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+///Chapter 9 Law Of Thermodynamics
+///Example 9.9 Page No:171
+///Find Net heat transfer in 1st cycle
+///Input data
+clc;
+clear;
+Qab=-6500;  //Heat transferred in 1st process KJ/min
+Qbc=0;      //Heat transferred in 2nd process 
+Qcd=-10200; //Heat transferred in 3rd process KJ/min
+Qda=32600;  //Heat transferred in 4th process KJ/min
+Wab=-1050;  //Heat transferred in 1st process KJ
+Wbc=-3450;  //Heat transferred in 2nd process KJ
+Wcd=20400;  //Heat transferred in 3rd process KJ
+Wda=0;      //Heat transferred in 4th process
+
+///Calculator
+dQ=Qab+Qbc+Qcd+Qda; //Net heat transfer in 1st cycle
+dW=Wab+Wbc+Wcd+Wda; //Net work done in 1st cycle
+dW1=dW/60;          //Net work done in 1st cycle
+DeltaUab=Qab-Wab;   //ab process
+DeltaUbc=Qbc-Wbc;   //bc processes
+DeltaUcd=Qcd-Wcd;   //cd processes
+DeltaUda=Qda-Wda;   //dc processes
+
+///Output
+printf('Net heat transfer in 1st cycle= %f KJ/min \n',dQ);
+printf('Net work done in 1st cycle= %f KJ/min \n',dW);
+printf('Net work done in 1st cycle=%f KW \n ',dW1);
+printf('ab process= %f KJ/min \n',DeltaUab);
+printf('bc processes= %f KJ/min \n ',DeltaUbc);
+printf('cd processes= %f KJ/min \n ',DeltaUcd);
+printf('dc processes= %f KJ/min \n',DeltaUda);