140 likes | 394 Views
Island Recharge Problem. L. y. 2L. ocean. ocean. well. R= 0.00305 ft/d T= 10,000 ft 2 /day. x. ocean. 1D approximation used by C.E. Jacob. R. ocean. ocean. x = - L. x = 0. x = L. at x =0. h(L) = 0. C.E. Jacob’s Model. Governing Eqn. at x =0. Boundary conditions. h(L) = 0.
E N D
Island Recharge Problem L y 2L ocean ocean well R= 0.00305 ft/d T= 10,000 ft2/day x ocean
1D approximation used by C.E. Jacob R ocean ocean x = - L x = 0 x = L at x =0 h(L) = 0
C.E. Jacob’s Model Governing Eqn. at x =0 Boundary conditions h(L) = 0 Analytical solution for 1D “confined” version of the problem h(x) = R (L2 – x2) / 2T Forward solution R = (2 T) h(x) / ( L2 – x2) Inverse solution for R
Inverse solution for R R = (2 T) h(x) / ( L2 – x2) Solve for R with h(x) = h(0) = 20 ft. You will find that R 0.00305 ft/d L Inverse solution for T Re-arrange eqn to solve for T, given value for R and h(0) = 20 ft.
Targets used in Model Calibration • Head measured in an observation well is known as a target. • 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.
Island Recharge Problem L = 12,000 ft y 2L ocean ocean well Solve the forward problem: R= 0.00305 ft/d T= 10,000 ft2/day x ocean
Water Balance IN = Out IN = R x AREA Out = outflow to the ocean
Top 4 rows Head at a node is the average head in the area surrounding the node. Red dots represent specified head cells, which are treated as inactive nodes. Note that the well node (not shown in this figure) is an active node.
Top 4 rows L 2L IN =R x Area = R (L-x/2) (2L - y/2) Also: IN = R (2.5)(5.5)(a2)
Top 4 rows Qy x/2 OUT = Qy + Qx Qy = K (x b) (h/y) or Qy = T h Qx = K (y b)(h/x) or Qx = T h Qx
Bottom 4 rows y/2 Well
Unconfined version of the Island Recharge Problem R groundwater divide h 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