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Solution Type. Head at center of island (ft). 1D confined*. 2D confined. 20.00. R = (2 T) h(x) / ( L 2 – x 2 ). Inverse solution for R. Island Recharge Problem Analytical Solutions (with R = 0.00305 ft/day). h(x) = R (L 2 – x 2 ) / 2T. 21.96.
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Solution Type Head at center of island (ft) 1D confined* 2D confined 20.00 R = (2 T) h(x) / ( L2 – x2) Inverse solution for R Island Recharge Problem Analytical Solutions (with R = 0.00305 ft/day) h(x) = R (L2 – x2) / 2T 21.96 *1D confined inverse solution with h(0)= 20, gives R = 0.00278 ft/day.
2D Confined Numerical Solution Gauss-Seidel Iteration Formula
Island Recharge Problem 4 X 7 Grid (a = 4000 ft)
Solution Type Head at center of island (ft) 1D confined analytical 2D confined numerical (a = 4000 ft) 19.87 2D confined analytical 20.00 2D confined numerical (a = 1000 ft) ? Island Recharge Problem Solutions (with R = 0.00305 ft/day) 21.96 Too low; error = 0.13 ft Same conceptual model
Sensitivity of solution to grid spacing 1. The approximation to the derivative improves as grid spacing zero: as x 0 2. The calculation of the water budget improves with smaller grid spacing.
IN increases as grid spacing decreases. x, y IN 4000 ft 6710E 2 1000 8243E 2 500 8511E 2 250 8647E 2 L y/2 For full area (L x 2L) IN = 8784E 2 ft3/day x/2 2L
Unconfined version of the Island Recharge Problem R groundwater divide ocean ocean b x = - L x = 0 x = L
Seepage face Derivation of 2D unconfined equation with the Dupuit Assumptions • No vertical flow • No seepage face h Impermeable rock
h Unconfined version of the Island Recharge Problem R groundwater divide ocean ocean b datum x = - L x = 0 x = L Let b = 100 ft & K = 100 ft/d so that at x=L, Kb= 10,000 ft2/day 1. Could define an unconfined “transmissivity”: Tu = Kh
2. Re-write the equation as follows: General Approaches for Unconfined Aquifers 1.Define an unconfined “transmissivity”: Tu = Kh Update Tx = Kxh and Ty = Kyh during the solution. MODFLOW uses this approach.
h Homogeous & isotropic conditions Unconfined version of the Island Recharge Problem R groundwater divide ocean ocean b datum x = - L x = 0 x = L Let b = 100 ft & K = 100 ft/d so that at x=L, Kb= 10,000 ft2/day
This equation is linear in h2. Let v = h2 Poisson Equation Solve the finite difference equations for v and then solve for h as v
Unconfined Poisson Equation Confined Poisson Equation
Unconfined aquifer Poisson eqn solved for v h= SQRT(v)
Outflow terms for the Water Balance of the Unconfined Problem with x = y Qx = -K v /2 Also: Qy = -K v /2
2D confined numerical solution for the 4x 7 grid: h = 19.87 ft Fluxes calculated using v… Qx = K v /2
Solution Type Head at center of island (ft) 1D confined* 1D unconfined 2D confined 20.00 Island Recharge Problem Analytical Solutions (with R = 0.00305 ft/day) 21.96 ?
h at x =0 h(L) = hL = 100 Unconfined version of the Island Recharge Problem R groundwater divide ocean ocean b datum x = - L x = 0 x = L Let b = 100 ft & K = 100 ft/d so that at x=L, Kb= 10,000 ft2/day
at x =0 h(L) = hL = 100 Analytical solution for 1D unconfined version of the problem h2(x) = R (L2 -x2 )/K + (hL)2 h(x) = [R (L2 -x2 )/K] + (hL)2 1D solution for unconfined aquifer Governing Eqn. Boundary conditions
Analytical solution for 1D “confined” version of the problem h(x) = R (L2 – x2) / 2T Analytical solution for 1D “unconfined” version of the problem h(x) = [R (L2 -x2 )/K] + (hL)2
ho datum h Island Recharge Problem R groundwater divide ocean ocean b datum x = - L x = 0 x = L Let b = 100 ft & K = 100 ft/d so that at x=L, Kb= 10,000 ft2/day At groundwater divide: confined aquifer h = ho unconfined aquifer h=b+ ho
h(x) = [R (L2 -x2 )/K] + (hL)2 “unconfined” at x = 0; h = b + ho & hL = b h(x) = R (L2 – x2) / 2T “confined” at x = 0; h = ho ho = R L2 / 2T R = 2 Kb ho / L2 b hL (b + ho)2 = [R L2 /K] + b2 0 L R = (2 Kb ho / L2)+ (ho2 K / L2) To maintain the same head (ho) at the groundwater divide as in the confined system, the 1D unconfined system requires that recharge rate, R, be augmented by the term shown in blue.