Page 226 - Solutions Manual to accompany Electric Machinery Fundamentals

E n
A 
E Ao n o
 E 
n    E  Ao  n
A
o
(a) If V = 120 V, then E = 120 V, and
A A
 120 V 
   n  1200 r/min  598 r/min
 241 V 

(b) If V = 180 V, then E = 180 V, and
A A
 180 V 
   n  1200 r/min  986 r/min
 241 V 

(c) If V = 240 V, then E = 240 V, and
A A
 240 V 
   n  1200 r/min  1195 r/min
 241 V 
8-9. For the separately excited motor of Problem 8-8:

(a) What is the maximum no-load speed attainable by varying both V and R adj ?
A
(b) What is the minimum no-load speed attainable by varying both V and R adj ?
A
(c) What is the motor’s efficiency at rated conditions? [Note: Assume that (1) the brush voltage drop
is 2 V; (2) the core loss is to e 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

(a) The maximum speed will occur with the maximum V A and the maximum R . The field current
adj
when R = 400  is:
adj
V 240 V 240 V
I  T    0.505 A
R  adj R F 400   75  475 
F
From Figure P8-1, this field current would produce an internal generated voltage E of 167 V at a speed
Ao
n of 1200 r/min. At no-load conditions, the maximum internal generated voltage E  V = 240 V.
o A A
Therefore, the speed n with a voltage of 240 V would be
E A  n
E Ao n o
 E   240 V 
n   A  n o    1200 r/min  1725 r/min
 E Ao   167 V 
(b) The minimum speed will occur with the minimum V A and the minimum R . The field current
adj
when R = 100  is:
adj
V 240 V 240 V
I  T    1.37 A
R  F adj R F 100   75  175 

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