Biology, asked by samayasiri, 7 months ago

In a two-stage chemostat system, the volumes of the first and second reactors are V1 ¼500 L and V2 ¼300 L, respectively. The first reactor is used for biomass production and the second is for a secondary metabolite formation. The feed flow rate to the first reactor is Q¼100 L/h, and the glucose concentration in the feed is S0 ¼5.0 g/L. Use the following constants for the cells: μmax ¼0:3 h 1,KS ¼0:1 g=L,YFX=S ¼0:4 g-cells=g-glucose (a) Determine cell and glucose concentrations in the effluent of the first stage. (b) Assume that growth is negligible in the second stage and the specific rate of product formation is μP ¼0.02 g-P/(g-cell h), and YFP/S ¼0.6 g-P/g-S. Determine the product and substrate concentrations in the effluent of the second reactor.

Answers

Answered by chhibbershivansh10
6

Answer:

it is question of back excercise please reffer the book q 20 to 26 in middle

Similar questions