Question

A battery of cylindrical membranes shown below is used for an extractive bioconversion. Extractive bloconversion means that fermentation and extraction of product occur at the same time. inner tube product) annular space A 10% glucose in water solution is passed through the annular space at a rate of 70 kg/h. An organic solvent enters the inner tube at a rate of 50 kg/h which cannot penetrate the membrane. On the other hand, glucose and water do not enter the inner tube to an appreciable extent (can be neglected). Yeast cells- immobilized in the membrane- convert glucose to ethanol according to the equation Glucose Ethanol Ethanol is soluble in the solvent; it diffuses into the inner tube and is carried out of the system. CO2 gas exits from the membrane unit through an escape valve. In the aqueous stream leaving the annular space, the concentration of unconverted glucose is 0.2% and the concentration of ethanol is 0.5%, and the rest is water. If the system operates at steady state, What is the concentration of ethanol in the product stream leaving the system? b. a. What is the amount of CO2 leaving the system? Hint: Product only includes ethanol and solvent. Molecular Weights: Glucose 180, Ethanol- 46, CO2- 44 Off-gas Feed 70 kg Aqueous stream Membrane System Solvent 50kg Product

Answer 1

The system operates in a steady state, i.e. at any point of time the concentration of unconverted glucose will be 0.2% and the concentration of ethanol will be 0.5% in the water exiting the annular tube.

Let us consider the system for 1 hour.

In 1 hour, the amount of 10% glucose solution entered the annular tube=70 kg

the amount of glucose entered the annular tube=7 kg = 7000 g

Molar mass of glucose is 180 g/mol.

Number of moles of glucose entering the annular tube in 1 hour= (7000/180) mol

The system operates in a steady state, i.e. at any point of time the concentration of unconverted glucose will be 0.2% and the concentration of ethanol will be 0.5% in the water exiting the annular tube. Let us consider the system for 1 hour. In 1 hour, the amount of 10% glucose solution entered the annular tube = 70 kg the amount of glucose entered the annular tube = 7 kg = 7000 g Molar mass of glucose is 180 g/mol. Number of moles of glucose entering the annular tube in 1 hour = (7000/180) mol In 1 hour the glucose solution entered the annular tube = 70 kg The solution is 10% in glucose that is the amount of glucose in 70 kg solution = 7kg = 7000 g number of moles of glucose entered = 7000 g/180 g/mol = 350/9 mol if 10 % corresponds to 7000 g Then 0.2 % will correspond to 7000/10 times 0. 2 g = 140 g and 0. 5 % will

a. The concentration of ethanol in the product stream leaving the system is 6.31%

b. The amount of CO₂ leaving the system is 3.35 kg/h