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+// FUNDAMENTALS OF ELECTICAL MACHINES 

+// M.A.SALAM 

+// NAROSA PUBLISHING HOUSE 

+// SECOND EDITION

+

+// Chapter 10 : SYNCHRONOUS MOTOR

+// Example : 10.2

+

+clc;clear; // clears the console and command history 

+

+// Given data

+kVA = 1200   // kVA ratings

+V = 14*10^3  // supply voltage in V

+R_r = 4.8    // per phase resistance in ohm

+X_r = 35     // syncronous reactance in ohm

+pf = 0.95    // leading power factor

+

+// caclulations

+phi = acosd(pf)

+Z_s = R_r+%i*X_r           // impedance per phase ohm

+I_a = kVA*10^3/(sqrt(3)*V) // armature current in A

+E_r = I_a*Z_s              // resultant voltage due to impedance in V

+V_t = V/sqrt(3)            // terminal voltage per phase in V

+b = atand(X_r/R_r)         // beta value

+E_f = sqrt(V_t^2+abs(E_r)^2-2*V_t*abs(E_r)*cosd(b-phi)) // excitation voltage per phase in V

+teta = sind(64)

+D = (E_r*teta/E_f)         // torque angle

+delta = asind(abs(D))

+

+// display the result  

+disp("Example 10.2 solution"); 

+printf(" \n Excitation voltage per phase \n E_f = %.2f V \n", E_f );

+printf(" \n Torque angle at 0.95 power factor lagging \n delta = %.2f degree \n", delta );