General governing equation for steady-state, heterogeneous, anisotropic conditions 2D Laplace Eqn. --Homogeneous and isotropic aquifer without a sink/source.

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Presentation transcript:

General governing equation for steady-state, heterogeneous, anisotropic conditions 2D Laplace Eqn. --Homogeneous and isotropic aquifer without a sink/source term. --2D flow in a profile; (Unconfined aquifer with a water table boundary condition; recharge occurs as a result of the boundary condition.)

Mathematical Model of the Toth Problem h = c x + z o Unconfined aquifer

b = 1 m xx z zz xx b Aquifer b Toth problem x z

2D horizontal flow in an aquifer with constant thickness, b. Aquifer b x y

Figure from Hornberger et al b unconfined aquifer b is not constant confined aquifer

2D horizontal flow in an aquifer with constant thickness, b. Aquifer b x y with recharge

with a source/sink term Poisson Equation 2D horizontal flow; homogeneous and isotropic aquifer with constant aquifer thickness, b, so that T=Kb. 2D horizontal flow

Map of Long Island, N.Y. South Fork Charles Edward Jacob ( ) Consultant to the Town of Southampton, NY December 1968

bocean groundwater divide C.E. Jacob’s Conceptual Model of the South Fork of Long Island R x = 0x = Lx = - L We can simulate this system assuming horizontal flow in a “confined” aquifer if we assume that T= Kb. hdatum water table

1D approximation used by C.E. Jacob h(L) = 0 at x =0 ocean R x = 0x = Lx = - L Governing Eqn. Boundary Conditions

h(x) = R (L 2 – x 2 ) / 2T Analytical solution for 1D “confined” version of the problem C.E. Jacob’s Model h(L) = 0 at x =0 Governing Eqn. Boundary conditions R = (2 T) h(x) / ( L 2 – x 2 ) Forward solution Inverse solution for R

Rearrange eqn to solve for T, given value for R and h(0) = 20 ft. Inverse solution for T L R = (2 T) h(x) / ( L 2 – x 2 ) Inverse solution for R Solve for R with h(x) = h(0) = 20 ft. Observation well on the groundwater divide

x y ocean T= 10,000 ft 2 /day L = 12,000 ft 2L L Island Recharge Problem ocean well

Head measured in an observation well is known as a head target. Targets used in Model Calibration The simulated head at the node representing the observation well is compared with the measured head. During model calibration, parameter values (e.g., R and T) are adjusted until the simulated head matches the observed value. Model calibration solves the inverse problem.

x y ocean Solve the forward problem: Given R= ft/d T= 10,000 ft 2 /day Solve for h at each nodal point 2L L = 12,000 ft Island Recharge Problem ocean well

Gauss-Seidel Iteration Formula for 2D Poisson Equation with  x =  y = a Write the finite difference approximation:

Island Recharge Problem 4 X 7 Grid

Water Balance IN = Out IN = R x AREA Out = outflow to the ocean

Top 4 rows Red dots represent specified head cells, which are treated as inactive nodes. Black dots are active nodes. (Note that the nodes along the groundwater divides are active nodes.) Head at a node is the average head in the area surrounding the node.

Top 4 rows IN =R x Area = R (L-  x/2) (2L -  y/2) 2L L Also: IN = R (2.5)(5.5)(a 2 )

Top 4 rows  x/2 xx xx

Top 4 rows OUT =  Qy +  Qx Qy = K (  x b) (  h/  y) Note:  x =  y  Qy = T  h Qx = K (  y b)(  h/  x) or Qx = T  h Qx Qy yy Qy = (T  h) /2

Well  y/2 Bottom 4 rows Qx = (T  h)/2

Island Recharge Problem 4 X 7 Grid

Water Budget Error