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Published byCory Wheeler Modified over 9 years ago
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Midterm Review
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Calculus Derivative relationships d(sin x)/dx = cos x d(cos x)/dx = -sin x
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Calculus Approximate numerical derivatives d(sin)/dx ~ [sin (x + x) – sin (x)]/ x
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Calculus Partial derivatives h(x,y) = x 4 + y 3 + xy The partial derivative of h with respect to x at a y location y 0 (i.e., ∂h/∂x| y=y 0 ), Treat any terms containing y only as constants –If these constants stand alone they drop out of the result –If the constants are in multiplicative terms involving x, they are retained as constants Thus ∂h/ ∂x| y=y 0 = 4x 3 + y 0
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Ground Water Basics Porosity Head Hydraulic Conductivity Transmissivity
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Porosity Basics Porosity n (or ) Volume of pores is also the total volume – the solids volume
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Porosity Basics Can re-write that as: Then incorporate: Solid density: s = M solids /V solids Bulk density: b = M solids /V total b s = V solids /V total
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Cubic Packings and Porosity http://members.tripod.com/~EppE/images.htm Simple Cubic Body-Centered Cubic Face-Centered Cubic n = 0.48 n = 0. 26 n = 0.26
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FCC and BCC have same porosity Bottom line for randomly packed beads: n ≈ 0.4 http://uwp.edu/~li/geol200-01/cryschem/ Smith et al. 1929, PR 34:1271-1274
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Effective Porosity
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Porosity Basics Volumetric water content ( ) –Equals porosity for saturated system
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Ground Water Flow Pressure and pressure head Elevation head Total head Head gradient Discharge Darcy’s Law (hydraulic conductivity) Kozeny-Carman Equation
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Multiple Choice: Water flows…? Uphill Downhill Something else
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Pressure Pressure is force per unit area Newton: F = ma –F force (‘Newtons’ N or kg m s -2 ) –m mass (kg) –a acceleration (m s -2 ) P = F/Area (Nm -2 or kg m s -2 m -2 = kg s -2 m -1 = Pa)
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Pressure and Pressure Head Pressure relative to atmospheric, so P = 0 at water table P = gh p – density –g gravity –h p depth
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P = 0 (= P atm ) Pressure Head (increases with depth below surface) Pressure Head Elevation Head
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Elevation Head Water wants to fall Potential energy
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Elevation Head (increases with height above datum) Elevation Head Elevation Head Elevation datum
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Total Head For our purposes: Total head = Pressure head + Elevation head Water flows down a total head gradient
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P = 0 (= P atm ) Total Head (constant: hydrostatic equilibrium) Pressure Head Elevation Head Elevation Head Elevation datum
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Head Gradient Change in head divided by distance in porous medium over which head change occurs dh/dx [unitless]
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Discharge Q (volume per time) Specific Discharge/Flux/Darcy Velocity q (volume per time per unit area) L 3 T -1 L -2 → L T -1
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Darcy’s Law Q = -K dh/dx A where K is the hydraulic conductivity and A is the cross-sectional flow area www.ngwa.org/ ngwef/darcy.html 1803 - 1858
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Darcy’s Law Q = K dh/dl A Specific discharge or Darcy ‘velocity’: q x = -K x ∂h/∂x … q = -K grad h Mean pore water velocity: v = q/n e
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Intrinsic Permeability L T -1 L2L2
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Kozeny-Carman Equation
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Transmissivity T = Kb
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Potential/Potential Diagrams Total potential = elevation potential + pressure potential Pressure potential depends on depth below a free surface Elevation potential depends on height relative to a reference (slope is 1)
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Darcy’s Law Q = -K dh/dl A Q, q K, T
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Mass Balance/Conservation Equation I = inputs P = production O = outputs L = losses A = accumulation
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Derivation of 1-D Laplace Equation Inflows - Outflows = 0 (q| x - q| x+ x ) y z = 0 q| x – (q| x + x dq/dx) = 0 dq/dx = 0 (Continuity Equation) xx yy qx|xqx|x q x | x+ x zz (Constitutive equation)
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General Analytical Solution of 1-D Laplace Equation
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Particular Analytical Solution of 1-D Laplace Equation (BVP) BCs: - Derivative (constant flux): e.g., dh/dx| 0 = 0.01 - Constant head: e.g., h| 100 = 10 m After 1 st integration of Laplace Equation we have: Incorporate derivative, gives A. After 2 nd integration of Laplace Equation we have: Incorporate constant head, gives B.
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Finite Difference Solution of 1-D Laplace Equation Need finite difference approximation for 2 nd order derivative. Start with 1 st order. Look the other direction and estimate at x – x/2:
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Finite Difference Solution of 1-D Laplace Equation (ctd) Combine 1 st order derivative approximations to get 2 nd order derivative approximation. Set equal to zero and solve for h:
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2-D Finite Difference Approximation
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Matrix Notation/Solutions Ax=b A -1 b=x
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Toth Problems Governing Equation Boundary Conditions
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Recognizing Boundary Conditions Parallel: –Constant Head –Constant (non-zero) Flux Perpendicular: No flow Other: –Sloping constant head –Constant (non-zero) Flux
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Internal ‘Boundary’ Conditions Constant head –Wells –Streams –Lakes No flow –Flow barriers Other
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Poisson Equation Add/remove water from system so that inflow and outflow are different R can be recharge, ET, well pumping, etc. R can be a function of space and time Units of R: L T -1
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Poisson Equation (q x | x+ x - q x | x ) yb -R x y = 0
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Dupuit Assumption Flow is horizontal Gradient = slope of water table Equipotentials are vertical
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Dupuit Assumption (q x | x+ x h x | x+ x - q x | x h x | x ) y - R x y = 0
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Capture Zones
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Water Balance and Model Types
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X 0 2x2x1x1x 2y2y 1y1y 0 Y Effective outflow boundary Only the area inside the boundary (i.e. [(i max -1) x] [(j max -1) y] in general) contributes water to what is measured at the effective outflow boundary. In our case this was 23000 11000, as we observed. For large i max and j max, subtracting 1 makes little difference. Block-centered model
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X 0 2x2x1x1x 2y2y 1y1y 0 Y Effective outflow boundary An alternative is to use a mesh-centered model. This will require an extra row and column of nodes and the constant heads will not be exactly on the boundary. Mesh-centered model
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Summary In summary, there are two possibilities: –Block-centered and –Mesh-centered. Block-centered makes good sense for constant head boundaries because they fall right on the nodes, but the water balance will miss part of the domain. Mesh-centered seems right for constant flux boundaries and gives a more intuitive water balance, but requires an extra row and column of nodes. The difference between these models becomes negligible as the number of nodes becomes large.
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Dupuit Assumption Water Balance h1h1 h2h2 Effective outflow area (h 1 + h 2 )/2
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Water Balance Given: –Recharge rate –Transmissivity Find and compare: –Inflow –Outflow
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Water Balance Given: –Recharge rate –Flux BC –Transmissivity Find and compare: –Inflow –Outflow
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