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// FUNDAMENTALS OF ELECTICAL MACHINES
// M.A.SALAM
// NAROSA PUBLISHING HOUSE
// SECOND EDITION
// Chapter 3 : TRANSFORMER AND PER UNIT SYSTEM
// Example : 3.17
clc;clear; // clears the console and command history
// Given data
S_b1 = 100 // base apparent power
V_bT11 = 220 // voltage of 1st transformer in kV
V_bT12 = 132 // voltage of 1st transformer in kV
X_T1 = 0.02 // impedance of 1st transformer in pu
S_b2 = 50 // base apparent power
V_bT21 = 132 // voltage of 2nd transformer in kV
V_bT22 = 66 // voltage of 2nd transformer in kV
X_T2 = 0.05 // impedance of 2nd transformer in pu
X_L = 4 // line impedance in ohm
P = 50 // power absorded in MW
pf = 0.6 // lagging power factor from transmission line
Z_p = 0.32*%i //Reactance of transformer in ohm
// caclulations
S_b = S_b1 //Base power(MW)
V_b = V_bT11 //Base voltage(kV)
a = V_bT11/V_bT12 // turn ratio for 1st transformer
Vb_line = (V_bT11/a) // base voltage of line in kV
Zb_line = Vb_line^2/S_b1 // base impedance of line in ohm
Xpu_line = X_L/Zb_line // per unit reactance of line
Xpu_T1 = X_T1*(V_bT11/V_b)^2*(S_b/S_b1) // 1st grid transformer ,the per unit reactance
Vb_load = (V_bT12/(V_bT12/V_bT22)) // load side base voltage in kV
Xpu_load = X_T2*(V_bT22/Vb_load)^2*(S_b/S_b2) // second load transformer ,the per unit reactance
I_b = S_b*1000/(sqrt(3)*Vb_load) // base current
I_L = S_b2*1000/(sqrt(3)*V_bT22*pf) // actualcurrent in load in A
I_Lpu = I_L/I_b // per unit value of the load
V_L = V_bT22/V_bT22 //per unit value of the voltage at the load terminal(bus4)
V_gb = I_Lpu*exp(%i*acos(pf))*Z_p + 1 // per unit value of bus voltage
V_gba = abs(V_gb)*V_bT11 // actual value of grid to bus voltage
// display the result
disp("Example 3.17 solution");
printf(" \n Actual value of grid to bus voltage \n V_gba = %.2f kV \n", V_gba);
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