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.
Find. The column length L required for 80% conversion under each of the three kinetic/mass-transfer
regimes (i)–(iii).
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.
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.
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):
$$Q(S_0-S)=v_{max}A\,L_1 \quad\Rightarrow\quad L_1=\frac{Q(S_0-S)}{v_{max}A}$$
$$L_1=\frac{(5\times10^{-4})(100-20)}{(7.45\times10^{-2})(0.070686)}=\frac{0.0400}{5.2661\times10^{-3}}
\boxed{=7.60\ \text{m}}$$
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).
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
Case
Controlling resistance
Column length L
(i)
Zero order, no MT limitation
7.60 m
(ii)
First order, external MT limited
122.2 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).