Applications of Darcy’s Law

Vertical Flow is in the “z” direction.z

xy

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.

Steps in the application of Darcy’s Law

• Define a Reference Elevation

• Determine two points where H is known

• Calculate the Gradient

Horizontal Column

1 2x2 - x1 = column length L

2

1

Vertical Column

z2 - z1 = column length L

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)

In the generic ----

ZR = 0

p1 p2

1

2L

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

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

ZR = 0

p1 = 10 cm

1

2L = 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

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/dNegative, 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)

Vertical Flow (Up):

Jw = Ksp1/L - Ks

p = 110 cm

L = 100 cmKs = 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.

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

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 “-”.

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

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

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

Water Flow in a Layered Soil

b

k5

k4

k3

k2

k1

k6

Jw

L

1

54

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)

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

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

place to place

3-dimensional

H/I = 1K H/AI = 3K IH/I = 2K IH/AI = 6K