1. Suppose that we have available a bacterium that grows in accordance with the

Question

 1. Suppose that we have available a bacterium that grows in accordance with the Monod model with a very small death rate.   The minimum doubling time (based on the maximum specific growth rate) for the bacterium is 0.7 hr, with a saturation constant of 0.1 g/l, and a maximum cell growth yield (Equation 6.11) of 0.6 grams of cell per gram of substrate.    Suppose we use the bacteria in an ideal chemostat that will be operated at steady state.   The dilution rate will be chosen to be 10% less than the maximum possible value for steady state operation.    What will be the space-time value for the substrate?   What will be the concentration of substrate in the exit stream from the chemostat?   

Explanation / Answer

FROM THE PROBLEM WE GET ;

MINIMUM DOUBLING TIME =0.7/hr

SATURATION CONSTANT = 0.1g/lt

maximum cells =0.6 cells /g of substrate .

In a chemostat ,Because the chemical environment is static, or at steady state. The fluid volume, concentration of nutrients, pH, cell density, and other parameters all are assumed to remain constant .

Dilution Rate = volume of nutrient/ per hour divided by the volume of the culture.

Space time :the time necessary to process one reactor volume of fluid based on entrance conditions

concentration of substrate in the exit stream from the chemostat;

Reactors in Series;The exit of one reactor is fed to the next one.

x= moles of susbtrate reacted upto point / moles of substrate reacted fed to first reactor. , by substituting thre values .we get

= 0.6/0.7 =0.857 concentration of substrate found in the exit stream of chemostat.

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