Question

c) Lactobacillus casei is propagated under essentially anaerobic conditions to provide a starter culture for manufacture of Swiss cheese. The culture produces lactic acid as a by-product of energy metabolism. The system has the following characteristics: Yx/s 0.23 kgkg Ks0.15 kg/m Hmax 0.35/h ms 0.135 kg/kg.h A stirred fermenter is operated in fed-batch mode at quasi-steady state witha feed flow rate of 3 m/h and feed substrate concentration of 60 kg/ m3. After 6 h, the liquid volume is 30 m3 What was the initial culture volume? i) ii) What is the concentration of substrate at quasi-steady state? ili) What is the concentration of cells at quasi-steady state?

Answer 1

1) Volume can be calculated using formula :

   V_t = V_o + Ft

Vt = 30 m³ /hr ( 1m³ = 1000 lt) = 30000 lts

Time, t = 6 hrs

Flow rate, F = 3 m³/hr = 3000 lt/hr

Vo + 3000 lt/hr x 6 hr = 30000 lt

Vo = 30000 lts -18000 lts = 12000 lts

2) According to Monod specific growth rate, $\mu$ = $\mu$max S/(Ks + S)

At quasi steady state, growth rate, $\mu$ = D = F/V

D = F/V = 3000 l/hr/30000 lts = 1/10 hr = 0.1 /hr

D = $\mu$max S/(Ks + S)

$\mu$max S = (Ks+S) D

$\mu$max S - SD = KsD

S($\mu$max - D) = KsD

S = KsD/($\mu$max - D)

Ks = 0.15 kg/m³ = 0.15 g/l

$\mu$max = 0.35/h

Substrate concentration at quasi steady state (S) : = KsD/($\mu$max - D) = (0.15 g/l x 1/10 hr)/(0.35/hr - 0.1 /hr) = 0.015 g/l/0.25 = 0.06 g/l

3) Concentration of cells at quasi steady state :

FSo - V$\mu$X/Yx/s = SdV/dt

F(So -S) = V$\mu$X/Yx/s

(So -S) F/V = $\mu$X/Yx/s

D = F/V

(So-S) D = $\mu$X/Yx/s

X = Yx/s ((So-S) D/$\mu$

At quasi steady state , $\mu$= D

X = Yx/s (So-S)

Yx/s = 0.23 kg cells/ kg substrate = 0.23 g cells/g substrate

So = 60 Kg/m³ = 60 g/l

S = 0.06 g/l

X = 0.23 g cells/g substrate (60 g/l - 0.06 g/l) = 13.79 g cells/l

Concentration of cells at quasi steady state (X) = 13.79 g cells/l