Chemostat Washout and Critical Dilution Rate
Washout in a chemostat occurs when the rate at which cells are removed by the flow ($D \times X$) exceeds the rate at which they are produced by growth ($\mu \times X$). The critical dilution rate ($D_c$) is the highest dilution rate at which a steady state for biomass can be maintained. Above $D_c$, the biomass concentration approaches zero (washout).
Monod Kinetics and $D_c$
For a microbe exhibiting Monod growth kinetics and operating in a chemostat with sterile feed and no maintenance term, the specific growth rate is given by: $ \mu = \frac{\mu_{max} S}{K_s + S} $ The critical dilution rate ($D_c$) is equal to the maximum growth rate achievable given the inlet substrate concentration ($S_0$): $ D_c = \frac{\mu_{max} S_0}{K_s + S_0} $
Growth Parameters
The problem provides the following values:
- Maximum specific growth rate, $\mu_{max} = 0.5 \text{ h}^{-1}$
- Monod constant, $K_s = 0.1 \text{ mg L}^{-1}$
- Chemostat volume, $V = 5 \text{ L}$
- Inlet substrate concentration, $S_0 = 10 \text{ g L}^{-1}$
Substrate Concentration Conversion
To ensure consistent units for $S_0$ and $K_s$, convert $S_0$ from $\text{g L}^{-1}$ to $\text{mg L}^{-1}$:
$ S_0 = 10 \text{ g L}^{-1} \times \frac{1000 \text{ mg}}{1 \text{ g}} = 10000 \text{ mg L}^{-1} $
Critical Dilution Rate ($D_c$) Calculation
Substitute the values into the formula for $D_c$:
$ D_c = \frac{(0.5 \text{ h}^{-1}) \times (10000 \text{ mg L}^{-1})}{0.1 \text{ mg L}^{-1} + 10000 \text{ mg L}^{-1}} $
$ D_c = \frac{5000}{10000.1} \text{ h}^{-1} $
$ D_c \approx 0.499995 \text{ h}^{-1} $
Minimum Flow Rate ($F_{min}$) Calculation
The dilution rate is related to the flow rate ($F$) and volume ($V$) by $D = \frac{F}{V}$. The minimum inlet feed flow rate for washout ($F_{min}$) is:
$ F_{min} = D_c \times V $
$ F_{min} \approx 0.499995 \text{ h}^{-1} \times 5 \text{ L} $
$ F_{min} \approx 2.499975 \text{ L h}^{-1} $
Rounded Answer
Rounding the result to one decimal place gives:
$ F_{min} \approx 2.5 \text{ L h}^{-1} $
Important Questions from Batch Fed Batch and Continuous Processes
- Under complete cell washout condition in a chemostat with sterile feed, which of the following statements is/are correct?
A fed batch process is running at quasi-steady state with respect to substrate and biomass concentration. At $2 \text{ h}$, the culture volume is $500 \text{ L}$ with a constant sterile inlet feed at $50 \text{ L } h^{-1}$ of glucose. The culture kinetic parameters $ \mu_m$ and $K_s$ are $0.2 \text{ } h^{-1}$ and $0.1 \text{ } g \text{ } L^{-1}$, respectively.
The substrate concentration in the reactor will be ________ $g \text{ } L^{-1}$ (rounded off to one decimal place).
The following schematic diagram shows a chemostat with cell recycle
where $F_0$ and $F_r$ are the volumetric flow rates (in $L.h^{-1}$) of feed and recycle streams, respectively. $X_1$, $X_0$ and $X$ are the cell concentrations (in $g.L^{-1}$) in the reactor, recycle-stream and product-stream, respectively. If $\frac{X_0}{X_1}=1.5$, $\frac{F_r}{F_0}=0.7$ and $X_1$ is $7.3 g.L^{-1}$, the value of $X$ (in $g.L^{-1}$, rounded off to one decimal place) is ________
A $2 \text{ L}$ bioreactor is being operated as a chemostat, at a flow rate of $0.8 \text{ L/h}$ and sterile feed of $10 \text{ g/L}$ substrate. The bacterial growth follows Monod kinetics at a maximum specific growth rate of $0.6 \text{ h}^{-1}$ with a Monod constant of $0.5 \text{ g/L}$ and a biomass yield coefficient of $0.4 \text{ g/g}$. The exit biomass concentration is __________ $\text{g/L}$.
(Round off to one decimal place)
- The amount of biomass in a reactor at the end of the batch process is 50 g. Fed- batch operation is initiated by feeding the substrate solution at a constant rate of $1 \text{ L h}^{-1}$. The concentration of substrate in the feed is $50 \text{ g L}^{-1}$. The maximum biomass yield ($Y_{XS}^M$) is $0.4 \frac{\text{g biomass}}{\text{g substrate}}$. Assuming the system is at quasi-steady state, the maximum amount of biomass after 5 h of feeding is ________________ g.
