20-Bio-A5 Systems Analysis & Control · Undated paper
Question 3 of 6: External-Mass-Transfer-Limited Immobilized-Enzyme Plate Reactor
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams / EGBC — May 2019 — 04-Bio-A5 Enzyme and Microbial Kinetics. Three-hour open-book examination; any non-communicating calculator is permitted. Six questions constitute a complete paper (10 + 15 + 20 + 20 + 15 + 20 = 100 marks). Content spans a rapid-equilibrium mixed-inhibition mechanism, Lineweaver–Burk inhibitor-type determination from initial-rate data, external-mass-transfer-limited immobilized-enzyme kinetics on a flat plate, aerobic-growth stoichiometry with an elemental-balance yield calculation, logistic growth-associated product-formation kinetics in a batch fermenter, and continuous tubular-sterilizer design (vitamin retention and probability of successful sterilization).
Reference texts: Bailey & Ollis, Biochemical Engineering Fundamentals (2nd ed.) — enzyme inhibition mechanisms and kinetics (Ch. 3), immobilized-enzyme diffusion-reaction (Ch. 8), stoichiometry and yields (Ch. 6), unstructured growth/product kinetics (Ch. 7), and batch sterilization design (Ch. 9). All quantities are used exactly as printed on the exam.
Check — four points were resolved before authoring: (1) the printed marking scheme lists six separate questions (10, 15, 15+5, 15+5, 15, 10+10 marks) — Questions 1 and 2 are separate questions, treated as such here to match the marking scheme exactly. (2) Question 1 asks for the “rapid equilibrium approach,” not the QSSA. (3) Question 2’s data table has four rows, not three — including the row I = 0, S = 0.02, V = 0.9 mmol/(mL·min), which is essential (it is the only other no-inhibitor point). (4) Question 6’s printed sub-parts really are only “b)” and “c)” — part “a)” is genuinely absent from the paper; the marking scheme’s (a) 10 marks / (b) 10 marks are mapped onto the two printed sub-parts b) and c) respectively.
Given. Well-mixed batch liquid of volume $V_L$; enzyme immobilized as a monolayer on plates of total area $A_T$; external film mass-transfer coefficient $k_L$; pseudo-first-order surface reaction $(-r_s')=V_m'S_WA_T$.
Find. (a) the two governing differential/flux-balance equations linking $S_L(t)$ (bulk), $S_W$ (surface) and the two rate processes in series; (b) the closed-form $S_L(t)$.
Fig. 3 — two resistances in series: external-film mass transfer ($k_L$, bulk $S_L$ to surface $S_W$) feeding a pseudo-first-order surface reaction ($V_m'$, per unit area $A_T$).
Approach. Write the unsteady bulk-phase substrate balance driven by the film flux, and the pseudo-steady surface balance (film flux = surface reaction rate, since the film itself holds negligible substrate inventory); combine the two resistances in series into one overall first-order rate constant to get $S_L(t)$.
(a) Bulk-phase unsteady balance. The well-mixed bulk liquid loses substrate only by transfer across the film to the plate surface: $$V_L\frac{dS_L}{dt}=-k_LA_T\big(S_L-S_W\big).$$
(a) Surface (film) balance. The film itself stores negligible substrate, so at every instant the flux arriving at the surface equals the rate consumed by the immobilized enzyme (pseudo-steady-state at the surface): $$k_LA_T\big(S_L-S_W\big)=V_m'S_WA_T\ \ \Rightarrow\ \ k_L\big(S_L-S_W\big)=V_m'S_W.$$
Eliminate $S_W$. Solving Step 2 for the surface concentration: $$S_W=\frac{k_L}{k_L+V_m'}S_L.$$
(b) Combine resistances into an overall rate constant. Substituting $S_W$ back into Step 1: $$V_L\frac{dS_L}{dt}=-k_LA_T\left(S_L-\frac{k_L}{k_L+V_m'}S_L\right)=-A_T\underbrace{\left(\frac{k_LV_m'}{k_L+V_m'}\right)}_{k_{\text{eff}}}S_L,$$ i.e. film resistance and surface-reaction resistance add like resistors in series, $1/k_{\text{eff}}=1/k_L+1/V_m'$.