Permeability

• Soils have interconnected voids through which

water can flow from points of high energy to
points of low energy.
• Estimating the quantity of underground seepage
under various hydraulic conditions.
• For investigating problems involving the pumping
of water for underground construction.
• Making stability analyses of earth dams and
earth-retaining structures that are subject to
seepage forces.
 Hydraulic Conductivity

Bernoulli’s Equation:

Where: h = total head

u = pressure
v = velocity
g = acceleration due to gravity
γw = unit weight of water
Z = is the vertical distance of a given point
above or below a datum plane.
Pressure, elevation, and total heads for flow of water
through soil
Because the seepage velocity is small:
u
h= +z
w
The head loss, Δh, can be expressed in a
nondimensional form as:
h
i=
L
Where: i = hydraulic gradient
L = distance between points A and B –
that is, the length of flow over which the loss of
head occurred
Darcy’s Law:
v = ki
Where: v = discharge velocity, which is the
quantity of water flowing in unit time through a
unit gross cross-sectional area of soil at right angles
to the direction of flow
k = hydraulic conductivity (otherwise
known as the coefficient of permeability)
Given, q, the quantity of water flowing through the
soil in unit time:
q = vA = Avvs
Where: vs = seepage velocity
Av = area of voids in the cross section of
the specimen

Also:
A = Av + As
Where: As = area of soil solids in the cross
section of the specimen
 q = v( Av + As ) = Av vs
v( Av + As ) v( Av + As )L v(Vv + Vs )
vs = = =
Av Av L Vv

Where: Vv = volume of voids in the specimen

Vs = volume of soil solids in the
specimen
  Vv   Where: e = void ratio
1 +    n = porosity
  V s   1 + e  v
vs = v = v =
 Vv   e  n
 V 
 s 
Hydraulic Conductivity
The hydraulic conductivity of a soil is also related to the
properties of the fluid flowing through it by the equation:

w
k= K

Where :
 w = unit weigh t of water
 = viscosity of water
K = absolute permeabili ty
Example Problem:

A permeable soil layer is underlain by an

impervious layer as shown. With k = 5.3 x 10-3
cm/sec for the permeable layer, calculate the rate
of seepage through it in m3/hr/m width if H = 3 m
and α = 8˚.
 ∆ℎ
𝑞 = 𝑣𝐴 = 𝑘𝑖𝐴 = 𝑘 𝐴
𝐿′
−3
𝑐𝑚
𝑘 = 5.3 𝑥 10
𝑠𝑒𝑐
∆ℎ = 𝐿(𝑡𝑎𝑛 𝛼) = 𝐿 tan 8 °
𝐿 𝐿
𝐿′= =
cos 𝛼 cos 8°
𝐴 = (3 𝑚) (cos 𝛼) 1 𝑚 = (3 𝑚2 )(cos 8°)
𝑐𝑚 1𝑚 3600 𝑠𝑒𝑐 𝐿 tan 8°
𝑞 = 0.0053 3 𝑚2 cos 8°
𝑠𝑒𝑐 100 𝑐𝑚 1 ℎ𝑟 𝐿
cos 8°
𝑞 = 0.0789 𝑚3 /ℎ𝑟/𝑚
Example Problem:

The hydraulic conductivity of a clayey soil is 3 x 10-7

cm/sec. The viscosity of water at 25°C is 0.0911 x 10-4 g
sec/cm2. Calculate the absolute permeability of the soil.
 𝛾𝑤
𝑘= ഥ
𝐾
𝜂
𝑘 = 3 𝑥 10−7 𝑐𝑚/𝑠𝑒𝑐
𝛾𝑤 = 1 𝑔/𝑐𝑚3
−4
𝑠𝑒𝑐
𝜂 = 0.0911 𝑥 10 𝑔 −
𝑐𝑚2
𝑔
𝑐𝑚 1 3
3 𝑥 10 −7 = 𝑐𝑚 ഥ
𝐾
𝑠𝑒𝑐 0.0911 𝑥 10−4 𝑔 − 𝑠𝑒𝑐
𝑐𝑚2
ഥ = 2.733 𝑥 10−12 𝑐𝑚2
𝐾
Laboratory Determination of Hydraulic
Conductivity

1. Constant Head Test (For Coarse-Grained Soils)

2. Falling Head Test (For Fine-Grained Soils)
Constant Head Test:
Q = Avt = A(ki )t
Where: A = area of cross
section of the soil specimen
t = duration of
water collection
h Where: L = length
i=
L of the specimen
 h QL
Q = A k t k=
 L Aht
Example Problem:

A constant head laboratory permeability test on a fine

sand gives the following values:
Length of specimen = 254 mm
Diameter of specimen = 63.5 mm
Head difference = 457 mm
Water collected in 2 min = 0.51 cm3

Determine these values:

a. Hydraulic conductivity, k, of the soil (cm/sec)
b. Discharge velocity (cm/sec)
c. Seepage velocity (cm/sec)

The void ratio of the soil specimen is 0.46.

𝐿 = 254 𝑚𝑚 = 25.4 𝑐𝑚
𝐷 = 63.5 𝑚𝑚 = 6.35 𝑐𝑚
𝜋 6.35 2
𝐴= = 31.67 𝑐𝑚2
4
ℎ = 457 𝑚𝑚 = 45.7 𝑐𝑚
𝑡 = 2 𝑚𝑖𝑛 = 120 𝑠𝑒𝑐
𝑄 = 0.51 𝑐𝑚3
𝑄𝐿
𝑎. ) 𝑘 =
𝐴ℎ𝑡
𝑘
0.51 𝑐𝑚3 25.4 𝑐𝑚
=
31.67 𝑐𝑚2 45.7 𝑐𝑚 120 𝑠𝑒𝑐
−5
𝑐𝑚
𝑘 = 7.46 𝑥 10
𝑠𝑒𝑐
 ℎ
𝑏. ) 𝑣 = 𝑘𝑖 = 𝑘
𝐿
𝑐𝑚 45.7 𝑐𝑚
𝑣 = 7.46 𝑥 10−5
𝑠𝑒𝑐 25.4 𝑐𝑚
−4
𝑐𝑚
𝑣 = 1.342 𝑥 10
𝑠𝑒𝑐
𝑣 𝑣
𝑐. ) 𝑣𝑠 = = 𝑒
𝑛
1+𝑒
𝑒 = 0.46
1.342 𝑥 10−4 𝑐𝑚/𝑠𝑒𝑐
𝑣𝑠 =
0.46
1 + 0.46
−4
𝑐𝑚
𝑣𝑠 = 4.259 𝑥 10
𝑠𝑒𝑐
Example Problem:

For a constant head permeability test, the

following values are given:
L = 300 mm
A = specimen area = 32 cm2
k = 0.0244 cm/sec

The head difference was slowly changed in steps to

800, 700, 600, 500, and 400 mm. Calculate and plot
the rate of flow, q, through the specimen, in
cm3/sec, against the head difference.
𝐿 = 300 𝑚𝑚 = 30.0 𝑐𝑚
𝐴 = 31.67 𝑐𝑚2
𝑘 = 0.0244 𝑐𝑚/𝑠𝑒𝑐
ℎ = 80 𝑐𝑚, 70 𝑐𝑚, 60 𝑐𝑚, 50 𝑐𝑚, 40 𝑐𝑚
𝑄𝐿 𝑞𝐿
𝑘= =
𝐴ℎ𝑡 𝐴ℎ h (cm) q (cm3/sec)
𝑘𝐴ℎ 80 2.06
𝑞=
𝐿
70 1.80
60 1.55
50 1.29
40 1.03
Falling Head Test:
h dh
q = k A = −a
L dt

Where: a = cross-
sectional area of the
standpipe
A = cross-
sectional area of the soil
specimen
 aL  dh 
dt =  − 
Ak  h 

aL h1
t= log e
Ak h2

aL h1
k = 2.303 log10
At h2
Example Problem:

For a variable head permeability test, the following

values are given: length of specimen = 380 mm,
area of specimen = 19.4 cm2, and k = 2.92 x 10-3
cm/sec. What should be the area of the standpipe
for the head to drop from 640 to 320 mm in 8 min?
𝐿 = 380 𝑚𝑚 = 38.0 𝑐𝑚
𝐴 = 19.4 𝑐𝑚2
𝑘 = 0.00292 𝑐𝑚/𝑠𝑒𝑐
ℎ1 = 640 𝑚𝑚 = 64.0 𝑐𝑚
ℎ2 = 320 𝑚𝑚 = 32.0 𝑐𝑚
𝑡 = 8 𝑚𝑖𝑛 = 480 𝑠𝑒𝑐
𝑎𝐿 ℎ1
𝑘 = 2.303 𝑙𝑜𝑔10
𝐴𝑡 ℎ2
𝑐𝑚 𝑎 38 𝑐𝑚 64 𝑐𝑚
0.00292 = 2.303 2
𝑙𝑜𝑔10
𝑠𝑒𝑐 19.4 𝑐𝑚 480 𝑠𝑒𝑐 32 𝑐𝑚
𝑎 = 1.032 𝑐𝑚2
Empirical Relations for Hydraulic Conductivity

Hazen:
k (cm/sec) = cD 2
10

Where: c = a constant that varies from 1.0 to

1.5.
D10 = the effective size (mm).
Casagrande:
k = 1.4e k0.85
2

Where: k = hydraulic conductivity at a void ratio

e.
k0.85 = the corresponding value at a void
ratio of 0.85.
Kozeny-Carman:
e3 Where: k = hydraulic
k = C1
1 + e conductivity at a void ratio e.
C1 = a constant.
Samarasinghe, Huang, and Drnevich:
en
k = C3
1+ e
Where: C3 and n are constants to be determined
experimentally

The equation can be rewritten as follows:

logk (1 + e ) = log C3 + n log e
Example Problem:

The hydraulic conductivity of a sand at a void ratio

of 0.5 is 0.02 cm/sec. Estimate its hydraulic
conductivity at a void ratio of 0.65.
𝑒1 = 0.5, 𝑘1 = 0.02 𝑐𝑚/𝑠𝑒𝑐
𝑒2 = 0.65, 𝑘2 = ?
𝑒13
𝑘1 𝐶1 1 + 𝑒1
=
𝑘2 𝑒23
𝐶1
1 + 𝑒2
0.53
0.02
= 1 + 0.5
𝑘2 0.653
1 + 0.65
𝑘2 = 0.04 𝑐𝑚/𝑠𝑒𝑐
Example Problem:

For a normally consolidated clay soil, the following

values are given:
Void ratio k (cm/sec)
1.1 0.302 x 10-7
0.9 0.12 x 10-7

Estimate the hydraulic conductivity of the clay at a

void ratio of 0.75. Use Huang, et. al.’s empirical
equation.
 −7
𝑐𝑚
𝑒1 = 1.1, 𝑘1 = 0.302 𝑥 10
𝑠𝑒𝑐
−7
𝑐𝑚
𝑒2 = 0.9, 𝑘2 = 0.12 𝑥 10
𝑠𝑒𝑐
𝑒 = 0.75, 𝑘 = ?
𝑒1𝑛
𝑘1 𝐶3 1 + 𝑒1 5.1
= 1.1
𝑘2 𝑒2𝑛 0.302 𝑥 10−7 = 𝐶3
𝐶3 1 + 1.1
1 + 𝑒2
𝐶3 = 3.9 𝑥 10−8 𝑐𝑚/𝑠𝑒𝑐
1.1 𝑛
0.302 𝑥 10−7 0.755.1
= 1 + 1.1 𝑘 = 3.9 𝑥 10−7
0.12 𝑥 10−7 0.9 𝑛 1 + 0.75
1 + 0.9 𝑐𝑚
𝑘 = 0.514 𝑥 10 −7
𝑛 = 5.1 𝑠𝑒𝑐
Equivalent Hydraulic Conductivity in Stratified Soil
Example Problem:
𝐻 =2+3+4=9𝑚
1
𝑘𝐻 𝑒𝑞 = 𝑘𝐻1 𝐻1 + 𝑘𝐻2 𝐻2 + 𝑘𝐻3 𝐻3
𝐻
1
𝑘𝐻 𝑒𝑞 = 10−4 2 + 3.2 𝑥 10−2 3 + 4.1 𝑥 10−5 4
9
𝑘𝐻 𝑒𝑞 = 0.0107𝑐𝑚/𝑠𝑒𝑐
𝐻
𝑘𝑉 𝑒𝑞 =
𝐻1 𝐻2 𝐻3
+ +
𝑘𝑉1 𝑘𝑉2 𝑘𝑉3
9
𝑘𝑉 𝑒𝑞 =
2 3 4
+ +
10−4 3.2 𝑥 10−2 4.1 𝑥 10−5
𝑘𝑉(𝑒𝑞) = 7.65 𝑥 10−5 𝑐𝑚/𝑠𝑒𝑐
𝑘𝐻(𝑒𝑞) 0.0107
= = 139.87
𝑘𝑉(𝑒𝑞) 7.65 𝑥 10−5