Batch Esterification — Ethyl Acetate Reactor Sizing: 23-Chem-A4 Chemical Reactor Engineering, May 2015, Q1

Question 1: Batch Esterification — Ethyl Acetate Reactor Sizing (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. Reversible liquid-phase esterification $A+B\rightleftharpoons C+D$ in a constant-density isothermal batch reactor at 100 °C. Charge composition per m³: ethanol 500, acetic acid 250, water 295 kg (density 1045 kg/m³). Acetic acid (A) is the limiting reactant, discharged at $X_A=0.30$; turnaround time 30 min; target 10,000 kg/day ethyl acetate.

Quantity	Value
$C_{A0}$ acetic acid $=250/60.05$	4.163 kmol/m³
$C_{B0}$ ethanol $=500/46.07$	10.853 kmol/m³
$C_{D0}$ water $=295/18.015$	16.375 kmol/m³
$\theta_B=C_{B0}/C_{A0}$ / $\theta_D=C_{D0}/C_{A0}$	2.607 / 3.933
$k_f$ / $k_r$ (m³/kmol·s)	$8.0\times10^{-6}$ / $2.7\times10^{-6}$
Turnaround / production target	30 min / 10,000 kg·day⁻¹

Find. The batch-reactor volume $V$ (m³) needed to make 10,000 kg/day of ethyl acetate, allowing for reaction time plus the 30-min turnaround per cycle.

Figure 1 — Isothermal batch reactor charged with the ethanol / acetic-acid / water mixture; it is filled, reacted to $X_A=0.30$, then discharged, cleaned and recharged (30-min turnaround) before the next batch.

Approach. Write the constant-volume batch design equation for the reversible second-order rate in terms of the single conversion $X$, integrate numerically to the discharge conversion, add the turnaround time to get the cycle time, then size the reactor from the required daily output.

Express all concentrations through the limiting-reactant conversion. With A limiting, $C_A=C_{A0}(1-X)$, $C_B=C_{A0}(\theta_B-X)$, $C_C=C_{A0}X$, $C_D=C_{A0}(\theta_D+X)$. Substituting into $-r_A=k_fC_AC_B-k_rC_CC_D$ gives $-r_A=C_{A0}^2\big[k_f(1-X)(\theta_B-X)-k_rX(\theta_D+X)\big].$
Batch design equation. For constant volume, $t=C_{A0}\displaystyle\int_0^{X}\frac{dX}{-r_A}=\frac{1}{C_{A0}}\int_0^{0.30}\frac{dX}{k_f(1-X)(\theta_B-X)-k_rX(\theta_D+X)}.$ The integrand rises modestly toward the (distant) equilibrium conversion, so Simpson quadrature is accurate.
Reaction time. Evaluating the integral numerically and dividing by $C_{A0}=4.163$ kmol/m³ gives $t_\text{rxn}=5.02\times10^{3}\ \text{s}=83.7\ \text{min}.$ Adding the 30-min turnaround, the cycle time is$$t_\text{cycle}=83.7+30=\boxed{113.7\ \text{min}}.$$
Ester made per batch and batches per day. Ester produced per m³ of charge is $C_{A0}X\,M_C=4.163(0.30)(88.11)=110.0$ kg/m³. The number of batches per day is $N=1440/113.7=12.67.$
Size the reactor. The daily output is $N\,V\,(C_{A0}XM_C)=10{,}000$ kg/day, so$$V=\frac{10{,}000}{12.67\times110.0}=\boxed{7.2\ \text{m}^3}.$$A reactor of about 7.2 m³ working volume (allowing freeboard, ≈8 m³ vessel) meets the production target.

Quantity	Result
Reaction time to $X_A=0.30$	83.7 min
Cycle time (with turnaround)	113.7 min (12.67 batches/day)
Ester per m³ of charge	110.0 kg/m³
Required reactor volume	≈ 7.2 m³

Concept & Theory

Topic. Constant-volume batch reactor sizing for a reversible second-order liquid-phase reaction, including turnaround (dead) time.

Key relations. Batch design equation $t=C_{A0}\int_0^X dX/(-r_A)$; reversible rate $-r_A=k_fC_AC_B-k_rC_CC_D$; cycle time $t_\text{cycle}=t_\text{rxn}+t_\text{down}$; output $=N\,V\,C_{A0}XM_C$.

Why this works. With constant density the volume cancels from the rate law, so the design equation is a single integral in $X$. Expressing every species through $X$ and $\theta_i$ collapses the reversible rate to one function of $X$; numerical integration then gives the reaction time directly. The turnaround time is added because it consumes the same reactor-day budget as reaction.

Common pitfall. Forgetting the water already present in the charge ($C_{D0}\ne0$) — it feeds the reverse reaction and lengthens the required time; or omitting the 30-min turnaround, which would undersize the reactor by ~25%.

Source. Fogler, Elements of Chemical Reaction Engineering §5 (isothermal batch design, reversible reactions); Levenspiel, Chemical Reaction Engineering §3–5 (batch reactor, reversible kinetics).

Question 1 of 5: Batch Esterification — Ethyl Acetate Reactor Sizing

Question 1: Batch Esterification — Ethyl Acetate Reactor Sizing (20 marks)