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authorpriyanka2015-06-24 15:03:17 +0530
committerpriyanka2015-06-24 15:03:17 +0530
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+//Exa 5.14
+clc;
+clear;
+close;
+format('v',6);
+
+//Given Data :
+m=3;//Kg
+p1=3;//bar
+T1=450;//Kelvin
+Q=600;//KJ
+Cv=0.81;//KJ/Kg
+T0=300;//Kelvin
+T=1500;//Kelvin
+deltaSsource=Q/T;//KJ/K
+//Q=m*Cv*(T2-T1)
+T2=Q/m/Cv+T1;//Kelvin
+A1=Q-T0*deltaSsource;//KJ
+deltaSg=m*Cv*log(T2/T1);//KJ/K
+A2=Q-T0*deltaSg;//KJ
+Loss=A1-A2;//KJ
+disp(Loss,"Loss in available energy due to heat transfer in KJ : ");