Continuous Fermenter — Unsteady-State Cell Mass Balance: 04-BS-13, December 2016, Q2

Question 2: Continuous Fermenter — Unsteady-State Cell Mass Balance (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. Constant-volume ($V$), constant-flow ($F$ in = $F$ out) fermenter; sterile feed (no cells entering); cell growth rate $r_x=k_1x$; initial condition $x(0)=x_0$.

Find. (a) the unsteady-state cell balance ODE; (b) the $F$–$k_1$–$V$ relationship at steady state; (c) $x(t)$.

Approach. Write accumulation = in $-$ out $+$ generation on the cell population, non-dimensionalise by the constant volume $V$, then set the accumulation term to zero for steady state and separate variables to integrate the general (non-steady) case.

Part (a): unsteady-state cell balance. With sterile feed ($x_{in}=0$), constant $V$, and cell growth as an internal generation term: $$\frac{d(Vx)}{dt}=F\cdot 0 - Fx + r_xV = -Fx+k_1xV.$$ Since $V$ is constant, $d(Vx)/dt=V\,dx/dt$, so dividing through by $V$: $$\boxed{\frac{dx}{dt}=\left(k_1-\frac{F}{V}\right)x}.$$ This is the governing unsteady-state cell mass balance — net specific growth rate is the intrinsic growth rate $k_1$ minus the dilution rate $D=F/V$ (washout term from the continuous outflow).
Part (b): steady-state relationship. At steady state $dx/dt=0$. For any non-trivial (non-washed-out) population $x\neq0$, this forces the bracket itself to vanish: $$k_1-\frac{F}{V}=0 \;\Rightarrow\; \boxed{F=k_1V}\quad\text{(equivalently, the dilution rate }D=F/V\text{ must equal }k_1\text{).}$$ Physically: the fermenter can only sustain a non-zero, non-growing, non-shrinking cell population if the rate at which broth (and hence cells) is washed out exactly matches the rate at which cells are being produced by growth.
Part (c): solve the ODE. The ODE from part (a) is separable (both $k_1$ and $F/V$ are constants): $$\frac{dx}{x}=\left(k_1-\frac{F}{V}\right)dt \;\Rightarrow\; \ln x = \left(k_1-\frac{F}{V}\right)t + \ln x_0.$$ Exponentiating and applying $x(0)=x_0$: $$\boxed{x(t)=x_0\exp\!\left[\left(k_1-\frac{F}{V}\right)t\right]}.$$ Note the consistency with part (b): if $F=k_1V$ exactly, the exponent is zero and $x(t)=x_0$ for all $t$ — a true steady state, as required.

Result	Expression
Unsteady-state cell balance	$dx/dt=(k_1-F/V)\,x$
Steady-state condition	$F=k_1V$ (i.e. $D=F/V=k_1$)
Cell concentration vs. time	$x(t)=x_0\exp[(k_1-F/V)t]$

Topic: Unsteady-state mass balance on a continuous (chemostat-type), constant-volume stirred fermenter.

Key relations: accumulation = in $-$ out $+$ generation, $d(Vx)/dt=F x_{in}-Fx+r_xV$; dilution rate $D=F/V$; net specific growth rate $=k_1-D$.

Why this works: because the outflow carries cells away at the same volumetric rate as the inflow (no accumulation of liquid), the cell concentration's fate is a pure competition between intrinsic growth ($k_1$) and washout ($D$) — whichever is larger determines whether $x$ grows, decays, or (in the exact balance case) holds steady, which is exactly the operating principle of a chemostat.

Common pitfall: forgetting that $x_{in}=0$ for a sterile feed (a fed culture entering with its own cells would add an $Fx_{in}$ inflow term), or setting $dx/dt=0$ and concluding $x=0$ is the only steady state — the washout ($x=0$) state is always a trivial steady state, but the physically interesting one requires the bracket $(k_1-F/V)$ itself to vanish.

Source: Shuler & Kargi, Bioprocess Engineering: Basic Concepts, Ch. 6 (chemostat/continuous-culture mass balances).

Question 2 of 8: Continuous Fermenter — Unsteady-State Cell Mass Balance

Question 2: Continuous Fermenter — Unsteady-State Cell Mass Balance (20 marks)