1. Derivation of the Dupuit Equation - Unconfined Flow

2. Dupuit Assumptions
For unconfined ground water flow Dupuit developed a theory that allows for a simple solution based off the following assumptions:
1) The water table or free surface is only
slightly inclined
2) Streamlines may be considered horizontal
and equipotential lines, vertical
3) Slopes of the free surface and hydraulic
gradient are equal

3. Derivation of the Dupuit Equation
Darcy’s law gives one-dimensional flow per unit width as:
q = -Kh dh/dx
At steady state, the rate of change of q with distance is zero, or
d/dx(-Kh dh/dx) = 0
OR (-K/2) d2h2/dx2 = 0
Which implies that,
d2h2/dx2 = 0

4. Dupuit Equation Integration of d2h2/dx2 = 0 yields h2 = ax + b
Where a and b are constants. Setting the boundary
     condition h = ho at x = 0, we can solve for b
b = ho2
Differentiation of h2 = ax + b allows us to solve for a,
a = 2h dh/dx
And from Darcy’s law,
hdh/dx = -q/K

5. Dupuit Equation So, by substitution h2 = h02 – 2qx/K
Setting h = hL2 = h02 – 2qL/K
Rearrangement gives
q = K/2L (h02- hL2) Dupuit Equation
Then the general equation for the shape of the parabola is
h2 = h02 – x/L(h02- hL2) Dupuit Parabola
However, this example does not consider recharge to the aquifer.

6. Cross Section of Flow q

7. Adding Recharge W - Causes a Mound to Form
Divide

8. Dupuit Example Example: 2 rivers 1000 m apart K is 0.5 m/day
average rainfall is 15 cm/yr
evaporation is 10 cm/yr
water elevation in river 1 is 20 m
water elevation in river 2 is 18 m
Determine the daily discharge per meter width into each
River.

9. Example L = 1000 m Dupuit equation with recharge becomes
h2 = h02 + (hL2 - h02) + W(x - L/2)
If W = 0, this equation will reduce to the parabolic
Equation found in the previous example, and
q = K/2L (h02- hL2) + W(x-L/2)
Given:
L = 1000 m
K = 0.5 m/day
h0 = 20 m
hL= 28 m
W = 5 cm/yr = x 10-4 m/day

10. Example For discharge into River 1, set x = 0 m
q = K/2L (h02- hL2) + W(0-L/2)
= [(0.5 m/day)/(2)(1000 m)] (202 m2 – 18 m2 ) +
(1.369 x 10-4 m/day)(-1000 m / 2)
q = – 0.05 m2 /day
The negative sign indicates that flow is in the opposite direction
From the x direction. Therefore,
q = 0.05 m2 /day into river 1

11. Example For discharge into River 2, set x = L = 1000 m:
q = K/2L (h02- hL2) + W(L-L/2)
= [(0.5 m/day)/(2)(1000 m)] (202 m2 – 18 m2 ) +
(1.369 x 10-4 m/day)(1000 m –(1000 m / 2))
q = m2/day into River 2
By setting q = 0 at the divide and solving for xd, the
water divide is located m from the edge of
River 1 and is 20.9 m high

12. Flow Nets - Graphical Flow Tool
Q = KmH / n
n = # head drops
m= # streamtubes
K = hyd cond
H = total head drop

13. Flow Net in Isotropic Soil
Portion of a flow net is shown below Y
Stream tube F
Curvilinear Squares

14. Flow Net Theory Streamlines Y and Equip. lines  are .
Streamlines Y are parallel to no flow boundaries.
Grids are curvilinear squares, where diagonals cross at right angles.
Each stream tube carries the same flow.

15. Seepage Flow under a Dam