Midterm Review
Calculus
• Derivative
relationships
• d(sin x)/dx = cos x
• d(cos x)/dx = -sin x
1.5


1
0.5
0
0
-0.5
-1
-1.5
1
2
3
4
5
6
7
Sin(x)
Cos(x)
Calculus
• Approximate numerical derivatives
• d(sin)/dx ~
[sin (x + Dx) – sin (x)]/ Dx
1.5


1
0.5
0
0
-0.5
-1
-1.5
1
2
3
4
5
6
7
Sin(x)
Cos(x)
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)
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
Cubic Packings and Porosity
Simple Cubic
n = 0.48
Body-Centered Cubic
Face-Centered Cubic
n = 0. 26
http://members.tripod.com/~EppE/images.htm
n = 0.26
FCC and BCC have same porosity
http://uwp.edu/~li/geol200-01/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
Vwater
q
Vtotal
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-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
• 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 grad h
• Mean pore water velocity:
v = q/ne
Intrinsic Permeability
rw g
K k

L
T-1
L2
Kozeny-Carman Equation
3
2
m
n
d
k
2
1  n  180
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   P  O   L   A
•
•
•
•
•
I = inputs
P = production
O = outputs
L = losses
A = accumulation
 I  O  0
Derivation of 1-D Laplace Equation
qx|x
•
•
•
•
Dz
Inflows - Outflows = 0
(q|x - q|x+Dx)DyDz = 0
q|x – (q|x +Dx dq/dx) = 0
dq/dx = 0 (Continuity Equation)
h
q  K
x
(Constitutive equation)
h 

d K 
x 

0
dx
qx|x+Dx
Dx
Dy
h

0
2
x
2
General Analytical Solution of 1-D
Laplace Equation
h
0
2
x
2
h
 x 2 x   0x
2
h
A
x
h

x

A

x
 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
Incorporate derivative, gives A.
After 2nd integration of Laplace Equation
we have:
h  Ax  B
Incorporate constant head, gives B.
Finite Difference Solution of 1-D
Laplace Equation
h/x|x+Dx/2
h|x
x
h|x+Dx
Estimate here
x +Dx
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

x  Dx / 2
h x  h x Dx
x  x  Dx 

h x  h x Dx
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
Combine 1st order derivative approximations to get 2nd order derivative approximation.
 h

2
x
2
h
x

x  Dx / 2
Dx
h
x
h x  Dx  h x
x  Dx / 2

Set equal to zero and solve for h:
Dx
h x  h x Dx

Dx
Dx
hx 

h x  Dx  2h x  h x Dx
Dx 2
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
Matrix Notation/Solutions
4h2, 2  h2,3  h1,2  h2,1  h3, 2
 h2, 2  4h2,3  h1,3  h2, 4  h3,3
 4  1 h2, 2  h1, 2  h2,1  h3, 2 
  1 4  h   h  h  h 

  2,3   1,3 2, 4 3,3 
• Ax=b
• A-1b=x
Toth Problems
• Governing Equation
h h
 2 0
2
x
y
• Boundary Conditions
19
17
15
13
11
9
7
5
3
2
1
2
S1
S2
S3
S4
S5
S6
S7
S8
S9
S10
S11
10.09-10.1
10.08-10.09
10.07-10.08
10.06-10.07
10.05-10.06
10.04-10.05
10.03-10.04
10.02-10.03
10.01-10.02
10-10.01
Recognizing Boundary Conditions
S1
• Parallel:
S3
S5
– Constant Head
– Constant (non-zero) Flux
S7
S9
S11
• Perpendicular: No flow
• Other:
S13
S15
S17
S19
– Sloping constant head
– Constant (non-zero) Flux
S21
S23
S25
S29
13
10
7
4
1
S27
Internal ‘Boundary’ Conditions
• No flow
S11
S3
S5
S7
S9
S11
S13
S15
S17
S19
S21
S23
S25
S27
S29
4
7
10
– Wells
– Streams
– Lakes
13
• Constant head
S1
S3
S5
– Flow barriers
S7
S9
• Other
S11
S13
S15
S17
S19
15
13
11
9
7
5
3
1
S21
S23
Poisson Equation
R
• Add/remove water
from system so that q |
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
xx
qx|x+Dx
b
Dx
Dy
Poisson Equation
R
(qx|x+Dx - qx|x)Dyb -RDxDy = 0
h
q  K
x

h
 K x

x  Dx
 h
 x

qx|x
qx|x+Dx
b
Dx
Dy
h 
K
Dyb  RDxDy

x x 
h 



x
R
x  Dx
x

Dx
T
 2h
R

2
x
T
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
h
q  K
x


h
h
hx Dx  K
hx  Dy  RDxDy
 K x
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
Capture Zones
Water Balance and Model
Types
Block-centered model
Effective
outflow
boundary
2Dy
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.
1Dy
0
0
1Dx
X
2Dx
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
2Dy
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.
1Dy
0
0
1Dx
X
2Dx
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
h1
(h1 + h2)/2
h2
Water Balance
h
0
y x ,1000
• Given:
– Recharge rate
– Transmissivity
h
0
x 1000, y
h
0
x 0, y
• Find and compare:
– Inflow
– Outflow
h x ,0  0
Water Balance
• Given:
– Recharge rate
– Flux BC
– Transmissivity
• Find and compare:
– Inflow
– Outflow