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+clear;
+clc;
+printf("\t\t\tProblem Number 3.16\n\n\n");
+// Chapter 3 : The First Law Of Thermodynamics
+// Problem 3.16 (page no. 116) 
+// Solution
+
+//In problem 3.15,
+p1=100; //Unit:psia //Initial pressure
+t1=950; //Unit:Fahrenheit //Temperature at pressure p1
+p2=76; //Unit:psia //Final pressure
+t2=580; //Unit:Fahrenheit //Temperature at pressure p2
+v1=4; //Unit:ft^3/LBm //Specific volume at inlet conditions
+v2=3.86; //Unit:ft^3/LBm //Specific volume at outlet conditions
+Cv=0.32; //Unit:Btu/(LBm*R) //Specific heat for constant volume process
+
+T1=t1+460; //Unit:R //Temperature at pressure p1
+T2=t2+460; //Unit:R //Temperature at pressure p2
+J=778; //J=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
+
+//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
+
+//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
+//In 3.15, the elevation of the pipe at section 1 makes Z1 = 0
+// Also no work crosses the boundaries of the system, the energy equation is reduced to
+//u1 + ((p1*v1)/J) + q = u2 + ((p2*v2)/J) + ((Z2/J)*(g/gc))
+//In problrm 3.16,
+Z2=100; //Given //Unit:ft //Outlet position
+//u2-u1=Cv*(T2-T1) //For a constant volume process //u2-u1=Chnage in internal energy
+//So,
+q=Cv*(T2-T1) + (p2*v2*144)/J - (p1*v1*144)/J + ((Z2/J)*(g/gc)) ;  //q=heat transfer //1 ft^2=144 in^2 //Unit:Btu/LBm
+printf("%f Btu/LBm heat is transferred from the gas \n",q);
+//For this problem , neglecting the elevation term leads to an insignificant error