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Applications of Darcys Law

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Steps in the application of Darcy's Law. Define a Reference Elevation ... Apply Darcy's Law over the entire column and compute Jw and Ks ... – PowerPoint PPT presentation

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Title: Applications of Darcys Law


1
Applications of Darcys Law
Vertical Flow is in the z direction.
z
Horizontal Flow is in the x direction.
x
y
In addition to the sign convention for flow
direction, we must also consider the soil as a
3-dimensional system.
Flow in the y direction is in the 3rd dimension.
2
  • Steps in the application of Darcys Law
  • Define a Reference Elevation
  • Determine two points where H is known
  • Calculate the Gradient

3
Horizontal Column
1
2
x2 - x1 column length L
2
Vertical Column
z2 - z1 column length L
1
4
Horizontal Flow Jw - Ks ?H - Ks (H2 H1)/(x2
x1) Vertical Flow Jw - Ks ?H - Ks (H2
H1)/(z2 z1) Where H p z (Driving Force)
5
In the generic ----
Jw - Ks ?H - Ks (?H/L) - Ks (H2 - H2)/L
p1
p2
ZR 0
Ks
1
2
L
H1 p1 z1 p1 0 p1 H2 p2 z2 p2 0
p2 ?H H2 - H1 p2 p1 ?H (p2 - p1)/L Jw
- Ks (p2 - p1)/L
6
Jw - Ks ?H - Ks (?H/L) - Ks (H2 - H2)/L
p
2
H1 p1 z1 0 0 0 H2 p2 z2 p2 L ?H
H2 - H1 (p2 L) - 0 p2 L ?H (p2 L)/L
p2/L 1
Ks
L
1
ZR 0
Jw - Ks (p2/L 1)
p atmosphere
7
Horizontal Flow
H1 p1 z1 10 cm 0 10 cm H2 p2 z2
0 0 0 ?H H2 - H1 0 - 10 cm -10 cm
?H 10 cm/100 cm - 0.10
p1 10 cm
ZR 0
Ks 100 cm/d
1
2
L 100 cm
Jw - Ks (H2 - H1)/L Jw - (100 cm/d)(- 0.10)
10 cm/d Positive so confirms flow is to the
right.
8
Vertical Flow (Down)
p 10 cm
H1 p1 z1 0 0 0
H2 p2 z2 10 cm 100 cm 110 cm ?H H2 -
H1 110 cm - 0 cm 110 cm ?H 110 cm/100 cm
1.1
2
Ks 100 cm/d
L 100 cm
Jw - Ks (H2 H1)/L Jw - (100 cm/d)(1.1)
-110 cm/d Negative, so confirms flow is downward,
and is greater than that of horizontal flow by
the magnitude of Ks.
1
ZR 0
p atmosphere
9
Vertical Flow (Up) Jw Ksp1/L - Ks
H1 p1 z1 110 cm 0 cm 110 cm H2 p2
z2 0 100 cm 100 cm ?H H2 - H1 100 cm -
110 cm - 10 cm ?H ?H/L -10cm/100cm - 0.1
2
Ks 100 cm/d
L 100 cm
1
ZR 0
p 110 cm
Jw - (100 cm/d) (- 0.10) 10 cm/d Positive
value so we know that flow is upward.
10
Measurement of Ks Constant Head Permeameter
Measure Q/t A pr2 Compute Jw Solve for Ks Jw
Q/t(pr2) - Ks?H Ks - Q/(tpr2?H) Ks -
(Jw/?H) Where Jw is negative for downward flow
p b
pr2
Ks
L
ZR 0
Jw
11
Jw - Ks (H2 - H1)/L H1 p1 z1 0 H2
p2 z2 b L Jw - Ks (bL)/L Ks -
JwL/(bL) where Jw is negative (-)(-) NOTE
Ks cannot be -.
p b
pr2
Ks
L
ZR 0
Jw
12
Falling Head Permeameter
?H (H2 - H1)/L (b(t)L)/L Jw(t) db/dt -Ks
(b(t)L)/L db/(b(t)L) -Ks/L dt where b b0
at t0 Integrating for t0 to tti for biltb0 Ks
L/t1 ln (b0L)/(b1L)
b(t)
pr2
Ks
L
ZR 0
Jw(t)
13
Hydrostatic Pressure at point z
b
Apply Darcys Law over the entire column and
compute Jw and Ks At steady-state water flow Jw
is constant throughout the column
pr2
Ks
L
z
ZR 0
Jw
14
Hydrostatic Pressure at point z
b
At point z (2) H1 p1 z1 0 H2 p
z Jw -Ks (pz)/z
pr2
Ks
L
z
ZR 0
Jw
-Ks(pz)z -Ks(bL)L (pz)/z (bL)/L p
(bL)/L z - z p ((b/L)1)z z p
bz/L p decreases linearly over the column
15
Water Flow in a Layered Soil
b
6
k6
Ke is the total resistance to water flow and is
the sum of the individual hydraulic resistivites
over the column length Ke SLi/S(Li/Ki)
k5
5
k4
4
L
Ke
k3
3
k2
2
k1
1
Jw
At steady-state water flow, Jw must be the same
through all layers such that Jw -k1(H2-H1)L1
-k2(H3-H2)/L2 -kn(Hn1-Hn)/Ln
16
b 10 cm
First compute Ke Apply Darcys Law over the
entire column to determine Jw Do the same example
but place the layer of lower conductivity at the
top instead of the bottom of the column
K2 25 cm/d
L2 75cm
p ?
K1 5 cm/d
L1 25cm
Jw ?
17
Intrinsic Permeability Ks k(?lg)/? where k is
a property of the medium itself, and is not
influenced by the liquid Homogeneity and
Isotrophy
place to place
each dimension
18
3-dimensional
place to place
H/I 1K H/AI 3K IH/I 2K IH/AI 6K
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