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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)
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