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20-Bio-A5 Systems Analysis & Control · Undated paper

Question 6 of 6: Continuous Tubular Steam Sterilizer — Vitamin Retention and Sterilization Probability

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.

Question 6: Continuous Tubular Steam Sterilizer — Vitamin Retention and Sterilization Probability (a. 10 marks; b. 10 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. Tubular (plug-flow) sterilizer, pipe length $L$, cross-section $A$, feed flow rate $F$; feed vitamin concentration $C_{V,0}$, spore concentration $N_0$; first-order Arrhenius degradation constants $k_{D,\text{vitamin}}(T)$, $k_{D,\text{spores}}(T)$.

Find. (b) $C_V$ as a function of $F$ and $T$; (c) the probability of a successful (contamination-free) sterilization as a function of $F$, $T$ and $t_{\text{oper}}$.

Feed, F $C_{V,0}$, $N_0$ steam (T) tubular sterilizer — length L, area A, T residence time τ = AL / F $C_V$, $N$ to fermenter plug flow: every fluid element spends the same time τ at temperature T — a batch-equivalent first-order exposure
Fig. 6 — continuous tubular sterilizer: steam injection brings the feed to temperature $T$; each fluid element is a plug spending the same residence time $\tau=AL/F$ in the hot zone.

Approach. Treat the pipe as an ideal plug-flow reactor: every fluid element experiences the same residence time $\tau=AL/F$ at temperature $T$, which is exactly equivalent to a first-order batch exposure of duration $\tau$ — apply that to vitamin retention directly, then to the expected number of surviving spores accumulated over the whole run, and use the Poisson (rare-event) result to get the probability of zero survivors.

  1. Residence time in the sterilizer. $$\tau=\frac{V_{\text{pipe}}}{F}=\frac{AL}{F}.$$
  2. (b) First-order vitamin degradation over one pass. Each element degrades as a first-order batch reaction for time $\tau$: $$\frac{dC_V}{dt}=-k_{D,\text{vitamin}}C_V\ \Rightarrow\ \boxed{C_V=C_{V,0}\exp\!\big(-k_{D,\text{vitamin}}(T)\,\tau\big)=C_{V,0}\exp\!\left(-\alpha_{\text{vit}}e^{-E_{a,\text{vit}}/RT}\frac{AL}{F}\right).}$$
  3. Spore survival fraction over one pass. By the same argument, $$\frac{N}{N_0}=\exp\!\big(-k_{D,\text{spores}}(T)\,\tau\big)=\exp\!\left(-\alpha_{\text{spores}}e^{-E_{a,\text{spores}}/RT}\frac{AL}{F}\right)\equiv e^{-\text{Da}},$$ where $\text{Da}=k_{D,\text{spores}}\tau$ is the sterilization Damköhler ("Del") factor.
  4. Total spores fed and expected survivors over the whole run. Over the total operating time $t_{\text{oper}}$, the total number of spores entering is $N_0\,F\,t_{\text{oper}}$ (concentration × total volume processed); the expected number of surviving spores is $$n_s=N_0\,F\,t_{\text{oper}}\,e^{-\text{Da}}.$$
  5. (c) Probability of a successful (contamination-free) run. Rare survival events over a large number of spores are Poisson-distributed, so the probability of exactly zero survivors is $$P_{\text{success}}=e^{-n_s}=\boxed{\exp\!\left[-N_0\,F\,t_{\text{oper}}\exp\!\left(-\alpha_{\text{spores}}e^{-E_{a,\text{spores}}/RT}\frac{AL}{F}\right)\right].}$$
ResultExpression
Residence time$\tau=AL/F$
(b) $C_V$$C_{V,0}\exp(-k_{D,\text{vitamin}}AL/F)$
Del factor$\text{Da}=k_{D,\text{spores}}AL/F$
(c) $P_{\text{success}}$$\exp[-N_0Ft_{\text{oper}}e^{-\text{Da}}]$
the source page prints only sub-parts b) and c); part a) is genuinely absent from the exam paper. The marking scheme's two mark buckets, (a) 10 marks and (b) 10 marks, are mapped onto the two sub-questions that are actually printed, b) and c) respectively, since no third sub-part exists to answer. Both parts assume ideal plug flow (no axial dispersion) and that $F$, $T$ are held constant through a full run of duration $t_{\text{oper}}$.
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