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 & R) Requires curvilinear square grids for solution After Philip Bedient Rice University
Flow Net in a Corner: Streamlines are at right angles to equipotential lines
Flow to Pumping Center:Contour map of the piezometric surface near Savannah, Georgia,1957, showing closed contours resulting from heavy localgroundwater pumping (after USGS Water-Supply Paper 1611).
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 OrphanagePeacock Pond
Travel time Luthy Lake Contamination in Luthy Lake h = 100 m wi Qi K hi b 4 4 4 4 li li ni li li l1 ttotal ti n 1 vi qi qi 1 1 1 1 Qi K hi K h qi l2 2 wi b li li 4 4 n ttotal ti li2 l3 3 1 K h 1 l1 500m Note: because l terms are squared, accuracy in estimating l is l2 250m l4 4 important. l3 100m n 0.25, h 2m, K 3.6 x10 3 m/s, h = 92 m l4 50m Travel time is only 131 days!ltotal 900m Peacock Pond
2 ltotal nltotal nltotal nltotal t total simple 80days 131days Supposed we used averages? v average KJ average K(4 h /ltotal ) K(4 h) Travel time Luthy Lake Contamination in Luthy Lake h = 100 m wi Qi K hi b li l1 1 l2 2 Co exp( t) Co exp ln 2 t 100exp( ln 2 131 ) 78 gL 1 l3 3 t1 365 2 l1 500m l2 250m l4 4 l3 100m n 0.25, h 2m, K 3.6 x10 3 m/s, h = 92 m l4 50m Travel time is only 131 days!ltotal 900m Peacock Pond
Two Layer Flow System with Sand Below Ku / Kl = 1 / 50
Two Layer Flow System with Tight Silt BelowFlow nets for seepage from one side of a channel through two differentanisotropic two-layer systems. (a) Ku / Kl = 1/50. (b) Ku / Kl = 50. Source: Todd &Bear, 1961.
Flow nets in anisotropic media SZ2005 Fig. 5.11
Flownets in Anisotropic MediaSo far we have only talked about flownets inisotropic material. Can we draw flownets foranisotropic circumstances? For steady-state anisotropic media, with x and y aligned with Kx and Ky, we can write the flow equation: 2 2 h h Kx Ky 0 x2 y2 2 2 Kx h h dividing both sides by Ky: 0 Ky x2 y2
Flownets in Anisotropic Media Next, we perform an extremely cool transformation of the coordinates: 1 Ky 2 1 Kx 1 x X Kx X2 Ky x 2This transforms our governing equation to: 2 2 h h 0 Laplace’s Eqn! X2 y2
Flownets in Anisotropic MediaSteps in drawing an anisotropic flownet:1. Determine directions of max/min K. Rotate axes so that x aligns with Kmax and y with Kmin2. 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
Flownets in Anisotropic MediaExample: Ky Kx 1 1 Ky 2 1 2 Kx = 15Ky 0.26 Kx 15
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
Flow Nets: an examplePosit ion: A B C D E F G H I JDist ance 0 3 22 3 7 .5 50 6 2 .5 75 86 94 100f romf ront t oe( f t)n 1 6 .5 9 8 7 6 5 4 3 2 1.2The flow net is drawn with: m = 5 head drops = 17 After Philip Bedient Rice University
Flow Nets: the solution• Solve for the flow per unit width: total change in head, H q=mK number of head drops = (5)(150)(35/17) = 1544 ft3/day per ft After Philip Bedient Rice University
Flow Nets: An ExampleThere is an earthen dam 13 meters acrossand 7.5 meters high.The Impounded water is6.2 meters deep, while the tailwater is 2.2meters deep. The dam is 72 meters long. Ifthe hydraulic conductivity is 6.1 x 10-4centimeter per second, what is the seepagethrough the dam if the number of head dropsis = 21 K = 6.1 x 10-4cm/sec = 0.527 m/day After Philip Bedient Rice University
Flow Nets: the solutionFrom the flow net, the total head loss, H, is 6.2 -2.2 = 4.0 meters.There are (m=) 6 flow channels and21 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
Aquifer Pumping TestsWhy 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 transportHow 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
Steady Radial Flow Toward a WellAquifer Equation, based on assumptions becomes an ODE for h(r) : 2 (Laplace’s -steady flow in a homogeneous, isotropic aquifer h 0 equation) 2 h 2 h (Laplace’s -fully penetrating pumping well & horizontal, 0 equation confined aquifer of uniform thickness, thus x2 y2 in x,y 2D) essentially horizontal groundwater flow (Laplace’s 1 d dh equation in -flow symmetry: radially symmetric flow r 0 radial r dr dr coordinates)Boundary conditions -2nd order ODE for h(r) , need two BC’s at two r’s, say r1 and r2 -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.
Island with a confined aquifer in a lake Q b h2 h1 We need to have a confining layer for our 1-D (radial) flow equation toapply—if the aquifer were unconfined, the water table would slope down to the well, and we would have flow in two dimensions.
Steady Radial Flow Toward a Well dh 1 d dh Constant C1: Qw (2 r)TODE r 0 dr r dr dr dh Qw r C1Multiply both sides by r dr dr 2 T dh Separate variables: d r 0 Qw dr dr dh 2 T r dhIntegrate r C1 Integrate again: dr r2 h2 Qw drEvaluate constant C1 using continuity. dh The Qw pumped by the well must equal 2 T r1 r h1 the total Q along the circumference for every radius r. We will solve Darcys Law for r(dh/dr). dh dh Qw Q(r) KA K(2 r)b dr dr dh dh (2 r)Kb (2 r)T dr dr
Steady-State Pumping Tests Q r2 r 1 bh2 r2 h2 h1 Qw dr dh 2 T r1 r h1
Steady-State Pumping TestsWhy is the equation logarithmic?This came about during the switch to radial coordinates. What is the physical rationale forh the shape of this curve (steep at small r, flat at high r)? r
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 dh must increase. dl(Driscoll, 1986)
The Thiem equation tells us there is a logarithmic relationship betweenhead and distance from the pumping well. 10r1 Q ln r1 T 2 hlc hlc one log cycle (Schwartz and Zhang, 2003)
The Thiem equation tells us there is a logarithmic relationship betweenhead and distance from the pumping well. 10r1 Q ln r1 T 2 hlc hlc ln x 2.3 log x ln 10 2.303 2 .3 Q T 2 hlc one log cycle If T decreases, slope increases (Schwartz and Zhang, 2003)
Steady-State Pumping TestsCan we determine S from a Thiem analysis? No—head isn’t changing!