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//CHAPTER 7- SINGLE PHASE TRANSFORMER
//Example 35
disp("CHAPTER 7");
disp("EXAMPLE 35");
//VARIABLE INITIALIZATION
va=200000; //apparent power
v1=11000; //primary voltage in Volts
v2=230; //secondary voltage in Volts
Woc=1600; //watts also equals core losses
Wc=2600; //watts, also equals cu losses
f=50;
//no load parameters
//day cycle given
h1=8;
load1=160000;
pf1=0.8;
h2=6;
load2=100000;
pf2=1;
h3=10;
load3=0;
pf3=0;
//SOLUTION
//24 hr energy output
Pout=load1*h1*pf1+load2*h2*pf2+load3*h3*pf3;
Pc24=Woc*24; // 24 hours Pc loss
//cu loss= hours.(kva output/kva rated)^2.Full load Cu loss
Pcu24=h1*(load1/va)^2*Wc+h2*(load2/va)^2*Wc+h3*(load3/va)^2*Wc;
Pin=Pout+Pc24+Pcu24;
eff=Pout*100/Pin;
//disp(sprintf("The value Pout is %f",Pout));
//disp(sprintf("The value Pc is %f",Pc24));
//disp(sprintf("The value Pcu is %f",Pcu24));
disp(sprintf("The percent efficiency at full load is %f",eff));
disp(" ");
//
//END
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