Page 252 - Solutions Manual to accompany Electric Machinery Fundamentals

. The adjustable resistance in the field circuit R adj may be varied over the range from 0 to 200  and is
currently set to 100 . Armature reaction may be ignored in this machine. The magnetization curve for
this motor, taken at a speed of 1000 r/min, is given in tabular form below:
E , V 5 78 95 112 118 126 130
A
I , A 0.00 0.80 1.00 1.28 1.44 2.88 4.00
F
(a) What is the speed of this motor when it is running at the rated conditions specified above?

(b) The output power from the motor is 10 hp at rated conditions. What is the output torque of the
motor?

(c) What are the copper losses and rotational losses in the motor at full load (ignore stray losses)?
(d) What is the efficiency of the motor at full load?
(e) If the motor is now unloaded with no changes in terminal voltage or R adj , what is the no-load

speed of the motor?
(f) Suppose that the motor is running at the no-load conditions described in part (e). What would
happen to the motor if its field circuit were to open? Ignoring armature reaction, what would the final
steady-state speed of the motor be under those conditions?
(g) What range of no-load speeds is possible in this motor, given the range of field resistance
adjustments available with R adj ?

Note: An electronic version of this magnetization curve can be found in file
prob8_21_mag.dat, which can be used with MATLAB programs. Column
1 contains field current in amps, and column 2 contains the internal generated
voltage E A in volts.

SOLUTION
(a) If R = 100 , the total field resistance is 140 , and the resulting field current is
adj
V 120 V
I  T   0.857 A
R  F F R adj 100 40   

This field current would produce a voltage E of 82.8 V at a speed of n o = 1000 r/min. The actual E
A
Ao
is
E  A V  T I R  A A 120 V    70 A 0.12   111.6 V
so the actual speed will be

E 111.6 V
n  A n   1000 r/min  1348 r/min
E Ao o 82.8 V

(b) The output power is 10 hp and the output speed is 1000 r/min at rated conditions, therefore, the
torque is

P  10 hp 746 W/hp 

 out     2 rad    1 min   71.2 N m
out

m
 1000 r/min  1 r     60 s  
(c) The copper losses are

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