Page 192 - Solutions Manual to accompany Electric Machinery Fundamentals

  3261 V 2
 
max  2 2 
  188.5 rad/s 0.559         0.559  0.332    0.368   


  373 N m
max
(c) The starting torque of this motor is the torque at slip s = 1. It is

2
3VR s
/
 ind  TH 2
 sync    TH  2 / R 2  R s  TH  X 2  X 2  
  3 261 V 2   0.077 

 ind   93.3 N m
 188.5 rad/s    0.559   2   0.077    0.332 0.368 2  
(d) To determine the starting code letter, we must find the locked-rotor kVA per horsepower, which is
equivalent to finding the starting kVA per horsepower. The easiest way to find the line current (or
armature current) at starting is to get the equivalent impedance Z of the rotor circuit in parallel with
F
jX at starting conditions, and then calculate the starting current as the phase voltage divided by the sum
M
of the series impedances, as shown below.
I A,start
R 1 jX 1 jX F R F

+ 0.33  j0.42 

V 

-

The equivalent impedance of the rotor circuit in parallel with jX at starting conditions (s = 1.0) is:
M
1 1

Z    0.0838  j 0.3782 0.385 79.0   
F ,start 1  1 1  1
jX M Z 2 j 18  0.077  j 0.386
The phase voltage is 460/ 3 = 266 V, so line current I is
L ,start

V 266 0 V
 I  I  
R  1 jX  1 R  F jX F 0.058   j 0.32   0.0838   j 0.3782 
L ,start A

 I  I 373   78.5 A
L ,start A
Therefore, the locked-rotor kVA of this motor is
S  3 V T I L ,rated    3 460 V 373 A  297 kVA

and the kVA per horsepower is
297 kVA
kVA/hp   3.96 kVA/hp
75 hp

This motor would have starting code letter C, since letter C covers the range 3.55 – 4.00.
6-24. Answer the following questions about the motor in Problem 6-21.

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