 # Derivation of the Dupuit Equation Unconfined Flow Dupuit

• Slides: 15 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) d 2 h 2/dx 2 = 0 Which implies that, d 2 h 2/dx 2 = 0 Dupuit Equation Integration of d 2 h 2/dx 2 = 0 yields h 2 = ax + b Where a and b are constants. Setting the boundary condition h = ho at x = 0, we can solve for b b = ho 2 Differentiation of h 2 = ax + b allows us to solve for a, a = 2 h dh/dx And from Darcy’s law, hdh/dx = -q/K Dupuit Equation So, by substitution h 2 = h 02 – 2 qx/K Setting h = h. L 2 = h 02 – 2 q. L/K Rearrangement gives q = K/2 L (h 02 - h. L 2) Dupuit Equation Then the general equation for the shape of the parabola is h 2 = h 02 – x/L(h 02 - h. L 2) 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: 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. Example Dupuit equation with recharge becomes h 2 = h 02 + (h. L 2 - h 02) + W(x - L/2) If W = 0, this equation will reduce to the parabolic Equation found in the previous example, and q = K/2 L (h 02 - h. L 2) + W(x-L/2) Given: L = 1000 m K = 0. 5 m/day h 0 = 20 m h. L= 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/2 L (h 02 - h. L 2) + W(0 -L/2) = [(0. 5 m/day)/(2)(1000 m)] (202 m 2 – 18 m 2 ) + (1. 369 x 10 -4 m/day)(-1000 m / 2) q = – 0. 05 m 2 /day The negative sign indicates that flow is in the opposite direction From the x direction. Therefore, q = 0. 05 m 2 /day into river 1 Example For discharge into River 2, set x = L = 1000 m: q = K/2 L (h 02 - h. L 2) + W(L-L/2) = [(0. 5 m/day)/(2)(1000 m)] (202 m 2 – 18 m 2 ) + (1. 369 x 10 -4 m/day)(1000 m –(1000 m / 2)) q = 0. 087 m 2/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 = Km. H / n n = # head drops m= # streamtubes K = hyd cond H = total head drop Flow Net in Isotropic Soil Portion of a flow net is shown below Y Stre am t ube F Curvilinear Squares Flow Net Theory 1. Streamlines Y and Equip. lines are . 2. Streamlines Y 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