Page 136 - Solutions Manual to accompany Electric Machinery Fundamentals

5-5. Plot the V-curves ( I A versus I F ) for the synchronous motor of Problem 5-4 at no-load, half-load, and
full-load conditions. (Note that an electronic version of the open-circuit characteristics in Figure P5-1 is
available at the book’s Web site. It may simplify the calculations required by this problem.)
Note: This problem can be greatly simplified if the armature resistance and copper
losses are ignored. This solution does not ignore them. Instead, it tries to
estimate the copper losses by first getting an estimate of the armature current.

SOLUTION The input power at no-load, half-load and full-load conditions is given below. Note that we are
assuming that R A is negligible in each case.
P IN,nl  30 kW  20 kW  50 kW

P IN,half  500 hp 746 W/hp  30 kW 20 kW    423 kW

P IN,full  1000 hp 746 W/hp  30 kW 20 kW    796 kW

If the power factor is adjusted to unity, then armature currents will be

P 50 kW
I A ,nl    12.6 A

3 V T PF  V 1.0 
 3 2300
P 423 kW
I    106 A

A
,half
3 V T PF   3 2300 V 1.0 
P 796 kW
I A ,full    200 A

3 V T  PF  3 2300 V 1.0 
The copper losses due to the armature currents are
P  3I 2  R   3 12.6 A 2   0.3   0.14 kW
CU,nl A A

P  3I 2 R   3 106 A 2   0.3   10.1 kW
CU,half A A
P  3I 2 R   3 200 A 2   0.3   36.0 kW
CU,full A A
Therefore, a better estimate of the input power at the various loads is

P IN,nl  30 kW 20 kW + 0.14 kW  50.1 kW
  P 500 hp 746 W/hp    30 kW 20 kW + 10.1 kW   433 kW
IN,half
  P 1000 hp 746 W/hp    30 kW 20 kW + 36 kW 832 kW


IN,full
and a better estimate of the line and phase current at unity power factor is
P 50.1 kW
I    12.6 A

,nl
A
3 V T  PF  3 2300 V 1.0 
P 433 kW
I    109 A

,half
A
3 V T PF   3 2300 V 1.0 
P 832 kW
I  I    209 A

L
A
,full
3 V PF  V 1.0 
 3 2300
T
The corresponding internal generated voltages at unity power factor are:
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