CSTRs in Series vs. a Single CSTR — First-Order Liquid Reaction: 23-Chem-A4 Chemical Reactor Engineering, May 2013, Q5

Question 5: CSTRs in Series vs. a Single CSTR — First-Order Liquid Reaction (20 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. Isothermal first-order liquid reaction, $k=0.020$ min$^{-1}$, constant density. Design (a): two 100 L CSTRs in series, feed 0.7884 L/min. Design (b): one 200 L CSTR at the same conversion.

Quantity	Value
Rate constant $k$	0.020 min$^{-1}$
Design (a) vessels	two × 100 L in series
Design (a) feed rate	0.7884 L/min
Design (b) vessel	single 200 L

Find. (a) conversion leaving the second CSTR; (b) feed rate for a single 200 L CSTR at the same conversion; (c) which design converts more A.

Figure 6 — Design (a): two identical 100 L CSTRs in series for a first-order liquid reaction ($k=0.020$ min⁻¹).

Approach. Use the tanks-in-series formula for first-order CSTRs to get $X_2$, invert the single-CSTR relation for the equal-conversion feed rate, then compare throughput $v\,C_{A0}\,X$.

(a) Space time and conversion for two equal CSTRs in series. Each tank has space time $\tau=V/v=100/0.7884=126.8$ min, so $k\tau=(0.020)(126.8)=2.54$. For $N$ identical first-order CSTRs in series, $$\frac{C_{A,N}}{C_{A0}}=\frac{1}{(1+k\tau)^N}\;\Rightarrow\; \frac{C_{A2}}{C_{A0}}=\frac{1}{(1+2.54)^2}=\frac{1}{12.5}=0.0799,$$ so the conversion leaving the second CSTR is $$X_2 = 1-0.0799=\boxed{0.920\;(92.0\%)}.$$
(b) Single 200-L CSTR for the same conversion. One CSTR gives $C_A/C_{A0}=1/(1+k\tau_s)$, hence $1+k\tau_s=1/(1-X)=1/0.0799=12.5$, so $k\tau_s=11.51$ and $\tau_s=11.509/0.020=575.4$ min. With $V=200$ L the required feed rate is $$v_b=\frac{V}{\tau_s}=\frac{200}{575.4}=\boxed{0.348\ \text{L/min}}.$$
(c) Which design converts more A? The molar rate of A converted is $\dot n_{conv}=v\,C_{A0}\,X$. Both designs reach the same conversion (0.920) from the same feed concentration, so the comparison reduces to the feed rate: $$\frac{(\dot n_{conv})_a}{(\dot n_{conv})_b}=\frac{v_a}{v_b}=\frac{0.7884}{0.348}=2.27.$$ Design (a), the two CSTRs in series, processes about $\boxed{2.3\times}$ the throughput at the same conversion, so it converts the larger quantity of A. Two tanks in series approximate plug flow and are more volume-efficient than a single stirred tank of the same total volume.

Quantity	Result
(a) Conversion, two CSTRs in series	$X_2 = 0.920$ (92.0%)
(b) Feed rate, single 200 L CSTR	$v_b = 0.348$ L/min
(c) Larger quantity converted	Design (a) — $\sim2.3\times$ the throughput

Concept & Theory

Topic. Tanks-in-series model; comparison of CSTR configurations; throughput vs. fractional conversion.

Key relations. Space time $\tau=V/v$; $N$ equal first-order CSTRs $C_{A,N}/C_{A0}=(1+k\tau)^{-N}$; single CSTR $C_A/C_{A0}=(1+k\tau)^{-1}$; molar conversion rate $\dot n_{conv}=v\,C_{A0}\,X$.

Why this works. Splitting a fixed total volume into several CSTRs in series keeps each tank at a higher average reactant concentration than one large well-mixed tank, so the series train behaves more like a plug-flow reactor and is more volume-efficient. At a fixed conversion, that efficiency shows up as a higher tolerable feed rate, hence a larger quantity of A converted per unit time.

Common pitfall. Confusing ‘larger conversion’ with ‘larger quantity converted’: here both designs have the same fractional conversion, so the throughput (feed rate) decides. Also, each tank’s $\tau$ uses its own volume, not the total.

Source. Fogler, Elements of Chemical Reaction Engineering, Ch. 2 & Ch. 5 (CSTRs in series); Levenspiel, Chemical Reaction Engineering, Ch. 6 (comparison of single and multiple reactors).

Question 5 of 5: CSTRs in Series vs. a Single CSTR — First-Order Liquid Reaction

Question 5: CSTRs in Series vs. a Single CSTR — First-Order Liquid Reaction (20 marks)