Bioprocess Engineering Questions and Answers – Ideal Reactor Operation

This set of Bioprocess Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Ideal Reactor Operation”.

1. Aerobic reactions are not batch operations.
a) True
b) False
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2. In a perfectly mixed reactor _________
a) The output composition is different from input composition
b) The output composition is identical from input composition
c) Both output and input composition are constant
d) Both output and input composition are not constant
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3. Convert v_max = 2.5 mmol m^-3 s^-1 into mM h^-1.
a) 2500
b) 900
c) 25
d) 9
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Answer: d
Explanation:

4. Convert v_max = 7 mmol m^-3 s^-1 into mM h^-1.
a) 25
b) 20
c) 25.20
d) 20.25
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Answer: c
Explanation:

5. In batch reaction time with enzyme deactivation, calculate the first-order deactivation rate constant.
(Given – s_o = 12 mM; v_max = 9 mM h^-1; K_m = 8.9 mM; s_f = 1.2 mM; t_h = 4.4 h).
a) 0.150 h^-1
b) 0.158 h^-1
c) 0.155 h^-1
d) 0.154 h^-1
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Answer: b
Explanation:

6. Given the batch reaction time with enzyme deactivation, calculate the batch time tb. (Given: k_d = 0.158 h^-1, s_o = 12 mM, v_max = 9 mM h⁻¹^-1, K_m = 8.9 mM, s_f = 1.2 mM, t_h = 4.4 h).
a) 5.0 h
b) 10.0 h
c) 15.0 h
d) 20.0 h
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Answer: a
Explanation:

7. Zymomonas mobilis is used to convert glucose to ethanol in a batch fermenter under anaerobic conditions. The yield of biomass from substrate is 0.06 g g^-1; Y_PX is 7.7 g g^-1. The maintenance coefficient is 2.2 g g^-1 h^-1; the specific rate of product formation due to maintenance is 1.1 h^-1. The maximum specific growth rate of Z. mobilis is approximately 0.3 h^-1.5 g bacteria are inoculated into 50 litres of medium containing 12 g l^-1 glucose. Determine batch culture times required to produce 10 g biomass.
a) 3.1 h
b) 3.3 h
c) 3.5 h
d) 3.7 h
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8. What is the batch culture time required to achieve 90% substrate conversion in a fermenter with the following conditions: Y_xs = 0.06 g/g, Y_px = 7.7 g/g, μ_max = 0.3 h^-l, maintenance coefficient (m_s) = 2.2 g/g/h, initial substrate concentration (s_o) = 12 g/l, and initial biomass concentration (x_o) = 0.1 g/l?
a) 5.1 h
b) 5.3 h
c) 5.5 h
d) 5.7 h
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Answer: d
Explanation:

9. What is the batch culture time required to produce 100 g of ethanol in a fermenter with the following conditions: μ_max = 0.3 h^-l, Y_px = 7.7 g/g, initial substrate concentration (s_₀) = 12 g/l, and initial biomass concentration (x_₀) = 0.1 g/l?
a) 3.2 h
b) 3.4 h
c) 3.6 h
d) 3.8 h
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Answer: b
Explanation:

10. The Zymomonas mobilis cells are used for chemostat culture in a 60 m³ fermenter. The feed contains 12 g l^-1 glucose; K_s for the organism is 0.2 g l^-1. What flow rate is required for a steady-state substrate concentration of 1.5 g l^-1?
a) 15.6 m³ h^-1
b) 15.8 m³ h^-1
c) 15.4 m³ h^-1
d) 15.2 m³ h^-1
View Answer

Answer: a
Explanation: Y_xs = 0.06 g g^-1; Y_px = 7.7 g g^-1;μ_max = 0.3 h^-1; K_s = 0.2 g^-1; m_s = 2.2 g g^-1 h^-l; s_i = 12 g 1^-1; V=60 m³. q_p = 3.4 h^-1, Y_PS = Y_px Y_xs = 0.46 g g^-1.
s = 1.5 g l^-1

11. A chemostat fermenter operates at a flow rate of 15.6 m³/h. Given the following parameters: initial substrate concentration s₀ = 12 g/l, steady-state substrate concentration s = 1.5 g/l, maximum specific growth rate μ_max = 0.3 h^-1, yield of biomass from substrate Y_xs = 0.06 g/g, and maintenance coefficient m_s = 2.2 g/g/h. What is the cell density (x) at steady state?
a) 0.42 g l^-1
b) 0.44 g l^-1
c) 0.46 g l^-1
d) 0.48 g l^-1
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Answer: a
Explanation: When synthesis of product is coupled with energy metabolism as for ethanol, x is evaluated. Therefore:

12. In a chemostat operated at a flow rate of 15.6 m³/h with a substrate concentration of 1.5 g/l and a cell density of 0.42 g/l, what is the concentration of ethanol produced at steady state? Assume the maximum ethanol production rate Y_px = 7.7 g/g and the maintenance coefficient for the product formation m _p = 1.1 h^-1.
a) 5.1 g l^-1
b) 5.3 g l^-1
c) 5.5 g l^-1
d) 5.7 g l^-1
View Answer

Answer: c
Explanation: Assuming ethanol is not present in the feed, P_I = 0. Steady-state product concentration is given by:

13. Which of the following type is of the perfusion culture?
a) Batch
b) Conc. Batch
c) Continuous
d) Conc. Fed-Batch
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14. Immobilised lactase is used to hydrolyse lactose in dairy waste to glucose and galactose. Enzyme is immobilised in resin particles and packed into a 0.5 m³ column. The total effectiveness factor for the system is close to unity; K_m for the immobilised enzyme is 1.32 kg m^-3; V_max is 45 kg m^-3 h^-1. The lactose concentration in the feed stream is 9.5 kg m^-3; a substrate conversion of 98% is required. The column is operated with plug flow for a total of 310 d per year. At what flow rate should the reactor be operated?
a) 1.56 m³ h^-1
b) 1.58 m³ h^-1
c) 1.50 m³ h^-1
d) 1.54 m³ h^-1
View Answer

Answer: a
Explanation: For 98% substrate conversion, s_f = (0.02 s_I) = 0.19 kg m^-3

15. An immobilized lactase enzyme is used in a column reactor to hydrolyze lactose to glucose and galactose. The enzyme is immobilized in resin particles, and the reactor operates under plug flow. Given the following conditions: reactor volume = 0.5 m³, initial lactose concentration s₀ = 9.5 kg/m³, final substrate concentration s_f = 0.19 kg/m³ (98% conversion), the maximum velocity V_max = 45 kg/m³/h, and K_m = 1.32 kg/m³. The reactor operates for 310 days per year. What is the total amount of glucose produced per year in tonnes?
a) 56.3 tonnes yr^-1
b) 56.6 tonnes yr^-1
c) 56.7 tonnes yr^-1
d) 56.5 tonnes yr^-1
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Answer: c
Explanation: The rate of lactose conversion is equal to the difference between inlet and outlet mass flow rates of lactose:
F(s_i – s_f) = 1.56 m³ h^-1 (9.5 – 0.19) kg m^-3 = 14.5 kg h^-1
Converting this to an annual rate based on 310 d per year and a molecular weight for lactose of 342:

The enzyme reaction is:
lactose + H₂0 → glucose + galactose.
Therefore, from reaction stoichiometry, 315 kg mol glucose are produced per year. The molecular weight of glucose is 180; therefore:

Sanfoundry Global Education & Learning Series – Bioprocess Engineering.

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