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// ELECTRICAL MACHINES
// R.K.Srivastava
// First Impression 2011
// CENGAGE LEARNING INDIA PVT. LTD
// CHAPTER : 7 : SPECIAL MOTORS AND INTRODUCTION TO GENERALIZED MACHINE THEORY
// EXAMPLE : 7.8
clear ; clc ; close ; // Clear the work space and console
// GIVEN DATA
V = 220; // supply voltage in Volts
f = 50; // Frequency in Hertz
p = 4; // Number of poles
Xm = 60; // Mutual reactance in Ohms
Rs = 1.0; // Resistance of stator windings in Ohms
Xs = 6.0; // Leakage reactance of stator windings in Ohms
Ra = 2.5; // Resistance of Armature windings in Ohms
Xa = 6.0; // Leakage reactance of armature windings in Ohms
P_hp = 1; // Output power in HP
N = 1400; // Motor running speed in RPM
alpha = 15; // Brush displacement from the low-impedance position in degree
// CALCULATIONS
Ns = (120*f)/p; // Synchronous speed in RPM
s = N/Ns; // Speed ratio
I = V / (Rs + %i*(Xs+Xm) + (%i*Xm^2*cosd(alpha))*(s*sind(alpha)-(%i*cosd(alpha)))/(Ra+%i*(Xa+Xm))); // Curent in Amphere
pf = cosd(atand(imag(I),real(I))); // Power factor lagging
// DISPLAY RESULTS
disp("EXAMPLE : 7.8: SOLUTION :-");
printf("\n (a) Currents, I = %.2f < %.2f A \n",abs(I),atand(imag(I),real(I)))
printf("\n (b) Power factor = %.4f lagging \n",pf)
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