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.

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

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

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 =

  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 =

 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

the equation above shows how temperature f, measured in degrees fahrenheit, relates to a temperature c, measured in degrees celsius. based on the equation, which of the following must be true?

a temperature increase of 1 degree fahrenheit is equivalent to a temperature increase of

5

9

degree celsius.

a temperature increase of 1 degree celsius is equivalent to a temperature increase of 1.8 degrees fahrenheit.

a temperature increase of

5

9

degree fahrenheit is equivalent to a temperature increase of 1 degree celsius.

a) i only

b) ii only

c) iii only

d) i and ii only

answer:

i dont know the answer but good luck

explanation: