Page 215 - Solutions Manual to accompany Electric Machinery Fundamentals

P   
conv ind
7-7. An eight-pole, 25-kW, 120-V dc generator has a duplex lap-wound armature which has 64 coils with 10
turns per coil. Its rated speed is 3600 r/min.
(a) How much flux per pole is required to produce the rated voltage in this generator at no-load
conditions?
(b) What is the current per path in the armature of this generator at the rated load?

(c) What is the induced torque in this machine at the rated load?
(d) How many brushes must this motor have? How wide must each one be?
(e) If the resistance of this winding is 0.011  per turn, what is the armature resistance R A of this
machine?

SOLUTION
ZP
(a) E  K   

A
2 a
In this machine, the number of current paths is
a  mP    16
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The number of conductor is

Z  64 coils 10 turns/coil 2 conductors/turn  1200
The equation for induced voltage is

ZP
E  
A

2 a
so the required flux is
1200 cond 8 poles   2 rad    1 min 
120 V    3600 r/min    
2  16 paths  1 r   60 s 
120 V 36,000 
  0.00333 Wb

(b) At rated load, the current flow in the generator would be
25 kW
I  A  208 A
120 V
There are a = m P = (2)(8) = 16 parallel current paths through the machine, so the current per path is

I 208 A
I  A   13 A
a 16
(c) The induced torque in this machine at rated load is

ZP
  I 
ind

2 a A
1200 cond 8 poles 


    0.00333 Wb 208 A
ind
2  paths 16

 ind  66.1 N m
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