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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')