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+// Electric Machinery and Transformers

+// Irving L kosow 

+// Prentice Hall of India

+// 2nd editiom

+

+// Chapter 13: RATINGS,SELECTION,AND MAINTENANCE OF ELECTRIC MACHINERY

+// Example 13-9

+

+clear; clc; close; // Clear the work space and console.

+

+// Given data

+// single phase alternator

+V = 500 ; // Rated voltage of the alternator in volt

+P = 20 ; // Rated power of the alternator in kVA

+I = 40 ; // Rated current of the alternator in A

+R = 2 ; // Armature resistance in ohm

+X = 15 ; // Armature reactance in ohm

+

+// Calculations

+// case a

+V_b = V ; // base voltage in volt

+I_b = I ; // base current in A

+R_pu = (R*I_b)/V_b ; // per-unit value of armature resistance in p.u

+

+// case b

+jX_pu = (X*I_b)/V_b ; // per-unit value of armature reactance in p.u

+

+// case c

+// subscript 1 indicates method 1 for finding Z_p.u

+Z_pu1 = R_pu + %i*(jX_pu); // per-unit value of armature impedance in p.u

+Z_pu1_m = abs(Z_pu1);//Z_pu1_m = magnitude of Z_pu1 in p.u

+Z_pu1_a = atan(imag(Z_pu1) /real(Z_pu1))*180/%pi;//Z_pu1_a=phase angle of Z_pu1 in degrees

+

+// subscript 2 indicates method 2 for finding Z_p.u

+Z_pu2 = (R + %i*X)*(I/V); // per-unit value of armature impedance in p.u

+Z_pu2_m = abs(Z_pu2);//Z_pu2_m = magnitude of Z_pu2 in p.u

+Z_pu2_a = atan(imag(Z_pu2) /real(Z_pu2))*180/%pi;//Z_pu2_a=phase angle of Z_pu2 in degrees

+

+// Display the results

+disp("Example 13-9 Solution : ");

+

+printf(" \n a: Armature resistance per unit value\n    R_p.u = %.2f p.u \n",R_pu);

+

+printf(" \n b: Armature reactance per unit value\n    jX_p.u in p.u = ");disp(%i*jX_pu);

+

+printf(" \n c: Armature impedance per unit value\n");

+printf(" \n    (method 1)\n    Z_p.u in p.u = ");disp(Z_pu1);

+printf(" \n    Z_p.u = %.3f <%.1f p.u \n",Z_pu1_m,Z_pu1_a );

+

+printf(" \n    (method 2)\n    Z_p.u in p.u = ");disp(Z_pu2);

+printf(" \n    Z_p.u = %.3f <%.1f p.u \n",Z_pu2_m,Z_pu2_a );