Determining the recycle ratio in a PFR

[MichelV]

Homework Statement

For an irreversible first-order liquid-phase reaction (C_A,0 = 10 mol/L) conversion is 90% in a plug flow reactor. If two-thirds of the stream leaving the reactor is recycled to the reactor entrance, and if the throughput to the whole reactor-recycle system is kept unchanged, what does this do to the concentration of reactant leaving the system?

Homework Equations

For a first-order reaction where $\epsilon_A = 0$ without a recycle:
C_A / C_A,0 = e^-k$\tau$ (i)

For a first-order reaction where $\epsilon_A = 0$ with a recycle:
$k \tau = (R+1) ln \left[ \frac{C_{A,0} + R C_{A,f}} {(R+1) C_{A,f}}\right]$ (ii)

The Attempt at a Solution

Rewriting equation (i) gives:
$k\tau = ln \left( \frac {C_{A,0}} {C_A} \right) = ln \left( \frac {10} {1} \right) = ln (10)$

And if I were to know the value of R i could fill in equation (ii) except for C_A,f and combining equation (i) and (ii) would give me the opportunity to solve for C_A,f and determine the conversion in the PFR with recycle to answer the question.

I think determining the value of R should have something to do with this line from the problem:
"If two-thirds of the stream leaving the reactor is recycled to the reactor entrance, and if the throughput to the whole reactor-recycle system is kept unchanged."

It's just that I don't really get what they mean by this. Could someone point me in the right direction?

Thanks a lot!

 

[BvU]

So for every m3 of feed (feeds to the whole system are the same in the original case and in the recyclce case) the recylce adds 2 m3 and sends the 3 m3 to the reactor input. At the output 2 m3 is recycled and 1 m3 is product.

You use symbols without explaining them and without differentiating them ($\tau, R$). What is $\tau$ and is it the same in both cases ?

 

[tnich]

Homework Statement

For an irreversible first-order liquid-phase reaction (C_A,0 = 10 mol/L) conversion is 90% in a plug flow reactor. If two-thirds of the stream leaving the reactor is recycled to the reactor entrance, and if the throughput to the whole reactor-recycle system is kept unchanged, what does this do to the concentration of reactant leaving the system?

Homework Equations

For a first-order reaction where $\epsilon_A = 0$ without a recycle:
C_A / C_A,0 = e^-k$\tau$ (i)

For a first-order reaction where $\epsilon_A = 0$ with a recycle:
$k \tau = (R+1) ln \left[ \frac{C_{A,0} + R C_{A,f}} {(R+1) C_{A,f}}\right]$ (ii)

The Attempt at a Solution

Rewriting equation (i) gives:
$k\tau = ln \left( \frac {C_{A,0}} {C_A} \right) = ln \left( \frac {10} {1} \right) = ln (10)$

And if I were to know the value of R i could fill in equation (ii) except for C_A,f and combining equation (i) and (ii) would give me the opportunity to solve for C_A,f and determine the conversion in the PFR with recycle to answer the question.

I think determining the value of R should have something to do with this line from the problem:
"If two-thirds of the stream leaving the reactor is recycled to the reactor entrance, and if the throughput to the whole reactor-recycle system is kept unchanged."

It's just that I don't really get what they mean by this. Could someone point me in the right direction?

Thanks a lot!

The way I read this is that the reactant concentration of the output is 10% that of the input. So draw a flow diagram and write the flow balance equation. Then solve for the concentration of the output.