Page 134 - Solutions Manual to accompany Electric Machinery Fundamentals

(c) If the field current were increased by 5 percent, what would the new value of the armature current
be? What would the new power factor be? How much reactive power is being consumed or supplied
by the motor?

(d) What is the maximum torque this machine is theoretically capable of supplying at unity power
factor? At 0.8 PF leading?

Note: An electronic version of this open circuit characteristic can be found in file
p51_occ.dat, which can be used with MATLAB programs. Column 1
contains field current in amps, and column 2 contains open-circuit terminal
voltage in volts.
SOLUTION

(a) At full load, the input power to the motor is
P  P OUT  P mech  P core  P CU
IN
We can’t know the copper losses until the armature current is known, so we will find the input power and
armature current ignoring that term, and then correct the input power after we know it.
P  IN  1000 hp 746 W/hp    30 kW 20 kW   796 kW

Therefore, the line and phase current at unity power factor is
P 796 kW
I  A I  L   200 A

3 V T  PF  3 2300 V 1.0 
The copper losses due to a current of 200 A are

P  CU 3I A 2 A R   3 200 A 2   0.3   36.0 kW

Therefore, a better estimate of the input power at full load is


P   1000 hp 746 W/hp    30 kW 20 kW + 36 kW 832 kW
IN
and a better estimate of the line and phase current at unity power factor is
P 832 kW
I  A I  L   209 A

3 V T  PF  3 2300 V 1.0 
The phasor diagram of this motor operating a unity power factor is shown below:
I V
A 

jX I
S A

E
A R I
A A
The phase voltage of this motor is 2300 / 3 = 1328 V. The required internal generated voltage is
E  V R I  jX I
A  A A S A



 E 1328 0  V  0.3   209 0  A   j 2.5   209 0  A 
A

 E 1370   22.44 V
A
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