User's guide of Proceessing Modflow 5.0

280 Processing Modflow
6.50. Except the four fixed-head cells at the right-hand side of the dam, the initial hydraulic
head for all cells are 10 m.
Q ' K @ B @
(h 2
1
& h 2
2 )
2L
' K @ h 1 % h 2
2
6.5.3 Seepage Surface through a Dam
Fig. 6.50 Model grid and the boundary conditions
The first step in solving this problem is to carry out a steady-state flow simulation with these
data. Fig. 6.51 shows the calculated hydraulic heads. By comparing the calculated heads with
the elevation of the cell bottom, we can easily find that the hydraulic heads of some of the cells
at the upper-right corner of the model are lower than the cell bottom. This means that these cells
went dry. In the second step, these dry cells will be defined as inactive cells by setting IBOUND
= 0 and a steady-state flow simulation will be carried out again. Now, it is possible that some of
the calculated heads are higher than the top elevation of the highest active cell. In this case,
these cells will be defined as active and a steady-state flow simulation will be performed again.
This iterative solution will be repeated until the water table remains unchanged between two
iteration steps. Fig. 6.52 shows the calculated head distribution and the form of the seepage
-5 3
surface. The seepage rate is about 4.8 × 10 m /s/m and the total seepage rate through the dam
-3 3
(lenght 100 m) is 4.8 × 10 m /s.
The analytical solution of the seepage rate after the Dupuit assumption is
@ B @ h 1 &h 2
L
(6.2)
where B is the length of the dam, L is the thickness of the dam, K is the hydraulic conductivity,
h and h are the heads at the uptream and downstream sides of the dam, respectively. The
1 2
modified form of the analytical solution is Darcy’s Law with a mean transmissivity of
K (h + h )/2. For this example with h = 10 m, h = 2 m, L = 10 m, B = 100 m and K = 1 × 10
1 2 1 2
-3 3
m/s, the seepaga rate Q is exactly equal to 4.8 × 10 m /s.
-5