Page 289 - Solutions Manual to accompany Electric Machinery Fundamentals

V
I 
R  1 jX  1 0.5Z  F 0.5Z B
1

220 0 V
I  1  j 1.30  0.5 19.12   2.01  j  0.5 5.654  j 1.979  0.854   17.32  27.3 A 

(b) The stator power factor is

PF cos 27.3   0.889 lagging
(c) The i put power is
n

P  IN VI cos  220 V 17.32 A cos 27.3  3386 W
(d) The air-gap power is

P I 2   0.5R    17.32 A 2   9.56   2868 W
AG,F 1 F
P I 2   0.5R    17.32 A 2   0.427   128 W
AG,B 1 B

P AG  P AG,F  P AG,B  2868 W 128 W  2740 W
(e) The power converted from electrical to mechanical form is

P conv,F    1 s P AG,F   1 0.05 2868 W  2725 W
P conv,B    1 s P AG,B   1 0.05 128 W  122 W

P conv  P conv,F  P conv,B  2725 W 122 W  2603 W
(f) The output power is

P OUT  P conv  P  rot 2603 W 291 W  2312 W
(g) The synchronous speed for a 6 pole 50 Hz machine is 1000 r/min, so induced torque is

P 2740 W
 ind  AG   26.17 N m

 sync  1000 r/min  2 rad    1 min 
  1 r     60 s  
(h) The load torque is
P 2312 W
  OUT   23.24 N m

 m  0.95 1000 r/min   2 rad    1 min 
load

  1 r     60 s  
(i) The overall efficiency is
P 2312 W
  OUT  100%   100% 68.3%

P IN 3386 W
9-6. Find the induced torque in the motor in Problem 9-5 if it is operating at 5 percent slip and its terminal
voltage is (a) 190 V, (b) 208 V, (c) 230 V.
The impedances Z and Z are:
B
F
s
R /  jX  jX 
Z  R 2 / s  jX  2 jX M
F
2
The slip s  0.05 , so R 2 2  / s  M 1.73  / 0.05   20.6

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