Page 235 - Solutions Manual to accompany Electric Machinery Fundamentals

P 5595 W

  OUT  100%   100% 83.3%
P IN 6720 W
(b) If the armature current is 40 A, then the input power to the motor will be

P  IN V I  T L 120 V 40 A  4800 W
The internal generated voltage at this condition is

E A 2  V  T I A R  A R S  120 V 40 A   0.10   0.08   112.8 V
and the internal generated voltage at rated conditions is


E A 2  V  T I A R  A R S  120 V   56 A 0.10   0.08   109.9 V
The final speed is given by the equation
E A 2  K  2  E Ao ,2 n 2

2

E A 1 K  2 E Ao ,1 n 1
2
/
since the ratio E Ao ,2 / E Ao ,1 is the same as the ratio   . Therefore, the final speed is
2
1
E E
n  A 2 Ao ,1 n
2
E A 1 E Ao ,2 1
From Figure P8-5, the internal generated voltage E for a current of 40 A and a speed of n = 1200
Ao ,2 o
r/min is E = 120 V, and the internal generated voltage E for a current of 56 A and a speed of n =
Ao ,2 Ao ,1 o
1200 r/min is E = 133 V.
Ao ,1
E E  112.8 V   133 V 
n  A 2 Ao ,1 n   1050 r/min  1195 r/min
2 1    
E A 1 E Ao ,2  109.9 V   120 V 
The power converted from electrical to mechanical form is

P conv  E I  AA 112.8 V 40 A  4512 W

The core losses in the motor are 220 W, and the mechanical losses in the motor are 230 W at a speed of
1050 r/min. The mechanical losses in the motor scale proportionally to the cube of the rotational speedm
so the mechanical losses at 1326 r/min are
 n  3  1195 r/min  3
P mech  2    230 W     230 W  339 W
 n 1   1050 r/min 
Therefore, the output power is



P OUT  P conv  P mech  P core  4512 W 339 W 220 W  3953 W
and the efficiency is
P 3953 W
  OUT  100%   100% 82.4%

P IN 4800 W
(c) A MATLAB program to plot the torque-speed characteristic of this motor is shown below:

% M-file: prob9_13.m
% M-file to create a plot of the torque-speed curve of the
% the series dc motor in Problem 9-13.

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