Michaelis–Menten Enzyme Reaction in a CSTR — Design-Equation Derivation & Reactor Volume for Glucose–Fructose Isomerization: 23-Chem-B4 Biochemical Engineering, December 2019, Q2

Question 2: Michaelis–Menten Enzyme Reaction in a CSTR — Design-Equation Derivation & Reactor Volume for Glucose–Fructose Isomerization (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.

(i) Derivation of the CSTR design equation

Approach. Write a steady-state substrate mass balance on the well-mixed CSTR (in − out − consumed by reaction = 0), express the consumption rate with Michaelis–Menten kinetics, and recast the result in terms of conversion X and residence time τ.

Steady-state substrate balance. For a CSTR of volume V fed at volumetric flow rate F with inlet substrate concentration S₀ and outlet (=tank) concentration S, $$FS_0-FS-rV=0,\qquad r=\frac{v_{max}S}{K_m+S}$$ where r is the (Michaelis–Menten) volumetric substrate-consumption rate at the tank's own outlet concentration S — the defining feature of a CSTR is that the reaction proceeds everywhere at the exit condition, not the inlet condition.
Introduce conversion and residence time. Define conversion X=(S₀−S)/S₀ (so S=S₀(1−X)) and residence time τ=V/F. Dividing the balance by F, $$S_0-S=r\tau=\frac{v_{max}S}{K_m+S}\,\tau \;\Rightarrow\; S_0X=\frac{v_{max}S_0(1-X)}{K_m+S_0(1-X)}\,\tau$$
Clear the denominator and simplify. Multiplying both sides by [K_m+S₀(1−X)], $$S_0X\big[K_m+S_0(1-X)\big]=v_{max}S_0(1-X)\,\tau$$ Dividing through by S₀ and expanding, $$K_mX+S_0X(1-X)=v_{max}(1-X)\tau$$
Divide by (1−X) to isolate the two conversion-dependent terms on the left: $$\frac{K_mX}{1-X}+S_0X=v_{max}\tau$$ $$\boxed{K_m\left(\frac{X}{1-X}\right)+S_0X=v_{max}\tau}$$ — exactly the given design equation.

Term	Meaning
K_m	Michaelis constant (g/L) — substrate concentration at half V_max
X	Fractional conversion of substrate, X=(S₀−S)/S₀
S₀	Feed (inlet) substrate concentration
v_max (V_max)	Maximum reaction velocity at enzyme saturation
τ	Residence time, V/F

Assumptions. Steady-state, well-mixed (perfectly back-mixed) CSTR so the tank concentration equals the exit concentration and the reaction rate throughout the vessel equals its rate at S (not S₀); constant volumetric flow rate and density (no volume change on reaction); enzyme concentration and activity constant over the residence time (no significant deactivation); the free/dissolved-enzyme (homogeneous) Michaelis–Menten rate law applies with no product or substrate inhibition.

(ii) Reactor volume for the glucose–fructose isomerization

Given.

Quantity	Symbol	Value
Feed glucose concentration	S₀	100 g/L
Desired conversion	X	0.40
Michaelis constant	K_m	0.26 g/L
Maximum velocity	v_max	10 g/(L·h)
Feed flow rate	F	100 L/h

Find. The CSTR volume V needed to achieve X=0.40.

Approach. Substitute the given values into the Part (i) design equation to solve for τ, then recover V=Fτ.

Evaluate the conversion term. $$\frac{X}{1-X}=\frac{0.40}{0.60}=0.6667$$
Substitute into the design equation and solve for τ. $$K_m\left(\frac{X}{1-X}\right)+S_0X=v_{max}\tau$$ $$(0.26)(0.6667)+(100)(0.40)=0.1733+40.00=40.173\ \text{g/L}$$ $$\tau=\frac{40.173}{v_{max}}=\frac{40.173\ \text{g/L}}{10\ \text{g/(L}\cdot\text{h)}}=4.017\ \text{h}$$
Recover the reactor volume. $$V=F\tau=(100\ \text{L/h})(4.017\ \text{h})$$ $$\boxed{V\approx 401.7\ \text{L}}$$

Quantity	Value
X/(1−X)	0.6667
K_mX/(1−X) + S₀X	40.17 g/L
Residence time, τ	4.017 h
Reactor volume, V	≈ 401.7 L

Check

Assumes the free-enzyme homogeneous Michaelis–Menten rate law applies with no immobilization/mass-transfer resistance (unlike packed-bed problems) and that K_m, v_max are constant over the whole reactor (no enzyme deactivation, no product inhibition from fructose).

Topic: CSTR sizing for a homogeneous Michaelis–Menten enzyme reaction.

Key relations: Steady-state balance FS₀−FS=rV; design equation K_mX/(1−X)+S₀X=v_maxτ; V=Fτ.

Why this works: A CSTR's defining feature — instantaneous, complete mixing — means the reaction proceeds everywhere at the single exit concentration S, so the balance uses r(S) rather than an integral over a concentration profile as a PFR would require; substituting the Michaelis–Menten rate law directly into the algebraic CSTR balance and recasting in terms of X gives a design equation solvable for τ (and hence V) with pure algebra, no integration.

Common pitfall: Using the PFR (integrated, logarithmic) Michaelis–Menten sizing equation $\tau=\big(K_m\ln(1/(1-X))+S_0X\big)/v_{max}$ for a CSTR problem, or vice versa — the two reactor types give materially different volumes for the same X because a CSTR runs at the (slower) exit rate throughout, while a PFR's rate varies continuously along its length.

Source: Bailey & Ollis, Biochemical Engineering Fundamentals, 2nd ed., Ch. 3 (enzyme reactor design); Shuler & Kargi, Bioprocess Engineering, 2nd ed., Ch. 4.

Question 2 of 5: Michaelis–Menten Enzyme Reaction in a CSTR — Design-Equation Derivation & Reactor Volume for Glucose–Fructose Isomerization

Question 2: Michaelis–Menten Enzyme Reaction in a CSTR — Design-Equation Derivation & Reactor Volume for Glucose–Fructose Isomerization (20 marks)

(i) Derivation of the CSTR design equation

(ii) Reactor volume for the glucose–fructose isomerization