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+// Calculating the per unit leakage reactance

+clc;

+disp('Example 5.14, Page No. = 5.90')

+// Given Data

+// 2000/400 V, single phase shell type transformer

+Q = 100;// kVA rating

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

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

+Tp = 200;// h.v winding turns

+Lmt = 1.5;// Length of mean turn (in meter)

+W = 0.12;// Width of winding (in meter)

+a = 0.016;// Width of duct between h.v and l.v. windings (in meter)

+bp = 0.04;// Width of h.v. winding (in meter)

+bs = 0.036;// Width of l.v. winding (in meter)

+// Calculation of the per unit leakage reactance

+Xp = %pi*f*u0*Tp*Tp/2*Lmt/W*(a+(bp+bs)/6);// Leakage reactance referred to the primary side (ohm)

+I_hv = Q*10^(3)/2000;// H.V. winding current at full load (in ampere)

+Xp_pu = Xp*I_hv/2000;// Per unit leakage reactance

+disp(Xp_pu,'Per unit leakage reactance=');

+//in book answer is 0.0353.  The answers vary due to round off error