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Sidana [21]
3 years ago
6

Why is the drawdown cone for a well completed in a low permeability aquifer narrower and deeper than a drawdown cone for a well

in a high permeability'aquifer? a) Low permeability aquifers typically produce water at a higher discharge rate b) The pump in a low permeability well casing typically has a higher capacity c) It takes a greater hydraulic head to drive the groundwater laterally to the well casing in the lower permeability aquifer d) The radius of influence of a high permeability well is typically shorter than that of a low permeability well
Engineering
1 answer:
densk [106]3 years ago
5 0

Answer:

c) It takes a greater hydraulic head to drive the groundwater laterally to the well casing in the lower permeability aquifer

Explanation:

The groundwater are contains under the rock and in the open spaces within the rocks and the unconsolidated sediments. Aquifer refers to the underground layers of the permeable sand or rocks that transmits the groundwater below water table which provides a sufficient supply of water to the well. Groundwater is present everywhere where there is porosity in the rocks and it depends on the permeability of the rocks to allow them flow.

A drawdown cone is completed in the lower permeable aquifer deeper and narrower than the high permeable aquifer as it takes more amount hydraulic head or energy to drive groundwater to the well casing which is in the lower permeable aquifer.

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A soil is at a void ratio e = 0.90 with a specific gravity of the solid particles Gs = 2.70.
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Answer:

The correct answers are:

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when S = 100%, it means that the soil is fully saturated and this will give the upper limit of water content.

when S < 100%, the soil is partially saturated, and this will give the lower limit of water content.

Note; S = 0% means that the soil is perfectly dry. Hence, when s = 1 will give the lowest limit of water content.

To get the relationship between water content and saturation, we will manipulate the equations above;

w =  \frac{M_w}{Ms}

Recall; mass = Density × volume

w = \frac{V_wP_w}{V_sP_s} ------(5)

From eqn. (2)  G_{s} = \frac{P_s}{P_w}

∴ \frac{1}{G_s} = \frac{P_w}{P_s} ------(6)

putting eqn. (6) into (5)

w = \frac{V_w}{V_sG_s} -----(7)

Again, from eqn (1)

V_s = \frac{V_v}{e}

substituting into eqn. (7)

w = \frac{V_w}{\frac{V_v}{e}{G_s} } = \frac{V_w e}{V_vG_s} \\ but \frac{V_w}{V_v}  = S

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Given, G_{s} = 2.7, e= 0.9

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0.50 = \frac{V_w}{0.9* 100}  = 45cm^{3}

mass of water = P_wV_w= 1 * 45 = 45_{g}

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