Geo-environmental CIVI 469 lab evaluate different slurry reactor configuration f

Question

Geo-environmental CIVI 469 lab

evaluate different slurry reactor configuration for the following case:
subsurface soil at a site is contaminated with diesel fuel at a concentration of 1800 mg/kg. treatment system is required to handle slurry flow of 0.04m^3/min. the required final diesel concentration in the soil is 100mg/kg. the reaction is first order with a rate constant of 0.1/min as determined from bench-scale study tests. four different configurations of slurry bio reactors in plug flow (PFR) mode are considered. determine the final effluent concentration from each of these arrangements and if it meets the cleanup requirements.
a.one 4-m^3 reactor
b> two 2-m^3 reactor in series
c. one 1-m^3 reactor followed by one 3-m^3 reactor
d. one 3-m^3 reactor followed by one 1-m^3 reactor
use the following formula
Cn/C0=(Cn/Cn-1)(Cn-1/Cn-2)....(C2/C1)(C1/C0)=e^(-kntn)e^(-kn-1 tn-1)...(e-K1t1)

Explanation / Answer

a. For the 4-m3 reactor, the residence time = V/Q = 4 m3/(0.04 m3/min)
= 100 minutes.

Use Eq. IV.3.6 to find the final effluent concentration

Cout/Cin= Cout/1800= 1/ (1)+(0.1)+(100)

Cout = 164 mg/kg. (It exceeds the cleanup level.)

b. For the two 2-m3 reactors, the residence time = V/Q = 2 m3/(0.04
m3/min) = 50 minutes each.

Use Eq. IV.5.1 to find the final effluent concentration

C2/Co= (C2/1800)=(C2/C1)(C1/Co)=[(1)/ (1)+(0.1)+(50)] [(1)/ (1)+(0.1)+(50)]

Cout = 50 mg/kg. (It is below the cleanup level.)

c. The residence time of the first reactor = 1 m3/(0.04 m3/min) = 25
minutes. The residence time of the second reactor = 3 m3/(0.04
m3/min) = 75 minutes.

Use Eq. IV.5.1 to find the final effluent concentration

C2/Co= (C2/1800)=(C2/C1)(C1/Co)=[(1)/ (1)+(0.1)+(25)] [(1)/ (1)+(0.1)+(75)]

Cout = 60.5 mg/kg. (It is below the cleanup level.)

d. The residence time of the first reactor = 3 m3/(0.04 m3/min) = 75
minutes. The residence time of the second reactor = 1 m3/(0.04
m3/min) = 25 minutes.

C2/Co= (C2/1800)=(C2/C1)(C1/Co)=[(1)/ (1)+(0.1)+(75)] [(1)/ (1)+(0.1)+(25)]

Cout = 60.5 mg/kg. (It is below the cleanup level.)

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