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Vilka [71]
2 years ago
6

Two piezometers have been placed along the direction of flow in a confined aquifer that is 30.0 m thick. The piezometers are 280

m apart. The difference in piezometric head between the two is 1.4 m. The aquifer hydraulic conductivity is 50 m · day−1, and the porosity is 20%. Estimate the travel time for water to flow between the two piezometers.
Engineering
1 answer:
Dafna11 [192]2 years ago
8 0

Answer:

time = 224 days

Explanation:

given data

thick = 30 m

piezometers =  280 m

head between  two = 1.4 m

aquifer hydraulic conductivity = 50 m

porosity =  20%

solution

we get here average pole velocity that is get by using Darcy law that is

Va = \frac{k}{\eta } \times \frac{\Delta h}{L}   ................1

here Va is average pole velocity and k is hydraulic conductivity and \eta is porosity  

here v is = k \times  \frac{dh}{dl}   ...........2

v = 50 × \frac{1.4}{280}

v = 0.25 m/day

and here average linear velocity Va will be

Va = \frac{v}{\eta }  

Va = \frac{0.25}{0.2}  

Va = 1.25 m/day  

travel time for water will be

time = \frac{280}{1.25}  

time = 224 days

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Explanation:

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6 0
3 years ago
At a point on the free surface of a stressed body, the normal stresses are 20 ksi (T) on a vertical plane and 30 ksi (C) on a ho
victus00 [196]

Answer:

The principal stresses are σp1 = 27 ksi, σp2 = -37 ksi and the shear stress is zero

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The expression for the maximum shear stress is given:

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The principal stress is:

\sigma _{x}+\sigma _{y} =\sigma _{p1}+\sigma _{p2}

Where

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σp2 = -30 ksi

\sigma _{p1}  +\sigma _{p2}=-10 ksi (equation 1)

\tau _{M} =\frac{\sigma _{p1}-\sigma _{p2}}{2} \\\sigma _{p1}-\sigma _{p2}=2\tau _{M}\\\sigma _{p1}-\sigma _{p2}=32*2=64ksi equation 2

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3 years ago
Consider the expansion of a gas at a constant temperature in a water-cooled piston-cylinder system. The constant temperature is
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Answer:

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Explanation:

According to the first thermodynamic law, the energy must be conserved so:

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This equation can be solved by integration between an initial and a final state:

(1) \int\limits^1_2 {} \, dQ = \int\limits^1_2 {} \, dU - \int\limits^1_2 {} \, dW

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\int\limits^1_2 {- P} \, dV = \int\limits^1_2 {- \frac{n*R*T}{V}} \, dV

In this particular case as the systems is closed and the temperature constant, n, R and T are constants:

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from fick's first law

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Take C_A as point  on which nitrogen concentration is 2 kg/m^3

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