Derivation of the Dupuit Equation - Unconfined Flow

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

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

Dupuit Equation

Integration of d2h2/dx2 = 0 yieldsh2 = 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

Dupuit Equation

So, by substitution

h2 = h02 – 2qx/K

Setting h = hL2 = h0

2 – 2qL/KRearrangement gives

q = K/2L (h02- hL

2) Dupuit Equation

Then the general equation for the shape of the parabola is

h2 = h02 – x/L(h0

2- hL2) Dupuit Parabola

However, this example does not consider recharge to the aquifer.

Cross Section of Flow

q

Adding Recharge W - Causes a Mound to Form

Divide

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.

ExampleDupuit equation with recharge becomes

h2 = h02 + (hL

2 - h02) + W(x - L/2)

If W = 0, this equation will reduce to the parabolicEquation found in the previous example, and

q = K/2L (h02- hL

2) + W(x-L/2)Given:

L = 1000 m K = 0.5 m/day

h0 = 20 m

hL= 28 m

W = 5 cm/yr = 1.369 x 10-4 m/day

Example

For discharge into River 1, set x = 0 m

q = K/2L (h02- hL

2) + 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 /dayThe negative sign indicates that flow is in the opposite direction

From the x direction. Therefore,

q = 0.05 m2 /day into river 1

Example

For discharge into River 2, set x = L = 1000 m:

q = K/2L (h02- hL

2) + 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 = 0.087 m2/day into River 2

By setting q = 0 at the divide and solving for xd, the

water divide is located 361.2 m from the edge of

River 1 and is 20.9 m high

Flow Nets - Graphical Flow Tool

Q = KmH / n

n = # head dropsm= # streamtubesK = hyd condH = total head drop

Flow Net in Isotropic Soil

Portion of a flow net is shown below

Stream tube

Curvilinear Squares

Flow Net Theory1. Streamlines and Equip. lines are .2. Streamlines are parallel to no flow

boundaries.3. Grids are curvilinear squares, where

diagonals cross at right angles.4. Each stream tube carries the same

flow.

Seepage Flow under a Dam