Page 188 - Solutions Manual to accompany Electric Machinery Fundamentals

The phas e voltage is 460/ 3 = 266 V, so line current I is
L
V 266 0 V

I  I   
R  1 jX  1 R  F jX F 0.54   j 2.093   0.435   j 2.90 
L A
I  I  52.3   79 A
L A
The start ing kVA of the motor is

S start  3V  A  I   3 266 V 52.3 A   41.7 kVA

The locked rotor kVA/hp is
41.7 kVA
kVA/hp   4.17
10 hp

Therefore this motor is Starting Code Letter D.
6-22. Plot the torque-speed characteristic of the motor in Problem 6-21. What is the starting torque of this
motor?
SOLUTION A MATLAB program to calculate the torque-speed characteristic of this motor is shown below:

% M-file: prob6_22.m
% M-fil e create a plot of the torque-speed curve of the
% induction motor of Problem 6-21.

% First, initialize th e values needed in this program.
r1 = 0.54; % Stator resistance
x1 = 2.093; % Stator r eactance
r 2 = 0.488; % Rotor res istance
x2 = 3.209; % Rotor reactance
xm = 51.12; % Magnetization bran ch react ance
v_phase = 460 / sqrt(3); % Phase voltage
n_sync = 1800; % Synchronous speed (r/min)
w_sync = 188.5; % Synchronous spee d (rad/s)

% Calculate the Thevenin voltage and impedan ce from Equations
% 6-41a and 6-43.
v_th = v_phase * ( xm / sqrt(r1^2 + (x1 + xm)^2) );
z _th = ((j*xm) * (r1 + j*x1)) / (r1 + j*(x1 + xm));
r_th = real(z_th);
x_th = imag(z_th);

% Now calculate the torque-speed characteristic for m any
% slips between 0 a nd 1. Note that the first slip va lue
% is set to 0.001 in stead of exactly 0 to avoid divide-
% by-zero problems.
s = (0:0.1:50) / 50; % Slip
s(1) = 0.001;
nm = (1 - s) * n_sync; % Mechanical speed

% Calculate torque versus speed
for ii = 1:len gth(s)
t_ind(ii) = (3 * v_th^2 * r2 / s(ii)) / ...
(w_sync * ((r_th + r2/s(ii))^2 + (x_th + x2)^2) );
end

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