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//CHAPTER 7- SINGLE PHASE TRANSFORMER
//Example 37
clc;
disp("CHAPTER 7");
disp("EXAMPLE 37");
//500 kVA 3300/500 V 50 hz single phase transformer
//VARIABLE INITIALIZATION
va=500000; //apparent power
v1=3300; //primary voltage in Volts
v2=500; //secondary voltage in Volts
f=50;
//loads
pf=1; //power factor unity
eff=0.97; // at 3/4 full load at unity pf
pf2=0.8; //power factor
//
//SOLUTION
I1=va/v1;
loss=(1-eff)*va*(3/4)*pf/eff; //=Pc+Pcu losses at 3/4 load
//since the eff value is maximum, Pcu=Pc; therefore, 2*Pc=loss
Pc=loss/2;
//(3/4)^2*Pcu=Pc;
f=(3/4)^2; //3/4 load
//Pcu=Pc/f
Pcu=Pc/f;
//disp(sprintf("The Pc is %f W",Pc));
//disp(sprintf("The Pcu is %f W",Pcu));
//
R_e1=Pcu/I1^2;
disp(sprintf("The value of Re1 is %.3f W",R_e1));
//10% impedance
Z_e1=v1*0.1/I1;
X_e1=sqrt(Z_e1^2-R_e1^2);
phi=acos(0.8);
%reg=(I1*R_e1*cos(phi)+I1*X_e1*sin(phi))*100/v1;
disp(sprintf("The percent regulation at full load 0.8 pf is %.2f W",%reg));
disp(" ");
//
//END
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