Saccharomyces cerevisiae is cultured in a chemostat (continuous fermentation) at a dilution rate of $0.5 \ h^{-1}$. The feed substrate concentration is $10 \ g.L^{-1}$. The biomass concentration in the chemostat at steady state will be ____________________ $g.L^{-1}$. Assumptions: Feed is sterile, maintenance is negligible and maximum biomass yield with respect to substrate is 0.4 (g biomass per g ethanol). Microbial growth kinetics is given by $ \mu = \frac{\mu_m s}{K_s+s}$ where $ \mu$ is specific growth rate ($h^{-1}$), $ \mu_m = 0.7 \ h^{-1}$, $K_s = 0.3 \ g.L^{-1}$ and $s$ is substrate concentration ($g.L^{-1}$).
Chemostat Steady-State Dynamics
For a chemostat operating at steady state, the specific growth rate ($ \mu $) of microorganisms is equal to the dilution rate ($D$), assuming no cell death and continuous substrate supply. The growth follows the Monod kinetics model:
$ \mu = \frac{\mu_m s}{K_s+s} $
Key parameters provided:
- Dilution rate, $D = 0.5 \ h^{-1}$
- Maximum specific growth rate, $ \mu_m = 0.7 \ h^{-1}$
- Half-saturation constant, $K_s = 0.3 \ g.L^{-1}$
At steady state, $ \mu = D$, leading to:
$ \frac{\mu_m s}{K_s+s} = D $
Calculating Steady-State Substrate ($s$)
To find the substrate concentration ($s$) in the chemostat at steady state, plug in the given values:
$ \frac{0.7 \ s}{0.3+s} = 0.5 $
Solve this equation for $s$:
$ 0.7 s = 0.5 (0.3+s) $
$ 0.7 s = 0.15 + 0.5 s $
$ 0.7 s - 0.5 s = 0.15 $
$ 0.2 s = 0.15 $
$ s = \frac{0.15}{0.2} = 0.75 \ g.L^{-1} $
The steady-state substrate concentration is $0.75 \ g.L^{-1}$.
Biomass Concentration ($X$) Calculation
The biomass concentration ($X$) is calculated using the yield coefficient and the difference between feed and effluent substrate concentrations. With negligible maintenance, the substrate balance equation is:
$ S_f = s + \frac{X}{Y_{X/S}} $
Given:
- Feed substrate concentration, $S_f = 10 \ g.L^{-1}$
- Steady-state substrate concentration, $s = 0.75 \ g.L^{-1}$
- Maximum biomass yield ($Y_{X/S}$) is stated as 0.4 g biomass per g ethanol. Assuming this refers to the yield from the limiting substrate, $Y_{X/S} = 0.4 \ g.g^{-1}$.
Rearrange the equation to solve for $X$:
$ X = Y_{X/S} (S_f - s) $
Substitute the values:
$ X = 0.4 \ g.g^{-1} \times (10 \ g.L^{-1} - 0.75 \ g.L^{-1}) $
$ X = 0.4 \times 9.25 \ g.L^{-1} $
$ X = 3.7 \ g.L^{-1} $
The biomass concentration in the chemostat at steady state is $3.7 \ g.L^{-1}$.
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.
