23-Chem-A4 Chemical Reactor Engineering · December 2019
Question 1 of 5: Sizing a PFR and an MFR from Batch Kinetic Data
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. Chemical Engineering — 23-Chem-A4 Chemical Reactor Engineering, December 2019. Open-book, 3 hours. Five questions; answering any four constitutes a complete paper (each worth 25 marks). All five are solved below.
Reference texts. O. Levenspiel, Chemical Reaction Engineering, 3rd ed. (Wiley, 1999); H. S. Fogler, Elements of Chemical Reaction Engineering, 5th ed. (Pearson, 2016); Perry's Chemical Engineers' Handbook, 9th ed. Gas constant $R = 0.082057\ \text{L}\,\text{atm}\,\text{mol}^{-1}\text{K}^{-1} = 8.314\ \text{J}\,\text{mol}^{-1}\text{K}^{-1}$.
Question 1: Sizing a PFR and an MFR from Batch Kinetic Data (25 marks: a 15, b 10)
Given. Constant-volume batch decomposition $2A \rightarrow R + S$ at $T = 373.15$ K; the partial-pressure history of $A$ above; a continuous feed of $A$ with 20% inerts ($y_{A0} = 0.80$) at $P = 1$ atm, $100\,{}^\circ$C; required conversion $X = 0.95$.
Find. The reaction order and rate constant from the batch data, then the volumes of a PFR (a) and an MFR (b) delivering $X = 0.95$.
Check (feed basis). The printed phrase “treat 120 moles of $A$ per litre” is dimensionally inconsistent with a reactor-sizing question (a concentration cannot be “treated”). It is read here as a molar feed rate $F_{A0} = 120\ \text{mol/min}$, the only interpretation that yields a finite reactor volume. Reactor volume scales linearly with $F_{A0}$, so a different feed rate rescales both answers by the same factor; if the intended basis were 120 mol/h, both volumes fall by a factor of 60, to $V_{PFR} \approx 417$ L and $V_{MFR} \approx 2.64\times10^{3}$ L.
Figure 1.1 — The $\ln p_A$–$t$ data fall on a straight line, the signature of a first-order reaction; a second-order test ($1/p_A$) would curve upward.
Approach. Recognise the reaction is mole-conserving so $p_A \propto C_A$ at constant volume; diagnose the order by which linear plot is straight; then apply the constant-density design equations ($\varepsilon_A = 0$) for a PFR and an MFR.
Establish that $p_A$ tracks concentration. Because $2A \rightarrow R + S$ converts 2 mol of gas into 2 mol of gas, the total mole count is constant; at fixed $V$ and $T$ the total pressure never changes and the expansion factor is $\varepsilon_A = 0$. Hence $C_A = p_A/RT$ and the tabulated $p_A$ is a direct proxy for $C_A$.
Diagnose the reaction order. Testing a first-order model means checking whether $\ln p_A$ is linear in $t$. Regressing the eleven points, $$\ln p_A = -k\,t, \qquad k = 9.17\times10^{-3}\ \text{s}^{-1}\ (R^2 = 0.996).$$ The plot (Fig. 1.1) is straight, whereas $1/p_A$ curves — so the decomposition is first order. In per-minute units $$\boxed{k = 0.550\ \text{min}^{-1}.}$$
Feed concentration and volumetric flow. At inlet conditions $$C_{A0} = \frac{y_{A0}P}{RT} = \frac{(0.80)(1)}{(0.082057)(373.15)} = 0.02613\ \text{mol/L},$$ so the volumetric feed is $v_0 = F_{A0}/C_{A0} = 120/0.02613 = 4.59\times10^{3}\ \text{L/min}.$
(a) Plug-flow reactor. For a first-order reaction with no volume change, the design equation is $k\tau = -\ln(1-X)$. Thus $$\tau_{PFR} = \frac{-\ln(1-0.95)}{0.550} = \frac{2.996}{0.550} = 5.44\ \text{min},$$ and $$\boxed{V_{PFR} = v_0\,\tau_{PFR} = (4593)(5.44) = 2.50\times10^{4}\ \text{L} \approx 25\ \text{m}^3.}$$
(b) Mixed-flow reactor. For a first-order reaction in a CSTR, $k\tau = X/(1-X)$, so $$\tau_{MFR} = \frac{0.95}{(0.550)(0.05)} = 34.5\ \text{min},$$ giving $$\boxed{V_{MFR} = v_0\,\tau_{MFR} = (4593)(34.5) = 1.59\times10^{5}\ \text{L} \approx 159\ \text{m}^3.}$$
Compare. The CSTR needs $V_{MFR}/V_{PFR} = 34.5/5.44 = 6.3$ times the plug-flow volume — a first-order reaction pushed to 95% conversion penalises back-mixing heavily, because a CSTR runs entirely at the low exit concentration where the rate is slowest.
