Objective: Determine the required input sucrose concentration ($S_{0, \text{new}}$) to achieve a specific biomass output rate in a chemostat.
Step 1: Calculate Current Volumetric Flow Rate ($F$)
The volumetric flow rate is calculated using the dilution rate ($D$) and the working volume ($V$).
- Given: $D = 0.55 \text{ h}^{-1}$ and $V = 60 \text{ m}^3$.
- Formula: $F = D \times V$
- Calculation: $F = 0.55 \text{ h}^{-1} \times 60 \text{ m}^3 = 33 \text{ m}^3 \text{ h}^{-1}$.
Step 2: Determine Required New Biomass Concentration ($X_{\text{new}}$)
The target biomass output rate dictates the necessary biomass concentration in the chemostat, assuming the flow rate remains constant.
- Target Biomass Output Rate = $45 \text{ Kg h}^{-1}$.
- Formula: $X_{\text{new}} = \frac{\text{Target Biomass Output Rate}}{F}$
- Calculation: $X_{\text{new}} = \frac{45 \text{ Kg h}^{-1}}{33 \text{ m}^3 \text{ h}^{-1}} = \frac{15}{11} \text{ Kg m}^{-3}$.
Step 3: Calculate Yield Coefficient ($Y_{xs}$)
The yield coefficient ($Y_{xs}$) relates the biomass produced to the substrate consumed. It is calculated from the initial steady-state conditions.
- Steady-state sucrose mass balance: $F \times S_0 = F \times S_{\text{res}} + \frac{F \times X}{Y_{xs}}$
- Simplified balance: $S_0 = S_{\text{res}} + \frac{X}{Y_{xs}}$
- Given initial values: $S_0 = 10.0 \text{ Kg m}^{-3}$, $S_{\text{res}} = 2.0 \text{ Kg m}^{-3}$, $X = 4.5 \text{ Kg m}^{-3}$.
- Rearranging for $Y_{xs}$: $Y_{xs} = \frac{X}{S_0 - S_{\text{res}}}$
- Calculation: $Y_{xs} = \frac{4.5 \text{ Kg m}^{-3}}{10.0 \text{ Kg m}^{-3} - 2.0 \text{ Kg m}^{-3}} = \frac{4.5}{8.0} = 0.5625 \text{ Kg biomass / Kg sucrose}$.
Step 4: Calculate New Input Sucrose Concentration ($S_{0, \text{new}}$)
Using the steady-state mass balance with the new biomass concentration and assuming $S_{\text{res}}$ and $Y_{xs}$ remain constant:
- Formula: $S_{0, \text{new}} = S_{\text{res}} + \frac{X_{\text{new}}}{Y_{xs}}$
- Values: $S_{\text{res}} = 2.0 \text{ Kg m}^{-3}$, $X_{\text{new}} = \frac{15}{11} \text{ Kg m}^{-3}$, $Y_{xs} = 0.5625 \text{ Kg/Kg}$.
- Calculation: $S_{0, \text{new}} = 2.0 \text{ Kg m}^{-3} + \frac{15/11 \text{ Kg m}^{-3}}{0.5625 \text{ Kg/Kg}}$ $S_{0, \text{new}} = 2.0 + \frac{1.3636...}{0.5625}$ $S_{0, \text{new}} = 2.0 + 2.4242...$ $S_{0, \text{new}} \approx 4.424 \text{ Kg m}^{-3}$
This value closely matches option B.
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.
