1 reply to this topic

[philipxenakis]

I am currently doing a design project and the following reactions taking place in an isothermal CSTR:

k1
A + B -> C

k2
D + B -> C

k3
C + A -> E


I have the inlet concentrations for A, B and D; I need to find the reactor volume. Would I be correct in saying that:

rate of formation of C = k1 [A] [B] + k2 [D] [B] - k3 [C] [A] ?

Then I could say that, since rate of formation of C = moles of C / volume of CSTR, that

moles of C / volume of CSTR = k1 [A] [B] + k2 [D] [B] - k3 [C] [A]

in order to work out the volume of the CSTR?

I've seen reaction schemes with parallel reactions, and ones with consecutive reactions, but not both together as seen here for component C. Have had a look in some reaction engineering books including Froment and Bischoff, didn't find any reaction schemes specifically like this. Any help would be much appreciated.

[rana680]

I am currently doing a design project and the following reactions taking place in an isothermal CSTR:

k1
A + B -> C

k2
D + B -> C

k3
C + A -> E


I have the inlet concentrations for A, B and D; I need to find the reactor volume. Would I be correct in saying that:

rate of formation of C = k1 [A] [B] + k2 [D] [B] - k3 [C] [A] ?

Then I could say that, since rate of formation of C = moles of C / volume of CSTR, that

moles of C / volume of CSTR = k1 [A] [B] + k2 [D] [B] - k3 [C] [A]

in order to work out the volume of the CSTR?

I've seen reaction schemes with parallel reactions, and ones with consecutive reactions, but not both together as seen here for component C. Have had a look in some reaction engineering books including Froment and Bischoff, didn't find any reaction schemes specifically like this. Any help would be much appreciated.

You are correct in saying r_C = k1 [A] [B] + k2 [D] [B] - k3 [C] [A] if all the above reactions follow elementary rate laws.
For a CSTR V=(F_c0-F_c)/-r_c
where F_c0= nu₀C_C0 AND F_C=nuC_c
usually nu₀=nu
which gives V=(nu₀C_C0-nuC_c)/-r_c

regards
Rahul

Edited by rana680, 22 February 2010 - 03:01 PM.