 Rsh

The shunt field current Ish is small compared with IL so may be neglected

Generator input = VIL+ IL2Ra + Wc

Now efficiency η iquest outputinput = VI L

VIL+ Ia 2Ra+Wc

Efficiency is maximum when the denominator of the above equation is minimumie

ddIL ( VI L

VIL+ Ia 2Ra+Wc ) = 0

Which gives IL2 Ra = WC

ie Variable loss = constant lossThe load current for maximum efficiency is given by

IL iquest radicWcRa

Fig 127 Variation of efficiency with load

Hence the efficiency is maximum for that load current at which the variable loss is equal to the constant loss

131 APPLICATIONS

Separately excited generators

These types of generators have special applications such as electroplating electro refining of materials because separate supply is required for field excitation

Shunt generators

These types of generators are used in battery charging and ordinary lighting purposes

Series generators

This type of generators is used commonly as boosters on DC feeders as current generators for welding and arc lamp

Cumulatively compounded generators

This type of generators is used for domestic lighting purposes and to transmit energy over long distances

Differential compound generators

This type of generators is rarely used They are used for special applications such as electric arc welding

132 NUMERICAL EXCERCISES

1 The emf induced in the armature of a 450 kW 250 volt shunt generator is

2588volt when the field current is 200 amp and the generator is supplying power to a load at rated terminal voltage The armature circuit resistance is 0005 ohm Determine (i) load current (ii) power generated (iii) power output and (iv) electrical efficiency Neglect brush contact drop

Solution

Let the load current be I amp Given Ish = 20 A

Armature current = Load current + field current or Ia = I + 20 Induced emf = Terminal voltage + Ia Ra drop

2588 = 250 + (I + 20) times 0005

So (i) Load current I = 1740 amp

(ii) Power generated E Ia = 2588 (1740 + 20) watts = 455488 kW (iii) Power output V I = 250 times 1740 watts

= 435 kW

(iv) Electrical efficiency = 435455488 times 100 = 955

2 A 4 pole wave wound dc shunt generator delivers a load of 45 kW at a terminal voltage of 220 volts Its armature has 150 single turn coils and has a resistance of 001 ohm The air gap flux per pole is 002 weber Shunt field resistance is 50 ohm Calculate the speed at which it is being driven Neglect brush contact drop

Solution Load current I = W

V = 45000220 = 20455 amp

Field Current Ish = V

R s h = 220

50 = 44 amp

Armature current Ia = I +I sh = 20455 +44 = 20895 amp

Armature drop I aRa = 20895 X 001 = 20895 volts Induced emf Eg = V + Ia Ra = 220 + 20895 = 2220895 volts

But induced emf

So N = Eg60 AΦ ZP = 2220895times 60 times2

002 times150 times2 times 4 = 111045 rpm

3 The following data refer to the OCC of a DC separately excited generator at 1000 rpm

Field current in A 00 02 03 04 05 06 07 08 09 10 11 12Armature voltage in V 5 40 75 100 124 145 162 178 188 195 200 205

The machine is now connected as a shunt generator with a total field resistance of 200 Ω and driven at 1000 rpm Estimate graphically (i) the voltage to which the generator will build up at no load (ii) the armature field and load currents when the terminal voltage is found to be 150 V Neglect the effect of armature reaction and brush drop and assume armature resistance Ra to be 08 Ω (iii) Finally estimate the steady state armature current when the machine terminals are shorted

Solution

First of all the OCC of the generator is plotted in a graph paper as shown in figure

(i) The total field circuit resistance is given to be Rf = 200 Ω The Rf line is now drawn on the same graph paper passing through the origin The point of intersection of the Rf line and the OCC decides the final no load voltage and can be read from the graph as 192 V and the corresponding field current is 096 A (ii) Since the terminal voltage V is 150 V (= BM) field current If

is OM = 077 A Generated voltage E is given by AM But we know E = IaRa + V Hence IaRa= E ndash V = AM ndash BM = AB Now from the graph AB = 25 V there4Ia = 2508 = 3125 A So load current IL= Ia ndash If= 3125 ndash 077 = 3048 A

ADDITIONAL PROBLEMS

1 A 4 pole lap wound 115 kW 230 volts dc shunt generator has armatureresistance of 02 ohm and field resistance of 100 ohms Calculate (i) emf

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES

Fig 125 Eddy current loss Fig 126 Core laminations

Eddy current loss is biven by the formula Pe= Ke Bmax

2 f2t2 V Watts

Where Ke = Constant depending upon electrical resistace of core Bmax = Maximum flux density in wbm2

f = frequency f magnetic reversals in Hz t = thickness of laminations in m and V = volume of core in m3

It may be noted that eddy current loss depends upon the square of lamination thickness For this reason lamination thickness should be kept as small as possible

3MECHANICALLOSSES

These losses are due to friction and windage(i) Friction losses eg bearing friction brush friction etc(ii) Windage loss ie air friction of rotating armature

These losses depend upon the speed of the machine But for a given speed they are practically constant

Note Iron losses and mechanical losses together are called stray losses

CONSTANT AND VARIABLE LOSSES

The losses in a dc generator (or dc motor) may be sub-divided into

(i) constant losses (ii) variable losses

(i) Constant losses

Those losses in a dc generator which remain constant at all loads are known as constant losses The constant losses in a dc generator are(a) iron losses(b) mechanical losses(c) shunt field losses(ii) Variable lossesThose losses in a dc generator which vary with load are called variable lossesThe variable losses in a dc generator are (a) Copper loss in armature winding (Ia

2 Ra) (b) Copper loss in series field winding (Ise

2Rse)

Total losses = Constant losses + Variable lossesNote Field Cu loss is constant for shunt and compound generators

128 POWER FLOW DIAGRAM

The power flow in a dc generator may be represented as

The different power stages are shown as A B and CA ndash The mechanical power input which is the power output of the prime mover is given as Pin = 2πNT

60 watts

B - The power developed in armature is Pa = Eg Ia watts C- The electrical power output is Pout = VIL watts

129 EFFICIENCY Efficiency of a machine is the ratio of its output power to the input power

Efficiency η = Output powerInput power

= Input powerminusLossesInput power = Output power

Output power+Losses

In case of a dc generator depending upon the power stages three different efficiencies can be defined as-

1 Mechanical Efficiency(ηm)

BA = Electrical power developedisinarmature

Mechanical power input2 Electrical Efficiency(ηe)

CB = Electrical power output

Electrical power developedisinthe armature 3 Overall or commercial efficiency (η)

CA = Electrical power output

Mechanical power input = ηmtimes ηe

130 CONDITION FOR MAXIMUM EFFICIENCY The efficiency of a dc generator is not constant but varies with load Fig 127shows the variation of efficiency with load current Consider a dc shunt generator with terminal voltage V delivering a load current of IL amp Generator output = V IL

Generator input = Output + Losses = V IL + constant losses + Variable losses = V IL + Wc+ Ia

2Ra

= V IL + Wc + (IL+ Ish)2 Rsh

The shunt field current Ish is small compared with IL so may be neglected

Generator input = VIL+ IL2Ra + Wc

Now efficiency η iquest outputinput = VI L

VIL+ Ia 2Ra+Wc

Efficiency is maximum when the denominator of the above equation is minimumie

ddIL ( VI L

VIL+ Ia 2Ra+Wc ) = 0

Which gives IL2 Ra = WC

ie Variable loss = constant lossThe load current for maximum efficiency is given by

IL iquest radicWcRa

Fig 127 Variation of efficiency with load

Hence the efficiency is maximum for that load current at which the variable loss is equal to the constant loss

131 APPLICATIONS

Separately excited generators

These types of generators have special applications such as electroplating electro refining of materials because separate supply is required for field excitation

Shunt generators

These types of generators are used in battery charging and ordinary lighting purposes

Series generators

This type of generators is used commonly as boosters on DC feeders as current generators for welding and arc lamp

Cumulatively compounded generators

This type of generators is used for domestic lighting purposes and to transmit energy over long distances

Differential compound generators

This type of generators is rarely used They are used for special applications such as electric arc welding

132 NUMERICAL EXCERCISES

1 The emf induced in the armature of a 450 kW 250 volt shunt generator is

2588volt when the field current is 200 amp and the generator is supplying power to a load at rated terminal voltage The armature circuit resistance is 0005 ohm Determine (i) load current (ii) power generated (iii) power output and (iv) electrical efficiency Neglect brush contact drop

Solution

Let the load current be I amp Given Ish = 20 A

Armature current = Load current + field current or Ia = I + 20 Induced emf = Terminal voltage + Ia Ra drop

2588 = 250 + (I + 20) times 0005

So (i) Load current I = 1740 amp

(ii) Power generated E Ia = 2588 (1740 + 20) watts = 455488 kW (iii) Power output V I = 250 times 1740 watts

= 435 kW

(iv) Electrical efficiency = 435455488 times 100 = 955

2 A 4 pole wave wound dc shunt generator delivers a load of 45 kW at a terminal voltage of 220 volts Its armature has 150 single turn coils and has a resistance of 001 ohm The air gap flux per pole is 002 weber Shunt field resistance is 50 ohm Calculate the speed at which it is being driven Neglect brush contact drop

Solution Load current I = W

V = 45000220 = 20455 amp

Field Current Ish = V

R s h = 220

50 = 44 amp

Armature current Ia = I +I sh = 20455 +44 = 20895 amp

Armature drop I aRa = 20895 X 001 = 20895 volts Induced emf Eg = V + Ia Ra = 220 + 20895 = 2220895 volts

But induced emf

So N = Eg60 AΦ ZP = 2220895times 60 times2

002 times150 times2 times 4 = 111045 rpm

3 The following data refer to the OCC of a DC separately excited generator at 1000 rpm

Field current in A 00 02 03 04 05 06 07 08 09 10 11 12Armature voltage in V 5 40 75 100 124 145 162 178 188 195 200 205

The machine is now connected as a shunt generator with a total field resistance of 200 Ω and driven at 1000 rpm Estimate graphically (i) the voltage to which the generator will build up at no load (ii) the armature field and load currents when the terminal voltage is found to be 150 V Neglect the effect of armature reaction and brush drop and assume armature resistance Ra to be 08 Ω (iii) Finally estimate the steady state armature current when the machine terminals are shorted

Solution

First of all the OCC of the generator is plotted in a graph paper as shown in figure

(i) The total field circuit resistance is given to be Rf = 200 Ω The Rf line is now drawn on the same graph paper passing through the origin The point of intersection of the Rf line and the OCC decides the final no load voltage and can be read from the graph as 192 V and the corresponding field current is 096 A (ii) Since the terminal voltage V is 150 V (= BM) field current If

is OM = 077 A Generated voltage E is given by AM But we know E = IaRa + V Hence IaRa= E ndash V = AM ndash BM = AB Now from the graph AB = 25 V there4Ia = 2508 = 3125 A So load current IL= Ia ndash If= 3125 ndash 077 = 3048 A

ADDITIONAL PROBLEMS

1 A 4 pole lap wound 115 kW 230 volts dc shunt generator has armatureresistance of 02 ohm and field resistance of 100 ohms Calculate (i) emf

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES

(i) constant losses (ii) variable losses

(i) Constant losses

Those losses in a dc generator which remain constant at all loads are known as constant losses The constant losses in a dc generator are(a) iron losses(b) mechanical losses(c) shunt field losses(ii) Variable lossesThose losses in a dc generator which vary with load are called variable lossesThe variable losses in a dc generator are (a) Copper loss in armature winding (Ia

2 Ra) (b) Copper loss in series field winding (Ise

2Rse)

Total losses = Constant losses + Variable lossesNote Field Cu loss is constant for shunt and compound generators

128 POWER FLOW DIAGRAM

The power flow in a dc generator may be represented as

The different power stages are shown as A B and CA ndash The mechanical power input which is the power output of the prime mover is given as Pin = 2πNT

60 watts

B - The power developed in armature is Pa = Eg Ia watts C- The electrical power output is Pout = VIL watts

129 EFFICIENCY Efficiency of a machine is the ratio of its output power to the input power

Efficiency η = Output powerInput power

= Input powerminusLossesInput power = Output power

Output power+Losses

In case of a dc generator depending upon the power stages three different efficiencies can be defined as-

1 Mechanical Efficiency(ηm)

BA = Electrical power developedisinarmature

Mechanical power input2 Electrical Efficiency(ηe)

CB = Electrical power output

Electrical power developedisinthe armature 3 Overall or commercial efficiency (η)

CA = Electrical power output

Mechanical power input = ηmtimes ηe

130 CONDITION FOR MAXIMUM EFFICIENCY The efficiency of a dc generator is not constant but varies with load Fig 127shows the variation of efficiency with load current Consider a dc shunt generator with terminal voltage V delivering a load current of IL amp Generator output = V IL

Generator input = Output + Losses = V IL + constant losses + Variable losses = V IL + Wc+ Ia

2Ra

= V IL + Wc + (IL+ Ish)2 Rsh

The shunt field current Ish is small compared with IL so may be neglected

Generator input = VIL+ IL2Ra + Wc

Now efficiency η iquest outputinput = VI L

VIL+ Ia 2Ra+Wc

Efficiency is maximum when the denominator of the above equation is minimumie

ddIL ( VI L

VIL+ Ia 2Ra+Wc ) = 0

Which gives IL2 Ra = WC

ie Variable loss = constant lossThe load current for maximum efficiency is given by

IL iquest radicWcRa

Fig 127 Variation of efficiency with load

Hence the efficiency is maximum for that load current at which the variable loss is equal to the constant loss

131 APPLICATIONS

Separately excited generators

These types of generators have special applications such as electroplating electro refining of materials because separate supply is required for field excitation

Shunt generators

These types of generators are used in battery charging and ordinary lighting purposes

Series generators

This type of generators is used commonly as boosters on DC feeders as current generators for welding and arc lamp

Cumulatively compounded generators

This type of generators is used for domestic lighting purposes and to transmit energy over long distances

Differential compound generators

This type of generators is rarely used They are used for special applications such as electric arc welding

132 NUMERICAL EXCERCISES

1 The emf induced in the armature of a 450 kW 250 volt shunt generator is

2588volt when the field current is 200 amp and the generator is supplying power to a load at rated terminal voltage The armature circuit resistance is 0005 ohm Determine (i) load current (ii) power generated (iii) power output and (iv) electrical efficiency Neglect brush contact drop

Solution

Let the load current be I amp Given Ish = 20 A

Armature current = Load current + field current or Ia = I + 20 Induced emf = Terminal voltage + Ia Ra drop

2588 = 250 + (I + 20) times 0005

So (i) Load current I = 1740 amp

(ii) Power generated E Ia = 2588 (1740 + 20) watts = 455488 kW (iii) Power output V I = 250 times 1740 watts

= 435 kW

(iv) Electrical efficiency = 435455488 times 100 = 955

2 A 4 pole wave wound dc shunt generator delivers a load of 45 kW at a terminal voltage of 220 volts Its armature has 150 single turn coils and has a resistance of 001 ohm The air gap flux per pole is 002 weber Shunt field resistance is 50 ohm Calculate the speed at which it is being driven Neglect brush contact drop

Solution Load current I = W

V = 45000220 = 20455 amp

Field Current Ish = V

R s h = 220

50 = 44 amp

Armature current Ia = I +I sh = 20455 +44 = 20895 amp

Armature drop I aRa = 20895 X 001 = 20895 volts Induced emf Eg = V + Ia Ra = 220 + 20895 = 2220895 volts

But induced emf

So N = Eg60 AΦ ZP = 2220895times 60 times2

002 times150 times2 times 4 = 111045 rpm

3 The following data refer to the OCC of a DC separately excited generator at 1000 rpm

Field current in A 00 02 03 04 05 06 07 08 09 10 11 12Armature voltage in V 5 40 75 100 124 145 162 178 188 195 200 205

The machine is now connected as a shunt generator with a total field resistance of 200 Ω and driven at 1000 rpm Estimate graphically (i) the voltage to which the generator will build up at no load (ii) the armature field and load currents when the terminal voltage is found to be 150 V Neglect the effect of armature reaction and brush drop and assume armature resistance Ra to be 08 Ω (iii) Finally estimate the steady state armature current when the machine terminals are shorted

Solution

First of all the OCC of the generator is plotted in a graph paper as shown in figure

(i) The total field circuit resistance is given to be Rf = 200 Ω The Rf line is now drawn on the same graph paper passing through the origin The point of intersection of the Rf line and the OCC decides the final no load voltage and can be read from the graph as 192 V and the corresponding field current is 096 A (ii) Since the terminal voltage V is 150 V (= BM) field current If

is OM = 077 A Generated voltage E is given by AM But we know E = IaRa + V Hence IaRa= E ndash V = AM ndash BM = AB Now from the graph AB = 25 V there4Ia = 2508 = 3125 A So load current IL= Ia ndash If= 3125 ndash 077 = 3048 A

ADDITIONAL PROBLEMS

1 A 4 pole lap wound 115 kW 230 volts dc shunt generator has armatureresistance of 02 ohm and field resistance of 100 ohms Calculate (i) emf

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES

129 EFFICIENCY Efficiency of a machine is the ratio of its output power to the input power

Efficiency η = Output powerInput power

= Input powerminusLossesInput power = Output power

Output power+Losses

In case of a dc generator depending upon the power stages three different efficiencies can be defined as-

1 Mechanical Efficiency(ηm)

BA = Electrical power developedisinarmature

Mechanical power input2 Electrical Efficiency(ηe)

CB = Electrical power output

Electrical power developedisinthe armature 3 Overall or commercial efficiency (η)

CA = Electrical power output

Mechanical power input = ηmtimes ηe

130 CONDITION FOR MAXIMUM EFFICIENCY The efficiency of a dc generator is not constant but varies with load Fig 127shows the variation of efficiency with load current Consider a dc shunt generator with terminal voltage V delivering a load current of IL amp Generator output = V IL

Generator input = Output + Losses = V IL + constant losses + Variable losses = V IL + Wc+ Ia

2Ra

= V IL + Wc + (IL+ Ish)2 Rsh

The shunt field current Ish is small compared with IL so may be neglected

Generator input = VIL+ IL2Ra + Wc

Now efficiency η iquest outputinput = VI L

VIL+ Ia 2Ra+Wc

Efficiency is maximum when the denominator of the above equation is minimumie

ddIL ( VI L

VIL+ Ia 2Ra+Wc ) = 0

Which gives IL2 Ra = WC

ie Variable loss = constant lossThe load current for maximum efficiency is given by

IL iquest radicWcRa

Fig 127 Variation of efficiency with load

Hence the efficiency is maximum for that load current at which the variable loss is equal to the constant loss

131 APPLICATIONS

Separately excited generators

These types of generators have special applications such as electroplating electro refining of materials because separate supply is required for field excitation

Shunt generators

These types of generators are used in battery charging and ordinary lighting purposes

Series generators

This type of generators is used commonly as boosters on DC feeders as current generators for welding and arc lamp

Cumulatively compounded generators

This type of generators is used for domestic lighting purposes and to transmit energy over long distances

Differential compound generators

This type of generators is rarely used They are used for special applications such as electric arc welding

132 NUMERICAL EXCERCISES

1 The emf induced in the armature of a 450 kW 250 volt shunt generator is

2588volt when the field current is 200 amp and the generator is supplying power to a load at rated terminal voltage The armature circuit resistance is 0005 ohm Determine (i) load current (ii) power generated (iii) power output and (iv) electrical efficiency Neglect brush contact drop

Solution

Let the load current be I amp Given Ish = 20 A

Armature current = Load current + field current or Ia = I + 20 Induced emf = Terminal voltage + Ia Ra drop

2588 = 250 + (I + 20) times 0005

So (i) Load current I = 1740 amp

(ii) Power generated E Ia = 2588 (1740 + 20) watts = 455488 kW (iii) Power output V I = 250 times 1740 watts

= 435 kW

(iv) Electrical efficiency = 435455488 times 100 = 955

2 A 4 pole wave wound dc shunt generator delivers a load of 45 kW at a terminal voltage of 220 volts Its armature has 150 single turn coils and has a resistance of 001 ohm The air gap flux per pole is 002 weber Shunt field resistance is 50 ohm Calculate the speed at which it is being driven Neglect brush contact drop

Solution Load current I = W

V = 45000220 = 20455 amp

Field Current Ish = V

R s h = 220

50 = 44 amp

Armature current Ia = I +I sh = 20455 +44 = 20895 amp

Armature drop I aRa = 20895 X 001 = 20895 volts Induced emf Eg = V + Ia Ra = 220 + 20895 = 2220895 volts

But induced emf

So N = Eg60 AΦ ZP = 2220895times 60 times2

002 times150 times2 times 4 = 111045 rpm

3 The following data refer to the OCC of a DC separately excited generator at 1000 rpm

Field current in A 00 02 03 04 05 06 07 08 09 10 11 12Armature voltage in V 5 40 75 100 124 145 162 178 188 195 200 205

The machine is now connected as a shunt generator with a total field resistance of 200 Ω and driven at 1000 rpm Estimate graphically (i) the voltage to which the generator will build up at no load (ii) the armature field and load currents when the terminal voltage is found to be 150 V Neglect the effect of armature reaction and brush drop and assume armature resistance Ra to be 08 Ω (iii) Finally estimate the steady state armature current when the machine terminals are shorted

Solution

First of all the OCC of the generator is plotted in a graph paper as shown in figure

(i) The total field circuit resistance is given to be Rf = 200 Ω The Rf line is now drawn on the same graph paper passing through the origin The point of intersection of the Rf line and the OCC decides the final no load voltage and can be read from the graph as 192 V and the corresponding field current is 096 A (ii) Since the terminal voltage V is 150 V (= BM) field current If

is OM = 077 A Generated voltage E is given by AM But we know E = IaRa + V Hence IaRa= E ndash V = AM ndash BM = AB Now from the graph AB = 25 V there4Ia = 2508 = 3125 A So load current IL= Ia ndash If= 3125 ndash 077 = 3048 A

ADDITIONAL PROBLEMS

1 A 4 pole lap wound 115 kW 230 volts dc shunt generator has armatureresistance of 02 ohm and field resistance of 100 ohms Calculate (i) emf

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES

Generator input = VIL+ IL2Ra + Wc

Now efficiency η iquest outputinput = VI L

VIL+ Ia 2Ra+Wc

Efficiency is maximum when the denominator of the above equation is minimumie

ddIL ( VI L

VIL+ Ia 2Ra+Wc ) = 0

Which gives IL2 Ra = WC

ie Variable loss = constant lossThe load current for maximum efficiency is given by

IL iquest radicWcRa

Fig 127 Variation of efficiency with load

Hence the efficiency is maximum for that load current at which the variable loss is equal to the constant loss

131 APPLICATIONS

Separately excited generators

These types of generators have special applications such as electroplating electro refining of materials because separate supply is required for field excitation

Shunt generators

These types of generators are used in battery charging and ordinary lighting purposes

Series generators

This type of generators is used commonly as boosters on DC feeders as current generators for welding and arc lamp

Cumulatively compounded generators

This type of generators is used for domestic lighting purposes and to transmit energy over long distances

Differential compound generators

This type of generators is rarely used They are used for special applications such as electric arc welding

132 NUMERICAL EXCERCISES

1 The emf induced in the armature of a 450 kW 250 volt shunt generator is

2588volt when the field current is 200 amp and the generator is supplying power to a load at rated terminal voltage The armature circuit resistance is 0005 ohm Determine (i) load current (ii) power generated (iii) power output and (iv) electrical efficiency Neglect brush contact drop

Solution

Let the load current be I amp Given Ish = 20 A

Armature current = Load current + field current or Ia = I + 20 Induced emf = Terminal voltage + Ia Ra drop

2588 = 250 + (I + 20) times 0005

So (i) Load current I = 1740 amp

(ii) Power generated E Ia = 2588 (1740 + 20) watts = 455488 kW (iii) Power output V I = 250 times 1740 watts

= 435 kW

(iv) Electrical efficiency = 435455488 times 100 = 955

2 A 4 pole wave wound dc shunt generator delivers a load of 45 kW at a terminal voltage of 220 volts Its armature has 150 single turn coils and has a resistance of 001 ohm The air gap flux per pole is 002 weber Shunt field resistance is 50 ohm Calculate the speed at which it is being driven Neglect brush contact drop

Solution Load current I = W

V = 45000220 = 20455 amp

Field Current Ish = V

R s h = 220

50 = 44 amp

Armature current Ia = I +I sh = 20455 +44 = 20895 amp

Armature drop I aRa = 20895 X 001 = 20895 volts Induced emf Eg = V + Ia Ra = 220 + 20895 = 2220895 volts

But induced emf

So N = Eg60 AΦ ZP = 2220895times 60 times2

002 times150 times2 times 4 = 111045 rpm

3 The following data refer to the OCC of a DC separately excited generator at 1000 rpm

Field current in A 00 02 03 04 05 06 07 08 09 10 11 12Armature voltage in V 5 40 75 100 124 145 162 178 188 195 200 205

The machine is now connected as a shunt generator with a total field resistance of 200 Ω and driven at 1000 rpm Estimate graphically (i) the voltage to which the generator will build up at no load (ii) the armature field and load currents when the terminal voltage is found to be 150 V Neglect the effect of armature reaction and brush drop and assume armature resistance Ra to be 08 Ω (iii) Finally estimate the steady state armature current when the machine terminals are shorted

Solution

First of all the OCC of the generator is plotted in a graph paper as shown in figure

(i) The total field circuit resistance is given to be Rf = 200 Ω The Rf line is now drawn on the same graph paper passing through the origin The point of intersection of the Rf line and the OCC decides the final no load voltage and can be read from the graph as 192 V and the corresponding field current is 096 A (ii) Since the terminal voltage V is 150 V (= BM) field current If

is OM = 077 A Generated voltage E is given by AM But we know E = IaRa + V Hence IaRa= E ndash V = AM ndash BM = AB Now from the graph AB = 25 V there4Ia = 2508 = 3125 A So load current IL= Ia ndash If= 3125 ndash 077 = 3048 A

ADDITIONAL PROBLEMS

1 A 4 pole lap wound 115 kW 230 volts dc shunt generator has armatureresistance of 02 ohm and field resistance of 100 ohms Calculate (i) emf

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES

Shunt generators

These types of generators are used in battery charging and ordinary lighting purposes

Series generators

This type of generators is used commonly as boosters on DC feeders as current generators for welding and arc lamp

Cumulatively compounded generators

This type of generators is used for domestic lighting purposes and to transmit energy over long distances

Differential compound generators

This type of generators is rarely used They are used for special applications such as electric arc welding

132 NUMERICAL EXCERCISES

1 The emf induced in the armature of a 450 kW 250 volt shunt generator is

2588volt when the field current is 200 amp and the generator is supplying power to a load at rated terminal voltage The armature circuit resistance is 0005 ohm Determine (i) load current (ii) power generated (iii) power output and (iv) electrical efficiency Neglect brush contact drop

Solution

Let the load current be I amp Given Ish = 20 A

Armature current = Load current + field current or Ia = I + 20 Induced emf = Terminal voltage + Ia Ra drop

2588 = 250 + (I + 20) times 0005

So (i) Load current I = 1740 amp

(ii) Power generated E Ia = 2588 (1740 + 20) watts = 455488 kW (iii) Power output V I = 250 times 1740 watts

= 435 kW

(iv) Electrical efficiency = 435455488 times 100 = 955

2 A 4 pole wave wound dc shunt generator delivers a load of 45 kW at a terminal voltage of 220 volts Its armature has 150 single turn coils and has a resistance of 001 ohm The air gap flux per pole is 002 weber Shunt field resistance is 50 ohm Calculate the speed at which it is being driven Neglect brush contact drop

Solution Load current I = W

V = 45000220 = 20455 amp

Field Current Ish = V

R s h = 220

50 = 44 amp

Armature current Ia = I +I sh = 20455 +44 = 20895 amp

Armature drop I aRa = 20895 X 001 = 20895 volts Induced emf Eg = V + Ia Ra = 220 + 20895 = 2220895 volts

But induced emf

So N = Eg60 AΦ ZP = 2220895times 60 times2

002 times150 times2 times 4 = 111045 rpm

3 The following data refer to the OCC of a DC separately excited generator at 1000 rpm

Field current in A 00 02 03 04 05 06 07 08 09 10 11 12Armature voltage in V 5 40 75 100 124 145 162 178 188 195 200 205

The machine is now connected as a shunt generator with a total field resistance of 200 Ω and driven at 1000 rpm Estimate graphically (i) the voltage to which the generator will build up at no load (ii) the armature field and load currents when the terminal voltage is found to be 150 V Neglect the effect of armature reaction and brush drop and assume armature resistance Ra to be 08 Ω (iii) Finally estimate the steady state armature current when the machine terminals are shorted

Solution

First of all the OCC of the generator is plotted in a graph paper as shown in figure

(i) The total field circuit resistance is given to be Rf = 200 Ω The Rf line is now drawn on the same graph paper passing through the origin The point of intersection of the Rf line and the OCC decides the final no load voltage and can be read from the graph as 192 V and the corresponding field current is 096 A (ii) Since the terminal voltage V is 150 V (= BM) field current If

is OM = 077 A Generated voltage E is given by AM But we know E = IaRa + V Hence IaRa= E ndash V = AM ndash BM = AB Now from the graph AB = 25 V there4Ia = 2508 = 3125 A So load current IL= Ia ndash If= 3125 ndash 077 = 3048 A

ADDITIONAL PROBLEMS

1 A 4 pole lap wound 115 kW 230 volts dc shunt generator has armatureresistance of 02 ohm and field resistance of 100 ohms Calculate (i) emf

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES

So (i) Load current I = 1740 amp

(ii) Power generated E Ia = 2588 (1740 + 20) watts = 455488 kW (iii) Power output V I = 250 times 1740 watts

= 435 kW

(iv) Electrical efficiency = 435455488 times 100 = 955

2 A 4 pole wave wound dc shunt generator delivers a load of 45 kW at a terminal voltage of 220 volts Its armature has 150 single turn coils and has a resistance of 001 ohm The air gap flux per pole is 002 weber Shunt field resistance is 50 ohm Calculate the speed at which it is being driven Neglect brush contact drop

Solution Load current I = W

V = 45000220 = 20455 amp

Field Current Ish = V

R s h = 220

50 = 44 amp

Armature current Ia = I +I sh = 20455 +44 = 20895 amp

Armature drop I aRa = 20895 X 001 = 20895 volts Induced emf Eg = V + Ia Ra = 220 + 20895 = 2220895 volts

But induced emf

So N = Eg60 AΦ ZP = 2220895times 60 times2

002 times150 times2 times 4 = 111045 rpm

3 The following data refer to the OCC of a DC separately excited generator at 1000 rpm

Field current in A 00 02 03 04 05 06 07 08 09 10 11 12Armature voltage in V 5 40 75 100 124 145 162 178 188 195 200 205

The machine is now connected as a shunt generator with a total field resistance of 200 Ω and driven at 1000 rpm Estimate graphically (i) the voltage to which the generator will build up at no load (ii) the armature field and load currents when the terminal voltage is found to be 150 V Neglect the effect of armature reaction and brush drop and assume armature resistance Ra to be 08 Ω (iii) Finally estimate the steady state armature current when the machine terminals are shorted

Solution

First of all the OCC of the generator is plotted in a graph paper as shown in figure

(i) The total field circuit resistance is given to be Rf = 200 Ω The Rf line is now drawn on the same graph paper passing through the origin The point of intersection of the Rf line and the OCC decides the final no load voltage and can be read from the graph as 192 V and the corresponding field current is 096 A (ii) Since the terminal voltage V is 150 V (= BM) field current If

is OM = 077 A Generated voltage E is given by AM But we know E = IaRa + V Hence IaRa= E ndash V = AM ndash BM = AB Now from the graph AB = 25 V there4Ia = 2508 = 3125 A So load current IL= Ia ndash If= 3125 ndash 077 = 3048 A

ADDITIONAL PROBLEMS

1 A 4 pole lap wound 115 kW 230 volts dc shunt generator has armatureresistance of 02 ohm and field resistance of 100 ohms Calculate (i) emf

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES

3 The following data refer to the OCC of a DC separately excited generator at 1000 rpm

Field current in A 00 02 03 04 05 06 07 08 09 10 11 12Armature voltage in V 5 40 75 100 124 145 162 178 188 195 200 205

The machine is now connected as a shunt generator with a total field resistance of 200 Ω and driven at 1000 rpm Estimate graphically (i) the voltage to which the generator will build up at no load (ii) the armature field and load currents when the terminal voltage is found to be 150 V Neglect the effect of armature reaction and brush drop and assume armature resistance Ra to be 08 Ω (iii) Finally estimate the steady state armature current when the machine terminals are shorted

Solution

First of all the OCC of the generator is plotted in a graph paper as shown in figure

(i) The total field circuit resistance is given to be Rf = 200 Ω The Rf line is now drawn on the same graph paper passing through the origin The point of intersection of the Rf line and the OCC decides the final no load voltage and can be read from the graph as 192 V and the corresponding field current is 096 A (ii) Since the terminal voltage V is 150 V (= BM) field current If

is OM = 077 A Generated voltage E is given by AM But we know E = IaRa + V Hence IaRa= E ndash V = AM ndash BM = AB Now from the graph AB = 25 V there4Ia = 2508 = 3125 A So load current IL= Ia ndash If= 3125 ndash 077 = 3048 A

ADDITIONAL PROBLEMS

1 A 4 pole lap wound 115 kW 230 volts dc shunt generator has armatureresistance of 02 ohm and field resistance of 100 ohms Calculate (i) emf

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES

(i) The total field circuit resistance is given to be Rf = 200 Ω The Rf line is now drawn on the same graph paper passing through the origin The point of intersection of the Rf line and the OCC decides the final no load voltage and can be read from the graph as 192 V and the corresponding field current is 096 A (ii) Since the terminal voltage V is 150 V (= BM) field current If

is OM = 077 A Generated voltage E is given by AM But we know E = IaRa + V Hence IaRa= E ndash V = AM ndash BM = AB Now from the graph AB = 25 V there4Ia = 2508 = 3125 A So load current IL= Ia ndash If= 3125 ndash 077 = 3048 A

ADDITIONAL PROBLEMS

1 A 4 pole lap wound 115 kW 230 volts dc shunt generator has armatureresistance of 02 ohm and field resistance of 100 ohms Calculate (i) emf

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES

generated if the brush contact drop is one volt per brush and (ii) flux per pole if the machine is driven at 1000 rpm and 1200 conductors are in the armature slots with generator supply rated load at rated voltage

Ans (i) 24246 volts (ii) 0012123 weberspole

2 A long shunt compound generator delivers a load current of 50 A at 500 Vand has armature series-field and shunt field resistances of 005 W 003 W and 250 W respectively Calculate the generated emf and the armaturecurrent Allow 10 volt per brush for contact drop Ans 50616 volts 52 amp

3 (a) The OCC of a dc generator having Ra = 08Ω and driven at 500 rpm is given below Field current (A) 20 30 40 50 60 70 80 90 Armature voltage (V) 110 155 186 212 230 246 260 271

The voltage induced due to residual field may be neglected The machine is now connected as shunt generator and driven at 500 rpm (i) What should be the field circuit resistance in order to have no load terminal voltage to be 230V Also calculate the critical field resistance (ii) What maximum current can be supplied to the load and at what terminal voltage Assume the speed to remain constant at 500 rpm (iii) What should be the range of variation of field circuit resistance in order to have a terminal voltage of 230V from no load to the full load condition the full load armature current being 20A The speed drops to 450 rpm at full load condition

4 The OCC of a shunt generator running at 850 rpm is given below Field current (A) 08 16 24 32 4 48 56 Armature voltage (V) 28 55 75 82 100 108 115

(i) Find the open circuit induced emfs for field resistances of 22Ω and 33Ω (ii) What should be the field resistance so that the open circuit induced emf at 850 rpm is 100 V (iii) Find the critical speed for the field resistance found in (ii) and (iv) find the critical field resistance at 850 rpm

• 131 Yoke (Frame)
• 132 Pole cores and pole shoes
• 134 Armature Core
• 135 Armature Winding
• 136 Commutator
• Fig16 Commutator segments
• 137 Brushes
• 138 Bearing
• 131 APPLICATIONS
• 132 NUMERICAL EXCERCISES