Page 172 - Solutions Manual to accompany Electric Machinery Fundamentals

2
VR
P  I R  2 L
 S L R   X  S X L   R
L L 2 2
  R  2  X  X 2  V  2 2 2  V R  R 
R 
 P   S L  R L  S  L  S L 
 R L   R  2   R X  X 2  2
 S L L  S 

To find the point of maximum power supplied to the load, set P / R = 0 and solve for R .
L L
  R  2   R X  X 2  V  2 2 2  V R  R  R   0
 S L L  S  L  S L 

  R  2   R X  X 2  2R   R
R 
 S L L  S  L S L
R  S 2 2RR  S L R  L 2  X  S X L   2 2RR  S L 2R
2
L
R  S 2 R  L 2  X  S X L   2 2R
2
L
R  S 2  X  S X L   2 R
2
L
Therefore, for maximum power transfer, the load resistor should be

R L R  S 2   S X L  X  2

6-15. A 460-V 60-Hz four-pole Y-connected induction motor is rated at 25 hp. The equivalent circuit
parameters are
R = 0.15  R 2 = 0.154  X M = 20 
1
X = 0.852  X 2 = 1.066 
1
P = 400 W P = 150 W P = 400 W
F&W misc core
For a slip of 0.02, find
(a) The line current I L

(b) The stator power factor
(c) The rotor power factor

(d) The rotor frequency
(e) The stator copper losses P SCL

(f) The air-gap power P AG

(g) The power converted from electrical to mechanical form P conv

(h) The induced torque 
ind
(i) The load torque 
load
(j) The overall machine efficiency 
(k) The motor speed in revolutions per minute and radians per second
(l) What is the starting code letter for this motor?

SOLUTION The equivalent circuit of this induction motor is shown below:
166