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

A steady-state mass balance on the reactor is

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

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

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

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

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

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

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