Download - Midterm Review. Calculus Derivative relationships d(sin x)/dx = cos x d(cos x)/dx = -sin x
Calculus
• Derivative relationships
• 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)
Calculus
• Approximate numerical derivatives• d(sin)/dx ~
[sin (x + x) – sin (x)]/ x
-1.5
-1
-0.5
0
0.5
1
1.5
0 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
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
Cubic Packings and Porosity
http://members.tripod.com/~EppE/images.htm
Simple Cubic Body-Centered Cubic Face-Centered Cubic n = 0.48 n = 0. 26 n = 0.26
FCC and BCC have same porosity
• Bottom line for randomly packed beads: n ≈ 0.4
http://uwp.edu/~li/geol200-01/cryschem/
Smith et al. 1929, PR 34:1271-1274
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 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 = 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
Elevation
Head
Ele
vatio
n H
ead
(incr
ease
s w
ith h
eigh
t ab
ove
datu
m)
Eleva
tion
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 (= Patm)
Tot
al H
ead
(con
stan
t: h
ydro
stat
ic e
quili
briu
m)
Pressure Head
Eleva
tion
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)• 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
www.ngwa.org/ ngwef/darcy.html
1803 - 1858
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
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)
Mass Balance/Conservation Equation
• I = inputs
• P = production
• O = outputs
• L = losses
• A = accumulation
ALOPI
0OI
Derivation of 1-D Laplace Equation
• Inflows - Outflows = 0
• (q|x - q|x+x)yz = 0
• q|x – (q|x +x dq/dx) = 0
• dq/dx = 0 (Continuity Equation)
x
hKq
x y
qx|x qx|x+xz
0
dxxh
Kd0
2
2
x
h(Constitutive equation)
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
22/2/
2
2 2
x
hhh
xx
hh
x
hh
x
x
h
x
h
x
h xxxxx
xxxxxx
xxxx
Set equal to zero and solve for h:
2xxxx
x
hhh
2-D Finite Difference Approximation
h|x,y h|x+x,y
x, y
y +y
h|x-x,y
x -x x +x
h|x,y-y
h|x,y+y
4,,,,
,
yyxyyxyxxyxx
yx
hhhhh
Matrix Notation/Solutions
• Ax=b
• A-1b=x
3,34,23,13,22,2
2,31,22,13,22,2
4
4
hhhhh
hhhhh
3,34,23,1
2,31,22,1
3,2
2,2
41
14
hhh
hhh
h
h
Toth Problems
• Governing Equation
• Boundary Conditions
1 3 5 7 9
11
13
15
17
19
S 1
S 2
S 3
S 4
S 5
S 6
S 7
S 8
S 9
S 10
S 11
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
10-10.0102
2
2
2
y
h
x
h
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
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
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+x - qx|x)yb -Rxy = 0
x
hKq
yxRybx
hK
x
hK
xxx
T
R
x
xh
xh
xxx
T
R
x
h
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+x hx|x+x- qx|x hx|x)y - Rxy = 0
x
hKq
yxRyhx
hKh
x
hK x
xxx
xx
K
R
x
xh
xh
xxx
2
22
x
hh
x
h
22
X
0
2x1x
2y
1y
0
Y
Effective outflow boundary
Only the area inside the boundary (i.e. [(imax -1)x] [(jmax -1)y] in general) contributes water to what is measured at the effective outflow boundary.
In our case this was 23000 11000, as we observed. For large imax and jmax, subtracting 1 makes little difference.
Block-centered model
X
0
2x1x
2y
1y
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.
Water Balance
• Given: – Recharge rate – Transmissivity
• Find and compare:– Inflow– Outflow
0,1000
yx
h0
,0
yx
h
01000,
xy
h
00,
xh