10Lecture16ApplicationsDarcysLaw

1. Applications of Darcy’s Law
Vertical Flow is in the “z”
direction.
z
x
y
In addition to the sign convention
for flow direction, we must also
consider the soil as a 3-
dimensional system.
Horizontal Flow is in the “x”
direction.
Flow in the “y”
direction is in the
3rd dimension.

2. Steps in the application of Darcy’s
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
1
Vertical Column
z2 - z1 = column length L

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 ----
ZR = 0
p1
p2
1 2
L
Jw = - Ks ▼H = - Ks (∆H/L)
= - Ks (H2 - H2)/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
Ks

6. p
1
2
ZR = 0
Jw = - Ks ▼H = - Ks (∆H/L)
= - Ks (H2 - H2)/L
p = atmosphere
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
L
Jw = - Ks (p2/L + 1)
Ks

7. ZR = 0
p1 = 10 cm
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.
Ks = 100 cm/d
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
Horizontal Flow

8. p = 10 cm
1
2
ZR = 0
p = atmosphere
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
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.
Ks
= 100
cm/d
Vertical Flow (Down)

9. Vertical Flow (Up):
Jw = Ksp1/L - Ks
p = 110 cm
L = 100 cm
Ks =
100
cm/d
1
2
ZR = 0
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
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
p = b
Ks
Jw
L
πr2
ZR = 0
Measure Q/t
A = πr2
Compute Jw
Solve for Ks
Jw = Q/t(πr2) = - Ks▼H
Ks = - Q/(tπr2▼H)
Ks = - (Jw/▼H)
Where Jw is negative for
downward flow

11. p = b
Ks
Jw
L
πr2
ZR = 0
Jw = - Ks (H2 - H1)/L
H1 = p1 + z1 = 0
H2 = p2 + z2 = b + L
Jw = - Ks (b+L)/L
Ks = - JwL/(b+L) where Jw
is negative (-)(-) = +
NOTE: Ks cannot be “-”.

12. b(t)
Ks
Jw(t)
L
πr2
ZR = 0
▼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 t=0
Integrating for t=0 to t=ti for
bi<b0
Ks = L/t1 * ln [(b0+L)/(b1+L)]
Falling Head Permeameter

13. b
Ks
Jw
L
πr2
ZR = 0
Hydrostatic Pressure at point z’
z’
Apply Darcy’s Law over the
entire column and compute Jw
and Ks
At “steady-state” water flow Jw
is constant throughout the
column

14. b
Ks
Jw
L
πr2
ZR = 0
Hydrostatic Pressure at point z’
z’
At point z’ (2):
H1 = p1 + z1 = 0
H2 = p’ + z’
Jw = -Ks (p’+z’)/z’
-Ks(p’+z’)z’ = -Ks(b+L)L (p’+z’)/z’ = (b+L)/L
p’ = (b+L)/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
k5
k4
k3
k2
k1
k6
Jw
L
1
5
4
3
2
6
Ke
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(Hn+1-Hn)/Ln
Ke is the total resistance
to water flow and is the
sum of the individual
hydraulic resistivities
over the column length
Ke = ΣLi/Σ(Li/Ki)

16. b =10 cm
K2 = 25
cm/d
K1 = 5
cm/d
Jw = ?
L1 = 25cm
L2 = 75cm
p = ?
First compute Ke
Apply Darcy’s 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

17. Intrinsic Permeability
Ks = k(ρlg)/η
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. place to place
3-dimensional
H/I = 1K H/AI = 3K IH/I = 2K IH/AI = 6K