Question

Compare the reactor volume required to achieve 95% contaminant removal efficiency for a CMFR vs. a PFR for the following conditions: steady-state, first-order reaction, flow rate 14 m3/d, and reaction rate coefficient 0.05 d. Your answers should represent the dimensionless ratio VCMER/VPR
It's not 840, 5320, or 6.34

Answer 1

According to the given data following is the answer to the problem. If it isn't what required there will be something wrong with the given data.

Assumptions:

1. Steady State condition
2. First order reaction
3. No outside gain or loss due to percolation or evaporation
4. Non conservative compound

Given:

Q = 14 m³/day

k = 0.05 /day

n = 95%

Answer:

n = [ 1- ( C₀/Cin) ] * 100

= 95 %

Therefore,

C₀/Cin = 1 - (95/100)

= 0.05

For CMFR,

C₀/Cin = [1/ ( 1+K*a )]

0.05 = [ 1/ ( 1+ 0.05 * a) ]

a = 380 days

V_CMFR = a * Q

= 380 days * 14 m³/day

= 5320 m³

For PFR,

C₀/Cin = e ^{- (ka)}

0.05 = e ^{- 0.05 * a}

    a = 59.915 days

V_PFR = a * Q

   = 59.915 days * 14 m³/day

= 838.81 m³

Dimensionless Ratio,

V_CMFR / V_PFR = 5320 / 838.81

= 6.342

Hence the required ratio is 6.342