Application of the Riccati Equation to Solve Adsorption Problems
Authors: Housam Binous and Ahmed Bellagi
Application of the Riccati Equation to Solve Adsorption Problems
Authors: Housam Binous and Ahmed Bellagi
Authors: Housam Binous and Ahmed Bellagi
We solve an adsorption problem (adsorption of acetone in air using clean charcoal) using the solution of a Riccati equation . We find that approximately 3.45 plates are needed to achieve the desired product purity (0.0202 Kg acetone/Kg air) . We also compute the number of stages with the classical approach by stepping off stages using the equilibrium curve (Langmuir relationship) and the operating line . With this second approach, we find 3.36 plates in close agreement with the Riccati equation method .
Acknowledgement :
The Riccati equation obtained when solving a binary distillation problem can be found in
Applied Mathematics and Modeling for Chemical Engineers by Rice and Do, Wiley, 1995.
The Riccati equation obtained when solving a binary distillation problem can be found in
Applied Mathematics and Modeling for Chemical Engineers by Rice and Do, Wiley, 1995.
Solution of the Riccati equation
In[]:=
sol=RSolve[x[n+1]x[n]+Adx[n+1]+Bdx[n]+Cd0,x[n],n]//FullSimplify
Out[]=
x[n]-Bd-4Cd+(-Bd(Ad+Bd)+2Cd)+Bd-4Cd-(Bd(Ad+Bd)-2Cd)(-Ad+Bd)+-4Cd-(Ad-Bd)--4Cd
n
4
n
-
1
Ad-Bd+-4Cd
2
(Ad+Bd)
2
(Ad+Bd)
n
4
n
-
1
Ad-Bd+-4Cd
2
(Ad+Bd)
2
(Ad+Bd)
n
Ad-Bd+-4Cd
2
(Ad+Bd)
AdBd-Cd
1
n
Ad-Bd+-4Cd
2
(Ad+Bd)
AdBd-Cd
1
n
4
n
-
1
Ad-Bd+-4Cd
2
(Ad+Bd)
n
4
n
-
1
Ad-Bd+-4Cd
2
(Ad+Bd)
2
(Ad+Bd)
n
Ad-Bd+-4Cd
2
(Ad+Bd)
AdBd-Cd
1
2
(Ad+Bd)
n
Ad-Bd+-4Cd
2
(Ad+Bd)
AdBd-Cd
1
Expressions for Ad, Bd and Cd for the adsorption problem
In[]:=
Ad=
VY1-LX0-V
L
Out[]=
-V-LX0+VY1
L
In[]:=
Bd=
1
K
Out[]=
1
K
In[]:=
Cd=
VY1-LX0
LK
Out[]=
-LX0+VY1
KL
Air rate V=3.5 Kg/s
In[]:=
V=3.5;
Clean Charcoal (X0=0) enters the system at L=2.5 Kg:s
In[]:=
X0=0;L=2.5;
Remove acetone from air such that:
YN+1=0.222 Kg acetone/ Kg air
and Y1=0.0202 Kg acetone/ Kg air
YN+1=0.222 Kg acetone/ Kg air
and Y1=0.0202 Kg acetone/ Kg air
In[]:=
Y1=0.0202;
Equilibrium constant K=0.5
In[]:=
K=0.5;
Values of the constants Cd, Bd and Ad
In[]:=
Cd
Out[]=
0.05656
In[]:=
Bd
Out[]=
2.
In[]:=
Ad
Out[]=
-1.37172
Exit charcoal contains 0.28525 Kg acetone/Kg charcoal
In[]:=
LX0+VY[N+1]-VY1
L
Out[]=
0.28252
We need exactly n=3.45 plates to obtain air with Y1 Kg acetone/ Kg air from YN+1 Kg acetone/ Kg air
In[]:=
FindRoot[{First[(x[n]/.sol/.C[1]C1/.n0)]X0,First[(x[n]/.sol/.C[1]C1)]0.28252},{C1,-300},{n,3}]
Out[]=
{C1-6.09164,n3.45327}
Equilibrium function
In[]:=
equilb=Function[With[{a=0.5},(a#/(1+(a)#))]]
Out[]=
With{a=0.5},&
a#1
1+a#1
Operating line
In[]:=
OpRect=Function[With[{L=2.5,V=3.5,X0=0},L/V#+Y1-LX0/V]]
Out[]=
With{L=2.5,V=3.5,X0=0},+Y1-&
L#1
V
LX0
V
Stepping off stages, we find a similar value as above (i.e. 3.36 plates)
In[]:=
x[0]=X0;i=0;y[0]=Y1;
In[]:=
While[x[i]<0.5,{Print[x[i]," ",i," ",y[i]],v=FindRoot[y[i]equilb[u],{u,0.5}][[1,2]],t=OpRect[v],If[v<0.5,{x[i+1]=v,y[i+1]=t}],i++}]
0 0 0.0202
0.0412329 1 0.0496521
0.104492 2 0.0948374
0.209548 3 0.169877
0.409282 4 0.312544
In[]:=
(0.28252-0.20954786)/(0.4092816-0.20954786)
Out[]=
0.365347