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+// To determine all day efficiency
+
+clc;
+clear;
+
+p=15*(10^3);
+t1=12;
+t2=6;
+t3=6;
+
+pf1=0.5;
+pf2=0.8;
+pf3=0.9;
+
+x=poly([0 1],'x','c');
+
+nm=0.98; // Max Efficiency
+
+y=(nm*(p+(2*x)))-p;
+
+x=roots(y); // To find the iron loss or copper loss at unity p.f for maximum efficiency
+
+Pil=x; // Iron loss
+
+Pc=x; // Copper Loss at unity p.f for maximum efficiency
+
+deff('a=culoss(b,c)','a=b*Pc*((c/(p/1000))^2)');
+
+Pc1=culoss(12,(2/pf1)); // Total Copper Loss for 12hrs - 2 kW at p.f 0.5
+Pc2=culoss(6,(12/pf2)); // Total Copper Loss for 6hrs - 12 kW at p.f 0.8
+Pc3=culoss(6,(18/pf3)); // Total Copper Loss for 6hrs - 18 kW at p.f 0.9
+
+Po=((12*2)+(6*12)+(6*18))*(10^3);// Power Output
+
+eff=Po*100/(Po+(Pc1+Pc2+Pc3)+(24*Pil));
+
+// Note the iron loss has to be considered to calculate the Efficiency, Text Error
+
+printf('The all day effciency = %f percent \n',eff)
+