Page 34 - Materials Chemistry, Second Edition

Site Assessment and Remedial Investigation 17

Example 2.8: Bulk Densities, Water Content, and
Degree of Water Saturation
The average specific gravity of soil grains at a site is 2.65; porosity is equal
to 0.40; and water content (weight of moisture/weight of dry soil) is 0.12.
Determine the (dry) bulk density, total bulk density, volumetric water con-
tent, and degree of water saturation of the soil.

Solution:
(a) Basis: soil volume = 1 m of soil
3
Total pore volume of the soil = (soil volume) × (porosity of soil)
= (1)(0.4) = 0.4 m 3
Volume occupied by the soil grains
= (soil volume) – (its pore volume)
= 1 – 0.4 = 0.6 m 3
Mass of the dry soil = (volume occupied by the soil grains)
× (density of the soil grains)
= (0.6 m )(2,650 kg/m ) = 1,590 kg
3
3
Bulk density (dry) = (mass of the dry soil) ÷ (soil volume)
= 1,590 kg ÷ 1 m 3
= 1,590 kg/m = 1.59 g/cm (= 99.2 lb/ft )
3
3
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(b) Mass of water/moisture in soil
= (water content) × (mass of the dry soil)
= (0.12)[(1,590 kg/m )(1 m )] = 190.8 kg
3
3
Mass of the wet soil
= (mass of water) + (mass of the dry soil)
= 190.8 + 1,590 = 1,781 kg
Total bulk density = (mass of the wet soil) ÷ (soil volume)
= 1,781 kg ÷ 1 m = 1,781 kg/m 3
3
= 1.78 g/cm (= 111.1 lb/ft )
3
3
(c) Volume of water = (mass of water) ÷ (density of water)
= (190.8 kg) ÷ (1,000 kg/m ) = 0.19 m 3
3
Volumetric water content = (volume of water) ÷ (soil volume)
= (0.19 m ) ÷ 1.0 m = 19%
3
3
(d) Degree of water saturation
= (volume of water) ÷ (total pore volume)
= (0.19 m ) ÷ (0.4 m ) = 47.5%
3
3