GATE BT

Batch, fed-batch and continuous bioreactors, and scale-up

A bioreactor can be run as a batch, as a fed-batch or continuously (a chemostat). All three are described by the same mass balances on cells, substrate, product and oxygen; only the inflow and outflow terms change. The topic is mostly quantitative: you solve a question by writing the balance that fits the mode of operation and knowing the conditions behind each standard result. That covers steady states and washout in a chemostat, feed rates in fed-batch, $k_{L} a$ from oxygen data, how speed and power change on scale-up, and the residence time an enzyme reactor needs for a given conversion.

24 previous-year questions · 23 solved · 7 practice questions

Batch, fed-batch and continuous operation

Every bioreactor calculation starts from one balance on the liquid in the vessel: accumulation = in − out + generation − consumption. The three modes differ only in the flow terms. In a batch run nothing enters or leaves after inoculation (apart from air, exhaust gas and small additions of acid, base or antifoam), so the volume $V$ is constant while $X$ , $S$ and $P$ change with time. In fed-batch medium is added but nothing is withdrawn, so $V$ rises. In continuous operation medium enters and broth leaves at the same flow $F$ , so $V$ is constant and, after start-up, the culture settles to a steady state.

Conventions used throughout: concentrations in g/L (biomass $X$ as dry cell weight), time in h, specific growth rate $μ$ in h $^{- 1}$ , dilution rate $D = F / V$ in h $^{- 1}$ with $V$ the working (liquid) volume, yield $Y_{X / S} = - d X / d S$ in g cells per g substrate, $K_{s}$ in g/L, and natural logarithms. $S_{0}$ is the initial substrate concentration in a batch and the feed concentration in a chemostat or fed-batch. Oxygen concentrations are in mg/L or mmol/L (1 mmol O $_{2}$ = 32 mg). A few letters are reused and are defined where they appear: $P$ is product concentration in the balances, total pressure in the oxygen section and shaft power in the agitation section; $α$ is the recycle ratio under cell recycle and the growth-associated constant in product formation; $C$ alone is the recycle concentration factor, while $C_{L}$ and $C^{*}$ are dissolved-oxygen concentrations; $k_{d}$ is a first-order rate constant for cell death or for enzyme deactivation.

MONOD KINETICS

μ = \frac{μ _{m} S}{K _{s} + S}

$μ_{m}$ = maximum specific growth rate (h $^{- 1}$ ); $K_{s}$ = saturation constant, the substrate concentration at which $μ = μ_{m} /2$ (g/L). For $S ≫ K_{s}$ , $μ \approx μ_{m}$ . Every result below that contains $K_{s}$ assumes this form, with a single growth-limiting substrate.

Feature	Batch	Fed-batch	Continuous (chemostat)
Liquid flows during the run	None after inoculation	Feed in, nothing out	Feed in = broth out (flow $F$ )
Volume	Constant	Rises: $V = V_{0} + \int F d t$	Constant
Specific growth rate	Changes through the lag, exponential ( $μ \approx μ_{m}$ ), deceleration, stationary and death phases	Set by the feed: $μ \approx F / V$ at quasi-steady state, or held constant by exponential feeding	Chosen by the operator: $μ = D$ at steady state
Substrate concentration	Starts high, falls to near zero	Kept low by the feed rate	Constant and low, $\overset{ˉ}{S}$
Strengths	Simple; each run is a fresh, contained lot; suits products formed after growth stops	High cell density; avoids substrate inhibition, overflow metabolism and catabolite repression; caps O $_{2}$ demand and heat load	Usually the highest volumetric productivity for growth-associated products; steady product quality; growth studies at a chosen $μ$
Weaknesses	Down-time between runs lowers average productivity	Needs feed control; the run ends when the vessel is full	Contamination and strain-instability risk over long runs; washout at high $D$

EXPONENTIAL GROWTH AND DOUBLING TIME

X = X_{0} e^{μ_{m} t}, t_{d} = \frac{ln 2}{μ _{m}}

Exponential phase of a batch culture, where $S ≫ K_{s}$ so that $μ \approx μ_{m}$ ; no lag and no death.

BATCH YIELD BALANCE

X - X_{0} = Y_{X / S} (S_{0} - S)

Constant yield, substrate used only for growth. When the substrate is exhausted ( $S = 0$ ) the biomass reaches its maximum, $X_{m} = X_{0} + Y_{X / S} S_{0}$ .

BATCH GROWTH TIME

t_{b} = \frac{1}{μ _{m}} ln \frac{X _{m}}{X _{0}}

Time to grow from $X_{0}$ to $X_{m}$ if the cells grow at $μ_{m}$ the whole time ( $K_{s}$ negligible, lag ignored).

BATCH TIME WITH MONOD KINETICS (INTEGRATED)

μ_{m} t = (1 + \frac{K _{s} Y _{X / S}}{X _{0} + Y _{X / S} S _{0}}) ln \frac{X}{X _{0}} + \frac{K _{s} Y _{X / S}}{X _{0} + Y _{X / S} S _{0}} ln \frac{S _{0}}{S}

Exact result with Monod kinetics and constant yield, with no lag and no death; $X$ and $S$ are linked by the yield balance. When $K_{s} Y_{X / S}$ is negligible beside $X_{0} + Y_{X / S} S_{0}$ it reduces to $μ_{m} t = ln (X / X_{0})$ , the exponential-growth result. With $K_{s} > 0$ , reaching exactly $S = 0$ takes infinite time, so the formula is used for a stated final $S$ or $X$ .

BATCH PRODUCTIVITY

Q_{X, b} = \frac{X _{m} - X _{0}}{t _{b} + t _{L}}

Averaged over the whole cycle. $t_{L}$ = turnaround time per cycle: harvesting, cleaning, sterilising, filling and lag.

EXAM TIP

Batch productivity is averaged over the WHOLE cycle, including the turnaround time $t_{L}$ ; leaving $t_{L}$ out overstates it. The log in $t_{b} = (1/ μ_{m}) ln (X_{m} / X_{0})$ is the natural log. Using $lo g_{10}$ gives a time 2.303 times too small.

The chemostat: balances, washout and optimum dilution rate

A chemostat is a well-mixed vessel fed with sterile medium that contains the limiting substrate at $S_{0}$ , at flow rate $F$ . Broth is removed at the same $F$ , so $V$ stays constant. Because the contents are well mixed, the outflow carries the reactor concentrations $X$ , $S$ and $P$ . Dividing the balances by $V$ brings in the dilution rate $D = F / V$ , whose reciprocal is the mean residence time.

CHEMOSTAT BIOMASS BALANCE

\frac{d X}{d t} = μ X - D X

No cells in the feed (sterile), cell death neglected. At steady state with $X \neq = 0$ this gives $μ = D$ : the feed rate sets the growth rate.

CHEMOSTAT SUBSTRATE BALANCE

\frac{d S}{d t} = D (S_{0} - S) - \frac{μ X}{Y _{X / S}}

Substrate used only for growth (no maintenance term), with $Y_{X / S}$ constant.

STEADY-STATE SUBSTRATE CONCENTRATION

\overset{ˉ}{S} = \frac{K _{s} D}{μ _{m} - D}

Monod kinetics with $μ = D$ . It does not depend on $S_{0}$ and holds only for $D < D_{c}$ .

STEADY-STATE BIOMASS CONCENTRATION

\overset{ˉ}{X} = Y_{X / S} (S_{0} - \overset{ˉ}{S}) = Y_{X / S} (S_{0} - \frac{K _{s} D}{μ _{m} - D})

From the substrate balance at steady state. Proportional to $Y_{X / S}$ ; it rises linearly with $S_{0}$ , and nearly in proportion to it when $S_{0} ≫ \overset{ˉ}{S}$ .

WASHOUT (CRITICAL) DILUTION RATE

D_{c} = \frac{μ _{m} S _{0}}{K _{s} + S _{0}} \approx μ_{m} (S_{0} ≫ K_{s})

Found by setting $\overset{ˉ}{S} = S_{0}$ (so $\overset{ˉ}{X} = 0$ ). For $D \geq D_{c}$ cells leave faster than they can grow and the culture washes out. Without recycle, $D_{c} < μ_{m}$ for any finite $S_{0}$ .

FINDING THE MONOD CONSTANTS FROM CHEMOSTAT DATA

\frac{1}{D} = \frac{1}{μ _{m}} + \frac{K _{s}}{μ _{m}} \frac{1}{S ˉ}

Monod kinetics with $μ = D$ at each steady state. Measure $\overset{ˉ}{S}$ at several dilution rates: a plot of $1/ D$ against $1/ \overset{ˉ}{S}$ is a straight line with intercept $1/ μ_{m}$ and slope $K_{s} / μ_{m}$ .

VOLUMETRIC BIOMASS PRODUCTIVITY

Q_{X} = D \overset{ˉ}{X}

Grams of cells per litre of reactor per hour. For a product, $Q_{P} = D \overset{ˉ}{P}$ .

DILUTION RATE FOR MAXIMUM PRODUCTIVITY

D_{o pt} = μ_{m} (1 - \frac{K _{s}}{K _{s} + S _{0}})

From $d (D \overset{ˉ}{X}) / d D = 0$ with Monod kinetics, no maintenance and no death. It lies below $D_{c}$ , and close to it when $S_{0} ≫ K_{s}$ .

MAXIMUM CHEMOSTAT PRODUCTIVITY

(D \overset{ˉ}{X})_{ma x} = Y_{X / S} μ_{m} (S_{0} + K_{s} - K_{s})^{2}

Productivity at $D_{o pt}$ . The substrate concentration there is $\overset{ˉ}{S} = K_{s} (K_{s} + S_{0}) - K_{s}$ .

How the steady state moves with $D$ : at low $D$ , $\overset{ˉ}{S}$ is small and $\overset{ˉ}{X} \approx Y_{X / S} S_{0}$ . As $D$ approaches $D_{c}$ , $\overset{ˉ}{S}$ rises steeply toward $S_{0}$ and $\overset{ˉ}{X}$ falls to zero. The productivity $D \overset{ˉ}{X}$ rises almost linearly, peaks at $D_{o pt}$ and drops to zero at $D_{c}$ . Because $D_{o pt}$ lies close to $D_{c}$ , where $\overset{ˉ}{X}$ changes steeply with $D$ , a small rise in $F$ or a small fall in $μ_{m}$ (from a change in temperature or pH) can push a culture run at $D_{o pt}$ toward washout, so in practice a chemostat is run a little below $D_{o pt}$ . When maintenance or cell death is included, $\overset{ˉ}{X}$ also falls at very low $D$ .

CONTINUOUS VS BATCH PRODUCTIVITY (ESTIMATE)

\frac{( D X ˉ ) _{ma x}}{Q _{X, b}} \approx ln \frac{X _{m}}{X _{0}} + μ_{m} t_{L}

Assumes $D_{o pt} \approx μ_{m}$ and $\overset{ˉ}{X} \approx X_{m}$ (both need $S_{0} ≫ K_{s}$ ), $X_{m} ≫ X_{0}$ , and batch growth at $μ_{m}$ . The ratio is usually well above 1. That is why continuous culture gives higher volumetric productivity for growth-associated products, as long as sterility and strain stability can be maintained over a long run.

WORKED EXAMPLE · CHEMOSTAT STEADY STATE AND OPTIMUM

A 2.0 L chemostat is fed sterile medium with $S_{0} = 20$ g/L of glucose at 0.60 L/h. The organism follows Monod kinetics with $μ_{m} = 0.50$ h $^{- 1}$ , $K_{s} = 0.20$ g/L and $Y_{X / S} = 0.50$ g/g (no maintenance, no death). Find (a) the steady-state $\overset{ˉ}{S}$ and $\overset{ˉ}{X}$ , (b) the biomass productivity, (c) the washout dilution rate, and (d) the maximum possible productivity and the $D$ at which it occurs.

Solution

$D = F / V = 0.60/2.0 = 0.30$ h $^{- 1}$ . This is below $μ_{m}$ , so a steady state is possible (confirmed in step 5).
$\overset{ˉ}{S} = K_{s} D / (μ_{m} - D) = (0.20) (0.30) / (0.50 - 0.30) = 0.30$ g/L.
$\overset{ˉ}{X} = Y_{X / S} (S_{0} - \overset{ˉ}{S}) = 0.50 (20 - 0.30) = 9.85$ g/L.
$Q_{X} = D \overset{ˉ}{X} = 0.30 \times 9.85 = 2.96$ g/(L h).
$D_{c} = μ_{m} S_{0} / (K_{s} + S_{0}) = (0.50) (20) /20.2 = 0.495$ h $^{- 1}$ , so $D = 0.30$ h $^{- 1}$ is safely below washout.
$K_{s} / (K_{s} + S_{0}) = 0.20/20.2 = 0.0995$ , so $D_{o pt} = 0.50 (1 - 0.0995) = 0.4502 \approx 0.450$ h $^{- 1}$ .
$(D \overset{ˉ}{X})_{ma x} = Y_{X / S} μ_{m} (S_{0} + K_{s} - K_{s})^{2} = (0.50) (0.50) (4.494 - 0.447)^{2} = 0.25 \times 16.38 = 4.10$ g/(L h).
Check: at $D = 0.4502$ h $^{- 1}$ , $μ_{m} - D = 0.0498$ h $^{- 1}$ , so $\overset{ˉ}{S} = (0.20) (0.4502) /0.0498 = 1.81$ g/L, $\overset{ˉ}{X} = 0.50 (20 - 1.81) = 9.10$ g/L and $D \overset{ˉ}{X} = 0.4502 \times 9.10 = 4.10$ g/(L h).

Answer: $\overset{ˉ}{S} = 0.30$ g/L, $\overset{ˉ}{X} = 9.85$ g/L, $Q_{X} \approx 2.96$ g/(L h), $D_{c} \approx 0.495$ h $^{- 1}$ , $D_{o pt} \approx 0.450$ h $^{- 1}$ with $(D \overset{ˉ}{X})_{ma x} \approx 4.10$ g/(L h).

EXAM TIP

Chemostat traps: (1) $\overset{ˉ}{S}$ has no $S_{0}$ in it. A stronger feed raises $\overset{ˉ}{X}$ , not $\overset{ˉ}{S}$ (and raises $D_{c}$ slightly). (2) $D$ uses the working liquid volume, not the vessel's total volume. (3) Check $D < D_{c}$ before using the steady-state formulas. If $D \geq D_{c}$ the answer is washout ( $\overset{ˉ}{X} = 0$ , $\overset{ˉ}{S} = S_{0}$ ), not a negative concentration. (4) Keep units consistent: if $F$ is in L/min, $D$ is in min $^{- 1}$ and must match the units of $μ_{m}$ .

Chemostat variants: cell recycle, maintenance, products, stages and the turbidostat

Cell recycle. A separator placed after the reactor (a settler, a centrifuge or a membrane) returns part of the cells to the vessel. Convention: the recycle ratio $α$ is the recycle flow divided by the fresh feed flow $F$ . The concentration factor $C$ is the cell concentration in the recycle stream divided by $X$ in the reactor ( $C > 1$ ). The separator is assumed not to change the substrate concentration, the fresh feed is sterile, and $D = F / V$ is based on the fresh feed.

GROWTH RATE WITH CELL RECYCLE

μ = D (1 + α - α C)

From the steady-state biomass balance on the reactor (no death). A balance on the separator shows that $1 + α - α C = X_{e} / \overset{ˉ}{X}$ , where $X_{e}$ is the cell concentration in the effluent that leaves the system, so the factor lies between 0 and 1: cells grow more slowly than the liquid is diluted.

STEADY STATE WITH CELL RECYCLE

\overset{ˉ}{S} = \frac{K _{s} D ( 1 + α - α C )}{μ _{m} - D ( 1 + α - α C )}, \overset{ˉ}{X} = \frac{Y _{X / S} ( S _{0} - S ˉ )}{1 + α - α C}

Monod kinetics. Same form as the simple chemostat, with $D$ replaced by $D (1 + α - α C)$ in $\overset{ˉ}{S}$ . $\overset{ˉ}{X}$ is raised by the factor $1/ (1 + α - α C)$ .

WASHOUT WITH CELL RECYCLE

D_{c} = \frac{μ _{m} S _{0}}{( K _{s} + S _{0} ) ( 1 + α - α C )}

Recycle raises the washout dilution rate, so with enough recycle ( $α (C - 1)$ close to 1) $D$ can exceed $μ_{m}$ without washout.

Effect of recycle: at a given $D$ the reactor holds more cells and leaves less substrate unconverted. It can also run at a higher dilution rate, which raises volumetric productivity. The net biomass leaving the system per unit reactor volume per hour is $D (1 + α - α C) \overset{ˉ}{X} = μ \overset{ˉ}{X}$ . The same idea underlies activated-sludge treatment and perfusion culture of animal cells.

MAINTENANCE (PIRT) RELATION

\frac{1}{Y _{X / S}^{o b s}} = \frac{1}{Y _{G}} + \frac{m _{s}}{μ}

$Y_{G}$ = true growth yield; $m_{s}$ = maintenance coefficient (g substrate per g cells per h). In a chemostat $μ = D$ and $Y_{X / S}^{o b s} = \overset{ˉ}{X} / (S_{0} - \overset{ˉ}{S})$ , so a plot of $1/ Y^{o b s}$ against $1/ D$ is a straight line with slope $m_{s}$ and intercept $1/ Y_{G}$ .

STEADY-STATE BIOMASS WITH MAINTENANCE

\overset{ˉ}{X} = \frac{D Y _{G} ( S _{0} - S ˉ )}{D + m _{s} Y _{G}}

Applies when substrate is also consumed for maintenance ( $μ = D$ , no death). $\overset{ˉ}{S}$ is still $K_{s} D / (μ_{m} - D)$ . $\overset{ˉ}{X}$ falls as $D \to 0$ .

PRODUCT IN A CHEMOSTAT

D \overset{ˉ}{P} = q_{P} \overset{ˉ}{X}, q_{P} = α μ + β (μ = D)

Product-free feed, no product degradation. $q_{P}$ = specific product formation rate (g product per g cells per h), written in the Luedeking–Piret form with growth-associated $α$ and non-growth-associated $β$ (this $α$ is unrelated to the recycle ratio). So $\overset{ˉ}{P} = α \overset{ˉ}{X} + β \overset{ˉ}{X} / D$ .

CHEMOSTAT WITH CELL DEATH

μ = D + k_{d}, \overset{ˉ}{S} = \frac{K _{s} ( D + k _{d} )}{μ _{m} - D - k _{d}}, \overset{ˉ}{X} = \frac{D Y _{X / S} ( S _{0} - S ˉ )}{D + k _{d}}

First-order death at rate $k_{d}$ (h $^{- 1}$ ) acting on the viable cells $X$ ; dead cells take up no substrate, and $Y_{X / S}$ is the growth yield. Washout now occurs at $D_{c} = μ_{m} S_{0} / (K_{s} + S_{0}) - k_{d}$ .

SECOND STAGE OF TWO CHEMOSTATS IN SERIES

μ_{2} = D (1 - \frac{X _{1}}{X _{2}}), D (S_{1} - S_{2}) = \frac{μ _{2} X _{2}}{Y _{X / S}}

Equal volumes, all fresh feed to stage 1, $D = F / V$ for each stage, no extra cells or substrate added to stage 2. Stage 1 behaves as a simple chemostat. Combining the two equations gives $X_{2} = X_{1} + Y_{X / S} (S_{1} - S_{2})$ , and with Monod kinetics $μ_{2} = μ_{m} S_{2} / (K_{s} + S_{2})$ closes the set. Stage 2 grows more slowly ( $μ_{2} < D$ ), which suits products made at low growth rates.

Turbidostat. Here the biomass concentration (read as optical density) is held at a set value, and a controller varies $F$ to keep it there. The steady-state balances are the same ( $μ = D$ ), but $X$ is the set variable and $D$ follows from it. A turbidostat is stable at high dilution rates close to $μ_{m}$ , the region near $D_{c}$ where a chemostat drifts toward washout. It is used to keep cells growing near their maximum rate or to select fast-growing variants. It needs a reliable online biomass sensor, which is hard to achieve at high cell density or with wall growth.

Chemostat vs turbidostat

Feature	Chemostat	Turbidostat
Set by the operator	Feed rate $F$ , and so $D$	Biomass concentration $X$ (via optical density)
Control	No feedback loop: a constant-rate pump sets $F$ , and the limiting substrate regulates growth by itself	Feedback loop that adjusts $F$
Stable operating range	Low to moderate $D$	High $D$ , close to $μ_{m}$
What limits growth	A limiting nutrient at low $\overset{ˉ}{S}$	Nutrients in excess; $μ$ near $μ_{m}$

Fed-batch operation

A fed-batch run starts as a batch. Once the initial substrate is nearly used up, a concentrated feed ( $S_{0}$ , far higher than any concentration in the reactor) is added at a controlled rate $F (t)$ , with no outflow. The reasons: keep $S$ low to avoid substrate inhibition, overflow metabolism (acetate in E. coli, ethanol from the Crabtree effect in baker's yeast) and catabolite repression; hold the oxygen demand and heat load to what the vessel can supply; and reach high cell densities.

VOLUME IN FED-BATCH

\frac{d V}{d t} = F, V = V_{0} + F t (constant F)

Liquid is only added, never withdrawn.

FED-BATCH BALANCES ON TOTAL AMOUNTS

\frac{d ( X V )}{d t} = μ X V, \frac{d ( S V )}{d t} = F S_{0} - \frac{μ X V}{Y _{X / S}}

No biomass in the feed, nothing withdrawn, substrate used only for growth. Write fed-batch balances on total amounts ( $X V$ , $S V$ ), because $V$ changes.

FED-BATCH BALANCES AS CONCENTRATIONS

\frac{d X}{d t} = (μ - D) X, \frac{d S}{d t} = D (S_{0} - S) - \frac{μ X}{Y _{X / S}}, D = \frac{F}{V ( t )}

Obtained from the total-amount balances using $d V / d t = F$ . They look like the chemostat equations, but here $D = F / V (t)$ changes with time; under constant feed it falls as $V$ grows.

TOTAL BIOMASS UNDER CONSTANT FEED

\frac{d ( X V )}{d t} = F Y_{X / S} S_{0} \Rightarrow X_{t} = X_{t 0} + F Y_{X / S} S_{0} t

Holds once substrate is used as fast as it is fed ( $S$ small and steady, $S ≪ S_{0}$ ). $X_{t} = X V$ is the total mass of cells, and under constant feed it rises linearly, not exponentially.

SPECIFIC GROWTH RATE UNDER CONSTANT FEED

μ = \frac{F Y _{X / S} S _{0}}{X V}

Applies when all the fed substrate is consumed as it arrives. It falls continuously as the total biomass $X V$ grows.

QUASI-STEADY STATE (CONSTANT FEED)

μ \approx D = \frac{F}{V _{0} + F t}, X \approx Y_{X / S} S_{0}

Approached as the run proceeds: with constant feed $X V = X_{t 0} + F Y_{X / S} S_{0} t$ , so $X$ tends to $Y_{X / S} S_{0}$ , after which $d X / d t \approx 0$ and $d S / d t \approx 0$ . Because $V$ keeps rising, $D$ and so $μ$ fall with time, and the residual $S \approx K_{s} D / (μ_{m} - D)$ falls too.

EXPONENTIAL FEEDING FOR CONSTANT GROWTH RATE

F (t) = \frac{μ _{se t} X _{0} V _{0}}{Y _{X / S} ( S _{0} - S )} e^{μ_{se t} t}

Holds $μ$ at a chosen $μ_{se t} < μ_{m}$ , with $S$ kept constant at $K_{s} μ_{se t} / (μ_{m} - μ_{se t})$ . $X_{0} V_{0}$ is the total biomass when exponential feeding starts ( $t = 0$ ). It follows from $X V = X_{0} V_{0} e^{μ_{se t} t}$ and the substrate balance with $S$ constant. If maintenance matters, replace $μ_{se t} / Y_{X / S}$ by $μ_{se t} / Y_{G} + m_{s}$ . If $S ≪ S_{0}$ , use $S_{0}$ in the denominator.

PRODUCT BALANCE IN FED-BATCH

\frac{d ( P V )}{d t} = q_{P} X V

No product in the feed, no outflow.

Feeding strategies

Constant feed: the simplest. Once the quasi-steady state is reached, $μ \approx F / V$ falls as $V$ grows, and total biomass rises linearly.
Linearly increasing or stepped feed: partly keeps up with the growing biomass.
Exponential feed: holds $μ$ constant, typically below the growth rate at which overflow metabolism starts.
Feedback feeding (DO-stat, pH-stat): feed is added when the dissolved oxygen rises or the pH shifts, both signs that the substrate has run out.

WORKED EXAMPLE · CONSTANT-FEED FED-BATCH

A fed-batch culture holds 2.0 L of broth with 20 g/L biomass when constant feeding starts: $F = 0.10$ L/h of medium with $S_{0} = 300$ g/L glucose, $Y_{X / S} = 0.50$ g/g. Glucose is consumed as fast as it is fed ( $S \approx 0$ ). After 10 h, find the volume, the total biomass, the biomass concentration, $μ$ and $D$ .

Solution

$V = V_{0} + F t = 2.0 + 0.10 \times 10 = 3.0$ L.
Biomass formed per hour $= F Y_{X / S} S_{0} = 0.10 \times 0.50 \times 300 = 15$ g/h.
Initial total biomass $= 20 \times 2.0 = 40$ g, so $X_{t} = 40 + 15 \times 10 = 190$ g and $X = 190/3.0 = 63.3$ g/L.
$μ = (d (X V) / d t) / (X V) = 15/190 = 0.079$ h $^{- 1}$ . At the start it was $15/40 = 0.375$ h $^{- 1}$ , so $μ$ falls continuously under constant feed. (The assumption $S \approx 0$ needs $μ_{m}$ above 0.375 h $^{- 1}$ ; otherwise glucose would accumulate at the start.)
$D = F / V = 0.10/3.0 = 0.033$ h $^{- 1}$ . Here $μ > D$ , so $X$ is still rising toward $Y_{X / S} S_{0} = 150$ g/L, where $μ$ would equal $D$ (the true quasi-steady state).

Answer: $V = 3.0$ L, $X_{t} = 190$ g, $X \approx 63.3$ g/L, $μ \approx 0.079$ h $^{- 1}$ , $D \approx 0.033$ h $^{- 1}$ .

EXAM TIP

Under constant feed the specific growth rate is NOT constant: at quasi-steady state $μ = F / V$ , which falls as $V$ grows. Only exponential feeding holds $μ$ fixed. Do not confuse the concentration $X$ (g/L) with the total biomass $X V$ (g): the linear-growth law is for $X V$ .

Oxygen transfer: Henry's law, OTR, OUR and kLa

Oxygen dissolves poorly in water, so in aerobic cultures it is usually the first nutrient to become limiting. On its way from bubble to cell it meets several resistances in series. For a sparingly soluble gas the liquid film around the bubble dominates, so the transfer rate is written with the liquid-side volumetric coefficient $k_{L} a$ : $k_{L}$ (m/s) times the interfacial area per unit liquid volume $a$ (m $^{2}$ /m $^{3}$ ). $k_{L} a$ therefore has units of s $^{- 1}$ or h $^{- 1}$ .

HENRY'S LAW

p_{O_{2}} = H C^{*}, C^{*} = \frac{y _{O_{2}} P}{H}

$C^{*}$ = dissolved O $_{2}$ concentration in equilibrium with the gas; $H$ = Henry's constant for O $_{2}$ in the medium at the operating temperature (pressure per concentration, for example atm L/mmol); $y_{O_{2}}$ = mole fraction of O $_{2}$ in the gas (0.21 for air); $P$ = total pressure. $C^{*}$ rises in proportion to the O $_{2}$ partial pressure (O $_{2}$ -enriched air, higher head pressure). It falls with temperature and with dissolved salts and sugars. Air-saturated pure water at 25 °C and 1 atm holds only about 8 mg/L (about 0.26 mmol/L), and less at higher temperatures.

OXYGEN TRANSFER RATE (OTR)

OTR = k_{L} a (C^{*} - C_{L})

$C_{L}$ = dissolved O $_{2}$ in the broth. With $C$ in mmol/L and $k_{L} a$ in h $^{- 1}$ , OTR is in mmol/(L h). The maximum OTR is $k_{L} a C^{*}$ , reached when $C_{L} = 0$ .

OXYGEN UPTAKE RATE (OUR)

OUR = q_{O_{2}} X

$q_{O_{2}}$ = specific O $_{2}$ uptake rate (mmol O $_{2}$ per g cells per h). It does not depend on $C_{L}$ above the critical dissolved-oxygen concentration $C_{cr i t}$ , and falls below it.

DISSOLVED-OXYGEN BALANCE

\frac{d C _{L}}{d t} = k_{L} a (C^{*} - C_{L}) - q_{O_{2}} X

Well-mixed liquid. At steady state OTR = OUR.

OXYGEN-LIMITED DESIGN

(k_{L} a)_{min} = \frac{q _{O_{2}} X}{C ^{*} - C _{cr i t}}, X_{ma x} = \frac{k _{L} a ( C ^{*} - C _{cr i t} )}{q _{O_{2}}}

The smallest $k_{L} a$ that keeps $C_{L} \geq C_{cr i t}$ at a given cell density, or the largest cell density a given $k_{L} a$ can support.

Dynamic gassing-out method (with respiring cells)

With the culture at a steady dissolved-oxygen (DO) level, switch the air off. DO falls in a straight line with $d C_{L} / d t = - q_{O_{2}} X$ (OUR constant while $C_{L} > C_{cr i t}$ ; surface aeration neglected), so the slope gives OUR.
Switch the air back on before DO falls below $C_{cr i t}$ , so that respiration is not disturbed.
While DO recovers, at every point $d C_{L} / d t = k_{L} a (C^{*} - C_{L}) - q_{O_{2}} X$ .
Rearrange: $C_{L} = C^{*} - \frac{1}{k _{L} a} (\frac{d C _{L}}{d t} + q_{O_{2}} X)$ . A plot of $C_{L}$ against $(d C_{L} / d t + q_{O_{2}} X)$ is a straight line with slope $- 1/ k_{L} a$ and intercept $C^{*}$ .
The method assumes the DO probe responds much faster than $1/ k_{L} a$ .

STATIC GASSING-OUT (NO OXYGEN UPTAKE)

ln \frac{C ^{*} - C _{L 0}}{C ^{*} - C _{L}} = k_{L} a t

Cell-free medium: strip the O $_{2}$ with N $_{2}$ , then aerate starting from $C_{L 0}$ . The slope of $ln (C^{*} - C_{L})$ against $t$ is $- k_{L} a$ . Valid when the DO probe responds much faster than $1/ k_{L} a$ .

Sulphite oxidation. In $SO_{3}^{2 -} + \frac{1}{2} O_{2} \to SO_{4}^{2 -}$ , catalysed by Cu $^{2 +}$ or Co $^{2 +}$ , the reaction is fast enough to keep dissolved O $_{2}$ at essentially zero, so OTR $= k_{L} a C^{*}$ . The remaining sulphite is titrated (iodometrically) at intervals, and the O $_{2}$ transfer rate is half the sulphite consumption rate. The test is run in a sulphite solution, not in broth. Its ionic strength changes bubble coalescence, so the $k_{L} a$ found can differ considerably from the value in the real medium.

KLA FROM SULPHITE OXIDATION

k_{L} a = \frac{OTR}{C ^{*}} = \frac{\frac{1}{2} ( - d [ SO _{3}^{2 -} ] / d t )}{C ^{*}}

2 mol of sulphite consume 1 mol of O $_{2}$ , and $C_{L} \approx 0$ throughout.

OUR FROM AN EXHAUST-GAS BALANCE

OUR = \frac{Q _{g}}{V} \frac{P}{R T} (y_{O_{2}, in} - y_{O_{2}, out})

Steady-state gas-phase balance: $Q_{g}$ = gas volumetric flow at pressure $P$ and temperature $T$ ; $R$ = gas constant; $V$ = liquid volume. It neglects the small change in total molar gas flow (an inert-gas balance corrects for it). Then $k_{L} a = OUR / (C^{*} - C_{L})$ , with $C_{L}$ read from the DO probe.

WORKED EXAMPLE · DYNAMIC GASSING-OUT

A culture with $X = 5.0$ g/L sits at a steady dissolved-oxygen level of 5.0 mg/L. The air is switched off and the DO falls linearly from 5.0 to 3.0 mg/L in 40 s. The saturation concentration in the medium is $C^{*} = 7.0$ mg/L. Find (a) the OUR and $q_{O_{2}}$ , (b) $k_{L} a$ , and (c) the highest cell concentration this vessel could support at the same $k_{L} a$ if DO must stay above a critical value of 1.0 mg/L (assume $q_{O_{2}}$ does not change).

Solution

With the air off, $d C_{L} / d t = - q_{O_{2}} X$ , so OUR $= (5.0 - 3.0) /40 = 0.050$ mg/(L s) $= 0.050 \times 3600 = 180$ mg O $_{2}$ /(L h), which is $180/32 = 5.63$ mmol/(L h).
$q_{O_{2}} = OUR / X = 180/5.0 = 36$ mg O $_{2}$ /(g h), or 1.13 mmol/(g h).
Before the air was switched off the DO was steady, so OTR = OUR: $k_{L} a (7.0 - 5.0) = 180$ , giving $k_{L} a = 90$ h $^{- 1}$ ( $= 0.025$ s $^{- 1}$ ).
$X_{ma x} = k_{L} a (C^{*} - C_{cr i t}) / q_{O_{2}} = 90 (7.0 - 1.0) /36 = 15$ g/L.

Answer: OUR = 180 mg O $_{2}$ /(L h), $q_{O_{2}}$ = 36 mg O $_{2}$ /(g h), $k_{L} a$ = 90 h $^{- 1}$ , $X_{ma x}$ = 15 g/L.

EXAM TIP

(1) While the air is off, the slope is $- q_{O_{2}} X$ (the OUR), with no $k_{L} a$ in it. $k_{L} a$ comes only from the re-aeration curve or from a steady-state balance. (2) In sulphite oxidation, 2 mol of sulphite consume 1 mol of O $_{2}$ ; forgetting the ½ doubles $k_{L} a$ . (3) $k_{L} a$ in s $^{- 1}$ × 3600 gives h $^{- 1}$ ; O $_{2}$ in mg/L ÷ 32 gives mmol/L. (4) Raising $C^{*}$ (O $_{2}$ -enriched air, higher back-pressure) increases the driving force $C^{*} - C_{L}$ without, to a first approximation, changing $k_{L} a$ .

Agitation power and scale-up

Agitation mixes the broth, breaks gas into small bubbles (raising $a$ and so $k_{L} a$ ), keeps cells suspended and helps heat transfer. Take a stirred tank with impeller diameter $D_{i}$ turning at $N$ (rev/s), holding a liquid of density $ρ$ and viscosity $μ_{L}$ (the subscript avoids a clash with the growth rate). Two dimensionless groups govern the power it draws:

IMPELLER REYNOLDS NUMBER

Re_{i} = \frac{ρN D _{i}^{2}}{μ _{L}}

$N$ in rev/s (not rad/s, not rpm). In a baffled tank, flow is laminar below about 10 and fully turbulent above about $1 0^{4}$ .

POWER NUMBER

N_{P} = \frac{P}{ρ N ^{3} D _{i}^{5}}

$P$ = ungassed shaft power (W, with $ρ$ in kg/m $^{3}$ , $N$ in rev/s and $D_{i}$ in m). For a given impeller and baffled-tank geometry, $N_{P}$ depends only on $Re_{i}$ .

LAMINAR REGIME

N_{P} = \frac{k _{1}}{Re _{i}} \Rightarrow P = k_{1} μ_{L} N^{2} D_{i}^{3}

Power does not depend on density ( $k_{1}$ is a geometry constant).

TURBULENT REGIME

N_{P} = constant \Rightarrow P = N_{P} ρ N^{3} D_{i}^{5}

Fully turbulent, baffled tank: power does not depend on viscosity. For a six-blade Rushton turbine $N_{P}$ is about 5 to 6. Doubling $N$ multiplies $P$ by 8; doubling $D_{i}$ multiplies it by 32.

GASSED POWER

P_{g} = c (\frac{P ^{2} N D _{i}^{3}}{Q _{g}^{0.56}})^{0.45}

Empirical correlation (Michel–Miller form) for the gassed power $P_{g}$ ; $Q_{g}$ is the gas volumetric flow and $c$ depends on the geometry and on the units used. Gas collects in cavities behind the blades and lowers their drag, so $P_{g} < P$ at the same $N$ . $P_{g} / P$ falls as the aeration number $N_{a} = Q_{g} / (N D_{i}^{3})$ rises, until the impeller floods: gas then rises past the impeller without being dispersed.

SUPERFICIAL GAS VELOCITY, AERATION RATE AND TIP SPEED

u_{s} = \frac{Q _{g}}{π D _{T}^{2} /4}, vvm = \frac{Q _{g}}{V}, tip speed = π N D_{i}

$D_{T}$ = tank diameter; vvm = volume of gas per volume of liquid per minute, so $Q_{g}$ in the vvm expression is per minute. Tip speed is a common index of the highest shear, which occurs near the impeller.

KLA CORRELATION (GENERAL FORM)

k_{L} a = K (\frac{P _{g}}{V})^{a} u_{s}^{b}

Empirical: $K$ , $a$ and $b$ depend on geometry and on whether the medium is coalescing. Holding $P_{g} / V$ and $u_{s}$ constant on scale-up is a common way to aim for constant $k_{L} a$ .

MIXING TIME, TURBULENT REGIME

N t_{m} = constant

Geometrically similar tanks, fully turbulent: mixing time is inversely proportional to impeller speed. Circulation time scales as $V / (N D_{i}^{3})$ .

Scale-up. Under geometric similarity every length scales as $L_{2} / L_{1} = (V_{2} / V_{1})^{1/3}$ , so $D_{i}$ and $D_{T}$ grow as the cube root of volume. In the turbulent regime $P \propto N^{3} D_{i}^{5}$ and $V \propto D_{i}^{3}$ . Each scale-up criterion holds one quantity fixed, and that fixes how $N$ must change:

GENERAL SCALE-UP RULE FOR IMPELLER SPEED

\frac{N _{2}}{N _{1}} = (\frac{D _{i 1}}{D _{i 2}})^{n}

$n = 2/3$ for constant $P / V$ ; $n = 1$ for constant tip speed ( $N D_{i}$ ); $n = 2$ for constant $Re_{i}$ ; $n = 0$ for constant mixing time (constant $N$ ). Turbulent regime at both scales, geometric similarity, ungassed power.

CONSTANT POWER PER UNIT VOLUME

\frac{P}{V} \propto N^{3} D_{i}^{2} \Rightarrow N_{2} = N_{1} (\frac{D _{i 1}}{D _{i 2}})^{2/3}

Turbulent regime, geometric similarity. Tip speed then rises as $D_{i}^{1/3}$ and mixing time as $D_{i}^{2/3}$ .

Scale-up by 5 times in impeller diameter (125 times in volume), fully turbulent at both scales

Held constant	$N_{2} / N_{1}$	$P_{2} / P_{1}$	$(P / V)_{2} / (P / V)_{1}$	Tip speed ratio	$Re_{i}$ ratio	Mixing time ratio
$P / V$	0.342	125	1	1.71	8.55	2.92
$N$ (mixing time)	1	3125	25	5	25	1
Tip speed $N D_{i}$	0.200	25	0.200	1	5	5
$Re_{i}$	0.0400	0.200	0.0016	0.200	1	25

Scale-up trade-offs

No single criterion keeps everything the same, so scale-up is always a compromise.
Constant $P / V$ (a common starting point for aerobic cultures, together with constant $u_{s}$ to hold $k_{L} a$ ): tip speed rises and mixing time lengthens, so a large vessel has gradients of substrate, pH and DO that the small one did not.
Constant tip speed protects shear-sensitive cells but cuts $P / V$ as $1/ D_{i}$ , which lowers $k_{L} a$ .
Constant mixing time would need $P / V$ to rise as $D_{i}^{2}$ , which is impractical at production scale.
Constant $Re_{i}$ makes $P / V$ collapse (as $D_{i}^{- 4}$ ); it is not used for aerobic cultures.
Heat: the surface-to-volume ratio falls as $1/ D_{T}$ while heat generation scales with $V$ , so a jacket alone stops being enough and internal coils are needed.
Gas: at constant vvm, $u_{s}$ rises in proportion to $D_{T}$ (flooding risk); at constant $u_{s}$ , vvm falls. Hydrostatic pressure at the bottom of a tall vessel raises $C^{*}$ there but also raises the dissolved CO $_{2}$ .

WORKED EXAMPLE · SCALE-UP AT CONSTANT P/V

A 10 L fermenter (impeller diameter 0.080 m, six-blade turbine, $N_{P} = 5.0$ ) runs at 600 rpm with a broth of density 1000 kg/m $^{3}$ and viscosity 0.001 Pa s. It is scaled up to a geometrically similar 10 000 L vessel. Find (a) the ungassed power and $P / V$ at small scale, (b) the large-scale impeller speed, power and tip speed at constant $P / V$ , and (c) the speed if constant tip speed were used instead, with the resulting $P / V$ .

Solution

$N_{1} = 600/60 = 10$ rev/s. $Re_{i} = ρN D_{i}^{2} / μ_{L} = (1000) (10) (0.0064) /0.001 = 6.4 \times 1 0^{4}$ , which is turbulent, so $N_{P}$ is constant.
$P_{1} = N_{P} ρ N^{3} D_{i}^{5} = 5.0 \times 1000 \times 1 0^{3} \times (0.080)^{5} = 5 \times 1 0^{6} \times 3.277 \times 1 0^{- 6} = 16.4$ W, so $P_{1} / V_{1} = 16.4/0.010 = 1640$ W/m $^{3}$ .
The volume grows 1000 times, so every length grows 10 times: $D_{i 2} = 0.80$ m.
Constant $P / V$ : $N_{2} = N_{1} (D_{i 1} / D_{i 2})^{2/3} = 10 \times (0.1)^{2/3} = 10 \times 0.2154 = 2.15$ rev/s $\approx 129$ rpm. Large-scale $Re_{i} = (1000) (2.154) (0.64) /0.001 = 1.4 \times 1 0^{6}$ , still turbulent, so $N_{P}$ is unchanged.
$P_{2} = P_{1} \times 1000 = 16.4$ kW. Check: $5.0 \times 1000 \times (2.154)^{3} \times (0.80)^{5} = 5000 \times 10.0 \times 0.3277 \approx 16.4$ kW.
Tip speed: small $π \times 10 \times 0.080 = 2.51$ m/s; large $π \times 2.154 \times 0.80 = 5.41$ m/s, up by $1 0^{1/3} = 2.15$ times. Mixing time goes up by $N_{1} / N_{2} = 4.64$ times.
Constant tip speed: $N_{2} = N_{1} D_{i 1} / D_{i 2} = 1.0$ rev/s $= 60$ rpm ( $Re_{i} = 6.4 \times 1 0^{5}$ , turbulent). $P / V \propto N^{3} D_{i}^{2}$ changes by $(0.1)^{3} \times 1 0^{2} = 0.1$ , so $P / V = 164$ W/m $^{3}$ .

Answer: (a) $P_{1} \approx 16.4$ W, $P / V \approx 1.64$ kW/m $^{3}$ . (b) $N_{2} \approx 129$ rpm, $P_{2} \approx 16.4$ kW, tip speed 5.41 m/s (from 2.51 m/s). (c) 60 rpm, with $P / V$ falling to about 164 W/m $^{3}$ .

EXAM TIP

(1) Use $N$ in rev/s in $N_{P}$ and $Re_{i}$ : rpm ÷ 60. (2) At constant $P / V$ , $N$ falls but tip speed RISES, a common option trap. (3) Power and power per unit volume behave differently: at constant tip speed $P$ still rises (as $D_{i}^{2}$ ) while $P / V$ falls (as $1/ D_{i}$ ).

Microbial and enzyme reactors

Reactor configurations. The balances above assume an ideal, well-mixed stirred tank. Other configurations are chosen for shear-sensitive cells, for immobilized cells or enzymes, or to avoid moving parts:

Bioreactor types

Type	How it works	Strengths and limits
Stirred tank	Impeller on a shaft, baffles and a gas sparger; contents well mixed	The most versatile; high $k_{L} a$ and copes with viscous broths better than gas-driven vessels; costs agitation power, and the impeller creates shear
Bubble column	Tall vessel; gas sparged at the base provides both aeration and mixing; no moving parts	Simple and low shear; mixing and gas dispersion weaken in viscous broths
Airlift	Gas sparged into a riser only; the gas-free downcomer is denser, so liquid circulates between them (internal draft tube or external loop)	Low shear, no moving parts and a defined circulation path; suits shear-sensitive cells and low-viscosity broths
Packed bed	Immobilized cells or enzyme on particles in a fixed bed; liquid flows through, close to plug flow	High catalyst loading and no need to separate catalyst from product; pressure drop, channelling, and poor removal of gas and heat
Fluidized bed	Upflow of liquid (and gas) keeps the immobilized particles suspended	Lower pressure drop and better mixing and mass transfer than a packed bed; the flow must stay within the range that fluidizes the particles without carrying them out
Membrane reactor	A membrane (ultrafiltration, hollow fibre) holds back the enzyme or cells while product passes through	Continuous use of a soluble enzyme; perfusion culture; membrane fouling limits run length

Enzyme reactors use either free enzyme (in a batch, or continuously with an ultrafiltration membrane to keep the enzyme in) or immobilized enzyme (in a packed bed or a stirred tank). The kinetics are Michaelis–Menten, $v = V_{ma x} S / (K_{m} + S)$ , with $V_{ma x} = k_{c a t} E_{0}$ in concentration per time (for example mM/min, based on the reactor volume) and $K_{m}$ in concentration units. The conversion is $x = (S_{0} - S) / S_{0}$ , with $S_{0}$ the initial or inlet concentration. The design equations for the three ideal reactors, with no inhibition and constant enzyme activity, are:

BATCH ENZYME REACTOR (INTEGRATED MICHAELIS–MENTEN)

V_{ma x} t = K_{m} ln \frac{S _{0}}{S} + (S_{0} - S)

Constant enzyme activity, no inhibition. In conversion form: $V_{ma x} t = S_{0} x - K_{m} ln (1 - x)$ .

PLUG-FLOW (PACKED-BED) ENZYME REACTOR

V_{ma x} τ = K_{m} ln \frac{S _{0}}{S} + (S_{0} - S), τ = \frac{V}{F}

The batch equation with residence time $τ$ in place of $t$ (ideal plug flow). Use $V$ and $V_{ma x}$ on the same basis. For an immobilized enzyme limited by diffusion, the local rate is $η$ times the intrinsic rate; because $η$ varies with $S$ , this integrated form then holds only where $η$ is roughly constant along the bed.

CSTR ENZYME REACTOR

V_{ma x} τ = \frac{( S _{0} - S ) ( K _{m} + S )}{S}

Well mixed, with the enzyme kept in the vessel; the whole reactor runs at the outlet concentration $S$ . In conversion form: $V_{ma x} τ = S_{0} x + K_{m} x / (1 - x)$ .

FIRST-ORDER LIMIT (S MUCH SMALLER THAN KM)

ln \frac{S _{0}}{S} = \frac{V _{ma x}}{K _{m}} τ (PFR), \frac{S _{0}}{S} = 1 + \frac{V _{ma x}}{K _{m}} τ (CSTR)

Natural log. In the zero-order limit ( $S ≫ K_{m}$ ), $S_{0} - S = V_{ma x} τ$ for both reactors.

ENZYME DEACTIVATION

V_{ma x} (t) = V_{ma x, 0} e^{- k_{d} t}, t_{1/2} = \frac{ln 2}{k _{d}}

First-order loss of active enzyme; $k_{d}$ in h $^{- 1}$ or min $^{- 1}$ .

BATCH REACTOR WITH FIRST-ORDER DEACTIVATION

K_{m} ln \frac{S _{0}}{S} + (S_{0} - S) = \frac{V _{ma x, 0}}{k _{d}} (1 - e^{- k_{d} t})

As $t \to \infty$ the right side tends to $V_{ma x, 0} / k_{d}$ , which caps the conversion that can ever be reached.

CSTR vs PFR for an enzyme reaction

Feature	CSTR (stirred tank)	PFR / packed bed
Substrate concentration seen by the enzyme	Uniform, equal to the low outlet value	Falls from $S_{0}$ at the inlet to $S$ at the outlet
Residence time for a given conversion (Michaelis–Menten)	Longer	Shorter
Substrate inhibition	Favoured: the enzyme works at low $S$	Worse: enzyme near the inlet sees high $S$
Product inhibition	Worse: the whole vessel is at the outlet product concentration	Favoured
Operation	Easy pH and temperature control; suits slurries and insoluble substrates; enzyme easy to top up	Pressure drop, channelling and bed compaction; flow is lowered over time to hold conversion as the enzyme deactivates

With immobilized enzyme the observed rate is $η$ times the intrinsic rate. The effectiveness factor $η$ accounts for diffusion inside the pores and through the liquid film around the particle; for Michaelis–Menten kinetics $η \leq 1$ , and it rises toward 1 as the substrate concentration at the particle surface rises well above $K_{m}$ . Smaller particles, faster flow past the particles and lower enzyme loading all raise $η$ .

EFFECTIVENESS FACTOR, FIRST-ORDER REACTION IN A SPHERICAL PARTICLE

ϕ = R \frac{k _{1}}{D _{e}}, η = \frac{3}{ϕ ^{2}} (ϕ coth ϕ - 1)

Internal diffusion only, first-order limit ( $S ≪ K_{m}$ , so $k_{1} = V_{ma x} / K_{m}$ with $V_{ma x}$ per unit particle volume); $R$ = particle radius, $D_{e}$ = effective diffusivity of the substrate inside the particle. $η \to 1$ as $ϕ \to 0$ , and $η \approx 3/ ϕ$ for large $ϕ$ (strong diffusion limitation). The modulus is also defined as $(R /3) k_{1} / D_{e}$ , which changes the form of $η$ , so check which definition a question uses.

EXAM TIP

For Michaelis–Menten kinetics, a CSTR always needs a longer residence time (a larger volume) than a PFR for the same conversion, and the gap widens as the conversion rises. A stirred tank can still be the better choice under substrate inhibition, where working at low $S$ can make it the smaller reactor, and for autocatalytic reactions such as cell growth, where the stirred tank keeps the cells that seed the reaction.

PREVIOUS-YEAR QUESTIONS

24 questions from the BT papers

Try each question first, then open the answer. One shows the official answer only. We placed each question under a topic ourselves; the exam paper does not label topics.

See these papers in full: GATE BT 2026 · GATE BT 2025 · GATE BT 2024 · GATE BT 2023 · GATE BT 2022 · GATE BT 2020 · GATE BT 2019

GATE 2026 BT Q.44

GATE 2026 · BTQ.44 · 2 marks · MSQ

A microbial culture is being grown in a bioreactor with continuous aeration. The dissolved oxygen probe provided with the system shows a reading of zero. In this context, which of the following inferences is/are correct?

(A)
The culture has died due to absence of oxygen
(B)
The culture has run out of carbon source
(C)
The culture growth becomes oxygen limited
(D)
The culture growth is independent of dissolved oxygen

Official answer

Official answer: CSource: Official answer key, GATE 2026 BT (IIT Guwahati)

See the question in the paper

Batch, fed-batch and continuous bioreactors, and scale-up

Batch, fed-batch and continuous operation

The chemostat: balances, washout and optimum dilution rate

Chemostat variants: cell recycle, maintenance, products, stages and the turbidostat

Fed-batch operation

Oxygen transfer: Henry's law, OTR, OUR and kLa

Agitation power and scale-up

Microbial and enzyme reactors

24 questions from the BT papers

7 questions to test yourself