Question

Glucose is converted to ethanol by immobilized yeast cells entrapped in gel beads. The specific rate of ethanol production is: \(q_{P}=0.2 \mathrm{~g}\) ethanol/g-cell-h. The effectiveness factor for an average bead is \(0.8\). Each bead contains \(50 \mathrm{~g} / \mathrm{L}\) of cells. The voids volume in the column is \(40 \%\). Assume growth is negligible (all glucose is converted into ethanol). The feed flow rate is \(F=400 \mathrm{l} / \mathrm{h}\) and glucose concentration in the feed is \(S_{0 \mathrm{i}}=150 \mathrm{~g}\) glucose/l. The diameter of the column is \(1 \mathrm{~m}\) and the yield coefficient is about \(0.49 \mathrm{~g}\) ethanol/g glucose. The column height is \(4 \mathrm{~m}\). a. What is the glucose conversion at the exit of the column? b. What is the ethanol concentration in the exit stream?

Short answer

a. 51.28% glucose conversion at the exit b. 37.68 g/L ethanol concentration in the exit stream.

Step-by-step solution

Calculate the volumetric flow rate in the feed

The volumetric flow rate in the feed is given as: $$ F = 400 \text{ L/h} $$

Calculate the total volume of the column

The total volume of the column can be calculated using its diameter and height: $$ V = \text{Cross sectional area} \times \text{Height}\V = \frac{\text{π}d^2}{4} \times h \V = \frac{\text{π}(1 \text{ m})^2}{4} \times 4 \text{ m}\V = 3.14 \text{ m}^3 $$ or 3140 L

Calculate the effective volume occupied by the liquid

The effective volume occupied by the liquid is determined by the void volume percentage: $$ V_{\text{liquid}} = V \times (\text{void fraction}) \V_{\text{liquid}} = 3140 \text{ L} \times 0.4\V_{\text{liquid}} = 1256 \text{ L} $$

Calculate the number of beads

The number of beads needed can be determined from the total volume and the volume occupied by the liquid: $$ \text{Volume occupied by beads} = (1 - \text{void fraction}) \times V \ = (1 - 0.4) \times 3140 \text{ L}\ = 0.6 \times 3140\ = 1884 \text{ L}$$

Calculate the total cell mass in the column

The total cell mass in the column can be calculated using cell concentration per bead volume: $$ \text{Cell concentration} = 50 \text{ g/L}\ \text{Total cell mass} = 50 \text{ g/L} \times \text{Volume occupied by beads} \= 50 \text{ g/L} \times 1884 \text{ L}\ = 94200 \text{ g} \= 94.2 \text{ kg} $$

Calculate the ethanol production rate

The ethanol production rate can be calculated using the specific rate of ethanol production and the effectiveness factor: $$ q_P = 0.2 \text{ g/g-cell-h} \ \text{Effectiveness factor} = 0.8\ \text{Effective production rate} = q_P \times \text{Effectiveness factor}\ = 0.2 \text{ g/g-cell-h} \times 0.8\ = 0.16 \text{ g/g-cell-h} \ \text{Ethanol production rate} = \text{Effective production rate} \times \text{Total cell mass}\ = 0.16 \text{ g/g-cell-h} \times 94200 \text{ g}\ = 15072 \text{ g/h}\ = 15.072 \text{ kg/h} $$

Calculate glucose consumption rate

Using the yield coefficient to determine the glucose consumption rate: $$ Y_{P/S} = 0.49 \text{ g ethanol/g glucose}\ \text{Glucose consumption rate} = \frac{\text{Ethanol production rate}}{Y_{P/S}} \ = \frac{15.072 \text{ kg/h}}{0.49 \text{ g ethanol/g glucose}}\ = 30.769 \text{ kg glucose/h} $$

Determine glucose conversion and ethanol concentration

Calculate total glucose input per hour and use it to find the conversion and ethanol concentration at the exit: $$ \text{Glucose feed rate} = S_{0 \text{i}} \times F\ = 150 \text{ g/L} \times 400 \text{ L/h}\ = 60000 \text{ g/h}\ = 60 \text{ kg/h}\ \text{Glucose conversion} = \frac{\text{Glucose consumption rate}}{\text{Glucose feed rate}}\ = \frac{30.769 \text{ kg/h}}{60 \text{ kg/h}}\ = 0.5128 \text{ or } 51.28 \text{ %}$$ For ethanol concentration:$$ \text{We know the production rate is 15.072 kg/h and the exit flow rate is the same as feed rate which doesn't change}\ \text{Ethanol concentration at exit} = \frac{\text{Ethanol production rate}}{\text{Feed flow rate}} \ = \frac{15.072 \text{ kg/h}}{400 \text{ L/h}}\ = 0.03768 \text{ kg/L}\ = 37.68 \text{ g/L} $$

Key concepts

Glucose to Ethanol Conversion

The conversion of glucose to ethanol is a fundamental process in bioprocess engineering, especially in the production of biofuels and alcoholic beverages. This process is mainly facilitated through fermentation by microorganisms like yeast. In a controlled environment, such as a bioreactor, glucose is metabolized by yeast to produce ethanol and carbon dioxide. Understanding the specifics of this biochemical reaction is vital. The balanced chemical equation for this conversion is: \(C_6H_{12}O_6 \rightarrow 2C_2H_5OH + 2CO_2\). This simple equation encapsulates the transformation of one glucose molecule into two ethanol and two carbon dioxide molecules. However, the real-world application involves various factors that influence the efficiency and rate of conversion, such as the type of microorganisms used, temperature, pH levels, and the presence of inhibitors. Moreover, theoretical yield is often affected by practical constraints, making it essential to optimize conditions in industrial applications.

Immobilized Yeast Cells

Yeast cells are key players in the fermentation process, but their performance can be significantly enhanced by immobilization. Immobilized yeast cells are entrapped or encapsulated in a matrix such as gel beads, allowing them to be used repeatedly and enabling continuous processing. The main advantages include:

• Increased stability and reusability of cells
• Higher cell density in the reactors
• Easier separation of cells from the product

The entrapped state allows these cells to remain in the bioreactor for extended periods, continuously catalyzing the conversion of glucose to ethanol. This leads to more efficient and cost-effective production. Also, the gel beads provide a protective environment for the yeast, improving their tolerance to stress conditions such as high ethanol concentration and temperature variations.

Effectiveness Factor

The effectiveness factor is a numerical indicator of how efficiently an immobilized enzyme or cell catalyzes a reaction compared to its free counterpart. For immobilized yeast cells, the effectiveness factor accounts for limitations imposed by the immobilization matrix, including mass transfer restrictions. It is mathematically defined as:
\( \text{Effectiveness Factor } (\text{η}) = \frac{\text{Observed Rate}}{\text{Intrinsic Rate}} \)
In our exercise, the effectiveness factor of 0.8 indicates the reaction proceeds at 80% of the rate it would if the yeast cells were free and unimmobilized. This factor is crucial because it helps adjust calculations for actual ethanol production rates, providing a more realistic estimation of process performance. Understanding and optimizing the effectiveness factor are critical to achieving maximum productivity in industrial bioprocesses.

Ethanol Production Rate

The rate of ethanol production is a measure of how much ethanol is produced per unit of time. This is influenced by several factors, including cell density, specific ethanol production rate, and effectiveness factor. From our exercise:

• Specific ethanol production rate: 0.2 g ethanol/g-cell-h
• Effectiveness factor: 0.8
• Total cell mass: 94.2 kg

These parameters combine to give an ethanol production rate of 15.072 kg/h. The calculation is straightforward:
\( \text{Effective production rate} = \text{Specific rate} \times \text{Effectiveness factor} = 0.2 \times 0.8 = 0.16 \text{ g/g-cell-h} \)
\( \text{Ethanol production rate} = 0.16 \text{ g/g-cell-h} \times 94200 \text{ g} = 15072 \text{ g/h} = 15.072 \text{ kg/h} \)
Tracking the ethanol production rate is vital for assessing the efficiency of the bioprocess and ensuring that the production meets desired targets.

Glucose Consumption Rate

The glucose consumption rate is directly related to the production of ethanol, guided by the yield coefficient. The yield coefficient (Y_P/S) indicates the amount of ethanol generated per unit of glucose consumed. In this scenario, with Y_P/S of 0.49 g ethanol/g glucose and an ethanol production rate of 15.072 kg/h:

• The glucose consumption rate = Ethanol production rate / Y_P/S
• \( \text{Glucose consumption rate} = \frac{15.072 \text{ kg/h}}{0.49 \text{ g ethanol/g glucose}} = 30.769 \text{ kg glucose/h} \)

This calculation is crucial because it provides insights into how efficiently the feedstock (glucose) is being converted into the desired product (ethanol). Monitoring glucose consumption helps optimize feed rates and ensures that the bioprocess operates at peak efficiency.