Question

An enzyme acts as a catalyst in the fermentation of A to form product R. An aqueous feed stream containing the enzyme and compound A flows into a CSTR at 25 L/min, and the initial concentration of A is 2 mol*L^-1 . Determine the volume of the CSTR (in Liters, expressed to the nearest integer ) needed to achieve 98% conversion of reactant A. You may assume that the enzyme concentration and volumetric flowrates are constant. The maximum rate of destruction of the substrate is 0.4 mol/(L min). When the substrate concentration is 0.5 mol/L, the rate of destruction of the substrate is 0.2 mol/(L min), i.e. half of the maximum rate.

Answer 1

The design equation for a CSTR is:

V / (F C_Ao) = X_A / r

where;

V = volume of CSTR

F = volumetric flowrate = 25 L / min

C_Ao = initial concentration of A = 2 mol / L

r = rate of reaction

X_A = conversion of reactant A = 0.98

The enzymatic reactions follow the Michaelis-Menten Kinetics which is given by:

r = v_max C_A / (K_M + C_A)

where;

v_max = maximum rate of destruction of substrate = 0.4 mol / (L min)

K_M = substrate concentration when rate of destruction of substrate is half of the maximum value = 0.5 mol/L

We know;

C_A = C_Ao (1 - X_A)

Using this in the rate expression:

r = v_max C_Ao (1 - X_A) / (K_M + C_Ao (1 - X_A))

Substituting the values of all the variables:

r = 0.4 X 2 X (1 - 0.98) / (0.5 + 2(1 - 0.98))

which on solving gives:

r = 0.0296 mol / L.min

Substituting this in the CSTR design equation:

V / (F C_Ao) = X_A / r

V / (25 X 2) = 0.98 / 0.0296

V / 50 = 33.11

V = 33.11 X 50

V = 1655 L

which is the volume of the CSTR.