Electric Machines I - Philadelphia Universityย ยท 2018. 1. 22.ย ยท For Short Shunt Cumulatively...
Transcript of Electric Machines I - Philadelphia Universityย ยท 2018. 1. 22.ย ยท For Short Shunt Cumulatively...
Electric Machines I DC Machines - DC Generators
1
Dr. Firas Obeidat
2
Table of contents
1 โข Construction of Simple Loop Generator
2 โข Working of Simple Loop Generator
3 โข Types of DC Generators
4 โข The Terminal Characteristic of a Separately Excited DC Generator
5 โข The Terminal Characteristic of a Self Excited Shunt DC Generator
6 โข The Terminal Characteristic of a Self Excited Series DC Generator
7 โข The Terminal Characteristic of Cumulatively Compound DC Generator
8 โข E.M.F. Equation of DC Generator
9 โข Total Loss in a DC Generator
10 โข Power Stages and Efficiency
11 โข Voltage Regulation
12 โข 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 โaโ and โbโ 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 โarmatureโ.
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=Ndฯ/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 ฮธ=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
โaโ is connected to brush 1 and
segment โbโ is connected to brush 2,
while in the second half revolution
segment โbโ is connected to brush 1
and segment โaโ 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
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Types of DC Generators
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The Terminal Characteristic of a Separately Excited DC Generator
๐ผ๐ด = ๐ผ๐ฟ
๐๐ = ๐ธ๐ด โ ๐ผ๐ด ๐ ๐ด
๐๐น = ๐ผ๐น ๐ ๐น
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
ฯ: is the flux
๐m: is the rotor angular speed
๐ธ๐ด = ๐ฯ๐๐
The terminal voltage can be controlled by:
1. Change the speed of rotation: If ๐
increases, then ๐ธ๐ด=๐ฯ๐๐ increases, so
๐๐ = ๐ธ๐ด โ ๐ผ๐ด ๐ ๐ด increases as well.
2. Change the field current. If RF is
decreased. then the field current
increases (๐๐น = ๐ผ๐น ๐ ๐น ). Therefore, the
flux in the machine increases. As the
flux rises, ๐ธ๐ด=๐ฯ๐๐ must rise too, so
๐๐ = ๐ธ๐ด โ ๐ผ๐ด ๐ ๐ด 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
๐ผ๐ด = ๐ผ๐น + ๐ผ๐ฟ
๐๐ = ๐ธ๐ด โ ๐ผ๐ด ๐ ๐ด
๐๐ = ๐ผ๐น ๐ ๐น
For Self Excited Shunt DC Generator
The terminal voltage can be controlled by:
1. Change the speed of rotation: If ๐
increases, then ๐ธ๐ด=๐ฯ๐๐ increases, so
๐๐ = ๐ธ๐ด โ ๐ผ๐ด ๐ ๐ด increases as well.
2. Change the field current. If RF is
decreased. then the field current
increases (๐๐น = ๐ผ๐น ๐ ๐น ). Therefore, the
flux in the machine increases. As the
flux rises, ๐ธ๐ด=๐ฯ๐๐ must rise too, so
๐๐ = ๐ธ๐ด โ ๐ผ๐ด ๐ ๐ด increases.
EA
-
+ RA
IA IL
-
+
VT
RF
LF
IF
๐ธ๐ด = ๐ฯ๐๐
12 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
The Terminal Characteristic of a Self Excited Series DC Generator
๐ผ๐ด = ๐ผ๐ = ๐ผ๐ฟ
๐๐ = ๐ธ๐ด โ ๐ผ๐ด(๐ ๐ด+๐ ๐ )
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.
๐ธ๐ด = ๐ฯ๐๐
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
๐ผ๐ด = ๐ผ๐น + ๐ผ๐ฟ
๐๐ = ๐ธ๐ด โ ๐ผ๐ด(๐ ๐ด+๐ ๐ )
For Long Shunt Cumulatively Compound DC Generator
๐ธ๐ด = ๐ฯ๐๐
EA
-
+ RA
IA IL
-
+
VT
Rs Ls RF
LF
IF
For Short Shunt Cumulatively Compound DC Generator
๐๐ = ๐ผ๐น ๐ ๐น
๐ผ๐ด = ๐ผ๐น + ๐ผ๐ฟ
๐๐ = ๐ธ๐ด โ ๐ผ๐ด๐ ๐ด โ๐ผ๐ฟ๐ ๐
๐ธ๐ด = ๐ฯ๐๐ 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 ๐ increases, then ๐ธ๐ด=๐ฯ๐๐ increases, so
๐๐ = ๐ธ๐ด โ ๐ผ๐ด ๐ ๐ด increases as well.
2. Change the field current. If RF is decreased. then the field current increases
(๐๐น = ๐ผ๐น ๐ ๐น). Therefore, the flux in the machine increases. As the flux rises, ๐ธ๐ด=๐ฯ
๐๐ must rise too, so ๐๐ = ๐ธ๐ด โ ๐ผ๐ด ๐ ๐ด 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 50ฮฉ and 0.3 ฮฉ respectively. Calculate emf.
๐ผ๐ =230
50= 4.6๐ด
๐ผ๐ด = ๐ผ๐น + ๐ผ๐ฟ = 4.6 + 450 = 454.6๐ด
๐ธ๐ด = ๐๐ + ๐ผ๐ด๐ ๐ด = 230 + 454.6 ร 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
0.05ฮฉ, 0.03ฮฉ and 250ฮฉ respectively. Calculate the generated voltage and the
armature current. Allow 1V per brush for contact drop.
๐ผ๐น =500
250= 2๐ด
๐ผ๐ด = ๐ผ๐น + ๐ผ๐ฟ = 2 + 50 = 52๐ด
Voltage drop across series winding=๐ผA๐ s=52ร0.03=1.56V
Armature voltage drop=๐ผA๐ A=52ร0.05=2.6V
Drop at brushes=2ร1=2V
๐ธ๐ด = ๐๐ + ๐ผ๐ด๐ ๐ด + ๐ ๐๐๐๐๐ ๐๐๐๐ + ๐๐๐ข๐ โ๐๐ ๐๐๐๐ = 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
0.05ฮฉ, 0.3ฮฉ and 200ฮฉ respectively. Calculate the induced emf and the
armature current. Allow 1V per brush for contact drop.
๐ผ๐น =229
200= 1.145๐ด
Armature voltage drop=๐ผ๐ด๐ ๐ด = 31.145 ร 0.05 = 1.56V
Voltage drop across series winding=๐ผL๐ s
=30 ร0.3=9V
Drop at brushes=2ร1=2V
๐ธ๐ด = ๐๐ + ๐ผ๐ด๐ ๐ด + ๐ ๐๐๐๐๐ ๐๐๐๐ + ๐๐๐ข๐ โ๐๐ ๐๐๐๐ = 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
0.032ฮฉ, 0.015ฮฉ and 92ฮฉ 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
๐ผ๐น =230
92= 2.5๐ด
๐ผ๐ด = ๐ผ๐น + ๐ผ๐ฟ = 2.5 + 150 = 152.5๐ด
Voltage drop across series winding=
๐ผA๐ s=152.5 ร0.015=2.2875V
Armature voltage drop=๐ผA๐ A=152.5 ร0.032=4.88V
Total power generated by the armature=๐ธ๐ด๐ผ๐ด=237.1675ร152.5=36168.04375W
๐ธ๐ด = ๐๐ + ๐ผ๐ด๐ ๐ด + ๐ผA๐ s = 230 + 2.2875 + 4.88 = 237.1675V
(i)
(ii)
(iii) Power lost in armature=๐ผ๐ด2๐ ๐ด=152.52ร0.032=744.2W
Power dissipated in shunt winding=๐๐๐ผ๐น=230ร2.5=575W
Power dissipated in series winding=๐ผ๐ด2๐ ๐ =152.52ร0.015=348.84375W
Power delivered to the load=๐๐๐ผ๐ฟ=230ร150=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
ฯ: flux/pole in weber.
Z: total number of armature conductors
Z=number of slots ร 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=dฯ/dt volt
Flux cut/conductor in one revolution dฯ=ฯP Wb
Number of revolutions /second=N/60
Time for one revolution dt=60/N second
E.M.F. generated/conductor= dฯ/dt= ฯPN/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
๐ธ. ๐. ๐น. ๐๐๐๐๐๐๐ก๐๐/๐๐๐กโ(๐ธ๐ด) =๐๐๐
60ร
๐
2=
๐๐๐๐
120๐ฃ๐๐๐ก
For simplex lap-wound generator
Number of parallel paths=P
Number of conductors (in series) in one path=Z/P
๐ธ. ๐. ๐น. ๐๐๐๐๐๐๐ก๐๐/๐๐๐กโ(๐ธ๐ด) =๐๐๐
60ร
๐
๐=
๐๐๐
60๐ฃ๐๐๐ก
In general
๐ธ๐ด =๐๐๐
60ร
๐
๐ด๐ฃ๐๐๐ก
where
A=2 for simplex wave-winding
A=P for simplex lap-winding
๐ธ๐ด =1
2ฯร
2ฯ๐
60ร ๐๐ ร
๐
๐ด=
๐๐
2ฯ๐ด๐๐๐ ๐ฃ๐๐๐ก Where ๐๐ =
2ฯ๐
60
For a given DC machine Z,P and A are constant
๐ธ๐ด = ๐๐๐๐ ๐ฃ๐๐๐ก Where ๐ =๐๐
2ฯ๐ด
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?
๐ธ๐ด =๐๐๐
60ร
๐
๐ด=
7 ร 10โ3 ร 51 ร 20 ร 1500
60ร
4
2= 357๐ฃ๐๐๐ก
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?
๐ธ๐ด =๐๐๐
60ร
๐
๐ด=
0.05 ร 500 ร 1200
60ร
8
8= 500๐ฃ๐๐๐ก
With lap-wound, P=A=8
With wave-wound, P=8, A=2
๐ธ๐ด =๐๐๐
60ร
๐
๐ด=
0.05 ร 500 ร ๐
60ร
8
2= 500 โ ๐ = 300๐๐๐
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 0.2ฮฉ and a field resistance of 250ฮฉ.
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
๐ธ๐ด =๐๐๐
60ร
๐
๐ด=
๐ ร 120 ร 1000
60ร
4
4= 258.2๐ฃ๐๐๐ก โ ๐ = 129.1๐๐๐
With lap-wound, P=A=4
๐ผ๐ฟ =5000 + 2500
250= 30๐ด
๐ผ๐น =250
250= 1๐ด
๐ผ๐ด = ๐ผ๐ฟ + ๐ผ๐น = 30 + 1 = 31๐ด
๐ธ๐ด = ๐๐ + ๐ผ๐ด๐ ๐ด + ๐๐๐ข๐ โ๐๐ ๐๐๐๐ = 250 + 31 ร 0.2 + 2 ร 1 = 258.2V
EA
-
+
RA=0.2ฮฉIA IL
-
+
LF
IF
VT =
250V
RF=250ฮฉ
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 0.04ฮฉ and
brush drop 2V.
๐ธ๐ด2 = ๐๐2 + ๐ผ๐ด2๐ ๐ด + ๐๐๐ข๐ โ๐๐ ๐๐๐๐
๐ธ๐ด1 = ๐๐1 + ๐ผ๐ด๐ ๐ด + ๐๐๐ข๐ โ๐๐ ๐๐๐๐ = 125 + 200 ร 0.04 + 2 = 135V
๐๐ด1 = 1000๐๐๐
๐ธ๐ด2 = ๐ธ๐ด1
๐๐ด2
๐๐ด1= 135
800
1000= 108๐
๐ ๐๐๐๐ =125
200= 0.625ฮฉ
108 = ๐ผ๐ด2 ร 0.625 + ๐ผ๐ด2 ร 0.04 + 2
๐๐2 = ๐ผ๐ด2๐ ๐๐๐๐
๐ผ๐ด2 =108 โ 2
0.625 + 0.04= 159.4A
๐๐2 = ๐ผ๐ด2๐ ๐๐๐๐ = 159.4 ร 0.625 = 99.6๐
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, ๐พ๐ โ ๐ฉ๐๐๐๐.๐๐
(ii) Eddy Current Loss, ๐พ๐ โ ๐ฉ๐๐๐๐๐๐
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
๐๐ =๐๐๐ก๐๐ ๐ค๐๐ก๐ก๐ ๐๐๐๐๐๐๐ก๐๐ ๐๐ ๐๐๐๐๐ก๐ข๐๐
๐๐๐โ๐๐๐๐๐๐ ๐๐๐ค๐๐ ๐ ๐ข๐๐๐๐๐๐ร 100% =
๐ธ๐ด๐ผ๐ด
๐๐ข๐ก๐๐ข๐ก ๐๐ ๐๐๐๐ฃ๐๐๐ ๐๐๐๐๐๐ร 100%
Electrical Efficiency
๐๐ =๐๐๐ก๐ก๐ ๐๐ฃ๐๐๐๐๐๐๐ ๐๐ ๐๐๐๐ ๐๐๐๐ข๐๐ก
๐๐๐ก๐๐ ๐ค๐๐ก๐ก๐ ๐๐๐๐๐๐๐ก๐๐ ๐๐ ๐๐๐๐๐ก๐ข๐๐ร 100% =
๐๐ผ๐ฟ
๐ธ๐ด๐ผ๐ดร 100%
Overall or Commercial Efficiency
๐๐ = ๐๐ ร ๐๐ =๐๐๐ก๐ก๐ ๐๐ฃ๐๐๐๐๐๐๐ ๐๐ ๐๐๐๐ ๐๐๐๐ข๐๐ก
๐๐๐โ๐๐๐๐๐๐ ๐๐๐ค๐๐ ๐ ๐ข๐๐๐๐๐๐ร 100% =
๐๐ผ๐ฟ
๐๐ข๐ก๐๐ข๐ก ๐๐ ๐๐๐๐ฃ๐๐๐ ๐๐๐๐๐๐ร 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 0.02ฮฉ and 50ฮฉ 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) ๐ผ๐ =250
50= 5๐ด
๐ผ๐ด = ๐ผ๐น + ๐ผ๐ฟ = 5 + 195 = 200๐ด
๐ธ๐ด = ๐๐ + ๐ผ๐ด๐ ๐ด = 250 + 200 ร 0.02 = 254V
(b) ๐ด๐๐๐๐ก๐ข๐๐ ๐ถ๐ข ๐๐๐ ๐ = ๐ผ๐ด2๐ ๐ด = 2002 ร 0.02 = 800๐
๐โ๐ข๐๐ก ๐ถ๐ข ๐๐๐ ๐ = ๐ผ๐2๐ ๐ = 52 ร 50 = 1250๐
๐๐๐ก๐๐ ๐ถ๐ข ๐๐๐ ๐ = 800 + 1250 = 2050๐
(c) Stray losses=950W
Total losses=950+2050=3000W
๐บ๐๐๐๐๐๐ก๐๐ ๐๐ข๐ก๐๐ข๐ก = ๐๐ผ๐ฟ = 250 ร 195 = 48750๐
๐๐ข๐ก๐๐ข๐ก ๐๐ ๐กโ๐ ๐๐๐๐๐ ๐๐๐ก๐๐ = ๐บ๐๐๐๐๐๐ก๐๐ ๐๐๐๐ข๐ก
26 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
Power Stages and Efficiency
๐บ๐๐๐๐๐๐ก๐๐ ๐๐๐๐ข๐ก = ๐บ๐๐๐๐๐๐ก๐๐ ๐๐ข๐ก๐๐ข๐ก + ๐ก๐๐ก๐๐ ๐๐๐ ๐ ๐๐ = 48750 + 3000 = 51750๐
๐๐ข๐ก๐๐ข๐ก ๐๐ ๐กโ๐ ๐๐๐๐๐ ๐๐๐ก๐๐ = 51750๐
๐๐ =๐ธ๐ด๐ผ๐ด
๐๐ข๐ก๐๐ข๐ก ๐๐ ๐๐๐๐ฃ๐๐๐ ๐๐๐๐๐๐ร 100% =
50800
51750ร 100% = 98.2%
๐๐ =๐๐ผ๐ฟ
๐ธ๐ด๐ผ๐ด=
48750
50800ร 100% = 95.9%
๐๐ =๐๐ผ๐ฟ
๐๐ข๐ก๐๐ข๐ก ๐๐ ๐๐๐๐ฃ๐๐๐ ๐๐๐๐๐๐ร 100% =
48750
51750ร 100% = 94.2%
(c) ๐บ๐๐๐๐๐๐ก๐๐ ๐๐๐๐๐ก๐๐๐๐๐ ๐๐๐ค๐๐(๐ธ๐ด๐ผ๐ด) = ๐บ๐๐๐๐๐๐ก๐๐ ๐๐๐๐ข๐ก โ ๐ ๐ก๐๐๐ฆ ๐๐๐ ๐
๐บ๐๐๐๐๐๐ก๐๐ ๐๐๐๐๐ก๐๐๐๐๐ ๐๐๐ค๐๐(๐ธ๐ด๐ผ๐ด) = 51750 โ 950 = 50800๐
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 55ฮฉ. If it has full load
efficiency of 88%, find the armature resistance.
๐โ๐ข๐๐ก ๐ถ๐ข ๐๐๐ ๐ = ๐ผ๐๐ = 4 ร 220 = 880๐
Constant losses=Shunt Cu losses+stray losses=880+720=1600W
๐บ๐๐๐๐๐๐ก๐๐ ๐๐ข๐ก๐๐ข๐ก = ๐๐ผ๐ฟ = 220 ร 196 = 43120๐
๐๐ =๐๐ผ๐ฟ
๐ธ๐ด๐ผ๐ดร 100% = 88% โ ๐ธ๐ด๐ผ๐ด = 43120 รท 0.88 = 49000๐
๐๐๐ก๐๐ ๐๐๐ ๐ ๐๐ = 49000 โ 43120 = 5880๐
๐ผ๐ = 220 รท 55 = 4๐ด
๐ผ๐ด = ๐ผ๐ฟ + ๐ผ๐ = 195 + 4 = 199๐ด
Total losses=Armature losses + Constant losses=๐ผ๐ด2๐ ๐ด+1600=5880
๐ผ๐ด2๐ ๐ด = 5880 โ 1600 = 4280๐
๐ ๐ด = 4280 รท 1992 = 0.108ฮฉ
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 10ฮฉ. If
the armature resistance is 0.5 ฮฉ and the shunt field resistance is 50 ฮฉ,
calculate the induced emf and the efficiency of the machine. Allow a drop of
1V per brush.
๐๐๐๐๐๐๐๐ ๐๐๐๐ก๐๐๐ = ๐ผ๐ฟ๐ = 20 ร 10 = 200๐
๐๐ =๐๐ผ๐ฟ
๐ธ๐ด๐ผ๐ดร 100% =
200 ร 20
214 ร 24ร 100% = 77.9%
๐ผ๐ = 200 รท 50 = 4๐ด ๐ผ๐ด = ๐ผ๐ฟ + ๐ผ๐ = 20 + 4 = 24๐ด
๐ผ๐ด๐ ๐ด = 24 ร 0.5 = 12๐
๐ธ๐ด = ๐ผ๐ด๐ ๐ด + ๐ + ๐๐๐ข๐ โ ๐๐๐๐ = 12 + 200 + 2 = 214๐
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,
๐๐๐๐ก๐๐๐ ๐ ๐๐๐ข๐๐๐ก๐๐๐ ๐๐ =๐๐๐ฟ โ ๐๐น๐ฟ
๐๐น๐ฟร 100% =
๐ธ๐ด โ ๐๐น๐ฟ
๐๐น๐ฟร 100%
๐๐ =๐ธ๐ด โ ๐๐น๐ฟ
๐๐น๐ฟร 100% =
214 โ 200
200ร 100% = 7%
29 Dr. Firas Obeidat Faculty of Engineering Philadelphia University
Uses of DC Generators
โข 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
โข Series generators are used as boosters in a certain types of distribution systems particularly in railway service.
Series Generators
โข 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.
โข 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
30