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galina1969
8 days ago
12

A permeability test was run on a compacted sample of dirty sandy gravel. The sample was 175 mm long and the diameter of the mold

175 mm. In 90 s the discharge under a constant head of 38 em was 405 cm3 The sample had a dry mass of 4950 g and its Ps was 2710 kg!m3 Calculate (a) the coefficient of permeability, (b) the seepage velocity, and (c) the discharge velocity during the test.
Physics
1 answer:
Maru [2.3K]8 days ago
8 0

Answer:

(a). The coefficient of permeability is 8.6\times10^{-3}\ cm/s.

(b). The seepage velocity is 0.0330 cm/s.

(c). The discharge velocity during the test is 0.0187 cm/s.

Explanation:

Given that,

Length = 175 mm

Diameter = 175 mm

Time = 90 sec

Volume= 405 cm³

We need to calculate the discharge

Using formula of discharge

Q=\dfrac{V}{t}

Insert the values into the formula

Q=\dfrac{405}{90}

Q=4.5\ cm^3/s

(a). We will compute the coefficient of permeability

Applying the formula for permeability

Q=kiA

k=\dfrac{Q}{iA}

k=\dfrac{Ql}{Ah}

Where, Q=discharge

l = length

A = cross-sectional area

h=constant head that causes flow

Plugging the value into the formula

k=\dfrac{4.5\times175\times10^{-1}}{\dfrac{\pi(175\times10^{-1})^2}{4}\times38}

k=8.6\times10^{-3}\ cm/s

The coefficient of permeability measures as 8.6\times10^{-3}\ cm/s.

(c). To ascertain the discharge velocity during the testing phase

Utilizing the discharge velocity formula

v=ki

v=\dfrac{kh}{l}

Substituting the values into the equation

v=\dfrac{8.6\times10^{-3}\times38}{17.5}

v=0.0187\ cm/s

The discharge velocity during the test measures 0.0187 cm/s.

Thus, (a). The coefficient of permeability is 8.6\times10^{-3}\ cm/s.

(b). The seepage velocity computes at 0.0330 cm/s.

(c). The observed discharge velocity during the test equals 0.0187 cm/s.

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