Problem 1. A fermentation process involving a single stage chemostat is used to
ID: 51692 • Letter: P
Question
Problem 1.
A fermentation process involving a single stage chemostat is used to produce the amino acid lysine. The same process has been modified to include cell recycle, with a recycle ratio of 0.3 and a cell concentration factor of 2.5. Assuming that all other conditions are identical, how much will cell concentration change due to the use of cell recycle?
Problem 2.
A two-stage fermentation process is being used to grow fungal cells. The feed stream is at a flowrate of 50 L/min and contains 220 g/l Glucose (which is the limiting nutrient). The volume of the first fermenter is 400 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 fermentation parameters are Yx/s=0.65, µ2=0.04 min-1.
Determine the following: a) µ1, b) Volume of fermenter 2, c) Steady-state substrate concentrations in fermenter 1 and fermenter 2.
Problem 3.
A fermentation process is being used to produce yeast cells. Using the following conditions, determine Productivity for both batch and continuous fermentation processes at growth rates of µ = 0.05, 0.1, 0.2, 0.4, 0.8, and 1 hr-1. Assume that the continuous system is a standard single stage continuous reactor with no recycle. Xo = 2.0 g/l, Xf =3.5 g/l Total batch turnaround time, t = 1.5 hrs (includes lag, delay, and turnaround times).
Make a graph of Productivity vs Growth Rate comparing both systems.
What conclusions can be made regarding productivity in batch and continuous systems as a function of growth rate?
Explanation / Answer
The critical dilution rate with cell cycle is increased by a factor of 1/(1 + - ) relative to the simple chemostat.
Where =recycle ratio, = cell concentration factor
1/(1+0.3-0.3x2.5)= 1/1(1+0.3-0.75)=1/1(0.55)=1.81
biomass productivity is greater in recycle systems by the same factor that is 1.8 times of initial.
2.
µ= (F . Yx/s) / x , F is the substrate feed rate, x= steady state biomass concentration
therefore,
µ= (F . Yx/s) / x , F is the substrate feed rate, x= steady state biomass concentration
therefore, 0.04=110x0.65/x
x=1787.5
similarly for 2,
on puting values in equation we get value of
µ2=140x0.65/1787.5=0.05
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