# Exercise: CSTR design I - Given inlet and outlet conditions

A steady-state mass balance on the reactor is

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

or

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

A mole-rate balance on species \(E\) is

\[\dot n_{E, in} = \dot n_{E, out} + r_{\text{consumption, E}}\]

which is expressed more conveniently for this problem as

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

The conversion of \(E\) is defined by

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

As noted in the problem statement, the fluid density is constant,
so the mass balance equation reduces to
\(\dot V_{in} = \dot V_{out} = \dot V\).

Therefore,

\[X_{E} = \frac{c_{E, in} - c_{E, out}}{c_{E, in}}
= \frac{\SI{0.95}{M} - \SI{0.14}{M}}{\SI{0.95}{M}}
= 0.853\]

The mole-rate balance then takes the form

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

or

\[\dot V \, X_{E} \, c_{E, in} = r_{\text{consumption, E}}\]

We also know that

\[r_{\text{consumption, E}} = r_{reaction, E} \, V\]

The reaction rate equation is

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

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

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

Putting all of this together, we get

\[\begin{split}\begin{align*}
V &= \frac{r_{\text{consumption, E}}}{r_{reaction, E}} \\
&= \frac{\dot V \, X_{E} \, c_{E, in}}{k \, c_{E, out} \, c_{F, out}^{2}} \\
&= \frac{(\SI{432}{L/min}) (0.853) (\SI{0.95}{gmol/L}) (\SI{1}{min}/\SI{60}{s})}
{(\SI{6.1}{L^{2}/gmol^{2}.s}) \, (\SI{0.14}{gmol/L}) \, (\SI{0.38}{gmol/L})^{2}} \\
&= \SI{47.3}{L}
\end{align*}\end{split}\]

The mean residence time is then

\[\tau = \frac{V}{\dot V}
= \frac{\SI{47.3}{L}}{\SI{432}{L/min}}
= \SI{0.109}{min} = \SI{6.54}{s}\]

**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:

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

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

\[\alpha = (\SI{6.1}{L^{2}/gmol^{2}.s}) \, (\SI{0.38}{gmol/L})^{2} \,
(\SI{6.54}{s}/2) = 2.893\]

and

\[X_{\text{E, new}} = \frac{\alpha}{1 + \alpha}
= \frac{2.893}{1 + 2.893}
= 0.743\]