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

Question

Suppose that we have available a bacterium that grows in accordance with the Monod model with a very small death rate, and a saturation constant of 0.2 g/l.    The bacteria are in an ideal chemostat of volume 10 liters that will be operated at steady state.   The feed flow rate is 4 liters/hr, the substrate concentration in the feed stream to the reactor is 2.0 g/l, the substrate concentration in the product stream is 0.1 g/l. The reactor cell productivity is 0.5 grams of cell per liter per hour.

What is the value of the cell yield (Equation 6.11)?

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 2%, holding substrate concentration in the feed constant. What should be the new feed flow rate to the reactor?

Explanation / Answer

form the question , we get ;

SATURATION CONSTSTNT =0.2g/L

CHEMOSTAT VOLUME =10 L

FEED FLOW RATE =4 L /hr

SUBSTRATE CONCENTRATION =2.0G/L

SUBSTRATE CONCENTRATION IN THE PRODUCT =0.1g/l

REACTOR CELL PRODUCTIVITY = 0.5gCELLS /L/HR.

IDEAL CHEMOSTAT;

mixed continuous-flow, stirred-tank reactor (CFSTR).Control elements: pH, dissolved oxygen, temperaturE- Fresh sterile medium is fed to the completely mixed and aerated (if required) reactor. Suspension is removed at the same rate. Liquid volume in the reactor is kept constant.

CELL GROWTH; value of the cell yield

SPECIFIC GROWTH RATE = Cell growth is the primary response of viable cells tosubstrates and nutrients.Substrates/nutrients + cells ? products + more cells.

Specific growth rate (h-1),??1/X *dX/dt,

x cell mass concentration g/ L

t = time /hr.

by substituting values we get ;

= 1/0.5 *0.5/60

=2*0.008=0.016 cells / hr = 0.16 g/10lt /hr , 0.16 cells per ten liter volume per hour .

minimum doubling time for the bacteria;

for x to be doubled x/x 0 = 2,

2= e?t=  0.32 cells / hr

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.