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

Question 5 of 6: Batch Fermenter — Time to 50% of Maximum Product Concentration

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 5: Batch Fermenter — Time to 50% of Maximum Product Concentration (15 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.

QuantitySymbolValue
Initial substrate conc.$S_0$2000 mg/L
Initial biomass conc.$X_0$50 mg/L
Growth-rate constant$k$ (in $\mu_g=kS$)0.002 hr$^{-1}$ per mg/L
True growth yield$Y_{X/S}^T$0.3 g dw/g substrate
Product yield$Y_{P/X}$0.2 g product/g cells

Find. The time $t^*$ at which $P(t^*)=0.5\,P_{\max}$.

t (hr) X (mg/L) Xmax = 650 mg/L X(t*) = 350 mg/L t* = 0.609 hr (36.5 min) X0 = 50
Fig. 5 — logistic biomass growth $X(t)$ (substrate-limited via the mass-balance link $Y_{X/S}^T=-dX/dS$); since product formation is growth-associated, $P(t)$ tracks $X(t)$ linearly, so $P=0.5P_{\max}$ occurs exactly where $X$ reaches the corresponding intermediate value $X(t^*)=350$ mg/L.

Approach. Use the constant-yield mass-balance link $Y_{X/S}^T=-dX/dS$ to eliminate $S$ from $\mu_g=kS$, turning the growth equation into the closed-form logistic equation; find $X_{\max}$ and $P_{\max}$ from the yields, express $P(t)$ via the growth-associated relation $dP/dt=Y_{P/X}\,dX/dt$, then solve the logistic solution for the time at which $X$ (equivalently $P$) reaches the 50%-of-maximum value.

  1. Eliminate $S$ via the yield mass balance. $Y_{X/S}^T=-dX/dS$ (constant) integrates to $X=X_0+Y_{X/S}^T(S_0-S)$, so $S=S_0-(X-X_0)/Y_{X/S}^T$. Substrate is fully consumed ($S\to0$) at $$X_{\max}=X_0+Y_{X/S}^T S_0=50+0.3(2000)=650\ \text{mg/L}.$$
  2. Logistic growth equation. $$\frac{dX}{dt}=\mu_g X=kSX=k\left[\frac{X_{\max}-X}{Y_{X/S}^T}\right]X=r\,X\left(1-\frac{X}{X_{\max}}\right),\qquad r=\frac{kX_{\max}}{Y_{X/S}^T}=\frac{0.002(650)}{0.3}=4.333\ \text{hr}^{-1}.$$
  3. Logistic solution. $$X(t)=\frac{X_{\max}}{1+\left(\dfrac{X_{\max}-X_0}{X_0}\right)e^{-rt}}.$$
  4. Product concentration and its maximum. Growth-associated kinetics: $dP/dt=Y_{P/X}\,dX/dt\ \Rightarrow\ P(t)=Y_{P/X}\big(X(t)-X_0\big)$, so $$P_{\max}=Y_{P/X}(X_{\max}-X_0)=0.2(650-50)=120\ \text{mg/L}.$$
  5. Target biomass concentration. $P(t^*)=0.5P_{\max}=60$ mg/L requires $$X(t^*)-X_0=\frac{60}{0.2}=300\ \Rightarrow\ X(t^*)=350\ \text{mg/L}.$$
  6. Solve the logistic equation for $t^*$. $$\frac{X_{\max}}{X(t^*)}-1=\frac{X_{\max}-X_0}{X_0}e^{-rt^*}\ \Rightarrow\ e^{-rt^*}=\left(\frac{650}{350}-1\right)\frac{50}{600}=0.85714(0.083333)=0.071429,$$ $$t^*=\frac{-\ln(0.071429)}{4.333}=\frac{2.6391}{4.333}=\boxed{0.609\ \text{hr}\ (\approx36.5\ \text{min}).}$$
ResultValue
$X_{\max}$650 mg/L
Logistic rate constant $r$4.333 hr$^{-1}$
$P_{\max}$120 mg/L
$X(t^*)$ for 50% $P_{\max}$350 mg/L
$t^*$0.609 hr (36.5 min)
Check: the printed growth-rate expression $\mu_g=0.002\,S\ \text{hr}^{-1}$ (S in mg/L) is used exactly as given even though it makes the initial specific growth rate very fast ($\mu_g(S_0)=4\ \text{hr}^{-1}$, a ~10-minute doubling time) — this is the source's own simplified numbers. The closed-form logistic result was cross-checked against a direct RK4 numerical integration of $dX/dt=kSX$ with $S=S_0-(X-X_0)/Y_{X/S}^T$; both give $t^*=0.6090$ hr.