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+clear;
+clc;
+printf("\t\t\tProblem Number 3.14\n\n\n");
+// Chapter 3 : The First Law Of Thermodynamics
+// Problem 3.14 (page no. 114) 
+// Solution
+
+//In problem 3.13 ,
+p1=150; //Unit:psia //Initial pressure
+T1=1000; //Unit:R //Temperature at pressure p1
+p2=15; //Unit:psia //Final pressure
+T2=600; //Unit:R //Temperature at pressure p2
+Cp=0.24; //Unit:Btu/(LBm*R) //Specific heat for constant pressure process
+v1=2.47; //Unit:ft^3/LBm //Specific volume at inlet conditions
+v2=14.8; //Unit:ft^3/LBm //Specific volume at outlet conditions
+
+//For a non-flow,constant pressure process
+//w/J=deltah=h2-h1=Cp(T2-T1) //deltah=change in enthalpy
+//W=w/J
+W=Cp*(T1-T2); //W=Work output //Unit:Btu/LBm //h2-h1
+printf("In problem 3.13,The work output of the turbine per pound of working fluid is %f Btu/LBm \n \n",W);
+
+//Now,In problem 3.14 , 
+q=1.1; //Unit:Btu/LBm //Heat losses
+//For a non-flow,constant pressure process
+//q-w/J=deltah=h2-h1=Cp(T2-T1) //deltah=change in enthalpy
+//W1=w/J
+W1=-q+W; //W=Work output //Unit:Btu/LBm //W=h2-h1 //Because q  is out of the system,it is a negative quantity
+printf("In problem 3.14,heat loss equal to 1.1 Btu/LBm,\n");
+printf("The work output of the turbine per pound of working fluid is %f Btu/LBm \n",W1);