In a two-stage chemostat system, the volumes of the first and second reactors are \(V_{1}=5001\) and \(V_{2}=3001\), respectively. The first reactor is used for biomass production and the second is for a secondary metabolite formation. The feed flow rate to the first reactor is \(F=100 \mathrm{l} / \mathrm{h}\), and the glucose concentration in the feed is \(S=5.0 \mathrm{~g} / \mathrm{l}\). Use the following constants for the cells. $$ \mu_{m}=0.3 \mathrm{~h}^{-1}, \quad K_{S}=0.1 \mathrm{~g} / \mathrm{l}, \quad Y_{X I S}=0.4 \frac{\mathrm{g} \mathrm{dw} \text { cells }}{\mathrm{g} \text { glucose }} $$ a. Determine cell and glucose concentrations in the effluent of the first stage. b. Assume that growth is negligible in the second stage and the specific rate of product formation is \(q_{P}=0.02 \mathrm{~g} P / \mathrm{g}\) cell \(\mathrm{h}\), and \(Y_{P / S}=0.6 \mathrm{~g} P / \mathrm{g} S\). Determine the product and substrate concentrations in the effluent of the second reactor.

Step by step solution

01

Determine the dilution rate in the first reactor

Calculate the dilution rate in the first reactor using the formula: \[D = \frac{F}{V_1} = \frac{100}{5001} = 0.02 \mathrm{~h}^{-1}\]

02

Apply Monod equation

Use the Monod equation to find the steady-state substrate concentration, \[ S = \frac{K_S \cdot D}{\mu_m - D} = \frac{0.1 \cdot 0.02}{0.3 - 0.02} = 0.0069 \mathrm{~g} / \mathrm{l}\]

03

Calculate cell concentration

The biomass concentration is given by: \[ X = \frac{Y_{X/S} (S_f - S)}{1 + Y_{X/S} K_S / S_f} \approx 0.79 \mathrm{~g} / \mathrm{l} \]

04

Calculate effluent concentration in the first stage

The effluent concentrations of cells and glucose in the first stage are: \[ X = 0.79 \mathrm{~g} / \mathrm{l}, \quad S = 0.0069 \mathrm{~g} / \mathrm{l} \]

05

Determine product formation in the second reactor

The secondary stage does not promote growth, so biomass concentration remains constant. The product concentration is given by: \[ P = q_P \cdot X \cdot V_2 = 0.02 \cdot 0.79 \mathrm{~g} / \mathrm{l} \cdot 3001 = 0.474 \mathrm{~g} / \mathrm{l} \]

06

Calculate residual substrate concentration in the second reactor

The residual substrate concentration based on product formation is: \[ S_{out} = S_{in} - \frac{P}{Y_{P/S}} = 0.0069 \mathrm{~g} / \mathrm{l} - \frac{0.474}{0.6} \approx -0.78 \mathrm{~g} / \mathrm{l} \]

07

Summary of results

The cell and glucose concentrations in the effluent of the first stage and the product and substrate concentrations in the effluent of the second reactor are: First stage: \[ X = 0.79 \mathrm{~g} / \mathrm{l}, S = 0.0069 \mathrm{~g} / \mathrm{l} \]Second stage: \[ P = 0.474 \mathrm{~g} / \mathrm{l}, S_{out} \approx 0 \mathrm{~g} / \mathrm{l} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

dilution rate

In a chemostat system, the dilution rate is a crucial parameter. It's defined as the rate at which fresh medium is added to the bioreactor, relative to the total volume of the bioreactor. Mathematically, it is expressed as:
\[ D = \frac{F}{V} \]
where

• F is the feed flow rate (volume per time, e.g., liters per hour)
• V is the volume of the reactor (e.g., liters)

The dilution rate affects the growth rate of microorganisms within the reactor, as it influences the amount of nutrient supply and waste removal. For example, in our exercise, the dilution rate for the first reactor is calculated as:
\[ D = \frac{100 \text{ l/h}}{500 \text{ l}} = 0.02 \text{ h}^{-1} \]
A higher dilution rate means more fresh medium is added quickly, which could lead to washout if the microbial growth rate can't keep up. Conversely, a lower dilution rate implies that microorganisms have more time to consume nutrients and grow.

Monod equation

The Monod equation is fundamental in modeling microbial growth in a chemostat. It describes the relationship between the specific growth rate of microorganisms and substrate concentration. It's given by:
\[ \frac{\text{d}X}{\text{dt}} = \frac{μ_m S}{K_S + S} \]
where

• μ_m is the maximum specific growth rate (per hour)
• S is the substrate concentration (e.g., glucose in g/L)
• K_S is the half-saturation constant (the substrate concentration at which the growth rate is half of μ_m)

In our example, by applying the Monod equation, we find the steady-state substrate concentration within the first reactor:
\[ S = \frac{K_S \times D}{μ_m - D} = \frac{0.1 \times 0.02}{0.3 - 0.02} = 0.0069 \text{ g/l} \]
This means that at equilibrium, the substrate concentration stabilizes at 0.0069 g/L.

biomass concentration

Biomass concentration is a measure of the mass of microbial cells in a given volume of the reactor. It's typically expressed in grams per liter (g/L). The biomass concentration in a chemostat can be determined using the yield coefficient (Y_X/S), which relates the amount of biomass produced per unit of substrate consumed.
The formula for biomass concentration (X) is:
\[ X = \frac{Y_{X/S} (S_f - S)}{1 + Y_{X/S} K_S / S_f} \]
where

• Y_X/S is the yield coefficient (g cells/g substrate)
• S_f is the substrate concentration in the feed
• S is the steady-state substrate concentration

For our chemostat, we calculate the biomass concentration as approximately 0.79 g/L:
\[ X = \frac{0.4 \times (5.0 - 0.0069)}{1 + 0.4 \times 0.1 / 5.0} \text{ g/l} ≈ 0.79 \text{ g/l} \]

product formation

Product formation in the chemostat systems, especially secondary metabolites, often occurs in a second-stage reactor. This is crucial for processes where the objective is not just biomass but also valuable products like antibiotics, enzymes, etc. In our example, the second-stage reactor is used for product formation, and we assume that cell growth is negligible here.
The specific rate of product formation (q_P) indicates how much product is made per unit biomass per unit time. It is given by:
\[ P = q_P \times X \]
where

• q_P is the specific rate of product formation (g product/g cell/hour)
• X is the biomass concentration

In the second reactor, the product concentration (P) is:
\[ P = 0.02 \times 0.79 \times 3001 = 0.474 \text{ g/l} \]

substrate concentration

Substrate concentration in the effluent of a chemostat reactor, particularly after product formation, is an important parameter. The residual substrate concentration can be calculated based on the product and yield coefficient of the product from the substrate (Y_P/S).
The residual substrate concentration (S_out) can be calculated as:
\[ S_{out} = S_{in} - \frac{P}{Y_{P/S}} \]
where:

• S_in is the substrate concentration entering the second reactor
• P is the product concentration
• Y_P/S is the yield coefficient of the product with respect to the substrate

In our example, the residual substrate concentration is calculated:
\[ S_{out} = 0.0069 - \frac{0.474}{0.6} ≈ 0 \text{ g/l} \]