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+// Electric Machinery and Transformers

+// Irving L kosow 

+// Prentice Hall of India

+// 2nd editiom

+

+// Chapter 6: AC DYNAMO VOLTAGE RELATIONS-ALTERNATORS

+// Example 6-6

+

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

+

+// Given data

+// 3-phase Y-connected alternator

+E_L = 11000 ; // Line voltage generated in volt

+kVA = 165000 ; // kVA rating of the alternator

+R_p = 0.1 ; // Armature resistance in ohm/per phase

+Z_p = 1.0 ; // Synchronous reactance/phase

+Z_r = 0.8 ; // Reactor reactance/phase

+

+// Calculations

+E_p = E_L / sqrt(3); // Rated phase voltage in volt

+I_p = (kVA * 1000)/(3*E_p); // Rated current per phase in A

+

+// case a

+I_max_a = E_p / R_p ; // Maximum short-circuit current in A (case a)

+overload_a = I_max_a / I_p ; // Overload (case a)

+

+// case b

+I_steady = E_p / Z_p ; // Sustained short-circuit current in A

+overload_b = I_steady / I_p ; // Overload (case b)

+

+// case c

+Z_t = R_p + %i*Z_r ; // Total reactance per phase

+I_max_c = E_p / Z_t ; // Maximum short-circuit current in A (case b)

+I_max_c_m=abs(I_max_c);//I_max_c_m=magnitude of I_max_c in A

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

+overload_c = I_max_c_m / I_p ; // Overload (case a)

+

+// Display the results

+disp("Example 6-6 Solution : ");

+printf("\n root 3 value is taken as %f , so slight variations in the answer.\n", sqrt(3));

+printf(" \n a: I_max = %d A ", I_max_a );

+printf(" \n    overload = %.1f * rated current \n", overload_a );

+

+printf(" \n b: I_steady = %d A ", I_steady );

+printf(" \n    overload = %.2f * rated current \n", overload_b );

+

+printf(" \n c: Rectangular form :\n   I_max = "); disp(I_max_c);

+printf(" \n    Polar form :");

+printf(" \n    I_max = %d <%.2f A ", I_max_c_m , I_max_c_a );

+printf(" \n    where %d is magnitude and %.2f is phase angle\n",I_max_c_m,I_max_c_a);

+printf(" \n    overload = %.3f * rated current \n", overload_c );