Transcript Calculus 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
14
12
10
8
y
y = axn + bxp + cxq + …
6
a
n
b
p
c
q
4
3
3
5
2
5
0
2
0
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
0.2
0.4
0.6
0.8
1
x
25
20
dy/dx = a n x(n-1) +
b p x(p-1) +
c q x(q-1) + …
y
15
10
5
0
-5
-1
-0.8
-0.6
-0.4
-0.2
0
x
Numerical Derivatives
• ‘finite difference’ approximation
• slope between points
• dy/dx ≈ Dy/Dx
Derivative of Sine and Cosine
1.5
p
p
1
0.5
0
0
1
2
3
4
5
-0.5
-1
-1.5
•
•
•
•
sin(0) = 0
period of both sine and cosine is 2p
d(sin(x))/dx = cos(x)
d(cos(x))/dx = -sin(x)
6
7
Sin(x)
Cos(x)
Partial Derivatives
3
2.5
2
1.5
1
0.5
0
-0.5
S19
-1
S13
19
13
16
X
10
S7
7
4
1
-1.5
Y
S1
• Functions of more than one variable
• Example: h(x,y) = x4 + y3 + xy
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
3
2.5
2
1.5
1
0.5
0
-0.5
S19
-1
S13
19
Example:
h(x,y)
= x4 + y3 + x2y+ xy
∂h/∂x
= 4x3
+ 2xy + y
∂h/∂x|y=y0 = 4x3
+ 2xy0+ y0
16
X
10
7
S7
13
•
•
•
•
4
1
-1.5
S1
Y
WHY?
Gradients
• del h (or grad h)
h
h
h  i
j
x
y
• Darcy’s Law:
q   Kh
Equipotentials/Velocity Vectors
Capture
Zones
Watersheds
http://www.bsatroop257.org/Documents/Summer%20Camp/Topographic%20map%20of%20Bartle.jpg
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 f)
n
• Volume of pores is
also the total volume
– the solids volume
Vpores
Vtotal
Vtotal  Vsolids
n
Vtotal
•
Porosity Basics
Vtotal  Vsolids
Can re-write that as:
n
Vtotal
• Then incorporate:
• Solid density: rs
= Msolids/Vsolids
• Bulk density: rb
= Msolids/Vtotal
• rb/rs = Vsolids/Vtotal
Vsolids
n  1
Vtotal
rb
n  1
rs
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
– F force (‘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 = rghp
– r density
– g gravity
– hp depth
Pressure Head
(increases with depth below surface)
Elevation
P = 0 (= Patm)
Head
Elevation Head
• Water wants to fall
• Potential energy
Head
Elevation
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
Elevation datum
Head
Total Head
(constant: hydrostatic equilibrium)
Elevation
P = 0 (= Patm)
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 crosssectional flow area
• Transmissivity T = Kb
– b = aquifer thickness
• Q = T dh/dx L
– L = width of flow field
1803 - 1858
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
rwg
K k

L
T-1
L2
More on gradients
2.445659
3.399753
4.067833
4.549766
4.902074
5.160327
5.348374
5.482701
5.574732
5.63216
5.659738
1
2.445659
3.399754
4.067834
4.549768
4.902077
5.160329
5.348377
5.482704
5.574736
5.632163
5.659741
5.659741
2
2.937225
3.685772
4.253985
4.679399
4.99614
5.230543
5.402107
5.52501
5.609349
5.662024
5.68733
5.68733
3
3.61747
4.152128
4.582937
4.917709
5.172544
5.363601
5.504502
5.605886
5.675635
5.719259
5.740232
5.740232
4
4.380528
4.722335
5.007931
5.235958
5.412733
5.546819
5.646422
5.718404
5.768053
5.799151
5.814114
5.814114
5
5.182307
5.348756
5.490497
5.605464
5.695616
5.764526
5.815968
5.853259
5.879029
5.895187
5.902965
5.902965
6
5.999944
5.99989
5.999838
5.999789
5.999745
5.999705
5.999672
5.999644
5.999623
5.999608
5.999601
5.999601
7
6.817582
6.651023
6.509179
6.394115
6.303874
6.234885
6.183375
6.146028
6.120216
6.10403
6.096237
6.096237
S1
S2
S3
S4
S5
S6
S7
S8
S9
S10
S11
1
2
3
4
5
6
7
8
9
S12
10 11
8
7.619361
7.277444
6.991744
6.76362
6.586756
6.452591
6.35292
6.280883
6.231191
6.200064
6.185087
6.185087
9
8.382418
7.847651
7.416737
7.081868
6.826944
6.635808
6.494838
6.393399
6.323607
6.279955
6.258968
6.258968
10
9.062663
8.314006
7.745689
7.320177
7.003347
6.768864
6.597232
6.474273
6.389891
6.337188
6.311867
6.311867
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
11
9.554228
8.600023
7.931838
7.449806
7.097408
6.839075
6.650959
6.516576
6.424502
6.367045
6.339453
6.339453
9.554228
8.600023
7.931838
7.449806
7.097408
6.839075
6.650959
6.516576
6.424502
6.367045
6.339453
6.339453
More on gradients
• Three point problems:
h
h
h
More on gradients
h = 10m
• Three point problems:
– (2 equal heads)
h = 10m
h = 9m
• Gradient = (10m9m)/CD
• CD?
– Scale from map
– Compute
More on gradients
h = 11m
• Three point problems:
– (3 unequal heads)
Best guess
for h = 10m
h = 10m
h = 9m
• Gradient = (10m9m)/CD
• CD?
– Scale from map
– Compute
Types of Porous Media
Isotropic
Anisotropic
Homogeneous
Heterogeneous
Hydraulic
Conductivity
Values
K
(m/d)
8.6
0.86
Freeze and Cherry, 1979
Layered media
(horizontal conductivity)
M
Q4
Kh 
Q3
Q2
Q1
b K
i 1
M
b
i 1
Q = Q1 + Q2 + Q 3 + Q4
Flow
K2
b2
K1
b1
i
i
hi
Layered media
(vertical conductivity)
Q4
R4
Q3
R3
Q2
Q1
Flow
R2
Controls
flow
R1
Q ≈ Q 1 ≈ Q2 ≈ Q3 ≈ Q4
The overall resistance is controlled
by the largest resistance:
R = R1 + R2 + R3 + R4
K2
b2
K1
b1
M
Kv 
b
i 1
i
M
b / K
i 1
i
The hydraulic resistance is b/K
hi
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
T2 (or K2)
b2 (or h2)
i=2
k2
d2
T1
b1
k1
d1
i=1
Pumped Aquifer Heads
T2 (or K2)
b2 (or h2)
i=2
k2
d2
T1
b1
k1
d1
i=1
Heads
h2
h2 - h1
T2 (or K2)
b2 (or h2)
i=2
h1
k2
d2
T1
b1
k1
d1
i=1
Flows
h2
h2 - h1
h1
T2 (or K2)
b2 (or h2)
i=2
qv
k2
d2
T1
b1
k1
d1
i=1
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+Dx)DyDz = 0
h
q  K
x
qx|x


h
h
 K x   K x
 DyDz  0
x
x  Dx 

 h
h 

 x


x
x
 x  Dx
0
Dx
Dz
qx|x+Dx
Dx
Dy
h
0
2
x
2
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
h
0
2
x
2
h
 x 2 x   0x
h
 x x   Ax
2
h
A
x
h  Ax  B
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:
h
A
x
After 2nd integration of Laplace Equation
we have:
h  Ax  B
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.
h xDx  h x h xDx  h x
h


x xDx / 2 x  Dx  x
Dx
Look the other direction and estimate at x
– Dx/2:
h x  h xDx h x  h xDx
h


x xDx / 2 x  x  Dx 
Dx
h/x|x+Dx/2
h|x
x
Estimate here
h|x+Dx
x +Dx
Finite Difference
Solution of 1-D
Laplace Equation
(ctd)
h|x-Dx
h/x|x-Dx/2
x -Dx Estimate here
h|x
h/x|x+Dx/2
x
Estimate here
h|x+Dx
x +Dx
2h/x2|x
Estimate here
1st
Combine
order derivative
approximations to get 2nd order derivative
approximation.
h
h
h x  Dx  h x h x  h x  Dx


h x  Dx  2h x  h x  Dx
 2h x x  Dx / 2 x x  Dx / 2
D
x
D
x



0
2
2
x
Dx
Dx
Dx
Solve for h:
hx 
h x  Dx  h x Dx
2
2-D Finite Difference Approximation
y +Dy
h|x-Dx,y
x -Dx
h|x,y+Dy
h|x,y
h|x+Dx,y
x, y
x +Dx
h|x,y-Dy
h x, y 
h xDx, y  h xDx, y  h x, y Dy  h x, y Dy
4
Poisson Equation
R
qx|x
qx|x+Dx
b
Dx
Dy
• 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
Derivation of Poisson Equation
R
(qx|x- qx|x+Dx)Dyb + RDxDy =0
h
q  K
x
qx|x
qx|x+Dx
b
Dx
Dy


h
h
 K x   K x
 Dyb   RDxDy
x
x  Dx 

 h
 x

h 



x
R
x  Dx
x

Dx
T
 2h
R

2
x
T
General Analytical Solution of 1-D
Poisson Equation
h
R

2
x
T
h
 R

 x x    T x  Ax
h
R
 x 2 x    T x
R 2
h
x  Ax  B
2T
2
2
h
R
 x A
x
T
R 2
h
x  Ax  B
2T
Water balance
R 2
h
x  Ax  B
2T
qx|x
R
qx|x+Dx
b
Dx
•
•
•
•
•
Dy
Qin + RDxDy – Qout = 0
qin bDy + RDxDy – qout bDy = 0
-K dh/dx|in bDy + RDxDy – -K dh/dx|out bDy = 0
-T dh/dx|in Dy + RDxDy – -T dh/dx|out Dy = 0
-T dh/dx|in + RDx +T dh/dx|out = 0
Dupuit Assumption
• Flow is horizontal
• Gradient = slope of water table
• Equipotentials are vertical
Dupuit Assumption
(qx|x hx|x - qx|x+Dx h|x+Dx)Dy + RDxDy = 0
q  K
h
x


h
h
hx  Dx  Dy   RDxDy
 K x hx   K x
x
x  Dx


h
h
 2h
x
x
2
 h 2

 x
h 2 



x
R
x  Dx
x

2Dx
K
h
2R

2
x
K
2 2
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 2002
Dr. 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
qx|x
qx|x+Dx
b
Dx
• DVw = DxDy S Dh
(qx|x - qx|x+Dx)Dyb = SDxDyh/t
dq
h

Dyb  SDxDy
dx
t
h
q  K
x
h 

   K 
x  S h


x
b t
K  K (x)
 h S h

2
x
T t
2
Finite Difference Solution
• First order spatial derivative:
C/x|x+Dx/2
h|x
h|x+Dx
x
Estimate here
x +Dx
h x Dx  h x h x Dx  h x
h


x x Dx / 2 x  Dx  x
Dx
Second order spatial derivative
h|x-Dx
x -Dx
h/x|x-Dx/2
h|x
Estimate here
x
h/x|x+Dx/2
Estimate here
h|x+Dx
x +Dx
2h/x2|x
Estimate here
 2h

2
x
h
x

x  Dx / 2

x  Dx / 2
Dx
h x  Dx  h x

h
x
Dx

Dx
h x  h x  Dx
Dx
h x  Dx  2 h x  h x  Dx
Dx 2
Finite Difference Solution
• Temporal devivative
h|x, t-Dt
t-Dt
x +Dx
x -Dx
C/t|t-Dt/2
Estimate here
h|x, t
t
x
h Dh hx ,t  hx ,t Dt


t Dt
Dt
All together:
 h S h

2
x
T t
2
h x Dx , y ,t Dt  2h x ,t Dt  h x Dx ,t Dt
Dx 2
hx ,t  hx ,t Dt
S hx ,t  hx , y ,t Dt

T
Dt
TDt  h x  Dx ,t Dt  2h x ,t Dt  h x Dx ,t Dt 



2
S 
Dx

• Stability criterion (Mesh Ratio):
TDt/(S(Dx)2) < ½.
Diffusion
Dz
jx|x
Jx|x+Dx
Dy
x
x + Dx
C
 jx x  jx xDx DyDz  t DxDyDz
• Fick’s Law:

C
C
  D
D
x x
x


C
C
  D
D
x x
x

Dx
C
j  D
x

C
DyDz 
DxDyDz
t
x  Dx 


C
x  Dx 

t
• Heat/Diffusion Equation:
 C 



x
  C
D 
x
t
 2 C C
D 2 
t
x
Temporal Derivative
C|x, t-Dt
t-Dt
x +Dx
x -Dx
C/t|t-Dt/2
Estimate here
C|x, t
t
C DC C x ,t  C x ,t Dt


t
Dt
Dt
x
All together:
 C C
D 2 
t
x
2
D
C xDx,t Dt  2C x,t Dt  C xDx,t Dt
(Dx)
Cx ,t  Cx ,t Dt
2


Cx,t  Cx,t Dt
Dt
DDt

C x  Dx ,t Dt  2C x ,t Dt  C x Dx ,t Dt
2
(Dx)

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:
C
j  D
x
Notice that if ∂C/∂x = 0, then there is no flux. The finite
difference expression we developed for ∂C/∂x is
C
x

x  Dx / 2
C x  Dx  C x
Dx
• Setting this to 0 is equivalent to
C x  Dx  C x
Convection-Dispersion Equation
ja  qC
C
jd  nD
x

C
C
 qC x  nD
 qC x  Dx  nD
x x
x


C
DyDz 
nDxDyDz
t
x  Dx 
•Key difference from diffusion here!
• Convective flux

C
C
 vC x  D
 vC x  Dx  D
x x
x

Dx


C
x  Dx 

t
CDE

C
v
x
 C x  Dx
Dx

 C
C
 

x x x

D
Dx


C
x  Dx 

t
 C 


C
x  C

v
D

x
x
t
C
 C C
v
D 2 
x
x
t
2
Finite Difference: Spatial
C|x-Dx
x -Dx
C/x|x-Dx/2
Estimate here
C|x
C/x|x+Dx/2
x
Estimate here
2C/x2|x
C|x+Dx
x +Dx
 2C

2
x
C
x
C

x
x  Dx / 2
Dx
x  Dx / 2
Estimate here
C x  Dx  C x


Dx

C x  C x  Dx
Dx
Dx
C x  Dx  2C x  C x  Dx
(Dx) 2
Finite Difference: Temporal
t-Dt
x +Dx
x -Dx
C/t|t-Dt/2
Estimate here
C|x, t
t
x
C DC C x ,t  C x ,t Dt


t
Dt
Dt
Centered Finite Difference
• For first order spatial derivative:
C
x

x  Dx / 2
C x  Dx  C x
Dx
• Worked for estimating second order
derivative (estimate ended up at x).
• Need centered derivative approximation
C xDx  C xDx
C

x x
2Dx
All together:
C
 2C C
v
D 2 
x
x
t
-v
C xDx,t Dd  C xDx,t Dt
2Dx
Cx ,t  Cx ,t Dt 

D
C xDx,t Dt  2C x,t Dt  C xDx,t Dt
(Dx)
2


Cx,t  Cx,t Dt
Dt

DDt
vDt
C

2
C

C

C x  Dx ,t Dt  C x Dx ,t Dt
2
x  Dx ,t  Dt
x ,t  Dt
x  Dx ,t  Dt
(Dx)
2Dx
• prone to numerical instabilities depending
on the values of the factors DDt/(Dx)2 and
vDt/2Dx (CFL)

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
C
j  qC  nD
x
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
C
x

x  Dx / 2
C x  Dx  C x
Dx
• Setting this to 0 is equivalent to
C x  Dx  C x
Fitting the CDE
1.2
Relative Concentration
1
0.8
Model
0.6
Data
0.4
0.2
0
0
500
1000
1500
2000
2500
Time (sec)
3000
3500
4000
4500
Adsorption Isotherm
Cs
• Linear: Cs = Kd C
C
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
Acenaphthene
mean Koc (L/kg)
5,028
Compound
1,4-Dichlorobenzene(p)
mean Koc (L/kg)
687
Compound
Methoxychlor
mean Koc (L/kg)
80,000
Aldrin
48,686
1,1-Dichloroethane
54
Methyl bromide
9
Anthracene
24,362
1,2-Dichloroethane
44
Methyl chloride
6
1,1-Dichloroethylene
65
Methylene chloride
trans-1,2-Dichloroethylene
38
Naphthalene
1,231
1,166,733
1,2-Dichloropropane
47
Nitrobenzene
141
76
1,3-Dichloropropene
27
Pentachlorobenzene
36,114
25,604
Pyrene
70,808
84
Styrene
912
Benz(a)anthracene
Benzene
Benzo(a)pyrene
Bis(2-chloroethyl)ether
Bis(2-ethylhexyl)phthalate
Bromoform
Butyl benzyl phthalate
Carbon tetrachloride
Chlordane
Chlorobenzene
Chloroform
459,882
66
114,337
126
14,055
158
51,798
Dieldrin
Diethylphthalate
10
Di-n-butylphthalate
1,580
1,1,2,2-Tetrachloroethane
Endosulfan
2,040
Tetrachloroethylene
272
Toluene
145
Endrin
11,422
260
Ethylbenzene
207
57
Fluoranthene
49,433
DDD
45,800
Fluorene
DDE
86,405
DDT
792,158
Toxaphene
1,2,4-Trichlorobenzene
79
95,816
1,783
8,906
1,1,1-Trichloroethane
139
Heptachlor
10,070
1,1,2-Trichloroethane
77
Hexachlorobenzene
80,000
Trichloroethylene
97
Dibenz(a,h)anthracene
2,029,435
a-HCH (a-BHC)
1,835
o-Xylene
241
1,2-Dichlorobenzene(o)
390
b-HCH (b-BHC)
2,241
m-Xylene
204
g-HCH (Lindane)
1,477
p-Xylene
313
Retardation
• Incorporate adsorbed solute mass
R  1
r b Kd
n
V
R
Vs
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
C
C
C
v
D 2 R
x
x
t
2
Aquifer Tests
• Theis
Matching 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/(4ps) W(u)
S = 4Ttu/r2.
Ghyben-Herzberg
h
z
Pf = Ps
rg(h+z) = rsgz
r(h+z) = rsz
rh = (rs-r z→ h r/(rs-r = z
z
Ghyben-Herzberg
•
•
•
•
Seawater: 1.025 g cm-3
h r/(rs-r = z
h 1/(1.05 - 1 = z
h 40 = z
Major Cations and Anions
• Cations:
– Ca2+, Mg2+, Na+, K+
• Anions:
– Cl-, SO42-, HCO3-
Chemical Concentration
Conversions
• Usually given ML-3 (e.g. g L-1 mg L-1)
• Convert to mol L-1:

molCa 
2
1
10mgL Ca 

2
.
5

10
molL

3

40

10
mgCa


1
2
• Convert to mol (+/-) L-1:
 2m ol( ) 
2
1
2.5  10 m olCa L 

5

10
m
ol
(

)
L
2 
 m olCa 
2
2 1
Charge Balance
(  )   ( )

Charge Balance 
 (  )   ( )
Piper Diagrams
Miami Beach GW and 1-7% Miami Beach sea water
100
EXPLANATION
1378.352602
+C
l
0
2+
2+
g
SO
4
+M
• Convert to
% mol (+/-)
2+
Ca
2-
-
35800
0
Mg
100 0
2-
SO4
100 0
100
0
100
Ca
2+
100
0
+
+
Na + K
0
100
100
0
2-
-
CO3 + HCO3
Cl
-
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)