Maximum specific growth rate ($\mu_m$) = $1 \text{ h}^{-1}$
Saturation constant ($K_s$) = $100 \text{ mg L}^{-1}$
Cell death rate ($k_d$) = $0.01 \text{ h}^{-1}$
Assuming that the bioreactor operates under 'chemostat' mode, the working volume required for this process is __________ L (rounded off to the nearest integer).
Phenol Wastewater Bioreactor Volume Calculation
Input Parameters
- Influent Phenol Concentration ($S_0$): $10,000 \text{ ppm}$
- Effluent Phenol Concentration ($S$): $10 \text{ ppm}$
- Wastewater Feed Rate ($Q$): $80 \text{ L h}^{-1}$
- Maximum Specific Growth Rate ($\mu_m$): $1 \text{ h}^{-1}$
- Saturation Constant ($K_s$): $100 \text{ mg L}^{-1}$
- Cell Death Rate ($k_d$): $0.01 \text{ h}^{-1}$
- Reactor Mode: Chemostat (Aerobic)
Chemostat Steady-State Principles
For a chemostat operating at steady state with cell death, the dilution rate ($D$) is equal to the net specific growth rate ($\mu_{net}$). The net specific growth rate accounts for cell death.
- Dilution Rate ($D$) = Net Specific Growth Rate: $D = \mu - k_d$
- Specific growth rate ($\mu$) follows Monod kinetics: $\mu = \frac{\mu_m S}{K_s + S}$
- Dilution rate is also defined by flow rate ($Q$) and working volume ($V$): $D = Q/V$
Step-by-Step Calculation
1. Calculate Dilution Rate ($D$)
Combine the steady-state and kinetic equations. Substitute the Monod equation for $\mu$ into the $D = \mu - k_d$ equation:
$D = \left( \frac{\mu_m S}{K_s + S} \right) - k_d$
Plug in the given values. Note that $\text{ppm}$ and $\text{mg L}^{-1}$ are numerically equivalent for dilute solutions.
$D = \left( \frac{(1 \text{ h}^{-1}) \times (10 \text{ ppm})}{ (100 \text{ mg L}^{-1}) + (10 \text{ ppm})} \right) - 0.01 \text{ h}^{-1}$
$D = \left( \frac{10}{110} \right) \text{ h}^{-1} - 0.01 \text{ h}^{-1}$
$D = \frac{1}{11} \text{ h}^{-1} - \frac{1}{100} \text{ h}^{-1}$
$D = \frac{100 - 11}{1100} \text{ h}^{-1} = \frac{89}{1100} \text{ h}^{-1}$
2. Calculate Reactor Volume ($V$)
Rearrange the dilution rate definition ($D = Q/V$) to solve for the working volume $V$:
$V = \frac{Q}{D}$
Substitute the calculated $D$ and the given $Q$:
$V = \frac{80 \text{ L h}^{-1}}{\frac{89}{1100} \text{ h}^{-1}}$
$V = 80 \times \frac{1100}{89} \text{ L}$
$V = \frac{88000}{89} \text{ L} \approx 988.76 \text{ L}$
3. Final Answer Rounding
The question requires rounding the working volume to the nearest integer.
$V \approx 989 \text{ L}$
This calculated volume falls within the specified range of 970 L to 1010 L.
Important Questions from Kinetics of Cell Growth Substrate Utilization and Product Formation
If the rate at which $E. coli$ divides is $0.5 \text{ h}^{-1}$, then its doubling time is _______________ h.
- Which of the following factors can affect the growth of a microbial culture in a batch cultivation process?
Let $y(t)$ be a bacterial population whose growth is given by
$ \frac{dy}{dt} = \lambda(y + 2) $
where $ \lambda $ is the growth rate constant. If $y(0) = 1$ and $y(1) = 4$, then the value of $ \lambda $ is
If the doubling time of a bacterial population is 3 hours, then its average specific growth rate during this period is _________ $h^{-1}$.
(Round off to two decimal places)
- A microorganism is grown in a batch culture using glucose as a carbon source. The apparent growth yield is $0.5 \frac{\text{g biomass}}{\text{g substrate}}$. The initial concentrations of biomass and substrate are $2 \text{ g L}^{-1}$ and $200 \text{ g L}^{-1}$, respectively. Assuming that there is no endogenous metabolism, the maximum biomass concentration that can be achieved is ________ $\text{g L}^{-1}$.
