Question

A poorly educated engineer designs a chlorine disinfection tank as a CSTR rather than as a PFR. It has a detention time of 30 min.

(a) Calculate the reduction in bacteria [1-(COUT/CIN)]*100% that would be achieved using an ideal PFR with a 30 minute detention time operating at the design chlorine dose. The first-order rate constant K = 0.23/min.
(b) What reduction in bacteria does the CSTR achieve?
(c) The first-order rate of kill depends also on the chlorine concentration, k = K (Cl2), where k = new rate constant, K = constant at the design chlorine dose, and (Cl2) = multiple of the design chlorine concentrations.
How much would the chlorine have to be increased (by what multiple) to increase the rate so that the CSTR would reduce the bacteria from 100,000 cells/ml to 100 cells/ml? Assume K is linearly related to chlorine concentration.

Answer 1

(a). 99.89% reduction

(). 87.34% reduction

(c). (Cl₂) = 14478.3

Step-by-step explanation:

given in the question that a CSTR tank is designed instead of a PFR with a time of detention T given as 30 min.

(a). from question , the first order rate constant K is 0.23/min.

the reduction in bacteria is calculated with a PFR by

-dCA/dt = KCA

integrating both sides we have

CA/CAo = e∧-Kt

1 - C out/ C in = 1 - e∧-Kt

inputing value we have

1 - C out/ C in = 1 - e∧-(0.23)(30)

= 1 - e∧-6.9 = 0.9989

to cal. in percent

∴ [1 - C out/ C in] × 100 = 0.9989 × 100 = 99.89%

(b). using CSTR in the bacteria reduction we have

T = CAoXA/-rA = CAo-CA/KCA

= 1/K [CAo/Ca - 1]

TK = CAo/CA -1

CA/CAo = 1/1+KT

1 - CAo/CA = 1 - 1/1+KT

= 1 - 1/1+KT

= 1 - 1/1+(0.23×30)

= 0.8734 which in percent gives 87.34%

(c). we have from the question that CAin = 100,000 cells.ml

while CAout = 100 cells/ml

using a CSTR gives;

T = CAin - CAout / -rA = 100000 - 100 / -rA

where -rA = K(Cl₂)

∴ T = 100000 - 100 / K(Cl₂)

100000 - 100 = 30×0.23(Cl₂)

∴ (Cl₂) = 14478.3