Calculate the CSTR volume required for 98% degradation of an organic compound. The inflow rate into the reactor is 75 L/s, the contaminant concentration is 0.05 mol/L, and the degradation rate constant (K) is 0.10/s. also calculate the PFR volume required to achieve the same level of degradation for the preceding problem.

Answer: CSTR volume required = 367500 Litres

PER volume required = 29340 Litres

Explanation: From units of the rate constant, (0.1 / s), it is evident that the reaction is first order with respect to the contaminant.

Starting from the overall material balance

Input = output + disappearance by reaction + accumulation

Input of reactant in moles/time = Fo

Output of reactant in moles/time = F = Fo (1 - X)

Disappearance by reaction = (-r) V

In a CSTR with constant volume; it is assumed that accumulation = 0

r = rate of reaction = - KC (first order reaction)

V = volume of the reactor

Fo = flow rate into the reactor in moles/s

F = flow out of the reactor.

K = rate constant

C = concentration of contaminant at any time = Co (1-X)

(Fo) X = (-r) V

(Fo) X = (KC) V = KV (Co) (1-X)

The performance equation for a CSTR and first order reaction is;

(K (Co) V) / Fo = X / (1 - X)

V = (X (Fo)) / (K (Co) (1-X))

X = 0.98, Fo = 75 L/s * 0.05 mol/L = 3.75 mol/s, K = 0.01/s, Co = 0.05 mol/L,

V = (0.98 * 3.75) / (0.01 * 0.05 * (1-0.98))

V = 367500 L

b) for PFR,

(Fo) dX = (-r) dV

(Fo) dX = (KC) dV

(Fo) dX = KCo (1-X) dV

dX / (1-X) = (K (Co) / (Fo)) dV

On integrating,

V = ((Fo) / ((Co) K) (In (1/1-X))

V = (3.75 / (0.05*0.01) (In (1/1-0.98))

V = 29340 L

QED!