It’s Totally Tubular, Man!

DOFPro Team

Chem. Reactor Example

The engineering team has designed a process for converting species \(\mathrm{A}\) into a valuable product, species \(\mathrm{P}\), according to the reaction

\[ \mathrm{A+2B \rightarrow 2P} \]

A kinetic study has shown that the reaction is 1st order in species \(\mathrm{A}\) and in species \(\mathrm{B}\), so

\[ -r_\mathrm{A} = k C_\mathrm{A} C_\mathrm{B} \]

where

\[ k = k_0 e^{\frac{-E_a}{RT}} = 1 \times 10^{12} \mathrm{\frac{m^3}{mol \cdot s}} e^{\frac{-12,000}{T}} \]

Chem. Reactor Example (cont.)

The plant is designed to feed the reactor with an aqueous solution of \(\mathrm{A}\) and \(\mathrm{B}\), with \(C_\mathrm{A_0} = 1000\ \mathrm{mol/m^3}\) and \(C_\mathrm{B_0} = 4000\ \mathrm{mol/m^3}\) at a feed rate of \(200\ \mathrm{kg/min}\).

Over the range of possible concentrations for \(\mathrm{A}\), \(\mathrm{B}\), and \(\mathrm{P}\), the density of the aqueous solution is \(1000\ \mathrm{kg/m^3}\).

The constraints from the separators and the recycle loop require that \(f_\mathrm{A}\) be \(80\%\), and that the reactor temperature be \(60\ ^\circ \mathrm{C}\).

Design and compare a tubular reactor (PFR) and a mixed reactor (CSTR) to process the required flow and achieve the required conversion. For completeness, design a batch processing scheme with the same feed rate and temperature, where the batch switchover time (the time from one batch ending to the next batch starting) is 2 minutes.

Kinetics Solution

We need

\[ k = 1 \times 10^{12} \mathrm{\frac{m^3}{mol \cdot s}} e^{\frac{-12,000}{T}} \]

\[ = 1 \times 10^{12} e^{\frac{-12,000}{(60 +273.15)}} = 2.274 \times 10^{-4}\ \mathrm{\frac{m^3}{mol \cdot s}} \]

\[ -r_\mathrm{A} = k C_\mathrm{A} C_\mathrm{B} = k C_\mathrm{A_0}(1-f_\mathrm{A})C_\mathrm{B_0}(1-f_\mathrm{B}) \]

Since \(C_\mathrm{B_0} = 4 C_\mathrm{A_0}\), then \(\dot{n}_\mathrm{B_0} = 4 \dot{n}_\mathrm{A_0}\). The moles of \(\mathrm{B}\) reacted or converted, \(\dot{n}_\mathrm{B_0} - \dot{n}_\mathrm{B}\) equals twice the moles of \(\mathrm{A}\) converted from stoichiometry, so

Kinetics Solution (cont.)

\[ \dot{n}_{\mathrm{B}_0} - \dot{n}_\mathrm{B} = 2(\dot{n}_{\mathrm{A}_0} - \dot{n}_\mathrm{A}) \]

\[ \frac{\dot{n}_{\mathrm{B}_0} - \dot{n}_\mathrm{B}}{\dot{n}_{\mathrm{B}_0}} = 2\frac{(\dot{n}_{\mathrm{A}_0} - \dot{n}_\mathrm{A})}{4 \dot{n}_\mathrm{A_0}} \]

\[ f_\mathrm{B} = \frac{1}{2} f_\mathrm{A} \]

\[ -r_\mathrm{A} = 4 k C_\mathrm{A_0}^2(1-f_\mathrm{A})(1-\frac{f_\mathrm{A}}{2}) \]

Also \[ \dot{V}_0 = \frac{\dot{m}}{\rho} = \frac{200\ \mathrm{\frac{kg}{min}}}{1000\ \mathrm{\frac{kg}{m^3}}} = 0.200\ \mathrm{\frac{m^3}{min}} \]

Totally Tubular Solution

The design equation for the PFR is

\[ \tau = C_\mathrm{A_0} \int_0^{f_\mathrm{A}} \frac{df_\mathrm{A}}{(-r_\mathrm{A})} = C_\mathrm{A_0} \int_0^{f_\mathrm{A}} \frac{df_\mathrm{A}}{4 k C_\mathrm{A_0}^2(1-f_\mathrm{A})(1-\frac{f_\mathrm{A}}{2})} \]

\[ = \frac{1}{4 k C_\mathrm{A_0}} \int_0^{f_\mathrm{A}} \frac{df_\mathrm{A}}{(1-f_\mathrm{A})(1-\frac{f_\mathrm{A}}{2})} = \frac{1}{2 k C_\mathrm{A_0}} \ln \frac{2-f_\mathrm{A}}{2(1-f_\mathrm{A})} \]

For \(f_\mathrm{A} = 0.8\),

\[ \tau = \frac{1}{2 k C_\mathrm{A_0}} \ln \frac{2-f_\mathrm{A}}{2(1-f_\mathrm{A})} = \frac{1}{2 \cdot 2.274 \times 10^{-4} \cdot 1000} \ln \frac{2-0.8}{2(1-0.8)} \]

\(\tau = 2.416\ \mathrm{min}\)

\[ \tau = \frac{V_R}{\dot{V_0}},\ \ \ V_R = \tau \dot{V_0} = 2.416\ \mathrm{min} \cdot 0.2\ \frac{\mathrm{m^3}}{\mathrm{min}} = 0.4831\ \mathrm{m^3} \]

Wishing Upon a CSTR

The design equation for a CSTR is


\[ \tau = \frac{C_\mathrm{A_0}f_\mathrm{A}}{{-r_\mathrm{A}}} = \frac{C_\mathrm{A_0}f_\mathrm{A}}{4 k C_\mathrm{A_0}^2(1-f_\mathrm{A})(1-\frac{f_\mathrm{A}}{2})}= \frac{f_\mathrm{A}}{4 k C_\mathrm{A_0}(1-f_\mathrm{A})(1-\frac{f_\mathrm{A}}{2})} \]


\[ \tau = \frac{0.8}{4 \cdot 2.274 \times 10^{-4} \cdot 1000 \cdot (1-0.8)(1 - 0.4)} = 7.329\ \mathrm{min} \]


\[ \tau = \frac{V_R}{\dot{V}},\ \ \ V_R = \tau \dot{V} = 7.329\ \mathrm{min} \cdot 0.2\ \frac{\mathrm{m^3}}{\mathrm{min}} = 1.466\ \mathrm{m^3} \]

Batch Reactor

Since the feed and flow conditions are the same, the batch time will equal the space time for the PFR, so

\[ t = 2.416\ \mathrm{min} \]

The total processing time is the sum of the batch time and the switchover time

\[ t_\mathrm{total} = t + t_\mathrm{switchover} = 2.416 + 2 = 4.416\ \mathrm{min} \]

We have to process \(V = \dot{V} t_\mathrm{total} = 0.2 \cdot 4.416 = 0.8832\ \mathrm{m^3}\) in the total time. We could build one reactor with that volume, or we could build two reactors with 1/2 that volume, and switch back and forth between them. Economics would determine the final choice based on capital costs and labor costs. The single reactor would probably, but not necessarily be the more economical.

Comparison

  • The PFR had a small space time and a smaller volume for the processing requirements than the CSTR. For any positive-order overall kinetics, the PFR will always have a smaller space time. The advantage grows as the fractional conversion increases.
  • The PFR required calculus, but the CSTR just needed algebra. This is a general result.
  • Because of the switchover time, and switchover labor, batch reactors are almost always more expensive than continuous reactors.
  • The extent of reaction could be calculated several different ways:
    • From \(\dot{n}_\mathrm{A_0}\) and \(\dot{n}_\mathrm{A}\)
    • From \(\dot{n}_\mathrm{B_0}\) and \(\dot{n}_\mathrm{B}\)
    • From \(\dot{n}_\mathrm{P}\)
    • From the fractional conversions
    • As \(-r_\mathrm{A} V_R\) for the CSTR

The Takeaways

  1. Usually the PFR had a smaller space time and a smaller volume for the processing requirements than the CSTR.
  2. CSTR calculations don’t require calculus.
  3. Batch reactors require accounting for switchover times.






Thanks for watching!
The previous video in the series is in the link in the upper left. The next video in the series, is in the upper right. To learn more about Chemical and Thermal Processes, visit the website linked in the description.