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CIV3248 Flow Net Workshop1 CIV3248 Groundwater,seepage and environmental engineering Workshop (4) Flownet sketching Keith H McKenry

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CIV3248 Flow Net Workshop2 Intention: Mathematical insight. Practise sketching skills. Flow net compared with SEEPW output. Interpretation of flow nets

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CIV3248 Flow Net Workshop3 Mathematics. Steady state: conservation of mass, continuity equation, stream function ( ), streamlines. Energy balance: Potential function ( ), total head lines, irrotational flow (flow around a curve by distortion rather than rotation). Laplaces Equation for & .

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CIV3248 Flow Net Workshop4 Streamlines Seepage velocity vector is tangent to streamline. Streamlines cannot cross. Commence on inflow boundary, finish at outflow boundary. Streamlines have constant stream function value.

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CIV3248 Flow Net Workshop5 Streamlines Seepage velocity stream function value dQ = Stream tube dQ Unconfined flow: is a boundary streamline

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CIV3248 Flow Net Workshop6 Streamlines and stream function (x,y), Two streamlines define a streamtube. Streamtube width is inversely proportional to local seepage velocity. Difference in in stream function value equals the discharge between two streamlines defining the stream tube.

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CIV3248 Flow Net Workshop7 Total head lines Total head = - / k Lines of constant value of potential function. Zero hydraulic gradient zero flow orthogonal to streamlines (cross at right angles).

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CIV3248 Flow Net Workshop8 Equipotential / Total head lines equipotential value = - H*k m 3 /s/m

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CIV3248 Flow Net Workshop9 Flownet definition…1 Selection of streamlines and total head lines. Constant difference in stream function values between all streamlines. Constant difference in potential function values between all total head lines. Same difference value for both stream function and potential function.

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CIV3248 Flow Net Workshop10 Flownet definition…2 Streamlines Equipotential lines (constant Total Head) dq = 3

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CIV3248 Flow Net Workshop11 Flownet features Orthogonal net. Inscribed circles. Diagonals to curvilinear squares form another orthogonal net. Ratio of number of streamtubes to number of potential drops is constant for a given flow boundary.

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CIV3248 Flow Net Workshop12 Flownet Examples Based on explicit functions for total head and stream function. Prior to the advent of cheap computers much mathematical effort was expended on techniques to allow explicit equations to fit a wide range of boundary geometries.

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CIV3248 Flow Net Workshop13 Basis For a given boundary geometry, only ONE true flownet exists. Find the true flow net by “trial and correct”.

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CIV3248 Flow Net Workshop14 Strategy Identify the two streamlines and two total total head lines that define the problem. For 4 or 5 STREAM TUBES draw the 3 or 4 internal streamlines. Draw TOTAL HEAD lines to attempt to form curvilinear squares (a true flownet). Adjust the trial lines to get better curvilinear squares.

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CIV3248 Flow Net Workshop15 Confined Flow example “Impermeable” “Pervious” Boundary streamlines Boundary total head lines Total head difference

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CIV3248 Flow Net Workshop16 Cross-section to scale: 10 mm = 1 m. W1 A C DE F B Datum RL 60m

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CIV3248 Flow Net Workshop17 Cross-section to scale: 10 mm = 1 m. W2 A C DE F B Datum RL 60m

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CIV3248 Flow Net Workshop18 Cross-section to scale: 10 mm = 1 m. W3 A C DE F B Datum RL 60m

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CIV3248 Flow Net Workshop19 Cross-section to scale: 10 mm = 1 m. W4 A C DE F B Datum RL 60m

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CIV3248 Flow Net Workshop20 Flownet Interpretataion Discharge (m 3 /sec/m) Pore pressure Hydraulic gradient.

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CIV3248 Flow Net Workshop21 Discharge per unit width (Q m 3 /sec/m) Q = k.H L.(N /N ) k - coefficient of permeability. H L - drop in Total Head across flow. N - number of stream tubes. N - number of potential drops. Note N might not be an integer.

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CIV3248 Flow Net Workshop22 H L = 4 m N = 3 N = 8 Q = k.H L.(N /N ) = 7 x 10 -5. 4.(3/8) = 10.5 x 10 -5 m 3 /s/m K = 7 x 10 -5 m/s

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CIV3248 Flow Net Workshop23 Pore Pressure “u” at a point Identify total head value of all constant total head lines relative to datum. Find total head at point by interpolation “H”. Identify elevation head of point “z”. u = (H - z) * w ( w = unit weight of water)

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CIV3248 Flow Net Workshop24 RL95m RL91m RL72m H=95m H=94.5m H=94m H=93.5m H=93mH=92.5m H=92m H=91.5m H=91m Total head datum = RL 0m RL80m RL94m RL83m RL92.5m A B u = (H - z) * w u A =(94 – 80)*9.8 = 137 kPa u BD =(94 – 75)*9.8 = 186 kPa u B =(92.5 – 83)*9.8 = 93 kPa D RL75m

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CIV3248 Flow Net Workshop25 RL93m RL91m RL72m H=93m H=92.75m H=92.5m H=92.25m H=92mH=91.75m H=91.5m H=91.25m H=91m Total head datum = RL 0m RL80m RL92.5m RL83m RL91.75m A B u = (H - z) * w u A =(92.5 – 80)*9.8 = 123 kPa u B =(91.75 – 83)*9.8 = 86 kPa

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CIV3248 Flow Net Workshop26 Hydraulic gradient at a point - “i” Interpolate streamline through the point. Determine difference in total head between constant total head lines either side of point “dH”. Measure real distance between these constant total head lines along the interpolated streamline “dL” i = dH /dL

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CIV3248 Flow Net Workshop27 RL95m RL91m RL72m H=95m H=94.5m H=94m H=93.5m H=93mH=92.5m H=92m H=91.5m H=91m Total head datum = RL 0m C dL = 4.5m dH = 0.5m (-)i C = dH /dL = 0.5 / 4.5 = 0.11

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CIV3248 Flow Net Workshop28 Upward seepage flow: Critical hydraulic gradient at exit - Icrit Liquifaction of soil may occur if the exit hydraulic gradient exceeds a critical value such that effective stress equals zero. Icrit = (G s - 1)/(e +1) G s - Specific gravity of soil particles e - Void Ratio Icrit ~ 1.0

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CIV3248 Flow Net Workshop29 Seepw solution

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CIV3248 Flow Net Workshop30 Conclusion: flow net sketching Limited to isotropic soils, homogeneous flow Quick, simple equipment, Analysis of multiple geometry for design. Check on computer packages.

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