Borrow soil is used to fill a 75,000m3 depression. The borrow soil has the following characteristics. Density: 1540kg/m3, water content: 8%, specific gravity of the solids: 2.66. The final in-place dry density should be 1790 kg/m3 and the final water content should be 13%. a) How many m3 of borrow soil are needed to fill the depression? b) Assuming no evaporation loss, what water mass is needed to achieve 13% moisture?

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wagonbelleville wagonbelleville · Answer 1 · 2020-02-05T10:19:28+01:00

Answer

given,

Volume of Depression, V = 75000 m³

borrow soil

Density of soil,γ = 1540 Kg/m³

water content,w = 8 % = 0.08

Specific gravity of solid,G = 2.66

Final in-place

dry density = 1790 kg/m³

water content = 13 % = 0.13

a) Volume of borrow soil require to fill the depression

Mass of solid solid,m= dry density of inplace soil x Volume of depression

= 1790 x 75000

m = 1.3425 x 10⁸ Kg

dry density of the borrow pit

[tex]\gamma_d = \dfrac{\gamma}{1 + w}[/tex]

[tex]\gamma_d = \dfrac{1540}{1 + 0.08}[/tex]

[tex]\gamma_d = 1425.93\ kg/m^3[/tex]

Volume of borrow soil required

[tex]V = \dfrac{m}{\gamma_d}[/tex]

[tex]V = \dfrac{1.3425\times 10^8}{1425.93}[/tex]

V = 94149 m³

b) Water required

[tex]W = 1790\times 75000\times 0.13 - 1425.93\times 94149\times 0.08[/tex]

W = 6.71 x 10⁶ Kg

Water required to achieve 13% moisture is equal to W = 6.71 x 10⁶ Kg