4
ELEN3441FundamentalsofPowerEngineering
Spring2011
Homework Solutions
4.1. (25 pt.) a) At no-load conditions, EA = VT = 240 V. The field current is given by
VT
240
IF =
=
= 0.96 A
Radj + RF 175 + 75
From Figure 5-1, this field current would produce an internal generated voltage EAo of 277 V at a
speed no of 1200 r/min. Therefore, the speed n with a voltage of 240 V would be
En
EA
240 1200
n
=
n= A 0 =
= 1040 rpm
277
E A0 n0
E A0
b) The minimum speed will occur when Radj = 100 , and the maximum speed will occur when
Radj = 400 . The field current when Radj = 100 is:
VT
240
IF =
=
= 1.37 A
Radj + RF 100 + 75
From the magnetization curve, this field current would produce an internal generated voltage EAo
of 289 V at a speed no of 1200 r/min. Therefore, the speed n with a voltage of 240 V would be
En
EA
240 1200
n
=
n= A 0 =
= 997 rpm
289
E A0 n0
E A0
The field current when Radj = 400 is:
VT
240
=
= 0.505 A
Radj + RF 400 + 75
From the magnetization curve, this field current would produce an internal generated voltage EAo
of 207 V at a speed no of 1200 r/min. Therefore, the speed n with a voltage of 240 V would be
En
EA
240 1200
n
=
n= A 0 =
= 1391 rpm
207
E A0 n0
E A0
IF =
c) The starting current of this machine (ignoring the small field current) is
V
240
I L , start = T =
= 1263 A
RA 0.19
The rated current is 110 A, so the starting current is 11.5 times greater than the full-load current.
This current is extremely likely to damage the motor.
4.2. (25 pt.) a) The output power of this motor at full load is
Pout = 15 746 = 11190 W
The input power is
Pin = VT I L = 120 115 = 13800 W
Therefore the efficiency is
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ELEN3441FundamentalsofPowerEngineering
=
Spring2011
Pout
11190
100% =
100% 81%
13800
Pin
b) If the armature current is 50 A, then the input power to the motor will be
Pin = VT I L = 120 50 = 6000 W
The internal generated voltage at this condition is
E A 2 = VT I A ( RA + RS ) = 120 50 ( 0.1 + 0.08 ) = 111V
and the internal generated voltage at rated conditions is
E A1 = VT I A ( RA + RS ) = 120 115 ( 0.1 + 0.08 ) = 99.3 V
The final speed is given by the equation
since the ratio EAo,2 /EAo,1 is the same as the ratio 2/1. Therefore, the final speed is
EE n
n = A 2 Ao ,1 1
E A1 E Ao ,2
The internal generated voltage EAo,2 for a current of 50 A and a speed of n0 = 1200 r/min is EAo,2
= 129V, and the internal generated voltage EAo,1 for a current of 115 A and a speed of n0 = 1200
rpm cannot be determined from the curve but will be approximated as EAo,1 = 155 V.
E E n 111 155 1050
= 1410 rpm
n = A 2 Ao ,1 1 =
99.3 129
E A1 E Ao ,2
The power converted from electrical to mechanical form is
Pconv = E A I A = 111 50 = 5550 W
The core losses in the are motor 420 W, and the mechanical losses in the motor are
3
Pmech
3
n
1410
= 2 460 =
460 = 1114 W
1050
n1
Therefore, the output power is
Pout = Pconv Pmech Pcore = 5550 1114 420 = 4016 W
and the efficiency is
=
Pout
4016
100% =
100% 67%
6000
Pin
4.3. (25 pt.) These computations must be performed for armature currents of 25.3 A, 50.7 A, 76
A, and 101.3 A.
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ELEN3441FundamentalsofPowerEngineering
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If IA = 25.3 A, then
E A = VT I A ( RA + RS ) = 240 V 25.3 A ( 0.09 + 0.06 ) = 236.2 V
The mmf is = NI A = 33 25.3 A = 835 A turn that produces a voltage EA0 of 134 V at n0 = 900
rpm. Therefore, the speed of the motor at these conditions is
n=
EA
236.2 V
n0 =
900 = 1586 rpm
E A0
134 V
The power converted from electrical to mechanical form is
Pconv = E A I A = 236.2 V 25.3 A = 5976 W
Since the rotational losses are ignored, this is also the output power of the motor. The induced
torque is
ind =
Pconv
m
=
5976 W
= 36 N m
1586 rpm 2 60
Similar computations for other armature currents lead to the following results:
IA (A)
25.3
50.7
76
101.3
n (rpm)
1586
1062
899
803
Pconv (W)
5976
11780
17370
22770
ind (Nm)
36
106
185
271
4.4. (25 pt.) The rated line current of this motor is 60 A, and the rated armature current is IA = IL
IF = 60 A 6 A = 54 A. The maximum desired starting current is (2.5)(54 A) = 135 A.
Therefore, the total initial starting resistance must be
240
RA + Rstart ,1 =
= 1.778
135
Rstart ,1 = 1.778 0.12 = 1.658
The current will fall to rated value when EA rises to
E A = 240 1.658 54 = 150.5 V
At that time, we want to cut out enough resistance to get the current back up to 135 A. Therefore,
240 150.5
RA + Rstart ,2 =
= 0.663
135
Rstart ,2 = 0.663 0.12 = 0.543
With this resistance in the circuit, the current will fall to rated value when EA rises to
E A = 240 0.543 54 = 210.7 V
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ELEN3441FundamentalsofPowerEngineering
Spring2011
At that time, we want to cut out enough resistance to get the current back up to 185 A. Therefore,
240 210.7
RA + Rstart ,3 =
= 0.217
135
Rstart ,3 = 0.217 0.12 = 0.097
With this resistance in the circuit, the current will fall to rated value when EA rises to
E A = 240 0.217 54 = 228.3 V
If the resistance is cut out when EA reaches 224.6 V, the resulting current is
240 228.3
IA =
= 97.5 A < 135 A
0.12
so there are only three stages of starting resistance. The three stages of starting resistance can be
found from the resistance in the circuit at each state during starting.
Rstart ,1 = R1 + R2 + R3 = 1.658
Rstart ,2 = R2 + R3 = 0.543
Rstart ,3 = R3 = 0.097
Therefore, the starting resistances are
R1 = 1.115
R2 = 0.446
R3 = 0.097
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