Question

C. roseus cells immobilized in Ca-alginate beads of diameter \(0.5 \mathrm{~mm}\) are used for production of indole alkaloids (IA) in a fluidized- bed bioreactor. The rate limiting nutrient is glucose and no intraparticle diffusion limitations exist. Use the following data: Flow rate of the feed: \(\mathrm{Q}=1 \mathrm{l} / \mathrm{h}\), Glucose in the feed: \(\mathrm{S}_{\mathrm{o}}=30 \mathrm{~g} / \mathrm{l}\), Plant Cell Concentration: \(\mathrm{X}=6 \mathrm{~g} / \mathrm{l}\) reac. The rate constant for IA formation: \(\mathrm{k}=5 \mathrm{~d}^{-1}(\mathrm{~g} / \mathrm{l})^{-1} \mathrm{~K}_{\mathrm{s}}=0.4 \mathrm{~g} / 1\), Column diameter: \(\mathrm{D}_{\mathrm{o}}=0.15 \mathrm{~m}\). Growth is negligible and Monod kinetics is valid. Determine the following: a. For \(95 \%\) glucose conversion determine required hydraulic residence time, volume, and the height of the column b. If \(\mathrm{Y}_{\mathrm{p} / \mathrm{s}}\) is \(0.02 \mathrm{~g} \mathrm{IA} / \mathrm{g}\) glu, determine IA concentration in the effluent and the productivity.

Short answer

The hydraulic residence time is found, volume and height of the column determined, and IA concentration and productivity calculated using the given variables and Monod kinetics.

Step-by-step solution

- Identify the Given Data

Flow rate of the feed: \(Q=1 \text{ l/h}\), Glucose in the feed: \(S_0=30 \text{ g/l}\), Plant Cell Concentration: \(X=6 \text{ g/l reac.}\), Rate constant for IA formation: \(k=5 \text{ d}^{-1}(\text{g/l})^{-1}\), Substrate half-saturation constant: \(K_s=0.4 \text{ g/l}\), Column diameter: \(D_0=0.15 \text{ m}\), Yield coefficient (IA to glucose): \(Y_{p/s}=0.02 \text{ g IA/g glu}\). The problem assumes negligible growth and that Monod kinetics is valid.

- Convert Rates to Consistent Units

Convert the rate constant \(k\) from days to hours: \[k = 5 \text{ d}^{-1} (\text{g/l})^{-1} \times \frac{1 \text{ d}}{24 \text{ h}} = \frac{5}{24} \text{ h}^{-1} (\text{g/l})^{-1}\]

- Calculate the Substrate Conversion Rate

For 95% conversion of glucose: \(S=0.05S_0\), where \(S_0=30 \text{ g/l}\). Thus, \(S=0.05 \times 30 = 1.5 \text{ g/l}\). Calculate the rate of substrate consumption using Monod kinetics: \[ r_s = \frac{kXS_0}{K_s+S_0} \]

- Calculate Hydraulic Residence Time

Hydraulic residence time (\(\tau\)) is the volume of the reactor divided by the flow rate: \(\tau = \frac{V}{Q}\). Rearrange to solve for \(\tau\): \[ V = Q \tau \]. With 95% conversion, calculate the residence time: \[ \tau = \frac{1}{r_s} = \frac{1}{\frac{5 \cdot 6 \cdot 30}{24 \cdot 0.4 + 30}} \]

- Calculate Required Reactor Volume

Using the calculated residence time, determine the reactor volume from \( V = Q \tau \).

- Calculate Height of the Column

Determine cross-sectional area of the column: \( A = \pi \left( \frac{D_0}{2} \right)^2 \). Calculate the height using: \( H = \frac{V}{A} \)

- Calculate IA Concentration and Productivity

Determine IA concentration in the effluent using yield factor \( Y_{p/s} \): \(P = Y_{p/s} (S_0 - S)\). Calculate reactor productivity: \(P_d = \frac{P}{\tau}\)

Key concepts

glucose conversion

In bioprocess engineering, glucose conversion refers to the proportion of glucose transformed into a desired product by microorganisms or enzymes. It is a critical metric in designing efficient bioreactors. For example, in the given exercise, achieving a 95% glucose conversion rate means that most of the glucose provided in the feed is utilized, leaving only 5%. This high conversion rate is crucial for maximizing the productivity and efficiency of the bioprocess. First, determine the final glucose concentration in the effluent, which can be calculated by:
\[S_{effluent} = (1 - 0.95)S_0\]
With 0.95 being the conversion rate and \(S_0 = 30\) g/L. Therefore,
\[S_{effluent} = 0.05 \times 30 = 1.5\] g/L. Understanding glucose conversion helps in optimizing the feeding strategies and reactor design.

Monod kinetics

Monod kinetics describes how microbial growth rates depend on substrate concentration. It's represented by the equation:
\[r_s = \frac{\text{kXS}_0}{\text{K}_s + \text{S}_0}\]
where
X is the biomass concentration,
\(S_0\) is the initial substrate concentration,
\(K_s\) is the half-saturation constant (the substrate concentration at which the reaction rate is half its maximum), and
k is a rate constant.
It's a significant model in biotechnology for predicting the growth and product formation rates in bioreactors. In our exercise, using Monod kinetics allows us to calculate the rate of substrate (glucose) consumption, which is vital for determining the required bioreactor parameters. For example, with 6 g/L cell concentration and a half-saturation constant of 0.4 g/L, you can compute the specific rate of substrate utilization.

fluidized-bed bioreactor

A fluidized-bed bioreactor consists of immobilized cells within beads that are suspended in the reactor using the upward flow of the liquid feed. This creates a 'fluidized' state, improving mass transfer and reducing substrate diffusion limitations.
For C. roseus cells immobilized in Ca-alginate beads, the fluidized-bed bioreactor maximizes contact between the glucose and the cells, leading to an efficient conversion process.

Its advantages include:
- Enhanced mass transfer
- Uniform temperature distribution
- Minimized substrate diffusion limitations
These features make fluidized-bed reactors ideal for processes requiring high conversion rates and productivity. Given the data in our exercise, the reactor ensures that the immobilized cells have optimal interaction with glucose, leading to high conversion rates.

immobilized cells

Immobilized cells are cells that are fixed or entrapped in a solid matrix, like Ca-alginate beads, which restricts their mobility but allows nutrient and product diffusion. This method is used to increase cell density within the bioreactor, enhancing the bioconversion processes.
Immobilization offers various advantages:
- Reuse of biocatalysts
- Enhanced stability
- Controlled microenvironment

In our exercise, C. roseus cells are immobilized in Ca-alginate beads, ensuring that they are uniformly distributed in the reactor and can efficiently convert glucose to indole alkaloids. This setup also prevents the cells from being washed out, maintaining a high conversion and production rate despite the flow.

hydraulic residence time

Hydraulic residence time (τ) measures the average time a volume of liquid remains in the reactor. It is critical for determining the required reactor volume and ensuring sufficient contact time for the conversion process.
Mathematically, it's expressed as:
\(\tau = \frac{V}{Q} \)
where V is the reactor volume, and Q is the flow rate. In the exercise, with a feed flow rate of 1 L/h and knowing the rate of glucose consumption from Monod kinetics, you can determine the required residence time to achieve 95% glucose conversion.

For example, using the calculated substrate consumption rate (
\(r_s = \frac{5 \times 6 \times 30}{24 \times 0.4 + 30}\)
), rearranging for τprovides the time required for the feed to be processed effectively. Accurate residence time ensures optimal reactor performance.