CSTR + PFR in Series — Best Ordering for Maximum Conversion: 23-Chem-A4 Chemical Reactor Engineering, December 2015, Q3

Question 3: CSTR + PFR in Series — Best Ordering for Maximum Conversion (25 marks)

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. Second-order liquid-phase decomposition, $-r_A=kC_A^2$, $k=2\times10^{-3}$ L·mol⁻¹·s⁻¹; $C_{A0}=5$ mol/L; $v_0=0.02$ L/s; two 2-L reactors (one CSTR, one PFR). Constant density (liquid).

Quantity	Value
Rate constant, $k$	$2\times10^{-3}$ L·mol⁻¹·s⁻¹
$C_{A0}$ / $v_0$	5 mol/L / 0.02 L/s
Each reactor volume / space time $\tau=V/v_0$	2 L / 100 s
$k\tau$ (per L·mol⁻¹)	0.20

Find. The reactor ordering (CSTR-then-PFR vs PFR-then-CSTR) that maximizes overall conversion of A, and that maximum conversion.

Figure 3 — The conversion-maximizing arrangement: PFR first (operating at high $C_A$, where a second-order rate is fastest), CSTR second to mop up the remainder.

Approach. Apply each ideal-reactor design equation for a second-order rate in sequence for both orderings, propagating the exit concentration of the first unit into the second, and compare final conversions.

Design equations (second order, constant density). PFR: $\dfrac{1}{C_\text{out}}-\dfrac{1}{C_\text{in}}=k\tau$. CSTR: $\tau=\dfrac{C_\text{in}-C_\text{out}}{kC_\text{out}^2}$, i.e. $k\tau\,C_\text{out}^2+C_\text{out}-C_\text{in}=0$. Here $\tau=100$ s and $k\tau=0.20$ L/mol for each vessel.
Case A — CSTR then PFR. CSTR first: solve $0.20\,C^2+C-5=0\Rightarrow C_1=3.09$ mol/L. PFR second: $\dfrac{1}{C_2}=\dfrac{1}{3.09}+0.20\Rightarrow C_2=1.91$ mol/L, giving $X_A=\dfrac{5-1.91}{5}=0.618.$
Case B — PFR then CSTR. PFR first: $\dfrac{1}{C_1}=\dfrac{1}{5}+0.20\Rightarrow C_1=2.50$ mol/L. CSTR second: $0.20\,C_2^2+C_2-2.50=0\Rightarrow C_2=1.83$ mol/L, giving $X_A=\dfrac{5-1.83}{5}=0.634.$
Compare and conclude. $$X_{\text{PFR}\to\text{CSTR}}=0.634>X_{\text{CSTR}\to\text{PFR}}=0.618.$$ The maximum conversion is $$\boxed{X_{\max}=63.4\%\quad(\text{PFR first, then CSTR}).}$$

Arrangement	Exit $C_A$ (mol/L)	Conversion
CSTR → PFR	1.91	61.8%
PFR → CSTR (best)	1.83	63.4%

Concept & Theory

Topic. Sequencing of ideal reactors for reaction orders greater than one.

Key relations. Second-order PFR $1/C_\text{out}-1/C_\text{in}=k\tau$; second-order CSTR $k\tau C_\text{out}^2+C_\text{out}-C_\text{in}=0$; conversion $X=(C_{A0}-C_\text{out})/C_{A0}$.

Why this works. For $n>1$ the rate falls steeply as $C_A$ drops. A PFR keeps concentration high along its length, so it does its best work at the reactor inlet; placing it first — where $C_A$ is largest — extracts the most conversion per unit volume, leaving the CSTR (which always runs at its low exit concentration) to finish. Reversing the order forces the PFR to work on an already-diluted stream and loses about 1.6 conversion points.

Common pitfall. Assuming order never matters because the two volumes are equal — true only for first-order kinetics, where series arrangements are order-independent. For $n>1$ put the PFR (or the smaller CSTR, in a CSTR train) first.

Source. Levenspiel, Chemical Reaction Engineering §6 (reactor combinations, best arrangement for $n>1$); Fogler §2 (reactors in series).

Question 3 of 5: CSTR + PFR in Series — Best Ordering for Maximum Conversion

Question 3: CSTR + PFR in Series — Best Ordering for Maximum Conversion (25 marks)