NivaarExam PrepOfficial exam papers ↗

23-Chem-B4 Biochemical Engineering · May 2018

Question 1 of 5: Immobilized-Enzyme Packed-Bed Column — Zero-Order, External- and Internal-Diffusion-Limited Sizing

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

Notes on this paper

National Exam 16-Chem-B4, Biochemical Engineering — May 2018. 3 hours, Closed-Book Exam (any non-communicating Casio or Sharp calculator permitted). Per the exam notes, FIVE (5) questions constitute a complete paper and all five must be answered; most require a short-essay-format answer, and clarity/organization of the answer are explicitly marked.

Reference texts: Shuler & Kargi, Bioprocess Engineering: Basic Concepts, 2nd ed.; Bailey & Ollis, Biochemical Engineering Fundamentals, 2nd ed.; Madigan et al., Brock Biology of Microorganisms, 13th ed.

Question 1: Immobilized-Enzyme Packed-Bed Column — Zero-Order, External- and Internal-Diffusion-Limited Sizing (20 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
Maximum reaction ratevmax$7.45\times10^{-2}$ mol/m³/s
Michaelis constantKm8 mol/m³
External mass-transfer coefficient (lumped, ksa)ks$9.3125\times10^{-5}$ s−1
Column diameterD0.30 m
Feed substrate concentrationS₀100 mol/m³
Volumetric feed flow rateQ0.03 m³/min = $5\times10^{-4}$ m³/s
Target conversionX80% (S = 20 mol/m³ at exit)
Internal Thiele modulus (case iii, given)φ80

Find. The column length L required for 80% conversion under each of the three kinetic/mass-transfer regimes (i)–(iii).

Feed: S₀ = 100 mol/m³Q = 0.03 m³/minProduct: S = 20 mol/m³(80% conversion)D = 30 cmL = ?Immobilized-enzymepacked bedPlug flow ↓
Fig. 1 — immobilized-enzyme packed-bed column: feed enters at S₀=100 mol/m³, plug flow through the packed bed, product exits at S=20 mol/m³ (80% conversion). Column length L is the unknown solved for in each of the three cases.

Approach. In each regime the intrinsic Michaelis–Menten rate collapses to a simpler rate expression (zero order, or pseudo-first order with an effective rate constant set by whichever resistance — reaction, external film, or internal pore diffusion — controls); substitute that expression into the plug-flow design equation $Q\,dS=-r(S)\,dV=-r(S)\,A\,dz$, integrate from the inlet to 80% conversion, and solve for L.

  1. Column cross-section and target exit concentration. $$A=\frac{\pi D^2}{4}=\frac{\pi(0.30)^2}{4}=0.070686\ \text{m}^2$$ At 80% conversion, $S=S_0(1-X)=100(1-0.80)=20\ \text{mol/m}^3$. The intrinsic first-order rate constant (valid whenever $S\ll K_m$) will be needed for cases (ii) and (iii): $$k_1=\frac{v_{max}}{K_m}=\frac{7.45\times10^{-2}}{8}=9.3125\times10^{-3}\ \text{s}^{-1}$$ Note that the given $k_s=9.3125\times10^{-5}\ \text{s}^{-1}$ is exactly $k_1/100$ — i.e. external mass transfer in case (ii) is deliberately 100× slower than the intrinsic reaction, confirming that case (ii) is genuinely external-transfer-controlled.
  2. Case (i) — zero order, no mass-transfer limitations. Treating the reaction as zero order ($v\approx v_{max}$, independent of S, the stated idealization for this case) the plug-flow balance $Q\,dS=-v_{max}A\,dz$ integrates directly (constant rate, no S-dependence to carry through the integral): $$Q(S_0-S)=v_{max}A\,L_1 \quad\Rightarrow\quad L_1=\frac{Q(S_0-S)}{v_{max}A}$$ $$L_1=\frac{(5\times10^{-4})(100-20)}{(7.45\times10^{-2})(0.070686)}=\frac{0.0400}{5.2661\times10^{-3}} \boxed{=7.60\ \text{m}}$$
  3. Case (ii) — first order, external mass-transfer limitations only. With $S\ll K_m$ the intrinsic kinetics are pseudo-first order, but here the external film is the slow step, so the observed rate is controlled entirely by the (given, already area-lumped) external coefficient: $r=k_sS$. The plug-flow balance $Q\,dS/S=-k_sA\,dz$ integrates to a log form: $$Q\ln\!\left(\frac{S_0}{S}\right)=k_sA\,L_2 \quad\Rightarrow\quad L_2=\frac{Q\ln(S_0/S)}{k_sA}$$ $$L_2=\frac{(5\times10^{-4})\ln(100/20)}{(9.3125\times10^{-5})(0.070686)} =\frac{(5\times10^{-4})(1.6094)}{6.5826\times10^{-6}} \boxed{=122.2\ \text{m}}$$ The column is far longer than case (i) because the film coefficient is deliberately the bottleneck (100× slower than the intrinsic rate constant found in Step 1).
  4. Case (iii) — first order, internal mass-transfer limitations only. Internal pore diffusion knocks the observed rate down by the effectiveness factor η, evaluated from the given Thiele modulus with the standard spherical-particle relation: $$\eta=\frac{3}{\varphi^2}\left(\varphi\coth\varphi-1\right)=\frac{3}{80^2}\big(80\coth 80-1\big) =\frac{3}{6400}(80-1)=0.03703$$ (coth 80 ≈ 1 to more than 60 significant figures, since $\varphi=80$ is deep in the strongly diffusion-limited asymptote). The observed pseudo-first-order rate constant is then $k_{obs}=\eta k_1$, and the same integrated log form as case (ii) applies with $k_{obs}$ in place of $k_s$: $$k_{obs}=\eta k_1=(0.03703)(9.3125\times10^{-3})=3.448\times10^{-4}\ \text{s}^{-1}$$ $$L_3=\frac{Q\ln(S_0/S)}{k_{obs}A}=\frac{(5\times10^{-4})(1.6094)}{(3.448\times10^{-4})(0.070686)} \boxed{=33.0\ \text{m}}$$ Internal diffusion ($\eta\approx0.037$, an effective ~27× rate penalty) is a less severe bottleneck here than the externally-imposed 100× penalty of case (ii), so $L_3
CaseControlling resistanceColumn length L
(i)Zero order, no MT limitation7.60 m
(ii)First order, external MT limited122.2 m
(iii)First order, internal MT limited (η=0.0370)33.0 m

All three lengths satisfy the same 80% conversion target from the same feed, so the spread illustrates how strongly the identity of the rate-limiting step drives reactor size: removing (or reducing) whichever resistance dominates — larger external film coefficient, smaller/more porous immobilization beads — would shrink the required column substantially below the mass-transfer-limited cases.

Check: Case (i)'s "zero order" treatment is only a good approximation while $S\gg K_m$; near the 80%-conversion exit ($S=20$ vs $K_m=8$) the true Michaelis–Menten rate has already fallen measurably below $v_{max}$, so $L_1=7.60$ m is a lower-bound idealization, exactly as the question's "assuming zero order" instruction intends (integrating the full Michaelis–Menten form, $L=Q[K_m\ln(S_0/S)+(S_0-S)]/(v_{max}A)$, gives 8.82 m). Case (iii) assumes spherical beads and takes the given φ=80 as the modulus $\varphi=R\sqrt{k_1/D_e}$ (Bailey & Ollis), since no bead radius or effective diffusivity is given. Texts that define it as $\varphi=(R/3)\sqrt{k_1/D_e}$ (Shuler & Kargi) use $\eta=\frac{1}{\varphi}\left[\frac{1}{\tanh 3\varphi}-\frac{1}{3\varphi}\right]=0.01245$, which gives $L_3\approx98.2$ m — still shorter than case (ii). Case (ii) treats the film as the only resistance; adding the intrinsic reaction resistance in series ($1/k=1/k_1+1/k_s$) lengthens it by only 1% (123.5 m).
← Paper overview