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+//Chapter 5
+//Example 5.9
+//page 165
+//to determine power,voltage,compensating equipment rating
+clear;clc;
+A=0.85;
+B=200;
+
+//case(a)
+Vs=275000;
+Vr=275000;
+a=5;b=75; //alpha and beta
+Qr=0;
+//from equation 5.62
+d=b-asind((B/(Vs*Vr))*(Qr+(A*Vr^2*sind(b-a)/B))); //delta
+Pr=(Vs*Vr*cosd(b-d)/B)-(A*Vr^2*cosd(b-a)/B);
+printf('\n\ncase(a)\nPower at unity powerfactor that can be received =%0.1f MW',Pr/10^6);
+
+//case(b)
+Pr=150*10^6;
+d=b-acosd((B/(Vs*Vr))*(Pr+(A*Vr^2*cosd(b-a)/B))); //delta
+Qr=(Vs*Vr*sind(b-d)/B)-(A*Vr^2*sind(b-a)/B);
+Qc=-Qr;
+printf('\n\ncase(b)\nRating of the compensating equipment = %0.2f MVAR',Qc/10^6);
+printf('\ni.e the compensating equipment must feed positive VARs into the line');
+
+
+//case(c)
+Pr=150*10^6;
+Vs=275000;
+//by solving the two conditions given as (i) and (ii), we get
+Vr=244.9*10^3;
+printf('\n\ncase(c)\nReceiving end voltage = %0.1f kV',Vr/1000);