Question

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.

Answer 1

Answer:

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 2

The hydraulic conductivity and transmissivity of the aquifer are respectively; 0.0032 m/s and 0.11 m³/s

What is the hydraulic conductivity?

A) We are given;

Pump rate; Q = 0.5 m³/s

thickness of quifer; b = 34 m

depth 1; r₁ = 50 m

depth 2; r₂ = 100 m

distance 1; S₁ = 0.9 m

distance 2; S₂ = 0.4 m

Formula for the pump rate is;

Q = 2π × b × k × (S₁ - S₂)/(In r₂/r₁)

making k the subject gives;

k = Q(In r₂/r₁)/(2π × b × (S₁ - S₂))

k = 0.5(In 100/50)/(2π × 34 × (0.9 - 0.4))

Solving for K gives;

Hydraulic conductivity is; k = 0.0032 m/s

Transmissivity is;

T = K * b

T = 0.0032 * 34

T = 0.11 m³/s

B) Formula for radius of incfluence is;

S_w = S₁ - [(Q/2π × b × k) In (r_w/r₁)]

Plugging in the relevant values gives;

S_w = 4.338 m

C) Formula for expected drawdown is;

R = r₁ e^(2πbk(S_w - S₁)/Q)

R = 100 * e^(2π*34*0.0032(-78.9)/0.5)

R = 147.7 m

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