electric machines i - philadelphia university · 2018. 1. 22. · for short shunt cumulatively...
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Electric Machines I DC Machines - DC Generators
1
Dr. Firas Obeidat
2
Table of contents
1 ⢠Construction of Simple Loop Generator
2 ⢠Working of Simple Loop Generator
3 ⢠Types of DC Generators
4 ⢠The Terminal Characteristic of a Separately Excited DC Generator
5 ⢠The Terminal Characteristic of a Self Excited Shunt DC Generator
6 ⢠The Terminal Characteristic of a Self Excited Series DC Generator
7 ⢠The Terminal Characteristic of Cumulatively Compound DC Generator
8 ⢠E.M.F. Equation of DC Generator
9 ⢠Total Loss in a DC Generator
10 ⢠Power Stages and Efficiency
11 ⢠Voltage Regulation
12 ⢠Uses of DC Generators
Dr. Firas Obeidat Faculty of Engineering Philadelphia University
3 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
Construction of Simple Loop Generator
A single turn rectangular copper ABCD rotating about its own axis in a
magnetic field provided by either permanent magnet or electromagnet.
The two ends of the coil are joined to slip ring âaâ and âbâ which are
insulated from each other and from the central shaft. Two collecting
brushes press against the slip rings; their function is to collect the
current induced in the coil and to convey it to external load resistance.
The rotating coil is called the âarmatureâ.
4 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
Working of Simple Loop Generator Imagine the coil to be rotating in clockwise
direction. As the coil assumes successive
positions in the field, the flux linked with it
changes. An emf is induced in it which is
proportional to the rate of change of flux
linkages (e=NdÏ/dt).
When the plane of coil is in position 1, then
flux linked with the coil is maximum but rate
of change of flux linkage is minimum. Hence,
there is no induced emf in the coil.
As the coil continues rotating further, the rate
of change of flux linkages (and hence induced
emf in it) increases, till position 3 is reached
where Ξ=90o. The coil plane is horizontal
(parallel to the lines of flux). The flux linked
with the coil is minimum but rate of change of
flux linkage is maximum. Hence, maximum
emf is induced in the coil when in this position.
In the second half revolution, the direction of
the current flow is DCMLBA. Which is just
the reverse of the previous direction of flow.
5 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
Working of Simple Loop Generator In the next quarter revolution (from 90o
to 180o), the flux linked with the coil
gradually increases but the rate of change
of flux linkages decreases. Hence the
induced emf decreases gradually till in
position 5 of the coil, it reduces to zero
value.
In the first half revolution of the coil, no
emf is induced in it when in position 1,
maximum when in position 3 and no emf
when in position 5. In this half revolution,
the direction of the current flow is
ABMLCD. The current through the load
resistor R flows from M to L during the
first half revolution of the coil.
In the next half revolution (from 180o to
360o), the variations in the magnitude of
emf are similar to those in the first half
revolution. Its value is maximum when
the coil is in position 7 and minimum
when it in position 1.
6 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
Working of Simple Loop Generator
The current which is obtained from
such a simple generator reverses its
direction after every half
revolution, this current is known as
alternating current. To make the
flow of current unidirectional in
the external circuit, the slip rings
are replaced by split rings.
In the first half revolution segment
âaâ is connected to brush 1 and
segment âbâ is connected to brush 2,
while in the second half revolution
segment âbâ is connected to brush 1
and segment âaâ is connected to
brush 2. In this case the current
will flow in the resistor from M to
L in the two halves of revolution.
The resulting current is
unidirectional but not continuous
like pure direct current.
7 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
Types of DC Generators
Generators are usually classified according to the way in which their fields
are excited
A. Separately Excited Generators: are those whose field magnets are
energized from an independent external source of DC current.
B. Self Excited Generators: are those whose field magnets are energized by
current produced by the generators themselves. There are three types of
self excited generators named according to the manner in which their
field coils are connected to the armature.
i. Shunt Wound: the field windings are connected across or in parallel
with the armature conductors and have the full voltage of the generator
applied across them.
ii. Series Wound: the field windings are joined in series with the armature
conductors
iii.Compound Wound: it is a combination of a few series and a few shunt
windings and can be either short-shunt or long-shunt. In compound
generator, the shunt field is stronger than the series field. When series
field aids the shunt field, generator is said to be commutatively-
compound. In series field oppose the shunt field, the generator is said to
be differentially compounded.
8 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
Types of DC Generators
Separately Excited Generators Shunt Wound Generators Series Wound Generators
Short Shunt Generators Long Shunt Generators
9 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
Types of DC Generators
10 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
The Terminal Characteristic of a Separately Excited DC Generator
ðŒðŽ = ðŒð¿
ðð = ðžðŽ â ðŒðŽ ð ðŽ
ðð¹ = ðŒð¹ ð ð¹
For Separately Excited DC Generator
Where
IA: is the armature current
IL: is the load current
EA: is the internal generated voltage
VT: is the terminal voltage
IF: is the field current
VF: is the field voltage
RA: is the armature winding resistance
RF: is the field winding resistance
Ï: is the flux
ðm: is the rotor angular speed
ðžðŽ = ðÏðð
The terminal voltage can be controlled by:
1. Change the speed of rotation: If ð
increases, then ðžðŽ=ðÏðð increases, so
ðð = ðžðŽ â ðŒðŽ ð ðŽ increases as well.
2. Change the field current. If RF is
decreased. then the field current
increases (ðð¹ = ðŒð¹ ð ð¹ ). Therefore, the
flux in the machine increases. As the
flux rises, ðžðŽ=ðÏðð must rise too, so
ðð = ðžðŽ â ðŒðŽ ð ðŽ increases.
EA
-
+ RA
IA IL
-
+
VT
RF
LF
-
+
VF
IF
11 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
The Terminal Characteristic of a Self Excited Shunt DC Generator
ðŒðŽ = ðŒð¹ + ðŒð¿
ðð = ðžðŽ â ðŒðŽ ð ðŽ
ðð = ðŒð¹ ð ð¹
For Self Excited Shunt DC Generator
The terminal voltage can be controlled by:
1. Change the speed of rotation: If ð
increases, then ðžðŽ=ðÏðð increases, so
ðð = ðžðŽ â ðŒðŽ ð ðŽ increases as well.
2. Change the field current. If RF is
decreased. then the field current
increases (ðð¹ = ðŒð¹ ð ð¹ ). Therefore, the
flux in the machine increases. As the
flux rises, ðžðŽ=ðÏðð must rise too, so
ðð = ðžðŽ â ðŒðŽ ð ðŽ increases.
EA
-
+ RA
IA IL
-
+
VT
RF
LF
IF
ðžðŽ = ðÏðð
12 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
The Terminal Characteristic of a Self Excited Series DC Generator
ðŒðŽ = ðŒð = ðŒð¿
ðð = ðžðŽ â ðŒðŽ(ð ðŽ+ð ð )
For Self Excited Series DC Generator
At no load, there is no field current, so VT is
reduced to a small level given by the
residual flux in the machine. As the load
increases, the field current rises, so EA rises
rapidly The IA(RA+ Rs) drop goes up too,
but at first the increase in EA goes up more
rapidly than the IA(RA+ Rs) drop rises, so
VT increases. After a while, the machine
approaches saturation, and EA becomes
almost constant. At that point, the resistive
drop is the predominant effect, and VT
starts to fall.
ðžðŽ = ðÏðð
EA
-
+ RA
IA IL
-
+
VT
Rs Ls
Is
13 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
The Terminal Characteristic of Cumulatively Compound DC Generator
ðŒðŽ = ðŒð¹ + ðŒð¿
ðð = ðžðŽ â ðŒðŽ(ð ðŽ+ð ð )
For Long Shunt Cumulatively Compound DC Generator
ðžðŽ = ðÏðð
EA
-
+ RA
IA IL
-
+
VT
Rs Ls RF
LF
IF
For Short Shunt Cumulatively Compound DC Generator
ðð = ðŒð¹ ð ð¹
ðŒðŽ = ðŒð¹ + ðŒð¿
ðð = ðžðŽ â ðŒðŽð ðŽ âðŒð¿ð ð
ðžðŽ = ðÏðð EA
-
+ RA
IA IL
-
+
VT
Rs LsRF
LF
IF
The terminal voltage Cumulatively Compound DC Generator can be controlled by:
1. Change the speed of rotation: If ð increases, then ðžðŽ=ðÏðð increases, so
ðð = ðžðŽ â ðŒðŽ ð ðŽ increases as well.
2. Change the field current. If RF is decreased. then the field current increases
(ðð¹ = ðŒð¹ ð ð¹). Therefore, the flux in the machine increases. As the flux rises, ðžðŽ=ðÏ
ðð must rise too, so ðð = ðžðŽ â ðŒðŽ ð ðŽ increases.
14 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
Examples
Example: A shunt DC generator delivers 450A at 230V and the resistance of
the shunt field and armature are 50Ω and 0.3 Ω respectively. Calculate emf.
ðŒð =230
50= 4.6ðŽ
ðŒðŽ = ðŒð¹ + ðŒð¿ = 4.6 + 450 = 454.6ðŽ
ðžðŽ = ðð + ðŒðŽð ðŽ = 230 + 454.6 à 0.3 = 243.6V
Example: A long shunt compound DC generator delivers a load current of
50A at 500V and has armature, series field and shunt field resistances of
0.05Ω, 0.03Ω and 250Ω respectively. Calculate the generated voltage and the
armature current. Allow 1V per brush for contact drop.
ðŒð¹ =500
250= 2ðŽ
ðŒðŽ = ðŒð¹ + ðŒð¿ = 2 + 50 = 52ðŽ
Voltage drop across series winding=ðŒAð s=52Ã0.03=1.56V
Armature voltage drop=ðŒAð A=52Ã0.05=2.6V
Drop at brushes=2Ã1=2V
ðžðŽ = ðð + ðŒðŽð ðŽ + ð ððððð ðððð + ððð¢ð âðð ðððð = 500 + 2.6 + 1.56 + 2 = 506.16V
EA
-
+ RA
IA IL=450A
-
+ VT =
230V
RF
LF
IF
EA
-
+ RA
IA
-
+Rs Ls RF
LF
IF
VT=
500V
IL=50A
15 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
Examples
Example: A short shunt compound DC generator delivers a load current of
30A at 220V and has armature, series field and shunt field resistances of
0.05Ω, 0.3Ω and 200Ω respectively. Calculate the induced emf and the
armature current. Allow 1V per brush for contact drop.
ðŒð¹ =229
200= 1.145ðŽ
Armature voltage drop=ðŒðŽð ðŽ = 31.145 à 0.05 = 1.56V
Voltage drop across series winding=ðŒLð s
=30 Ã0.3=9V
Drop at brushes=2Ã1=2V
ðžðŽ = ðð + ðŒðŽð ðŽ + ð ððððð ðððð + ððð¢ð âðð ðððð = 220 + 9 + 1.56 + 2 = 232.56V
Voltage across shunt winding=220 + 9=229V
EA
-
+ RA
IA
-
+Rs LsRF
LF
IF
VT =
220V
IL=30A
16 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
Examples
Example: A long shunt compound DC generator delivers a load current of
150A at 230V and has armature, series field and shunt field resistances of
0.032Ω, 0.015Ω and 92Ω respectively. Calculate (i) induced emf (ii) total
power generated and (iii) distribution of this power.
EA
-
+ RA
IA
-
+Rs Ls RF
LF
IF
VT =
230V
IL=150A
ðŒð¹ =230
92= 2.5ðŽ
ðŒðŽ = ðŒð¹ + ðŒð¿ = 2.5 + 150 = 152.5ðŽ
Voltage drop across series winding=
ðŒAð s=152.5 Ã0.015=2.2875V
Armature voltage drop=ðŒAð A=152.5 Ã0.032=4.88V
Total power generated by the armature=ðžðŽðŒðŽ=237.1675Ã152.5=36168.04375W
ðžðŽ = ðð + ðŒðŽð ðŽ + ðŒAð s = 230 + 2.2875 + 4.88 = 237.1675V
(i)
(ii)
(iii) Power lost in armature=ðŒðŽ2ð ðŽ=152.52Ã0.032=744.2W
Power dissipated in shunt winding=ðððŒð¹=230Ã2.5=575W
Power dissipated in series winding=ðŒðŽ2ð ð =152.52Ã0.015=348.84375W
Power delivered to the load=ðððŒð¿=230Ã150=34500W
Total power generated by the armature=744.2 + 575 + 348.843 + 34500=36168.04375W
17 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
E.M.F. Equation of DC Generator
Let
Ï: flux/pole in weber.
Z: total number of armature conductors
Z=number of slots à number of conductors/slot
A: number of parallel paths in armature
N: armature rotation in rpm
E: emf induced in any parallel path in armature
Generated emf EA=emf generated in any one of the parallel paths
Average emf generated/conductor=dÏ/dt volt
Flux cut/conductor in one revolution dÏ=ÏP Wb
Number of revolutions /second=N/60
Time for one revolution dt=60/N second
E.M.F. generated/conductor= dÏ/dt= ÏPN/60 volt
18 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
E.M.F. Equation of DC Generator
For simplex wave-wound generator
Number of parallel paths=2
Number of conductors (in series) in one path=Z/2
ðž. ð. ð¹. ððððððð¡ðð/ððð¡â(ðžðŽ) =ððð
60Ã
ð
2=
ðððð
120ð£ððð¡
For simplex lap-wound generator
Number of parallel paths=P
Number of conductors (in series) in one path=Z/P
ðž. ð. ð¹. ððððððð¡ðð/ððð¡â(ðžðŽ) =ððð
60Ã
ð
ð=
ððð
60ð£ððð¡
In general
ðžðŽ =ððð
60Ã
ð
ðŽð£ððð¡
where
A=2 for simplex wave-winding
A=P for simplex lap-winding
ðžðŽ =1
2ÏÃ
2Ïð
60Ã ðð Ã
ð
ðŽ=
ðð
2ÏðŽððð ð£ððð¡ Where ðð =
2Ïð
60
For a given DC machine Z,P and A are constant
ðžðŽ = ðððð ð£ððð¡ Where ð =ðð
2ÏðŽ
19 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
E.M.F. Equation of DC Generator
Example: A four pole generator, having wave wound armature winding has 51
slots, each slot containing 20 conductor. What will be the voltage generated in
the machine when driven at 1500 rpm assuming the flux per pole to be
7mWb?
ðžðŽ =ððð
60Ã
ð
ðŽ=
7 Ã 10â3 Ã 51 Ã 20 Ã 1500
60Ã
4
2= 357ð£ððð¡
Example: An 8 pole Dc generator has 500 armature conductors, and a useful
flux of 0.05Wb per pole. what will be the emf generated if it is lap-connected
and runs at 1200 rpm? What must be the speed at which it is to be driven
produce the same emf if it is wave-wound?
ðžðŽ =ððð
60Ã
ð
ðŽ=
0.05 Ã 500 Ã 1200
60Ã
8
8= 500ð£ððð¡
With lap-wound, P=A=8
With wave-wound, P=8, A=2
ðžðŽ =ððð
60Ã
ð
ðŽ=
0.05 Ã 500 Ã ð
60Ã
8
2= 500 â ð = 300ððð
20 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
E.M.F. Equation of DC Generator
Example: A four pole lap-connected armature of a DC shunt generator is
required to supply the loads connected in parallel:
(a) 5kW Geyser at 250 V and (b) 2.5kW lighting load also at 250V.
The generator has an armature resistance 0.2Ω and a field resistance of 250Ω.
The armature has 120 conductors in the slots and runs at 1000 rpm. Allowing
1V per brush for contact drops, find
(1) Flux per pole, (2) armature current per parallel path
ðžðŽ =ððð
60Ã
ð
ðŽ=
ð Ã 120 Ã 1000
60Ã
4
4= 258.2ð£ððð¡ â ð = 129.1ððð
With lap-wound, P=A=4
ðŒð¿ =5000 + 2500
250= 30ðŽ
ðŒð¹ =250
250= 1ðŽ
ðŒðŽ = ðŒð¿ + ðŒð¹ = 30 + 1 = 31ðŽ
ðžðŽ = ðð + ðŒðŽð ðŽ + ððð¢ð âðð ðððð = 250 + 31 à 0.2 + 2 à 1 = 258.2V
EA
-
+
RA=0.2ΩIA IL
-
+
LF
IF
VT =
250V
RF=250Ω
5k
W G
eyser
2.5
kW
ligh
ting
(2)
(1)
Armature current per parallel path=31/4=7.75A
21 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
E.M.F. Equation of DC Generator
Example: A separately excited DC generator, when running at 1000 rpm
supplied 200A at 125V. What will be the load current when the speed drops to
800 rpm if IF is unchanged? Given that the armature resistance 0.04Ω and
brush drop 2V.
ðžðŽ2 = ðð2 + ðŒðŽ2ð ðŽ + ððð¢ð âðð ðððð
ðžðŽ1 = ðð1 + ðŒðŽð ðŽ + ððð¢ð âðð ðððð = 125 + 200 à 0.04 + 2 = 135V
ððŽ1 = 1000ððð
ðžðŽ2 = ðžðŽ1
ððŽ2
ððŽ1= 135
800
1000= 108ð
ð ðððð =125
200= 0.625Ω
108 = ðŒðŽ2 à 0.625 + ðŒðŽ2 à 0.04 + 2
ðð2 = ðŒðŽ2ð ðððð
ðŒðŽ2 =108 â 2
0.625 + 0.04= 159.4A
ðð2 = ðŒðŽ2ð ðððð = 159.4 à 0.625 = 99.6ð
EA2
-
+ RA
IL=159.4A
-
+RF
LF
-
+
VF
IF
VT
2 =99.6
V
EA1
-
+ RA
IL=200A
-
+RF
LF
-
+
VF
IF
VT
1 =125V
22 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
Total Loss in a DC Generator
(A) Copper Losses
(i) Armature copper losses=Ia2Ra
This loss is about 30-40% of full load losses.
(ii) Field copper loss:
In case of shunt generator, field copper losses=IF2RF
In case of shunt generator, field copper losses=IL2Rs
This loss is about 20-30% of full load losses.
(iii) The loss due to brush contact resistance.
(B) Magnetic (Iron or Core) Losses
(i) Hysteresis Loss, ðŸð â ð©ðððð.ðð
(ii) Eddy Current Loss, ðŸð â ð©ðððððð
These losses are practically constant for shunt and compound wound
generators, because in their case, field current is approximately constant.
This loss is about 20-30% of full load losses.
(C) Mechanical Losses
(i) Friction Loss at bearing and commutator. (ii) Air Friction or Windage Loss of rotating armature
This loss is about 10-20% of full load losses.
23 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
Total Loss in a DC Generator
Tota
l L
oss
es
Copper Losses
Armature Cu Loss
Shunt Cu Loss
Series Cu Loss
Iron Losses
Hysteresis Loss
Eddy Current Loss
Mechanical Losses
Friction Loss
Air Friction or Windage Loss
Stray Losses
Iron and mechanical losses are collectively known as Stray (Rotational) losses.
Constant or Standing Losses
Field Cu losses is constant for shunt and compound generators. Stray losses
and shunt Cu loss are constant in their case. These losses are together known
as Constant or Standing Losses (Wc).
24 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
Power Stages and Efficiency
Mechanical Efficiency
ðð =ððð¡ðð ð€ðð¡ð¡ð ððððððð¡ðð ðð ððððð¡ð¢ðð
ðððâðððððð ððð€ðð ð ð¢ððððððà 100% =
ðžðŽðŒðŽ
ðð¢ð¡ðð¢ð¡ ðð ðððð£ððð ððððððà 100%
Electrical Efficiency
ðð =ððð¡ð¡ð ðð£ððððððð ðð ðððð ðððð¢ðð¡
ððð¡ðð ð€ðð¡ð¡ð ððððððð¡ðð ðð ððððð¡ð¢ððà 100% =
ððŒð¿
ðžðŽðŒðŽÃ 100%
Overall or Commercial Efficiency
ðð = ðð à ðð =ððð¡ð¡ð ðð£ððððððð ðð ðððð ðððð¢ðð¡
ðððâðððððð ððð€ðð ð ð¢ððððððà 100% =
ððŒð¿
ðð¢ð¡ðð¢ð¡ ðð ðððð£ððð ððððððà 100%
25 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
Power Stages and Efficiency
Example: A shunt generator delivers 195A at terminal voltage of 250V. The
armature resistance and shunt field resistance are 0.02Ω and 50Ω respectively.
The iron and friction losses equal 950W. Find
(a) emf generated (b) Cu losses (c) output of the prime motor
(d) commercial, mechanical and electrical efficiencies.
(a) ðŒð =250
50= 5ðŽ
ðŒðŽ = ðŒð¹ + ðŒð¿ = 5 + 195 = 200ðŽ
ðžðŽ = ðð + ðŒðŽð ðŽ = 250 + 200 à 0.02 = 254V
(b) ðŽðððð¡ð¢ðð ð¶ð¢ ððð ð = ðŒðŽ2ð ðŽ = 2002 à 0.02 = 800ð
ðâð¢ðð¡ ð¶ð¢ ððð ð = ðŒð2ð ð = 52 à 50 = 1250ð
ððð¡ðð ð¶ð¢ ððð ð = 800 + 1250 = 2050ð
(c) Stray losses=950W
Total losses=950+2050=3000W
ðºðððððð¡ðð ðð¢ð¡ðð¢ð¡ = ððŒð¿ = 250 à 195 = 48750ð
ðð¢ð¡ðð¢ð¡ ðð ð¡âð ððððð ððð¡ðð = ðºðððððð¡ðð ðððð¢ð¡
26 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
Power Stages and Efficiency
ðºðððððð¡ðð ðððð¢ð¡ = ðºðððððð¡ðð ðð¢ð¡ðð¢ð¡ + ð¡ðð¡ðð ððð ð ðð = 48750 + 3000 = 51750ð
ðð¢ð¡ðð¢ð¡ ðð ð¡âð ððððð ððð¡ðð = 51750ð
ðð =ðžðŽðŒðŽ
ðð¢ð¡ðð¢ð¡ ðð ðððð£ððð ððððððà 100% =
50800
51750Ã 100% = 98.2%
ðð =ððŒð¿
ðžðŽðŒðŽ=
48750
50800Ã 100% = 95.9%
ðð =ððŒð¿
ðð¢ð¡ðð¢ð¡ ðð ðððð£ððð ððððððà 100% =
48750
51750Ã 100% = 94.2%
(c) ðºðððððð¡ðð ððððð¡ððððð ððð€ðð(ðžðŽðŒðŽ) = ðºðððððð¡ðð ðððð¢ð¡ â ð ð¡ðððŠ ððð ð
ðºðððððð¡ðð ððððð¡ððððð ððð€ðð(ðžðŽðŒðŽ) = 51750 â 950 = 50800ð
27 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
Power Stages and Efficiency
Example: A shunt generator has a full load current of 196 A at 220V. The stray
lassos are 720W and the shunt field coil resistance is 55Ω. If it has full load
efficiency of 88%, find the armature resistance.
ðâð¢ðð¡ ð¶ð¢ ððð ð = ðŒðð = 4 à 220 = 880ð
Constant losses=Shunt Cu losses+stray losses=880+720=1600W
ðºðððððð¡ðð ðð¢ð¡ðð¢ð¡ = ððŒð¿ = 220 à 196 = 43120ð
ðð =ððŒð¿
ðžðŽðŒðŽÃ 100% = 88% â ðžðŽðŒðŽ = 43120 ÷ 0.88 = 49000ð
ððð¡ðð ððð ð ðð = 49000 â 43120 = 5880ð
ðŒð = 220 ÷ 55 = 4ðŽ
ðŒðŽ = ðŒð¿ + ðŒð = 195 + 4 = 199ðŽ
Total losses=Armature losses + Constant losses=ðŒðŽ2ð ðŽ+1600=5880
ðŒðŽ2ð ðŽ = 5880 â 1600 = 4280ð
ð ðŽ = 4280 ÷ 1992 = 0.108Ω
28 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
Voltage Regulation
Example: A 4-pole shunt DC generator is delivering 20A to a load of 10Ω. If
the armature resistance is 0.5 Ω and the shunt field resistance is 50 Ω,
calculate the induced emf and the efficiency of the machine. Allow a drop of
1V per brush.
ðððððððð ðððð¡ððð = ðŒð¿ð = 20 à 10 = 200ð
ðð =ððŒð¿
ðžðŽðŒðŽÃ 100% =
200 Ã 20
214 Ã 24Ã 100% = 77.9%
ðŒð = 200 ÷ 50 = 4ðŽ ðŒðŽ = ðŒð¿ + ðŒð = 20 + 4 = 24ðŽ
ðŒðŽð ðŽ = 24 à 0.5 = 12ð
ðžðŽ = ðŒðŽð ðŽ + ð + ððð¢ð â ðððð = 12 + 200 + 2 = 214ð
The voltage regulation (VR) is defined as the difference between the no-load
terminal voltage (VNL) to full load terminal voltage (VFL) and is expressed as
a percentage of full load terminal voltage. It is therefore can be expressed as,
ðððð¡ððð ð ððð¢ððð¡ððð ðð =ððð¿ â ðð¹ð¿
ðð¹ð¿Ã 100% =
ðžðŽ â ðð¹ð¿
ðð¹ð¿Ã 100%
ðð =ðžðŽ â ðð¹ð¿
ðð¹ð¿Ã 100% =
214 â 200
200Ã 100% = 7%
29 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
Uses of DC Generators
⢠Shunt generators with field regulators are used for ordinary lighting and power supply purposes. They are also used for charging batteries because their terminal voltages are almost constant.
Shunt Generators
⢠Series generators are used as boosters in a certain types of distribution systems particularly in railway service.
Series Generators
⢠The cumulatively compound generator is the most used DC generator because its external characteristics can be adjusted for compensating the voltage drop in the line resistance. Cumulatively compound generators are used for motor driving which require DC supply at constant voltage, for lamp loads and for heavy power service such as electric railways.
⢠The differential compound DC generator has an external characteristic similar to that of shunt generator but with large demagnetization armature reaction. Differential compound DC generators re widely used in arc welding where larger voltage drop is desirable with increase in current.
Compound Generators
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