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authorprashantsinalkar2017-10-10 12:27:19 +0530
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+///Chapter 9 Law Of Thermodynamics
+///Example 9.1 Page No:165
+/// Find Work interaction during the 4th processes
+///Input data
+clc;
+clear;
+Qab=720; //Heat transfer of 1st processes in KJ
+Qbc=-80; //Heat transfer of 2nd processes in KJ
+Qcd=40; //Heat transfer of 3rd processes in KJ
+Qda=-640; //Heat transfer of 4th processes in KJ
+Wab=-90; //Work transfer of 1st processes in KJ
+Wbc=-50; //Work transfer of 2nd processes in KJ
+Wcd=130; //Work transfer of 3rd processes in KJ
+
+
+///Calculation
+///From the 1st law of thermodynamic for close system undergoing a cycle.
+
+//Work interaction during the 4th processes
+Wda=((Qab+Qbc+Qcd+Qda)-(Wab+Wbc+Wcd));
+
+///Output
+printf('Work interaction during the 4th processes= %f KJ \n",Wda);