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23-Chem-A4 Chemical Reactor Engineering · December 2014

Question 4 of 5: Parallel Reactions — Reactor Choice for Selectivity

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams / EGBC — December 2014 — 04-Chem-A4 Chemical Reactor Engineering. Three-hour open-book exam; the designated Fogler textbook (any edition), unit-conversion / mathematical tables and a non-communicating programmable calculator are permitted. Five questions are printed and any four constitute a complete paper (each worth 20 marks); all five are solved below for completeness. No credit is given for re-deriving standard rate expressions, so the batch / PFR / CSTR design equations are quoted and applied. These are Levenspiel-style graphical / analytical reactor problems (enzyme kinetics, rate-data analysis, autocatalytic staging, parallel-reaction selectivity, and CSTR kinetic determination).

Reference texts: O. Levenspiel, Chemical Reaction Engineering (3rd ed., Wiley) — batch/MFR/PFR design equations, the graphical “$1/(-r_A)$ vs $C_A$” area method, autocatalytic optimum staging, product distribution for parallel reactions, and constant-volume CSTR kinetic analysis; H. S. Fogler, Elements of Chemical Reaction Engineering (4th/5th ed., Prentice Hall) — the designated open-book text: stoichiometric tables with the expansion factor $\varepsilon$, gas-phase reactions with change in moles, and reactor sizing; supporting property data from Perry’s Chemical Engineers’ Handbook (9th ed.).

Question 4: Parallel Reactions — Reactor Choice for Selectivity (20 marks: a 8, 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. Three simultaneous parallel decompositions of A with orders 0.5, 1 and 2. Feed $C_{A0}=10$ kmol/m³, $v_0=1$ m³/min. Total consumption rate $-r_A=r_R+r_S+r_T=16C_A^{0.5}+12C_A+C_A^{2}$.

PathRate law (kmol/m³·min)OrderFavoured by
A → R$16C_A^{0.5}$0.5low $C_A$
A → S$12C_A$1—
A → T$C_A^{2}$2high $C_A$

Find. (a) $C_R$ and the CSTR volume when R is the target; (b) $C_T$ and the PFR volume when T is the target. Each unit is designed for 99% conversion of A (exit $C_A=0.1$ kmol/m³), the practical near-complete-conversion point.

0.021.683.355.016.678.341000.1670.3330.50.6670.8331phi_R (order 1/2)phi_S (order 1)phi_T (order 2)concentration C_A (kmol/m3)instantaneous fractional yield phiSelectivity vs C_A: R wins at low C_A, T wins at high C_A
Figure 4 — Instantaneous fractional yields $\varphi_i=r_i/(-r_A)$. The lowest-order product R dominates at low $C_A$ (favouring a CSTR run to high conversion); the highest-order product T dominates at high $C_A$ (favouring a PFR that keeps $C_A$ high). S is never the majority product.

Approach. The instantaneous fractional yield $\varphi_i=r_i/(-r_A)$ sets the selectivity; since R is lowest-order and T highest-order, run the low-$C_A$-favouring reaction in a CSTR and the high-$C_A$-favouring reaction in a PFR, then evaluate the mixing rule for $C_i$ and size from $-r_A$.

  1. Selectivity ranking. $\varphi_R=\dfrac{16C_A^{0.5}}{16C_A^{0.5}+12C_A+C_A^{2}}$ rises as $C_A$ falls (R is order ½), whereas $\varphi_T=\dfrac{C_A^{2}}{-r_A}$ rises as $C_A$ increases (T is order 2). Hence a CSTR — which holds the stream at the low exit $C_A$ — maximises R, and a PFR — which keeps $C_A$ high over most of its length — maximises T (Figure 4).
  2. R in a CSTR (part a). A CSTR mixes to the exit composition, so its overall yield equals the instantaneous yield there. At $C_A=0.1$: $r_R=16(0.1)^{0.5}=5.06$, $-r_A=5.06+1.20+0.01=6.27$, giving $\varphi_R=0.807$. Thus$$C_R=\varphi_R\,(C_{A0}-C_A)=0.807\,(9.9)=\boxed{7.99\ \text{kmol/m}^3}.$$The volume follows from the CSTR mole balance $V=\dfrac{v_0(C_{A0}-C_A)}{-r_A}=\dfrac{1(9.9)}{6.27}=\boxed{1.58\ \text{m}^3}$.
  3. T in a PFR (part b). A PFR passes through every concentration, so $C_T=\displaystyle\int_{C_A}^{C_{A0}}\varphi_T\,dC_A=\int_{0.1}^{10}\frac{C_A^{2}}{16C_A^{0.5}+12C_A+C_A^{2}}\,dC_A$. Numerical integration gives$$\boxed{C_T=1.97\ \text{kmol/m}^3}.$$
  4. PFR volume (part b). $$V=v_0\displaystyle\int_{C_A}^{C_{A0}}\frac{dC_A}{-r_A}=1\int_{0.1}^{10}\frac{dC_A}{16C_A^{0.5}+12C_A+C_A^{2}}=\boxed{0.148\ \text{m}^3}.$$The PFR is far smaller than the CSTR because the fast high-$C_A$ region (large $-r_A$) is swept quickly, whereas the CSTR runs its whole volume at the slow exit rate.

Check: The question fixes the reactor type but not the target conversion; both units are designed here for 99% conversion of A ($C_A=0.1$ kmol/m³) as the practical near-complete-conversion point. Since $\varphi_R$ keeps rising as $C_A\to0$, pushing beyond 99% yields only marginally more R for a rapidly growing CSTR volume (the CSTR volume $\propto 1/\sqrt{C_A}$ becomes infinite at complete conversion). The PFR behaves differently: $1/(-r_A)\approx1/(16\sqrt{C_A})$ is integrable at $C_A\to0$, so a PFR taken to complete conversion has a finite volume, $V=0.148+0.035=0.183$ m³, while $C_T$ is unchanged at 1.97 kmol/m³ (the last 1% of A forms almost no T). Either design basis is acceptable if stated; the conclusion — PFR for T, $C_T\approx1.97$ kmol/m³ — does not depend on it.

Target / reactorProduct conc.Volume
(a) R in CSTR$C_R=7.99$ kmol/m³1.58 m³
(b) T in PFR$C_T=1.97$ kmol/m³0.148 m³