final review exam cumulative: incorporate complete midterm review

Final Review

• Exam cumulative: incorporate complete midterm review

Calculus Review

Derivative of a polynomial

• In differential Calculus, we consider the slopes of curves rather than straight lines

• For polynomial y = axn + bxp + cxq + …, derivative with respect to x is:

• dy/dx = a n x(n-1) + b p x(p-1) + c q x(q-1) + …

Example

a 3 n 3 b 5 p 2 c 5 q 0

0

2

4

6

8

10

12

14

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

x

y

y = axn + bxp + cxq + …

-5

0

5

10

15

20

25

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1

x

y

dy/dx = a n x(n-1) + b p x(p-1) + c q x(q-1) + …

Numerical Derivatives

• ‘finite difference’ approximation

• slope between points

• dy/dx ≈ y/x

Derivative of Sine and Cosine

• sin(0) = 0 • period of both sine and cosine is 2• d(sin(x))/dx = cos(x) • d(cos(x))/dx = -sin(x)

-1.5

-1

-0.5

0

0.5

1

1.5

0 1 2 3 4 5 6 7

Sin(x)

Cos(x)

Partial Derivatives

• Functions of more than one variable• Example: h(x,y) = x4 + y3 + xy

1 4 7

10 13 16 19S1

S7

S13

S19

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

3

X

Y

2.5-3

2-2.5

1.5-2

1-1.5

0.5-1

0-0.5

-0.5-0

-1--0.5

-1.5--1

Partial Derivatives

• Partial derivative of h with respect to x at a y location y0

• Notation ∂h/∂x|y=y0

• Treat ys as constants• If these constants stand alone, they drop

out of the result• If they are in multiplicative terms involving

x, they are retained as constants

Partial Derivatives

• Example: • h(x,y) = x4 + y3 + x2y+ xy • ∂h/∂x = 4x3 + 2xy + y

• ∂h/∂x|y=y0 = 4x3 + 2xy0+ y0

1 4 7

10 13 16 19

S1

S7

S13

S19

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

3

X

Y

Gradients

• del h (or grad h)

• Darcy’s Law:

y

h

x

hh

ji

hKq

Equipotentials/Velocity Vectors

Capture Zones

Watersheds

http://www.bsatroop257.org/Documents/Summer%20Camp/Topographic%20map%20of%20Bartle.jpg

Capture Zones

Water (Mass) Balance

• In – Out = Change in Storage– Totally general– Usually for a particular time interval– Many ways to break up components– Different reservoirs can be considered

Water (Mass) Balance

• Principal components:– Precipitation– Evaporation– Transpiration– Runoff

• P – E – T – Ro = Change in Storage

• Units?

Ground Water (Mass) Balance

• Principal components:– Recharge– Inflow– Transpiration– Outflow

• R + Qin – T – Qout = Change in Storage

Ground Water Basics

• Porosity

• Head

• Hydraulic Conductivity

Porosity Basics

• Porosity n (or )

• Volume of pores is also the total volume – the solids volume

total

pores

V

Vn

total

solidstotal

V

VVn

Porosity Basics

• Can re-write that as:

• Then incorporate:• Solid density: s

= Msolids/Vsolids

• Bulk density: b

= Msolids/Vtotal • bs = Vsolids/Vtotal

total

solidstotal

V

VVn

total

solids

V

Vn 1

s

bn

1

Ground Water Flow

• Pressure and pressure head

• Elevation head

• Total head

• Head gradient

• Discharge

• Darcy’s Law (hydraulic conductivity)

• Kozeny-Carman Equation

Pressure

• Pressure is force per unit area• Newton: F = ma

– Fforce (‘Newtons’ N or kg ms-2)– m mass (kg)– a acceleration (ms-2)

• P = F/Area (Nm-2 or kg ms-2m-2 =

kg s-2m-1 = Pa)

Pressure and Pressure Head

• Pressure relative to atmospheric, so P = 0 at water table

• P = ghp

– density– g gravity

– hp depth

P = 0 (= Patm)

Pre

ssur

e H

ead

(incr

ease

s w

ith d

epth

bel

ow s

urfa

ce)

Pressure Head

Ele

vati

on

Head

Elevation Head

• Water wants to fall

• Potential energy

Ele

vatio

n H

ead

(incr

ease

s w

ith h

eigh

t ab

ove

datu

m)

Eleva

tion

Head

Ele

vati

on

Head

Elevation datum

Total Head

• For our purposes:

• Total head = Pressure head + Elevation head

• Water flows down a total head gradient

P = 0 (= Patm)

Tot

al H

ead

(con

stan

t: h

ydro

stat

ic e

quili

briu

m)

Pressure Head

Eleva

tion

Head

Ele

vati

on

Head

Elevation datum

Head Gradient

• Change in head divided by distance in porous medium over which head change occurs

• A slope

• dh/dx [unitless]

Discharge

• Q (volume per time: L3T-1)

• q (volume per time per area: L3T-1L-2 = LT-1)

Darcy’s Law

• q = -K dh/dx– Darcy ‘velocity’

• Q = K dh/dx A– where K is the hydraulic

conductivity and A is the cross-sectional flow area

• Transmissivity T = Kb– b = aquifer thickness

• Q = T dh/dx L– L = width of flow field

www.ngwa.org/ ngwef/darcy.html

1803 - 1858

http://www.ngwa.org/ngwef/darcy.html

Mean Pore Water Velocity

• Darcy ‘velocity’:q = -K ∂h/∂x

• Mean pore water velocity:v = q/ne

Intrinsic Permeability

g

kK w

L T-1 L2

More on gradients1 2 3 4 5 6 7 8 9 10 11

2.445659 2.445659 2.937225 3.61747 4.380528 5.182307 5.999944 6.817582 7.619361 8.382418 9.062663 9.554228 9.5542283.399753 3.399754 3.685772 4.152128 4.722335 5.348756 5.99989 6.651023 7.277444 7.847651 8.314006 8.600023 8.6000234.067833 4.067834 4.253985 4.582937 5.007931 5.490497 5.999838 6.509179 6.991744 7.416737 7.745689 7.931838 7.9318384.549766 4.549768 4.679399 4.917709 5.235958 5.605464 5.999789 6.394115 6.76362 7.081868 7.320177 7.449806 7.4498064.902074 4.902077 4.99614 5.172544 5.412733 5.695616 5.999745 6.303874 6.586756 6.826944 7.003347 7.097408 7.0974085.160327 5.160329 5.230543 5.363601 5.546819 5.764526 5.999705 6.234885 6.452591 6.635808 6.768864 6.839075 6.8390755.348374 5.348377 5.402107 5.504502 5.646422 5.815968 5.999672 6.183375 6.35292 6.494838 6.597232 6.650959 6.6509595.482701 5.482704 5.52501 5.605886 5.718404 5.853259 5.999644 6.146028 6.280883 6.393399 6.474273 6.516576 6.5165765.574732 5.574736 5.609349 5.675635 5.768053 5.879029 5.999623 6.120216 6.231191 6.323607 6.389891 6.424502 6.424502

5.63216 5.632163 5.662024 5.719259 5.799151 5.895187 5.999608 6.10403 6.200064 6.279955 6.337188 6.367045 6.3670455.659738 5.659741 5.68733 5.740232 5.814114 5.902965 5.999601 6.096237 6.185087 6.258968 6.311867 6.339453 6.339453

5.659741 5.68733 5.740232 5.814114 5.902965 5.999601 6.096237 6.185087 6.258968 6.311867 6.339453 6.339453

1 2 3 4 5 6 7 8 9 10 11

S1

S2

S3

S4

S5

S6

S7

S8

S9

S10

S11

S12

10.5-11

10-10.5

9.5-10

9-9.5

8.5-9

8-8.5

7.5-8

7-7.5

6.5-7

6-6.5

5.5-6

5-5.5

4.5-5

4-4.5

3.5-4

3-3.5

2.5-3

2-2.5

1.5-2

More on gradients

• Three point problems:

h

h

h

400 m

412 m

100 m

More on gradients

• Three point problems:– (2 equal heads)

h = 10m

h = 10m

h = 9m

400 m

412 m

100 m

CD • Gradient = (10m-9m)/CD

• CD?– Scale from map– Compute

More on gradients

• Three point problems:– (3 unequal heads)

h = 10m

h = 11m

h = 9m

400 m

412 m

100 m

CD • Gradient = (10m-9m)/CD

• CD?– Scale from map– Compute

Best guess for h = 10m

Types of Porous Media

Homogeneous Heterogeneous

Isotropic

Anisotropic

Hydraulic Conductivity Values

Freeze and Cherry, 1979

8.6

0.86

K

(m/d)

Layered media (horizontal conductivity)

M

ii

M

ihii

h

b

KbK

1

1

Q1

Q2

Q3

Q4

Q = Q1 + Q2 + Q3 + Q4

K1

K2

b1

b2

Flow

Layered media(vertical conductivity)

M

ihii

M

ii

v

Kb

bK

1

1

/

Controls flow

Q1

Q2

Q3

Q4

Q ≈ Q1 ≈ Q2 ≈ Q3 ≈ Q4

R1

R2

R3

R4

R = R1 + R2 + R3 + R4

K1

K2

b1

b2

Flow

The overall resistance is controlled by the largest resistance: The hydraulic resistance is b/K

Aquifers

• Lithologic unit or collection of units capable of yielding water to wells

• Confined aquifer bounded by confining beds

• Unconfined or water table aquifer bounded by water table

• Perched aquifers

Transmissivity

• T = Kb

gpd/ft, ft2/d, m2/d

Schematic

i = 1

i = 2

d1

b1

d2

b2 (or h2)

k1

T1

k2

T2 (or K2)

Pumped Aquifer Heads

i = 1

i = 2

d1

b1

d2

b2 (or h2)

k1

T1

k2

T2 (or K2)

Heads

i = 1

i = 2

d1

b1

d2

b2 (or h2)

k1

T1

k2

T2 (or K2)

h1

h2

h2 - h1

Flows

i = 1

i = 2

d1

b1

d2

b2 (or h2)

k1

T1

k2

T2 (or K2)h1

h2 h2 - h1

qv

Terminology

• Derive governing equation:– Mass balance, pass to differential equation

• Take derivative:– dx2/dx = 2x

• PDE = Partial Differential Equation• CDE or ADE = Convection or Advection

Diffusion or Dispersion Equation• Analytical solution:

– exact mathematical solution, usually from integration• Numerical solution:

– Derivatives are approximated by finite differences

Derivation of 1-D Laplace Equation

• Inflows - Outflows = 0

• (qx|x- qx|x+x)yz = 0

x

hKq

x y

qx|x qx|x+xz

0

zyx

hK

x

hK

xxx

0

x

xh

xh

xxx

02

2

x

h

Governing Equation

Boundary Conditions

• Constant head: h = constant

• Constant flux: dh/dx = constant– If dh/dx = 0 then no flow– Otherwise constant flow

General Analytical Solution of 1-D Laplace Equation

Ax

h

xAxx

h

BAxh

02

2

x

h

xxx

h0

2

2

Particular Analytical Solution of 1-D Laplace Equation (BVP)

Ax

h

BAxh

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:

Incorporate derivative, gives A.

After 2nd integration of Laplace Equation we have:

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 – x/2:

x

hh

xxx

hh

x

h xxxxxx

xx

2/

x

hh

xxx

hh

x

h xxxxxx

xx

2/

h|x h|x+x

x x +x

h/x|x+x/2

Estimate here

Finite Difference Solution of 1-D

Laplace Equation (ctd)

Combine 1st order derivative approximations to get 2nd order derivative approximation.

h|x h|x+x

x x +x

h|x-x

x -x

h/x|x+x/2

Estimate here

h/x|x-x/2

Estimate here

2h/x2|x

Estimate here

02

22/2/

2

2

x

hhh

xx

hh

x

hh

xxh

xh

x

h xxxxx

xxxxxx

xxxx

Solve for h:

2xxxx

x

hhh

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• Units of R: L T-1

x y

qx|x qx|x+xb

R

x y

qx|x qx|x+x

x yx yx y

qx|x qx|x+xb

R

Derivation of Poisson Equation

x y

qx|x qx|x+xb

R

x y

qx|x qx|x+x

x yx yx y

qx|x qx|x+xb

R(qx|x- qx|x+x)yb + Rxy =0

x

hKq

yxRybx

hK

x

hK

xxx

T

R

x

xh

xh

xxx

T

R

x

h

2

2

General Analytical Solution of 1-D Poisson Equation

AxT

R

x

h

xAxT

Rx

x

h

BAxxT

Rh 2

2

T

R

x

h

2

2

xT

Rx

x

h2

2

BAxxT

Rh 2

2

Dupuit Assumption

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

Dupuit Assumption

K

R

x

h 22

22

(qx|x hx|x - qx|x+x h|x+x)y + Rxy = 0

x

hKq

yxRyhx

hKh

x

hK xx

xxx

x

K

R

x

xh

xh

xxx

2

22

x

hh

x

h

22

Transient Problems

• Transient GW flow

• Diffusion

• Convection-Dispersion Equation

• All transient problems require specifying initial conditions (in addition to boundary conditions)

Storage Coefficient/Storativity

• S is storage coefficient or storativity: The amount of water stored or released per unit area of aquifer given unit head change

• Typical values of S (dimensionless) are 10-5 – 10-3

• Measuring storativity: derived from observations of multi-well tests

• GEOS 4310/5310 Lecture Notes, Fall 2002Dr. T. Brikowski, UTD

http://www.utdallas.edu/~brikowi/Teaching/Geohydrology/LectureNotes/Regional_Flow/Storativity.html

1-D Transient GW Flow

1-D Transient GW Flow: Deriving the Governing PDE

• Vw = xy S h

x

bqx|x qx|x+x

(qx|x - qx|x+x)yb = Sxyh/t

t

hyxSyb

dx

dq

t

h

b

S

xxh

K

t

h

T

S

x

h

2

2

x

hKq

)(xKK

Finite Difference Solution

• First order spatial derivative:

h|x h|x+x

x x +x

C/x|x+x/2

Estimate here

x

hh

xxx

hh

x

h xxxxxx

xx

2/

Finite Difference Solution

h|x, t

x

x +x

C/t|t-t/2 Estimate here

t-t

t

x -x

t

hh

t

h

t

h ttxtx

,,

h|x, t-t

• Temporal devivative

All together:

t

hh

T

S

x

hhhttyxtxttxxttxttyxx

,,,

2,,,,

2

2,,,

,,

2

x

hhh

S

tThh ttxxttxttxx

ttxtx

t

h

T

S

x

h

2

2

• Stability criterion (Mesh Ratio): Tt/(S(x)2) < ½.

Diffusion

x + x

y

z

x

jx|x Jx|x+x

zyxt

Czyjj

xxxxx

• Fick’s Law:

x

CDj

zyxt

Czy

x

CD

x

CD

xxx

t

C

x

x

CD

x

CD

xxx

t

C

xx

C

D

t

C

x

CD

2

2

• Heat/Diffusion Equation:

Temporal Derivative

C|x, t

x

x +x


t-t

t

x -x

t

CC

t

C

t

C ttxtx

,,

C|x, t-t

All together:

ttxxttxttxxttxtx CCC

x

tCC

,,,2,, 2

)(

D

t

C

x

CD

2

2

t

CC

x

CCCD ttxtxttxxttxttxx

,,

2

,,,

)(

2

Boundary conditions

• Specify either– Concentrations at the boundaries, or – Chemical flux at the boundaries (usually zero)

• Fixed concentration boundary concept is simple. • Chemical flux boundary is slightly more difficult. We go

back to Fick’s law:

Notice that if ∂C/∂x = 0, then there is no flux. The finite difference expression we developed for ∂C/∂x is

• Setting this to 0 is equivalent to

x

CDj

x

CC

x

C xxx

xx

2/

xxxCC

Convection-Dispersion Equation

x

CnDjd

qCja

zyxnt

Czy

x

CnDqC

x

CnDqC

xxxx

xx

t

C

x

xC

DvCxC

DvCxx

xxx

x

•Key difference from diffusion here!

• Convective flux

CDE

t

C

x

xC

xC

Dx

CCv xxxxxx

t

C

xxC

Dx

Cv

t

C

x

CD

x

Cv

2

2

Finite Difference: Temporal

C|x, tx

x +x


t-t

t

x -x

t

CC

t

C

t

C ttxtx

,,

Centered Finite Difference

• For first order spatial derivative:

x

CC

x

C xxxx

x

2

x

CC

x

C xxx

xx

2/

• Worked for estimating second order derivative (estimate ended up at x).

• Need centered derivative approximation

All together:

• prone to numerical instabilities depending on the values of the factors Dt/(x)2 and vt/2x (CFL)

t

CC

x

CCC

x

CCttxtxttxxttxttxxttxxdtxx

,,

2

,,,,,

)(

2D

2v-

ttxxttxxttxxttxttxxttxtx CC

x

tCCC

x

tCC

,,,,,2,, 2

v2

)(

D

t

C

x

CD

x

Cv

2

2

Boundary conditions• Specify either

– Concentrations at the boundaries, or – Chemical flux at the boundaries

• Fixed concentration boundary concept is simple• Chemical flux boundary is slightly more difficult. We go back to the

flux

Notice that if ∂C/∂x = 0, then there is no dispersive flux but there can still be a convective flux. This would apply at the end of a soil column for example; the water carrying the chemical still flows out of the column but there is no more dispersion. One of the finite difference expressions we developed for ∂C/∂x is

• Setting this to 0 is equivalent to

x

CnDqCj

x

CC

x

C xxx

xx

2/

xxxCC

Fitting the CDE

0

0.2

0.4

0.6

0.8

1

1.2

0 500 1000 1500 2000 2500 3000 3500 4000 4500

Time (sec)

Rel

ativ

e C

on

cen

trat

ion

Model

Data

Adsorption Isotherm

• Linear: Cs = Kd C

C

Cs

Koc Values

• Kd = Koc foc

Organic Carbon Partitioning Coefficients for Nonionizable Organic Compounds. Adapted from USEPA, Soil Screening Guidance: Technical Background Document. http://www.epa.gov/superfund/resources/soil/introtbd.htm

Compound mean Koc (L/kg) Compound mean Koc (L/kg) Compound mean Koc (L/kg)

Acenaphthene 5,028 1,4-Dichlorobenzene(p) 687 Methoxychlor 80,000

Aldrin 48,686 1,1-Dichloroethane 54 Methyl bromide 9

Anthracene 24,362 1,2-Dichloroethane 44 Methyl chloride 6

Benz(a)anthracene 459,882 1,1-Dichloroethylene 65 Methylene chloride 10

Benzene 66 trans-1,2-Dichloroethylene 38 Naphthalene 1,231

Benzo(a)pyrene 1,166,733 1,2-Dichloropropane 47 Nitrobenzene 141

Bis(2-chloroethyl)ether 76 1,3-Dichloropropene 27 Pentachlorobenzene 36,114

Bis(2-ethylhexyl)phthalate 114,337 Dieldrin 25,604 Pyrene 70,808

Bromoform 126 Diethylphthalate 84 Styrene 912

Butyl benzyl phthalate 14,055 Di-n-butylphthalate 1,580 1,1,2,2-Tetrachloroethane 79

Carbon tetrachloride 158 Endosulfan 2,040 Tetrachloroethylene 272

Chlordane 51,798 Endrin 11,422 Toluene 145

Chlorobenzene 260 Ethylbenzene 207 Toxaphene 95,816

Chloroform 57 Fluoranthene 49,433 1,2,4-Trichlorobenzene 1,783

DDD 45,800 Fluorene 8,906 1,1,1-Trichloroethane 139

DDE 86,405 Heptachlor 10,070 1,1,2-Trichloroethane 77

DDT 792,158 Hexachlorobenzene 80,000 Trichloroethylene 97

Dibenz(a,h)anthracene 2,029,435 -HCH (-BHC) 1,835 o-Xylene 241

1,2-Dichlorobenzene(o) 390 -HCH (-BHC) 2,241 m-Xylene 204

-HCH (Lindane) 1,477 p-Xylene 313

http://www.epa.gov/superfund/resources/soil/introtbd.htm

Retardation

• Incorporate adsorbed solute mass

n

KdR b1

Vs

VR

Sample problem:

A tanker truck collision has resulted in a spill of 5000 L of the insecticide diazinon 2000 m from the City of Miami’s water supply wells. Use a rule of thumb to estimate the dispersivity for the plume that is carrying the contaminant from the spill site to the wells.

Sample problem:

The transmissivity determined from aquifer tests is 100,000 m2 d-1 and the aquifer thickness is 20 m. The head in wells 1000 m apart along the flow path is 3.1 and 3 m. What is the gradient? What is the mean pore water velocity and what is the dispersion coefficient?

Sample problem:

• You look up the Koc value of diazinon (290 ml/g). The aquifer material you tested has an foc of 0.0001. What is the Kd? If the porosity is 50% and the bulk density is 1.5 Kg L-1, what is R?

• Assume retarded piston flow and estimate the arrival time of the insecticide at the well field using the appropriate data from the preceding problems.

Retardation

t

CR

x

CD

x

Cv

2

2

Aquifer Tests

• TheisMatching aquifer test data to the Theis type curve has resulted in the match point coordinates 1/u = 10, W(u) = 1, t = 83.9 minutes, and s =0.217 m. The pumping rate is 1 m3 min-1 and the observation well is 100 m away from the pumping well. Compute the aquifer transmissivity and storativity. Be sure to

specify the units.

Hints:

T = Q/(4s) W(u) S = 4Ttu/r2.

Ghyben-Herzberg

zz

h

Pf = Ps

g(h+z) = sgz

(h+z) = sz

h = (s- z→ h /(s- = z

Ghyben-Herzberg

• Seawater: 1.025 g cm-3

• h /(s- = z

• h /(- = z

• h = z

Major Cations and Anions

• Cations:– Ca2+, Mg2+, Na+, K+

• Anions:– Cl-, SO4

2-, HCO3-

Chemical Concentration Conversions

• Usually given ML-3 (e.g. g L-1 mg L-1)

• Convert to mol L-1:

123

21 105.21040

10

molLmgCa

molCaCamgL

• Convert to mol (+/-) L-1:

122

122 )(105)(2

105.2

LmolmolCa

molLmolCa

Charge Balance

)()(

)()( BalanceCharge

Piper DiagramsMiami Beach GW and 1-7% Miami Beach sea water

Ca2+

Mg2+

Na+ + K

+CO3

2- + HCO3

-

SO42-

Cl-

SO4

2- + C

l- Ca 2+

+ Mg 2+

1000

0100

100 0

1000

0100

100 0

100

100

0 0

EXPLANATION

1378.352602

35800

• Convert to % mol (+/-)

Stiff Diagrams

http://water.usgs.gov/pubs/wri/wri024045/htms/report2.htm

Redox Reactions

• O2 (disappear)

• NO3- (disappear)

• Fe/Mn (appear in solution)

• SO42- (disappear)

• CH4 (appear)