Fed-Batch Fermenter — Total and Substrate Mass Balances: 04-BS-13, May 2016, Q2

Question 2 (main): Fed-Batch Fermenter — Total and Substrate Mass Balances (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. Feed enters at constant rate $F_i$ (no outflow — fed-batch); initial volume $V_0$; constant density $\rho$; feed glucose concentration $S_i$; first-order consumption kinetics $r_s=k_1s$.

Find. (a) the total mass balance and $V(t)$; (b) the ODE governing $s(t)$.

Approach. A fed-batch fermenter has an inlet stream but no outlet, so the overall (total) mass balance fixes how the liquid volume grows with time; substituting that result into the species (substrate) mass balance and applying the product rule to $d(Vs)/dt$ isolates $ds/dt$.

Part (a): total mass balance. With constant density $\rho$, mass $=\rho V$, and there is no outflow: $$\frac{d(\rho V)}{dt}=F_i\rho \;\Rightarrow\; \rho\frac{dV}{dt}=F_i\rho \;\Rightarrow\; \frac{dV}{dt}=F_i.$$ Integrating with $V(0)=V_0$ gives $$\boxed{V(t)=V_0+F_it}.$$
Part (b): substrate (species) mass balance. Accumulation = in $-$ consumed (no outflow): $$\frac{d(Vs)}{dt}=F_iS_i-r_sV=F_iS_i-k_1sV.$$
Expand the left side by the product rule. $$V\frac{ds}{dt}+s\frac{dV}{dt}=F_iS_i-k_1sV.$$ From part (a), $dV/dt=F_i$, so $s\,dV/dt=sF_i$: $$V\frac{ds}{dt}+sF_i=F_iS_i-k_1sV.$$
Solve for $ds/dt$. $$V\frac{ds}{dt}=F_i(S_i-s)-k_1sV \;\Rightarrow\; \boxed{\frac{ds}{dt}=\frac{F_i(S_i-s)}{V}-k_1s},$$ with $V=V_0+F_it$ from part (a) substituted in. This single ODE captures both the dilution effect of the incoming feed (first term, vanishes when $s=S_i$) and the biological consumption (second term).

Result	Expression
Total mass balance	$dV/dt=F_i$
Volume–time relation	$V(t)=V_0+F_it$
Substrate ODE	$\dfrac{ds}{dt}=\dfrac{F_i(S_i-s)}{V_0+F_it}-k_1s$

Topic: Unsteady-state mass balances on a fed-batch (single-inflow, no-outflow) bioreactor.

Key relations: total balance $d(\rho V)/dt=F_i\rho-F_o\rho$ (here $F_o=0$); species balance $d(VC)/dt=F_iC_i-F_oC_o\pm rV$; product rule to separate $dV/dt$ from $dC/dt$.

Why this works: in a fed-batch operation the working volume is itself a function of time (it only grows), so the species balance cannot be written directly in terms of $ds/dt$ alone — the product-rule expansion of $d(Vs)/dt$ is what correctly separates the "dilution by the growing volume" contribution from the true concentration-rate term.

Common pitfall: forgetting the $s\,dV/dt$ term and writing $ds/dt=F_iS_i/V-k_1s$ directly — this silently drops the dilution-by-incoming-feed effect and is dimensionally inconsistent with the correct balance once $V$ is time-varying.

Source: Shuler & Kargi, Bioprocess Engineering: Basic Concepts, Ch. 6 (fermenter mass balances, fed-batch operation).

Question 2 of 9: Fed-Batch Fermenter — Total and Substrate Mass Balances

Question 2 (main): Fed-Batch Fermenter — Total and Substrate Mass Balances (20 marks)