Bioreactor operation formulas

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

Exponential phase of a batch culture, where $S\gg K_s$ so that $\mu \approx {\mu }_m$; no lag and no death.

Constant yield, substrate used only for growth. When the substrate is exhausted ($S=0$) the biomass reaches its maximum, $X_m=X_0+Y_{X/S}{S}_0$.

Time to grow from $X_0$ to $X_m$ if the cells grow at ${\mu }_m$ the whole time ($K_s$ negligible, lag ignored).

Averaged over the whole cycle. $t_L$ = turnaround time per cycle: harvesting, cleaning, sterilising, filling and lag.

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

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

Monod kinetics with $\mu =D$. It does not depend on $S_0$ and holds only for $D<D_c$.

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 $S_0\gg \bar{S}$.

Found by setting $\bar{S}=S_0$ (so $\bar{X}=0$). For $D\ge D_c$ cells leave faster than they can grow and the culture washes out. Without recycle, $D_c<{\mu }_m$ for any finite $S_0$.

Monod kinetics with $\mu =D$ at each steady state. Measure $\bar{S}$ at several dilution rates: a plot of $1/D$ against $1/\bar{S}$ is a straight line with intercept $1/{\mu }_m$ and slope $K_s/{\mu }_m$.

Grams of cells per litre of reactor per hour. For a product, $Q_P=D\bar{P}$.

From $d(D\bar{X})/dD=0$ with Monod kinetics, no maintenance and no death. It lies below $D_c$, and close to it when $S_0\gg K_s$.

Productivity at $D_{opt}$. The substrate concentration there is $\bar{S}=\sqrt{K_s(K_s+S_0)}-K_s$.

From the steady-state biomass balance on the reactor (no death). A balance on the separator shows that $1+\alpha -\alpha C=X_e/\bar{X}$, 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.

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

$Y_G$ = true growth yield; $m_s$ = maintenance coefficient (g substrate per g cells per h). In a chemostat $\mu =D$ and $Y_{X/S}^{obs}=\bar{X}/(S_0-\bar{S})$, so a plot of $1/Y^{obs}$ against $1/D$ is a straight line with slope $m_s$ and intercept $1/Y_G$.

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 $\alpha$ and non-growth-associated $\beta$ (this $\alpha$ is unrelated to the recycle ratio). So $\bar{P}=\alpha \bar{X}+\beta \bar{X}/D$.

Liquid is only added, never withdrawn.

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

Holds once substrate is used as fast as it is fed ($S$ small and steady, $S\ll S_0$). $X_t=XV$ is the total mass of cells, and under constant feed it rises linearly, not exponentially.

Approached as the run proceeds: with constant feed $XV=X_{t0}+F{Y}_{X/S}{S}_{0}t$, so $X$ tends to $Y_{X/S}{S}_0$, after which $dX/dt\approx 0$ and $dS/dt\approx 0$. Because $V$ keeps rising, $D$ and so $\mu$ fall with time, and the residual $S\approx K_{s}D/({\mu }_m-D)$ falls too.

Holds $\mu$ at a chosen ${\mu }_{set}<{\mu }_m$, with $S$ kept constant at $K_s{\mu }_{set}/({\mu }_m-{\mu }_{set})$. $X_0{V}_0$ is the total biomass when exponential feeding starts ($t=0$). It follows from $XV=X_0{V}_0{e}^{{\mu }_{set}t}$ and the substrate balance with $S$ constant. If maintenance matters, replace ${\mu }_{set}/Y_{X/S}$ by ${\mu }_{set}/Y_G+m_s$. If $S\ll S_0$, use $S_0$ in the denominator.

$C^{\ast }$ = 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. $C^{\ast }$ 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.

$C_L$ = 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 $k_{L}a{C}^{\ast }$, reached when $C_L=0$.

$q_{O_2}$ = 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 $C_{crit}$, and falls below it.

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

The smallest $k_{L}a$ that keeps $C_L\ge C_{crit}$ at a given cell density, or the largest cell density a given $k_{L}a$ can support.

Cell-free medium: strip the O${}_2$ with N${}_2$, then aerate starting from $C_{L0}$. The slope of $\ln (C^{\ast }-C_L)$ against $t$ is $-k_{L}a$. Valid when the DO probe responds much faster than $1/k_{L}a$.

2 mol of sulphite consume 1 mol of O${}_2$, and $C_L\approx 0$ throughout.

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

$P$ = ungassed shaft power (W, with $\rho$ 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 ${\mathrm{Re}}_i$.

Power does not depend on density ($k_1$ is a geometry constant).

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.

$D_T$ = 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.

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

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

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

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

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

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

Natural log. In the zero-order limit ($S\gg K_m$), $S_0-S=V_{max}\tau$ for both reactors.

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