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 +  x) – sin (x)]/  x

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

Ground Water Basics Porosity Head Hydraulic Conductivity Transmissivity

Porosity Basics Porosity n (or  ) Volume of pores is also the total volume – the solids volume

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

Cubic Packings and Porosity Simple Cubic Body-Centered Cubic Face-Centered Cubic n = 0.48 n = n = 0.26

FCC and BCC have same porosity Bottom line for randomly packed beads: n ≈ Smith et al. 1929, PR 34:

Effective Porosity

Porosity Basics Volumetric water content (  ) –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 –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)

Pressure and Pressure Head Pressure relative to atmospheric, so P = 0 at water table P =  gh p –  density –g gravity –h p depth

P = 0 (= P atm ) Pressure Head (increases with depth below surface) Pressure Head Elevation Head

Elevation Head Water wants to fall Potential energy

Elevation Head (increases with height above datum) Elevation Head Elevation Head Elevation datum

Total Head For our purposes: Total head = Pressure head + Elevation head Water flows down a total head gradient

P = 0 (= P atm ) Total Head (constant: hydrostatic equilibrium) Pressure Head Elevation Head Elevation Head Elevation datum

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) L 3 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 ngwef/darcy.html

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

Intrinsic Permeability L T -1 L2L2

Kozeny-Carman Equation

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

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) xx yy qx|xqx|x q x | x+  x zz (Constitutive equation)

General Analytical Solution of 1-D Laplace Equation

Particular Analytical Solution of 1-D Laplace Equation (BVP) BCs: - Derivative (constant flux): e.g., dh/dx| 0 = 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.

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:

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:

2-D Finite Difference Approximation

Matrix Notation/Solutions Ax=b A -1 b=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 (q x | x+  x - q x | x )  yb -R  x  y = 0

Dupuit Assumption Flow is horizontal Gradient = slope of water table Equipotentials are vertical

Dupuit Assumption (q x | x+  x h x | x+  x - q x | x h x | x )  y - R  x  y = 0

Capture Zones

Water Balance and Model Types

X 0 2x2x1x1x 2y2y 1y1y 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  11000, as we observed. For large i max and j max, subtracting 1 makes little difference. Block-centered model

X 0 2x2x1x1x 2y2y 1y1y 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

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.

Dupuit Assumption Water Balance h1h1 h2h2 Effective outflow area (h 1 + h 2 )/2

Water Balance Given: –Recharge rate –Transmissivity Find and compare: –Inflow –Outflow

Water Balance Given: –Recharge rate –Flux BC –Transmissivity Find and compare: –Inflow –Outflow