Engineering, 03.03.2020 05:04 Wanna14ever

Prob. 4.2.1. A well that pumps at a constant rate of 0.5 m3/s fully penetrates a confined aquifer of 34-m thickness. After a long period of pumping, steady-state drawdowns are measured at two observation wells 50 and 100 m from the pumping well as 0.9 m and 0.4 m, respectively. Determine: (a) the hydraulic conductivity and transmissivity of the aquifer; (b) the radius of influence of the pumping well; (c) the expected drawdown in the pumping well if the radius of the well is 0.4 m.

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Answer from: lildanielmabien

Part (a)

K = 0.00406 m / s

T = 0.14 m2 / s

Part (b)

R = Radius of influence of Pumping well = 237.94 m

Part (c)

S = Drawdown at well = 3.68 m

Explanation:

General formula for wells at confined aquifer is

Q =

{2*3.1416*K*B*(S1-S2)} / {2.303 log (R/r)}

............ eq (A)

Where ,

Q = Discharge

K = Hydraulic conductivity

B = Thickness of aquifer

S1 = Draw-down at point 1

S2 = Draw-down at point 2

R = Radius of influence of well

r = Radius of well

Part (a)

We will use

Q =

{2*3.1416*K*B*(S1-S2)} / {2.303 log (r2/r1)}

......... eq (1)

Where,

r 1 = Distance of first observation well from main well

r2 = Distance of 2nd observation well from main well

Given data: Q = 0.5 m3/s B = 34 m r 1 = 50 m

r2 = 100 m S1 = 0.9 m S2 = 0.4 m

Put all these values in equation 1

0.5 = 2*3.1416*K*34*(0.9 - 0.5) / {2.303 log(100 / 50)}

Write Equation in terms of K = hydraulic conductivity

K = {0.5*2.303 log (100 / 50)} / {2*3.1416*34*(0.9 - 0.5)}

K = 0.00406 m / s

Now Transmissivity

T = K*B

T = 0.14 m2 / s

Part (b)

To calculate radius of influence, use equation (A)

Q = {2*3.1416*K*B*(S1-S2)} / {2.303 log (R/r)}

At distance R from main well drawdown (S2) = 0

use r = r1 = 50 m and S1 = 0.9

use ln ( R / r) = 2.303 log (R / r)

0.5 = 2* 3.1416 * 0.00406*34* (0.9 - 0) / ln(R / 50)

Write equation in terms of R

ln( R / 50) = 2* 3.1416 * 0.00406*34* (0.9 - 0) / 0.5

ln( R / 50) = 1.56

Take anti log (e) of above equation

R / 50 = 4.76

R = Radius of influence of Pumping well = 237.94 m

Part (c)

Use equation A

Q = {2*3.1416*K*B*(S1-S2)} / {2.303 log (R/r)}

S1 = ?

Put S2 = 0 R = 237.94 r = 0.4

0.5 = 2* 3.1416* 0.00406*34*(S1 - 0) / 2.303 log(237.94 / 0.4)

Write equation in terms of S1

S1 = 0.5* 2.303 log(237.94 / 0.4) / 2*3.1416*0.00406*34

S1 = Drawdown at well = 3.68 m

Answer from: Quest

no he didnt use it

explanation:

he said no

Answer from: Quest

how is this a question? in order for me to answer it i have to understand the question

explanation:

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