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USPshnik [31]
3 years ago
14

A well penetrates an unconfined aquifer. Prior to pumping, the water level (head) is 25 meters. After a long period of pumping a

t a constate rate of 0.05 cubic meters per second, the drawdowns at distances of 50 and 150 meters from the well were observed to be three and 1.2 meters respectively. Computer the hydraulic conductivity of the aquifer and the radius of influence of the pumping well.

Engineering
1 answer:
GarryVolchara [31]3 years ago
3 0

Answer:

Explanation:

Find attach the solution

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Find the differential and evaluate for the given x and dx: y=sin2xx,x=π,dx=0.25
Sedaia [141]

By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.

<h3>How to determine the differential of a one-variable function</h3>

Differentials represent the <em>instantaneous</em> change of a variable. As the given function has only one variable, the differential can be found by using <em>ordinary</em> derivatives. It follows:

dy = y'(x) · dx     (1)

If we know that y = (1/x) · sin 2x, x = π and dx = 0.25, then the differential to be evaluated is:

y' = -\frac{1}{x^{2}}\cdot \sin 2x + \frac{2}{x}\cdot \cos 2x

y' = \frac{2\cdot x \cdot \cos 2x - \sin 2x}{x^{2}}

dy = \left(\frac{2\cdot x \cdot \cos 2x - \sin 2x}{x^{2}} \right)\cdot dx

dy = \left(\frac{2\pi \cdot \cos 2\pi -\sin 2\pi}{\pi^{2}} \right)\cdot (0.25)

dy = \frac{1}{2\pi}

By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.

To learn more on differentials: brainly.com/question/24062595

#SPJ1

4 0
2 years ago
For methyl chloride at 100°C the second and third virial coefficients are: B = −242.5 cm 3 ·mol −1 C = 25,200 cm 6 ·mol −2 Calcu
bogdanovich [222]

Answer:

a)W=12.62 kJ/mol

b)W=12.59 kJ/mol

Explanation:

At T = 100 °C the second and third virial coefficients are

B = -242.5 cm^3 mol^-1

C = 25200 cm^6  mo1^-2

Now according isothermal work of one mole methyl gas is

W=-\int\limits^a_b {P} \, dV

a=v_2\\

b=v_1

from virial equation  

\frac{PV}{RT}=z=1+\frac{B}{V}+\frac{C}{V^2}\\   \\P=RT(1+\frac{B}{V} +\frac{C}{V^2})\frac{1}{V}\\

And  

W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V}  } \, dV

a=v_2\\

b=v_1

Now calculate V1 and V2 at given condition

\frac{P1V1}{RT} = 1+\frac{B}{v_1} +\frac{C}{v_1^2}

Substitute given values P_1\\ = 1 x 10^5 , T = 373.15 and given values of coefficients we get  

10^5(v_1)/8.314*373.15=1-242.5/v_1+25200/v_1^2

Solve for V1 by iterative or alternative cubic equation solver we get

v_1=30780 cm^3/mol

Similarly solve for state 2 at P2 = 50 bar we get  

v_1=241.33 cm^3/mol

Now  

W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V}  } \, dV

a=241.33

b=30780

After performing integration we get work done on the system is  

W=12.62 kJ/mol

(b) for Z = 1 + B' P +C' P^2 = PV/RT by performing differential we get  

         dV=RT(-1/p^2+0+C')dP

Hence work done on the system is  

W=-\int\limits^a_b {P(RT(-1/p^2+0+C')} \, dP

a=v_2\\

b=v_1

by substituting given limit and P = 1 bar , P2 = 50 bar and T = 373 K we get work  

W=12.59 kJ/mol

The work by differ between a and b because the conversion of constant of virial coefficients are valid only for infinite series  

8 0
3 years ago
A Pelton wheel is supplied with water from a lake at an elevation H above the turbine. The penstock that supplies the water to t
gayaneshka [121]

Answer:

Following are the proving to this question:

Explanation:

\frac{D_1}{D} = \frac{1}{(2f(\frac{l}{D}))^{\frac{1}{4}}}

using the energy equation for entry and exit value :

\to \frac{p_o}{y} +\frac{V^{2}_{o}}{2g}+Z_0  = \frac{p_1}{y} +\frac{V^{2}_{1}}{2g}+Z_1+ f \frac{l}{D}\frac{V^{2}}{2g}

where

\to p_0=p_1=0\\\\\to Z_0=Z_1=H\\\\\to v_0=0\\\\AV =A_1V_1 \\\\\to V=(\frac{D_1}{D})^2 V_1\\\\\to V^2=(\frac{D_1}{D})^4 V^{2}_{1}

         = (\frac{1}{(2f (\frac{l}{D} ))^{\frac{1}{4}}})^4\  V^{2}_{1}\\\\

         = \frac{1}{(2f (\frac{l}{D})  )} \  V^{2}_{1}\\

\to \frac{p_o}{y} +\frac{V^{2}_{o}}{2g}+Z_0  =\frac{p_1}{y} +\frac{V^{2}_{1}}{2g}+Z_1+ f \frac{l}{D}\frac{V^{2}}{2g} \\\\

\to 0+0+Z_0 = 0  +\frac{V^{2}_{1} }{2g} +Z_1+ f \frac{l}{D} \frac{\frac{1}{(2f(\frac{l}{D}))}\ V^{2}_{1}}{2g}   \\\\\to Z_0 -Z_1 = +\frac{V^{2}_{1}}{2g} \ (1+f\frac{l}{D}\frac{1}{(2f(\frac{l}{D}) )} )  \\\\\to H= \frac{V^{2}_{1}}{2g} (\frac{3}{2}) \\\\\to  \frac{V^{2}_{1}}{2g} = H(\frac{3}{2})

L.H.S = R.H.S

7 0
3 years ago
How large a force is required to accelerate a 1300 kg car from rest to a speed of 20 m/s in a distance of 80 m?
topjm [15]

F=m*a

F=80*20

F =1600 ans"

7 0
3 years ago
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What type of building project is most like an enhanced skills occupation?
enot [183]

Answer:

4.modifying a building to achieve an entirely new purpose

8 0
3 years ago
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