Page 170 - Solutions Manual to accompany Electric Machinery Fundamentals

P 23.4 kW

  conv   132.6 N m
ind
 m  1685 r/min  2 rad    1 min 
  1 r     60 s  
Alternately, the induced torque can be found as

P 25.0 kW
  AG   132.6 N m

ind
 sync  1800 r/min  2 rad    1 min 
  1 r     60 s  
(c) The output power of this motor is

P  out P conv  P mech  23,400 W 300 W  23,100 W

P 23.1 kW
 load     2 rad    1 min   130.9 N m

out

m
 1685 r/min  1 r     60 s  
6-13. Figure 6-18a shows a simple circuit consisting of a voltage source, a resistor, and two reactances. Find
the Thevenin equivalent voltage and impedance of this circuit at the terminals. Then derive the
expressions for the magnitude of V TH and for R TH given in Equations (6-41b) and (6-44).

SOLUTION The Thevenin voltage of this circuit is
jX
V TH  M V

R 
1  1 X M  j X 
The magnitude of this voltage is
X
V TH  M V

R  1 2  1 X M  X  2

2
If X M  X , then R  1 2  X  1 X M   2  X  1 X M  , so
1
X
V  M V
X  X M
TH 
1
The Thevenin impedance of this circuit is
jX R  jX 
Z TH  R  1 M  1 1 X 1 M  j X 

164