Page 29 - Solutions Manual to accompany Electric Machinery Fundamentals

V 100 V
v  B   200 m/s

Bl 0.5 T 1 m 
(c) With a load of 25 N opposite to the direction of motion, the steady-state current flow in the bar will
be given by

F app  F ind  ilB
F 25 N
i  app   50 A

Bl 0.5 T 1 m 
The induced voltage in the bar will be

e ind  V  B iR  100 V -  50 A 0.25    87.5 V
and the velocity of the bar will be
V 87.5 V
v  B   175 m/s

Bl 0.5 T 1 m 
The input power to the linear machine under these conditions is

P  in B  V i  100 V 50 A   5000 W
The output power from the linear machine under these conditions is

P  out B  V i  87.5 V 50 A   4375 W
Therefore, the efficiency of the machine under these conditions is

P 4375 W
  out  100%   100%  87.5%
P in 5000 W
1-22. A linear machine has the following characteristics:

B  0 .5 T into page R  0.25 
l  0.5 m V  120 V
B
(a) If this bar has a load of 20 N attached to it opposite to the direction of motion, what is the steady-state
speed of the bar?

(b) If the bar runs off into a region where the flux density falls to 0.45 T, what happens to the bar? What
is its final steady-state speed?

(c) Suppose V is now decreased to 100 V with everything else remaining as in part (b). What is the new
B
steady-state speed of the bar?
(d) From the results for parts (b) and (c), what are two methods of controlling the speed of a linear
machine (or a real dc motor)?

SOLUTION

(a) With a load of 20 N opposite to the direction of motion, the steady-state current flow in the bar will
be given by

F app  F ind  ilB
F 20 N
i  app   80 A
Bl 0.5 T  0.5 m

The induced voltage in the bar will be
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