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+// Implementation of example 4.3
+// Basic and Applied Thermodynamics by P.K.Nag
+// page 73-74
+
+clc
+clear
+
+Q=-170 // (sum of all heat transfers during a cycle in kJ)
+
+Qab=0 // (heat transfer in process a-b in kJ/min)
+Qbc=21000 // (heat transfer in process b-c in kJ/min)
+Qcd=-2100 // (heat transfer in process c-d in kJ/min)
+Wab=2170 // (work done in process a-b in kJ/min)
+Wbc=0 // (work done in process b-c in kJ/min)
+dEcd=-36600 // (change in internal energy in kJ/min)
+
+dEab=Qab-Wab;
+dEbc=Qbc-Wbc;
+Wcd=Qcd-dEcd;
+// The system completes 100 cycles per min
+Qda=(Q*100)-(Qab+Qbc+Qcd);
+// Now dE=0,since cyclic integral of any property is zero
+dEda=-(dEab+dEbc+dEcd);
+Wda=Qda-dEda;
+printf("change in internal energy in a-b is = %.2f kJ/min \n",dEab);
+printf("change in internal energy in b-c is = %.2f kJ/min \n",dEbc);
+printf("work done in c-d is = %.2f kJ/min \n",Wcd);
+printf("change in internal energy in  d-a is = %.2f kJ/min \n",dEda);
+printf("work done in d-a is = %.2f kJ/min \n",Wda);
+printf("heat transfer in d-a is = %.2f kJ/min \n",Qda);
+// end
+\ No newline at end of file