Page 37 - Materials Chemistry, Second Edition

20 Practical Design Calculations for Groundwater and Soil Remediation

Total bulk density of soil in situ = (1.8 g/cm )[(62.4 lb/ft )/(1 g/cm )]
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= (1.8)(62.4) = 112.32 lb/ft 3
Total bulk density of soil in stockpiles = (1.5)(62.4) = 93.6 lb/ft 3
Mass of soil excavated
= (19,461 ft )(112.32 lb/ft ) = 2,185,800 lb
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= (23,353 ft )(93.6 lb/ft ) = 2,185,800 lb = 1,093 tons
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Discussion:
1. The fluffy factor of 1.2 is to take into account the loosening of soil
after being excavated from subsurface (the in situ soil is usually
more compacted). A fluffy factor of 1.2 means that the volume of soil
increases 20% from in situ to the stockpiles. On the other hand, the
bulk density of soil in the stockpiles would be smaller than that of in
situ soil as the result of becoming “loose” after excavation.
2. The calculated mass of the excavated soil should be the same; regard-
less, the volume of soil in the tank pit or that in the stockpiles is used.
3. For the US customary system, one (short) ton = 2,000 lb, while in
SI, one (long) ton = 1,000 kg (which is equivalent to 2,200 lb).
4. Sizes of USTs at gasoline stations nowadays are typically larger
and in the neighborhood of 10,000 gallons.

Example 2.10: Mass and Concentration Relationship of Excavated Soil

A leaky 20 m underground storage tank was removed. The excavation
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resulted in a tank pit of 4 m × 4 m × 5 m (L×W×H), and the excavated soil
was stockpiled on site. Three samples were taken from the pile, and the total
petroleum hydrocarbon (TPH) concentrations were determined to be ND
(not detectable, <100), 1,500, and 2,000 ppm. What is the amount of TPH in
the pile? Express your answers in both kilograms and liters.

Solution:
Volume of the tank pit = (4)(4)(5) = 80 m 3
Volume of soil in the tank pit before excavation
= (volume of the tank pit) − (volume of the USTs)
= 80 − 20 = 60 m 3

Average TPH concentration = (100 + 1,500 + 2,000)/3 = 1,200 ppm
= 1,200 mg/kg
Mass of TPH in soil = [(60 m )(1,800 kg/m )](1,200 mg/kg)
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= (1.08 × 10 )(1,200) mg = 1.30 × 10 mg = 130 kg
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