Question

A field density test is performed on a compacted fill to check the dry density and water content. The QA technician reported that the total density of the fill was 1.9 Mg/m3, the water content was 18% and Gs = 2.67. Assume total volume 1 m3.

(a) Find porosity
(b) degree of saturation in percentage.

Answer 1

Answer: The porosity of the fill is 18% and the degree of saturation is 100%.

Explanation: To find the porosity, we need to determine the volume of water and the volume of solids in the fill. (a) To calculate the volume of water: Volume of water = Water content × Total volume Volume of water = 0.18 × 1 m^3 Volume of water = 0.18 m^3 To calculate the volume of solids: Volume of solids = Total volume - Volume of water Volume of solids = 1 m^3 - 0.18 m^3 Volume of solids = 0.82 m^3 Now, to find the porosity: Porosity = (Volume of voids ÷ Total volume) × 100 Since the volume of voids is equal to the volume of water: Porosity = (Volume of water ÷ Total volume) × 100 Porosity = (0.18 m^3 ÷ 1 m^3) × 100 Porosity = 18% (b) The degree of saturation represents the percentage of voids that are filled with water. It can be calculated as: Degree of saturation = (Volume of water ÷ Volume of voids) × 100 Since the volume of voids is equal to the volume of water: Degree of saturation = (Volume of water ÷ Volume of water) × 100 Degree of saturation = 100% Therefore, the porosity of the fill is 18% and the degree of saturation is 100%.