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. By taking the parameters of Q5, and the batch reaction time t_b.
a) 5.0 h
b) 10.0 h
c) 15.0 h
d) 20.0 h
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Answer: a

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. By taking the pararmeters of Q7 and Determine batch culture times required to achieve 90% substrate conversion.
a) 5.1 h
b) 5.3 h
c) 5.5 h
d) 5.7 h
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Answer: d

9. By taking the parameters question 8 and question 7, determine batch culture times required to produce 100 g ethanol.
a) 3.2 h
b) 3.4 h
c) 3.6 h
d) 3.8 h
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Answer: b

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. By taking Q10 and at the flow rate of Q10, what is the cell density?
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. By taking Q10 and Q11 into account and at the flow rate of Q10, what concentration of ethanol is produced?
a) 5.1 g l^-1
b) 5.3 g l^-1
c) 5.5 g l^-1
d) 5.7 g l^-1
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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. By taking the parameters of Q14 into consideration, estimate how many tonnes of glucose are produced per year?
a) 56.3 tonnes yr^-1
b) 56.6 tonnes yr^-1
c) 56.7 tonnes yr^-1
d) 56.5 tonnes yr^-1
View Answer

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:

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