Autocatalytic Reaction — PFR, MFR and Minimum-Size Train: 23-Chem-A4 Chemical Reactor Engineering, December 2014, Q3

Question 3: Autocatalytic Reaction — PFR, MFR and Minimum-Size Train (20 marks: a 7, b 6, c 7)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $-r_A=kC_AC_R=kC_A(C_0-C_A)$ with $k=1$ L/(mol·min), $C_0=1$ mol/L. Feed $C_{A0}=0.99$ mol/L ($C_{R0}=0.01$); target $C_A=0.10$ mol/L ($C_R=0.90$). The rate is maximal at $C_A=C_0/2=0.5$ mol/L.

Find. Holding (space) time $\tau$ for (a) PFR, (b) MFR, (c) the minimum-total-volume PFR+MFR combination.

Figure 3 — $1/(-r_A)$ is U-shaped with a minimum (fastest rate) at $C_A=C_0/2=0.5$. A mixed-flow reactor is best where the curve is high on the feed side; a plug-flow reactor sweeps the low-lying tail — hence the minimum-volume MFR-then-PFR train.

Approach. Autocatalytic rate vanishes at both ends and peaks in the middle, so $1/(-r_A)$ is U-shaped (Figure 3); the MFR is best where the curve is high and the PFR where it is low — the classic minimum-volume argument.

Plug-flow reactor (part a). $\tau_{PFR}=\displaystyle\int_{C_A}^{C_{A0}}\frac{dC_A}{kC_A(C_0-C_A)}=\frac{1}{kC_0}\ln\frac{C_A}{C_0-C_A}\Big|_{0.10}^{0.99}$. Evaluating, $\tau_{PFR}=\ln\frac{0.99}{0.01}-\ln\frac{0.10}{0.90}=4.595+2.197=\boxed{6.79\ \text{min}}$.
Mixed-flow reactor (part b). The whole vessel sits at the exit ($C_A=0.10$, $C_R=0.90$): $\tau_{MFR}=\dfrac{C_{A0}-C_A}{kC_AC_R}=\dfrac{0.99-0.10}{(1)(0.10)(0.90)}=\dfrac{0.89}{0.09}=\boxed{9.89\ \text{min}}$. The single MFR is larger than the PFR because it runs entirely at the slow, low-$C_A$ exit rate.
Minimum-size train (part c). Because $1/(-r_A)$ is minimum at $C_A=0.5$, the least-volume arrangement is an MFR first, taking the feed down to the rate maximum ($0.99\to0.5$), followed by a PFR that finishes the slow tail ($0.5\to0.10$). MFR part: $\dfrac{0.99-0.50}{(1)(0.5)(0.5)}=1.96$ min; PFR part: $\ln\frac{0.5}{0.5}-\ln\frac{0.10}{0.90}=2.20$ min. Total$$\boxed{\tau_{min}=1.96+2.20=4.16\ \text{min}},$$about 39% below the single PFR and 58% below the single MFR.

Configuration	Holding time $\tau$
(a) Plug-flow reactor	6.79 min
(b) Mixed-flow reactor	9.89 min
(c) Minimum-size MFR→PFR train	4.16 min

Concept & Theory

Topic. Autocatalytic reactions and the minimum-volume reactor combination.

Key relations. Rate $-r_A=kC_A(C_0-C_A)$; PFR $\tau=\frac{1}{kC_0}\ln[C_A/(C_0-C_A)]$ between limits; MFR $\tau=(C_{A0}-C_A)/(-r_A)_{exit}$; optimum split at the rate maximum $C_A=C_0/2$.

Why this works. Autocatalysis makes the rate zero at both extremes (no R at the start, no A at the end), so a single PFR wastes volume near the feed while a single MFR wastes it near the exit. Splitting at the peak lets a small MFR jump the slow-start region and a PFR polish the slow finish — the general Levenspiel result for any U-shaped $1/(-r_A)$ curve.

Common pitfall. Placing the PFR first (it crawls through the low-rate feed region) or splitting at the wrong concentration; the optimum always joins the two units exactly at the rate maximum.

Source. Levenspiel, Chemical Reaction Engineering §6 (design for single reactions; autocatalytic reactions and reactor combinations).

Question 3 of 5: Autocatalytic Reaction — PFR, MFR and Minimum-Size Train

Question 3: Autocatalytic Reaction — PFR, MFR and Minimum-Size Train (20 marks: a 7, b 6, c 7)