Question

2..Three formations, each 25 m thick, overlie one another. If a constant-velocity vertical flow field is set up across the set of formations with h = 120 m at the top and h = 100 m at the bottom, calculate h at the two internal boundaries. The hydraulic conductivity of the top formation is 0.0001 m/s, the middle formation 0.0005 m/s, and the bottom formation 0.0010 m/s.

Answer 1

The values of h at the two internal boundaries are :

• h₁ = 104.625 m
• h₂ = 101.55 m

Given data :

Z₁ = Z₂ = Z₃ = 25 m

h top = 120 m

h bottom = 100 m

K₁ = 0.0001 m/s

K₂ = 0.0005 m/s

K₃ = 0.0010 m/s

First step : Calculate the value of Keq

we will apply the formula below since flow is perpendicular to the bedding plane

Keq = $\frac{Z1 + Z2 + Z3 }{\frac{Z1}{K1}+\frac{Z2}{K2} + \frac{Z3}{K3} }$  ----- ( 1 )

Insert values given above into equation 1

Therefore ; Keq = 2.307 * 10⁻⁴ m/s

Next step : determine the hydraulic gradient

Hydraulic gradient ( Ieq ) = head loss / length

                                          = ( 120 - 100 ) / 3 * 25

                                   Ieq  = 0.266

Given that the flow is perpendicular to bedding plane

q1 = q2 = q3

V₁ = V₂ = V₃ = V

K₁i₁ = K₂i₂ = K₃i₃ = Keq * ieq

Hence :

V = Keq* Ieq

   = 2.307 * 10⁻⁴ * 0.266

   = 6.15 * 10⁻⁵ m/s .

Also;

K₁i₁ =  Keq * ieq = K₂i₂ = K₃i₃

therefore :

  i₁ = 0.615

  i₂ =  0.123

  i₃ = 0.0615

Final step : determine the value of h at the two internal boundaries

Pressure at point 1 ( i.e. pressure between first two formations )

h₁ = h top - i₁L₁

   = 120 - 0.615 * 25

   = 104.625 m

Pressure at point 2 ( i.e. pressure between the 2nd and 3rd formation )

h₂ = h₁ - i₂L₂

    = 104.625 - 0.123 * 25

    = 101.55 m

Therefore we can conclude that The values of h at the two internal boundaries are :  h₁ = 104.625 m , h₂ = 101.55 m

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