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Fed-Batch Cheese Culture Exam Analysis

Lactobacillus casei is grown under anaerobic conditions in a 40,000 L fed-batch fermenter operated at quasi steady-state over 6 hours with a feed rate of 4,000 L/hr and substrate concentration in the feed of 80 g/L. The initial culture volume is calculated to be 16,000 L. The substrate concentration at quasi steady-state after 6 hours is calculated to be 0.06 g/L. The cell concentration at quasi steady-state is calculated to be 18.39 g cells/L. The total mass of cells produced after 6 hours is calculated to be 729.3 kg.

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51 views3 pages

Fed-Batch Cheese Culture Exam Analysis

Lactobacillus casei is grown under anaerobic conditions in a 40,000 L fed-batch fermenter operated at quasi steady-state over 6 hours with a feed rate of 4,000 L/hr and substrate concentration in the feed of 80 g/L. The initial culture volume is calculated to be 16,000 L. The substrate concentration at quasi steady-state after 6 hours is calculated to be 0.06 g/L. The cell concentration at quasi steady-state is calculated to be 18.39 g cells/L. The total mass of cells produced after 6 hours is calculated to be 729.3 kg.

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CM4710 ; Biochemical Processes

2nd Mid-Term Examination


October 26, 2001

1. Fed-Batch Production of a Cheese Starter Culture (Doran, 1995). (54)


Lactobacillus casei is grown under essentially anaerobic conditions to provide a starter
culture for manufacture of Swiss cheese. The system has the following characteristics:

YX/S = 0.23 (g dry cell wt./L)/(g substrate/L)


KS = 0.15 g/L
µmax = 0.35 hr-1

A stirred fermenter is operated in fed-batch mode at quasi steady-state (concentrations do not


change with respect to time in the reactor) with a feed flow rate of 4,000 L/hr and feed
substrate concentration of 80 g/L. After 6 hr, the liquid volume is 40,000 L.

a) What is the initial culture volume?


b) What is the concentration of substrate at quasi steady-state at 6 hr?
c) What is the concentration of cells at quasi steady-state at 6 hr?
d) What mass of cells is produced after 6 hr of fed-batch culture, if the initial cell
concentration prior to fed-batch culturing is 18 g/L?

Fed -Batch Reactor

F = 4,000 L/hr

S o = 80 g /L
Xo = 0 g/L
S X

V = rea ctor liquid volume a t any time

Relevant Equations:
µ max S
Monod Specific Growth Rate: µ=
KS + S

CM4710 Biochemical Processees 1


dV
Volume: = F ⇒ V = Vo + Ft
dt

d(XV) dX dV
Biomass: FX o + VµX = = V +X
dt dt dt
dV 1 dV F
VµX = X ⇒ µ = = = D
dt V dt V
F F Do
µ = = =
V Vo + Ft 1 + Do t
V µX d(SV) dS dV
Substrate: FSo - = = V +S
YX / S dt dt dt
V µX dV
FSo - = S
YX / S dt

a) dV/dt = F, or V=Vo+Ft. Solving for Vo, Vo = V-Ft = 40,000 L – 4,000 L/hr (6 hr) = 16,000

b) µ = D at quasi steady-state
µ max S
= D ⇒ µmax S = KSD + SD
KS + S
K SD
S( µmax - D) = KSD ⇒ S =
µ max - D
4,000 L / hr
(0.15 g / L)( )
40,000 L
S = = 0.06 g / L
-1 4,000 L / hr
(0.35 hr - )
40,000 L

CM4710 Biochemical Processees 2


c)
VµX dV V µX
FSo - = S ⇒ F(So - S) =
YX /S dt YX /S

µX YX / SD(So - S)
D(So - S) = ⇒ X =
YX / S µ
but D = µ therefore
X = YX /S (So - S) = (.23 g cells / g substrate)(80 -.6) g substrate / L = 18.39 g cells / L

d) Xt = X(initial)Vo + X(V-Vo) = (18 g/L)(16,000 L) + (18.39 g cells/L)(40,000-16,000 L)

= 729,268 g cells or 729.3 kg cells

CM4710 Biochemical Processees 3

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