Rate that Passes Through a Maximum — Choosing CSTR vs PFR: 23-Chem-A4 Chemical Reactor Engineering, December 2016, Q1

Question 1: Rate that Passes Through a Maximum — Choosing CSTR vs PFR (25 marks: a 5, b 12, c 8)

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. Because $B$ is in large excess, $C_B \approx C_{B0}$ is constant and the rate collapses to a function of $C_A$ alone, $-r_A = \dfrac{k' C_A}{(1 + K_A C_A)^2}$ with lumped constant $k' = k\,C_{B0}$. The design comparison uses the exit-based mixed-flow rule against the plug-flow area on a Levenspiel plot.

Find. (a) the concentration at which $-r_A$ peaks; (b)–(c) which single ideal reactor gives the smaller volume for two different exit concentrations, with reasoning from the $1/(-r_A)$ curve.

Approach. Differentiate the lumped rate to locate the maximum, then plot $1/(-r_A)$ and apply the Levenspiel-plot rule — a CSTR volume is the rectangle $[X_{\text{exit}}]\times[1/(-r_A)]_{\text{exit}}$ while a PFR volume is the area under the curve; whichever is smaller wins.

Part (a) — location of the maximum

Differentiate the lumped rate. With $-r_A = k' C_A (1+K_A C_A)^{-2}$, the quotient/product rule gives $$\frac{d(-r_A)}{dC_A} = k'\,\frac{(1+K_A C_A) - 2 K_A C_A}{(1+K_A C_A)^3} = k'\,\frac{1 - K_A C_A}{(1+K_A C_A)^3}.$$
Set the derivative to zero. The denominator is always positive, so the numerator vanishes when $1 - K_A C_A = 0$, i.e. $$\boxed{\,C_{A,\text{max}} = \dfrac{1}{K_A}\,}.$$ The second derivative is negative here, confirming a maximum rather than a minimum. Physically the rate first rises with $C_A$ (more reactant) then falls as the $(1+K_A C_A)^2$ adsorption/inhibition term in the denominator takes over — classic Langmuir–Hinshelwood behaviour.

Part (b) — feed $1.5/K_A$, exit $0.5/K_A$

Levenspiel plot $1/(-r_A)$ vs $C_A$ (in units of $K_A/k'$). The curve dips to a minimum at $C_A = 1/K_A$ (where the rate peaks). Reading right→left is the direction of increasing conversion. For part (b) the reactor operates from $1.5/K_A$ down to $0.5/K_A$; the exit sits at the highest point of $1/(-r_A)$ on that interval.

Tabulate $1/(-r_A)$ at the key concentrations. Writing $u = K_A C_A$, $1/(-r_A) = \dfrac{K_A}{k'}\dfrac{(1+u)^2}{u}$. In units of $K_A/k'$: at the feed $u=1.5\Rightarrow 4.167$; at the rate peak $u=1\Rightarrow 4.000$ (the minimum of the curve); at the exit $u=0.5\Rightarrow 4.500$.
Apply the Levenspiel rule at the exit. The operating window $[0.5/K_A,\,1.5/K_A]$ straddles the peak, so the reaction first speeds up then slows down as $A$ is consumed. The exit value $1/(-r_A)=4.5$ is the largest on the interval. A CSTR must run entirely at that worst (exit) rate — its volume is the tall rectangle of height $4.5$ — whereas a PFR only pays the exit rate at the very end and enjoys the lower values (down to $4.0$) throughout the rest of the reactor. $$\boxed{\text{Use a PFR}}$$ Because the area under the curve is smaller than the exit rectangle, the PFR needs less volume.

Part (c) — feed $1.5/K_A$, exit $1/K_A$

Re-read the curve over the narrower window. Now $C_A$ runs from $1.5/K_A$ only down to $1/K_A$. Over $[1/K_A,\,1.5/K_A]$ the rate is monotonically increasing as $A$ is consumed (we approach the peak from the right), so $1/(-r_A)$ falls steadily from $4.167$ to its minimum $4.000$ at the exit.
Apply the Levenspiel rule again. Here the exit $1/(-r_A)=4.0$ is the smallest value on the interval. A CSTR runs the whole reactor at that best (lowest) exit rate — the rectangle of height $4.0$ — which is smaller than the area under the (higher) curve that a PFR integrates. $$\boxed{\text{Use a CSTR — the answer changes}}$$ Whenever the exit lies on a falling stretch of $1/(-r_A)$, mixed flow wins; whenever the exit lies on a rising stretch, plug flow wins.

Quantity	Result
(a) Concentration at maximum rate	$C_A = 1/K_A$
(b) Best single reactor, exit $0.5/K_A$	PFR (exit is the peak of $1/(-r_A)$)
(c) Best single reactor, exit $1/K_A$	CSTR (exit is the trough of $1/(-r_A)$)

Concept & Theory

Topic. Reactor selection from the shape of the $1/(-r_A)$–$C_A$ (Levenspiel) curve when the rate is non-monotonic.

Key relations. PFR: $\tau = C_{A0}\!\int_{X}\! dX/(-r_A)$ = area under $1/(-r_A)$; CSTR: $\tau = C_{A0} X/(-r_A)|_{\text{exit}}$ = rectangle to the exit point (Levenspiel, Ch. 5; Fogler, Ch. 2). Rate maximum from $d(-r_A)/dC_A = 0$.

Why this works. A CSTR operates uniformly at exit conditions, so its "cost" per unit conversion is set entirely by the exit rate. A PFR sweeps through every intermediate concentration, so it is charged the average of $1/(-r_A)$ over the interval. Comparing a rectangle (CSTR) with the area under the curve (PFR) instantly tells you which is smaller — no integration needed when you only must pick one.

Common pitfall. Assuming a PFR is "always better" for positive-order kinetics. That is only true when $1/(-r_A)$ rises toward the exit. Here the rate maximises inside the window, so the winner flips depending on where the exit falls relative to the peak.

Source. Levenspiel, Chemical Reaction Engineering, 3rd ed., Ch. 5–6; Fogler, Elements of Chemical Reaction Engineering, 5th ed., Ch. 2.

Question 1 of 5: Rate that Passes Through a Maximum — Choosing CSTR vs PFR

Question 1: Rate that Passes Through a Maximum — Choosing CSTR vs PFR (25 marks: a 5, b 12, c 8)

Part (a) — location of the maximum

Part (b) — feed $1.5/K_A$, exit $0.5/K_A$

Part (c) — feed $1.5/K_A$, exit $1/K_A$