Page 153 - Solutions Manual to accompany Electric Machinery Fundamentals

I E A,g
A
 g
 m jX I
S,g A
V

jX I
S,m A

E
A,m
Then by Kirchhoff’s Voltage Law,
 E  E ( jX  X )I
, Ag , A m , S g , S m A
E  E
or I  , Ag , A m
A
( jX  X )
, Sg , S m
Note that this combined phasor diagram looks just like the diagram of a synchronous motor, so we can
apply the power equation for synchronous motors to this system.
3EE
P  , Ag , A m sin
X , Sg  X , S m

where      . From this equation,
g m
 X  X  P  2.4   80 kW 
  sin  1 , Sg , S m  sin  1  38.0

3E , Ag E , A m   3 280V 371 V 
Therefore,
E  E 280 0 V 371  38.0 V




,

I  Ag , A m   95.3 3.1 A
A
( jX , Sg  X , S m ) j 2.4 
The phase voltage of the system would be


 V  E jX  I 280 0 V   j 0.4  95.3 3.1 A   285   7.7 V
 , Ag , S g A
The phase voltage of the system can also be calculated as

 V  E jX  I 371   38 V   j 2  95.3 3.1 A   284.6   7.7 V
 , Ag , S m A
These two calculations agree, as we would expect.
If we now make V the reference (as we usually do), all of the phases will shift by 7.7, and these

voltages and currents become:


 E 280 7.7 V
, Ag


 V 285 0 V


 E 371   30.3 V
, Am


 I 95.3 10.8 A
A
The new terminal voltage is V  T   3 285 V  494 V , so the system voltage has increased.
(c) The impedance angle is   10.8 (the opposite of the current angle). The power factor of the
motor is now PF cos  10.8   0.982 lead ing .
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