What is the influent and effluent glucose concentration (mM)? A completely stirr

Question

What is the influent and effluent glucose concentration (mM)? A completely stirred tank reactor is used for growing penicillin according to a zero-order reaction. The input, glucose (CoH120%), is converted to various organic yeasts which eventually produce the penicillin. The flow to the system is 20 liters per minute (LPM) and the glucose conversion rate is 4 mg/(min L). The influent glucose concentration is 800 mg/L and the effluent must be less than 100 mg/L. a. What is the influent and effluent glucose concentration (in mM)? b. What is the smallest reactor (in m) capable of facilitating this reaction?

Explanation / Answer

For a completely Stirred Tank Reactor we have,

Flow rate Q = 20 LPM

Glucose conversation rate = 4 mg/(min×L)

Influent glucose concentration Cin = 800 mg/L = 0.8 g/L

Maximum Effluent glucose concentration Cin = 100 mg/L = 0.1g/L

a.)

Molecular mass of glucose = 12×6+1×12+16×6 = 180 g per mole.

Thus, 180 grams of glucose is equal to one mole of glucose

1 gram of glucose = 1/180 moles

Influent concentration = 0.8/180 moles per litre = 0.004444 mole/L

= 4.44 mM

Effluent concentration = 0.1/180 moles per litre = 0.005555 mole/L

= 5.555 mM

b.)

Now as we know for CSTR

Cout/Cin = 1/(1+K×t)

t= Hydraulic retention time =(1/K)×{(Cin/Cout)-1}

t =(1/4)×{8-1} =1.75 minutes

As t = V/Q

V= Q×t =20×1.75 =35 litre

This is te minimum volume of tank required.

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