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+// Calculating the instantaneous radial force and instantaneous axial force on the h.v. winding under short circuit conditions

+clc;

+disp('Example 5.17, Page No. = 5.98')

+// Given Data

+// 7500/435 V, single phase core type transformer

+Q = 575;// kVA rating

+f = 50;// Frequency (in Hz)

+u0 = 4*%pi*10^(-7);

+Z_pu = 0.036;// Per unit impedance

+T = 190;// h.v winding turns

+Lmt = 1.25;// Length of mean turn (in meter)

+Lc = 0.35;// Height of coils (in meter)

+m = 1.8;// Doubling effect multiplier

+a = 0.015;// Width of duct (in meter)

+bp = 0.027;// Width of h.v. winding (in meter)

+bs = 0.023;// Width of l.v. winding (in meter)

+k = 0.05;// Since the h.v. winding is 5% shorter than the l.v. winding at one end

+// Calculation of the instantaneous radial force

+I_fl = Q*1000/7500;// Rms value of full load current (in Ampere)

+i = m*2^(1/2)*(1/Z_pu)*I_fl;// Instantaneous peak value of short circuit current (in Ampere)

+Fr = u0/2*(i*T)^(2)*Lmt/Lc;// Instantaneous radial force on the h.v. coil (in N)

+disp(Fr,'(a) Instantaneous radial force on the h.v. winding (N)=');//in book answer is 2380000 (N).  The answers vary due to round off error

+// Calculation of the instantaneous axial force

+Fa = u0/2*k*(i*T)^(2)*Lmt/(2*(a+bp+bs));// Total instantaneous radial force on the h.v. coil (in N)

+disp(Fa,'(b) Instantaneous axial force on the h.v. winding (N)=');

+//in book answer is 3200000 (N).  The provided in the textbook is wrong

+disp('This shows that there is a very large axial force, even though one of the winding is only 5% shorter than the other at one end')