Suppose that 100 L/min are drawn from a fermentation tank and passed through an

Question

Suppose that 100 L/min are drawn from a fermentation tank and passed through an extraction tank in which the fermentation product (in the aqueous phase) is mixed with an organic solvent, and then the aqueous phase is separated from the organic phase. The concentration of the desired enzyme (3-hydroxybutyrate dehydrogenase) in the aqueous feed to the extraction tank is 10.2 g/L. The pure organic extraction solvent runs into the extraction tank at the rate of 9.5 L/min. If the ratio of the enzyme in the exit product stream (the organic phase) from the extraction tank to the concentration of the enzyme in the exit waste stream (the aqueous phase) from the tank is D= 18.5 (g/L organic)/(g/L aqueous), what is the fraction recovery of the enzyme and the amount recovered per minute? Assume negligible miscibility between the aqueous and organic liquids in each other, and ignore any change in density on removal or addition of the enzyme to either stream.

Explanation / Answer

Volume passed from fermentation tank to extraction tank = 100 L/min

Concentration of enzyme in feed to the extraction tank = 10.2 g/L.

Rate of pure organic solvent to the extraction tank = 9.5 L/min.

Rate of enzyme addition = 10.2 g/L * 9.5 L/min

= 96.9 g/min

Let us calculate for 1 min.

Mass of solvent added = 96.9 g

Total volume in extraction tank = 100 + 9.5

= 109.5 L

Concentratoion of enzyme in tank = 96.9 / 109.5

= 0.8849 g/L

The ratio of the enzyme in the exit product stream from the extraction tank to the concentration of the enzyme in the exit waste stream, D= 18.5 (g/L organic)/(g/L aqueous).

So,

enzyme in exit stream / enzyme in exit waste stream = 18.5 (g/L organic)/(g/L aqueous)

0.8849 / enzyme in exit waste stream = 18.5 (g/L organic)/(g/L aqueous)

enzyme in exit waste stream = 0.8849 / 18.5

= 0.0478 g/L

fraction of enzyme recovered = 0.8849 - 0.0478 / 0.8849

= 0.8371 / 0.8849

= 0.946

= 94.6%

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