Batch Reactor Sizing and Two-Stage CSTR Conversion: 23-Chem-A4 Chemical Reactor Engineering, May 2018, Q5

Question 5: Batch Reactor Sizing and Two-Stage CSTR Conversion (25 marks: a 13, b 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. Liquid-phase reaction of order 1.7; a batch route (with turnaround downtime) and a two-CSTR-in-series alternative are compared at the same throughput.

Quantity	Value
Inlet concentration $C_{A0}$	0.5 lb·mol/ft$^3$
Throughput	25 ft$^3$/hr
Rate law	$-r_A=2.33\,C_A^{1.7}$
(a) Downtime	45 min = 0.75 hr per batch
(b) CSTR volume (each)	50 ft$^3$ $\times$ 2

Find. (a) The batch reactor volume for 90% conversion; (b) the conversion at the exit of each of the two CSTRs.

Figure 5 — The continuous alternative of part (b): two equal 50-ft$^3$ CSTRs in series. Each tank operates at its own (low) exit concentration, so staging recovers much of the efficiency a single mixed tank would lose for this $>$1-order reaction.

Approach. (a) Integrate the batch rate to a reaction time, add downtime, and multiply the cycle time by throughput. (b) Solve each CSTR design equation for the exit concentration and convert to conversion.

(a) Batch reaction time. Separating variables, $2.33\,t=\dfrac{1}{0.7}\big(C_A^{-0.7}-C_{A0}^{-0.7}\big).$ With $C_{A0}=0.5$ and $C_A=0.05$ (90% conversion), $2.33\,t=\dfrac{1}{0.7}(8.14-1.62)=9.31,$ so $t_{rxn}=4.00\ \text{hr}.$
(a) Cycle time and batch volume. Each cycle takes $t_{cycle}=4.00+0.75=4.75\ \text{hr}.$ To process 25 ft$^3$/hr, one batch must hold $$V_{batch}=25\times4.75=\boxed{119\ \text{ft}^3.}$$
(b) Residence time per tank. $\tau=\dfrac{V}{v_0}=\dfrac{50}{25}=2\ \text{hr}$ in each CSTR.
(b) Stage 1. The CSTR design equation $\tau=\dfrac{C_{A0}-C_{A1}}{2.33\,C_{A1}^{1.7}}$ (rate at the exit concentration) solves to $C_{A1}=0.199\ \text{lb}\cdot\text{mol/ft}^3,$ so $$\boxed{X_1=1-\tfrac{0.199}{0.5}=0.601\ (60.1\%).}$$
(b) Stage 2. Feeding $C_{A1}$ to the second tank, $\tau=\dfrac{C_{A1}-C_{A2}}{2.33\,C_{A2}^{1.7}}$ solves to $C_{A2}=0.103,$ giving $$\boxed{X_2=1-\tfrac{0.103}{0.5}=0.795\ (79.5\%).}$$

Quantity	Result
(a) Batch reaction time	4.00 hr
(a) Batch reactor volume	119 ft$^3$
(b) Conversion after stage 1	60.1% ($C_A=0.199$)
(b) Conversion after stage 2	79.5% ($C_A=0.103$)

Concept & Theory

Topic: Batch reactor sizing with turnaround time, and staged CSTRs for a greater-than-first-order reaction.

Key relations: batch integral $t_{rxn}=\dfrac{1}{2.33}\dfrac{C_A^{-0.7}-C_{A0}^{-0.7}}{0.7}$; effective throughput $V_{batch}=v_0(t_{rxn}+t_{down})$; CSTR design $\tau=\dfrac{C_{in}-C_{out}}{-r_A(C_{out})}$ (Levenspiel Ch. 3, 5–6).

Why this works: A batch reactor is idle during charging, discharging and cleaning, so the volume that meets a continuous demand must cover the whole cycle, not just the reaction. For a positive-order reaction a single CSTR (held at the low exit concentration) is inefficient; splitting into stages keeps the first tank at a higher concentration and therefore a faster rate, so the pair reaches higher conversion than one equal-total-volume mixer.

Common pitfall: Omitting downtime when sizing the batch (under-sizes by $\sim16\%$ here); and evaluating the CSTR rate at the inlet rather than the exit concentration — a CSTR reacts everywhere at its outlet condition.

Source: Levenspiel, Ch. 3 (batch), Ch. 5–6 (single and multiple ideal reactors); Fogler, Ch. 2, 5.

Question 5 of 5: Batch Reactor Sizing and Two-Stage CSTR Conversion

Question 5: Batch Reactor Sizing and Two-Stage CSTR Conversion (25 marks: a 13, b 12)