Question

A continuous culture system is being constructed. The fermentation tank is to be \(50,000 \mathrm{l}\) in size and the residence time is to be \(2 \mathrm{~h}\). A continuous sterilizer is to be used. The unsterilized medium contains \(10^{4}\) spores \(/ \mathrm{l}\). The value of \(k_{d}\) has been determined to be \(1 \mathrm{~min}^{-1}\) at \(121^{\circ} \mathrm{C}\) and \(61 \mathrm{~min}^{-1}\) at \(140^{\circ} \mathrm{C}\). For each temperature \(\left(121^{\circ} \mathrm{C}\right.\) and \(\left.140^{\circ} \mathrm{C}\right)\), determine the required residence time in the holding section so as to ensure that \(99 \%\) of the time four weeks of continuous operation can be obtained without contamination (due to contaminants in the liquid medium).

Short answer

For \(121^\text{°C}\), the residence time is 4.6 minutes. For \(140^\text{°C}\), it is 0.075 minutes.

Step-by-step solution

Define the sterilization requirement

Ensure that 99% sterilization is achieved, which means 99% of the spores are killed.

Calculate the remaining spores after sterilization

Since 99% sterilization is needed, only 1% of the spores will remain. Therefore, if we start with an initial concentration of \(10^4\) spores per liter, we need to reduce it to \(10^4 \times 0.01 = 10^2\) spores per liter.

Apply first-order kinetic model

Use the first-order kinetic model for sterilization, which is given by: \(N = N_0 \times e^{-k_d \times t}\). Here, - \(N\) is the number of spores remaining after sterilization - \(N_0\) is the initial number of spores - \(k_d\) is the death rate constant - \(t\) is the time (residence time)

Rearrange the equation to solve for time

Rearrange the equation \(N = N_0 \times e^{-k_d \times t}\) to \(t = -\frac{1}{k_d} \times ln(\frac{N}{N_0})\)

Calculate the residence time for 121°C

For \(121^\text{°C}\), \(k_d = 1 \) min\(^{-1}\), \(N_0 = 10^4\) and \(N = 10^2\). Plugging these values in, we get: \(t = - \frac{1}{1}\times ln(\frac{10^2}{10^4}) = 4.6\) minutes.

Calculate the residence time for 140°C

For \(140^\text{°C}\), \(k_d = 61 \) min\(^{-1}\), \(N_0 = 10^4\) and \(N = 10^2\). Plugging these values in, we get: \(t = - \frac{1}{61}\times ln(\frac{10^2}{10^4}) = 0.075\) minutes.

Key concepts

FERMENTATION TANK

A fermentation tank is a crucial component in biochemical engineering. It's essentially a large vessel where microbial fermentation processes occur. In our example, the fermentation tank has a capacity of 50,000 liters. This size is typical for industrial applications where large-scale production is required.
Inside the tank, microorganisms grow and produce desired products, such as alcohol, enzymes, or antibiotics. Maintaining the right environment inside the tank is essential for optimal microorganism activity. Factors such as temperature, pH, and nutrient availability need to be carefully controlled. The design of the tank also helps prevent contamination and ensures efficient mixing.
Ensuring good mixing is vital because it ensures that nutrients and oxygen are evenly distributed, and waste products are removed effectively. This helps in achieving a uniform culture and maximizing production efficiency.

STERILIZATION REQUIREMENT

Sterilization is a process designed to eliminate all forms of microbial life, including spores. In the context of a continuous culture system, sterilization ensures that the medium entering the fermentation tank is free from unwanted contaminants.
The exercise specifies that a continuous sterilizer will be used and that the original medium contains 10,000 spores per liter. To meet the requirement of 99% sterilization, only 1% of these spores can remain post-sterilization. This translates into reducing the spore concentration from 10,000 to 100 spores per liter.
Ensuring effective sterilization is critical as it minimizes the risk of contamination. Contaminants can compromise the quality of the product and affect the overall yield of the fermentation process. Therefore, precise control of sterilization parameters is necessary to meet the desired microbial safety levels.

FIRST-ORDER KINETIC MODEL

The first-order kinetic model is commonly used to describe the sterilization process. It assumes that the rate of microbial death is directly proportional to the number of microorganisms present.
This model is mathematically represented by the equation:
\(N = N_0 \times e^{-k_d \times t}\)
In this equation:

• \(N\) is the number of spores remaining after sterilization.
• \(N_0\) is the initial number of spores.
• \(k_d\) is the death rate constant.
• \(t\) is the residence time (or time of exposure).

The exercise provides the values of the death rate constant \(k_d\) at different temperatures, which allows us to calculate the residence time required for effective sterilization. By rearranging the equation to solve for \(t\), we can determine the length of time needed at a given temperature to achieve the desired reduction in spore concentration.

RESIDENCE TIME

Residence time is the duration for which the medium stays within the continuous sterilization system. Correctly calculating residence time is crucial for ensuring effective sterilization.
The exercise requires calculating the residence time at two different temperatures, 121°C and 140°C, to ensure 99% sterilization (i.e., achieving a 100-fold reduction in spore concentration).
Using the first-order kinetic equation, the residence time at 121°C (where \(k_d = 1 \text{ min}^{-1}\)) is calculated to be approximately 4.6 minutes. For 140°C (where \(k_d = 61 \text{ min}^{-1}\)), the residence time reduces drastically to about 0.075 minutes.
This shows how increasing the temperature significantly decreases the necessary residence time, highlighting the temperature's critical role in the effectiveness of the sterilization process.