Page 291 - Materials Chemistry, Second Edition

274 Practical Design Calculations for Groundwater and Soil Remediation

Determine (1) the amount of activated carbon in each 55-gal unit, (2) the
amount of xylene that each unit can remove before being exhausted, and (3)
the minimum number of the 55-gal units needed.

Solution:
(a) Volume of the activated carbon inside a 55-gal drum = (πr )(h)
2
= (π)[(1.5/2) ](3) = 5.3 ft 3
2
Amount of the activated carbon inside a 55-gal drum = (V)(ρ )
b
= (5.3 ft )(28 lb/ft ) = 148 lb
3
3
(b) Amount of xylene that can be retained by a drum before the
GAC becomes exhausted
= (amount of the GAC)(actual adsorption capacity)
= (148 lb/drum)(0.193 lb xylene/lb GAC) = 28.6 lb xylene/drum
(c) Assuming a design air flow velocity of 60 ft/min, the required
cross-sectional area for the GAC adsorption can be found by
using Equation (7.4) as:
Q 200
A GAC = = = 3.33 ft 2
Airflowvelocity 60
If the adsorption system is tailor-made, then a system with a cross-
sectional area of 3.33 ft will do the job. However, the off-the-
2
shelf 55-gal drums are to be used, so we need to determine
the number of drums that will provide the required cross-
sectional area.
Area of the activated carbon inside a 55-gal drum = (πr )
2
= (π)[(1.5/2) ] = 1.77 ft /drum
2
2
Number of drums in parallel to meet the required hydraulic
loading rate
= (3.33 ft ) ÷ (1.77 ft /drum) = 1.88 drums
2
2
So, use two drums in parallel to provide the required cross-
sectional area. The total cross-sectional area of two drums is
equal to 3.54 ft (= 1.77 × 2).
2
Discussion:
1. The bulk density of vapor-phase GAC is typically in the neigh-
borhood of 30 lb/ft . The amount of activated carbon in a 55-gal
3
drum is approximately 150 pounds.
2. The minimum number of 55-gal drums for this project is two
to meet the air flow velocity requirement. The actual number