Page 28 - Solutions Manual to accompany Electric Machinery Fundamentals

SOLUTION
The first step is to apply the following identity:
1
cos cos    cos       cos     
2
The result is

( ) 
pt v () ()  t i t 2VI cos t cos  t    )
 1 
() 
pt cos  2VI     t    cos t   t      t     
 2 
() VI 
pt   cos cos     2 t   
Now we must apply the angle addition identity to the second term:
 cos    cos cos   sin sin  

The result is
pt  cos  cos2 t cos   sin2 t sin  


( ) VI
Collecting terms yields the final result:
p VI cos  () t   1 cos2 t  VI sin sin2 t

 
1-21. A linear machine has a magnetic flux density of 0.5 T directed into the page, a resistance of 0.25 , a bar
length l = 1.0 m, and a battery voltage of 100 V.
(a) What is the initial force on the bar at starting? What is the initial current flow?

(b) What is the no-load steady-state speed of the bar?
(c) If the bar is loaded with a force of 25 N opposite to the direction of motion, what is the new steady-
state speed? What is the efficiency of the machine under these circumstances?

SOLUTION

(a) The current in the bar at starting is
V 100 V
i  B   400 A
R 0.25 

Therefore, the force on the bar at starting is
  F  i  l  B  400 A   1 m 0.5 T   200 N, to the right

(b) The no-load steady-state speed of this bar can be found from the equation

V  e ind  vBl
B

22