Fermentation Reactors

A continuous reactor at steady state contains 0.04 g.l^-1 of biomass and 0.02 g.l^-1 of phenol. The feed contained 0.1 g.l^-1 of phenol. The biomass yield would be

A:

0.4 g.g^-1

B:

0.5 g.g^-1

C:

1.0 g.g^-1

D:

5.0 g.g^-1

Answer: B

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The growth of an organism on glucose is described by the following Monod model parameters: μ_m = 0.5 h^-1 and Ks =0.1 g.l^-1, if the concentration of glucose in the feed is 10 g.l^-1 and the dilution rate is set to 0.4 h^-1, then the steady state concentration of glucose in the effluent will be

A:

0 g.l^-1

B:

0.5 g.l^-1

C:

1.0 g.l^-1

D:

10 g.l^-1

Answer: B

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Mixing per unit volume is observed to be poorest in

A:

continuous packed bed reactor

B:

continuous fluidized bed bioreactor

C:

continuous airlift bioreactor

D:

none of the above

Answer: A

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A chemostat has a liquid volume of 2 litres and is being fed at a rate of 4 litres per hour. Dilution rate for this reactor will be

A:

2 litres

B:

2 litres per hour

C:

2 h^-1

D:

4 litres per hour

Answer: C

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In fed-batch fermentation, mass of substrate utilized is calculated using the relationship

A:

initial mass of substrate in the reactor + final mass of substrate in the reactor + flow rate x [substrate in the feed]

B:

initial mass of substrate in the reactor - final mass of substrate in the reactor + flow rate x [substrate in the feed]

C:

initial mass of substrate in the reactor + final mass of substrate in the reactor - flow rate x [substrate in the feed]

D:

initial mass of substrate in the reactor + final mass of substrate in the reactor - flow rate

Answer: B

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Questions for: Fermentation Reactors