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Diffstat (limited to '1445/CH7/EX7.36')
| -rw-r--r-- | 1445/CH7/EX7.36/ch7_ex_36.sce | 45 |
1 files changed, 45 insertions, 0 deletions
diff --git a/1445/CH7/EX7.36/ch7_ex_36.sce b/1445/CH7/EX7.36/ch7_ex_36.sce new file mode 100644 index 000000000..2aa003af2 --- /dev/null +++ b/1445/CH7/EX7.36/ch7_ex_36.sce @@ -0,0 +1,45 @@ +//CHAPTER 7- SINGLE PHASE TRANSFORMER +//Example 36 + +disp("CHAPTER 7"); +disp("EXAMPLE 36"); + +//VARIABLE INITIALIZATION +va=100000; //apparent power +v1=440; //primary voltage in Volts +v2=11000; //secondary voltage in Volts +f=50; +//loads +pf=1; +eff1=0.985; // at full load at 0.8pf +eff2=0.99; //at half full load at unity pf +pf1=0.8; // +pf2=1; // +// +//SOLUTION +loss1=(1-eff1)*va*pf1/eff1; //=Pc+Pcu losses +loss2=(1-eff2)*va*(1/2)*pf2/eff2; //=Pc+Pcu losses +//simultaneous equation to be solved +//eq 1: Pc+Pcu=loss; +//fractipon of copper/ ohmic losses +f=(1/2)^2; // 60% of full load +//the 2nd equation is Pc+f*Pcu=loss +//now the matrix +M=[1,1;1,f]; //Pc+Pcu=loss1; Pc+(1/2)^2*Pcu=loss2: 1,1,; 1,f +A=[loss1,loss2]; +Mi=inv(M); +Ans=A*inv(M); +Pc=Ans(1,1); +Pcu=Ans(1,2); +disp(sprintf("The Pc is %f W",Pc)); +disp(sprintf("The Pcu is %f W",Pcu)); +// +//maximumefficiency at farction x times the full load;and then f.Pcu=Pc +x=sqrt(Pc/Pcu); +disp(sprintf("The maximum efficiency would occur at a load of %f VA",x*va)); +I1=va/v1; +I1maxEff=I1*x; +disp(sprintf("The current at maximum efficeincy is %f A",I1maxEff)); +disp(" "); +// +//END |