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20-Bio-A5 Systems Analysis & Control · May 2017

Question 6 of 6: Fed-Batch Reactor — Volume and Pseudo-Steady-State Substrate Expressions

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

Notes on this paper

National Exams / EGBC — May 2017 — 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 + 25 + 15 + 20 + 15 + 15 = 100 marks). Content spans quasi-steady-state derivation of a substrate-inhibited enzyme mechanism, diffusion-reaction in an immobilized-enzyme bead (Thiele modulus / effectiveness factor), the rapid-equilibrium integrated batch equation for substrate inhibition, a growth-associated-product batch fermentation, Contois-kinetics chemostat design, and fed-batch quasi-steady-state theory.

Reference texts: Bailey & Ollis, Biochemical Engineering Fundamentals (2nd ed.) — enzyme kinetics, substrate inhibition, immobilized-enzyme diffusion-reaction and effectiveness factors, Monod/Contois growth models, chemostat and fed-batch theory. All quantities are used exactly as printed on the exam (SI/CGS mixed units as given).

Question 6: Fed-Batch Reactor — Volume and Pseudo-Steady-State Substrate Expressions (a. 10 marks; b. 5 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. $V(0)=V_0$; feed rate $F(t)=F_0+\alpha t$ with feed substrate concentration $S_f$; growth kinetics $\mu_g=kS$; no cell death; true growth yield $Y_{X/S}^M$; no outflow (fed-batch).

Find. (a) $V(t)$; (b) $S(t)$ under the pseudo-steady-state (quasi-steady-state, QSS) approximation.

V0 (t = 0) V(t), rising Feed, F(t) = F0 + alpha t substrate conc. Sf X(t), S(t) no outflow (fed-batch) V(t) = V0 + F0 t + (alpha/2) t^2
Fig. 4 — fed-batch reactor: an increasing feed rate $F(t)=F_0+\alpha t$ raises the liquid volume with no outflow; biomass and substrate concentrations evolve inside.

Approach. Part (a) is a direct integration of the overall liquid-volume balance. Part (b) applies the pseudo-steady-state approximation ($dS/dt\approx0$, i.e. $S$ adjusts on a much faster timescale than $V$ or the total biomass $XV$ change) to the coupled substrate and biomass balances, together with the standard fed-batch simplification that the feed is concentrated ($S_f\gg S$ at quasi-steady state — the whole point of running fed-batch).

  1. (a) Overall liquid-volume balance. With no outflow, $dV/dt=F(t)=F_0+\alpha t$. Integrating from $V(0)=V_0$: $$\boxed{V(t)=V_0+F_0t+\frac{\alpha}{2}t^2.}$$
  2. (b) Overall biomass and substrate balances. With no death and no outflow: $$\frac{d(XV)}{dt}=\mu_g\,XV, \qquad \frac{d(SV)}{dt}=F\,S_f-\frac{\mu_g\,XV}{Y_{X/S}^M}.$$
  3. Apply the pseudo-steady-state approximation. Expanding $d(SV)/dt=V\,dS/dt+S\,dV/dt$ and setting $dS/dt\approx0$ gives $d(SV)/dt\approx S\,dV/dt=SF$. Substituting into the substrate balance: $$SF=FS_f-\frac{\mu_g XV}{Y_{X/S}^M} \ \Rightarrow\ \mu_g XV=Y_{X/S}^M F(S_f-S)\approx Y_{X/S}^M F S_f$$ (dropping $S$ next to $S_f$, consistent with $S$ staying small — that is what “quasi-steady” means here).
  4. Integrate the total-biomass balance. The biomass balance becomes $d(XV)/dt=\mu_g XV=Y_{X/S}^M S_f F(t)$, a known function of time, so $$XV(t)=X_0V_0+Y_{X/S}^M S_f\int_0^t F\,dt'=X_0V_0+Y_{X/S}^M S_f\big(V(t)-V_0\big).$$
  5. Solve for $S(t)$. Since $\mu_g=kS$ and $\mu_g XV=Y_{X/S}^M S_f F(t)$ from Step 3, $$kS\cdot XV(t)=Y_{X/S}^M S_f F(t) \ \Rightarrow\ \boxed{S(t)=\frac{Y_{X/S}^M S_f\big(F_0+\alpha t\big)}{k\Big[X_0V_0+Y_{X/S}^M S_f\big(V(t)-V_0\big)\Big]}.}$$
ResultExpression
(a) Reactor volume $V(t)$$V_0+F_0t+\alpha t^2/2$
Total biomass $XV(t)$$X_0V_0+Y_{X/S}^M S_f\big(V(t)-V_0\big)$
(b) Substrate conc. $S(t)$$Y_{X/S}^M S_f F(t)\big/\big[k\,XV(t)\big]$
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