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A fed batch culture was operated with intermittent addition of glucose solution at a flow rate of 200 ml h–1. The values of Ks, µm and D are 0.3 g l–1, 0.4 h–1 and 0.1 h–1, respectively. Determine the concentration of growth limiting substrate (gl–1) in the reactor at quasi-steady state. _______
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- 0.1
- 1
- 0.01
- 10
Correct Option: A
Given, flow rate for the glucose solution, F = 200 mL/hr
KS = 0.3 g/L; µm = 0.4 h-1; D = 0.1 h-1
Therefore, Volume of the reactor, V = F / D = 200 / 0.1 mL = 2000 mL = 2L
We have, from Monod’s equation of substrate consumption µ = (µmax * S)/ (KS + S)
where, µ = rate of substrate consumed per unit time.
µmax = maximum rate of substrate consumed.
KS = Monod’s constant.
S = steady state substrate concentration.
In steady state, mass balance equation is given by :-
Input – Output – Consumption = 0 ⇒(F / V)*SR – µ = 0
⇒D(S) – {(µmax*S)/(KS + S)} = 0
Therefore, D(S) = {(µmax*S)/(KS + S)} = {(µmax*S)/(KS + S)}
= {(0.4S) / (0.3 + S)}
⇒ DS = 0.4S / (0.3 + S)
⇒ 0.1 = 0.4S / (0.3 + S)
⇒ 0.4S = 0.03 + 0.1S
⇒ 0.3S = 0.03
⇒ S = 0.1
Therefore, the concentration of the limiting substrate = 0.1 g/L
