105328616-3-GW-Flow-Equations

					Equations of Groundwater Flow

Description of ground water flow is based on:

Darcy’s Law

Continuity Equation – describes conservation of fluid
mass during flow through a porous medium; results in a
partial differential equation of flow.


Laplace’s Eqn - most important in math
Derivation of 3-D GW Flow
Equation from Darcy’s Law
                        z



       ρx                               ∂
       V                           ρ x + ( V)
                                    V      ρx
                                        ∂x

               y
    Mass In - Mass Out = Change in Storage

      ∂     ∂    ∂
              ρ ) ρz) 0
     − (V − ( y − (V =
        ρx)   V
      ∂
      x     ∂
            y    ∂
                 z
Derivation of 3-D GW Flow
Equation from Darcy’s Law
 Replace Vx, Vy, and Vz with Darcy using Kx, Ky, and Kz


    ∂  ∂ ∂
        h     ∂ ∂
               h    ∂
                     h
             ρy + ρz
      ρx +  K   K  0
     K               =
    ∂
    x   ∂ ∂ ∂ ∂ ∂
        x  y   y  z  z

 Divide out constant ρ, and assume Kx= Ky= Kz = K

              ∂h ∂h ∂h
               2    2   2
                  + 2 + 2 =0
              ∂x ∂
                2
                   y   ∂z
             ∇h=0cldaa E.
              2
                  le p e n
                 a Llc q
Transient Saturated Flow
      ∂     ∂    ∂    ∂
              ρ ) ρz)
     − (V − ( y − (V = (n
        ρx)   V         ρ)
      ∂
      x     ∂
            y    ∂
                 z    ∂
                      t
 A change in h will produce change in ρ and n, replaced
  with specific storage Ss = ρg(α + nΒ). Note, α is the
  compressibility of aquifer and B is comp of water,
  therefore,

     ∂ ∂ ∂
         h    ∂ ∂ ∂
                h    h    ∂h
      x 
       K   + y 
              K          =s
                  + z  S
                     K
     ∂ ∂ ∂ ∂ ∂ ∂
     x   x  y   y  z   z    ∂
                            t
Solutions to GW Flow Eqns.
Solutions for only a few simple problems can be obtained
directly
Generally need to apply numerical methods to address
complex boundary conditions.
               ∂h ∂h ∂h
                2    2   2
                   + 2 + 2 =0
               ∂x ∂
                 2
                    y   ∂z
               ∇h=0cldaa E.
                2
                    le p e n
                   a Llc q

          h0                                h1
Transient Saturated Flow
Simplifying by assuming K = constant in all dimensions
And assuming that S = Ssb, and that T = Kb yields

                   h
           ∂ ∂  ∂ ∂  ∂ ∂  S∂
             h         h    s h
              +   +   =
           x ∂  y ∂  z ∂
           ∂  x ∂  y ∂ z K t∂

            ∂2h ∂2h ∂2h S ∂  s h
                 + 2+ 2 =
            ∂x 2
                  ∂y   ∂z   K∂ t
                  S∂h
            ∇ h=
             2
                      rm o,
                         a
                      fo Jcb     hs
                                  e
                                 Ti
                  T∂t
Steady State Flow to Well
Simplifying by assuming K = constant in all dimensions
and assuming that Transmissivity T = Kb and
Q = flow rate to well at point (x,y) yields



             ∂h ∂h
              2
                 + 2=−
                      2
                       ( )
                        x
                      Q ,y
             ∂x ∂
               2
                  y    T

				
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posted:10/7/2012
language:English
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Description: Drainage Engineering