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//CHAPTER 7- SINGLE PHASE TRANSFORMER
//Example 20
disp("CHAPTER 7");
disp("EXAMPLE 20");
//VARIABLE INITIALIZATION
va=4000; //apparent power
v1=200; //primary voltage in Volts
v2=400; //secondary voltage in Volts
f=50;
R_e1=0.15;
Pi=60; //core losses iron core
pf1=0.9;
pf2=0.8;
//SOLUTION
R_e2=(v2/v1)^2*R_e1;
I1=va/v1;
I2=va/v2;
Pcu=I2^2*R_e2; //cu losses
disp("SOLUTION (i)");
disp(sprintf("The value of Copper Losses at full load is %f W",Pcu));
//
Pout=va*pf1;
Pin=Pout+Pi+Pcu;
eff=Pout*100/Pin;
disp("SOLUTION (ii)");
disp(sprintf("The percent efficiency at full load %f PF is %f",pf1,eff));
//
//at half load
Pout=va*pf2/2;
Pin=Pout+Pi+Pcu*(1/2)^2;
eff=Pout*100/Pin;
disp("SOLUTION (ii)");
disp(sprintf("The percent efficiency at full load %f PF is %f",pf2,eff));
disp(" ");
//
//END
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