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 question asks for the substrate concentration in a fed batch reactor operating at quasi-steady state for biomass and substrate.
Fed Batch Biomass Quasi-Steady State
A quasi-steady state condition for biomass concentration ($\frac{dX}{dt} \approx 0$) in a fed batch reactor implies that the rate of biomass production through growth is balanced by the rate of biomass removal due to the feed flow.
The dilution rate ($D$) in a fed batch process is defined as $D = \frac{F}{V}$, where $F$ is the feed rate and $V$ is the culture volume. For biomass to be at quasi-steady state, the specific growth rate ($\mu$) must equal the dilution rate:
$ \mu = D $
Given:
- Feed rate, $F = 50 \text{ L } h^{-1}$
- Culture volume, $V = 500 \text{ L}$
Calculate the dilution rate (and thus the specific growth rate):
$ \mu = \frac{F}{V} = \frac{50 \text{ L } h^{-1}}{500 \text{ L}} = 0.1 \text{ } h^{-1} $
Substrate Concentration via Monod Kinetics
The specific growth rate ($\mu$) is related to the substrate concentration ($S$) by the Monod equation:
$ \mu = \frac{\mu_m S}{K_s + S} $
Given kinetic parameters:
- Maximum specific growth rate, $\mu_m = 0.2 \text{ } h^{-1}$
- Half-saturation constant, $K_s = 0.1 \text{ } g \text{ } L^{-1}$
We have determined $\mu = 0.1 \text{ } h^{-1}$. Substitute these values into the Monod equation to solve for $S$:
$ 0.1 = \frac{0.2 \times S}{0.1 + S} $
Rearrange the equation to solve for $S$:
$ 0.1 \times (0.1 + S) = 0.2 \times S $
$ 0.01 + 0.1 S = 0.2 S $
$ 0.01 = 0.2 S - 0.1 S $
$ 0.01 = 0.1 S $
$ S = \frac{0.01}{0.1} $
$ S = 0.1 \text{ } g \text{ } L^{-1} $
The substrate concentration in the reactor is $0.1 \text{ } g \text{ } L^{-1}$.
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?
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.
- 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 ________
