Respuesta :
Answer:
Part A
The thickness of the compacted soil is approximately 4.3467 × 10⁻¹ m
Part B
The weight of water to be added is approximately 19886.[tex]\overline{36}[/tex] kN, the volume of the water added is approximately 2,027.77 m³
Explanation:
The parameters of the soil are;
The volume of sol the excavator excavates, [tex]V_T[/tex] = 10,000 m³
The moist unit weight, W = 17.5 kN/m³
The moisture content = 10%
The area of the project, A = 20,000 m²
The required dry unit weight = 18.3 kN/m³
The required moisture content = 12.5%
Part A
Therefore, we have;
The moist unit weight = Unit weight = ([tex]W_s[/tex] + [tex]W_w[/tex])/[tex]V_T[/tex]
The moisture content, MC = 10% = ([tex]W_w[/tex]/[tex]W_s[/tex]) × 100
∴ [tex]W_w[/tex] = 0.1·[tex]W_s[/tex]
∴ The moist unit weight = 17.5 kN/m³ = ([tex]W_s[/tex] + 0.1·[tex]W_s[/tex])/(10,000 m³)
1.1·[tex]W_s[/tex] = 10,000 m³ × 17.5 kN/m³ = 175,000 kN
[tex]W_s[/tex] = 175,000 kN/1.1 = 159,090.[tex]\overline{09}[/tex] kN
For the required soil, we have;
The required dry unit weight = 18.3 kN/m³ = [tex]W_s[/tex]/[tex]V_T[/tex] = 159,090.[tex]\overline{09}[/tex] kN/[tex]V_T[/tex]
[tex]V_T[/tex] = 159,090.[tex]\overline{09}[/tex] kN/(18.3 kN/m³) ≈ 8,693.4923 m³
The total volume of the required soil ≈ 8,693.4923 m³
Volume [tex]V_T[/tex] = Area, A × Thickness, d
∴ d = [tex]V_T[/tex]/A
d = 8,693.4923 m³/(20,000 m²) ≈ 4.3467 × 10⁻¹ m
The thickness of the compacted soil ≈ 4.3467 × 10⁻¹ m
Part A
The moisture content, MC = 12.5% = ([tex]W_w[/tex]/[tex]W_s[/tex]) × 100
[tex]W_w[/tex] = [tex]W_s[/tex] × MC/100 = 159,090.[tex]\overline{09}[/tex] kN × 12.5/100 = 19886.[tex]\overline{36}[/tex] kN
The weight of water to be added, [tex]W_w[/tex] = 19886.[tex]\overline{36}[/tex] kN
Where the density of water, ρ = 9.807 kN/m³
Therefore, we have;
The volume of water, V = [tex]W_w[/tex]/ρ
∴ V = 19886.[tex]\overline{36}[/tex] kN/(9.807 kN/m³) ≈ 2027.77 m³
The volume of water, V ≈ 2027.77 m³
