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3 years ago

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The natural material in a borrow pit has a mass unit weight of 110.0 pcf and a water content of 6%, and the specific gravity of

the soil solids is 2.63. The specifications require that the soil be compacted to a dry unit weight of 122 lb. per cf and the water content be held to 5.5%.
A. How many cubic yards of borrow are required to construct an embankment having a net section volume of 245,000 cy?
B. How many gallons of water must be added per cubic yard of borrow material assuming no loss by evaporation?

1 answer:

enot [183]3 years ago

7 0

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, $G_s$ = 2.63

The desired dry unit weight, $\gamma _d$ = 122 pcf

The water content, w₁ = 5.5 %

The net section volume, $V_T$ = 245,000 cy = 6,615,000 ft³

A. $\gamma _d$ = $W_s$ / $V_T$

∴ $W_s$ = $V_T$ × $\gamma _d$ = 6,615,000 ft³ × 122 lb/ft³ = 807030000 lbs

w = ( $W_w$ / $W_s$ ) × 100

∴ $W_w$ = (w/100) × $W_s$ = (6/100) × 807030000 lbs = 48421800 lbs

The weight of solids

W = $W_s$ + $W_w$ = 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;

$W_{w1}$ = (w₁/100) × $W_s$ = (5.5/100) × 807030000 lbs = 44386650 lbs

The amount of water that must be added = $W_{w1}$ - $W_w$ = 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

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3 years ago

Given the potential field in cylindrical coordinates, V = 100/ (z2+1) rho cos φV, and point P at rho = 3m, φ = 60°, z = 2m, find

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Answer:

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Explanation:

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Find:

values at P for:

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e.) aN

f.) rhov in free space

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$V = \frac{100*3*cos(60)}{2^2 + 1}\\\\V = \frac{150}{5} = 30 V$

b)

To compute the Electric field from Volt potential we have the following relation:

E = - ∀.V

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Plug in the values for point P:

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The magnitude of the Electric Field is given by:

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E = √975.9824

E = 31.241 N/C

d)

The dV/dN is the Field Strength E in normal to the surface with vector N given by:

dV / dN = | - ∀.V |

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e)

a_N is the unit vector in the direction of dV/dN or the electric field strength E, as follows:

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f)

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