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// ELECTRICAL MACHINES
// R.K.Srivastava
// First Impression 2011
// CENGAGE LEARNING INDIA PVT. LTD
// CHAPTER : 4 : DIRECT CURRENT MACHINES
// EXAMPLE : 4.23
clear ; clc ; close ; // Clear the work space and console
// GIVEN DATA
N1 = 600; // Speed of the DC shunt Motor in RPM
Out_hp = 10; // Output of the DC shunt Motor in HP
V = 220; // Motor operating Volatge in Volts
Ra = 1.5; // Armature Resistance in Ohms
Rf = 250; // Field Resistance in Ohms
eta = 88/100; // Operating Efficiency of the Motor
Rf_a = 50; // Resistance inserted to the field circuit
// CALCULATIONS
out = Out_hp * 746 // Output of the DC Motor in watts
I = out/(V * eta); // Rated current in Amphere
If1 = V/Rf; // Field current in Amphere
Ia1 = I - If1; // Aramature current in Amphere
E1 = V - Ra*Ia1; // Back EMF in Volts
If2 = V/(Rf+Rf_a); // New Field current in Amphere after 50 Ohms Resistance inserted to the field circuit
// Refer page no. 217 we have T1 = K*If1*Ia1 proportional to 1/W1^2 and T1 = K*If2*Ia2 proportional to 1/W2^2 thus T1/T2 = (If1*Ia1)/(If2*Ia2) = (W2^2)/(W1^2) = (N2^2)/(N1^2), Ia2 = (If1*Ia1*W1^2)/(If1*W1^2) = (0.88*37.65*N2^2)/(0.733*600*600)
// Now New EMF E2 is E2 = V - Ia2*Ra, E1/E2 = (k*If1*N1)/(k*If2*N2), E2 = (0.733*N2)/(0.88*600) = 220 - (0.88*37.65*1.5*N2^2)/(0.733*600*600) Thus we have 0.001388*N2^2 = 220 - 1.833*10^-4*N2
N2 = poly ([-220 0.001388 1.833*10^-4],'x','coeff'); // Expression for the new speed of the motor in Quadratic form
r = roots (N2); // Value of the New speed of the motor in RPM
// DISPLAY RESULTS
disp("EXAMPLE : 4.23 : SOLUTION :-") ;
printf("\n (a) New speed of the motor, N2 = %.2f RPM nearly %.f RPM \n",r(2,1),r(2,1));
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