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+// Implementation of example 6.3
+// Basic and Applied Thermodynamics by P.K.Nag
+
+clc
+clear
+
+// T temperature,Nmax max efficiency,W work done,Q heat trnsfer,COP coefficient of performance of refrigerator..
+t1=600 
+T1=t1+273 // K
+t2=40 
+T2=t2+273 // K
+t3=-20
+T3=t3+273 // K
+Q1=2000 // kJ
+W=360 // kJ
+
+Nmax=1-(T2/T1);
+W1=Nmax*Q1;
+COP=T3/(T2-T3);
+W2=W1-W;
+Q4=COP*W2;
+Q3=Q4+W2;
+Q2=Q1-W1;
+printf("heat rejection to 40 degree celsius reservoir = %.2f kJ \n",Q2+Q3);
+// part b
+N=0.4*Nmax;
+W1=N*Q1;
+W2=W1-W;
+COP2=0.4*COP;
+Q4=W2*COP2;
+Q3=Q4+W2;
+Q2=Q1-W1;
+printf("heat rejection to 40 degree celsius reservoir with decreased efficiency= %.2f kJ \n",Q2+Q3);
+// end
+\ No newline at end of file