Respuesta :
Answer:
A. 288,030.91 cy
B. The amount of water that must be removed from the natural material is 483541.04254 gallons of water
Explanation:
The natural material in the barrow properties are;
The mass unit weight, γ = 110.0 pcf
The water content, w = 6%
The specific gravity of the soil solids, [tex]G_s[/tex] = 2.63
The desired dry unit weight, [tex]\gamma _d[/tex] = 122 pcf
The water content, w₁ = 5.5 %
The net section volume, [tex]V_T[/tex] = 245,000 cy = 6,615,000 ft³
A. [tex]\gamma _d[/tex] = [tex]W_s[/tex]/[tex]V_T[/tex]
∴ [tex]W_s[/tex] = [tex]V_T[/tex] × [tex]\gamma _d[/tex] = 6,615,000 ft³ × 122 lb/ft³ = 807030000 lbs
w = ([tex]W_w[/tex]/[tex]W_s[/tex]) × 100
∴ [tex]W_w[/tex] = (w/100) × [tex]W_s[/tex] = (6/100) × 807030000 lbs = 48421800 lbs
The weight of solids
W = [tex]W_s[/tex] + [tex]W_w[/tex] = 807030000 lbs + 48421800 lbs = 855451800 lbs
V = W/γ = 855451800 lbs/(110.0 lb/ft.³) = 7776834.54545 ft.³ = 288,030.91 cy
V = 288,030.91 cy
The amount of cubic yards of borrow required = 288,030.91 cy
B. The volume of water in the required soil is found as follows;
[tex]W_{w1}[/tex] = (w₁/100) × [tex]W_s[/tex] = (5.5/100) × 807030000 lbs = 44386650 lbs
The amount of water that must be added = [tex]W_{w1}[/tex] - [tex]W_w[/tex] = 44386650 lbs - 48421800 lbs = -4,035,150 lbs
Therefore, 4,035,150 lbs of water must be removed
The density of water, ρ = 8.345 lbs/gal
Therefore, V = 4,035,150 lbs/(8.345 lbs/gal) = 483541.04254 gal of water must be removed from the natural material
