A fed batch culture was operated with intermittent addition of glucose solution at a flow rate of $200$ ml $h^{-1}$. The values of $K_s$, $\mu_m$ and $D$ are $0.3$ g $l^{-1}$, $0.4$ $h^{-1}$ and $0.1$ $h^{-1}$, respectively. Determine the concentration of growth limiting substrate (g$l^{-1}$) in the reactor at quasi-steady state._______
Determining Growth Limiting Substrate Concentration
In a fed-batch culture operating at quasi-steady state with a constant dilution rate ($D$), the specific growth rate ($\mu$) of the microorganisms equals the dilution rate.
Quasi-Steady State Principle
The condition for quasi-steady state in this context is:
$ \mu = D $
Applying Monod Kinetics
The specific growth rate ($\mu$) is described by the Monod equation:
$ \mu = \frac{\mu_m S}{K_s + S} $
Where:
- $\mu_m$ is the maximum specific growth rate.
- $S$ is the concentration of the growth-limiting substrate (glucose).
- $K_s$ is the half-saturation constant.
Deriving Substrate Concentration
Equating the Monod equation with the quasi-steady state condition:
$ \frac{\mu_m S}{K_s + S} = D $
Rearranging the equation to solve for $S$:
$ \mu_m S = D (K_s + S) $
$ \mu_m S = D K_s + D S $
$ S (\mu_m - D) = D K_s $
$ S = \frac{D K_s}{\mu_m - D} $
Calculation
Given values:
- $D = 0.1$ $h^{-1}$
- $K_s = 0.3$ $g \text{ } l^{-1}$
- $\mu_m = 0.4$ $h^{-1}$
Substitute these values into the derived equation:
$ S = \frac{(0.1 \text{ } h^{-1}) \times (0.3 \text{ } g \text{ } l^{-1})}{(0.4 \text{ } h^{-1}) - (0.1 \text{ } h^{-1})} $
$ S = \frac{0.03 \text{ } g \text{ } l^{-1} \text{ } h^{-1}}{0.3 \text{ } h^{-1}} $
$ S = 0.1 \text{ } g \text{ } l^{-1} $
Conclusion
The concentration of the growth-limiting substrate (glucose) in the reactor at quasi-steady state is $0.1$ $g \text{ } l^{-1}$.
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?
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.
