A two-stage fermentation process is being used to grow fungal cells. The feed st

Question

A two-stage fermentation process is being used to grow fungal cells. The feed stream is at a flow rate of 50 L/min and contains 220 g/l Glucose ( which is the limiting nutrient). The volume of the first fermeneter is 400 L and the volume of the second is 1250 L. The steady state cell concentration in fermenter 1 is 110 g/l and the steady state cell concentration in fermenter 2 is 140 g/l. Other known parameters are: Yx/s = 0.65, u1 = .125min^-1 and u2 = .04min^-1.

Determine the steady-state substrate concentrations in fermenters 1 and 2.

Please show your work. Thank you.

Explanation / Answer

Feed stream is at a flow rate = 50 L/min

Glucose ( which is the limiting nutrient) = 220 g/l

volume of the first fermeneter = 400 L

volume of the second fermeneter = 1250 L

steady state cell concentration in fermenter 1 = 110 g/l

the steady state cell concentration in fermenter 2 = 140 g/l

Yx/s = 0.65, u1 = .125min^-1 and u2 = .04min^-1

steady-state substrate concentrations in fermenters 1 and 2 = ?

steady state cell concentration in fermenter 1 = 110 g/l ( rate = K x Concentration of substrate in fermenter1)

rate = 0.125 x Y x concentration in fermenter1

50 = 0.125 x 0.65 x 110/220 x concentration in fermenter1

concentration in fermenter1 = 50 / (0.125 x 0.65 x 0.5) = 1230.769 g/l

steady state cell concentration in fermenter 2 = 140 g/l ( rate = K x Concentration of substrate in fermenter1)

rate = 0.125 x Y x concentration in fermenter1

50 = 0.04 x 0.65 x 140/220 x concentration in fermenter2

concentration in fermenter2 = 50 / (0.04 x 0.65 x 0.636) = 3023.70 g/l

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