# Exercise: CSTR design II - Given stoichiometry

A steady-state mass balance on the reactor is

\[\dot m_{in} = \dot m_{out}\]

or

\[(\dot V \rho)_{in} = (\dot V \rho)_{out}\]

As noted in the problem statement, the fluid density is constant,
so the mass balance equation reduces to

\[\dot V_{in} = \dot V_{out} = \dot V\]

Mole-rate balance on species \(E\) and \(F\) are

\[ \begin{align}\begin{aligned}\dot n_{E, in} = \dot n_{E, out} + r_{\text{cons, E}}\\\dot n_{F, in} = \dot n_{F, out} + r_{\text{cons, F}}\end{aligned}\end{align} \]

which are expressed more conveniently for this problem as

\[ \begin{align}\begin{aligned}(\dot V c_{E})_{in} = (\dot V c_{E})_{out} + r_{\text{cons, E}}\\(\dot V c_{F})_{in} = (\dot V c_{F})_{out} + r_{\text{cons, F}}\end{aligned}\end{align} \]

We are given information related to the entering and leaving concentrations of species \(F\),
so we can immediately solve for the rate of consumption:

\[r_{\text{cons, F}} = (\dot V c_{F})_{in} - (\dot V c_{F})_{out} = \dot V (c_{F, in} - c_{F,out})\]

or, in terms of numerical values,

\[r_{\text{cons, F}} = (\SI{432}{L/min}) (\SI{2}{gmol/L} - \SI{0.2}{gmol/L}) = \SI{777.6}{gmol/min}\]

Based on the stoichiometry, we know that

\[\frac{r_{\text{cons, E}}} {r_{\text{cons, F}}} = \frac{1}{2}\]

So,

\[r_{\text{cons, E}} = \frac{1}{2} (\SI{777.6}{gmol/min}) = \SI{388.8}{gmol/min}\]

The outlet concentration of species \(E\) is unknown can be found with

\[(\dot V c_{E})_{out} = (\dot V c_{E})_{in} - r_{\text{cons, E}}\]

or

\[c_{E,out} = c_{E,in} - \frac{r_{\text{cons, E}}}{\dot V}\]

Substituting in numbers, we have

\[c_{E,out} = \SI{0.95}{gmol/L} - \frac{\SI{388.8}{gmol/min}}{\SI{432}{L/min}} = \SI{0.05}{gmol/L}\]

The conversion of \(E\) is defined by

\[X_{E} = \frac{\dot n_{E, in} - \dot n_{E, out}}{\dot n_{E, in}}
= \frac{(\dot V c_{E})_{in} - (\dot V c_{E})_{out}}{(\dot V c_{E})_{in}}
=\frac{c_{E, in} - c_{E, out}}{c_{E, in}}\]

Therefore,

\[X_{E} = \frac{\SI{0.95}{gmol/L} - \SI{0.05}{gmol/L}}{\SI{0.95}{gmol/L}}
= 0.947\]

To determine the reactor volume, we can make use of the reactor design equation:

\[r_{\text{cons, E}} = r_{rxn, E} \, V\]

Since we know the rate of consumption, we need the rate of reaction, which is given by

\[r_{rxn, E} = k \, c_{E} \, c_{F}^{2}\]

NOTE: This relationship is based on **concentrations inside the reactor**, which,
for a perfect CSTR, **are the same as those exiting the reactor**:

\[r_{rxn, E} = k \, c_{E, out} \, c_{F, out}^{2}\]

Substituting in numbers, we have

\[r_{rxn, E} = (\SI{6.1}{L^{2}.gmol^{-2}.s^{-1}})(\SI{0.05}{gmol/L})(\SI{0.2}{gmol/L})^{2}
= \SI{0.0122}{gmol/(L \cdot s)}\]

or, converting units,

\[r_{rxn, E} = \SI{0.732}{gmol/(L \cdot min)}\]

Putting all of this together, we get

\[\begin{split}\begin{align*}
V &= \frac{r_{\text{cons, E}}}{r_{rxn, E}} \\
&= \frac{\SI{388.8}{gmol/min}}{\SI{0.732}{gmol/(L \cdot min)}} \\
&= \SI{531.1}{L}
\end{align*}\end{split}\]

The mean residence time is then

\[\tau = \frac{V}{\dot V}
= \frac{\SI{531.1}{L}}{\SI{432}{L/min}}
= \SI{1.23}{min} = \SI{73.8}{s}\]

**Challenge**

Assuming the reactor volume is the same as above, the only way to change
the mean residence time is to change the volumetric flow rate for the reactor.

Our mole-rate balance is then as follows:

\[\dot V_{\text{new}} \, X_{\text{E, new}} \, c_{E, in}
= r_{\text{rxn, E}} \, V = k \, c_{E, out} \, c_{F, out}^{2} \, V\]

The outlet concentration can be computed using the conversion and inlet concentration:

\[c_{E, out} = (1 - X_{\text{E, new}}) \, c_{E, in}\]

Combining the above, we get

\[\dot V_{\text{new}} \, X_{\text{E, new}} \, c_{E, in} =
k \, (1 - X_{\text{E, new}}) \, c_{E, in} \, c_{F, out}^{2} \, V\]

Since

\[\tau_{\text{new}} = \frac{V}{\dot V_{\text{new}}}\]

we can recast the mole-rate balance in terms of conversion and residence time:

\[X_{\text{E, new}} =
k \, ({1 - X_{\text{E, new}}}) \, c_{F, out}^{2} \, \tau_{\text{new}}\]

This expression can be solved for the conversion to arrive at

\[X_{\text{E, new}} = \frac{\alpha}{1 + \alpha}\]

where

\[\alpha = k \, c_{F, out}^{2} \, \tau_{new} = k \, c_{F, out}^{2} \, (\tau_{orig}/2)\]

Substituting numbers into these latter equations, we get

\[\alpha = (\SI{6.1}{L^{2}/gmol^{2}.s}) \, (\SI{0.2}{gmol/L})^{2} \,
(\SI{73.8}{s}/2) = 9\]

and

\[X_{\text{E, new}} = \frac{\alpha}{1 + \alpha}
= \frac{9}{1 + 9}
= 0.9\]