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+clear;
+clc;
+printf("\t\t\tProblem Number 3.11\n\n\n");
+// Chapter 3 : The First Law Of Thermodynamics
+// Problem 3.11 (page no. 111) 
+// Solution
+
+//Given data
+//                    Inlet     Outlet
+//Pressure(psia)      1000      1
+//Temperature(F)      1000      101.74
+//Velocity(ft/s)      125       430
+//Inlet position(ft)  +10       0
+//Enthalpy(Btu/LBm)   1505.4    940.0
+//Steam flow rate of 150000 LBm/hr
+
+//From the table,
+Z1=10; V1=125; h1=1505.4; Z2=0; V2=430; h2=940.0;
+
+//Energy equation is given by
+//((Z1/J)*(g/gc)) + (V1^2/(2*gc*J)) + h1 + q = ((Z2/J)*(g/gc)) + (V2^2/(2*gc*J)) + h2 + w/J
+printf("Solution for (a) \n");
+q=0; //net heat
+J=778; //Conversion factor
+gc=32.174; //Unit: (LBm*ft)/(LBf*s^2) //gc is constant of proportionality
+g=gc; //Unit:ft/s^2 //g=The local gravity
+//W1=w/J;
+//Energy equation is given by
+W1=((Z1/J)*(g/gc)) + (V1^2/(2*gc*J)) + h1 + q - ((Z2/J)*(g/gc)) - (V2^2/(2*gc*J)) - h2; //Unit:Btu/LBm
+printf("If heat losses are negligible,\n");
+printf("Total work of the turbine is %f Btu/LBm\n",W1);
+printf("Total work of the turbine is %f Btu/hr\n",W1*150000); 
+//(W*150000*778)/(60*33000) //in terms of horsepower  //1 hr=60 min //1 hp=33000 (ft*LBf)
+printf("Total work of the turbine is %f hp \n",(W1*150000*778)/(60*33000)); 
+//1 hp =0.746 kW
+printf("Total work of the turbine is %f kW \n\n",((W1*150000*778)/(60*33000))*0.746);
+
+
+printf("\nSolution for (b) \n");
+//Heat losses equal 50,000 Btu/hr
+q=50000/150000; //Unit:Btu/LBm //Heat loss
+W2=((Z1/J)*(g/gc)) + (V1^2/(2*gc*J)) + h1 - q - ((Z2/J)*(g/gc)) - (V2^2/(2*gc*J)) - h2; //Unit:Btu/LBm
+printf("If heat losses equal 50,000 Btu/hr , Total work of the turbine is %f Btu/LBm\n",W2);
+