PowerPoint PresentationSlide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25WzhKzyhKyxhKxzyx)()()(General governing equationfor steady-state, heterogeneous, anisotropic conditions02222zhxh2D 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 Problem02222zhxh0xh0xh0zhh = c x + zoUnconfined aquiferb = 1 mxzzxb02222zhxhAquiferbTothproblemxz02222yhxh2D horizontal flow in an aquifer with constant thickness, b.AquiferbxyFigure from Hornberger et al. 1998bunconfined aquiferb is not constantconfined aquifer2D horizontal flow in an aquifer with constant thickness, b.Aquiferbxywith rechargeWzhKzyhKyxhKxzyx)()()(with a source/sink termTRyhxh2222Poisson Equation2D horizontal flow; homogeneous and isotropic aquiferwith constant aquifer thickness, b, so that T=Kb.RbWzhKzyhbKyxhbKxzyx)()()(2D horizontal flowMap of Long Island, N.Y.South ForkCharles Edward Jacob (1914-1970)Consultant to the Town of Southampton, NYDecember 1968boceanoceangroundwater divideC.E. Jacob’s Conceptual Model of theSouth Fork of Long IslandRx = 0 x = Lx = - LWe can simulate this system assuming horizontal flow in a “confined” aquifer if we assume that T= Kb.h datumwater table1D approximation used by C.E. JacobTRdxhd22h(L) = 0 0dxdhat x =0oceanoceanRx = 0 x = Lx = - LGoverning Eqn.Boundary Conditionsh(x) = R (L2 – x2) / 2TAnalytical solution for 1D “confined” version of the problemC.E. Jacob’s ModelTRdxhd22h(L) = 0 0dxdhat x =0Governing Eqn.Boundary conditionsR = (2 T) h(x) / ( L2 – x2)Forward solutionInverse solution for RRearrange eqn to solve for T,given value for R and h(0) = 20 ft.Inverse solution for TLR = (2 T) h(x) / ( L2 – x2)Inverse solution for RSolve for R with h(x) = h(0) = 20 ft.Observation wellon the groundwaterdividexyoceanoceanT= 10,000 ft2/dayL = 12,000 ft2LLTRyhxh2222Island Recharge Problemoceanwell• 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.xyoceanoceanSolve the forward problem:GivenR= 0.00305 ft/dT= 10,000 ft2/daySolve for h at each nodal point2LL = 12,000 ftTRyhxh2222Island Recharge ProblemoceanwellTRahhhhhmjimjimjimjimji44211,1,1,1,11,TRyhxh2222Gauss-Seidel Iteration Formula for 2D Poisson Equation with x = y = aWrite the finite difference approximation:Island Recharge Problem4 X 7 GridWater BalanceIN = OutIN = R x AREAOut = outflow to the oceanTop 4 rowsRed 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 theaverage head in the areasurrounding the node.Top 4 rowsIN =R x Area = R (L-x/2) (2L - y/2)2LLAlso: IN = R (2.5)(5.5)(a2)Top 4 rowsx/2xxx/2Top 4 rowsOUT = Qy +  QxQy = K (x b) (h/y)Note: x = y  Qy = T hQx = K (y b)(h/x) or Qx = T hQxQyyQy = (Th) /2Welly/2Bottom 4 rowsQx = (T h)/2Island Recharge Problem4 X 7 GridWater Budget