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// FUNDAMENTALS OF ELECTICAL MACHINES
// M.A.SALAM
// NAROSA PUBLISHING HOUSE
// SECOND EDITION
// Chapter 5 : DIRECT CURRENT MOTORS
// Example : 5.5
clc;clear; // clears the console and command history
// Given data
P = 6 // number of poles
V_t = 230 // supply to shunt motor in V
Z = 450 // number of conductors
R_a = 0.8 // armature resistance in ohm
I = 30 // supply current in A
P_0 = 5560 // out put power in W
I_F = 3 // current through field winding
phi = 25*10^-3 // flux per pole in Wb
// caclulations
A = 6 // for lap wond A=P
I_a = I-I_F // armature current in A
E_b = V_t-I_a*R_a // back emf in V
N = 60*A*E_b/(P*Z*phi) // speed in rpm
T_sh = 9.55*P_0/N // shaft torque in N-m
// display the result
disp("Example 5.5 solution");
printf(" speed \n N = %.1f rpm \n", N );
printf(" shaft torque \n T_sh = %.1f N-m \n", T_sh );
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