Page 241 - Solutions Manual to accompany Electric Machinery Fundamentals

V  440 V including brushes I  7.50 A
F
A
I  231. A n  863 r/min
A
What is this motor ’s efficiency at the rated conditions? [Note: Assume that (1) the brush voltage drop is 2
V; (2) the core loss is to be determined at an armature voltage equal to the armature voltage under full
load; and (3) stray load losses are 1 percent of full load.]
SOLUTION The armature resistance of this motor is

V 14.9 V
R  A ,br   0.0298 
A
I A ,br 500 A
Under no-load conditions, the core and mechanical losses taken together (that is, the rotational losses) of
this motor are equal to the product of the internal generated voltage E and the armature current I ,
A
A
since this is no output power from the motor at no-load conditions. The refore, the rotational losses t
a
rated speed can be found as
E  V  V   I R 440 V 2 V  23.1 A 0.0298   437.3 V

A A brush A A

P  P   E I  437.3 V 23.1 A  10.1 kW
rot conv AA
The input power to the motor at full load is

P  IN V I  T L 440 V 560 A   246.4 kW

The outp ut power from the motor at full load is
P OUT  P  IN P  CU P  rot P brush  P stray

The copp er losses are

P  CU I R  A 2 A V I  F F  560 A 2  0.0 298    440 V 7.52 A   12.65 kW
The brus h losses are

I 
P brush  V brush A  2 V 560 A    1.12 kW
Therefore ,
P OUT  P  IN P  CU P  rot P brush  P stray




P OUT  246.4 kW 14.1 kW 12.65 kW 10.1 kW 1.12 kW 2.46 kW   206 kW
The moto r’s efficiency at full load is
P 206 kW

  OUT  100%   100% 83.6%
P IN 246.4 kW

P roblems 8-16 to 8-19 refer to a 240-V 100-A dc motor which has both shunt and series windings. Its
characteristics are
R = 0.14  N = 1500 turns
F
A
R S = 0.05  N SE = 15 turns
R F = 200  n = 3000 r/min
m
R adj = 0 to 300 , currently set to 120 
This motor has c ompensating windings and interpoles. T he magnetization curve for this motor at 3000 r/min is
shown in Figure P8-6.

235