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

Question 2 of 6: Immobilized-Enzyme CSTR — Bead Volume and Diffusion Limitation

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 2: Immobilized-Enzyme CSTR — Bead Volume and Diffusion Limitation (a. 18 marks; b. 7 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
Reactor volume$V$1000 L
Feed flow rate$F$10 L/min
Feed substrate conc.$S_0$10 mol/L
Target effluent product conc.$P_{\text{out}}$5 mol/L
Bead radius$R$5 mm = 0.005 m
Substrate conc. at bead surface$C_s$1 mol/L
Intrinsic max. rate (per L particle)$V_m''$10 mol/(L·min)
Michaelis constant$K_M$100 mol/L
Substrate diffusivity$D_e$$1\times10^{-4}$ m$^2$/min

Stoichiometry $S+E\rightarrow E+2P$: one mole of substrate consumed yields two moles of product.

Find. (a) the total volume of immobilized-enzyme particles needed so the reactor delivers $P_{\text{out}}=5$ mol/L; (b) how the observed (diffusion-limited) rate in the particles compares, quantitatively, to the rate an equal mass of enzyme would give free in solution.

CSTRV = 1000 Limmobilized-enzymebeads (R = 5 mm)FeedF = 10 L/minS0 = 10 mol/LEffluentF = 10 L/minS, P = 5 mol/L
Fig. 2a — CSTR mass balance: 10 L/min of 10 mol/L feed leaves at 5 mol/L product, giving the total substrate-consumption rate the immobilized beads must supply.
bulk fluid S_bulk (well mixed) external film (resistance) C(R) = Cs = 1 mol/L agarose bead R = 5 mm, reaction + diffusion r (0 to R) C(r) C(0) Cs phi = 0.158 (small) so eta = 0.998 -- profile nearly flat, diffusion not limiting
Fig. 2b — substrate must cross an external film (surface conc. $C_s=1$ mol/L) then diffuse and react inside the bead; since $\phi$ is small the internal profile is nearly flat.

Approach. Because $C_s\ll K_M$ (1 vs. 100 mol/L), the intrinsic kinetics are pseudo-first-order everywhere inside the bead, so the classical first-order-sphere Thiele modulus and effectiveness factor apply; combine the effectiveness factor with the overall CSTR substrate/product balance to size the particle volume, then compare the diffusion-limited rate to the un-hindered (free-enzyme) rate at the same surface concentration.

  1. Part (a) — pseudo-first-order rate constant. Since $C_s=1\ll K_M=100$, $\ V_m''C/(K_M+C)\approx (V_m''/K_M)C$ throughout $0\le C\le C_s$: $$k_1=\frac{V_m''}{K_M}=\frac{10}{100}=0.1\ \text{min}^{-1}.$$
  2. Thiele modulus (sphere, first order). $$\phi=R\sqrt{\frac{k_1}{D_e}}=0.005\sqrt{\frac{0.1}{1\times10^{-4}}}=0.005\sqrt{1000}=0.1581.$$
  3. Effectiveness factor. For a sphere with first-order kinetics, $$\eta=\frac{3}{\phi^2}\big(\phi\coth\phi-1\big)=\frac{3}{0.1581^2}\big(0.1581\coth 0.1581-1\big)=0.998.$$ Diffusion barely limits the reaction here — the bead is small and, at $C_s\ll K_M$, the intrinsic reaction itself is slow relative to diffusion.
  4. Observed (diffusion-limited) volumetric rate. Using the exact Michaelis–Menten rate at the surface, $r(C_s)=V_m''C_s/(K_M+C_s)=10(1)/101=0.0990\ \text{mol}/(\text{L}\cdot\text{min})$, so $$r_{\text{obs}}=\eta\, r(C_s)=0.998(0.0990)=0.0988\ \text{mol}/(\text{L}_{\text{particle}}\cdot\text{min}).$$
  5. Required total substrate-consumption rate. From the CSTR product balance ($S+E\rightarrow E+2P$, one S per two P): $$\dot n_{S,\text{consumed}}=\frac{F\,P_{\text{out}}}{2}=\frac{10(5)}{2}=25\ \text{mol S/min}.$$
  6. Particle volume required. $$V_{\text{particles}}=\frac{\dot n_{S,\text{consumed}}}{r_{\text{obs}}}=\frac{25}{0.0988}=\boxed{253\ \text{L (}\approx 25\%\text{ of the 1000 L reactor)}}.$$
  7. Part (b) — immobilized vs. free enzyme. An equivalent mass of enzyme, dispersed freely and exposed to the same surface concentration $C_s=1$ mol/L, reacts with no internal diffusion resistance at rate density $r(C_s)=0.0990\ \text{mol}/(\text{L}\cdot\text{min})$ — that is exactly the definition of $\eta$: $$\frac{r_{\text{obs}}}{r_{\text{free}}}=\frac{\eta\,r(C_s)}{r(C_s)}=\eta=\boxed{0.998\ \ (\text{immobilized rate} \approx 99.8\%\text{ of the free-enzyme rate}).}$$
ResultValue
Pseudo-first-order constant $k_1$0.1 min$^{-1}$
Thiele modulus $\phi$0.158
Effectiveness factor $\eta$0.998
(a) Particle volume required253 L
(b) Immobilized/free rate ratio0.998 (≈99.8%)
Check: assumes the given $C_s=1$ mol/L (post-external-film) is uniform over every bead and constant at steady state, and that the pseudo-first-order approximation (exact to <1% here since $C_s/K_M=0.01$) is used for both $\phi$ and $\eta$. Back-checking the bulk phase: the CSTR product balance gives an outlet/bulk substrate concentration of $S_{\text{out}}=S_0-25/F=7.5$ mol/L, so the external film drops the concentration from 7.5 mol/L (bulk) to 1 mol/L (bead surface) — i.e. external mass-transfer resistance, not internal diffusion, is what actually limits this reactor.