Question 1 of 5: Immobilized-Enzyme CSTR — Haldane Kinetics, Thiele Modulus & Effectiveness Factor

Question 1: Immobilized-Enzyme CSTR — Haldane Kinetics, Thiele Modulus & Effectiveness Factor (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.

Find. (a) The feed flow rate F giving maximum productivity; (b) the Thiele modulus φ for the immobilized bead; (c) the effectiveness factor η.

Approach. Haldane (substrate-inhibited) kinetics has an intrinsic rate MAXIMUM (not a monotonic rise, as in ordinary Michaelis–Menten) at S*=√(K_mK_s); the given S* pins down the otherwise-unstated inhibition constant K_s. A CSTR at steady state always satisfies F(S₀−S)=v(S)V, so its volumetric productivity equals v(S) for whatever S the reactor settles at — therefore maximum productivity means choosing F so the reactor settles exactly at S=S*, the rate maximum. Parts (b)/(c) then assess the SEPARATE complication that the immobilized catalyst is diffusion limited: build a pseudo-first-order rate constant at the operating point, form the generalized Thiele modulus for a sphere, and read the effectiveness factor off the standard first-order-sphere relation.

1. Solve for K_s from the given rate-maximum condition. Haldane kinetics v=v_m/(1+K_m/S+S/K_s) has dv/dS=0 at S*=√(K_mK_s). Rearranging with the given S*=20 and K_m=20: $$K_s=\frac{S^{*2}}{K_m}=\frac{20^2}{20}=\boxed{20\ \text{mol/m}^3}$$
2. Evaluate the maximum intrinsic rate v* at S=S*. $$v^{*}=\frac{v_m}{1+K_m/S^{*}+S^{*}/K_s}=\frac{4.45\times10^{-3}}{1+1+1}=\boxed{1.483\times10^{-3}\ \text{mol}\cdot\text{m}^{-3}\text{s}^{-1}}\ (=v_m/3)$$
3. (a) CSTR steady-state balance → feed flow rate for maximum productivity. At steady state F(S₀−S)=v(S)V for any operating point, so productivity F(S₀−S)/V equals v(S) always; it is maximized by running the reactor at S=S* (the intrinsic-rate peak): $$F=\frac{v^{*}V}{S_0-S^{*}}=\frac{(1.483\times10^{-3})(0.001)}{100-20}=\boxed{1.854\times10^{-8}\ \text{m}^3/\text{s}}=0.0668\ \text{L/hr}$$
4. (b) Pseudo-first-order rate constant at the operating point → Thiele modulus. Using the rate/concentration ratio AT the operating point S* as an effective first-order rate constant, and the sphere's characteristic length V_p/S_p=R/3: $$k=\frac{v^{*}}{S^{*}}=\frac{1.483\times10^{-3}}{20}=7.417\times10^{-5}\ \text{s}^{-1}\qquad D_e=3.6\times10^{-6}\ \text{cm}^2/\text{s}=3.6\times10^{-10}\ \text{m}^2/\text{s}$$ $$\phi=\frac{R}{3}\sqrt{\frac{k}{D_e}}=\frac{0.08}{3}\sqrt{\frac{7.417\times10^{-5}}{3.6\times10^{-10}}}=\boxed{12.10}$$ A Thiele modulus this large (≫1) signals severe internal diffusion limitation — consistent with the question's own remark that these 80 mm beads are unusually large.
5. (c) Effectiveness factor (first-order kinetics in a sphere). $$\eta=\frac{3}{\phi^2}\left(\phi\coth\phi-1\right)=\frac{3}{12.10^2}(12.10\times1.0000-1)=\boxed{0.227}$$ Only about 23% of the bead's intrinsic capacity is realized: substrate is consumed near the surface before it can diffuse to the bead core, exactly as the concentration profile in Fig. 2 shows.