Question

Streptomycin is extracted from the fermentation broth using an organic solvent in a counter-current staged extraction unit. The distribution coefficient of streptomycin at pH = 4 is KD = Yi/Xi = 40, and the flow rate of the aqueous (H) phase is H= 150 L/min. If only five extraction units are available to reduce the streptomycin concentration from 10 g/L in the aqueous phase to 0.2 g/L, determine the required flow rate of the organic phase (L) in the extraction unit.

Answer 1

The required flow rate of the organic phase in the extraction unit is approximately 0.573 L/min.

Step-by-step explanation:

The required flow rate of the organic phase (L) in the extraction unit can be determined by using the formula:

L = (H x (C1 - C2)) / ((KD x C2) - (KD x C1))

Where:

• L is the flow rate of the organic phase
• H is the flow rate of the aqueous phase (150 L/min)
• C1 is the initial concentration of streptomycin (10 g/L)
• C2 is the desired concentration of streptomycin (0.2 g/L)
• KD is the distribution coefficient of streptomycin (40)

Plugging in the values:

L = (150 x (10 - 0.2)) / ((40 x 0.2) - (40 x 10))

L ≈ 0.573 L/min

Answer 2

6.75l l/min

Step-by-step explanation:

Streptomycin is extracted from the fermentation broth using an organic solvent in a counter current staged extraction unit.

for 5 stage counter current staged extraction unit for streptomycin

X5 = 0.2g/l

X0 = 10.0g/l

X5/X0 = 0.02

given

Xn/X0, E, and

n = 5

A estimate of E can be evaluated directly;

E = 1.8

= LKd/H

by substituting the given values; Kd = 40, H= 150l/min

then we have,

L;L = 1.8(150)/40

= 6.75l l/min