Maple Solutions to the Chemical Engineering Problem Set

f(re);
C ( re )
D
The next step is to create a Maple procedure to evaluate the force balance for different values of the
terminal velocity. There are many ways to do this; our example follows:
> eqn := proc(vt)
local C, re;
global params;
re:= subs(params,D[p]*vt*rho/mu);
C[D]:=f(re);
subs(params,vt-sqrt(4*g*(rho[p]-rho)*D[p]/(3*C[D]*rho)));
end:
Finally, we invoke Maple's built in floating point solver
> vt1:=fsolve('eqn'(vt),vt=0.00001..0.05);
vt1 := .01578618582
This is the terminal velocity of the particle under the given conditions. Note that eqn in the call to
fsolve is contained within quote marks. If this were not done then Maple would attempt to call the eqn
procedure and pass that result to fsolve. Needless to say, that would be a disaster in this case. the quote
marks force Maple to re-evaluate the equation every time it tries a new value of the terminal velocity.
The units here are m/s. The Reynolds number at this velocity is
> subs(v[t]=vt1,params,reeqn);
re = 3.656696458
and the drag coefficient is
> f(rhs("));
8.840561062
For part (b) we are asked for the terminal velocity in a centifugal separator where the acceleration is
30 g. The only changes needed are to revise the set of parameters (in this case only g is changed).
> params := {g=9.81*30,rho[p]=1800, D[p]=0.208e-3,rho=994.6,mu=8.931e-4};
params := { g = 294.30 , ρ = 994.6 , μ = .0008931 , D = .000208 ,
ρ = 1800}
p
p
> eqn := proc(vt)
local C, re;
global params;
params := {g=9.81*30,rho[p]=1800, D[p]=0.208e-3,rho=994.6,mu=8.931e-4};
re:= subs(params,D[p]*vt*rho/mu);
C[D]:=f(re);
subs(params,vt-sqrt(4*g*(rho[p]-rho)*D[p]/(3*C[D]*rho)));
end:
Again we call fsolve
> vt2:=fsolve('eqn'(vt),vt);
vt2 := .2060692237
The units here are m/s. The Reynolds number at this velocity is
> subs(v[t]=vt2,params,reeqn);
re = 47.73367101
and the drag coefficient is
> f(rhs("));
1.556428713
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