Problem 1 Estimate the daily carbon utilization to remove chlorobenzene from 1 million gal/day of groundwater saturated
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Problem 1 Estimate the daily carbon utilization to remove chlorobenzene from 1 million gal/day of groundwater saturated
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Problem 1 Estimate the daily carbon utilization to remove chlorobenzene from 1 million gal/day of groundwater saturated with chlorobenzene. Assume a chlorobenzene concentration of 5 mg/L is acceptable for discharge to a POTW.
Problem 2 Provide a preliminary design of a carbon adsorption system ...
9 -10. 200 mL of a solution with a para-xylene concentration of 500 mg/L is placed
in each of six containers with activated carbon and shaken for 24 hours. The
samples are filtered and the concentration of p-xylene measured, yielding the
following analyses:
REGRESSION ANALYSIS ON Log X/M vs. Log Cf:
1/n = Slope 0.698058 -0.129959 Constant Log K
Std Error 0.031947 0.0476593 Std Error of Constant
R-Squared 0.991692 0.0258649 Std Error of y
F 477.4435 4 d.f.
SS reg 0.319406 0.002676 SS residual
9 -11. Estimate the daily carbon utilization to remove chlorobenzene from 1.0 MGD of
ground water saturated with chlorobenzene. Assume a chlorobenzene concentration of
5 mg/L is acceptable for discharge to a POTW.
Given:
Contaminant = chlorobenzene [C6H5CI]
Final concentration of chlorobenzene (Cg) = 5 mg/L
Saturation = 500 mg/L (Appendix A)
Density = 1.107 at 20°
Initial concentration = 500 mg/L
From Appendix A, for chlorobenzene, K = 91, l/n = 0.99
X 1
= kC f n = 91C f
0 . 99
where Cf is the final concentration.
M
∴Freundlich equation:
Since Cf = 5 mglL
X = 91 x 5 0.99 = 447. 73 mg Chlorobenzene/g Carbon = 0. 448 g Chlorobenzene/g Carbon
M
∴Carbon required = 0.448 g/g = 0.448 lb/lb
gal × (500) mg × 10 − 6 L
∴ Chlorobenzene removed = 1 × 10 × 8.34 lb = 4128.3 lbChlorobenzene
6
day L mg gal d
The density of chlorobenzene is greater than that of water. Hence Chlorobenzene can be considered as a
DNAPL and hence might be considered as a continuous source of contamination.
9-12. Provide a preliminary design of a carbon adsorption system for removal of 2, 4, 6
trichlorophenol from 250,000 gal/day of water. The following data is provided:
Bohart-Adams Model: a = 2.3 days/ft; b = -10 days in laboratory. tests where trichlorophenol concentration was
reduced from 395 mg/L to 10 mg/L at a loading of 4.0 gal/ft2 • min. The adsorption zone was 19.0 feet.
Given:
Contaminant = 2, 4, 6 trichlorophenol
water, Q = 250,000 gal/day
Bohart-Adams Model:
a = 2.3 days/ft
b = -10 days
Cin = 395 mg/L
Cout = 10 mg/L
Loading = 4.0 gal/ft2 min = V
AZ = 19.0 ft
Solution:
1. Height of Adsorption Zone = 19.0 ft = 5.79 m = 5.8 m
2. Number and Size of Units:
AZ 5 .8
n = + 1 = + 1 = 3 . 52
d 2 . 3
∴No. of Units = 4 columns
Area of lab columns = 2.043 x 10-3 m2 (3.14in2)
Loading rate in laboratory columns = 4.09 gal/ft2 • min
Applying the same loading rate for the full scale units
gal −3 m3
×1day
Q 250,000 day × 3.7854×10 gal 24hrs
× hr
60min
A= = 2
= 4.03m2 = 43.37 ft2
V 4gal 3 1 ft
× 3.7854×10−3 m ×
ft2 ⋅ min gal 9.2903×10−2 m2
3. BDST equation for 90% removal
Slope a = 2.3 days/ft = 7.52 days/m
Intercept b = -10 days
Equations of line: t = 7.52 x -10
1
Velocity of absorption zone = = 0.133m / day = (0.434 ft / day)
a
1
Carbon Utilization = Area × × UnitWeight
a
= 4.03 x 0.133 x 481 = 257.81 kg/day (5681bg/day)
2001McGraw-Hill, Inc.
Page 3 of 22
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