Ex. 13.6-1 condensation of Steam on a Vertical Surface

using property data for steam from the Unix program "steam" for delH and rho and from Perry's for mu and k.

>  restart;
>  Ren:=(mu,dHvap)->Q/(Pi*D1*mu*dHvap); eq. 13.6-9 for the film 
Reynold's Number
>  Q:=92000*Btu/hr;
>  D1:=2/12*ft;
>  Ren200:=evalf(Ren(.7376*lbm/ft/hr,977.86*Btu/lbm)); 
Perry's gave .3049 cp for mu
>  Gam200:=.7376*lbm/ft/hr*%; Re*mu is Gamma in 13.6
>  Ren220:=evalf(Ren(.6829*lbm/ft/hr,965.20*Btu/lbm));
>  Gam220:=.6829*lbm/ft/hr*%;
>  hm:=(kf,rhof,muf,Gam)->(4/3)*(kf^3*rhof^2*g/(3*muf*Gam))^
(1/3); Eq. 13.6-4 for the mean heat transfer coef.
>  hm200:=hm(.393*Btu/hr/ft/F,60.11*lbm/ft^3,.7376*lbm/ft/hr,
Gam200);
>  g:=32.174*ft/s^2*(3600*s/hr)^2;
>  hm200:=simplify(evalf(hm200),assume=positive);
>  hm220:=hm(.3934*Btu/hr/ft/F,59.61*lbm/ft^3,.6829*lbm/ft/hr,
Gam220);
>  hm220:=simplify(evalf(hm220),assume=positive);
>  dT:=h->Q/(h*A);
>  Q; This is still set.
>  A:=10*ft*Pi*D1; L is 10ft and D1 is the pipe diameter
>  dT(hm220);
>  evalf(%);
>  evalf(dT(hm200)); Thus a T of 221 would be required.
Now let's look at the x axis of Figure 13.6-2. One would hope that 
this group is dimensionless, but that is not obvious from a casual 
inspection. Let's put in all the parameters shown for the group with 
their dimensions to make sure it is dimensionless and then check to 
see that we get about 1700 for its value as they did in their solution.
>  kf:=.3934*Btu/(hr*ft*F);
>  rhof:=59.61*lbm/ft^3;
>  g;
>  delT:=dT(hm220);
>  L:=10*ft;
>  mu:=0.6829*lbm/(ft*hr);
>  dHvap:=965.2*Btu/lbm;
>  x:=kf*rhof^(2/3)*g^(1/3)*delT*L/(mu^(5/3)*dHvap);
>  x:=simplify(evalf(%),assume=positive); This is close 
to 1700 particulary seeing where the experimental values scatter 
in the figure.