Holding Times for an Autocatalytic Reaction: 23-Chem-A4 Chemical Reactor Engineering, December 2019, Q2

Question 2: Holding Times for an Autocatalytic Reaction (25 marks: a 8, b 5, c 12)

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. Elementary autocatalytic reaction with $-r_A = k\,C_A C_R = k\,C_A(C_0-C_A)$, $k = 1\ \text{L}\,\text{mol}^{-1}\text{min}^{-1}$, $C_0 = 1\ \text{mol/L}$. Product specification $C_R = 0.9 \Rightarrow C_A = 0.1\ \text{mol/L}$. Feed is taken as $C_{A0} = 0.9$, $C_{R0} = 0.1\ \text{mol/L}$ (a trace of $R$ must be present to initiate an autocatalytic reaction).

ASSUMPTION (feed composition). The paper does not print the feed composition, only that $C_{A0}+C_{R0}=1$ mol/L. A feed of $C_{A0}=0.9$, $C_{R0}=0.1$ mol/L is adopted here. With a feed of 99% $A$ and 1% $R$ instead, the identical method gives $\tau_{PFR}=\ln 891=6.79$ min, $\tau_{MFR}=0.89/0.09=9.89$ min and a minimum combination (MFR to $C_A=0.5$, then PFR) of $1.96+2.20=4.16$ min. The ranking and the optimum arrangement are the same under either feed.

Find. The holding time $\tau = V/v_0$ for (a) a PFR, (b) an MFR, and (c) the minimum-total-volume series combination reaching $C_A = 0.1$.

Figure 2.1 — The $1/(-r_A)$ curve is U-shaped, with a minimum at $C_A = C_0/2 = 0.5$. The PFR time is the area under the curve; the MFR time is the shaded rectangle at the (high) exit ordinate; the minimum combination follows the lower envelope.

Approach. Build the Levenspiel $1/(-r_A)$–$C_A$ curve; a PFR integrates the area, a CSTR uses the exit-ordinate rectangle, and the optimum series arrangement follows whichever reactor lies lower at each concentration.

(a) Plug-flow reactor. The PFR holding time is $$\tau_{PFR} = \int_{C_A}^{C_{A0}}\frac{dC_A}{k\,C_A(C_0-C_A)} = \frac{1}{k\,C_0}\ln\!\frac{C_A}{C_0-C_A}\Big|_{0.1}^{0.9}.$$ Evaluating, $\ln\frac{0.9}{0.1} - \ln\frac{0.1}{0.9} = 2\ln 9$, so $$\boxed{\tau_{PFR} = 2\ln 9 = 4.39\ \text{min}.}$$
(b) Mixed-flow reactor. A CSTR operates at the exit rate $-r_A|_{exit} = k\,C_A(C_0-C_A) = (1)(0.1)(0.9) = 0.09\ \text{mol}\,\text{L}^{-1}\text{min}^{-1}$, so $$\tau_{MFR} = \frac{C_{A0}-C_A}{-r_A|_{exit}} = \frac{0.9-0.1}{0.09} = \boxed{8.89\ \text{min}.}$$ The CSTR is worse than the PFR here because it sits at the slow, low-$C_A$ tail rather than passing through the fast region near $C_A = 0.5$.
(c) Minimum-size combination. The rate is fastest at the curve minimum $C_A = C_0/2 = 0.5$. The optimum series with no recycle places a CSTR first, taking the feed from $C_{A0}=0.9$ down to $C_A = 0.5$ (operating at the maximum rate), followed by a PFR from $0.5$ to $0.1$: $$\tau_{MFR}^{(1)} = \frac{0.9-0.5}{(1)(0.5)(0.5)} = 1.60\ \text{min},\qquad \tau_{PFR}^{(2)} = \frac{1}{k C_0}\Big[\ln\tfrac{0.5}{0.5}-\ln\tfrac{0.1}{0.9}\Big] = \ln 9 = 2.20\ \text{min}.$$ The total is $$\boxed{\tau_{min} = 1.60 + 2.20 = 3.80\ \text{min},}$$ smaller than either single reactor.
Rank the three schemes. $\tau_{min}\,(3.80) < \tau_{PFR}\,(4.39) < \tau_{MFR}\,(8.89)$. The CSTR-then-PFR combination exploits the CSTR’s ability to jump the slow low-conversion start to the fast midpoint, then lets the PFR finish efficiently — the textbook optimum for an autocatalytic reaction.

Reactor scheme	Holding time (min)
(a) Plug flow	4.39
(b) Mixed flow	8.89
(c) Minimum combination (MFR to $C_A{=}0.5$, then PFR)	3.80

Concept & Theory

Topic. Autocatalytic kinetics and optimum reactor selection from the $1/(-r_A)$ curve.

Key relations. $-r_A = kC_A(C_0-C_A)$; PFR $\tau = \int dC_A/(-r_A)$; CSTR $\tau = (C_{A0}-C_A)/(-r_A|_{exit})$; rate maximum at $C_A = C_0/2$.

Why this works. Because the rate depends on the product of reactant and autocatalyst, it starts small (little $R$), peaks at $C_A = C_0/2$, and falls again near completion — giving a U-shaped $1/(-r_A)$ curve. A CSTR is efficient where the curve is high (it works at a single low ordinate), so it should cover the slow start up to the minimum; a PFR is efficient where the ordinate is small, so it finishes the job. This is the general rule: use a CSTR over rising-rate regions and a PFR over falling-rate regions.

Common pitfall. Assuming a pure PFR is always smallest — false for autocatalytic reactions; and forgetting that some $R$ must seed the feed, else the rate is identically zero and no reactor works.

Source. Levenspiel, Chemical Reaction Engineering, 3rd ed., Ch. 6 (design for single reactions; autocatalytic reactions).

Question 2 of 5: Holding Times for an Autocatalytic Reaction

Question 2: Holding Times for an Autocatalytic Reaction (25 marks: a 8, b 5, c 12)