Question 6 of 6: Batch, Chemostat, and Fed-Batch Culture Kinetics Compared

Question 6: Batch, Chemostat, and Fed-Batch Culture Kinetics Compared (15 marks — a–d, assumed weight; see paper-level check note)

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.

Find. (a) maximum cell density and the time to reach it (after the lag phase); (b) the dilution rate giving maximum chemostat productivity; (c) how the batch and continuous volumetric production rates compare; (d) the fed-batch biomass production rate.

Approach. (a) The yield coefficient links $S$ linearly to $X$, turning the Monod batch growth ODE into a separable "logistic-type" equation solved by partial fractions. (b) The classical chemostat productivity-optimum condition ($d(DX)/dD=0$) gives a closed-form $D_{\text{opt}}$. (c) Compare the resulting volumetric rates directly. (d) Fed-batch has no outflow, so under the quasi-steady-state approximation (feed becomes the growth-limiting supply) the total biomass production rate is constant and set entirely by the feed.

1. (a) Yield-linked substrate–biomass relation. $$S=S_0-\frac{X-X_0}{Y_{x/s}}=\frac{X_m-X}{Y_{x/s}}, \qquad X_m\equiv X_0+Y_{x/s}S_0=0.01+0.4(10)=4.01\ \text{g/L (theoretical zero-substrate maximum)}.$$
2. Maximum cell density. At the stated minimum residual substrate $S_{\min}=0.01$ g/L: $$X_{\max}=X_0+Y_{x/s}(S_0-S_{\min})=0.01+0.4(9.99)=\boxed{4.006\ \text{g/L}}.$$
3. Integrated Monod batch growth time. Substituting $S(X)$ into $dX/dt=\mu_mSX/(K_s+S)$ gives $dX/dt=\mu_mX(X_m-X)/\big(K_sY_{x/s}+X_m-X\big)$, which integrates by partial fractions (with $a\equiv K_sY_{x/s}=0.04$) to $$t=\frac{1}{\mu_m}\left[\frac{a+X_m}{X_m}\ln\frac{X}{X_0}+\frac{a}{X_m}\ln\frac{X_m-X_0}{X_m-X}\right].$$
4. Evaluate at $X=X_{\max}=4.006$ g/L. $$t=\frac{1}{0.7}\left[1.010\ln\frac{4.006}{0.01}+0.00998\ln\frac{4.00}{0.004}\right]=\frac{1}{0.7}[6.053+0.069]=\boxed{8.75\ \text{h after the lag phase}}.$$
5. (b) Chemostat productivity optimum. Maximizing volumetric productivity $D\,X(D)$ over the Monod chemostat steady state gives the classical closed-form optimum $$D_{\text{opt}}=\mu_m\left(1-\sqrt{\frac{K_s}{K_s+S_0}}\right)=0.7\left(1-\sqrt{\frac{0.1}{10.1}}\right)=\boxed{0.630\ \text{h}^{-1}}.$$
6. (c) Batch vs. continuous volumetric production rate. Batch (whole-cycle average): $$\bar r_{\text{batch}}=\frac{X_{\max}-X_0}{t}=\frac{4.006-0.01}{8.75}=0.457\ \text{g/(L}\cdot\text{h)}.$$ Chemostat at $D_{\text{opt}}$: $S_{\text{opt}}=K_sD_{\text{opt}}/(\mu_m-D_{\text{opt}})=0.905$ g/L, $X_{\text{opt}}=Y_{x/s}(S_0-S_{\text{opt}})=3.638$ g/L, so $$r_{\text{cont}}=D_{\text{opt}}X_{\text{opt}}=0.630(3.638)=\boxed{2.29\ \text{g/(L}\cdot\text{h)}}\approx 5.0\times \bar r_{\text{batch}}.$$ The batch average is dragged down by the slow finish as $S$ crawls toward $S_{\min}$; the chemostat sustains its (lower per-cell, but continuously applied) growth rate indefinitely.
7. (d) Fed-batch feed rate. Using the part-(b) optimum as the INITIAL dilution rate with $V_0=100$ L: $$F=D_{\text{opt}}V_0=0.630(100)=63.0\ \text{L/h},$$ held constant thereafter (so $D=F/V(t)$ falls below $D_{\text{opt}}$ as $V$ grows past 100 L in the 1000 L vessel).
8. Fed-batch quasi-steady-state production rate. Fed-batch has no outflow, so the total-biomass balance is $d(XV)/dt=\mu XV$ (growth only). Under the fed-batch quasi-steady-state approximation, the feed itself is the growth-limiting supply — $S$ stays low and roughly constant because essentially every increment of incoming substrate is consumed almost as fast as it arrives — so $$\mu XV\approx Y_{x/s}\,F\,S_0 \quad\Rightarrow\quad \boxed{\frac{d(XV)}{dt}\approx Y_{x/s}FS_0=0.4(63.0)(10)\approx 252\ \text{g DCW/h}},$$ a constant TOTAL production rate, in contrast to the batch (whose rate collapses to near zero as $S\to S_{\min}$) and the fixed-volume chemostat (whose rate is capped by $V$).