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