Analytical solution for 1D unconfined version of the problem

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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
 h
 h
2
2
R



2
2
T
x
y
Gauss-Seidel Iteration Formula
hi , j
m 1

hi 1, j m  hi 1, j m 1  hi , j 1m  hi , j 1m 1
4
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
ocean
b
x=-L
x=0
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
ocean
b
x=-L
h
x=0
Let b = 100 ft & K = 100 ft/d
so that
at x=L, Kb= 10,000 ft2/day
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
)   2R
x
x
y
y
Unconfined version of the Island Recharge Problem
R
groundwater
divide
ocean
ocean
b
x=-L
 2h2
x
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
 2h2
x 2

 2h2
y 2
2R

K
This equation is linear in h2.
Let v = h2
 v
2
 v
2
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
 v
2
 v
2
2R



2
2
K
x
y
Confined Poisson Equation
 h
2
 h
2
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 (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 (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
“confined”
h(x) = R (L2 – x2) / 2T
ho = R L2 / 2T
at x = 0; h = ho
R = 2 Kb ho / L2
“unconfined”
h (x) = [R (L2 - x2 )/K] + (hL)2
at x = 0; h = b + ho

& hL = b
(b + ho)2 = [R L2 /K] + b2
b
0
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.
hL
L
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