Q48: B 26 m³ h⁻¹
Calculation: D = μₘ·S/(Kₛ+S) = 0.3·1.5/(0.2+1.5) ≈ 0.2647 h⁻¹; F = D·V = 0.2647·100 ≈ 26 m³ h⁻¹.
Q49: A 2 g L⁻¹
Mass balance: X = Yₓ/ₛ·(Sᵢ – S) → 4 = 0.4·(12 – S) ⇒ 12 – S = 10 ⇒ S = 2 g L⁻¹.
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Sourav Pan🥈 SilverJuly 3, 2025
A culture of Rhizobium is grown in a chemostat (100 m³ bioreactor). The feed contains 12 g L⁻¹ sucrose, Ks for the organism is 0.2 g L⁻¹ and μₘ = 0.3 h⁻¹.
A culture of Rhizobium is grown in a chemostat (100 m³ bioreactor). The feed contains 12 g L⁻¹ sucrose, Ks for the organism is 0.2 g L⁻¹ and μₘ = 0.3 h⁻¹.
The flow rate required to result in steady‐state concentration of sucrose as 1.5 g L⁻¹ in the bioreactor will be
A 15 m³ h⁻¹
B 26 m³ h⁻¹
C 2.6 m³ h⁻¹
D 150 m³ h⁻¹
If Yₓ/ₛ = 0.4 g g⁻¹ for the above culture and steady‐state cell concentration in the bioreactor is 4 g L⁻¹, the resulting substrate concentration will be
A 2 g L⁻¹
B 8 g L⁻¹
C 4 g L⁻¹
D 6 g L⁻¹
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Sourav Pan🥈 SilverJuly 3, 2025
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