Example 15.4-1 The Cooling of an Ideal Gas

>  restart; Start by inserting lots of data:
>  MW:=28.8*lbm/lbmol; For air.
>  T1:=(300+459.67)*R; v1:=100*ft/s; p1:=30*psia;
>  T2:=459.67*R; dh2:=10*ft; p2:=15*psia;
>  Cp:=T->(6.39+9.8e-4*T/R-8.18e-8*(T/R)^2)*Btu/(lbmol*R);
>  Cp(T1);Cp(T2); Checking the two ends of the exchanger.
>  Jc:=778.16*ft*lbf/Btu; Joule's Constant from Appendix C
>  w:=200*lbm/hr; The flow rate.
>  gcun:=32.174*lbm*ft/(lbf*s^2); The gravitational conversion 
factor used in English units. It makes F=m*a/gc consistent. The symbol 
gc is reserved in Maple.
>  g0:=32.174*ft/s^2; The standard acceration of Gravity from 
Appendix C.
>  Q:=w*(int(Cp(T),T=T1...T2)/MW+(v2^2-v1^2)/(2*gcun*Jc)+
Phi2-Phi1); Eq. 15.4-3
>  Phi2:=Phi1+dh2/Jc; This had better come out with the units: 
Btu/lbm
>  Phi2:=Phi1+dh2*g0/(Jc*gcun); This is the correction for that 
problem.
>  rho1:=p1*MW/(Rgas*T1); rho2:=p2*MW/(Rgas*T2); One way to get 
v2.
>  v2:=rho1*v1/rho2;

>  simplify(-Q); BS&L got 14,400 Btu/hr