# Reactor design

1. Nov 1, 2014

### Maylis

1. The problem statement, all variables and given/known data
The elementary liquid phase reaction
$$A + B \rightarrow C$$
is carried out in a 500 L reactor. The entering concentrations of streams A and B are both 2 M and the specific reaction rate is 0.01 L/(mol min).

(a) Calculate the time to reach 90% conversion if the reactor is a batch reactor filled to the brim

Assuming a stoichiometric feed (10 mol A/min) to a continuous-flow reactor, calculate the reactor volume and space-time to achieve 90% conversion if the reactor is
(b) a CSTR (Ans.: V = 22,500 L)
(c) a PFR (Ans: V = 2,250 L)
(d) redo (a) through (c) assuming the reaction is first order in B and zero order in A with k = 0.01/min.
(e) Assume the reaction is reversible with $K_{e} = 2 \frac {L}{mol}$. Calculate the equilibrium conversion and the CSTR and PFR volumes necessary to achieve 98% of the equilibrium conversion.

2. Relevant equations

3. The attempt at a solution
(a)
I use the equation $t = N_{A0} \int_0^X \frac {dX}{-r_{A}V}$
$$t = C_{A0} \int_0^X \frac {dX}{-r_{A}}$$
I assume this is a first order reaction with respect to A, $-r_{A} = kC_{A}$, and I know for a liquid phase reaction, $C_{A} = C_{A0}(1 - X)$. Therefore, $-r_{A} = kC_{A0}(1-X)$.
$$t = C_{A0} \int_0^X \frac {dX}{kC_{A0}(1-X)}$$
$$t = \frac {1}{k} \int_0^X \frac {dX}{1-X}$$
$$t = - \frac {1}{k} ln(1 - X) = - \frac {1}{0.01} ln(1 - 0.9) = 230.3 \hspace{0.05 in} min$$

(b) This is where I get stuck, I use the equation $V = \frac {F_{A0}X}{-r_{A}}$
$$V = \frac {F_{A0}X}{kC_{A0}(1-X)}$$
$$V = \frac {10(0.9)}{0.01(2)(1-0.9)} = 4,500 \hspace{0.05 in} L$$
But this is not the answer given in the textbook, 22,500 L and I haven't figured out why.
For the space time,
$$\tau = \frac {V}{v_{0}}$$
where $v_{0}$ is the initial volumetric flow rate. $F_{A0} = C_{A0}v_{A0}$, so $10 = 2v_{A0}$, therefore the volumetric flow rate for A is 5 L/min, and since this is a stoichiometric feed, the total volumetric flow rate is 10 L/min, so the space time is 4,500 L/10 = 450 min.

Last edited: Nov 1, 2014
2. Nov 1, 2014

### Maylis

My problem was my assumption, I should have said that $-r_{A} = kC_{A}C_{B}$, since this reaction is elementary.

Last edited: Nov 1, 2014