Page 251 - Solutions Manual to accompany Electric Machinery Fundamentals

8-20. An automatic starter circuit is to be designed for a shunt motor rated at 20 hp, 240 V, and 75 A. The
armature resistance of the motor is 0.12 , and the shunt field resistance is 40 . The motor is to start
with no more than 250 percent of its rated armature current, and as soon as the current falls to rated value,
a starting resistor stage is to be cut out. How many stages of starting resistance are needed, and how big
should each one be?

SOLUTION The rated line current of this motor is 75 A, and the rated armature current is I  A I  L I F = 75
A – 6 A = 69 A. The maximum desired starting current is (2.5)(69 A) = 172.5 A. Therefore, the total
initial starting resistance must be

240 V
R  R   1.391 
A start,1
172.5 A
R  1.391   0.12   1.271 
start,1
The current will fall to rated value when E rises to
A
E  240 V  1.391  69 A  144 V
A
At that time, we want to cut out enough resistance to get the current back up to 172.5 A. Therefore,

240 V 144 V
R  R   0.557 
start,2
A
172.5 A
R start,2  0.557   0.12   0.437 
With this resistance in the circuit, the current will fall to rated value when E A rises to
E  A 240 V  0.557  69 A  201.6 V

At that time, we want to cut out enough resistance to get the current back up to 172.5 A. Therefore,

240 V 201.6 V
R  R   0.223 
A
start,3
172.5 A
R start,3  0.223   0.12   0.103 
With this resistance in the circuit, the current will fall to rated value when E A rises to
E  A 240 V  0.223  69 A  224.6 V

If the resistance is cut out when E reaches 224,6 V, the resulting current is
A

240 V 224.6 V
I  A 0.12   128 A  172.5 A ,
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.
R start,1  R  1 R  2 R  3 1.217 
R start,2  R  2 R  3 0.437 
R start,3  R  3 0.103 

Therefore, the starting resistances are
R  1 0.780 
R  2 0.334 
R  3 0.103 

8-21. A 10-hp 120-V 1000 r/min shunt dc motor has a full-load armature current of 70 A when operating at
rated conditions. The armature resistance of the motor is R A = 0.12 , and the field resistance R F is 40

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