Page 177 - Solutions Manual to accompany Electric Machinery Fundamentals

The slip at pullout torque is found by calcula ting the Thevenin equivalent of the input circuit from the
rotor back to the power supply, and then using that with the rotor circuit model.
jX R  jX   16.67j   0.15   j 0.710  
Z TH  M 1 1   0.138  j 0.6822   0.696 78.6   
R 
1  1 M  j X  0.15   j  X  16.67   0.710 
jX  16.67 j  


V TH  M V      212.5 0 V  203 0.49 V
R 
1  1 X M  j X  0.15   j   16.67   0.710 
The slip at pullout torque is
R
s max  2
R TH 2   TH  X 2  X 2
0.154 
s max   0.0976
 0.138  2   0.6822  0.890    2

The synchronous speed of this motor is
120 f 120 50 Hz 
n  e   1500 r/min
sync
P 4
 sync   1500 r/min   2 rad      1 min    157.1 rad/s
 1 r   60 s 

sp
This corre onds to a rotor speed of
  n  1 s   n 1 976 0.0 1500 r/min  1354 r/min
max max sync
The pullout torque of the motor is

3V 2
 max  TH
  sync    R TH R TH   TH  X 2  X 2 
2


  3 203 V 2
 max 
  157.1 rad/s 0.138        2   0.138  0.6822    0.890  2   

 max  229.3 N m

6-18. Plot the following quantiti es for the motor in Problem 6-15 as slip varies from 0% to 10%: (a)  ind (b)

P conv (c) P (d) Efficiency . At what slip does P equal the rated power of the machine?
out
out
SOLUTION This problem is ideally suited to solution w ith a MATLAB program. An appropriate program
is shown below. It follows the calculations perform for Problem 6-18, but repeats them at many values
ed
of slip, and then plots the results. Note that it plots all the specified values versus n , which varies from
m
1620 to 1800 r/min, corresponding to a range of 0 to 10% slip.

% M-file: prob6_18.m
% M-file create a plot of the induced torque, power
% converted, power o ut, and efficiency of the induction
% motor of Problem 6-15 as a function of slip.

% First, initialize the values needed in this progr am.
r1 = 0.015; % Stator resistance
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