Soil Permeability
1 Introduction
Natural soil
deposits invariably include water. Under certain conditions soil moisture or
water in the soil is not stationary but is capable of moving through the soil.
Movement of water through soil affects the properties and behaviour of the
soil, rather in a significant way. Construction operations and the performance
of completed construction could be influenced by soil water. Ground water is
frequently encountered during construction operations; the manner in which
movement of water through soil can occur and its effects are, therefore, of
considerable interest in the practice of geotechnical engineering.
2 Flow of Water Through Soil (Permeability)
It is necessary for a Civil Engineer to study the principles of
fluid flow and the flow of water through soil in order to solve problems
involving, – (a) The rate at which water flows through soil (for
example, the determination of rate of leakage through an earth dam) ; (b)
Compression (for example, the determination of the rate of settlement of a
foundation ; and (c) Strength (for example, the evaluation of factors of
safety of an embankment). The emphasis in this discussion is on the influence
of the fluid on the soil through which it is flowing ; in particular on the
effective stress.
Soil, being a particulate material, has many void spaces between
the grains because of the irregular shape of the individual particles; thus,
soil deposits are porous media. In general, all voids in soils are connected to
neighbouring voids. Isolated voids are impossible in an assemblage of spheres,
regardless of the type of packing; thus, it is hard to imagine isolated voids
in coarse soils such as gravels, sands, and even silts. As clays consist of
plate-shaped particles, a small percentage of isolated voids would seem
possible. Modern methods of identification such as electron micrography suggest
that even in clays all voids are interconnected.
Water can flow through the pore spaces in the soil and the soil
is considered to be ‘permeable’ ; thus, the property of a porous medium such as
soil by virtue of which water (or other fluids) can flow through it is called
its ‘permeability’. While all soils are permeable to a greater or a smaller
degree, certain clays are more or less ‘impermeable’ for all practical
purposes.
Permeability is one of the most important of soil properties.
The path of flow from one point to another is considered to be a straight one,
on a macroscopic scale and the velocity of flow is considered uniform at an
effective value ; this path, in a microscopic scale, is invariably a tortuous
and erratic one because of the random arrangement of soil particles, and the
velocity of flow may vary considerably from point to point depending upon the
size of the pore and other factors.
According to
fundamental hydraulics flowing water may assume either of two characteristic
states of motion—the ‘laminar flow’ and the ‘turbulent flow’. In laminar flow
each particle travels along a definite path which never crosses the path of
other particles; while, in turbulent flow the paths are irregular and twisting,
crossing and recrossing at random. Osborne Reynolds, from his classic
experiments on flow through pipes, established a lower limit of velocity at
which the flow changes from laminar to a turbulent one; it is called the ‘lower
critical velocity’. In laminar flow, the resistance to flow is primarily due to
the viscosity of water and the boundary conditions are not of much
significance; in turbulent flow, however, the boundary conditions have a major
influence and the effect of viscosity is insignificant
It is difficult to
study the conditions of flow in an individual soil pore; only average
conditions existing at any cross-section in a soil mass can be studied. Since
pores of most soils are small, flow through them is invariably ‘laminar’ ;
however, in the case of soils coarser than coarse sand, the flow may be
turbulent.
3 Darcey`s Law
In order to obtain a fundamental relation for the
quantity of seepage through a soil mass under a given condition, consider the
case. The cross-sectional area of the soil is equal to A and
the rate of seepage is Q. According to Bernoulli’s theorem, the total
head for flow at any section in the soil can be given by:
Total head =
elevation head + pressure head + velocity head
The velocity head for flow through soil is
very small and can be neglected. The total heads at sections A and B can thus
be given by:
The hydraulic
gradient i can be written as:
where L is
the distance between sections A and B.
Darcy published
a simple relation between the discharge velocity
and the
hydraulic gradient:
v = k. i
where v =
discharge velocity
i = hydraulic gradient
k = coefficient of permeability
Hence the rate
of seepage q can be given by:
q = k i A
3.1
Discharge velocity and seepage velocity
In any random cross section through a soil sample:
A= area of voids + area of solids
For a unit
of thickness:
Voids ratio = e =
= area of voids / area of solids
Porosity
= n =
= area of voids / total area A
q = A× v = Av× vs
Av = area of voids = n × A
v = n
vs
where: v : discharge velocity, vs : seepage
velocity
4 In Situ Permeability Tests (Pumping Tests)
It is sometimes difficult to obtain undisturbed
soil specimens from the field. For large construction projects it is advisable
to conduct permeability tests in situ and compare the results with those
obtained in the laboratory. Several techniques are presently available for
determination of the coefficient of permeability in the field, such as pumping
from wells.
(a)
Unconfined flow
Fig 6.7 shows a permeable layer underlain by
an impermeable stratum. The coefficient of permeability of the top permeable
layer can be determined by pumping from a well at a constant rate and observing
the steady-state water table in nearby observation wells. The steady-state is
established when the water levels in the test well and the observation wells
become constant. At steady state, the rate of discharge due to pumping can be expressed
as:
(a)
Confined flow
The coefficient of permeability for a confined
aquifer can also be determined from well pumping tests
5
Degree of Permeability
Considering the magnitude of the coefficient of
permeability, soils are classified into five categories as shown in Table 6.1.
Table 6.1
Degree of permeability
|
k (cm/sec)
|
Soil
|
High permeability
Medium permeability
Low permeability
Very low
permeability
Impermeable
|
> 10-1
10-1 to
10-3
10-3 to
10-5
10-5 to
10-7
< 10-7
|
Gravel and coarse sand
Med. And fine sand
Very fine sand
Silt
Clays
|
Solved Examples
1- A coarse
sand, 12 cm long and 7.5 cm in diameter, is tested in a constant head
permeameter under a head of 100 cm for 1 min and 12 sec. The quantity of
discharge is 5 litres. Determine the coefficient of permeability and then
determine the quantity of water would pass through a mass of soil 6o cm long
and 100 cm2 in cross section under a constant head of 20 cm.
Solution
Q = k i A; V/t = k H/L A
L = 60 cm, H =
100 cm, V = 5 litres = 5000 cm3, D = 7.5 cm, A=pD2/4 = =p(7.5)2 /4 = 44.16 cm2, t = 1(60) + 12 = 72 sec.
L = 12 cm, H =
100 cm, A= 100 cm2, t =
60×60 sec.
V = 22680 cm3 = 22.68 litres
2- The following results are obtained from a
falling head permeameter test on a sandy silt sample of 140 mm long and 70 mm
in diameter. The initial head is 1400 mm and the final head is 220 mm and the
time for fall in head is 80 sec. The stand pipe diameter is 6 mm. The constant
head test is carried out on the same compacted
level to the same void ratio, calculate the quantity of water falling
through the same sample in 10 min. if the head of water over a 100 mm length of
sample is 30 mm. The internal diameter of the permeameter is 90 mm.
Solution
Falling test:
L = 140 mm, D =
70 mm, A =p(70)2/4 = 3846.5 mm2,
h1=1400mm, h2=220mm, t = 80 sec., a = p(6)2 /4 = 28.26 mm2
Constant head
test:
T = 10×60 =
600 sec., L = 100 mm, H = 30 mm, A=p(90)2/4=6358.5
mm2,
V = 27239.8 mm3 = 27.24 cm3
3- Calculate the coefficient of permeability of a
soil sample, 8 cm in height and 10 cm in diameter if a quantity of water equal
to 450 cm3 passed in 10 min. and 20 sec. under an effective constant
head of 40 cm. The weight of oven dried sample is 495 gm. The unit weight of
solid particles is 2.65 gm/cm3. Calculate the flow velocity and
seepage velocity during the test.
Solution
L = 8 cm, H =
40 cm, V = 450 cm3, D = 10 cm, A=pD2/4 = =p(10)2 /4
= 78.5 cm2, t = 10(60) + 20 = 620 sec.
Vs
(volume of solids) = 186.79 cm3
n = (Vt
- Vs)/Vt Vt
n = (628 -
186.79) / 628 = 0.702
or: gd = Gsgw/(1+e) = Ws /Vt
e = 2.36, n =
e/(1+e) = 0.702
Discharge velocity = v = Q/A = V/tA = 450/(620×78.5)
= 0.0092 cm/sec.
Seepage velocity = vs
= v/n = 0.0092/0.702 = 0.013 cm/sec.
4- The data of a falling head permeability test on two soils are as
follows:
Stand pipe area = 75 mm2
Sample area = 2600 mm2
Sample height = 150 mm
Initial water head in stand
pipe = 1200 mm
Finial water head in stand
pipe = 300 mm
Time for decreasing the
water: Soil (1) = 400 sec.
Soil (2) = 60 sec.
Determine the coefficient of permeability of each soil in mm/sec. If
these two soils form adjacent layers the depth of the first one is 1.5 m thick
and the depth of the second is 2.3 m, calculate the average permeability in
vertical and horizontal directions.
Solution
Flow in
horizontal direction:
Flow in vertical direction:
5- A pumping
test is made in sand extending to a depth of 15 m where an impermeable stratum
is encountered. The initial ground water level is at the ground surface. Observation
wells are installed at distances of and
7.5 m from the pumping well. A steady state is established at about 20 hours then
the discharge is 3.8×106 mm3/sec. The drawdown at
the two observation wells are 1.5 and 0.35 m. Calculate the coefficient of
permeability.
Solution
See Fig. 6.9.
|
15m
|
|
r1=3m
|
|
r2=7.5m
|
|
1.5m
|
|
0.35m
|
|
Q
|
|
Fig. 6.9
|
|
Sand
|
6-. An
unconfined aquifer is known to be 32 m thick below the water table. A constant discharge
of 2 cubic metres per minute is pumped out of the aquifer through a tube well till
the water level in the tube well becomes steady. Two observation wells at
distances of 15 m and 70 m from the tube well show falls of 3 m and 0.7 m
respectively from their static water levels. Find the permeability of the
aquifer.
Solution
See Fig. 6.10.
Fig. 6.10 Unconfined aquifer example 6
7-A deposit of
cohesionless soil with a permeability of 3 × 10–2 cm/s has a depth of 10 m with
an impervious ledge below. A sheet pile wall is driven into this deposit to a
depth of 7.5 m. The wall extends above the surface of the soil and a 2.5 m
depth of water acts on one side. Sketch the flow net and determine the seepage
quantity per metre length of the wall.
Solution
The flow net is shown.
Number of flow channels, nf = 4
Number of equipotential
drops, nd = 14
Quantity of seepage per meter
length of wall
= 3 × 10–4 × 2.5 ×4/14 m3/sec/metre run
= 2.143 × 10–4 m3/sec/
meter run
= 214.3 ml/sec/metre run
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