Nusrat Jahan

Lecturer (Grade-I)
Department of Civil Engineering
Ahsanullah University of Science & Technology

Nusrat Jahan, Lecturer, Department of CE, AUST 1

References
 Principles of Geotechnical Engineering – Braja M. Das
 Soil Mechanics and Foundation Engineering – Dr. K.
R. Arora
 www.slideshare.net/JISMIJACOB/soil-permeability-
ppt

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What is Permeability
 Permeability is the measure of the soil’s ability to
permit water to flow through its pore or void.
 Soil consists of solid particles with interconnected
voids where water can flow from a point of high energy
to a point of low energy.

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Importance of Permeability
 Estimating the quantity of underground seepage.
 Making stability analysis of earth dams and earth
retaining structures.

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Seepage and Discharge velocity
 The discharge of water per unit cross sectional area A
is known as Discharge Velocity.
 The flow which occur through the voids is known as
Seepage Velocity

 It is found that the seepage velocity is always greater

than the discharge velocity.
 Seepage velocity = Discharge velocity/Porosity

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Laboratory Determination of Hydraulic
Conductivity
 Constant Head Test.
 Falling Head Test.

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Constant Head Test

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Constant Head Test (cont.)
 In constant head permeability test, the water supply at
the inlet is adjusted in such a way that the difference of
head between the inlet and outlet remains constant
during the test period.
 After a constant flow rate is established, water is
collected in a graduated flask for a known duration.
 The total volume of water collected may be expressed as
Q = Avt = A(ki)t
Where, Q = Volume of water collected
A = Area of cross section of the soil specimen
t = Duration of water collection

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Constant Head Test (cont.)
 Hydraulic Gradient, i = h/L
Where, L = Length of specimen
So, Q = Ak(h/L)t
Or, k= QL/Aht

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Falling Head test

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Falling Head test (cont.)
From Darcy’s Law, v = ki
Again, flow rate q = vA = kiA
Where, i=h/L (here, h is continuously changing)
A very short interval of time dt, head is dh.
Amount of water collected in dt time, Q = q.dt
Continuity Equation:
-volume of water dropped in standpipe in time dt =
volume of water collected in the jar in time dt

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Falling Head test (cont.)
→ -a.dh = q.dt
→ -a.dh = kiA.dt
→ -a.dh = k(h/L)A.dt
→ (kA/aL)dt = -(dh/h) [where,(kA/aL) is constant]
→

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Example 7.1
 For a constant head permeability test, the following values are
given: Length, L=30 cm; Area of the specimen, A=177 cm2 ;
Constant head difference, h=50 cm; Water collected in a
period of 5 minutes=350 cm3. Calculate the hydraulic
conductivity in cm/sec.
 Solution: For constant head permeability, Given,
→ k=QL/hAt Q= 350 cm3,
Or, k= (350x30)/ (50x177x300) L=30 cm,
-3 A=177 cm2,
k =3.95x10 cm/sec h=50 cm,
t=5 min
(5x60=300sec)
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 Example 7.2
 For a falling head permeability test, the following values are
given: Length of specimen=203 mm; Area of the soil
specimen=10.3 cm2 ; Area of the standpipe=0.39 cm2 ; Head
difference at time t=0=508 mm; Head difference at time t=180
sec=305 mm. Determine the hydraulic conductivity of the soil
in cm/sec.
Given,
 Solution: For falling head permeability, L=203 mm =20.3cm,
k= A=10.3 cm2,
a=0.39 cm2,
h1=508 mm,
h2=305 mm,
t=180 sec

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Example 7.3
 The hydraulic conductivity of a clayey soil is 3 x 10-7
cm/sec. The viscosity of water at 250 C is 0.0911 x10-4
g.sec/cm2. Calculate the absolute permeability of the soil.
 Solution:

Given,
k= 3 x 10-7 cm/sec
Or, 3 x 10-7 = [1/ (0.0911 x10-4)]
= 0.2733x10-11 cm2

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Example 7.4
 A permeable soil layer is underlain by an impervious layer
with k= 5.3 x 10-5 m/sec for the permeable layer,
calculate the rate of seepage through it in m3/hr/m width
if H = 3 m and 𝜶 = 80

 Solution:

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Hydraulic Gradient, i = head loss/length =

= sin𝜶
Here, k = 5.3x10-5 m/sec
A = (3 cos𝜶 x 1) m2
Flow rate, q = kiA = (5.3x10-5)(sin𝜶)(3 cos𝜶 x 1)(3600)
q = 0.0789 m3/hr/m

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Example 7.5
 Find the flow rate in m3/sec/m length through the permeable
soil layer. Given, H = 8 m, H1 = 3m, h = 4 m, L = 50 m, 𝜶 =
80, and k = 0.08 cm/sec.
 Solution:

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Hydraulic Gradient, i = head loss/length =

q = kiA
= k (h cos𝜶/L)(H1 cos𝜶 x1)
= (0.08x10-2 m/sec)(4 cos 80 /50)(3 cos 80 x1)
= 0.19x 10-3m3/sec/m

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Relationships for hydraulic conductivity (Granular
soil)
 k 𝜶 (e3/ 1+e)
 Example 7.6 : 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
 Solution: Given,
k1= 0.02 cm/sec
e1 = 0.5
e2 = 0.65
k2 = ?

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Permeability of stratified soil deposits
 A stratified soil deposit consists of a number of soil layers having
different permeabilities. The average permeability of the deposit
as a whole parallel to the planes of stratification and that normal
of the planes of stratification can be determined as expressed
below.
 (a) Flow parallel to planes of stratification
Let us consider a deposit consisting of two horizontal layers of
soil of thickness H1 and H2.
For flow parallel to the planes of stratification, the loss of head
(h) over a length L is the same for both the layer. Therefore, the
hydraulic gradient (i) for each layers is equal to the hydraulic
gradient of the entire deposit.

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 (b) Flow normal to planes of stratification
Let us consider a deposit consisting of two layers of soil
of thickness H1 and H2 in which flow occurs normal to
the plane of stratification.

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Example 7.11
 A layered soil is shown in the following figure. Given,

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 Solution:

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Example 7.12

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