9 replies to this topic

[alovera88]

Hi everyone, I've got a problem where I have a two CSTR's joined together by a pipe that can be considered to be a PFR. I know the volumes of the reactors, the rate constant and the overall conversion, but I'm just stumped how I place all the design equations together (to make an implicit equation) to find out the volumetric flow rate for the desired conversion.

Any hints would be greatly appreciated, as I feel I am almost there.

Thanks.

Alex

(p.s. it is a liquid phase, first order reaction of species A to B.

90% conversion from A to B and there is no B in the feedstock)

[Andree]

solve equations:
1) Q*cA0=k*cA1*V1+Q*cA1
2) cA2=cA1*exp[-k*(V2/Q)]
3) Q*cA2=k*cA3*V3+Q*cA3
4) (cA0-cA3)/cA0=0.9

4 equations and for unknown variables: cA1,cA2,cA3,Q - solve for Q

Q - volumetric flow
cAi - concentrations: 0-inlet, 1-after 1st CSTR, 2-after PFR, 3-outlet (after 2nd CSTR)
Vi - volumes of reactors
k - reaction rate constant

[alovera88]

solve equations:
1) Q*cA0=k*cA1*V1+Q*cA1
2) cA2=cA1*exp[-k*(V2/Q)]
3) Q*cA2=k*cA3*V3+Q*cA3
4) (cA0-cA3)/cA0=0.9

4 equations and for unknown variables: cA1,cA2,cA3,Q - solve for Q

Q - volumetric flow
cAi - concentrations: 0-inlet, 1-after 1st CSTR, 2-after PFR, 3-outlet (after 2nd CSTR)
Vi - volumes of reactors
k - reaction rate constant

Hi, thanks for your reply, will have a look at it when I'm back from a trip.

Much appreciated , Alex

[alovera88]

Hey Andree,

jus been trying to work through the equations, but keep getting stuck, maybe my algebra's just rusty.

(1) Q*cA0=k*cA1*V1+Q*cA1
(2) cA2=cA1*exp[-k*(V2/Q)]
(3) Q*cA2=k*cA3*V3+Q*cA3
(4) (cA0-cA3)/cA0=0.9

[Values: V1,V3 = 5m^3
V2 = 11m^3
k = 0.1s^-1 ]

Starting from (4) we know that: cA3 = 0.1 , we can then complete the other equtions:

(1*) Q = (0.5*cA1) / ( 1 - cA1) [or] (1**) cA1 = Q / (0.5 + Q)

(3*) Q = 0.05 / (cA2 - 0.1) [or] (3**) cA2 = (0.05 / Q) + 0.1

Now we put these into equation (2) (1** and 3**) and taking logs to remove the exponent, but i can't seem to get my head around the next bit. Sorry if this seems trivial to you, but it's starting to annoy me as i know it should be relatively easy, but I can't see the solution!

Thanks again for your time.

[Andree]

Starting from (4) we know that: cA3 = 0.1 , we can then complete the other equtions:

cA3/cA0 = 0.1

[Andree]

I suggest solving the system of eqns numerically - analytically can be difficult (maybe even impossible) as you will have "Q" in exponent... I recommend MathCAD, which is very convenient for this purpose.

[Andree]

approx. 0.9593 [m3/s] - it seems to be too much (not realistic value unless it is a student task)

[alovera88]

approx. 0.9593 [m3/s] - it seems to be too much (not realistic value unless it is a student task)

Yeah, it's a student question, but should be able to be answered using just the design equations and I'm sure there must be a simpler way

thanks for the answer though, atleast it gives me something to work towards

Alex

[Andree]

to avoid wrtiting a numerical code (if you have no access to any math software, i.e. Mathcad or Matlab), you can find solution using simple spreadsheet... put guess value of Q and check if l.h.s of final equation is equal to r.h.s. ;-)

Attached Files

•  reactors_in_series.xls   21KB   44 downloads

[chenblue]

Excel can be used for iterative calculation.

1. We can simplify Andree's 4 equ's to become 2 equ's, like

Q1=f(Q2)
Q2=f(Q1)

2. Type the two equ's into Excel's two cells.
(Note that Q1 cell and Q2 cell need each other)

3. check the menu command/...Options/Tools/Calculation tab/Iteration checkbox,
select the iterative number(default=100).

It is vary easy, you could try it yourself.