Midterm Review Calculus Derivative relationships dsin xdx cos

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 + Dx) – sin (x)]/ Dx

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 = 4 x 3 + y 0

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 n = 0. 48 Body-Centered Cubic Face-Centered Cubic n = 0. 26 http: //members. tripod. com/~Epp. E/images. htm n = 0. 26

FCC and BCC have same porosity http: //uwp. edu/~li/geol 200 -01/cryschem/ • Bottom line for randomly packed beads: n ≈ 0. 4 Smith et al. 1929, PR 34: 1271 -1274

Effective Porosity

Effective Porosity

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 – 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 = rghp – r density – g gravity – hp depth

Elevation Pr es su re H ea d Head Pressure Head (increases with depth below surface) P = 0 (= Patm)

Elevation Head • Water wants to fall • Potential energy

Head El at io ev n ad He Elevation Head (increases with height above datum) Elevation datum

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

at io ev n ad He Pr es su re H ea d Elevation datum Head Total Head (constant: hydrostatic equilibrium) El Elevation P = 0 (= Patm)

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 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 grad h • Mean pore water velocity: v = q/ne

Intrinsic Permeability L T-1 L 2

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 qx | x • • Dz Inflows - Outflows = 0 (q|x - q|x+Dx)Dy. Dz = 0 q|x – (q|x +Dx dq/dx) = 0 dq/dx = 0 (Continuity Equation) (Constitutive equation) qx|x+Dx Dx Dy

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 = 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.

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 – Dx/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 (qx|x+Dx - qx|x)Dyb -RDx. Dy = 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 - RDx. Dy = 0

Capture Zones

Water Balance and Model Types

Block-centered model Effective outflow boundary 2 Dy Y 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. 1 Dy 0 0 1 Dx X 2 Dx In our case this was 23000 11000, as we observed. For large imax and jmax, subtracting 1 makes little difference.

Mesh-centered model Effective outflow boundary 2 Dy An alternative is to use a mesh-centered model. Y This will require an extra row and column of nodes and the constant heads will not be exactly on the boundary. 1 Dy 0 0 1 Dx X 2 Dx

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 Effective outflow area h 1 (h 1 + h 2)/2 h 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