Exercise: CSTR design II - Given stoichiometry¶

A steady-state mass balance on the reactor is

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

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

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

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

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

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

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