Pilot Sterilization Holding Time Calculation
This problem requires calculating the holding time needed for a pilot-scale sterilization process to reduce a high initial spore concentration to an acceptable low level.
Key Parameters Identified
- Initial spore concentration: $N_0 = 10^8 \text{ spores/ml}$
- Vessel Volume: $V = 100 \text{ m}^3$
- Target final spore count: $N_f = 1 \text{ spore}$ (in the entire vessel)
- Specific death rate constant: $k = 2 \text{ min}^{-1}$ at $121 \text{ }^\circ C$
- Assumption: No spore death during heating/cooling phases.
Calculating Initial Total Spores
First, convert the vessel volume to milliliters (ml) to match the concentration units:
$V = 100 \text{ m}^3 = 100 \times (100 \text{ cm})^3 = 100 \times 10^6 \text{ cm}^3 = 10^8 \text{ cm}^3$
Since $1 \text{ cm}^3 = 1 \text{ ml}$, the volume is $10^8 \text{ ml}$.
Calculate the total initial number of spores in the vessel:
Initial Total Spores ($N_{0, \text{total}}$) $= N_0 \times V$ $N_{0, \text{total}} = (10^8 \text{ spores/ml}) \times (10^8 \text{ ml}) = 10^{16} \text{ spores}$
Applying Sterilization Kinetics
The death of microorganisms during sterilization follows first-order kinetics. The relationship between the initial number of spores ($N_{0, \text{total}}$), the final number of spores ($N_f$), the specific death rate constant ($k$), and the holding time ($t$) is given by:
$N_f = N_{0, \text{total}} e^{-kt}$
We need to find the holding time ($t$) required to reduce $10^{16}$ spores to 1 spore.
Determining Holding Time
Substitute the known values into the kinetic equation:
$1 = 10^{16} \times e^{-2t}$
Rearrange the equation to solve for $t$:
$\frac{1}{10^{16}} = e^{-2t}$
$10^{-16} = e^{-2t}$
Take the natural logarithm ($\ln$) of both sides:
$\ln(10^{-16}) = \ln(e^{-2t})$
$-16 \ln(10) = -2t$
Solve for $t$:
$t = \frac{16 \ln(10)}{2}$
$t = 8 \ln(10)$
Using the approximate value $\ln(10) \approx 2.3026$:
$t \approx 8 \times 2.3026$
$t \approx 18.4208 \text{ min}$
Final Answer Calculation
The holding time is calculated to be approximately $18.42 \text{ min}$. Rounding this to the nearest integer gives:
$t \approx 18 \text{ min}$
Important Questions from Sterilization of Air and Media
The decimal reduction time of a microbe during sterilization at $120 \text{ °C}$ with a first order thermal death rate constant of $1 \text{ min}^{-1}$ will be _______________ min (rounded off to 1 decimal place).
Decimal reduction time of a bacterial strain is $20$ min. Specific death rate constant in $min^{-1}$ (rounded off to two decimal places) is____.
Moist heat sterilization of spores at $121 \text{ } ^\circ C$ follows first order kinetics as per the expression:
$ \frac{dN}{dt} = -k_d N $
where, N is the number of viable spores, t is the time, $k_d$ is the rate constant and $ \frac{dN}{dt} $ is the rate of change of viable spores.
If $k_d$ value is $1.0 \text{ min}^{-1}$, the time (in minutes) required to reduce the number of viable spores from an initial value of $10^{10}$ to a final value of 1 is (up to two decimal places)______.
- Decimal reduction time of bacterial spores is 23 min at $121 \ °C$ and the death kinetics follow first order. One liter medium containing $10^9$ spores per mL was sterilized for 10 min at $121 \ °C$ in a batch sterilizer. The number of spores in the medium after sterilization (assuming destruction of spores in heating and cooling period is negligible) will be ____________________ $ \times 10^7$.
- An industrial fermentor containing $10,000 \text{ L}$ of medium needs to be sterilized. The initial spore concentration in the medium is $10^6 \text{ spores mL}^{-1}$. The desired probability of contamination after sterilization is $10^{-3}$. The death rate of spores at $121 \text{ °C}$ is $4 \text{ min}^{-1}$. Assume that there is no cell death during heating and cooling phases. The holding time of the sterilization process is __________ min (rounded off to the nearest integer).
