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+clear;
+clc;
+printf("\t\t\tProblem Number 3.17\n\n\n");
+// Chapter 3 : The First Law Of Thermodynamics
+// Problem 3.17 (page no. 117) 
+// Solution
+
+p1=1000; //Unit:psia //Initial pressure
+t1=100;  //Unit:Fahrenheit //Temperature at pressure p1
+p2=1000; //Unit:psia //Final pressure
+t2=1000; //Unit:Fahrenheit //Temperature at pressure p2
+// feed in 10,000 LBm/hr 
+h1=70.68  //Unit:Btu/LBm //Inlet enthalpy
+h2=1505.9 //Unit:Btu/LBm //Outlet enthalpy
+
+T1=t1+460; //Unit:R //Temperature at pressure p1
+T2=t2+460; //Unit:R //Temperature at pressure p2
+//Energy equation is given by
+J=778; //J=Conversion factor
+
+//Z1=Inlet position //Unit:m 
+//V1=Inlet velocity //Unit:m/s
+//Z2=Outlet position //Unit:m 
+//V2=Outlet velocity Unit:m/s 
+//u1=internal energy //energy in
+//u2=internal energy //energy out
+//h=enthalpy
+
+//Energy equation is given by
+//((Z1/J)*(g/gc)) + (V1^2/(2*gc*J)) + u1 + ((p1*v1)/J) + q = ((Z2/J)*(g/gc)) + (V2^2/(2*gc*J)) + u2 + ((p2*v2)/J) + w/J; //Unit:Btu/LBm
+
+//we can consider this system as a single unit with feed water entering ans steam leaving. 
+//It well designed,this unit will be thoroughly insulated,and heat losse will be reduced to a negligible amount
+//Alos,no work will be added to the fluid during the time it is passing through the unit, and kinetic energy differences will be assumed to be negligibly small
+//Differennces in elevation also be considered negligible
+//So,the energy equation is reduced to 
+//u1 + ((p1*v1)/J) + q = u2 + ((p2*v2)/J)
+//Because h=u+(p*v/J)
+q=h2-h1; //q=net heat losses //Unit:Btu/LBm
+printf("Net heat losses is %f Btu/LBm \n",q);
+printf("For 10000 LBm/hr,\n");
+printf("%f Btu/hr energy has been added to the water to convert it to steam",q*10000)
+