水文地質(zhì)學(xué)基礎(chǔ)：地下水流網(wǎng)

1、Company Logo地下水流網(wǎng)原理流線與等水頭線相互垂直.流線平行于隔水頭邊界.流網(wǎng)由曲線網(wǎng)格組成，其中對(duì)角線成直角交叉。相鄰兩條流線之間通過(guò)的流量相等流線總是從“源”指向“匯”。Company LogoCompany Logo均質(zhì)土體中的流網(wǎng)The equation for flow nets originates from Darcys Law. Flow Net solution is equivalent to solving the governing equations of flow for a uniform isotropic aquifer with well-defi

2、ned boundary conditions.Company Logo均質(zhì)土體中的流網(wǎng)Flow through a channel between equipotential lines 1 and 2 per unit width is: q = K(dm x 1)(h1/dl)dmDh1dlF1F3DqF2Dh2DqnmCompany Logo均質(zhì)土體中的流網(wǎng)Flow through equipotential lines 2 and 3 is:q = K(dm x 1)(h2/dl)The flow net has square grids, so the head drop is t

3、he same in each potential drop:h1 = h2If there are nd such drops, then: h = (H/n)where H is the total head loss between the first and last equipotential lines.Company Logo均質(zhì)土體中的流網(wǎng)Substitution yields:q = K(dm x dl)(H/n)This equation is for one flow channel. If there are m such channels in the net, th

4、en total flow per unit width is: q = (m/n)K(dm/dl)HCompany LogoSince the flow net is drawn with squares, then dm dl, and:q = (m/n)KH L2T-1where:q = rate of flow or seepage per unit widthm= number of flow channelsn= number of equipotential dropsh = total head loss in flow systemK = hydraulic conducti

5、vityCompany LogoGroundwater MovementDetermine flow direction from well data:NWell #2 4315 elevdepth to WT = 78(WT elev = 4237)Well #3 4397 elevdepth to WT = 95(WT elev = 4302)Well #1 4252 elevdepth to WT = 120(WT elev = 4132)424042604280430041404160418042004220424042604280418042004220Calculate WT el

6、evations.Interpolate contour intervals.Connect contours of equal elevation.Draw flow lines perpendicular to contours.S. Hughes, 2003Company Logo流網(wǎng)繪制方法1. Draw to a convenient scale the cross sections of the structure, water elevations, and aquifer profiles.2. Establish boundary conditions and draw on

7、e or two flow lines Y and equipotential lines F near the boundaries.Company Logo流網(wǎng)繪制方法3. Sketch intermediate flow lines and equipotential lines by smooth curves adhering to right-angle intersections and square grids. Where flow direction is a straight line, flow lines are an equal distance apart and

8、 parallel.4. Continue sketching until a problem develops. Each problem will indicate changes to be made in the entire net. Successive trials will result in a reasonably consistent flow net.Company Logo流網(wǎng)繪制方法 In most cases, 5 to 10 flow lines are usually sufficient. Depending on the no. of flow lines

9、 selected, the number of equipotential lines will automatically be fixed by geometry and grid layout. 6. Equivalent to solving the governing equations of GW flow in 2-dimensions. Company LogoSeepage Under DamsFlow nets for seepage through earthen dams Seepage under concrete damsUses boundary conditi

10、ons (L & R)Requires curvilinear square grids for solutionCompany LogoTwo Layer Flow System with Sand BelowKu / Kl = 1 / 50Company LogoTwo Layer Flow System with Tight Silt BelowFlow nets for seepage from one side of a channel through two different anisotropic two-layer systems. (a) Ku / Kl = 1/50. (

11、b) Ku / Kl = 50. Source: Todd & Bear, 1961.Company LogoEffects of Boundary Condition on Shape of Flow NetsCompany LogoRadial Flow:Contour map of the piezometric surface near Savannah, Georgia, 1957, showing closed contours resulting from heavy local groundwater pumping (after USGS Water-Supply Paper

12、 1611).Company Logo Flow Net in a Corner:Streamlines Y are at right angles to equipotential F linesCompany LogoFlow Nets: an exampleA 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 hydraul

13、ic conductivity of 150 ft/day, determine the rate of flow or seepage under the dam.Company LogoFlow Nets: an exampleThe flow net is drawn with: m = 5 n = 17Company LogoFlow Nets: the solutionSolve for the flow per unit width:q = (m/n) K h = (5/17)(150)(35) = 1544 ft3/day per ft Company LogoFlow Nets

14、: An ExampleThere 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 n = 21 K = 6.1

15、 x 10-4cm/sec= 0.527 m/dayCompany LogoFlow Nets: the solutionFrom the flow net, the total head loss, H, is 6.2 -2.2 = 4.0 meters.There are 6 flow channels (m) and 21 head drops along each flow path (n):Q = (KmH/n) x dam length= (0.527 m/day x 6 x 4m / 21) x (dam length)= 0.60 m3/day per m of dam = 4

16、3.4 m3/day for the entire 72-meter length of the damCompany LogoWater table contour lines are similar to topographic lines on a map. They essentially represent elevations in the subsurface. These elevations are the hydraulic head mentioned above.Water table contour lines can be used to determine the

17、 direction groundwater will flow in a given region. Many wells are drilled and hydraulic head is measured in each one. Water table contours (called equipotential lines) are constructed to join areas of equal head. Groundwater flow lines, which represent the paths of groundwater downslope, are drawn

18、perpendicular to the contour lines.A map of groundwater contour lines with groundwater flow lines is called a flow net.Remember: groundwater always moves from an area of higher hydraulic head to an area of lower hydraulic head, and perpendicular to equipotential lines.Groundwater Flow NetsS. Hughes,

19、 2003Company LogoGroundwater Flow NetsAquitard (granite)Qal10050QalWTA simple flow netCross-profile viewwellAquitardQal Effect of a producing well Notice the approximate diameter of the cone of depressionS. Hughes, 2003Company LogoGroundwater Flow Nets707080809090100100AquitardAquitardQalQalWater table contoursWater is flowing from Qal to graniteWater is flowing from granite to QalDistorted contours may occur due to anisotropic conditions (changes in aquifer properties).Area of high permeability (high conductivity)S.