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+// Electric Machinery and Transformers
+// Irving L kosow 
+// Prentice Hall of India
+// 2nd editiom
+
+// Chapter 14: TRANSFORMERS
+// Example 14-25
+
+clear; clc; close; // Clear the work space and console.
+
+// Given data
+kVA_1 = 500 ; // Power rating of the transformer 1 in kVA
+R_1_pu = 0.01 ; // per-unit value of resistance of the transformer 1
+X_1_pu = 0.05 ; // per-unit value of reactance of the transformer 1
+Z_1_pu = R_1_pu + %i*X_1_pu ; //per-unit value of impedance of the transformer 1
+
+PF = 0.8 ; // lagging power factor
+V_2 = 400 ; // Secondary voltage in volt
+S_load = 750 ; // Increased system load in kVA
+
+kVA_2 = 250 ; // Power rating of the transformer  2 in kVA
+R_pu_2 = 0.015 ; // per-unit value of resistance of the transformer 2
+X_pu_2 = 0.04 ; // per-unit value of reactance of the transformer 2
+
+// smaller transformer secondary voltageis same as larger transformer
+
+// Calculations
+// Preliminary calculations
+Z_pu_1 = R_pu_2 + %i*X_pu_2 ; // New transformer p.u. impedance
+
+// Calculations
+// case a
+V_b1 = 400 ; // base voltage in volt
+V_b2 = 400 ; // base voltage in volt
+Z_pu_2 = (kVA_1/kVA_2)*(V_b1/V_b2)^2 * (Z_pu_1); // New transformer p.u impedance
+Z_2_pu = Z_pu_2 ; //New transformer p.u impedance
+
+// case b
+cos_theta = PF ; // Power factor
+sin_theta = sqrt( 1 - (cos_theta)^2 );
+S_t_conjugate = (kVA_1 + kVA_2)*(cos_theta + %i*sin_theta); // kVA of total load
+
+// case c
+S_2_conjugate = S_t_conjugate * ( Z_1_pu /(Z_1_pu + Z_2_pu) ); // Portion of load carried by the smaller transformer in kVA
+S_2_conjugate_m = abs(S_2_conjugate);//S_2_conjugate_m=magnitude of S_2_conjugate in kVA
+S_2_conjugate_a = atan(imag(S_2_conjugate) /real(S_2_conjugate))*180/%pi;//S_2_conjugate_a=phase angle of S_2_conjugate in degrees
+
+// case d
+S_1_conjugate = S_t_conjugate * ( Z_2_pu/(Z_1_pu + Z_2_pu) ); // Portion of load carried by the original transformer in kVA
+S_1_conjugate_m = abs(S_1_conjugate);//S_1_conjugate_m=magnitude of S_1_conjugate in kVA
+S_1_conjugate_a = atan(imag(S_1_conjugate) /real(S_1_conjugate))*180/%pi;//S_1_conjugate_a=phase angle of S_1_conjugate in degrees
+
+// case e
+S_1 = S_1_conjugate_m ; 
+S_b1 = kVA_1 ; // base power in kVA of trancsformer 1
+LF1 = (S_1 / S_b1)*100 ; // Load fraction of the original transformer in percent
+
+// case f
+S_2 = S_2_conjugate_m ; 
+S_b2 = kVA_2 ; // base power in kVA of trancsformer 2
+LF2 = (S_2 / S_b2)*100 ; // Load fraction of the original transformer in percent
+
+// Display the results
+disp("Example 14-25 Solution : ");
+
+printf(" \n a: New transformer p.u impedance :\n    Z_p.u.2 in p.u = ");disp(Z_pu_2);
+
+printf(" \n b: kVA of total load :\n    S*_t in kVA = ");disp(S_t_conjugate);
+
+printf(" \n c: Portion of load carried by the smaller transformer in kVA :");
+printf(" \n    S*_2 in kVA =  ");disp(S_2_conjugate);
+printf(" \n    S*_2 = %.1f <%.2f kVA (inductive load)\n",S_2_conjugate_m,S_2_conjugate_a);
+
+printf(" \n d: Portion of load carried by the original transformer in kVA :");
+printf(" \n    S*_2 in kVA = ");disp(S_1_conjugate);
+printf(" \n    S*_2 = %.1f <%.2f kVA (inductive load)\n",S_1_conjugate_m,S_1_conjugate_a);
+
+printf(" \n e: Load fraction of the original transformer :\n    L.F.1 = %.1f percent\n",LF1);
+
+printf(" \n f: Load fraction of the original transformer :\n    L.F.2 = %.1f percent\n",LF2);
+
+printf(" \n g: Yes. Reduce the no-load voltage of the new transformer to some value ");
+printf(" \n    below that of its present value so that its share of the load is reduced.");
+
+
+
+