Bioreaction Engineering — Kinetics, Ideal Reactors, Cell Growth, Oxygen Transfer and Scale-up
1. Rate laws, zero- and first-order kinetics, and enzyme kinetics for design
| Kinetics | Rate | Integrated form | Half-life |
|---|---|---|---|
| Zero order | −dC/dt = k | C = C₀ − kt | C₀/2k — grows with C₀ |
| First order | −dC/dt = kC | ln(C₀/C) = kt | 0.693/k — independent of C₀ |
| Michaelis–Menten | −dS/dt = Vₘₐₓ·S/(Kₘ + S) | Vₘₐₓ·t = Kₘ ln(S₀/S) + (S₀ − S) | First order when S ≪ Km, zero order when S ≫ Km |
The inhibition kinetics of the enzyme chapter carry into design unchanged — a competitive inhibitor replaces Km by Km(1 + I/Kᵢ), a non-competitive one divides Vmax by (1 + I/Kᵢ), an uncompetitive one divides both — and two further forms matter for reactors. Substrate inhibition, v = VmaxS/(Km + S + S²/K_SI), passes through a maximum at S = √(Km·K_SI), so feeding substrate slowly (fed-batch) or running a CSTR at low S can beat a batch. Product inhibition makes a CSTR, which operates at the high outlet product concentration throughout, worse than a PFR.
2. Ideal reactors — batch, mixed flow and plug flow
A batch reactor is closed and uniform, and time is its coordinate. A mixed-flow reactor (CSTR) is uniform and equal to its outlet, so the balance is algebraic: τ = V/F = (C₀ − C)/(−r) evaluated at the outlet. A plug-flow reactor has no axial mixing, each fluid element behaving as a small batch, so τ = ∫dC/(−r) exactly as batch time. For first-order kinetics, CSTR: X = kτ/(1 + kτ) and PFR: X = 1 − e^(−kτ); at kτ = 2 these are 0.667 and 0.865. For Michaelis–Menten kinetics the CSTR gives τ = (S₀ − S)(Km + S)/(VmaxS) and the PFR gives the integrated batch equation with t replaced by τ.
3. Immobilised enzymes and diffusion effects — Thiele modulus, effectiveness factor and Damköhler number
Enzymes are immobilised by adsorption (weak, reversible), covalent binding (stable, some activity lost), entrapment in a gel such as calcium alginate or polyacrylamide, encapsulation in membranes and cross-linking with glutaraldehyde, to be reused, run continuously in packed beds and kept out of the product — glucose isomerase for high-fructose syrup is the classic industrial case. The cost is mass transfer. External transfer from the bulk to the particle surface is measured by the Damköhler number, Da = maximum reaction rate/maximum mass-transfer rate = V′max/(k_L·S_b): Da ≪ 1 means reaction-limited, Da ≫ 1 means film-diffusion-limited, so stirring faster raises the observed rate only when Da is large.
Internal diffusion inside a porous particle is measured by the Thiele modulus φ, the square root of reaction rate over diffusion rate. For a first-order reaction in a slab of half-thickness L, φ = L√(k/Dₑ) and the effectiveness factor η = observed rate/rate with no diffusion limitation = tanh φ/φ; for a sphere with φ defined as (R/3)√(k/Dₑ), η = (1/φ)[1/tanh(3φ) − 1/(3φ)]. Small φ gives η → 1; large φ gives η ≈ 1/φ, the particle’s interior being starved. At φ = 2 a slab has η = tanh 2/2 = 0.964/2 = 0.482. Smaller particles, lower enzyme loading or faster diffusion all reduce φ.
4. Kinetics of cell growth, substrate utilisation and product formation
The specific growth rate is μ = (1/X)(dX/dt); in exponential growth X = X₀e^(μt) and the doubling time is t_d = ln 2/μ, so μ = 0.35 h⁻¹ gives t_d = 1.98 h, and growing from 0.1 to 3.2 g/L (five doublings) takes ln 32/0.4 = 8.66 h at μ = 0.4 h⁻¹. The Monod equation, μ = μmax·S/(Ks + S), is the unstructured workhorse: Ks is the substrate concentration giving μmax/2. With μmax = 0.5 h⁻¹, Ks = 0.2 g/L and S = 0.6 g/L, μ = 0.5 × 0.6/0.8 = 0.375 h⁻¹. Variants add substrate inhibition (Andrews, μ = μmaxS/(Ks + S + S²/Kᵢ)), product inhibition, or cell-density dependence (Contois).
| Quantity | Definition | Use |
|---|---|---|
| Biomass yield Yx/s | −ΔX/ΔS, grams of cells per gram of substrate consumed | Final biomass X = X₀ + Yx/s(S₀ − S): 0.6 + 0.45 × 28 = 13.2 g/L |
| Product yield Yp/s; oxygen yield Yx/O₂ | Product formed, or cells made per unit O₂, per unit consumed | Stoichiometric limits; oxygen demand of a culture |
| Maintenance (Pirt) | qs = μ/Y_G + m_s, equivalently 1/Yx/s,obs = 1/Y_G + m_s/μ | Observed yield falls at low growth rate, as maintenance takes a larger share |
| Luedeking–Piret | qp = αμ + β | α: growth-associated (primary metabolites such as ethanol, lactic acid); β: non-growth-associated (secondary metabolites such as penicillin); both: mixed |
Unstructured models treat biomass as one variable — Monod is the example — and describe balanced growth well. Structured models divide the cell into compartments (RNA, protein, a key enzyme pool) and can follow transient changes such as the lag phase or a shift of substrate. A separate axis is segregated (cells differ from one another, as in population-balance models) against unsegregated (an average cell). Most GATE numericals are unstructured and unsegregated.
5. Batch, fed-batch and continuous processes
In a chemostat (a CSTR fed sterile medium at dilution rate D = F/V) a steady state requires μ = D. Monod then fixes the outlet substrate, S = Ks·D/(μmax − D), independent of the feed, and the biomass follows from the yield, X = Yx/s(S₀ − S). With μmax = 0.8 h⁻¹, Ks = 0.5 g/L, S₀ = 20 g/L, Yx/s = 0.5 and D = 0.4 h⁻¹: S = 0.5 × 0.4/0.4 = 0.5 g/L, X = 0.5 × 19.5 = 9.75 g/L, and biomass productivity DX = 3.9 g L⁻¹ h⁻¹. Raising D past the critical dilution rate Dc = μmax·S₀/(Ks + S₀) = 0.8 × 20/20.5 = 0.78 h⁻¹ washes the culture out. Productivity DX peaks just below washout, at D_opt = μmax[1 − √(Ks/(Ks + S₀))], which is why chemostats are run close to, but safely under, Dc.
- Batch is simple, flexible and least prone to contamination, but its productivity must include the turnaround time for emptying, cleaning, sterilising and filling, and cells spend much of the run outside exponential growth.
- Fed-batch adds substrate during the run without removing broth, so volume rises. Feeding keeps S low, which avoids substrate inhibition and overflow metabolism (the Crabtree effect in yeast, acetate in E. coli), allows very high cell densities for recombinant proteins, and holds a secondary metabolite such as penicillin in its production phase. An exponential feed F = μX₀V₀e^(μt)/(Yx/s·S_F) holds μ constant; a quasi-steady state with D ≈ μ develops.
- Continuous operation has the highest volumetric productivity and steady product quality, but risks contamination, washout and genetic drift of production strains over long runs; cell recycle raises the biomass above Yx/s(S₀ − S) and the usable D above μmax.
6. Microbial and enzyme reactors, oxygen transfer, optimisation, scale-up and case studies
Microbial reactors are mostly stirred tanks (baffled, with Rushton or axial impellers and a sparger), bubble columns and airlift reactors (no moving parts, low shear, used for shear-sensitive and very large aerobic processes); enzyme reactors are packed beds, fluidised beds and membrane reactors that retain a soluble enzyme. For aerobic cultures oxygen, sparingly soluble in water, is usually the limit. The oxygen transfer rate is OTR = k_La(C* − C_L), the oxygen uptake rate is OUR = q_O₂·X, and at steady state OTR = OUR. C_L must stay above the critical dissolved oxygen concentration, below which respiration slows. With k_La = 100 h⁻¹, C* = 7 mg/L and C_L held at 2 mg/L, OTR = 500 mg O₂ L⁻¹ h⁻¹, which supports at most X = 500/200 = 2.5 g/L of cells respiring at 200 mg O₂ g⁻¹ h⁻¹. k_La is raised by higher agitation power, gas flow, pressure or O₂ enrichment (which raises C*).
Scale-up keeps geometric similarity and holds one criterion constant, and no single criterion can hold all of them. In turbulent flow power is P = N_p·ρN³D⁵ with a constant power number, so constant P/V needs N³D² constant, N₂ = N₁(D₁/D₂)^(2/3): from 300 rpm with a 0.1 m impeller to a 0.5 m impeller, N₂ = 300 × (0.2)^(2/3) = 103 rpm. Constant impeller tip speed πND (to cap shear) needs N₂ = N₁D₁/D₂ = 60 rpm, and constant k_La (the usual choice for aerobic processes) sits in between; constant mixing time would need the same N and an impossible power. Optimisation means choosing medium, temperature, pH, dissolved oxygen and feeding policy to maximise titre, yield and productivity together. Case studies: penicillin from Penicillium chrysogenum is a fed-batch, non-growth-associated secondary metabolite, made after the growth phase with a slow sugar feed and phenylacetic acid as the side-chain precursor of penicillin G; ethanol from Saccharomyces cerevisiae is growth-associated and anaerobic, C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂, with a theoretical yield of 2 × 46.07/180.16 = 0.511 g per g glucose.
Key takeaways
- Batch enzyme time: Vmax·t = Km ln(S₀/S) + (S₀ − S); enzyme CSTR: τ = (S₀ − S)(Km + S)/(VmaxS); first order: CSTR X = kτ/(1 + kτ), PFR X = 1 − e^(−kτ).
- Da ≫ 1 is external-diffusion-limited; a large Thiele modulus starves the particle interior, η = tanh φ/φ for a slab and η ≈ 1/φ when φ is large.
- t_d = ln 2/μ; Monod μ = μmaxS/(Ks + S); X = X₀ + Yx/s(S₀ − S); Luedeking–Piret qp = αμ + β separates growth- and non-growth-associated products.
- Chemostat: μ = D, S = KsD/(μmax − D), X = Yx/s(S₀ − S), washout at Dc = μmaxS₀/(Ks + S₀); fed-batch keeps S low for high density and secondary metabolites.
- OTR = k_La(C* − C_L) must match OUR = q_O₂X; gassing out gives k_La from a log plot; constant P/V scales N as D^(−2/3), constant tip speed as D^(−1).
Practice questions (17)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
In the Monod equation μ = μmax·S/(Ks + S), the constant Ks is
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Answer: A — the substrate concentration at which μ = μmax/2
Putting S = Ks gives μ = μmax·Ks/(2Ks) = μmax/2, exactly as Km does for Michaelis–Menten. A small Ks means the organism grows near its maximum rate even at low substrate levels, which is why chemostat outlet substrate is often tiny.For an immobilised enzyme particle, a Damköhler number much greater than 1 indicates that the observed rate is
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Answer: A — limited by external mass transfer from the bulk to the surface
Da is the maximum reaction rate divided by the maximum film mass-transfer rate, V′max/(k_L·S_b). When it is large the enzyme could consume substrate far faster than the film can deliver it, so the surface concentration falls towards zero and transfer controls; faster stirring then helps. Da ≪ 1 is the reaction-limited case.A soluble enzyme with Km = 2 mM and Vmax = 1 mM/min acts on a substrate at S₀ = 10 mM in a batch reactor. The time required for 90% conversion is ______ min (round off to one decimal place).
Numerical answer — type the value.
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Answer: 13.6
Integrated Michaelis–Menten: Vmax·t = Km ln(S₀/S) + (S₀ − S) with S = 1 mM, so t = [2 ln 10 + 9]/1 = 4.61 + 9 = 13.6 min. Treating the reaction as zero order gives 9 min and as first order (k = Vmax/Km) gives 4.6 min — each keeps only one of the two terms.An enzyme with Vmax = 10 mM/h and Km = 5 mM is used in a CSTR fed with 50 mM substrate. The residence time required for 90% conversion is ______ h.
Numerical answer — type the value.
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Answer: 9
The outlet is S = 5 mM. The CSTR balance F(S₀ − S) = V·VmaxS/(Km + S) gives τ = (S₀ − S)(Km + S)/(VmaxS) = 45 × 10/(10 × 5) = 9 h. Evaluating the rate at the inlet concentration instead of the outlet is the classic CSTR error and gives far too short a time.A first-order reaction with k = 0.5 h⁻¹ is run at a residence time of 4 h. The fractional conversions in an ideal CSTR and an ideal PFR are, respectively,
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Answer: A — 0.667 and 0.865
kτ = 2. CSTR: X = kτ/(1 + kτ) = 2/3 = 0.667. PFR: X = 1 − e⁻² = 0.865. The PFR converts more at the same τ because, for a positive-order reaction, it keeps the concentration and rate high near the inlet; swapping the two is the tempting error.An enzyme immobilised in a slab-shaped gel follows first-order kinetics with Thiele modulus φ = 2. Using η = tanh φ/φ, the effectiveness factor is ______ (round off to three decimal places).
Numerical answer — type the value.
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Answer: 0.482
tanh 2 = 0.964, so η = 0.964/2 = 0.482: less than half the enzyme’s intrinsic capacity is used because substrate is consumed before reaching the slab’s centre. The large-φ approximation 1/φ gives 0.5 and would be accurate only for φ well above 3.A culture grows exponentially with a specific growth rate of 0.35 h⁻¹. Its doubling time is ______ h (round off to two decimal places).
Numerical answer — type the value.
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Answer: 1.98
t_d = ln 2/μ = 0.693/0.35 = 1.98 h. Taking t_d = 1/μ (2.86 h) omits the ln 2, and dividing by log₁₀ 2 instead of ln 2 confuses the natural-log definition of μ with the base-10 generation count.An organism has μmax = 0.5 h⁻¹ and Ks = 0.2 g/L. At a substrate concentration of 0.6 g/L its specific growth rate by the Monod model is ______ h⁻¹ (round off to three decimal places).
Numerical answer — type the value.
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Answer: 0.375
μ = μmax·S/(Ks + S) = 0.5 × 0.6/(0.2 + 0.6) = 0.30/0.80 = 0.375 h⁻¹. With S three times Ks the culture grows at three-quarters of μmax, the same shape as v = 0.75 Vmax at [S] = 3Km.A batch culture is inoculated at 0.6 g/L biomass with 30 g/L glucose. When the glucose has fallen to 2 g/L, the biomass concentration, for Yx/s = 0.45 g/g, is ______ g/L.
Numerical answer — type the value.
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Answer: 13.2
Substrate consumed = 30 − 2 = 28 g/L, biomass formed = 0.45 × 28 = 12.6 g/L, so X = 0.6 + 12.6 = 13.2 g/L. Forgetting the inoculum gives 12.6, and using the initial 30 g/L rather than the amount consumed gives 14.1.A chemostat is operated at D = 0.4 h⁻¹ with μmax = 0.8 h⁻¹, Ks = 0.5 g/L, S₀ = 20 g/L and Yx/s = 0.5 g/g (Monod kinetics, no maintenance). The steady-state biomass productivity is ______ g L⁻¹ h⁻¹.
Numerical answer — type the value.
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Answer: 3.9
At steady state μ = D, so S = KsD/(μmax − D) = 0.5 × 0.4/0.4 = 0.5 g/L. Then X = Yx/s(S₀ − S) = 0.5 × 19.5 = 9.75 g/L and productivity DX = 0.4 × 9.75 = 3.9 g L⁻¹ h⁻¹. Assuming all 20 g/L is consumed gives 4.0, and reporting X alone answers a different question.For the same chemostat (μmax = 0.8 h⁻¹, Ks = 0.5 g/L, S₀ = 20 g/L), the critical dilution rate above which washout occurs is ______ h⁻¹ (round off to two decimal places).
Numerical answer — type the value.
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Answer: 0.78
Washout begins when the highest growth rate the feed can support — Monod evaluated at S = S₀ — no longer matches D: Dc = μmax·S₀/(Ks + S₀) = 0.8 × 20/20.5 = 0.78 h⁻¹. Answering μmax = 0.8 ignores that the cells in the vessel never see more than S₀.In the Luedeking–Piret model qp = αμ + β, which of the following are correct?
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Answer: A — α is the growth-associated coefficient.; B — β is the non-growth-associated coefficient.; C — Penicillin production is largely non-growth-associated, so β dominates.
The αμ term scales with growth and the β term is produced by cells whether or not they grow; penicillin, a secondary metabolite made after growth slows, is the textbook β-dominated case. A purely growth-associated product has β = 0 and qp = αμ, so it is proportional to μ, not independent of it.In a gassing-out experiment on a cell-free vessel with C* = 8 mg/L, the dissolved oxygen reads 2 mg/L at t = 10 s and 6 mg/L at t = 40 s. The volumetric mass-transfer coefficient k_La is ______ h⁻¹ (round off to the nearest integer).
Numerical answer — type the value.
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Answer: 132
The driving forces are C* − C₁ = 6 and C* − C₂ = 2 mg/L, so ln[(C* − C₁)/(C* − C₂)] = ln 3 = 1.0986 = k_La × 30 s, giving k_La = 0.0366 s⁻¹, or × 3600 = 132 h⁻¹. Forgetting the conversion from s⁻¹ to h⁻¹ leaves 0.037; a linear slope (4 mg/L in 30 s divided by a driving force) has no meaning for this exponential approach.A fermenter has k_La = 100 h⁻¹ and C* = 7 mg/L, and the dissolved oxygen must be held at 2 mg/L. If the cells consume 200 mg O₂ per gram of biomass per hour, the maximum biomass concentration that oxygen transfer can support is ______ g/L.
Numerical answer — type the value.
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Answer: 2.5
OTR = k_La(C* − C_L) = 100 × (7 − 2) = 500 mg O₂ L⁻¹ h⁻¹; setting OTR = OUR = q_O₂·X gives X = 500/200 = 2.5 g/L. Using C* alone as the driving force (700) overestimates the capacity, because the culture must be kept above its critical oxygen level.A geometrically similar fermenter is scaled up from a 0.1 m to a 0.5 m impeller diameter in turbulent flow. The pilot runs at 300 rpm. To keep power per unit volume constant, the large impeller should run at about
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Answer: A — 103 rpm
With a constant power number, P ∝ N³D⁵ and V ∝ D³, so P/V ∝ N³D²; holding it constant gives N₂ = N₁(D₁/D₂)^(2/3) = 300 × (0.2)^(2/3) = 300 × 0.342 = 103 rpm. 60 rpm is the constant-tip-speed answer (N ∝ 1/D), 12 rpm is constant Reynolds number (N ∝ 1/D²), and 300 rpm would keep mixing time but need 25 times the power per unit volume.Which of the following are genuine advantages of fed-batch over batch operation?
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Answer: A — It avoids substrate inhibition by keeping substrate concentration low.; B — It limits overflow metabolism such as the Crabtree effect in yeast.; C — It allows much higher final cell densities.
Controlled feeding holds S low, which prevents inhibition, stops glucose-rich yeast from diverting carbon to ethanol and E. coli to acetate, and lets biomass accumulate to high densities. Volume is exactly what does not stay constant: fed-batch adds medium without withdrawal, so V rises through the run.A kinetic model that describes the cell by several intracellular pools, such as RNA, protein and a key enzyme, so that it can follow the lag phase after a substrate shift, is
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Answer: A — a structured model
Structured models give the biomass internal composition, and changes in that composition are what produce lags and transients. Unstructured models such as Monod treat biomass as a single lumped variable and describe balanced growth only; Luedeking–Piret is a product-formation relation, not a description of the cell.