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+//Transport Processes and Seperation Process Principles
+//Chapter 2
+//Example 2.7-5
+//Principles of Momentum Transfer and Overall Balances
+//given data
+//cgs units
+V=69.1;//volumetric flow rate in gallon/min
+fr1=V*(1/60)*(1/7.481);//converting flow rate in ft3/s
+A1=0.0233;//cross section area in ft2 of the pipe
+v2=fr1/A1;//vel in ft/s
+v1=0;//since the tank is very large
+p1=1;//atm pressure
+p2=p1;
+gc=32.174;
+z1=0;//datum
+g=32.174;
+z2=50;//length of discharge line in ft
+rho=114.8;//density of liq soln
+F1=10;//friction loss in ft-lb force/lb mass
+Ws=(z1*(g/gc))-(z2*(g/gc))+(v1*v1/(2*gc))-(v2*v2/(2*gc))+((p1-p2)/rho)-F1;//shaft work calculated by mechanical energy equation
+n=0.65;//efficiency
+Wp=(-Ws/n);
+m=fr1*rho;//mass flow rate in lbm/s
+P=m*Wp*(1/550);//pump horsepower
+A2=0.05134;//area of cross section
+v3=fr1/A2;
+v4=v2;
+F2=0;//friction losses in the second pipe is negligible
+Pdbyrho=(v3*v3/(2*gc))-(v4*v4/(2*gc))-Ws-F2;
+Pdiff=Pdbyrho*(rho/144);//pressure diff in lb force/in2
+mprintf("pressure developed by pump = %f psia",Pdiff)
+mprintf(" pump horsepower= %f hp",P)