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iVinArrow [24]
2 years ago
8

An aquifer pump test was conducted in a confined aquifer where the initial piezometric surface was at elevation 12.45 m, and wel

l logs indicate that the thickness of the aquifer is 14 m. The well was pumped at 28.24 L/s, and after 1 day the piezometric levels at 25 m and 75 m from the pumping well were measured as 10.89 and 11.45, respectively. Assuming steady state conditions, estimate the transmissivity and hydraulic conductivity of the aquifer.
Physics
1 answer:
murzikaleks [220]2 years ago
4 0

Answer:

T = 0.0088 m²/s

Explanation:

given,

initial piezometric elevation = 12.5 m

thickness of aquifer = 14 m

discharge = 28.24 L/s = 0.02824 m³/s

we know                              

k = \dfrac{qln(\dfrac{R_2}{R_1})}{2\pi D(H_2-H_1)}

k = \dfrac{0.02824 \times ln(\dfrac{75}{25})}{2\pi \times 14 (11.45-10.89)}                                                                        

k = 0.629 mm/sec

Transmissibilty

T = k × H                          

T = 0.629 × 14 × 10⁻³

T = 0.0088 m²/s

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consider east-west direction along X-axis  and north-south direction along Y-axis

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using the equation

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3 years ago
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Answer:

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In a second experiment, you decide to connect a string which has length L from a pivot to the side of block A (which has width d
Salsk061 [2.6K]

Answer:

The answer is in the explanation

Explanation:

A)

i) The blocks will come to rest when all their initial kinetic energy is dissipated by the friction force acting on them. Since block A has higher initial kinetic energy, on account of having larger mass, therefore one can argue that block A will go farther befoe coming to rest.

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B) The equation of motion for block A is

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Here, \nu_{s} is the coefficient of friction between the block and the surface of the table. Equation (1) can be easily integrated to get

v(t) = C-\nu_{s}gt \quad (2)

Here, C is the constant of integration, which can be determined by using the initial condition

v(t=0) = v_{0}\Rightarrow C = v_{0} \quad (3)

Hence

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Block A will stop when its velocity will become zero,i.e

0 = v_{0}-\nu_{s}gT\Rightarrow T = \frac{v_{0}}{\nu_{s}g} \quad (5)

Going back to equation (4), we can write it as

\frac{\mathrm{d} x}{\mathrm{d} t} = v_{0}-\nu_{s}gt\Rightarrow x(t) = v_{0}t-\nu_{s}g\frac{t^{2}}{2}+D \quad (6)

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The distance travelled by block A before stopping is

x(t=T) = v_{0}T-\nu_{s}g\frac{T^{2}}{2} = v_{0}\frac{v_{0}}{\nu_{s}g}-\nu_{s}g\frac{v_{0}^{2}}{2\nu_{s}^{2}g^{2}} = \frac{v_{0}^{2}}{2\nu_{s}g} \quad (8)

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D)

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ii) The forces acting on the block are

a) Tension: Acting in the radially inwards direction, hence it is always perpendicular to the velocity of the block, therefore it does not change the speed of the block.

b) Friction: Acting tangentially, in the direction opposite to the velocity of the block at any given time, therefore it decreases the speed of the block.

The speed decreases linearly with time in the same manner as derived in part (C), using the expression for tension in part (D)(i) we can see that the tension in the string also decreases with time (in a quadratic manner to be specific).

8 0
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Show all work.
lys-0071 [83]

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Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

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