transformer part 3 of 3

7/28/2019 Transformer Part 3 of 3

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CHAPTER 6THE TRANSFORMER(ili)THE TRANSDUCTOR

This chapter is devoted to the various types of transformer and!eactor. which have been developed for special ~ p p l i c a . t i 0 n s . Mentionis made of instrument t r a n s f o r m ~ r s , auto-transformers, both of fixedand variable ratio, saturable reactors or transductors and m'agneticamplifiers. Since the treat ment is mostly of a d e s c r i p t i v ~ n a t u r e , theo p p o r t u ~ t y is taken to i n t r o d u ~ miscellaneous e x a m l ? l e ~ of a,c.theory for revision and further instruction. It will be appreciated that,in this volume, attention is often given to matte.= which are deemedby the author to be essential. Here he may'b e at variahce with th eexamining authority in that no relevant,questions are asked in theexaminations for marine engineers. It must, however, be concededthat, if the reader haS a desire to increase his electrical engineeringknowledge 'then, he must be'introduced to items of equipment whichwill be encountered to an ever increasing extent as the size andcomplexity of the ship-board electrical system develops.

INSTRUMENT TRANSFORMEItSThis heading covers current transformers and voltage transformers:the latter are sometimes called potential transformers. Instrument 'transformers are used in conjunction with indicating or recordinginstruments, protective equipment, alarms and control gear. Their.. usage gives the following advantages.

1. The indicating and protective instrument movements can bestandardised e.g. 5 A for the current coils and 110 V for t he voltagecoils. The arrangement allows a cheap, workable system of instrumentation which is flexible and yet uses meters which are producedon a bulk basis and are c0mparatively inexpensiv'l.2. The indicating and protecti on equipment can bl,.. isoJatea from th::;high or medIum voltage side of a system. The main insulation betweenthe primary and ~ c o n d a r y systems can be ht'ilt into the instrumenttransformers and the m eter system can utilise this to advantage. Thb.sinstruments, relays, alarm units, etc, can be small, with the minimumof insulation and at a potential to earth which is considered to be safe.

THE TRANSFORMER (iii): THE TRANSDUCTOR 1753. The connections from the transformer secondaries can be made'"light". This allows cheaper and neater wiring and c0ntrol boards canbe located conveniently and a ~ s o m e distance from the heavy powerequipment such as the generators, switchboards, transformers andmotors.In relation to instrument transformers some new te'rms will beintroduced such as burden, class, terminal markings, phase-angle andaccuracy. These terms will be understood when explained in contextbut, since the secondaries feed into systems which are isolated fromthe main system, it is obvious that the total load of these systemsmust be known and that the transformers must be of a size capableof operating the systems efficiently, accurately and effectively. Theratio of the transformer is the prime requirement but its rating mustbe .suited to the load or burden which it is to carry. It can be",entioned that the term "burden" is special to instrument transformers.

SPECIALFUSESC.T.

II i~VFig. 85

THE CURRENT TRANSFORMER (C.T.)

LOAD3000/SA

The main requirements for a C.T. are ratio accuracy, small angularvariation from the 1800 phase displacement between primary andsecondary currents,sufficient rating to operate the connected burden,high insulation strength, mechanical rigidity and ability to withstandheavy momentary currents.


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176 REED'S ADVANCED ELECTROTECHNOLOGYIt will be seen from the diagram of Fig. 85 that the primary forms

part of the main series circuit. In this respect, the C.T. is dissimilar toall other transformers where the primary is connect ed acroSS the sys- .tern and is subjected to the supply voltage.

PRIMARV OFHEAVY GAUGECONDUCTOR(INSULATED) ~AND WOUNDON. INSULATINGS H E E T ~

LAMINATEDIRON CORE

Fig.86a

Fig.86b

THE T R A N S F O R M ~ R (iii): THE TRANSDUCTOR 177Because of the method of connection,.the primary must consist

of a conduct or section which is decided by the magnitude of the'main line current and not by the secondary line curlent. The ampereturns, necessary for the magnetic circuit are similarly decided by thiscurrent, and we have transformer constructions of the types shownin diagrams (a) and (b) of Fig. 86. The former uses a few turns ofthick conductor and is known as a "wound primary", whereas thelatter uses the main cable or busbar as the primary which, in itself,constitutes a single tum. This arrangement is known as a "bar primary". The t ransformers are available in variations of these two basic formsof construction and are usually built by instrument makers orspecialists rath er thim by power transformer manufacturers.THEORY OF OPERATION. From knowledge offundamental transfor,mer theory it known that 12N2 = l iN. and it) practice12N2 isapproximately equal to 11Ni . Reference to the phasor diagram (Fig.87), shows that this approximation is acceptable if 10 , is small. In thedesign of current transformers every attempt is made to minimise 10by reducing its components 1m and Iw '

Awe, .. . "

Fig. 87The iron used for curren t transformers is chosen for its excellentmagnetic characteristics. Thus Mu-metal may used instead ofStalloy since, at low flux-density values, its permeability value maybe seven times that of Stalloy. Modern technology has produced grainoriented filicon irons such as Unidi and Stantranis which have evenmore superior characteristics, and if tl,tin laminations are used withgood insulation between th e laminations, the iron losses can be keptto a m inimum since the flux density is always low for associatedreasons. These requirements will thus make for a small Iw value,


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179178 REED'S ADVANCED ELECTROTECHNOLOGYFor any magnetic circuit Magnetomotive Force ::i:: Flux x Reluctance..r Im N (the energismg ampere-tuins) = wi-A ::i:: BAI-A or. '

Bl BlIm N = .and 1m ==thus 1m is decided by the factors shown and for a C.T. the workingflux density is kept at 1 to 1'5 X 10-4 T as compared to the nonnalfigure of 14 x10-4 T for a: power transformer. The length of the ironcircuit is made as. short as possible and, as mentioI1fd above, iron witha high permeability is used. N should be as large as possible, but thisis not w a y s possible Since frequently a single-tum or "bar:primary"must be wed. .Other e a t u r ~ s ' 'Of construction are the result of the followitlgrequirements. Since the iron circuit must have low reluctance, for thebetter cJass C.Ts., the l a m i n ~ t i o n s have the minimum of joints Or can

be conplete rings. The windings are put on-by hand and leakagereactance is reduced b;y: superimposing or sandwiching the windings.PRECAUTIONS. Since the secondary low voltage system is isolatedfrom the "mains", it is possible to work on this circuit providedappropriate precautions are taken. One rule must, however, alwaysbe observed. Never open the secondary circuit of a C.T. whilst theprimary is energised. The reason forthe rule will become obvious ifreference is made to the phasor d i a g r a m ~

W i ~ h correct functioning 12N2 = ItNt' or the. primary ampereturns are nullified by the secondary a m p e r e - t u m s ~ the main fluxbeing produced by J.mN1 , whose value, by design, is kept to anacceptable figure for the re8'llOns already mentioned. Thuslm.maybe in the region of500 mA for a C.T. of ratio 500/5. Assume thesecondary to develop it terminal voltage of 3 V under correct loadingconditions. If the secondary circuit was to be opened and thedemagnetising turns removed, the primary, magnetising current would'500now become 500 A or = 1000 times greater than the

normal value. The primary magnetising ampere-turns would rise inproportion and, as the magnetic material of tlie core \8 operated ata very low value of flux density, it is capable of being raisedappreciably. The material is' chosen for its magnetic characteristic

THE TRANSFORMER (ill): THE TRANSDUcrORi.e. a steep graph andbjgh saturation value-. It is thus possible for theflux density of the transformer to rise in proportion with the newampere-tum value and the secondary voltage would rise to some3000 V for the example. being considered.It is now evident that a transformer designed to operate with 3 Vacross'its output terminals may well break-down due to insulation-failure. Again if one terminal is earthed then other parts of the secondary system, being subjected to an exceedingly high voltage. may bedamaged. The danger to an operator working on this circuit would beconsiderable - hence the rule regarding open,circuiting the secondaryofaC.T.Unlike the normal transformer, no danger is introduced if the secondary is short-circuited since the maximum current which can flowis 5 A or less, the actual value being decided by the relationSecondary A( = Primary At. Under emergency conditions i t is possible to take an instrument out of circuit by first shorting the terminalsto which the meter is connected. Thus by ensuring that secondarycurrent is still flowing, the instiument can be disconnected, leavingthe shorting link in position. Similarly when inserting the current-coilof a meter into circuit, connect it across the shorting link and thenopen the latter. 1\It should be apparent from the above that working on the secondary side of a C.T. can be hazardous and should not be undertaken unless one is quite confident of the operation. Marine engineers mayhave seen specialists undertake this operation during "setting up?' ortesting procedure and may be tempted to make similar checks.Proceed with caution.

It s appropriate, at this point, t9 note that a special current-circuitswitch, such as an ammeter switch, is frequently provided on a switchboard .This would be used to enable the current to be noted in thethree lines of a three-phase supply. The meter is used with three C.Ts.,one in each line, and the switch is special, in that it shorts out a lineC.T. and then disconnects the instrument. It next connects theammeter acrosS the shorting link of the next line and then breaks thelatter connection. It thus carries out the operation described abovein correct sequence.BURDEN AND CLASS. The burden of a current transformer is theoutput VA value at the rated secondary current. Thus for a 15 VAburden at 5 A, the terminal voltage would be 3 V and the impedanceof the instrument ,or relay circuit must not exceed 0'6 n, otherwiseerrors would be ititroduced into the readings and operation. The

I. appropriate B.S. specification No. 3938 covers current transformersI!


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18180 REED'S ADVANCED ELECTROTECHNOLOGYand sets out the various sizes of burden together with additional useful information.B.S.S. 3938 also sets out the various classes of current-transformers.Those to be used for precision and industrial instruments are listed,together with the permissible errots for each class. A study of thisspecification is advised if more detailed information about instrumenttransformers is required.TERMINAL MARKINGS. For simple current metering conditions them a r ~ of terminals is not important but for the measurement ofpower, power factor and, especially for three-phase instruments, thecorrect sequence of connecting up is essential. To ensure that connectionsare made in accordance with the requirementsof the instrumentS, the primary terminals of the transformer are marked PI andP2 whilst the secondary terminals are marked 81 and S2 The connections would then be such that at the instant when current is flowing in the primary from PI to P2 ,i t is also flowing in the secondaryfrom 81 through the external circuit to S2 A colour (red spot) issometimes used to mark PI and 81 For switchboard work the 82side of the secondary is often connected to earth.

PHASE ANGLE. The phasor diagram shows the secondary currentphasor to be almost anti-phase i.e. almost at 1800 to the primary current phasor. If he angle was 1800 , no phase-angle error would result.Components 1m and Iw cause the angle to be less than 1800 andmethods of minimlsing these have been considered. The angle between1'1 or 12 reversed and II is the phase angle. The angle is +ve if , whenreversed, secondary current leads the primary current. On a lowpower-factor Circuit, this angle could be -ve.For current measurement, 12 should be a defmite known fractionof II and phase-angle error is not important. For power measurements,. current ratio must be accurate and secondary current should be 1800out of phase with the primary current. As this condition does notexist, the phase-angle error may introduce an appreciable error intopower measurements if appropriate precautions are not taken by theinstrument and transformer makers.Example 51. For the barprimary typeof current transformer, shownin Fig. 85, estimate the secondary turns and the burdenif the impedances of the ammeter, wattmeter and power-factor indicatorcurrent-coils are 0'15. 03 and 0'3 ohms respectively.;. . 3000 600 .Smce the ratio of the transformer 18 -5- or -1- ' It follows

THE TRANSFORMER (iii): THE .TRANSDUCTORthat if the primaryAts are to equal the secondaryAts then 1 x 3000== N2 x 5 or N2 =600 turns (in inverse proportion to the currentratio).The 600 turns would be of insulated wire of section sufficient tocarry 5 A.With 5 A flowing, the voltage drop across the ammeter would be5 x 0'15 = 0-75 V. The meter burden would be 5 x 0'75 = 375 VA.Similarly the wattmeter and P.F.I. burdens wouldbe 5 x03 x 5= 7'5 VA. The total burden would be 3'75 + 7'5 + 7'5 = 18'75 VAand a C.T. of rating equal or greater than this would be required.Thus a 30 VA unit would be suitable. Note that this is a case whereinstrument burdens have been added arithmetically, since they aredue mainly to a resistance load, and any error introduced by theassumption is small.

If a suitably rated unit was not available from the standard range,it would be necessary to use two C.Ts. and two independent instrument circuits. Thus the ammeter could be energised from the firstC.T. togethe.. with another instrument like an overcurrent relay, and .the second C.T. could be used for the wattmeter and power-factorindicator. It is usual however, to keep instrument and protective-gearcircuits separate rather than mixed. . .THE VOLTAGE TRANSFORMER


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182 REED'S ADVANCED ELECTROTECHNOLOGYcomponents and Iw are thus reduced and 10 would be muchsmaller than that for a power transformer of equivalent rating.4. Sensitive instruments should be used since these require a smalleroperating current and, if 12 is small then, the internal voltage dropsale leaveN. 'V,

v;

I,

r-v ....

Fig. isThe point to note regarding the use of voltage transformers is that,unlike the C.T., the primaries are connected across the "mains" and ashort-circuit on the secondary would be a short referred to the primary.Short-circuit protection is therefore necessary and high,rupturingcapacity fuses are fitted in the primary circuit. Since this is a highvoltage supply, the fuses must be suitably insulated and a special toolis usually provided to allow their renewal without requiring the

"mains" to be switched off. The mid-point of the secondary circuitcan be earthed so that the maxiinum voltage to earth on this systemis 55 V. Since the instruments are all in ,parallel and not in series, asfor the C.T., the disconnection or reconnection of a suspect meterpresents no difficulty. There is no danger of a high induced voltage,as for the C.T., and short-circuiting of terminals must not be undertaken. 'BURDEN AND CLASS. B.S.S'. No. 3941 specifies a secondary voltageof 110 V when the rated primary voltage is applied. A range ofprimary voltages from 110 V to 220 kV and more (line to line) isspecified together with the classes and permissible errors. The standard

IIIIII

'I

~ ,

THE TRANSFORMER (iii); THE TRANSDUCTOR 183rated burdens, from 15 to some 200 VA, are also listed together withthe type of duty for each class. These are usually similar tei thosementioned for the current transformers. ' .TERMINAL MARKINGS. Single-phase. Primary terminals are markedA and B or N. Secondary terminals are marked a and b or n andthe polarity is such that, when A is positive with respect to B thena is + ve with respect to b .Three Phase. StaIJdard P.Ts. are star-connected. on the primaryand secondary sides. The neutrals may be brol,1ghtout, especially forthe secondary, and other arrangements may be used for special requirements. The primary terminals are marked A, Band C while the secondaries are a, b, and c. The primary neutral (if provided) is Nandthe secondl!l'y neutral is marked as n. The transformer polarity issuch that when A is +ve with respect to N, a is +ve with respect to n.PHASE ANGLE. Ukethe currenttransformer, a voltage transformercan introduce ali error into the metering. Such an error can be bothof magnitude and phase ill terms of the voltage. Ratio and phaseangle errors depend on the relative sizes of the impedance voltaged.rops, and mention has been made as to how these can be kept to aminimum. Note that phase-angle error is only important when themeasuring of power is required. .Example 52. If the voltage transformer shown in Fig. 85 is rated at15 VA, estimate whether a frequency meter can be added to theinstallation. The manufacturer's catalogue lists the burden of thismeter at 110 V, to be 2 VA and that of the voltmeter is 4 VA. Thevoltage windings of the wattmeter and power factor indicator arerated at 3'5 VA and 45 VA respectively at 11 0 V.Total burden of existing instrument voltage windings

= 4 + 3'5 + 4'5 12 VAThe transformer is suitable for a burden of 15 VA, fhus an additional15 12 = 3 VA can be added. It win be possible to add the frequency meter in the manner shown.

THE AUTO-TRANSFORMERFIXED RATIO TYPEThe auto-transformer is a unit employing only one winding which .is tapped to provide the appropriate voltage. From the theory of


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184 REED'S ADVANCED ELECTROTECHNOLOGYoperation set out below. it will be seen that tkere is no electrical difference between the action of a normal double.;.wound trimsformerand that of the autotransformer, and the related ratio of V and I tothe number of turns on the winding still holds good. Because there isonly one winding, it is cheaper than a double-wound unit of the same .kVA rating. Also, as it contains less copper, the copper loss is lowerand the efficiency is higher.The auto-transformer is used to advantage for certain applications,the chief of these being where the output voltage is nO.t greatly dissimilar from th& input voltage. In this form it.is operated as a voltagebooster and is connected into a transmission line at a point wherethe load terminal voltage would be below an acceptable value. In thisposition, its function would be to raise the voltage to the requiredvalue. For marine work, its main application is in relation to motorstarters. Some forms of starters for induction motors operate on theprinciple of lowering the voltage to a motor, in order to limit thestarting-current surge taken from the ship's mains. As the motor runsup to speed, the voltage is increased by tappings on the transformer,until the full supply is attained when it is switcp,ed out of circuit.Since the transformer is used for only a brief period, it s apparentthat, the unit should be as cheap as possible and the aut o constructionsatisfies this requirement ..The autotransformer has an obvious disadvantage in that, the secondary is not isolated from the primary and the benefits of a twovoltage system are not realised. This would apply to a marine installation where the main power system was of the order of 440 V and thesmaUpower system at 110 V. An earth fault on one line of eithersystem, would immediately be felt on the other. Of greater importance is the fact that if a break occurred on the common section ofthe winding, the high voltage would become evident on the low voltagesystem. This will be ~ e e n , if the diagram of Fig. 89b is studied and anoPen circuit imagined .at the point X.

'11 ~ bv '_". 12N,

o 0Fig.89a

I,I ~ N , - N 2VII 12-1

Fig.89b

,qIi;1:,

I

THE TRANSFORMER (iii)! THE TRANSDUCTOR 185THEORY OF OPERATION. Consider the diagrams (a) and (b) of Fig.89. It will be seen that for a normal transformer, if one primary andsecondary terminal is joined- using a pair which are always of thesame polarity as each other, then it is possible to fmd a point on theprimary winding which has the same potential as the remaining secondary terminal. These points could be connected and the secondarywinding dispensed with. Advantage can now be taken of the fact thatI'). and 11 are practically in phase opposition. If the secondary isformed at two tappings on the primary, then current in this section isI'J, - I I and the area of the winding conductor can be reduced, thereby effecting a further saving in cost. The extent of the saving incopper can be investigatt'd thus:SAVING OF COPPER. For both the double-wound and autotransformer it is known that

VI Nl-= V'l Nz !1.. = k= IIIt cantherefore be deduced thatFor the double-wound transformer

Weight of copper in Primary (X IINIWeight of copper in Secondary (X I'lN2Total weight of copper (X (IINI + I1N 2 )

For the auto-transformerThe top section of (Nl - Nz) turns carries 11 so weight of copper(X II (N1 - N z)Bottom section of Nz turns carries (/2 - Ii) so ..-eight of copptra: N1.(I'l _. Idor total weight of copper (X l INt - I I Nz + 11N1- I INz

ex II (N I - 2N1 ) + I").N").Th Wt. of Cu. in autotransformer J1 (Nt 2Nz)+12Nzus =Wi. of Cu. in double-wound transfr. hNl +12 N2


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186 187EED'S A'DVANCED ELECTROTECHNOLOGYDividing top and bottom by l iN,

-2N2 ) + 12 NI _ 2 + 12 N, I I "N'J. 11

:: =NI 12 Nl 12 - + - -+N2 II N2 II

k - 2 + k 2k - 2 k - 1 1 = :=k + k 2ft k k0(a) If the primary to secondary voltage is 10: 9 then -k = 19

9 and the ratio is =10 10 IO 9

Therefore Wt. of Cu. in auto: Wt. of Cu. in double-wound unit = 1: 10 . 1 or Wt of Cu. in auto 10. of that ri,lquired for a normal t r a n s f o ~ m e r .

IO(b) If the primary to secondary voltage is 10: 1 then k9 and the ratio is =10 10

Therefore Wt. of Cu. in auto: Wt. of Co. in double-wound unit = 9: 10 9 or Wt. of Cu. in auto is to of that required fora normal transformer.

Examples (a,.) and' (b) show that'saving is only effective when k isfar r e ~ o v e d from unity.

THE TRANSFORMER (iii): THE TRANSDUCTOR. ,Example 53. ASingle-phase, 400/440 V auto-transformer is used tostep up the voltage and supply a load of88 kV1\, operating at unitypower factor. Neglecting losses and themagnetising current, find theoutput current, the input current and the current in the commonsection of the winding of the transformer.

8800 20 A Load or output current = 440 8800. Input current ;: = 22 A400

Current in common section 22 20 = 2 A. Example 54. Determine the core area, the numberof turns and the position of the tapping point for a 500 kVA, 50 Hz, singlephase 6:6/50 kV auto-transformer, assuming the following values: e.m.f. per tum' = 8 V, maximum flux density = 13 T.leo

Using the transformer e.m.f. equation E = 4'44


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188 189REED'S ADVANCED ELECTROTECHNOLOGYVARIABLE RATIO TYPEThis unit is usually called a Variable Transformer or a Variac, theoriginal trade name, and consists of one winding wound on an ironcore. The core is in the fOFm of a cylinder, built up from annularlamination.stampings and insulated from each other in accordancewith good transformer practice. The winding consistsof enamelledwire put on in toroidal form, as illustrated by the diagram of Fig. 90.A movable carbon brush is arranged on a radial arm, to make contactwith a track formed on the winding by cleaning off the enamel insulation.

CARBON BRUSH

TRACK

ENAMELLEDWIRE

Fig. 90An evident weakness of construction is that, full voltage is appliedalong the length of the track where the insulation between adjacentturns is reduced to a minimum, Furthermore, the collection of grimeand carbon dust at this very point can lead to insulation break-downbetween turns. This results in a shorting of the turns with eventualburn out or an insulation failure along the whole length of the track.Modern techniques of construction include the use of durable synthetic

I

THE TRANSFORMER (ill): THE TRANSDUCTORenanlel and varnish impregnation or "potting", together with a carbonbrush of suitable type, grade and shaped tip. These refinements haveremoved the original causes of failure, and have resulted in this autotransformer being accepted as a reliable item of equipment, muchused in laboratories and the test-rooms of electrical factories. Theunit is built in,size.s up to 2'5 kVA for 230V working, and can bearranged in "ganged" form for 440 V working. In the smaller sizes, itis easily operated by a small sensitive motor and in this form can bebuilt into voltage regulators.' The author knows of no specific marineapplications where Variacs are used for power purposes but the probability of their use in small automatic control and supervisory systemsrnustbe expected.

Example 55. A250/200 V auto-transformer is rated at 2 kVA. Whatis the economy in copper compared with the equivalent doublewound transformer. Estimate the copper section of the, windings.From the d e d u ~ t i o n made alreadyWt. of Cu. in auto-transformer I V11 - - where k =-Wt. of Cu. in double wound transformer - k V2

250 _ 1 4 1- -=Here k = 200- 4 So ratio = -- =5 S 54

Therefore Wt. of Cu. in auto is 5 of that in the double-wound unit.making for a saving of 5 or 80 per cent in the copper required.

2000. 2000Input current = 250 = 8 A Output current = 200 = lOACurrent in common section = 10 - 8 = 2 A

Working on a current densityoft'55 Almm2 the common sectionmould be, of: conductor at least 1'29 mm 2 say 1'3 mm 2 Top sectionshould be 4 times.this i.e. S'2 mm 2


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190 191REED'S ADVANCED ELECTROTECHNOl.,QGYTHREE-PHASE TRANSFORMATIONThe production of a three-phase output supply from a three-phaseinput supply of the same frequency but different voltage value, canbe effected in one of two ways: (a) by the use of one composite threephase unit, (b) by interconnecting three similar bu t separate singlephase units.(a) THE THREE-PHASE TRANSFORMER. The magnetic circuit for thisunit is built up so as to provide three cores on which are wound theprimary and secondary of each phase. The diagram of Fig. 91 showsthe arrangement. It should be noted that the three windings are neverwound on one single limb, since the resultant f l u ~ caused by the threeprimaries would be zero because, being 1200 displaced from"oneanother, the fluxes would cancel. The construction technique is ~ i m -ilar to that for the singlephase unit except that the windings for eachphase al': confined to one core. The most usual arrangement favoursthe connection of either the primary or secondary in delta to enablethe passage of harmonic currents. .

PR*ARY ( j f t deltalA 8 C

p -

b . c oSECONDARY (in star)

a

S 5 !....- - --I- - - F - F l - F- -l - S I- 1--'5 1--'"5l- I- -- -- I- -- ~ ..... f!

I1Fig. 91

Although, up to now, no mention has been made of the generationof harmonics in transformers, the condition can be summarised brieflythus.Assume a sinusoidal applied voltage, then since 0: V. The fluxwaveform will also be sinusoidal bu t pot so the no-load current. This

I

,.

THE TRANSFORMER (iii): THE TRANSDUCTORwill be distort ed for two reasons. 1. Since the BIH curve for the corematerial is not a straight line then a "peaky" current wave results,especially iftheiron is worked near the saturation value. 2. Sincedue to hysteresis, the BIH curve is dissimilar on the ascending anddescending values then the curren t waveform will be dis torted stillfurther, being unsymmetrical about the maximum value. The resultis illustrated by the diagram of Fig. 92.

Fig. 92Fourier's theorem shows that, on analysis, the noload currentwaveform can b e considered to be made up of a fundament al waveform together with those of 3rd, 5th, 7th and 9t h harmonics. Theseare present indecteasing magnitude values and their effects can beminimised by taking appropriate precautions. A delta connection fora threephase winding enables the 3rd hannonic e.m.f.s - the most

important, as produced in each phase, to circulate 3l'd harmonic currents. These currents add to the fundamental to produce the necessarysinusoidal flux wave with the resultant sinusoidal induced e.m.f.s.(b) THREE SEPARATE SINGLE-PHASE UNITS. Here three similartransformers are interconnected into the three-phase primary andsecondary lines. This arrangement is favoured for marine practice,since only one spare unit need be carried. By a system of selectorswitches, the spare unit can be connected into the circuit in the eventof one of the other units failing. The rating of each of the single-phaseunits will only need to be one-third of that of the total output. SeeFig. 93 for the arrangement.


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192 REED'S ADVANCED ELEC'fROTECHNOLOGY

MAIN BUSBARS RyB

3 ~ VLf_J-_-_-_-_-::::::::.:::.:-:::;:::::. . :... .:.:-:::::..:::,---; POWER LOADS'bruaJ. 1m..Jho . .o . .J .STANDBY'__1__ i ~ - L ~ - : : . . _ - = = L f ~ j TRANSFORMERb b. r SWITCHED AS No.1MECHANICALLY INTERLOCKED FOR. SMALL POWERUSE IN ONE POSITION ONLY 3ph LOADS..-..-...

ryb

1ph LOADSFig. 93

METHODS OF CONNECTION1. Star/Star. Diagram (a), of Fig. 94, shows the arrangement and" Vconnection diagram. The phase windings have to withstand only V l

which means that less insulation material is required, a saving in spaceand easier design. The windings have to carry full line current and ifone phase winding should fail, the whole threephase arrangementwould be out of action. For this connection, third harmonic currentswill not flow in the line and distortion of secondary e.m.f. can beexpected. If a neutral line is run from the star point, third harmoniccurrent will flow along this wire an undesirable feature.2. Delta/Del ta. Diagram (b) of Fig.. 94 shows that with this connection, each phase winding must be insulated to full line Voltage value

1but the windings need only be designed to carry . Pf' times line. v 3

f4

,"

f",

fit," ,

i

THE TRANSFORMER (iii): THE TRANSDUCTOR 193current. In the event of one winding becoming faulty the unit neednot be shutdown since, by disconnecting the faulty winding, thesupply can be maintained but with a reduced output. This methodof working is known as "open-delta" or ''vee'', and is an arrangementfavoured by Continental and American marine practice. As no neutralpoint is available, a t h r e e ~ w i r e supply only is used. Any third harmoniccurrents, as set up by third harmonic e.m.f.s the phases, will cir..culate round the closed mesh and will not flow out into the lines. Asseen earlier, the connection avoids distortion of the secondary e.m.f.waveform by internally generated harmonics.3. Delta/Star or Star/Del ta. Diagram (c), of Fig. 94, will stress theimportant that, here the transformation ratio is not merely thetransformerturns ratio but involves the type of connection.

Thus for a delta/star arrangement, transformation ratio = ~= Vph!..j3Vph , Nl= VlN'J h N1 , he ha .were - IS t P se-turns ratio.N'JIrrespective of whether the delta/star or'star/delta connection isused, no trouble is experienced with third harmonic currents sincethey circulate round the mesh. /Ii star"point is available if a four-wirearrangement is required for earthing purposes: For marine works,star/delta working allows an acceptable arrangement since a star-pointon the alternator is usual and a neutral can be introduced at a laterdate, should this be comidered advantageous.

C B A~ ~oc EOB Eoc fOb

po l"'l l - ~ ~m2.2 1l2S

Cl 1 Qlc b a

Fig.94a


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194 REED'S ADVANCED ELECTR


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196 REED'S ADVANCED ELECTROTECHNOLOGYThe turns rati o is in inverse proportion to the currents i.e.

154682 4'43 i.e. 1 :4'43

440The transformation ratio is 3300 = 430

THE SATURABLE REACTOR

75 i.e. 1 :7'5

If the BIHcurve of the iron material, as used for a reactor core, isexamined (Fig. 95a), it will be seen that there is a comparativelystraight portion for which R a: H and the permeability fJ. = BIH islarge and substantially constant. This part of the curve is used by thereactor and transformer designer. At higher values of H as the "knee"of the curve is reached, the characteristic ceases to be linear and theB/H or fJ. value varies from point to point as the graph flattens out.The curve shows that the iron has passed into a state of saturationand in its final state becomes almost horizontal. The j.t value afterdecreasing rapidly has now reached a minimum value.

H_ H-+Fig.95a Fig.95b

The variation of the fJ. value is often referred t o as its incrementalpermeability and, if plotted to a base of H, gives a graph of the form

L..

THE TRANSFORMER (iii): THE TRANSDUCTOR 19 7shown in Fig. 95b. From the work done in Chapter 6, it was seen that

L = ~ " or for a given reactor, L ex fJ.. Thus for an iron-coredreactor, inductance varies with j.t and it follows that if inductance isplotted against H, the curve "'ould be similar to that shown for fJ..It could well happen that the flux density of a reactor core iscarried into the region of saturation by the energising coil being subjected to over-voltage. This could also occur for a ba dly designedtransformer when an increase of the no-load current would result,being accompanied by distortion of the output voltage. waveform .For both conditions, the reactance would be reduced if the inductanceis reduced by saturation and, in the case of the reactor, control of thechoking effect could be used to advantage if the incremental permeability or inductance could be controlled.

H-+Fig. 96

The usual method of achieving saturati on is by providing the ironwith an initial biasing flux-density value. This can be done by fittingthe reactor with an additional winding fed from a d.c. source. Theadditional ampere-turns or initial H value would result in biasing thevalue of B and can also be seen as a movement of the H axis upwards.Irrespective of the polarity or direction 01 the direct current, variation of this biasing flux density would allow the inductance of thereactor to be varied from a value approaching :lero to its maximumvalue. Thus, when the current of the biasing or control coil is at itsmaximum value, the reactor inductance would be a minimum and


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198 REED'S ADVANCED ELECTROTECHNOLOGYwhen the. former is zero, the inductance would have its normal value.This control can be illustrated by the "butterfly" curve shown byFig. 96. Note that the two curves are due to the hysteresis effect ofthe B/H.curve, whilst symmetry about the 11 aXis is due to the reversalof the direction of curfent in the d.c. control coil.If a saturable reactor.is connected as shown in the diagram (Fig.97), the variation of inductance by the control winding affects thereactance and hence the total impedance (Z) offered to the a . c . ~ supply source. The control winding thus produces, by a changing impedance, a change of load current and thus a variation of power assupplied to the load. The arrangement was first used for fading oftheatre lights but an obvious disadvantage was that an alternatingvoltage was induced. into the control circuit. by transformer action.Refinements have resulted from the development work done to makethe unit more'effective, and have produced the transductor, as it isnow more commonly called.

O.C CONT ROt'VOLTA


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200Bt

REED'S ADVANCED ELECTROTECHNOLOGY

t ~ X

'wi."toT--- ....._------------ .....I )'" --_ ...

H ---+

I _-I _I I __---I _-I I

t-II,I,I,---------"'-------....\I ,I _ _ _ ..I _----

Fig. 99I _-I __

......-+B

1"'1 ?

-H H+

.L J)B

Fig. 100

THE TRANSFORMER (iii): THE TRANSDUCTOR 201To continue, remember that the load current is only dependent

on th e resistance of the circuit once the device has saturated. Aswitch if timed to close at an appropriate instant after the start of acycle and open at the end of a half-cycle, would result in voltagebeing applied to the load as shown (Fig. 101). The shaded area showswhen the voltage is effective, and the mean value ofthe current passedby the load canthus be controlled byvarying the instant of closingthe switch.

VorIt

LOAD.J ~ I o ~SUPPLY I

Fig. lOla

Fig.lOtbThe similarity with transductor operation is illustrated by the diagrams of Figs. 98 and 99. With no control current the transductorsupports the full a.c. supply voltage and the only current evident at

the load is the small no-load magnetising current. With the applic ationof control current, the transductor is driven into saturation on the+ve half-cycle by the sum of the load current magnetising m.m.f. andthe control current m.m.f. The instant, at which this occurs, can bevaried by adjusting the level of the control current i.e. the length ofthe ordinate OA of Fig. 99. As stated, upon saturation,the devicecannot support the a.c. supply voltage and the load current is limited


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202 REED'S ADVANCED ELECTROTECHNOLOGYonly by the circuit resistance. On the -v e half-cycles the controlm.m.f. opposes saturation and the'transquc tor supports t he supplyvoltage.' .

The disadvantages of the saturable reactor, as already mentioned,are still present, namely: 1. Due to transformer ac.tion a voltage isinduced into the d.c. cont-rolciruict and because of this an uneven,pulsating direct current is the result. 2. Control is'not as effective asdesired since it occurs for half-cycles only. These defects are overcome by using two transductor elements connected as shown in thediagram of Fig. 1()'2. The connection allows the induced e.m.f.'s inthe control windings to cancel out, and the arrangement can befurther developed to give a characteristic nearer that required of anamplifier.THE MAGNETIC AMPLIFIERFor the arrangement of Fig. 102, the load windings are shownconnected in parallel even though the control windings are in series.The load windings could also be arranged in series bu t the parallelarrangement is easier to understand, and f o l l o ~ the usual practicalmethod.

A.C.SUPPLY

Fig. 102 D.C.::ONTROlII\lPUTThe load windings are connected in opposition so that on alternatehalf-cycles of the supply voltage, the contr"ol ampere-turns drive anopposite core into saturation. Each load winding is designed to sup

port the full supply voltage when the control current is zero.Consider core 1. With any given control current condition, for a+ve half-cycle of supply voltage, the flux increases until a point isreached when the core goes into saturation and the winding cannotthen support the voltage. Core 2 is, at the same time, driven awayfrom saturation and would support the supply voltage but for thefact that the winding of core 1 is across it. Since core I has gone intosaturation, its winding effectively short-circuits the load winding ofcore 2 with the result that the supply voltage is not carried by thetransductor and is impressed across the load, to result in a partialhalf-cycle of current.

THE T R : A N S F O R M ~ R (iii): THE TRANSDUcrOR 203For the -v e half-cyc1e,of the supply voltage the operation is-repeated. Core 1 comes out of saturation but its winding is shortcircuited by that of c()re 2 which has gone into saturation. Thus control is obtained over afull cycle by the d.c. contfolwinding .with itscurrent maintained continuously in one direction. Reversal of th!scontrol current would merely change the order in which the saturation

of the cores occurs, and the controlling effect would"be as it wasoriginally.The diagram of Fig. 103 shows the waveforms associated with themagnetic amplifier in its simple form. The output is alternating,though of a pulsing nature, and can be converted for d.c. operationby the use of a suitably rated rectifier. . .

I

/III

Fig. 103

I1

CONTROL CURRENT


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204 REED'S ADVANCEDELECTROTECHNOLOGYCURRENT GAIN. This is a measure of the amplifying property of thearrangement and can be simply deduced. Under the controlled condition i.e. when one core is saturated and the effect of the other is off-"set by the short-circuiting of its winding, there can be no change offlux. Any increase of load ampere-turns must be balanced by an equaland oppcsite increase of control At, so that the resultant m.m.. forthe core remains the same. Since high quality iron is used and workedat high permeability, the m.m.. necessary to cause saturation isrelatively small and can be neglected. Thus, under saturated conditions,we can write

IleNe = IN or Ie NeN where N" = the number of turns ofthe control winding and Ie the control current. N = the number ofturns of the main winding and I the load current. The relationshipshows the current gain to be proportional to the turns ratio of loadand control windings.

leN"Also since IN load current is dependent on the turns ratio

and the control current. It is independent of the supply voltages orload resistance and the amplifier can be regarded as a constant currentsource.Before leaving the current waveforms shown on the diagram ofFig. 103, it is pointed out that, since the control current is fed troma d.c. source its value might be assumed constant being decided bythe control circuit resistance. The above deduction however, showsthat if flux is constant, IeNe = IN and if I varies, Ie will vary.Hence a reflection of the load-current peaks is evident in the controlcurrent.When the cores are unsaturated, the impedance of the load windingis large and only ;l small magnetising current will flow, which lags thesupply v0ltage by nearly 90. When control is ineffective, I is deter

Supply voltagemined by . , and since we have the relation I kle,Load resIstancethe magnetising current being neglected, the characteristic shown byFig. 104 can be deduced. This indicates that the magnetic amplifiercan be used to advantage where a large load current is tel be variedin a linear. manner by a small control current.

THE TRANSFORMER (iii): THE TRANgDUCTOR 205

f...z:t" /0..JI

CONT 1t0L CURReNT

Fig. 104,D.C.INPUT

LOAD

Fig. 105

CVUENT=-1:

A.C.SUPPLY


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206 !tEED'S ADVANCED ELECTROTECHNOLOGYFEEDBACK. Another way of showing the construction and diagramoJ a magnetic amplifier is by Fig. 105. From this it is seen that, theinductance of the a.c. windings is controlled by a d:c. winding whichcan consist of two sections in series opposition to eliminate theinduced alternating voltage. As seen above, by suitably proportioningthe ratio of d.c. to a.c. turns, the amplifier is effective with an outputpractically independent of the load circuit resistance. When d.c. amplification is required, a rectifier is placed in series with the a.c. windings. The amplifier will not however, discriminate between thepolarity of d.c. inp ut signals. ,

To increase the gain of the device and to make it polarity sensitive,positive feedback is employed. For the type of construction justmentioned, a second winding can be placed on t he centre limb andconnected to the d.c. terminals of a further bridge rectifier ,in serieswith the load rectifier (Fig. 106). .D.C.INPUT

A.C.SUPPLY

Fig. 106

THE TRANSFORMER (iii): THE TRANSDUCTOR 207With the input signal in one direction the current throughfeedback winding will be aiding the control winding and increasingthe gain. With a reversed signal, the current through the feedbackwinding will remain in the same direction and will now oppose thecontrol current to reduce the gain. The new oharacteristicis shown

by Fig. 1,07. Current gains of up to 300 and power gains of 104 arepossible with this arrangement, which can be developed further forspecialised work. Thus a bias winding can be added to move the nosignal operating point and if two amplifiers are used in p u s h ~ p u l l , areversible output with no standing current is possible. The amplifierscan also be used in cascade for increased gain.

T--- ...~ ~ . " o r iD

CONTROL CURRENT +

Fig. 107

The magnetic amplifier has been used in various control schemesfor many marine applications. It' formed the basis of the B.T.H."Magnastat" and the G.E.C. "Accurex" automatic voltage regulatingsystems and can be found in associatims. It must be stressedhowever that recent developments in semiconductor technology indicate that usage of thy "Mag-amp" will decrease. Devices, such as theThyristor orS. C.R: (silicon controlled rectifier), can be developed tohave characteristics similar to a magnetic amplifier,and. since the disadvantages of the lat ter include its size and weight then, the electronicalternative is to be preferred once the cost of production is reducedand availability improves.


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208 REED'S ADVANCED ELECTROTECHNOLOGYMISCELL.A.NEOUS EXAMPLES

Example 57. A coil of unknown inductance and resistance is connected in series with a 25 n, non-inductive resistor across 250 V,50 Hz mains. The p.d. across the resistor is found to be 150 V andacross the coil 180 V. Calculate the value of coil resistance and inductance .. Find also the power factor of the coil.

r - 1 5 0 V ~ 1 8 0 V = 1~V y. __ , - - _ l ~ O YI -

IIIIIIIIIIIIIIIIIVC.i

Vx

Fig.108a

V-2S0VIIIIII,1I I .I -I I

I :I II II II II II I, II I


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21110 ~ D ' S ADVANCED ELECTROTECHNOLOGYImpedance of new coil on 50 Hz = ,11050 2 + 471'l

:;:: 100 y'10'52 +4'7P= l00y'I11 + 22'2

100 y'133'2 = 1155 n120 Current taken by coil = i155'5 = 0'104 A

220Required impedance on 220 V circuit = 0'104 = 2120 nThe reactance of the 220 V circuit would be= y'2-1202 -10502 = looy'ii-z2 '::"'1O'S2 = 100y'450-111

:;:: 100y'339 = 1840 nReactance of-new coil on 50 Hz is 471 n

6On 60 Hz it is 471 x "5'= 565'2 nNow the reactance of the required capacitor arrangement mustcancel this inductive reactance and provide the additional reactancefor the 220 V circuit, i.e. it must be 1840 + 565,2 = 24052 n.The catch in the problem is involved wit h this point. Note it. 1

106 The required value of capacitance is given by Xc = 21ffC 106 106 102orC = - - - - = =21ffXc 2 x 3 .14 x 60.x 2405 6'28 x 6 x 2'405100 = 1'1 F90'62 p

This could be made up from capacitors of IpF and 0-lpF in parallel.

. Example 59. A h r e e ~ p h a s e , 440 V, 50 Hz, 90 kW "circulating waterpump" motot has an efficiency of 82 per cent and operates at a:power factor 0[08 (lagging). Calculate (a) the kVA input to themotor. (b) the load current. If the motor is connected to the mainswitchboard by a three-core, 95 mm2 cable, 100 m long; calculatethe cable line-voltage drop. Take the resistivity of copper as 17 pO mm,

';!\I'i;fi!l

!\"

I\,

THE TRANSFORMER (iii): THE TRANSJ)UCfOR. . ('\1'\ 100Motor power Input == JfV x 82 :;= 109'76kWApparent power input =

kVALoad current = y'3kVResistance of cable core

kW 109'76== :;:: 137'2 kVA (a)cos


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212 REED'S ADVANCED ELEcrR.OTECHNOLOGY93Secondary phase current = Vl = 54 A

. 10_1_ =Primary phase current =S4x 13'5 A440

Primary line current = 135 x Vl = ,233 A

Supply transformer.. Secondary line current = 23'3 A

Secondary phase current = 23'3 A (since the Windi.ngs areconnected in star)440/Vl = 1'76 Arimary phase current = ,123'3. x 3300

Primary line current = 1'8 x Vl = 3'1 AAssuming no losses

kVA input from supply = kVA output from lighting transformerkW _ 15- 0.85 = 1765kVA.= cvs.

THE TRANSFORMER (iii): THE TRANSDUCTOR 213CHAPTER 6

PRACTICE EXAMPLES1. . A direct-reading wattmeter having a 5 A current-coil and a '

-110 V, voltage-coil is to be used to measure the power in asingle-phase, 66 kV circuit. ciurying a maximum current of100 A. State the appropriate ratios for the instrument tr;msformers and calculate the constant by which the wattmeterreading must be multiplied to measure the power consumed.2. A 75 kW, 415 V, three-phase induction motor has a full-load efficiency of 80 per cent. The input line currentis to be

m e a s u r ~ by' a C.T. operated ammeter. Suggest an appropriateratio fortheC.T. if the full-scale deflection of the ammeter is5 A. State the.expected ammeter reading. The power factor ofthe motor on full load is 0'87 (lagging).3. A single-phase wattmeter with 5 A and 250 V ranges is usedlJ1 conjunction with a 25/5 C.T. to measure the power of onephase of a balanced, three-phase, star-connected load. If theload absorbs 12 kW from a three-phase, 415V, 50 Hz, supplyand the power-factor is 0'8 (lagging), calculate (a) the wattmeter reading (b) the impedance per phaseof the load ( c) if aphase impedance consists of a resistor and reactance in series,calculate their phase values.4. The output power of a 415 V. three-phase alternator supplyinga balanced load is measured by one wattmeter. If the current. transformer has a 2515 ratio and the wattmeter reading is 7 kW,what is the total "Power?S. For recording the input to a three-phese, 7"5 kW inductionmotor, which is rated to take a line curren Ilf 14 A at 415 V,the following instruments are to be used.Ammeter OlO 5 A" of resistance 0-08 n. Voltmeter 0 to 120 V, of resistance 3636 n. Three-phase Wattmeter (two element type). Current coils oto 5 A, of resistance 0'1 n. Voltage coils 0 to 120 V, ofresistance 4000 n.Estimate the ratio and burden rating of suitable current andvoltage transformers.


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214 REED'S ADVANCED ELECTROTECHNOLOGY6. The primary and secondary voltages of an auto-transformerare 500 V and 400 V respectively. Find the current distributionin the windings when the secondary current is 100 A and calculate the economy of copper for this arrangement. .7. The possibility, is to be considered, of using a o u b l e - w o u n d ~

6 kVA, 250/150 V, single-phase transformer, as an autotransformer on 400 V mains, to supply a load of 12 kVA at250V. By checking the current loading of the windings and apractical connection method, determin,e the suita.bility of the. arrangement.8. Find the values of the currents flowing in the v ~ r i o u sbranches of a three-phase, star-connected auto-transformer loadedwith 400 kW at a power factor o f 0'8 (lagging) and having aratio of440/550 V. Neglect voltage drops, magnetising currentand all losses in the transformer.9. A440 V, three-phase induction motor is to be started by the use of a delta-connected auto-transformer provided with a 70 per cent tapping on each phaSe winding. Find the voltage applied across the motor terminals at the stage of starting when the transformer tappings are in circuit.

10. A three-phase, 440 V, 40 kW induction motor has anefficiency of82 per cent and operates at a power factor of0'85 (lagging). When direct-on started, the motor takes a. current of 6 x full-load current and produces a torque. of 1'5 x.fuU-loadtorque. Calculate the current taken from the supplyand the ratio of starting to fuU-load torque if the motor isstarted through an auto-transformer having a 75 per centtapping. This transformer is star-connected.

CHAPTER 7

THE A.C. GENERATOR 6)In Chapter 7 of Vol. 6 the reader was introduced to the basic lawsof electromagnetic induction, and the principles of e.m.f. generationby dynamic induction were also considered. The generation of analternating voltage was seen to be accomplished by a comparativelysimple arrangement but, if the a..c. generator was to be functional asa machine; distinct from the d.c. generator then, it \lIaS evident tqatmuch development work was necessary before it became a commercial proposition. The advantages of a.c. over d.c., from the generationand utilisation pqirtt of view, are many and wiI.l-be outlined later butif the generators are compared, it is apparent that the a.c. generator. scores because it has no commutator. Furthermore, since it is merelya matter of moving conductors with respect to the magnetic fieldthen it is immaterial as to whether the field moves and the conductorstpe stationary or vice verSa. Thus we have two methods of operationopen to; the machlne designer but the arrangement of a moving fieldtRd fixed armature conductors is usually favoured because of reasonstQ be mentioned. The basics of machine construction are thereforeconsidered before the main substance of a.c. generator theory isdetailed.mE L T E R N A T I ~ G - C U R R E N T GENERATOR, In the interests of standardisation the title of "synchronous a.c.pnerator" or more simply the "a.c. generator" is now s u g g ~ s t e d tocompare with that of "d.c. generator" but the term alternator will.,c:ontinue to be used by engineers whilst the new terminology is being

a c ~ p t e d and established. The main types of a.c. generators in generaluse are now c o n s i d ~ e d .ROTATING-ARMATURE TYPE..;.In the C?lementary generator, a.c.is always produced. The provisionof a reversing switch - the commutator, resUlts in d.c. but, if a.c. isrequired then slip-rings are necessary. A machine of this type is'constructed like a d.c. generator with a fIXed-field system and a .moving armature. It is known as a "rQtating-armature" alternator andis an acceptable arrangement for smlJlI machines up to about 40 kVAat a voltage of 450 V. It would be a cheaper machine than as a.c. generator constructed on the rotating-field.principle, because a d.c. gen


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546 REED'S ADVANCED ELECI'ROTECHNOLOGYand minimum value of rectified voltage = 5515 x cos.= 5515 cos 30= 5515 x 0-866= 4775 Vor voltage dropped = 5515 - 4 77 5 = 740 V peak to peak740=.-- x 100 = 13'4 per cent_5515

13-4As before, ripple voltage = = 6'7 per cent.2

CHAPTER 61. For tll.e voltage transformer. the appropriate ratio would be

6600 60 'V.T. ratio = 110 = T' The voltage will not vary appreciablyand this ratio should be suitable.

For the current transformer the appropriate ratio. to give100 20f u l l ~ deflection, would be C.T, ratio .,. T =I' Since

current is a variable quantity depending on the load, in theopi,nion of the author, it is always safer to allow for overload conditions. Thus the full-load reading should not occur at full scale and this can be achieved' by psing a larger C.T, ratio.,. In this case 1 or would be suitable. Normal reading would then occur at about two-thirds full scale . .By using a 30 C.T. ratio, the wattmeter. beingused on the1 .5 A and 110V ranges, must hJve its reading multiplied by thefactors of 30 for the current and 60 for the voltage.

The required constant is 30 x.6O = 1800. 2. Motor output = 75 kW

_ 75 X 100 93-75 kWMotor iriput - 8 0 ' 93'75 or = 107'76 kVA0'87

107760 107760The line ,current would be V3 x 415 71S'78__ 149-92 A


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549548 REED'S ADVANCED ELECTROTECHNOLOGYA 150/5 ratio current-transformer would be suitable bu t itis usual to allow a portion of the scale for noting starting andoverload conditions and a 250/5 ratio C.T. would be the better

149'92proposition. I f used, the meter reading would be ---SO' theI50 C.T. ratio being taken into account.Reading 3 A for normal running conditions.

123. (a) The three-phase apparent power 0.8 15 kVA15000 1000The line current ::=

.J3x415 1'732 x 27'67100047'82 "" 20'91 A

The wattmeter reading would be secondary C.T. current x phasevbitage x load power factor.25 5Since the C.T. ratio =: 5 or 1 20'91 415reading -- x - x Og = 800W5 .j 3

Note. The result could be arrived at, in a more direct fashion,12since the power per phase or 4000WkW-3

. 5The C.T. rabo reading 4000 800W5It is pointed out that, since the load is balanced and starconnected, the voltage-coil of the wattmeter is connectedbetween a line and the star-point. The wattmeter is therefore

415 subjected to ' volts, ie. a value within the rated range.

SOLlITIONS TO PRACTICE EXAMPLESI 415 239;6tb ) The impedance per phase = .j 3 x 20'91 20'91

= 11'48n(c) The equivalent resistance :::: 1148 x08 9'184 n

.The equivalent reactance == 11 '41$ x06 6'888 n.4. Since the power is being measured by one wattmeter, it isevident that a neutral or star-point is available or has been artificially created and the total power will be three times thepower measured - which is phase power.

Although phase voltage is applied to the instrument, reducedcurrent in the ratio of 2: or Tis being applied. :. The wattmeter constant for the arrangement would be five and phasepower would be 7 x 5 = 35 kW.

The total three-phase power == 3 x 35 = 105 kW.5. . Ratio of C.T.'s = 25/5 or 5/1

440Ratio ofV.T.'1 = 110 or 4/1Ammeter voltage drop at 5 A = 5 x 008 = 04 VAmmeter burden = 5 x 0-4 = 2 VAWattmeter voltage drop at 5 A = 5 x 0'1 :::: 0'5 VWattmeter burden per current coil =: 5 x 0'5 = 2'5 VA:. C.T. burden = 2 + 2'5 = 4'5 say 5 VA. Thi s is a minimum value, a tral1sformer capable of a larger burden say I 0 VAwould give greater accurac y on overload and would allow forextension of instrumentation should this be required.

3120 10 X 10 = 33 rnAVoltmeter current = =, 3636 303


24/25

550 REED'S ADVANCED ELECTROUCIINOLOGYVoltmeter burden : : ' 120 )( 33 X 10-3 : : 3;3.)C. 1'2 = 3-96 VA

_ ' i 20 _ 12 )( 103 Wattmeter voltase-coil eurren! - 4000 - 400:: lOrnA

Wattmeter burden per vo l " coil :: 30 )( 10-3 X 120 = 3 ,x 1:2 = 3-6 VA V.T. burden = 3-96 + 3-6 = 7'56 say 10 VA. Hereagain, a transformer rating greater than the bueden 11 usual.

500 56. Transformation J'lltio = = k = 4400 Secondary. current = l00A'and since primary kVA = ,' secondary kVA

500 x 400 x ioo 400 :: 400 x 100 or II = 1000 1000 500 5 SOA

Thus the single section of winding carries 80 A and thecommon section of winding carries (1 00 80). = 20 A,Wt of Cu in auto-transformer 1

Now Wt of Cu in double-wound transformer = k 4 1= 1 = 1 - - =5 5 5

4 4 . i.e. SOT 80 per cent of copper is saved.

551 6000

SOLUTIONS TO PRACTICE EXAMPLES7. Normal current rating of L.T. winding .!: 150 = 40 A

6000 Normal current rating of H.T. winding = 250 = 24 A With theL.T. winding connected in series with the H.T.winding so that the latter forms the common section, the outputcurrent would be 40 + 24. = 64 A. Thus the maximum loa


25/25

552 REED'S ADVANCED ELECTRO', ECHNOLOGY9. The most usual method ,of using a three-phase auto-tr ansformeris to have the phase windings connected in star. The catch in

R.

this problem is the fact that, a delta connection is used and forsuch an arrangement, the secondary line voltage is not 70 percent of the primary voltage, when the 70 per cent tapping isused. This is illustrated by the diagrams and solution

L'S",/O.7VRY I ' . . Q P Q Q g ~ 1 O,3VYa VI2l - t - - - - - - - - - = = ~ ~

The phasor diagram shows the output voltage between linesas given by the solutionVL 2 = {(0'72 + 0'3 2) (2 x0'7 x 0'3 x cos 60) } V

L1

Note here VL is the line voltage = Vay = VYB:. VJ-2 = (0.49 + 009) - (042 x 0.5) VL 2= (0'58 -0'21) VL 1or V 1-2 = ";0'37 VL = 061 VL (approx.)The voltage applied to the motor terminals would be

0'61 x 440 = 2684 V.JO. Motor power output == 40 000 W

. 40000Motor power mput = watts08240000

Motor truee-phase rating = 0'82 x 0,8'5 volt amperes

SOLunONS TO PRACTICE EXAMPLES 55340 000 = 75,3 AFull-load curren t = ..;3 x 0-82 x 0-85 x 440

Direct-on starting current = 75'3 x 6 = 452 AAs already noted for a star-connected arrangement, the out

put voltage is proportional to the transformer tapping. Thusthe output voltage = 75 per cent of the input voltage. and inconsequence the starting cutrent will be only 75 per c ent of thedirect-on starting value because of the reduced impressed voltage on the motor. .Thus starting current = 075 x 452 339 A

Agam due to transformer action and the fact thatSec kVA = Prim kJI'A, the primary or supply current is 75per cent of the secondary current = 0-75 x 339 amperes.. Thus supply current = 254 A

Note. This latter part of the question is better answered witha knowledge of induction motor theory and is fully dealt within the appropriate section of this book. Here we can say that,torque is proportional to the voltage squared and therefore, ifthe voltage is reduced to a value then, the torque will be

(3)2 9 9reduced to '4 16 of the starting value = 16 x 1-5of the fuUlo ad value = 0'562 5 x 1' 5 x full-load value= 0-844 x full-load value_

transformer part 3 of 3

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