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 25Slide 26Slide 27Slide 28Slide 29Slide 30Slide 31Slide 32Slide 33Slide 34Slide 35Conceptual ModelA descriptive representationof a groundwater system that incorporates an interpretation of the geological & hydrological conditions. Generally includes information about the water budget.a set of equations that describesthe physical and/or chemicalprocesses occurring in a system.Mathematical Model• Governing equation• Boundary conditions• Initial conditions for transient simulationsR x yQyxz1. Consider flux (q) through REV2. OUT – IN = - Storage3. Combine with: q = -KK grad hqDerivation of the Governing Equationdiv q = 0q = - K grad hSteady state mass balance eqn.Darcy’s lawgrad hq equipotential linegrad hqIsotropic AnisotropicKx = KzKx  Kzzxxzx’z’global localKxx Kxy KxzKyx Kyy KyzKzx Kzy KzzK’x 0 00 K’y 00 0 K’zbedding planesqxqyqz= - zhyhxhq = - K grad hzhKyhKxhKqzhKyhKxhKqzhKyhKxhKqzzzyzxzyzyyyxyxzxyxxxKxx Kxy KxzKyx Kyy KyzKzx Kzy KzzWthSzhKzyhKyxhKxszyx)()()(RthSyhTyxhTxyx)()(RthSyhhKyxhhKxyyx)()(2D confined:2D unconfined w/Dupuit assumptions:Storage coefficient (S) is either storativity or specific yield.S = Ss b & T = K bGeneral 3D equationHetergeneous, anisotropic, transient, sink/source termTypes of Boundary Conditions1. Specified head (including constant head)2. Specified flow (including no flow)3. Head-dependent flowFrom conceptual model to mathematical model…Toth ProblemLaplace Equation2D, steady state02222zhxh0xh0xh0zhh = c x + zoCross section through an unconfined aquifer.Water table formsWater table formsthe upper boundary conditionthe upper boundary condition0)()()( zhKzyhKyxhKxzyxGoverning Eqn. for TopoDrive2D, steady-state, heterogeneous, anisotropicbhoceanoceangroundwater divide“Confined” Island Recharge ProblemRx = 0 x = Lx = - LWe can treat this system as a “confined” aquifer if we assume that T= Kb.datumTRyhxh2222ArealviewWater table is the solution.Water table is the solution.Poisson’s Eqn.2D horizontal flow through an unconfined aquifer where T=Kb.bhoceanoceangroundwater divideRx = 0 x = Lx = - LdatumUnconfined version of the Island Recharge ProblemKRyhxh 2222222Water table is the solution.Water table is the solution.(Pumping can be accommodated by appropriatedefinition of the source/sink term.)2D horizontal flow through an unconfined aquifer underthe Dupuit assumptions.02222yhxh02222zhxhVertical cross section through an unconfinedaquifer with the water table as the upper boundary.2D horizontal flow in a confined aquifer; solutionis h(x,y), i.e., the potentiometric surface.02222yvxv2D horizontal flow in an unconfined aquiferwhere v= h2. Solution is h(x,y), i.e., the water table.All three governing equations are the LaPlace Eqn.t = 0t > 0BC:h (0, t) = 16 m; t > 0h (L, t) = 11 m; t > 0datum0 L = 100 mxIC: h (x, 0) = 16 m; 0 < x < L(represents static steady state)thTSxh22Reservoir Problem1D transient flow through a confined aquifer.confining bedSolution techniques…• Analytical solutions• Numerical solutions finite difference (FD) methods finite element (FE) methods• Analytic element methods (AEM)Toth ProblemzxAnalytical SolutionNumerical Solution02222zhxhh(x,z) = zo + cs/2 – 4cs/2 …hi,j = (hi+1,j + hi-1,j + hi,j+1 + hi,j-1)/4zxcontinuous solution discrete solution(eqn. 2.1 in W&A)0xh0xh0zhh = c x + zoMathematicalmodelh = 10011090 ftToth Problemmesh vs block centered grids another viewx = y = a = 20 ft200 ftGrid DesignThree options for solving the set of algebraic equationsthat result from applying the method of FD or FE:• Iteration• Direct solution by matrix inversion• A combination of iteration and matrix solutionNote: The explicit solution for the transient flow equation is anothersolution technique, but in practice is never used.Examples of Iteration methods include:Gauss-Seidel IterationSuccessive Over-Relaxation (SOR)02222yhxh02221,1,2,1,,1yhhhxhhhjiijjijijiji41,1,,1,1,jijijijijihhhhhLet x=y=a411,1,1,1,11,mjimjimjimjimjihhhhhGauss-Seidel Formula for 2D Laplace EquationGeneral SOR Formula)( ,1,,1,mjimjimjimjihhhh Relaxation factor= 1 Gauss-Seidel< 1 under-relaxation>1 over-relaxation, typically between 1 and 2TRahhhhhmjimjimjimjimji44211,1,1,1,11,Gauss-Seidel Formula for 2D Poisson EquationSOR Formula)( ,1,,1,mjimjimjimjihhhh Relaxation factor= 1 Gauss-Seidel< 1 under-relaxation>1 over-relaxation(Eqn. 3.7W&A)mm+1m+2m+3solution(Initial guesses)Iteration fora steady state problem.Iteration levelsnn+1n+2n+3Steady statetttInitial conditions(steady state)Transient Problemsrequire time steps.Time levelsthhTSxhhhninininini1211)(2Explicit ApproximationthTSxh22thhTSxhhhninininini1211111)(2Implicit Approximation211211111)(21()(2xhhhxhhhnininininini22xhwhere  = 1 for fully implicit  = 0.5 for Crank-Nicolson  = 0 for explicitIn general:• Explicit solutions do not require iteration but are unstable with large time steps.• We can derive the stability criterion by writingthe explicit approx. in a form that looks like the SORiteration formula and setting the terms in theposition occupied by omega equal to 1.• For the 1D governing equation used in the reservoirproblem, the stability criterion is:1)(22xStT<TxSt2)(5.0<orNote that critical t value is directlydependent on grid spacing,  x.Implicit solutions require iteration