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# Effects of Boundary Condition on Shape of Flow Nets

Effects of Boundary Condition on Shape of Flow Nets. ?. ?. ?. ?. ?. Seepage and Dams. Flow nets for seepage through earthen dams Seepage under concrete dams Uses boundary conditions (L &amp; R) Requires curvilinear square grids for solution. After Philip Bedient Rice University.

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## Effects of Boundary Condition on Shape of Flow Nets

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1. Seepage and Dams Flow nets for seepage through earthen dams Seepage under concrete dams Uses boundary conditions (L & R) Requires curvilinear square grids for solution After Philip Bedient Rice University

2. Flow Net in a Corner: Streamlines Y are at right angles to equipotential F lines

3. Flow to Pumping Center: Contour map of the piezometric surface near Savannah, Georgia, 1957, showing closed contours resulting from heavy local groundwater pumping (after USGS Water-Supply Paper 1611).

4. Example: Travel Time Luthy Lake h = 100 m The deadly radionuclide jaronium-121 was released in Luthy Lake one year ago. Half-life = 365 days Concentration in Luthy Lake was 100 g/L. If orphans receive jaronium over 50 g/L, they will poop their diapers constantly. How long will it take for the contaminated water to get to Peacock Pond and into the orphanage water supply? Will jaronium be over the limit? h = 92 m Peacock Pond Orphanage

5. Luthy Lake h = 100 m h = 92 m Peacock Pond Travel time Contamination in Luthy Lake 1 2 3 Note: because l terms are squared, accuracy in estimating l is important. 4 Travel time is only 131 days!

6. Luthy Lake h = 100 m h = 92 m Peacock Pond Supposed we used averages? Travel time Contamination in Luthy Lake 1 2 3 4 Travel time is only 131 days!

7. Two Layer Flow System with Sand Below Ku / Kl = 1 / 50

8. Two Layer Flow System with Tight Silt Below Flow nets for seepage from one side of a channel through two different anisotropic two-layer systems. (a) Ku / Kl = 1/50. (b) Ku / Kl = 50. Source: Todd & Bear, 1961.

9. Flow nets in anisotropic media SZ2005 Fig. 5.11

10. Flownets in Anisotropic Media So far we have only talked about flownets in isotropic material. Can we draw flownets for anisotropic circumstances? For steady-state anisotropic media, with x and y aligned with Kx and Ky, we can write the flow equation: dividing both sides by Ky:

11. Flownets in Anisotropic Media Next, we perform an extremely cool transformation of the coordinates: This transforms our governing equation to: Laplace’s Eqn!

12. Flownets in Anisotropic Media Steps in drawing an anisotropic flownet: Determine directions of max/min K. Rotate axes so that x aligns with Kmax and y with Kmin 2. Multiply the dimension in the x direction by (Ky/Kx)1/2 and draw flownet. 3. Project flownet back to the original dimension by dividing the x axis by (Ky/Kx)1/2

13. Flownets in Anisotropic Media Example: Ky Kx Kx = 15Ky

14. Flownets in Anisotropic Media Ky Kx Kx = 15Ky

15. Ky Kx Flownets in Anisotropic Media Kx = 15Ky

16. Flownets in Anisotropic Media Kx = 15Ky

17. Flownets in Anisotropic Media Kx = 15Ky

18. Flownets in Anisotropic Media Kx = 15Ky

19. 25% Flownets in Anisotropic Media Kx = 15Ky

20. 25% Flownets in Anisotropic Media Kx = 15Ky

21. Flownets in Anisotropic Media Kx = 15Ky

22. Flownets in Anisotropic Media Kx = 15Ky

23. Flownets in Anisotropic Media Kx = 15Ky

24. Flownets in Anisotropic Media Kx = 15Ky

25. Flownets in Anisotropic Media Kx = 15Ky

26. Flow Nets: an example • A dam is constructed on a permeable stratum underlain by an impermeable rock. A row of sheet pile is installed at the upstream face. If the permeable soil has a hydraulic conductivity of 150 ft/day, determine the rate of flow or seepage under the dam. After Philip Bedient Rice University

27. Flow Nets: an example The flow net is drawn with: m = 5 head drops = 17 After Philip Bedient Rice University

28. Flow Nets: the solution • Solve for the flow per unit width: q = m K = (5)(150)(35/17) = 1544 ft3/day per ft • total change in head, H • number of head drops After Philip Bedient Rice University

29. Flow Nets: An Example There is an earthen dam 13 meters across and 7.5 meters high.The Impounded water is 6.2 meters deep, while the tailwater is 2.2 meters deep. The dam is 72 meters long. If the hydraulic conductivity is 6.1 x 10-4 centimeter per second, what is the seepage through the dam if the number of head drops is = 21 K = 6.1 x 10-4cm/sec = 0.527 m/day After Philip Bedient Rice University

30. Flow Nets: the solution From the flow net, the total head loss, H, is 6.2 -2.2 = 4.0 meters. There are (m=) 6 flow channels and 21 head drops along each flow path: Q = (mKH/number of head drops) x dam length = (6 x 0.527 m/day x 4m / 21) x(dam length) = 0.60 m3/day per m of dam = 43.4 m3/day for the entire 72-meter length of the dam After Philip Bedient Rice University

31. Aquifer Pumping Tests Why do we need to know T and S (or K and Ss)? -To determine well placement and yield -To predict future drawdowns -To understand regional flow -Numerical model input -Contaminant transport How can we find this information? -Flow net or other Darcy’s Law calculation -Permeameter tests on core samples -Tracer tests -Inverse solutions of numerical models -Aquifer pumping tests

32. Steady Radial Flow Toward a Well • Aquifer Equation, based on assumptions becomes an ODE for h(r) : • steady flow in a homogeneous, isotropic aquifer • fully penetrating pumping well & horizontal, • confined aquifer of uniform thickness, thus • essentially horizontal groundwater flow • flow symmetry: radially symmetric flow (Laplace’s equation) (Laplace’s equation in x,y 2D) (Laplace’s equation in radial coordinates) • Boundary conditions • 2nd order ODE for h(r) , need two BC’s at two r’s, sayr1andr2 • Could be • one Dirichlet BC and one Neumann, say at well radius, rw , if we know the pumping rate, Qw • or two Dirichlet BCs, e.g., two observation wells.

33. Steady Radial Flow Toward a Well

34. Q b h2 h1 Island with a confined aquifer in a lake We need to have a confining layer for our 1-D (radial) flow equation to apply—if the aquifer were unconfined, the water table would slope down to the well, and we would have flow in two dimensions.

35. Steady Radial Flow Toward a Well Constant C1: ODE Multiply both sides by r dr Separate variables: Integrate Integrate again: • Evaluate constant C1 using continuity. • The Qw pumped by the well must equal • the total Q along the circumference for every • radius r. We will solve Darcys Law for r(dh/dr).

36. Steady-State Pumping Tests Q b h2 h1 r2 r1

37. Integrate Thiem Equation

38. Steady-State Pumping Tests h r Why is the equation logarithmic? This came about during the switch to radial coordinates. What is the physical rationale for the shape of this curve (steep at small r, flat at high r)?

39. Think of water as flowing toward the well through a series of rings. As we approach the well the rings get smaller. A is smaller but Q is the same, so must increase. (Driscoll, 1986)

40. hlc one log cycle The Thiem equation tells us there is a logarithmic relationship between head and distance from the pumping well. (Schwartz and Zhang, 2003)

41. The Thiem equation tells us there is a logarithmic relationship between head and distance from the pumping well. hlc one log cycle If T decreases, slope increases (Schwartz and Zhang, 2003)

42. Steady-State Pumping Tests Can we determine S from a Thiem analysis? No—head isn’t changing!

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