You are culturing a bacterium that n an ideal chemostat of v

You are culturing a bacterium that n an ideal chemostat of volume 5 liters that will be operated at steady state. The feed flow rate is set at 1.75 liters/hr, the substrate concentration in the feed stream to the reactor is 2.0 g/l, and the substrate concentration in the product stream is 0.3 g/l. The reactor cell productivity is 0. 4 grams of cell per liter per hour, (you can assume the bacteria growth follows the Monod model with a very small death rate. The saturation constant of 0.4 g/l.) What is the value of the apparent cell yield? What is the minimum doubling time for the bacteria? Suppose we wish to increase the concentration of cells in the product stream leaving the reactor by 5%, holding substrate concentration in the feed constant. What should be the new feed flow rate to the reactor?

Solution

Dilution rate = flow rate/culture volume

Dilution rate = 1.75liters/hr/5 liters

=> 1.75/5 = 0.34/hr = D

The specific growth rate (µ) of bacteria depends on the time (t_d) taken for the biomass to double.

µ = ln(2)/t_d

t_d = ln(2)/D

t_d = 0.693/0.34 = 2.038hrs

µ = 0.693/2.038 = 0.34 times/hr

The reaction is in the steady state as the dilution rate is equal to the specific growth rate

a)

The apparent cell yield is the µ_max

µ = µ_max [S/(Ks+S)]

0.34 = µ_max [2/(0.4+2)]

0.34*1.2 = µ_max

µ_max = 0.408 times/hr

b)

The doubling time for the bacteria is calculated above as approximately 2 hours.

c)

If the concentration of cells is increased by 5percent, µ_max is increased by 5 percent.

Therefore, (0.408/20) +0.408 = 0.428 = µ_max

Now, the critical dilution rate = µ_max [S/(Ks+S)]

= 0.428[2/2.4] = 0.428/1.2 = 0.356/hr

Now, the new feed flow rate = D * total volume = 0.356*5 = 1.78 liters/hr