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Midterm Review. Calculus. Derivative relationships d(sin x)/dx = cos x d(cos x)/dx = -sin x. Calculus. Approximate numerical derivatives d(sin)/dx ~ [sin (x + D x) – sin (x)]/ D x. Calculus. Partial derivatives h(x,y) = x 4 + y 3 + xy

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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**Calculus**• Approximate numerical derivatives • d(sin)/dx ~ [sin (x + Dx) – sin (x)]/ Dx**Calculus**• Partial derivatives • h(x,y) = x4 + y3 + xy • The partial derivative of h with respect to x at a y location y0 (i.e., ∂h/∂x|y=y0), • 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=y0 = 4x3 + y0**Ground Water Basics**• Porosity • Head • Hydraulic Conductivity • Transmissivity**Porosity Basics**• Porosity n (or f) • Volume of pores is also the total volume – the solids volume**Porosity Basics**• Can re-write that as: • Then incorporate: • Solid density: rs = Msolids/Vsolids • Bulk density: rb = Msolids/Vtotal • rb/rs = Vsolids/Vtotal**Cubic Packings and Porosity**Simple Cubic Body-Centered Cubic Face-Centered Cubic n = 0.48 n = 0. 26 n = 0.26 http://members.tripod.com/~EppE/images.htm**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**Porosity Basics**• Volumetric water content (q) • Equals porosity for saturated system**Ground Water Flow**• Pressure and pressure head • Elevation head • Total head • Head gradient • Discharge • Darcy’s Law (hydraulic conductivity) • Kozeny-Carman Equation**Multiple Choice:Water flows…?**• Uphill • Downhill • Something else**Pressure**• Pressure is force per unit area • Newton: F = ma • Fforce (‘Newtons’ N or kg m s-2) • m mass (kg) • a acceleration (m s-2) • P = F/Area (Nm-2 or kg m s-2m-2 = kg s-2m-1 = Pa)**Pressure and Pressure Head**• Pressure relative to atmospheric, so P = 0 at water table • P = rghp • r density • g gravity • hpdepth**P = 0 (= Patm)**Pressure Head Pressure Head (increases with depth below surface) Elevation Head**Elevation Head**• Water wants to fall • Potential energy**Elevation Head**(increases with height above datum) Elevation Elevation Head Elevation datum Head**Total Head**• For our purposes: • Total head = Pressure head + Elevation head • Water flows down a total head gradient**P = 0 (= Patm)**Pressure Head Total Head (constant: hydrostatic equilibrium) Elevation Elevation Head Elevation datum Head**Head Gradient**• Change in head divided by distance in porous medium over which head change occurs • dh/dx [unitless]**Discharge**• Q (volume per time) Specific Discharge/Flux/Darcy Velocity • q (volume per time per unit area) • L3 T-1 L-2→ L T-1**Darcy’s Law**• Q = -K dh/dx A where K is the hydraulic conductivity and A is the cross-sectional flow area 1803 - 1858 www.ngwa.org/ ngwef/darcy.html**Darcy’s Law**• Q = K dh/dl A • Specific discharge or Darcy ‘velocity’: qx = -Kx∂h/∂x … q = -K gradh • Mean pore water velocity: v = q/ne**Intrinsic Permeability**L2 L T-1**Transmissivity**• T = Kb**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)**Darcy’s Law**• Q = -K dh/dl A • Q, q • K, T**Mass Balance/Conservation Equation**• I = inputs • P = production • O = outputs • L = losses • A = accumulation**qx|x**Dz qx|x+Dx Dx Dy Derivation of 1-D Laplace Equation • Inflows - Outflows = 0 • (q|x - q|x+Dx)DyDz = 0 • q|x – (q|x +Dx dq/dx) = 0 • dq/dx = 0 (Continuity Equation) (Constitutive equation)**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 1st integration of Laplace Equation we have: After 2nd integration of Laplace Equation we have: Incorporate derivative, gives A. Incorporate constant head, gives B.**Finite Difference Solution of 1-D Laplace Equation**Need finite difference approximation for 2nd order derivative. Start with 1st order. Look the other direction and estimate at x – Dx/2:**Finite Difference Solution of 1-D Laplace Equation (ctd)**Combine 1st order derivative approximations to get 2nd order derivative approximation. Set equal to zero and solve for h:**Matrix Notation/Solutions**• Ax=b • A-1b=x**Toth Problems**• Governing Equation • Boundary Conditions**Recognizing Boundary Conditions**• Parallel: • Constant Head • Constant (non-zero) Flux • Perpendicular: No flow • Other: • Sloping constant head • Constant (non-zero) Flux**Internal ‘Boundary’ Conditions**• Constant head • Wells • Streams • Lakes • No flow • Flow barriers • Other**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**Poisson Equation**(qx|x+Dx - qx|x)Dyb -RDxDy = 0**Dupuit Assumption**• Flow is horizontal • Gradient = slope of water table • Equipotentials are vertical**Dupuit Assumption**(qx|x+Dx hx|x+Dx- qx|x hx|x)Dy - RDxDy = 0**2Dy**Y 1Dy 0 0 1Dx 2Dx Effective outflow boundary Block-centered model Only the area inside the boundary (i.e. [(imax -1)Dx] [(jmax -1)Dy] 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 imax and jmax, subtracting 1 makes little difference. X**Effective outflow boundary**Mesh-centered model 2Dy 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. Y 1Dy 0 0 1Dx 2Dx X**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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