Page 190 - Solutions Manual to accompany Electric Machinery Fundamentals

I A
R 1 jX 1 jX 2 R 2

+ 0.58 ? j0.32 ? j0.386 ? ??? ? I 2

 1  s
j18 ? jX M R  
 s 
V ? 2

-

The Thevenin equivalent of the input circuit is:
jX R  jX   18j   0.58   j 0.32  
Z  M 1 1   0.559  j 0.332   0.651 30.7  
TH
R  1  1 X M  j X  0.58   j   18   0.32 
jX  18 j  


V  M V     266 0 V  261 1.8 V
TH
R  1  1 X M  j X   0.58   j   18   0.32 
(a) If losses are neglected, the induced torque in a motor is equal to its load torque. At full load, the
output power of this motor is 75 hp and its slip is 1.2%, so the induced torque is
n  m  1 0.035 1800 r/min   1737 r/min
 
75 hp 746 W/hp 
 ind   load   307.6 N m

 1737 r/min   2 rad      1min  
 1 r   60 s 

The induced torque is given by the equation
2
3VR s
/
  TH 2
ind 2 2
 sync    TH  2 / R  R s  TH  X 2  X  
Substituting known values and solving for R s yields
/
2
2 3V / s R
  TH 2
ind 2 2
    / s  R  R  X  X 
sync  TH 2 TH 2 
  3261 V R / s
2
307.6 N m   2
 188.5 rad/s    2 / s  0.559 R 2     0.332 0.386 2  

205,932 R / s
57,983  2
   / s 0.559 R 2  0.516 
 2 

   / s 0.559 R 2  0.516   3.552 R / s
 2  2
 0.3125 1.118R  / s  / s R 2  0.516   3.552 R / s

 2 2  2

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