Page 160 - Solutions Manual to accompany Electric Machinery Fundamentals

P 51.3 kW
  conv   171.9 N m

ind
 m  2850 r/min  2 rad    1 min 
  1 r     60 s  
(e) The rotor frequency is

f  r e  sf  0.05 50 Hz   2.5 Hz

6-5. A 208-V four-pole 60-Hz Y-connected wound-rotor induction motor is rated at 30 hp. Its equivalent
circuit components are
R = 0.100  R 2 = 0.070  X M = 10.0 
1
X = 0.210  X 2 = 0.210 
1
P mech = 500 W P misc  0 P core = 400 W

For a slip of 0.05, find
(a) The line current I L

(b) The stator copper losses P SCL

(c) The air-gap power P AG
(d) The power converted from electrical to mechanical form P conv

(e) The induced torque  ind

(f) The load torque  load

(g) The overall machine efficiency
(h) The motor speed in revolutions per minute and radians per second

SOLUTION The equivalent circuit of this induction motor is shown below:

I A
R 1 jX 1 jX 2 R 2

+ j0.21 ? 0.07 ?
0.10 ? j0.21 ?

 1  s
j10 ? jX M R  
 s 
V ? 2
1.33 ?
-

(a) The easiest way to find the line current (or armature current) is to get the equivalent impedance Z F
of the rotor circuit in parallel with jX , and then calculate the current as the phase voltage divided by the
M
sum of the series impedances, as shown below.

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