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)
Chemostat Parameters Summary
The objective is to determine the exit biomass concentration ($X$) in a chemostat bioreactor at steady state. The given parameters are:
- Reactor Volume ($V$): $2 \text{ L}$
- Flow Rate ($F$): $0.8 \text{ L/h}$
- Feed Substrate Concentration ($S_f$): $10 \text{ g/L}$
- Maximum Specific Growth Rate ($\mu_{\text{max}}$): $0.6 \text{ h}^{-1}$
- Monod Constant ($K_s$): $0.5 \text{ g/L}$
- Biomass Yield Coefficient ($Y_{X/S}$): $0.4 \text{ g/g}$
Calculating Dilution Rate
The dilution rate ($D$) is crucial for chemostat operation. It's calculated as the volumetric flow rate ($F$) divided by the reactor volume ($V$).
$D = \frac{F}{V}$
$D = \frac{0.8 \text{ L/h}}{2 \text{ L}}$
$D = 0.4 \text{ h}^{-1}$
Determining Specific Growth Rate
For a chemostat operating at steady state with no cell death, the specific growth rate ($\mu$) of the biomass equals the dilution rate ($D$).
$\mu = D$
$\mu = 0.4 \text{ h}^{-1}$
Calculating Outlet Substrate Concentration
Monod kinetics describe the relationship between specific growth rate ($\mu$) and substrate concentration ($S$). We use this to find the steady-state substrate concentration in the effluent.
$\mu = \frac{\mu_{\text{max}} S}{K_s + S}$
Substituting the known values:
$0.4 \text{ h}^{-1} = \frac{0.6 \text{ h}^{-1} \times S}{0.5 \text{ g/L} + S}$
Solving for $S$:
$0.4 (0.5 + S) = 0.6 S$
$0.2 + 0.4 S = 0.6 S$
$0.2 = 0.2 S$
$S = 1.0 \text{ g/L}$
Calculating Exit Biomass Concentration
The exit biomass concentration ($X$) is calculated using the biomass yield coefficient ($Y_{X/S}$) and the difference between the feed substrate concentration ($S_f$) and the outlet substrate concentration ($S$).
$X = Y_{X/S} \times (S_f - S)$
Plugging in the values:
$X = 0.4 \text{ g/g} \times (10 \text{ g/L} - 1.0 \text{ g/L})$
$X = 0.4 \text{ g/g} \times 9.0 \text{ g/L}$
$X = 3.6 \text{ g/L}$
The exit biomass concentration is 3.6 g/L.
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 ________
- 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.
- In a chemostat with a dilution rate of $0.8 \text{ h}^{-1}$, the steady state biomass concentration and the specific product formation rate are $8 \text{ mol m}^{-3}$ and $0.2 \text{ (mol product) (mol biomass)}^{-1} \text{ h}^{-1}$, respectively. The steady state product concentration in $mol \text{ m}^{-3}$ is ________
