Question

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 1

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!!!