Page 41 - Materials Chemistry, Second Edition

24 Practical Design Calculations for Groundwater and Soil Remediation

Solution:
Thickness intervals for all areas of the plume are the same at 5 ft.
Volume of the impacted soil (using Equation [2.9])
= (5 ft)(350 ft ) + (5ft)(420 ft ) + (5 ft)(560 ft ) + (5 ft)(810 ft )
2
2
2
2
= (1,750 + 2,100 + 2,800 + 4,050) ft = 10,700 ft = 396 yd 3
3
3
or = (22.5 − 17.5)(350) + (27.5 − 22.5)(420) + (32.5 − 27.5)(560)
+ (37.5 − 32.5)(810) = 10,700 ft 3
Assuming the total bulk density of soil is 112 lb/ft , the mass of the
3
impacted soil = (10,700 ft )(112 lb/ft )
3
3
= 1,198,400 lb = 599 tons
Example 2.13: Determine the Amount of Impacted
Soil in the Vadose Zone
For the project described in Example 2.9, after the USTs were removed, five
soil borings were installed. Soil samples were taken every 5 ft below ground
surface (bgs). However, not all the samples were analyzed due to budget con-
straints. The areas of the plume at a few depths were determined as follows:

2
Depth (ft bgs) Area of the Plume at that Depth (ft )
15 0
20 350
25 420
35 810
40 0

Determine the volume and mass of the impacted soil left in the vadose zone.

Strategy:
The depth intervals given are not all the same; therefore, each plume
area represents a different depth interval. For example, the sample
taken at 25-ft depth represents a 7.5-ft interval, from 22.5 ft to 30 ft.

Solution:
Volume of the impacted soil (using Equation [2.9])
= (5)(350) + (7.5)(420) + (7.5)(810) ft 3
= 10,915 ft = 406 yd 3
3
or = (22.5 − 17.5)(350) + (30 − 22.5)(420) + (37.5 − 30)(810) = 10,915 ft 3
Assuming the total bulk density of soil is 112 lb/ft , the mass of the
3
impacted soil = (10,975 ft )(112 lb/ft )
3
3
= 1,229,200 lb = 615 tons