Permeability of Soils & Seepage Analysis

Permeability of Soils &
Seepage Analysis
Wasim Saikh
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Types of soil water
• Soil water may be in the forms of Gravitational
water (free water)‘’ and Held water.
• The gravity water is free to move through the
pore space of the soil mass under the influence
of gravity.
• Gravitational Water: ‘Gravitational water’ is the
water in excess of the moisture that can be
retained by the soil.
• It translocates as a liquid and can be drained by
the gravitational force.
• It is capable of transmitting hydraulic pressure.
• Gravitational water can be subdivided into (a)
free water (bulk water) and (b) Capillary water.
• Free water may be further distinguished as (i)
Free surface water and (ii) Ground water
Water
Gravitational
Water
free water
Free surface
water
Ground
water
Capillary
water
Held Water
Structural
water
Absorbed
water

Free Water
(a) Free water (bulk water). It has the usual properties of liquid water. It moves at all times
under the influence of gravity, or because of a difference in hydrostatic pressure head.
Free water is of two types: Free surface water and free ground water
Free surface water may be from
precipitation, run-off, flood water,
melting snow, water from certain
hydraulic operations.
Ground water is that water which fills
up the voids in the soil up to
the ground water table and
translocates through them.

Capillary wate
‘Held water’ is that water which is held in soil pores or void spaces because of certain forces of
attraction. It can be further classified as (a) Structural water and (b) Absorbed water.
Held Water
Water which is in a suspended condition, held by the forces of surface tension within the
interstices and pores of capillary size in the soil, is called ‘capillary water’.
Structural water: Water that is chemically
combined as a part of the crystal structure
of the mineral of the soil grains is called
‘Structural water’. Under the loading
encountered in geotechnical engineering,
this water cannot be separated by any
means.
Soils which appear quite dry contain,
nevertheless, very thin films of moisture
around the mineral grains, called adsorbed
water

Assumptions of Darcy’s Law
• Soil is homogeneous and isotropic
• The flow is laminar
• Soil is fully saturated

Darcy’s Law
• Darcy's law is an equation that describes the flow of a fluid through a porous medium. The law
was formulated by Henry Darcy based on results of experiments on the flow of water through beds
of sand, forming the basis of hydrogeology
Darcy’s Experiment
A
L
h
h
k
q 





 
 2
1
A
i
k
q .
.

i = The hydraulic gradient.
L
h
h
i 2
1 

A
v
q .


Superficial Velocity and Seepage Velocity ( Discharge velocity)
• Superficial Velocity: A is the total area of cross-section of the soil, same
as the open area of the tube above the soil, v is the average velocity of
downward movement of a drop of water.
• This velocity is numerically equal to ki. (Without assuming voids)
• In reality drop of water flows at a faster rate through the soil than this
approach velocity because the average area of flow channel through the
soil is reduced owing to the presence of soil grains.
• The actual velocity of flow is greater than Superficial Velocity.
• This velocity is called as Seepage velocity.
A
v
q .

n
v
v s 
Vs = Seepage velocity
n = Porosity of the soil

Limitations of Darcy’s Law
• Darcy’s law is generally valid only when the flow is laminar i.e when the reynold’s number is less
than on equals to 2300.
• Darcy’s law is valid only for clay, silt and sand and not for gravels, cobbles etc. This is because the
flow is always turbulant.
• The actual velocity of discharge is greater than the equation given by Darcy.
• It is only applicable to fully saturated soil.

Permeability of soil
• Soil permeability, or hydraulic conductivity, is the rate of the flow of water through soil materials,
and it is an essential characteristic of soil.
• It is denoted by symbol k.
• The permeability of a soil can be measured in either the laboratory or the field
• The following are some of the methods used in the laboratory.
1. Constant head permeameter (In Syllabus)
2. Falling or variable head permeameter (In Syllabus)
3. Direct or indirect measurement during an Oedometer test
4. Horizontal capillarity test.
The following are the methods used in the field to determine permeability.
1. Pumping out of wells
2. Pumping into wells

Constant-Head Permeameter
• The objective of this test is to determine permeability of the
soil.
• A simple set-up of the constant-head permeameter is shown
in Fig.
• The principle in this set-up is that the hydraulic head causing
flow is maintained constant.
• The quantity of water flowing through a soil specimen of
known cross-sectional area and length in a given time is
measured.
• If Q is the total quantity of water collected in the measuring
jar after flowing through the soil in an elapsed time t, from
Darcy’s law.
t
A
i
Q
k
.
.

A
i
k
t
Q .
.
/ 
t
A
L
h
Q
k
.


hAt
QL
k 

• In highly impervious soils the quantity of water that can be collected will be small and, accurate
measurements are difficult to make.
• Therefore, the constant head permeameter is mainly application cable to relatively pervious soils,
although, theoretically speaking, it can be used for any type of soil.
• The water should be collected only after a steady state of flow has been established.
Constant-Head Permeameter

• This is more accurate method of measuring permeability of soil.
• The water level in the stand-pipe falls continuously as water flows
through the soil specimen.
• Observations are taken after a steady state of flow has reached.
• If the head or height of water level in the standpipe above that in the
constant head chamber falls from h0 to h1, corresponding to elapsed
times t0 and t1, the coefficient of permeability, k, can be shown to be
• (Derivation is not required)
Falling or Variable Head Permeameter
)
/
(
log
)
(
303
.
2
1
0
10
0
1
h
h
t
t
A
aL
k 



The discharge of water collected from a constant head permeameter in a period of 15
minutes is 500 ml. The internal diameter of the permeameter is 5 cm and the measured
difference in head between two gauging points 15 cm vertically apart is 40 cm. Calculate
the coefficient of permeability.
If the dry weight of the 15 cm long sample is 4.86 N and the specific gravity of the solids is
2.65, calculate the seepage velocity. (May 15-10 marks)
Given Data:
Q = 500 ml ;
t = 15 × 60 = 900 s
d = 5
L = 15 cm
h = 40 cm
k = ? cm/sec
2
4
d
A


hAt
QL
k 
sec
/
06
.
1
40
900
14
.
3
25
.
6
15
500
cm
k 






Calculation of Seepage velocity, vs
n
v
v s 
At
Q
v 
)
1
( e
G w
d




V
W
d 

AL
V 
e
e
n


1

Flow Perpendicular to the Bedding Planes:
PERMEABILITY OF LAYERED SOILS











n
n
k
h
k
h
k
h
h
k
......
2
2
1
1

Flow Parallel to the Bedding Plane:
PERMEABILITY OF LAYERED SOILS





 


h
h
k
h
k
h
k
k n
n
......
2
2
1
1

A horizontal stratified soil deposit consists of three layers each uniform in itself. The permeabilities of these
layers are 8 × 10–4 cm/s, 52 × 10–4 cm/s, and 6 × 10–4 cm/s, and their thicknesses are 7, 3 and 10 m
respectively. Find the effective average permeability of the deposit in the horizontal and vertical directions.
k1 = 8 × 10–4 cm/s h1 = 7 m
k2 = 52 × 10–4 cm/s h2 = 3 m
k3 = 6 × 10–4 cm/s h3 = 10 m
Flow Perpendicular to the Bedding Planes:











n
n
v
k
h
k
h
k
h
h
k
......
2
2
1
1
Flow Parallel to the Bedding Plane:





 


h
h
k
h
k
h
k
k n
n
h
......
1
1
1
1
kh=13.6 × 10–3 mm/s
kv=7.7 × 10–3 mm/s.

December 2011- 10 marks
Given Data:
L = 0.17 m
A = 21.8 x10-4
h0 = 0.25 m h1 = 0.1 m
a = 0.0002 m2.
k1 = 0.00003 m/s h1 = 0.06 m
k2 = 0.00004 m/s h2 = 0.06 m
k3 = 0.00006 m/s h3 = 0.05 m
t1-t0 = ?












n
n
v
k
h
k
h
k
h
h
k
......
2
2
1
1
)
/
(
log
)
(
303
.
2
1
0
10
0
1
h
h
t
t
A
aL
kv 











00006
.
0
05
.
0
00004
.
0
06
.
0
00003
.
0
06
.
0
17
.
0
k

Determination of Permeability—Field Approach
• The average permeability of a soil deposit or stratum in
the field may be somewhat different from the values
obtained from tests on laboratory samples.
• A few terms must be understood before.
• ‘Aquifer’ is a permeable formation which allows a
significant quantity of water to move through it under
field conditions.
• Aquifers may be ‘Unconfined aquifers’ or ‘Confined
aquifers.
• Unconfined aquifer is one in which the ground water
table is the upper surface of the zone of saturation and it
lies within the test stratum.
• Confined aquifer is one in which ground water remains
entrapped under pressure greater than atmospheric, by
overlying relatively impermeable strata.

When a well is penetrated into a homogeneous aquifer, the water table in the well initially remains horizontal.
When water is pumped out from the well, the aquifer gets depleted of water, and the water table is lowered
resulting in a circular depression in the phreatic surface.
• In pumping-out tests, drawdowns corresponding to a steady
discharge are observed at number of observation wells.
• The analysis of flow towards such a well was given by Dupuit
(1863).
• There are two different analysis for Confined and un-confined
aquifer

Determination of Permeability—Field Approach
• The following assumptions are relevant to the discussion that would follow :
1. The aquifer is homogeneous with uniform permeability and is of infinite areal extent.
2. The flow is laminar and Darcy’s law is valid.
3. The flow is horizontal and uniform at all points in the vertical section.
4. The well penetrates the entire thickness of the aquifer.
5. Natural groundwater regime affecting the aquifer remains constant with time.

Analysis for Unconfined Aquifer (Very Important)
• r0 be radius of central well
• r1 and r2 be the radial distances from the central well to two of the
observation wells
• Z1 and Z2 be the corresponding heights of a drawdown curve above the
impervious boundary
• Z0 be the height of water level after pumping in the central well above
the impervious boundary
• d0, d1 and d2 be the depths of water level (Drawdowns) after pumping
from the initial level of water table
• h be the initial height of the water table above the impervious layer (h =
Z0 + d0, obviously) and,
• R be the radius of influence or the radial distance from the central well
of the point where the drawdown curve meets the original water table.

Analysis for Unconfined Aquifer (Not required for everyone to do the
derivation)
• Let r and z be the radial distance and height above the impervious
boundary at any point on the drawdown curve.
• By Darcy’s law, the discharge q is given by :
• Here, A is lateral surface area, A = 2πr.z
• The hydraulic gradient, i, is given by dz/dr by Dupuit’s assumption
A
i
k
q .
.

dr
dz
A
k
q .
.

dr
dz
rz
k
q .
2
. 


Analysis for Unconfined Aquifer
dr
dz
rz
k
q .
2
. 

r
dr
q
zdz
k 







2
.
Integrating between the limits r1 and r2 for r and z1 and z2 for z
  2
1
2
1
2
log
2
2
.
r
r
e
z
z
r
q
z
k 














 






2
1
2
1
2
.
r
r
z
z
r
dr
q
zdz
k


  2
1
2
1
2
log
2
2
.
r
r
e
z
z
r
q
z
k 













   
1
2
2
1
2
2 log
log
2
.
2
r
r
q
z
z
k
e
e 









  















1
2
2
1
2
2 log
.
r
r
q
z
z
k e

  

















1
2
2
1
2
2
log
.
r
r
z
z
q
k e

  


















1
2
10
2
1
2
2
log
303
.
2
.
r
r
z
z
q
k

  


















1
2
10
2
1
2
2
log
73
.
0
.
r
r
z
z
q
k

  


















1
2
10
2
1
2
2
log
73
.
0
.
r
r
z
z
q
k
  


















1
2
10
2
1
2
2
log
36
.
1
.
r
r
z
z
q
k
k can be evaluated if z1, z2, r1 and r2 are obtained from observations in the field
If the extreme limits z0 and h at r0 and R are applied,
  


















0
10
2
0
2
log
36
.
1
.
r
R
z
h
q
k

A pumping test was carried out in a soil bed of thickness 15 m and the following measurements were
recorded. Rate of pumping was 10.6 x 10-3m3 /s; drawdowns in observation wells located at 15 m and 30 m
from the center of the pumping well were 1.6 m and 1.4 m, respectively, from the initial groundwater level.
The initial groundwater level was located at 1.9 m below ground level. Determine k
r1 = 15 m
r2 = 30 m
q = 10.6 x10-3m3 /s
Z1 = 15-1.9-1.6 = 11.5
Z2 = 15-1.9-1.4 = 11.7
  


















1
2
10
2
1
2
2
log
36
.
1
.
r
r
z
z
q
k
  


















15
30
log
5
.
11
7
.
11
36
.
1
10
6
.
10
. 10
2
2
3
k
s
m
k /
10
05
.
5
. 4




h = 10 m
r1 = 20 m
r2 = 50 m
q = 19.72 m3 /hr
q = 5.47 x 10-3 m3 /sec
d1 = 1.9 m
d2 = ?
k = 3.8 x10-4 m /sec
  


















1
2
10
2
1
2
2
log
36
.
1
.
r
r
z
z
q
k
m
z 35
.
8
2 
m
d 65
.
1
2 

To determine drawdown at test well use extreme limit formula
  


















0
10
2
0
2
log
36
.
1
.
r
R
z
h
q
k
Here, R is the radius of influence and can be calculated using formula
k
d
R max
3000

NOTE: Here dmax is in m and k is in m/s
Here, dmax = d2
m
R 49
.
96

m
Z 52
.
8
0 

Factors affecting Permeability of soil.
• The following soil factors that influence on permeability:
1. Grain-size
2. Void ratio
3. Composition
4. Structural arrangement of particles
5. Degree of saturation
6. Presence of entrapped air and other foreign matter.

1) Grain-size
• The permeability varies with the square of particle diameter.
• It is logical that the smaller the grain-size the smaller the voids and thus the lower the permeability.
• Here, D is in mm
2
100 D
k 
2) Void Ratio
• Permeability of soil is directly proportional to void ratio
• Increase in the porosity leads to an increase in the
permeability of a soil for two distinct reasons.
• Firstly, it causes an increase in the percentage of cross-
sectional area available for flow.

• Increase in compaction reduces the permeability of soil.
• The influence of soil composition on permeability is generally of little significance in the case of gravels,
sands, and silts, unless mica and organic matter are present.
4) Structural Arrangement of Particles:
• Structural arrangement of particles is an important soil characteristic influencing permeability, especially of
fine-grained soils.
• At the same void ratio, it is logical to expect a soil in the most flocculated state will have the highest
permeability, and the one in the most dispersed state will have the lowest permeability.
3) Composition

• The higher the degree of saturation, the higher the permeability.
• In the case of certain sands the permeability may increase three-fold when the degree of saturation increases
from 80% to 100%.
5) Degree of Saturation

Rise of Water in Capillary Tubes
• Capillarity, rise or depression of a liquid in a small passage such
as a tube of small cross-sectional area, like the spaces between the
openings in a porous material such as soil.
• Capillarity is the result of surface, or interfacial, forces.
• The rise of water in a thin tube inserted in water is caused by
forces of attraction between the molecules of water and the glass
walls and among the molecules of water themselves.
c
w
s
c
d
T
h

4
 Here,
hc = Height of capilary rise
Ts = Surface tension
dc = diameter of the glass capillary

Rise of Water in Capillary Tubes
c
w
s
c
d
T
h

4

Here,
hc = Height of capilary rise
Ts = Surface tension
dc = diameter of the glass capillary
c
c
d
h
30

The value of Ts for water varies with temperature. At ordinary or
room temperature, Ts is nearly 7.3 dynes/mm or 73 × 10–6 N/mm
and γw may be taken as 9.81 × 10–6 N/mm3.
Note: hc and dc are in mm

To what height would water rise in a glass capillary tube of 0.01 mm diameter ? What is the water pressure just
under the meniscus in the capillary tube ?
Given Data:
dc = 0.01 mm
hc = ?
Ts = 73 × 10–6 N/mm (Assumed)
γw = 9.81 × 10–6 N/mm3. (Assumed)
c
w
s
c
d
T
h

4

01
.
0
10
81
.
9
10
73
4
6
6




 

c
h
m
mm
hc 3
3000 

What is the water pressure just under the meniscus in the
capillary tube ?
c
w
v h

 
6
10
81
.
9
3000 



v

2
/
02943
.
0 mm
N
v 

2
/
43
.
29 m
kN
v 


Permeability of Soils & Seepage Analysis

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