Chemostat Product Concentration Calculation
This problem involves calculating the steady-state product concentration ($P$) in a chemostat using the provided parameters: dilution rate ($D$), steady-state biomass concentration ($X$), and specific product formation rate ($q_p$).
Steady State Mass Balance
At steady state in a chemostat, the rate of product formation must equal the rate of product removal via the outflow. The rate of formation is given by the product of biomass concentration and the specific product formation rate ($q_p \times X$). The rate of removal is the product of the dilution rate and the product concentration ($D \times P$).
The mass balance equation at steady state is:
$ D \times P = q_p \times X $
Calculating Product Concentration
To find the product concentration ($P$), we rearrange the steady-state equation:
$ P = \frac{q_p \times X}{D} $
Now, substitute the given values:
- $D = 0.8 \text{ h}^{-1}$
- $X = 8 \text{ mol m}^{-3}$
- $q_p = 0.2 \text{ (mol product) (mol biomass)}^{-1} \text{ h}^{-1}$
Plugging these into the equation:
$ P = \frac{(0.2 \text{ (mol product) (mol biomass)}^{-1} \text{ h}^{-1}) \times (8 \text{ mol m}^{-3})}{0.8 \text{ h}^{-1}} $
First, calculate the numerator:
$ q_p \times X = 0.2 \times 8 = 1.6 \text{ (mol product) m}^{-3} \text{ h}^{-1} $
Now, divide by the dilution rate:
$ P = \frac{1.6 \text{ (mol product) m}^{-3} \text{ h}^{-1}}{0.8 \text{ h}^{-1}} $
$ P = 2 \text{ mol m}^{-3} $
The calculated steady-state product concentration is $2 \text{ mol m}^{-3}$.
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.
