Source: GEOTECHNICAL ENGINEERING

14 Seepage

14.1 GRAVITATIONAL FLOW OF WATER IN SOIL

14.1.1 The Overall Direction Is Down

Unlike capillary water that can move in any direction in response to suction or
negative pore water pressure, the net direction of gravitational flow is downward.
Even uphill flow in a siphon requires that the exit must be lower than the level of
the entrance.

The flow of water in soil in some ways is similar to flow in a pipe, but soil is a wild
array of tiny channels consisting of extremely irregular, angular crevices between
grains of the soil. Resisting the flow is viscous drag along these pores, and the
smaller the effective diameter of the pores, the larger the drag and the slower the
flow. The seepage rate in sand, other factors being the same, differs by a factor of
a thousand or more from that in clay.

14.1.2 Significance of Groundwater Flow

Any excavation that penetrates below the groundwater table becomes a target for
seepage, which unless controlled will flood the excavation up to the original
groundwater level. Because seepage exerts a viscous drag on the soil, it can affect
stability of the sides of the excavation. Groundwater elevations should be
measured as part of an exploration drilling program.

The groundwater table can be lowered locally with wells, and perimeter wells and
drains are commonly used for this purpose around a construction site. The flow of
water into an excavation also can be restricted by sheet-pile walls, but pumping
still is required to lower the water level and compensate for leaks.

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Seepage 319

The rate of groundwater flow affects the number and spacing of wells for site
dewatering, the size and number of pumps, and the means for disposal of the
extracted water. For this reason it is important that the flow rate be estimated
prior to construction dewatering. Sandy and gravelly soils that allow relatively
high flow rates are said to be aquifers, highly valued as sources for water but also
posing an inconvenience or danger if they are encountered unexpectedly in an
open excavation or tunnel.

Earth dams are designed to minimize seepage through, under, and around
the dam. Nevertheless some leakage is inevitable, and seepage through the
dam is directed into toe drains so that it will not exit and endanger the
integrity of the downstream face of the dam. Seepage force in an upward
direction, as can occur below an earth dam or near a levee, is the cause of
quicksand.

14.1.3 Seepage Forces

As soil restrains water from quickly draining out, water exerts an equal and
opposite drag on the soil in the direction of flow—for every action there is an
equal and opposite reaction. This becomes a matter of considerable importance in
the initiation of landslides, slope failures, and mudflows. It is no coincidence that
most slope failures occur after prolonged periods of rain that increase downslope
seepage of water within the soil.

14.1.4 Laminar Flow

Gravitational flow can be laminar or turbulent, depending on the fluid viscosity,
flow velocity, and the size, shape, and smoothness of the conduit or channel
through which the fluid is flowing. In soils the velocity of groundwater rarely, if
ever, becomes high enough to produce turbulence, so flow normally is laminar,
although exceptions can occur in underground caverns and possibly in rapidly
draining road base course materials.

Water in laminar flow moves in essentially parallel paths. The quantity of water
flowing past a fixed point in a stated period of time therefore equals the
cross-sectional area of the water multiplied by the average velocity of flow.
This relationship, referred to as a continuity condition, is expressed by the
formula

Q ¼ vA ð14:1Þ

where Q is the volume of flow per unit of time, v is the average flow velocity, and
A is the cross-sectional area of flowing water.

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320 Geotechnical Engineering

14.2 MECHANICS OF GRAVITATIONAL FLOW

14.2.1 Gravitational Potential

In hydraulics, gravitational potential is conveniently expressed in units of head.
Head includes three components: pressure head, elevation head, and velocity
head, all of which are expressed in terms of equivalent height above a datum.
Velocity head is negligible for seepage through soils where fluid velocities are low.

As shown in Fig. 14.1, in a static body of water there is a tradeoff between

pressure head and elevation head: as elevation decreases, pressure increases and
total head remains the same, so there is no head difference to drive gravitational
flow.

Water head is expressed as a height or equivalent height of water above a datum.

For example, if the datum is the ground surface, artesian water confined in a
vertical pipe will rise to the level indicated by its combined pressure and elevation
head.

14.2.2 Hydraulic Gradient

Hydraulic gradient is one of the factors that determines rate of flow, and is
defined as the reduction in total head divided by the distance over which it occurs.
By definition,
h
i¼ ð14:2Þ
d
where i is the hydraulic gradient, h is the change in head, and d is the distance over
which it occurs. For example, if an open channel is 1000 m long and declines 10 m
in that distance, the hydraulic gradient is 10/1000, or 0.01. If the decline is 100 m,

Figure 14.1
In standing water
such as in this
standpipe total
head everywhere
is the same, so
there is no
gravitational flow.

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Seepage 321

the hydraulic gradient is 0.1, and the water flow will be substantially faster.
Streams descend rapidly from mountains because of a high hydraulic gradient,
whereas rivers on floodplains lower their gradient by extending their channel
lengths through meandering, and have a much slower rate of flow.

14.2.3 Darcy’s Law

In 1856 a French engineer, Henry Darcy, measured the volume of water flowing
through saturated sand columns and discovered the relationship now known as
Darcy’s Law:
Q ¼ kiA ð14:3Þ
where Q is the volume of water flowing through the soil in a unit of time, k is a
proportionality constant, i is the hydraulic gradient, and A is the cross-sectional
area of the soil. This relationship is illustrated by flow through a container
filled with sand in Fig. 14.2. As the reduction in head occurs entirely within the
sand, it is possible to contour the loss with vertical lines called equipotential lines.
This relationship is general and may be applied to any flow that is laminar
in character.

The value of the proportionality constant k depends on soil properties,

in particular the size and number of soil pores. The equation simply means that

Figure 14.2
Diagram illustrating
the loss of head, h,
as water Q flows
through sand. The
hydraulic gradient,
i, equals h divided
by the flow
distance d.

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322 Geotechnical Engineering

the quantity of water flowing through a given cross-sectional area of soil is equal
to a constant multiplied by the hydraulic gradient.

14.2.4 Hydraulic Conductivity

The constant k in eq. (14.3) commonly is called the coefficient of permeability, but
since permeability also applies to gas flow, a more appropriate term for water flow
is coefficient of hydraulic conductivity. As k depends on the size and number of
voids in a unit cross-sectional area, it is indirectly related to the size, shape, and
packing of the soil grains. A clayey soil with very fine grains will have a much
lower coefficient than will a sand with relatively coarse grains, even though the
void ratio and the density of the two soils may be nearly the same. On the other
hand, when we consider the same soil in two different states of density, as density
increases due to compaction or consolidation, pore spaces are reduced in size,
resistance to flow is increased, and k decreases.

14.2.5 Approach Velocity

In the application of the Darcy Law and eq. (14.3), the cross-sectional area A is
the area of the soil including both solids and void spaces. Obviously, flow does not
occur through the solid fraction of a soil so the velocity, v, in eq. (14.1), or the
product ki in eq. (14.3), is a factitious or ‘‘made-up’’ velocity. It is referred to as
the ‘‘velocity of approach’’ or the ‘‘superficial velocity’’ of the water just before
entering or after leaving a soil mass. The true velocity can be approximated by
dividing v by the soil porosity.

As i in eq. (14.3) is dimensionless, k has the dimensions of a velocity, that is, a

distance divided by time. However, it should be emphasized that having units of
velocity does not mean that k is a velocity because the actual velocity also depends
on the hydraulic gradient. The value of k represents a ‘‘superficial velocity of
water flowing through soil under unit hydraulic gradient.’’ The actual flow
velocity of water through soil must be higher than the superficial velocity k,
because not only is the cross-section partly occupied by soil grains, but the flow is
not straight-line and follows tortuous paths around the soil grains.

14.2.6 Hagen-Poiseuille Equation

The Hagen-Poiseuille derivation assumes laminar flow through a straight tube of
uniform diameter and therefore does not take into account the irregular cross-
section and tortuosity of flow channels in soil. Nevertheless the derivation is
useful because it shows the relationship between flow velocity and an effective
pore diameter.

In Fig. 14.3 an axially symmetrical element of radius y is acted on at the ends by

uniformly distributed pressure heads h1 and h2, and on the sides by viscous shear.
Hydraulic head is expressed in terms of height of a water column, so to convert to
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Seepage 323

Figure 14.3
(a) Forces on a
fluid element in
laminar flow; (b)
distribution of
viscous shearing
resistance, which
is zero at the
center of the tube;
(c) distribution of
fluid velocity in the
tube.

pressure, h is multiplied by the unit weight of water,
w. Equating forces in the

x-direction gives

y2
w dh ¼ 2y dx

y
w dh
¼
ð14:4Þ
2 dx

where  is the viscous shearing resistance. Therefore if y ¼ 0 at the center of the

tube,  ¼ 0.

In an ideal Newtonian fluid, the shearing stress, , is proportional to the rate

of shearing, which can be represented as the change in velocity across a distance
y. If V is the velocity at any point within a capillary, the shearing rate is
dV/dy. Along the center axis of a tube where shearing stress is zero, the rate of
change of velocity with y also is zero, shown in Fig. 14.3(b). Since, according
to eq. (14.4), shearing stress is proportional to y, it increases linearly to a
maximum at the tube boundaries, where the velocity is zero.

Integration with respect to y indicates that the velocity distribution follows a

parabola across a tube with laminar flow (Fig. 14.3(c)). In a unit of time the
amount of fluid moved through a cylindrical tube is described by rotation of
a parabola about its central axis, which is a paraboloid. The volume q moved
per unit of time then is

D4 i
w
q¼ ð14:5Þ
128 
where D ¼ pore diameter
 ¼ fluid viscosity
i ¼ dh/dx, the hydraulic gradient or head loss per unit length

w ¼ unit weight of water.
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324 Geotechnical Engineering

Equation (14.5) emphasizes the important role of pore diameter, which is to

the fourth power. However, this is partly offset because a smaller pore
diameter means that there are more pores per unit cross-sectional area of
soil. As the number of pores n is approximately inverse to D2, and the flow per
unit area then is
Q
w
nq ¼ ¼ C1 D2 i ¼ ki ð14:6Þ
A 

w
k ¼ C1 D2 ð14:6aÞ

where C1 is a constant and k is the coefficient of hydraulic conductivity of the soil.
However, as the pore diameter D is not known and is not readily measured,
empirical investigations have for the most part attempted to relate fluid flow
characteristics of soils to their particle size and void ratio.

A commonly used empirical equation developed by Hazen is

k ¼ C2 ðD10 Þ2 ð14:7Þ
where D10 is the soil effective size, C2 ¼ 100, and k is in centimeters per second.

Hazen’s equation does not account for variability in void ratio or influences
from other grain sizes, and used alone is not considered reliable for
prediction. Other studies have shown that, as an approximation, with particle
size constant,

3 
e
k ¼ C3 ð14:8Þ
1þe
where C3 is a constant and e is the void ratio.

In summary, the coefficient of hydraulic conductivity, k, increases approx-

imately as the square of both the particle size and the void ratio. Equation (14.8)
is a simplified version of the Kozeny-Carman equation for flow through porous
media.

Typical ranges of k values are shown in Table 14.1. Traditional metric

units are in centimeters per second, but because of the extended range of
the values, the coefficient frequently is expressed in terms of those units and
an exponent to base 10. Therefore a coefficient of ‘‘minus 1’’ normally will
mean 10
1 cm/s.

In Table 14.1 it will be noted that k less than ‘‘minus 5’’ is considered low,
‘‘minus 3 to minus 4’’ is medium, and larger than ‘‘minus 2’’ is high. It should be
emphasized that the other major influence on k, density, is not included in this
table, and it is possible to convert a medium to a low coefficient simply by
compaction.
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Seepage 325

Descriptive term k, cm/s Soils Table 14.1

Representative
4100 Clean gravel
High 100 to 10
2 Gravel, clean coarse sand values of the
10
2 to 10
3 Well-graded sand, fine sand hydraulic
Medium 10
3 to 10
4 Silty sand conductivity
10
4 to 10
5 Dense silt, clayey silt coefficient, k
Low 10
5 to 10
6 Clay, silty clay
510
6 Clay

14.3 MEASURING HYDRAULIC CONDUCTIVITY OF SOIL

14.3.1 The Need to Know

Seepage affects water losses through and under earth dams and levees, from
irrigation ditches, and into open excavations. Seepage rates also affect pumping
capacity for dewatering, and spacing and depths of drains for lowering the
groundwater table. Seepage rates also influence settlement of structures as water is
forced out of the foundation soil, which is the subject of a later chapter.

14.3.2 Different Measurement Methods

Hydraulic conductivity measurements can be conducted in the field with various
pumping tests, or in a laboratory on disturbed or undisturbed samples. Each
procedure offers advantages for particular types of problems, and the method that
is most feasible and appropriate for a particular problem is the one that should be
used. For example, in order to predict seepage through an anticipated earth dam,
the most appropriate test would be conducted in the laboratory on samples of soil
compacted to represent the range in densities and moisture contents specified for
the dam. On the other hand, a field test of the soil in place is more appropriate for
studies related to dewatering of an excavation. In every case, the objective should
be to determine the conductivity of the soil either in its natural or an anticipated
condition and to do so as accurately as possible.

14.3.3 Precision of Measurements

Values of k vary by many orders of magnitude, depending on soil voids that
are highly sensitive to soil structure, gradation, and density. These variations
influence measurements that therefore are not precise and should not be reported
to more than 2 significant figures. Other factors, including remolding during
sampling, cracks, root holes, and perimeter leakage in the measuring chamber,
also can profoundly affect the data, which therefore in all cases must be examined
to determine if the numbers are reasonable. Anomalous tests should be repeated,
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326 Geotechnical Engineering

and if results are not consistent, the measurement setup and procedure should be
examined for possible sources of error.

For example, if a field permeability test gives k that is many times higher than
other tests in the vicinity, and the test is repeated and gives the same high value, it
could mean proximity to a fissure or an abandoned mine tunnel or cavern.

Another common cause of high k measurements is if the water pressure used for
testing opens fissures or even fractures the soil, a process called hydraulic
fracturing. If that happens, more realistic measurements will be obtained if the
tests are conducted at lower pressures. However, a pressure that can lead to
hydraulic fracturing also can be relevant. For example, it would be extremely
unwise for fluid pressures near or under a dam to be allowed to exceed those
required for hydraulic fracturing, and some major dam failures have been
attributed to this cause (Leonards, 1987).

The sensitivity of hydraulic conductivity to small changes in void ratio emphasizes

the importance of minimizing disturbance during sampling of soils for laboratory
tests. In addition, samples must be tight against walls of the test containers to
prevent seepage around the margins.

Another important factor influencing results is that natural soil deposits often are
nonisotropic with respect to flow, that is, the conductivity coefficient in the
vertical direction differs considerably from that in the horizontal direction, a
situation that is most evident in layered alluvial or lacustrine soil deposits. If such
a condition exists, horizontal and vertical conductivity measurements may be
conducted, and the ratio between the two is used to quantify flow in the field in
relation to the seepage direction.

Field measurements should simulate anticipated flow directions. For example,

flow into the sides of an open excavation will be horizontal, whereas flow up into
the bottom will be vertical. Flow rates in field tests should be reproducible within
5 percent, and repeated determinations of k should be within 25 percent of one
another.

14.4 LABORATORY MEASUREMENTS OF k

14.4.1 Constant-Head Permeameter

Coarse-grained soils can be tested in a laboratory constant-head permeameter
that uses an overflow to maintain a constant hydraulic head on the sample.
The hydraulic gradient then remains constant throughout the testing period.
A schematic diagram is shown in Fig. 14.4. The amount of water passing through
the soil sample during a period of time is measured, and appropriate values that

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Seepage 327

Figure 14.4
Schematic
diagram of a
constant-head
permeameter.

include dimensions of the apparatus and soil sample are substituted in eq. (14.3)
to obtain a value for the conductivity coefficient.

In preparation for testing, sand or silt soil is poured into the permeameter in layers
that are tamped in place to a predetermined relative density. Then to prevent air
blockage within the sample, air is removed from the soil by use of a vacuum
(ASTM Designation D-2434) or by purging with CO2 gas, which dissolves and is
removed by permeating water. Another potentially serious problem is if air that
initially is dissolved in the permeating water comes out of solution and forms
bubbles in the soil. This possibility is reduced by having both the inlet and outlet
pressures high to keep the air in solution, as shown in Fig. 14.4.

The test may be performed on the same sample with several values of h to deter-
mine linearity. If the flow rate is not linear with increasing h, this signals a
nonuniform response such as turbulent flow, air in the system, segregation of
fines, or development of channels, or piping. Piping also may be evident from
muddying of the exit water.

Example 14.1
A soil sample in a constant-head apparatus is 152 mm (6 in.) in diameter and 203 mm (8 in.)
long. The vertical distance from headwater to tailwater is 279 mm (11 in.). In a test run,
347 kg (766 lb) of water passes through the sample in 4 hr 15 min. Determine the coefficient
of hydraulic conductivity. Is this value considered high or low?
Answer: In this test, h ¼ 279 mm and d ¼ 203 mm, so i ¼ h/d ¼ 1.37. A ¼ P (15.2/2)2 ¼ 181 cm2.
Q ¼ 347 kg/255 min ¼ 1.36 kg/min ¼ 2.27 g/s ¼ 2.27 cm3/s. From eq. (14.3),

Q ¼ kiA
2:27 cm3 =s ¼ k  1:37  181 cm2
k ¼ 9:1  ð10Þ
3 cm=s

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328 Geotechnical Engineering

which is rounded to 10
2 cm/s. This in the medium range and representative of
sand.

In computing the value of the conductivity coefficient from data obtained in a test
of this type, as in all fluid flow problems, it is important to carry dimensions
through the calculations in order to ensure that they are dimensionally correct.
A relatively easy and sure way to do this is to decide in advance the final units
that are desired, and reduce the values of Q and A to those units before making
the computation, as was done in the example.

14.4.2 Falling-Head Permeameter

Fine-grained soils that require long testing times are most efficiently tested by
simply filling a permeameter and letting the head decrease as the water runs out.
The decline in head is measured in a small tube to increase precision. This
arrangement is shown in Fig. 14.5. In the conduct of the test, the water passing
through the soil sample causes water in the standpipe to drop from h0 to h1 in time
t1. Because the head is constantly decreasing, an integration is necessary.

Figure 14.5
Schematic
diagram of a
falling-head
permeameter.

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Seepage 329

The head on the sample at any time t between the start and finish of the test is h,
and in any increment of time dt there is a decrease in head dh. From these
definitions the following relationships may be written. The minus sign is required
as the head is decreasing with time.
h dh
k¼ A ¼
a ð14:9Þ
d dt
where A is the cross-sectional area of the soil specimen and a is that of the
standpipe. Then
Zt1 Zh1
A dh
k dt ¼
a ð14:10Þ
d h
t0 h0

from which
ad h0
k¼ ln ð14:11Þ
At1 h1

Example 14.2
A sample of clay soil having a cross-sectional area of 78.5 cm2 (12.15 in.2) and a height of
50 mm (1.97 in.) is placed in a falling-head permeameter in which the area of the standpipe
is 0.53 cm (0.082 in.2). In a test run, the head on the sample drops from 800 mm to 380 mm
(31.5 to 15.0 in.) in 1 hr 24 min 18 sec. What is the coefficient of hydraulic conductivity of
this soil? Is this value reasonable?
Answer: Substituting a ¼ 0.53 cm2, d ¼ 5.0 cm, A ¼ 78.5 cm2 and the appropriate values
for t1, h1, and h2 in eq. (14.11) gives

k ¼ [(0.53 cm2)(5.0 cm)  (78.5 cm2)(5058 s)] ln (800 mm/380 mm)

¼ 6.7(10)
6  0.744 ¼ 5(10)
6 cm/s

A similar procedure is used with English units, in which case the answer will be in inches/
sec. Note that by writing the units they cancel out and give the answer in cm/s. The value
for k should be carefully inspected to determine if it is reasonable. This value for k is very
low and indicates clay.

14.4.3 Flexible-Wall Permeameter

Perimeter leakage is reduced if the soil is contained in a rubber membrane and
placed inside a pressure cell, as shown in Fig. 14.6. Undisturbed samples of
fine-grained soils are trimmed to remove remolded soil around the edges, and
water pressure on the outside of the membrane pushes it into intimate contact
with the soil. Ideally the cell pressure should simulate lateral pressure existing in
the field. The sample can be de-aired with back-pressure on the water to dissolve
all air. Simultaneous testing of several samples is facilitated by using regulated
air pressure to drive the input.
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330 Geotechnical Engineering

Figure 14.6
Schematic
diagram of a
flexible-wall
permeameter.

Either constant or falling-head methods may be used, and specimen length can be
monitored to ensure that the soil does not consolidate as a result of seepage forces.
This test method is described in ASTM Designation D-5084.

14.4.4 Selecting Appropriate Test Conditions

Hydraulic conductivity is so variable and is influenced by so many factors that it is
important that soil samples have a structure, density, and lateral confining pressure
representative of field conditions. For example, air is present in all compacted soils
and eventually will dissolve in the seepage water, increasing conductivity. Therefore
compacted soil samples should be evacuated and de-aired prior to testing, or time
allowed for air under pressure in the sample to dissolve and be carried away in the
seepage water. In that case several changes of water are necessary, and the test
should be continued until the flow rate becomes constant. Water used in a test may
be de-aired by evacuating or boiling, and it is important that water be kept under
pressure in the sample to keep any air in solution.

Tap water ordinarily is used as a permeant, as distilled water becomes acidic from
capture of CO2 gas from air to create H2CO3. The surplus of Hþ ions can have
a dispersing effect on soil clay that will influence the soil structure and hydraulic
conductivity. Unsoftened the water also is preferred to avoid the dispersing effects
of Naþ ions in the water.

Another method for eliminating perimeter leakage is to test soil samples while
they are still inside the special thin-walled steel tubes that are used to acquire the
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Seepage 331

samples, cutting the tubes and the contained soil samples into lengths that fit into
a permeameter. In this case, conductivity will be reduced by remolding of soil
around the perimeter of the sample.

14.4.5 Temperature and Viscosity of Water

The coefficient of hydraulic conductivity is influenced by the viscosity of the
permeating water, which depends on its temperature. Permeability data therefore
are adjusted to a standard laboratory temperature of 208C. However, it will be
noted that that temperature does not represent most field conditions, as evidenced
by the temperature of groundwater in a well or spring, where the temperature
approximates that of the mean annual air temperature. Therefore a temperature
correction is required to simulate most field conditions—the cooler the water, the
higher the viscosity and the lower the hydraulic conductivity.

The coefficient of viscosity in SI is in Newton-seconds per square meter (N s/m2).

The coefficient of viscosity of water at 20.208C (68.368F) is 1mN s/m2, which also
equals 1 centipoise in the cgs metric system. Correction factors based on the
viscosity of water are presented in Table 14.2.

Example 14.3
Correct a coefficient of hydraulic conductivity k27 ¼ 12(10)
4 cm/s measured in the
laboratory at 278C to a k10 for a field temperature of 108C.
Answer: The first step is to adjust the laboratory data to the standard temperature of
208C, which, being lower than the measurement temperature, means that the water
will have a higher viscosity so the corrected k should be lower. The correction factor
for 278C is 0.85. Therefore
k20 ¼ 0:85  12ð10Þ
4 ¼ 10:2ð10Þ
4

The second step is to adjust k20 to k10, and again the corrected viscosity should be lower.
The correction factor is 1.3, which is divided into k20:

8C Correction Table 14.2

Correction factors
10 1.30
to adjust k to a
12 1.23
standard reference
14 1.16
temperature of 208C
16 1.11
18 1.05
20 1.00
22 0.953
24 0.910
26 0.869
28 0.832
30 0.797

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332 Geotechnical Engineering

k10 ¼ 10.2(10)
4  1.3 ¼ 7.8(10)
4 cm/s

As a result of the two-stage correction the final value for k is 35% lower than that
measured in the laboratory.

14.5 FIELD TESTS FOR HYDRAULIC CONDUCTIVITY

14.5.1 Advantages of Field Tests

Because most sampling is conducted vertically, laboratory tests normally use
vertical flow even though hydraulic conductivity often is higher in a horizontal
direction, sometimes by orders of magnitude. Field measurements therefore
are more reliable for predicting lateral seepage into excavations or under dams,
and in many other situations involving groundwater flow through natural soils.
Field measurements also avoid problems from sampling disturbance and
are conducted on a much larger scale that averages the effects of local variations
in the soil.

14.5.2 Drawdown Pumping Test

A relatively simple method for measuring hydraulic conductivity in the field
is to pump the water level down a test well and measure water levels in two or more
observation wells spaced at different distances from the test well. The test well has
a perforated casing and ideally is drilled through the water-bearing soil to an
underlying impervious stratum, but if no impervious stratum is present, the well is
drilled to a considerable depth below the water table so there will be little upward
flow of water into the well. The perforations in the well casing must be sufficiently
close to one another to permit the groundwater to flow into the well as fast as it
reaches the casing.

Observation wells are put down at various radial distances from the test well, the
number of such wells and the radial distances depending on site uniformity and
extensiveness of the investigations. If casing is required, it must be perforated or
equipped with a screen at the lower end. In some cases observation wells can be
installed by pushing in casing with a ‘‘sand point.’’

Elevations of the groundwater table are measured and recorded at each of the wells
before pumping begins. Pumping from the test well then proceeds and continued
until a steady flow is established, which is indicated when water levels in the test well
and observation wells become constant. The decrease in elevation of the
water table at each of the various observation wells, the radial distances out to
these wells, and the rate of discharge from the test well, provide the information
necessary for computing the coefficient of permeability of the soil within the zone of
influence of the test well.

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14.5.3 Calculation of Hydraulic Conductivity

With steady flow and uniform soil conditions, the groundwater surface assumes
the shape of an inverted hyperbolic cone. A radial section through the well and
two observation wells is shown in Fig. 14.7. The flow toward the test well is
assumed to be radial at the observation wells so the hydraulic gradient is dh/dr.
Water flows through a cylindrical area 2rh toward the test well where the
cylinder has a radius r and height h. Substituting these values into eq. (14.3) gives
dh
Q ¼ kIA ¼ 2rhk ð14:12Þ
dr
Integrating between r and h for the two observation wells gives
Zr1 Zh1
dr 2k
¼ h dh ð14:13Þ
r Q
r2 h2

from which
Q lnðr1 =r2 Þ
k¼  2  ð14:14Þ
 h1
h22

Example 14.4
A test well yields a steady discharge of 7.89 l/s (125 gal/min). Elevations of the
water table above an impervious layer (or above the bottom of the test well in case no
impervious layer exists) are 10.39 m (34.1 ft) and 10.58 m (34.7 ft), respectively, and these
wells are located 18.3 m (60 ft) and 30.5 m (100 ft) from the center of the test well.
Determine the coefficient of permeability of the soil.

Figure 14.7
Determination of k
from a drawdown
pumping test.

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334 Geotechnical Engineering

Answer: In SI, Q is expressed in m3/s; in English units in ft3/day. Note that

1 liter ¼ (10)
3 m3 and 1 gal ¼ 0.1337 ft3. Substitution of appropriate values in eq. (14.14)
gives

7:89ð10Þ
3 lnð30:5=18:3Þ
k¼ ¼ 320 mm=s ¼ 91 ft=day ¼ 3:2ð10Þ
3 cm=s
ð10:582
10:392 Þ
Again it should be emphasized that k does not represent a flow rate, which
depends on the hydraulic gradient.

14.5.4 Approximate Drawdown Method

If point A in Fig. 14.7 is taken at the circumference of the test well and B is on the
circle of zero drawdown of the water table, eq. (14.14) becomes
Q lnðr0 =r3 Þ
k¼  2  ð14:15Þ
 h0
h23
where r3 is the radius of the test well, r0 is the maximum radius of influence or
radius of the circle of zero drawdown, h0 is the height of the groundwater table
above an impervious layer or the bottom of the test well, and h3 ¼ h0
d where d is
the drawdown in the test well at a steady flow condition.

Since the radius of the circle of zero drawdown will always be several hundred
times the radius of the test well, the value of ln(r0/r3) varies over only a relatively
narrow range, so a value may be assumed for r0. It then is not necessary to observe
the drawdown in observation wells.

Example 14.5
The diameter of the test well of Fig. 14.7 is 0.61 m (24 in.) and the drawdown is 7.65 m
(25.1 ft) with a steady pumping discharge of 7.89 l/s (125 gal/min). The distance from the
impervious layer to the normal water table is 10.7 m (35.2 ft). Assume that the radius r0 of
zero drawdown is 150 m (500 ft). Determine the coefficient of permeability. (1 liter ¼
1000 cm3 ¼ 10
3 m3.)
Answer: Substitution of the appropriate values in eq. (14.15) gives
7:89ð10Þ
3 lnð150=0:305Þ m3 =s
k¼   ¼ 0:148 m=s ¼ 41:9 ft=day ¼ 15 cm=s
 ð10:72 Þ
ð10:7
7:65Þ2 m2

Questions: Is this answer reasonable for sand? What if r0 is assumed to be twice as

far, 300 m (1000 ft)?

Answers: The answer is high even for coarse sand. Recalculation with r0 ¼ 300 m
increases k by 11%, so the answers are within a range of acceptable accuracy
for k.

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14.5.5 Open-End Pumping Tests

Less accurate but faster and less costly is to simply pour water into an
open-ended drill casing that is bottomed in a test hole that extends below
the level of the groundwater table. In a procedure developed empirically by
the U.S. Bureau of Reclamation the depth to the water table is measured, then
casing is lowered and carefully cleaned out just to the bottom of the boring.
The rate at which clear water must be added to maintain a constant head
near the top of the casing is measured, and the hydraulic conductivity is
calculated from
Q
k¼ ð14:16Þ
5:5rh
in which Q is the flow rate for a constant head, r is the internal radius of the
casing, and h is the head of water above the water table. In this equation r and h
have units of length that must be consistent with the units of Q and k.

Example 14.6
An NX, 76 mm (3 in.) i.d. casing is set at 9.14 m (30 ft) depth, 4.57 m (15 ft) below the water
table. The casing is cleaned out, and it is found that by adding 0.0315 l/s (0.5 gal/min) of
water, the average water level is at the ground surface. Find k.

Answer:

31:5 cm3 =s
k¼ ¼ 0:33ð10Þ
3 cm=s
5:5ð3:8 cmÞð9:14
4:57Þ  1000 cm

Is this answer reasonable for a silty sand?

A disadvantage of open-end pumping tests is that only a very limited volume

of soil is tested, just below the end of the casing. The factor 5.5 in eq. (14.16)
was arrived at experimentally by use of an electrical analog where electrical
current, voltage, and resistance are analogous to Q, h, and the reciprocal of k,
respectively.

14.5.6 Packer Pumping Tests

Packer tests are conducted by partitioning off sections of a boring with
expandable packers (Fig. 14.8). These tests are commonly used to evaluate
foundation soils and rocks for dams, and have an advantage that
different depth intervals can be discretely tested. For best accuracy the
depth interval should be 5 and preferably 10 times the hole diameter. The
equation for k is
Q L
k¼ ln ð14:17Þ
2Lh r
where L is the length of hole tested, r is the radius of the hole, and h is the
pumping pressure expressed as hydraulic head.
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336 Geotechnical Engineering

Figure 14.8
Packer test for
determining
hydraulic
conductivity.

In eq. (14.17) it will be seen that if k is constant, Q is proportional to the pumping

pressure head, h. The test therefore may be repeated with different values of h to
observe the effects of hydraulic pressure and flow on k. A sharp increase in k
usually means that cracks are being opened, indicative of hydraulic fracturing.
This becomes likely if the water pressure exceeds the overburden pressure, as
water injects horizontally and lifts the soil. The pressure to cause hydraulic
fracturing also has been used to estimate lateral in-situ stress in the soil, but only
if that stress is lower than the overburden pressure, and other methods
are considered more reliable. Hydraulic fracturing can be deleterious if it occurs
during pressure grouting, which is injection of a fluid that then hardens and
stabilizes soil. This is discussed in Section 24.6.2.

14.5.7 Slug Tests

‘‘Slug tests’’ refer to either pouring a measured amount of water or ‘‘slug’’ into a
boring, or removing a measured amount of water from the boring, and observing
the rate of change of elevation of water in the hole (Fig. 14.9). Boast and Kirkham
(1971) showed that if the auger hole extends completely through a pervious
stratum to an impervious layer, the flow is subject to exact mathematical
analysis. In the absence of an impervious layer the results can present a good

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Figure 14.9
Example of a slug
test where water is
removed from an
auger hole and the
rate of re-entry is
determined.

approximation if the ratio of depth to diameter of the auger hole is large. In this
method, as in any method where water is taken from a boring, it is advisable
to pump water from the hole several times and allow the hole to refill, in order to
flush out soil particles from the remolded zone along the sides.

The formula for determining the k by this procedure is

r dh
k ¼ 0:617 ð14:18Þ
Sd dt
where r is the radius of the hole, d is the depth of the hole below the groundwater
table, dh/dt is the rate of rise of the water in the hole, and S is a coefficient that
depends on h/d where h is the depth of water in the hole at the time the rate of rise
is measured. Values of the coefficient S are given in Fig. 14.10.

In order to perform this test a hole is bored down to an impermeable layer or to a

depth below the water table that is at least 10 times the diameter of the hole.
Water in the hole is allowed to rise to a constant level, which is considered to be
the elevation of the water table. Water then is pumped from the hole and the
hole allowed to refill two or three times, more if the soil puddles readily.
The water is then pumped out of the hole again and the rate of rise of the water
level measured at several different elevations by measuring the rise in a short
period of time.

As flow into the auger hole is almost entirely horizontal, this method is especially
useful for determining horizontal permeability in nonisotropic soils. If the soil is
stratified, the flow into the auger hole is from all strata penetrated by the boring,
with the greatest inflow being through the most permeable layer.
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338 Geotechnical Engineering

Figure 14.10
Values of S for
use in eq. (14.18),
after Boast and
Kirkham (1971).

Example 14.7
A 4 in. diameter hole is augered to extend 2.5 ft below the groundwater table and is baled
several times to remove the water. When the water depth below the water table is 1.5 ft, the
rate of rise is 1 inch in 12 minutes. What is k?
Answer: h ¼ 1 ft and d ¼ 1.5 ft so h/d ¼ 0.67. r ¼ 2 in. so r/d ¼ 2/18 ¼ 0.11. From Fig. 14.10,
S ¼ 1.6. Then

k ¼ 0.617 (2 in./1.6  18 in.)  1 in./12 min ¼ 0.0036 in./min ¼ 0.43 ft/day ¼ 1.5 (10)
4 cm/s.

Is this a reasonable value? Yes, if the soil is fine-grained.

14.5.8 Dye Injection Tests

A dye or salt solution can be injected into a well and the time determined for it to
be detected at another well or seepage outlet that is downslope. This gives a direct
measurement of the true flow velocity through the pores. This method is most

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Seepage 339

applicable to granular soils because of the time required and because dyes and
salts may interact with and be adsorbed by clays. The hydraulic gradient i equals
the decrease in elevation of the groundwater table divided by the distance
downslope.

As discussed in Section 14.2.4, k defines a superficial flow velocity under a unit

hydraulic gradient in a soil that is all voids. The relation between flow velocity
measured by dye injection and k is
vs ne
k¼ ð14:19Þ
i
where vs is the seepage velocity, ne the porosity, and i the hydraulic gradient.

Example 14.8
Dye injected into a boring in sand is detected entering an open excavation 100 m
away after 85 minutes. The groundwater level in the boring is 4 m higher than in the
excavation, which is being dewatered to the capacity of the pumps. Estimate k and
recommend how much more pump capacity will be needed to lower the water level an
additional 3 m.
Answer: The seepage velocity ve ¼ 100/85 ¼ 1.18 m/min ¼ 2.2 cm/s. The hydraulic gradient
i ¼ 4 m/100 m ¼ 0.04. Porosity is not known and must be estimated; it is assumed to be
0.3. Then
k ¼ 2:2 cm=s  0:3=0:04 ¼ 16 cm=s:

This value of k is high but is reasonable for a sand aquifer. If k is constant, eq. (14.19)
indicates that the seepage velocity vs is proportional to i, so dewatering to a depth of 7 m
instead of 4 m will increase the required pump capacity by that ratio, other factors being
constant. However, the increased seepage also may cause slope instability, so that
possibility should be investigated. This problem can be avoided by reversing the seepage
direction with perimeter wells.

14.5.9 Tests in the Vadose Zone

‘‘Vadose zone’’ means a zone of unsaturated soils. The hydraulic conductivity in
such a zone normally is lower than if the soil were saturated because of partial
blocking of water flow by air. The rate of infiltration of rainwater into soil is in
part controlled by hydraulic conductivity in the vadose zone. Compacted soils
contain air, so they also constitute a vadose zone unless they later become
saturated with water.

Compacted clay layers used to seal the bottoms of ponds, lagoons, and solid
waste containments can contain air even though the soil is in contact with water,
so long as water drains out of the bottom as fast as or faster than it enters at the
top. However, a saturated coefficient often is used for design of such containers
to be on the safe side. The vadose zone also is a repository for insoluble
low-density pollutants such as gasoline that float on the groundwater table,
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340 Geotechnical Engineering

and geoenvironmental engineering procedures are available for sampling pore

fluids from the vadose zone.

14.5.10 Measuring Infiltration in the Vadose Zone: Overview

An indication of hydraulic conductivity in the vadose zone is obtained from
infiltration tests. The simplest of these is a borehole infiltration test or ‘‘perc test’’
that is widely used as an indicator of the suitability of a soil for a septic drain
field. In this test, water is poured into an auger hole and the rate of decline
of the water level is measured. Since the test is performed at shallow depth it
is strongly influenced by capillary attractions, so the hole should be kept
filled with water for 4 to 24 hours prior to measurement. Because infiltration
is both horizontal and vertical, results are considered empirical, and the
acceptability for drain fields is based on experience as reflected in local
building codes.

Auger hole ‘‘perc’’ and other infiltration tests differ from slug tests in that they do
not fully saturate the soil. Keeping an auger hole filled with water results in a
‘‘field-saturated’’ condition but the soil still contains air. The coefficient of
hydraulic conductivity therefore is designated by kfs (fs for field saturated),
and normally is only about 50 percent of the saturated k, or it may be as little
as 25 percent of the saturated k for fine-grained soils. A high conductivity is
required for operation of septic drain fields, so these results are on the safe side
for design.

14.5.11 Infiltrometers
More sophisticated infiltration tests are part of soil physics, which is a
branch of agronomic soil science. A condensed discussion of various test
methods is given in ASTM Designation D-5126. Only a brief treatment is
included here.

The most common method for testing is with an infiltrometer, which in its
simplest form is a metal ring that is driven a short distance into the ground surface
and filled with water. This creates a downward-penetrating wetting front that also
spreads laterally. A double-ring infiltrometer adds a concentric outer ring so that
the space between the rings also can be filled with water, which directs infiltration
from the inner ring downward, as shown in Fig. 14.11.

The driving force for infiltration includes both gravitational and matric or
capillary potential. Gravity head at the wetting front is zero, so the gravitational
head is measured from the surface of the ponded water down to the wetting
front. Therefore, as the wetting front advances, if the ponded water is
maintained at a constant elevation both the gravitational head and
flow distance increase. If the level of the ponded water approaches
the ground surface, the gravitational head h equals the thickness of the
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Figure 14.11
Schematic diagram
of a double-ring
infiltrometer. As
the water level
decreases, h ! l
and i ! 1.

partially saturated soil layer behind the wetting front. As this also equals the
flow distance l, the hydraulic gradient i ¼ h/l ¼ 1. The Darcy equation (14.3) then
becomes
Q=A ¼ kfs ð14:20Þ
where Q is the quantity of water infiltrating per unit of time, A is the infiltrating
area, and kfs is the coefficient of hydraulic conductivity in the field-saturated soil
behind the wetting front. This relationship can be used for sands or silts where the
wetting front has advanced sufficiently that h  hF, where hF is the matric
potential.

In clay, matric potential is the main driving force. A model developed by

J. R. Philip indicates that the horizontal infiltration rate without any influence of
gravity should go as the square root of time, whereas vertical infiltration that is
in response to both gravitational and matric potential goes as the square root of
time plus a time function (Jury and Horton, 2004). The constants in the required
equations are evaluated experimentally.

14.6 MANAGING THE GROUNDWATER TABLE

14.6.1 Drain Tiles

Perimeter drains are installed to lower the groundwater table prior to excavating,
and permanent drains are installed around building foundations to prevent
seepage into basements. Many natural wetlands have been drained with tile lines
to convert the land for agriculture, and soils at hazardous waste sites destined for
burning in kilns can be dewatered with drains and the water treated separately
instead of having to be burned off.

Drains draw down a groundwater table as shown in Fig. 14.12. A simplified

approach called the Dupuit-Forchheimer, or DF theory, defines the groundwater
surface between two tile lines as an ellipse with focal points located on the surface
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Figure 14.12
Recharge R from
infiltration of rain,
and elliptical
drawdown curve
to two parallel
drainage tile lines.

of an underlying impermeable stratum. The idealized ‘‘DF soil’’ is discussed by

Kirkham and Powers (1971), who show that the error from the assumptions in
most cases is relatively small. Then
S2 4k
¼ ð14:21Þ
b2
a2 R
where S is the tile line spacing, a and b are the depths to an impermeable
barrier under the tile and halfway between the tile, respectively, k is the coefficient
of hydraulic conductivity of the soil, and R is the infiltrating rainfall expressed
in the same units as k. A materials balance requires that the amount of
water going in equals the amount going out, so the quantity per unit of length
of tile is R  S. However, because low areas where drains are most likely to
be used collect runoff water from adjacent hillsides, R may exceed the annual
rainfall.

Example 14.9
A silty soil layer 5 m (16.4 ft) thick is to be drained so that the water table at its highest
point is 1 m below the ground surface. Assume k ¼ 5(10)
5 cm/s (1.6 m/yr) and R ¼ 0.76
m/yr (30 in./yr), and try two alternative tile depths and spacings.

Trial 1. Let the tile depth be 2 m (6.6 ft); a ¼ (5 – 2) ¼ 3 and b is unchanged:

  4  1:6
S2 ¼ 42
32  ¼ 59, and S ¼ 7:7 mð22ftÞ
0:76
Trial 2. Let the tile depth be 3 m (9.8 ft). Then a ¼ (5 – 3) ¼ 2 m and b ¼ (5 – 1) ¼ 4 m.
Therefore
  4  1:6
S2 ¼ 42
22  ¼ 100, and S ¼ 10 mð33 ftÞ
0:76
While the Trial 2 solution may appear more economical because it would reduce the length
of tile lines by 50 %, every solution that looks good on paper also must be checked for
feasibility or ‘‘buildability.’’ A failure to do so will lend to droll observations concerning
impractical solutions proposed by inexperienced engineers. In this case it should be noted
that the deeper the trench required to install the drains, the less stable are the trench walls,
and even the shallower trench might be unsafe with a seasonally high groundwater table.
Laws governing worker safety usually require that a rigid trench box be used to protect
workers inside such a trench.

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14.6.2 Average Drawdown Between Tile Lines

Drain tile lines are widely used to stabilize landslides. Because of the difficulties
and dangers of trenching in unstable, sliding soil, even with a trench box, the
drains often are installed by horizontal probing or drilling.

In order to analyze the effect of drainage on slope stability it is necessary

to approximate the percentage of the soil volume that will remain undrained,
which includes all soil below the level of the drain tile plus an upper part
that is defined by an ellipse extending between the drain tile, as shown in
Fig. 14.12. The problem is simplified if that part of the ellipse is approximated
by a parabola, in which case the cross-sectional area of soil above the level of
the tile is
2
A ¼ Sðb
aÞ ð14:22Þ
3
where b and a are as indicated in the figure and S is the tile spacing. Dividing by S
gives the average distance of the groundwater table above the tile line:
2
ðb
aÞav ¼ ðb
aÞ ð14:23Þ
3
Adding a to each side gives the average groundwater elevation above the
impermeable layer:
2b þ a
bav ¼ ð14:24Þ
3

Example 14.10
For each of the tile spacings indicated in the previous example calculate the percent of the
soil that will remain undrained, assuming that it initially was saturated to the ground
surface.
Answer: Trial 1: bav ¼ (2  4 þ 3)/3 ¼ 3.7 m; 100 (3.7/5) ¼ 73%.

Trial 2: bav ¼ (2  4 þ 2)/3 ¼ 3.3 m; 100 (3.3/5) ¼ 66%.

14.6.3 Emergency Procedures for Landslides

Landslides usually develop after periods of heavy rainfall that raise the
groundwater table, so steps may be taken to reduce the infiltration rate R so
that it can be accommodated by the natural soil drainage. Unfortunately, the
broken ground in landslides ponds water, and ground cracks funnel water
down into the soil. Ponded water should be drained and cracks closed by
spading in soil wedges. Slopes also can be temporarily covered with plastic
film (Fig. 14.13). Such measures obviously are only temporary, but may
delay sliding through a wet season to allow later use of more effective control
measures.

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344 Geotechnical Engineering

Figure 14.13
Temporary
landscaping with
plastic to reduce
infiltration R and
try and put some
brakes on a
landslide in
southern
California.

14.6.4 Groundwater Mounding under Lagoons

Unless steps are taken to prevent it, submerging soil under a lagoon will
increase the infiltration rate so that the groundwater level mounds up underneath
the lagoon, and eventually will come into contact with the polluted lagoon
water. Lagoons therefore are lined with clay or other material to decrease
infiltration. Even without mounding, seepage losses should be intercepted
before they enter the groundwater supply. A sand or permeable geotextile
layer may be placed underneath the clay liner and kept pumped out and the
seepage recycled. This procedure also is frequently used underneath sanitary
landfills.

Another alternative to keep lagoon water from entering the groundwater supply
was suggested by a soil physicist, Robert Horton. Horton’s tests showed
that adding clean water to the permeable layer so that the head is slightly
higher than that of the lagoon will reverse the flow direction through the clay liner
so that it is upward instead of downward, even if the liner has been punctured
and compromised.

Example 14.11
A sewage lagoon 55 ft deep is to be constructed with a clay liner 3 ft thick with
k ¼ (10)
6 cm/s. Underneath the clay is unsaturated sand. The mean annual rainfall is 35 in.
Should one anticipate groundwater mounding?

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Answer: From eq. (14.3), Q/A ¼ ki, where i ¼ 55/3 ¼ 18.3. Q/A therefore equals
18.3(10)
6 cm/s ¼ 230 in./year  35 in./year, so there should be groundwater mounding.

Question: What if the lagoon is 5.5 ft deep?

Answer: This would reduce i and ki by a factor of 10, so Q/A ¼ 23 in./year
5 35 in./year, and there should be no mounding—so long as there are no leaks.

14.7 FLOW NETS

14.7.1 Solving Two-Dimensional Flow Problems

The tile drain problem presented above is a two-dimensional flow
problem. A graphical scheme for solving such problems is called a flow net.
An example of a flow net is shown in Fig. 14.14 for a constant-head
permeameter. Horizontal lines represent flow paths, which are paths taken by
water in laminar flow, and vertical lines show contours of head, the equipotential
lines.

The space between adjacent flow lines is a flow path, which is analogous to a pipe
and must be continuous, having both an inlet and an outlet. This is obvious in
Fig. 14.14, but becomes less obvious when flow nets are drawn for more
complicated geometries.

Figure 14.14
Two-dimensional
flow net for the
permeameter of
Fig. 14.4, based on
the soil being
contained in a
box having a
rectangular
cross-section.

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346 Geotechnical Engineering

Because head decreases in the direction of flow, the intersections of equipotential

lines with flow lines must be at 908 even though the flow lines are curved. If the
intersections are not at 908, the flow net is incorrect.

14.7.2 Boundary Lines of a Flow Net

In Fig. 14.14 the boundary flow lines are along the outer walls of the
permeameter, but in field situations the boundaries may be less obvious
unless one remembers that they must be flow lines. For example, in the flow
net of Fig. 14.15 the flow lines dip sharply downward under the lower ends
of the sheet-pile cofferdam. The upper boundary is the flow path down
along the surface of the sheet pile and up along the surface on the other
side. The lower boundary is the contact between the soil and an imperme-
able layer. In other situations the upper boundary is at the level of the
groundwater table because that is the upper limit of saturation by gravitational
water.

14.7.3 Curved Flow Paths

In Fig. 14.15 the flow paths are sharply curved so the geometrical figures cannot
be true squares. However, it will be seen that the flow lines and equipotential lines
all intersect at right angles, and if the geometric figures were subdivided with more
flow paths and equipotential lines they would approach squares, in which case
the median dimensions of each figure will be equal.

Figure 14.15
Flow net for a
sheet-pile
cofferdam
showing flow
lines, flow paths,
and equipotential
lines. The
surfaces of the
sheet pile and of
the impervious
stratum are
boundary flow
lines.

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Seepage 347

14.7.4 Variable Hydraulic Gradient

The curved flow paths that are typical of flow nets drawn for field situations mean
that the hydraulic gradient, which is the head loss per unit of travel, is not the
same throughout the flow region, which is a reason for drawing a flow net. For
example, as shown in Fig. 14.15, the seepage path for water entering soil at b and
exiting at d travels a much shorter distance and has a higher hydraulic gradient,
and therefore a higher velocity, than that entering at a and exiting near e. As will
be shown later, this is important because a high emerging hydraulic gradient can
create quicksand.

14.7.5 Laplace’s Equation

While it is not necessary to comprehend the mathematical justification for
flow nets in order to use them, the derivation is not complicated and the
principles upon which it is based are important. From this it can be seen why
flow-net theory also applies, for example, to the flow of electricity, but not to
the flow of a gas, which is compressible, or of water in unsaturated soil,
because of the content of compressible air and suction derived from the matric
potential.

Pierre-Simon Laplace (pronounced with two ah’s) was an eighteenth/nineteeth-

century French mathematician and astronomer. The basis for Laplace’s equation
relating to flow nets is the principle of continuity, that the volume of water
flowing into a unit volume of saturated soil must equal the volume of water
flowing out. By dividing the flow into x and y components, this principle can be
represented as shown in Fig. 14.16, where the flow direction changes within the
element. The principle of continuity states

q ¼ qxe þ qye ð14:25Þ

where q is the total flow quantity and qxe and qye are flow quantities in the x and y
directions. This expression also applies to flow of electricity, which allows
two-dimensional flow nets to be modeled with an electrical analog such as a
resistance paper or even a layer of cardboard wet with salt water with electrodes
contacting the paper at the entrance and exit faces.

Equation (14.25) may be expressed in terms of fluid velocity V at the two entry
faces, since qxe ¼ Vx times area, and qye ¼ Vy times area. Then for a unit thickness,
as shown in the left-hand segment of Fig.14.16,
q ¼ Vx dy þ Vy dx ð14:26Þ

The velocities at the two departure faces usually are not the same as at the
corresponding entry faces because of a changing flow direction within the element.
This change can be represented by a rate of change with respect to distance,

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348 Geotechnical Engineering

Figure 14.16
Principle of
continuity for
derivation of
Laplace’s
equation:
entrance and exit
rates in a square
element must be
consistent
with volume
in ¼ volume out.

@Vx/@x and @Vy/@y, times the corresponding distances dx and dy. Then, as shown
by the right-hand segment of the figure, the departing velocities are
 
Vx þ ð@Vx =@xÞdx and Vy þ @Vy =@y dy ð14:27Þ
respectively. The departing volumes qxd and qyd are obtained by multiplying the
departure velocities times the corresponding areas. Then
 
q ¼ Vx dy þ ð@Vx =@xÞdxdy þ Vy dx þ @Vy =@y dydx ð14:28Þ
Equating (14.27) with (14.28) and canceling identities gives
 
ð@Vx =@xÞdxdy þ @Vy =@y dxdy ¼ 0
 
ð@Vx =@xÞ ¼ @Vy =@y ð14:29Þ
This is the equation of continuity for two-dimensional flow through a
homogeneous mass.

Darcy’s Law states that q ¼ kia or V ¼ ki, where k is the coefficient of hydraulic
conductivity and i is the hydraulic gradient. Hence,
Vx ¼ kix ¼
kð@h=@xÞ ð14:30aÞ
Vx y ¼ kiy ¼
kð@h=@yÞ ð14:30bÞ
where the (
) sign indicates that a flow velocity Vx or Vy corresponds to a
reduction in hydraulic head, @h/@x and @h/@y, respectively. Equations (14.30)
differentiated with respect to x and to y are

@Vx =@x ¼
kð@h=@xÞ and

ð14:31Þ
@Vy =@y ¼
kð@h=@yÞ

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Seepage 349

Substituting in eq. (14.28) gives

@2 h=@x2 ¼
k@2 h=@y2 ð14:32Þ

This is Laplace’s differential equation in two dimensions. It says that a rate of
change in head loss with respect to distance in one coordinate direction requires
an opposite rate of change in the other. A similar equation may be derived for
three-dimensional flow.

Laplace’s equation gives the mathematical justification for the existence of

nonparallel, curvilinear flow lines and equipotential lines. The rate of change in
head is a maximum in the direction of flow; hence equipotential lines cross flow
lines at right angles, this being one of the binding requirements for a flow net.
It also can be shown that the discharge quantity of water flowing between any
two adjacent flow lines is the same.

Laplace also was the first to suggest that randomness of measurements follows a
statistical normal or Gaussian distribution.

14.7.6 Determining Flow Quantities from a Flow Net

The total quantity of water Q flowing through a unit thickness of a soil mass
equals the sum of the quantities in all the flow paths of the flow net. Since a basic
requirement of a flow net is that every flow path must transmit the same quantity
of water,
Q ¼ F dQ ð14:33Þ
where F is the number of flow paths and dQ is the quantity in each path. Similarly,
the total head loss h is the sum of the drops in head in all equipotential spaces of
the flow net:
h ¼ N dh ð14:34Þ
where N is the number of equipotential spaces and dh is the reduction in head
across each space. In an arbitrarily selected square of length l, width w, and
thickness ¼ 1.0, the area normal to flow is w  1, and the hydraulic gradient
is i ¼ dh  l. Substituting these quantities in the Darcy Law for a single square
gives
Q ¼ kiA
dh ð14:35Þ
dQ ¼ k w
l
where dQ is the quantity in each flow path, dh is the head loss
between equipotential line, and w and l are dimensions of the square. Since
w/l ¼ 1,

dQ ¼ k dh ð14:36Þ

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350 Geotechnical Engineering

Substituting from eqs (14.33) and (14.34),

Q h
¼k
F N
ð14:37Þ
F
Q ¼ kh
N
This means that the total quantity of water that will seep through a unit thickness
(normal to the page) of a soil structure can be found by drawing a flow net on a
cross-section of the structure, and multiplying the coefficient of permeability k
times the total head loss h, times the ratio of the number of flow paths F to the
number of equipotential spaces N. To find the total seepage through the entire
structure, the flow through a unit dimension normal to the page is multiplied by
the total corresponding dimension of the prototype structure. For a dam or
similar structure this would be times the length of the dam.

The validity of the flow-net procedure for computing seepage quantities may be
demonstrated by comparing the result obtained by use of the Darcy equation
(14.3) with the result obtained by use of the flow-net equation (14.37) when both
are applied to the same flow situation such as represented in Fig. 14.14.

Example 14.12
Assume that the soil mass in Fig. 14.14 is 10 cm thick, 5 cm high, and 14 cm long; that the
head loss is 12 cm; and that k ¼ 10
3 cm/s. Determine the total seepage (a) by eq. (14.3) and
(b) by eq. (14.31).
Answer: (a) In the figure, there are 5 flow paths and 14 equipotential spaces, so according
to eq. (14.37),

Q ¼ (10)
3 cm/s  12 cm  (5  14) ¼ 0.043 cm2/s

which times the thickness of 10 cm gives 0.43 cm3/s.

(b) By eq. (14.3)

Q ¼ kiA ¼ (10)
3  (12/14)  (5 x 1) ¼ 0.043 cm2/s

which gives a total flow of 0.43 cm3/s, which is the same answer. A flow net obviously is not
necessary to solve a simple rectilinear flow problem, but becomes invaluable when flow
paths curve and change width.

14.8 HOW TO DRAW A FLOW NET

14.8.1 Why Engineers Draw Flow Nets

The Laplace solution means that there are many different ways to arrive at a
flow net, including mathematical solutions, electrical analogies, finite element
and finite difference solutions, and trial sketching. However, in soils the
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 Seepage

Seepage 351

most significant source of error in seepage problems is not the flow net but
measurement of k.

A useful exercise is to assign a seepage problem to a class in order to compare

results from independently sketched flow nets. If the flow nets meet
basic requirements of flow paths—that is, continuity, 908 intersections, and
quasi-square elements—the variation in F/N should not exceed about 10 percent,
compared with variations in k measurements that are many times that amount.
The time required to sketch a rudimentary flow net is minimal and may be less
than required to set up a problem on a computer where errors can be less obvious.
One advantage of a flow net is that mistakes generally are obvious. Many
engineering organizations therefore use trial sketching even though it may be
supplemented by electrical analogy or other methods.

14.8.2 Steps in the Trial-Sketching Method

1. The first step is to draw or trace a cross-section of the soil structure normal to
the direction of seepage flow.
2. Next, the two boundary flow lines should be established.
3. The entrance and exit boundaries should be established. These may or may
not be equipotential lines. In Fig. 14.15 both boundary lines are equipotential
lines along the ground surfaces, so flow lines must intersect ground surfaces at
right angles.

Question: What. if the ground surfaces on the two sides of the sheet pile in
Fig. 14.15 were sloping?
Answer: Submergence on the left side means that the head is constant even
though the elevation varies because of the compensating nature of pressure and
elevation head. However, on the right side the emerging water is at atmospheric
pressure, so the ground surface would not be an equipotential line, as the potential
would depend on the elevation.

4. In order to avoid discontinuous flow paths, the next step is to lightly sketch in
two or three trial flow lines so that flow paths do not pinch off or come to a
dead end. As a check, every flow path must link the entrance and exit
boundaries.
5. Now sketch trial equipotential lines with right-angle intersections and
essentially square figures, and adjust the positions of flow lines as
inconsistencies become apparent, such as squares that are rectangular or
nonperpendicular intersections. This can be done with pencil and paper, or
with a computer drawing program that allows lines to be shifted with a
mouse. Each inconsistency will indicate the direction and magnitude of a
necessary change.
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352 Geotechnical Engineering

Figure 14.17
Flow net for a
masonry dam.

Figure 14.18
Flow net for an
earth dam on an
impervious
stratum.

6. Adjustments are continued until the net appears to be a reasonable

representation. More flow lines and equipotential lines then can be added
during trial re-sketching for better precision, as F and N are whole numbers,
but ordinarily three to five flow lines will be sufficient.

Flow nets made by experts should be studied, leading to a sense of fitness

and intuition that will assist trial sketching. Some examples of flow nets
are shown in the Figs. 14.17 and 14.18. It will be noted that every
transition in a homogeneous soil is smooth and curvilinear, and flow paths
taper gradually.

14.8.3 Flow Nets when k Varies Depending on the

Direction of Flow
A common situation is where layering in soil causes horizontal permeability to be
higher, in some cases many times higher, than vertical permeability. This could
pose a difficult situation for analysis, but is readily taken into account in a flow
net by compressing the horizontal scale. The scale is compressed according to the
ratio of horizontal to vertical k, the flow net is drawn according to the
conventional rules, and the scale then is re-expanded back to its normal

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Seepage 353

Figure 14.19
Flow net where
kh 4 kv. (a) Flow
net constructed on
a transformed
section. (b) Section
re-expanded to
normal scale.

dimensions. This is easily and instantly accomplished with a computer drawing

program.

When the horizontal permeability is the greater, this transformation is made as

follows where x0 is the adjusted scale of x:
x
x0 ¼ pffiffiffiffiffiffiffiffiffiffiffi ð14:38Þ
kh =kv
A transformed flow net with right-angle intersections and nominally
square figures is shown in Fig. 14.19(a), and the flow net expanded to the
original scale is shown in Fig. 14.19(b), where the flow lines and equipotential
lines no longer intersect at right angles, nor are the figures squares. Because the
elements of the transformed net are squares, it can be shown that the geometric
mean of the two coefficients of permeability may be used to obtain an effective k0
to estimate the flow quantity:
pffiffiffiffiffiffiffiffiffi
k0 ¼ kh kv ð14:39Þ
Derivations of eqs. (14.38) and (14.39) will be found in Harr (1962).

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354 Geotechnical Engineering

14.8.4 Weighted Average Permeabilities

The cause of high horizontal permeabilities usually is horizontal layers of coarser,
more permeable soils. With normal vertical tube sampling it often is not practical
to measure horizontal permeabilities, but vertical permeabilities of samples from
individual layers can be measured and assumed to be the same as horizontal
permeabilities for the same layers. An adjusted mean can by obtained by use of an
electrical analogy.

A coefficient of hydraulic conductivity k is the equivalent of electrical

conductance, which is the reciprocal of resistance. The analog for parallel
flow of water along layers with different permeabilities is electricity
flowing through resistors with different resistances in parallel. The analog for
water transmitted across a series of soil layers is electrical resistors arranged in
series.

For an electrical circuit the total resistance of resistors in parallel is

1 1 1 1
¼ þ þ  ð14:40Þ
R R1 R2 Rn
where the R’s represent total and individual layer resistances in ohms. The
analogous equation for permeabilities is
t1 k1 þ t2 k2 þ    tn kn
kh ¼ ð14:41Þ
t
where kh is the coefficient for flow along the layers, which in this case is in the
horizontal direction, t represents the total thickness, and t1, . . . ,tn represent
thicknesses of layers with permeabilities k1, . . . ,kn.

In the case of vertical flow through horizontal layers of differing permeabilities,

the electrical analogy for resistance is
R ¼ R1 þ R2 þ    þ Rn ð14:42Þ
The analogy for fluid flow across layers is
t t1 t2 tn
¼ þ þ  ð14:43Þ
kv k1 k2 kn
where kv is the coefficient for flow across the layers, which in this case is in the
vertical direction, and the t’s are as above. Application of these equations is
simplified if all layers having the same permeability are grouped together. It will
be noted that the respective equations apply to flow along and across layers of
varying permeability and regardless of the layer orientation.

Example 14.13
A boring log in Fig. 14.26 shows 10 m of stratified soil that includes 6 silt layers with a total
thickness of 8 m and k ¼ 10
4 cm/s, and 2 sand seams with a total thickness of about 2 m
and k ¼ 10
2 cm/s. (a) Estimate the horizontal and vertical permeabilities on the basis of

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Seepage 355

these test data, and (b) recommend a horizontal compression factor for the flow net and (c)
an effective k’ for interpretation of flow net quantities.

Answer:\rm (a) From eq. (14.42) the horizontal permeability is

kh ¼ 8ð10
4 Þ þ 2ð10
2 Þ ¼ 2:1ð10
2 Þ cm=s

which is essentially the same k as that of the sand. From eq. (14.43) the vertical
permeability is obtained from

10 8 2
¼
4 þ
2 ¼ 8:02ð10Þ4
kv 10 10

kv ¼ 1:2ð10
4 Þ cm=s

which is approximately the same k as the silt.

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
(b) The horizontal compression factor for the flow net is 2:1ð10
2 =1:2ð10
4 Þ ¼ 13
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
(c) The effective k value is k0 ¼ ½ð2:1ð10
2 Þ  1:2ð10Þ
4  ¼ 1:6ð10
3 Þ cm=s:

14.9 HYDROSTATIC UPLIFT PRESSURE ON A STRUCTURE

Any structure that extends below a water surface, whether it is a free water surface
or the groundwater table, is subjected to uplift pressure from the water. The uplift
pressure below a free water surface is simply the depth times the unit weight of
water, which is the principle of Archimedes.

Uplift pressures from water seeping through soil are easily calculated
from equipotential lines in a flow net. Uplift pressure is important because it
in effect partially floats a structure, which reduces sliding friction between the
structure and the soil. For example, a dam that is subjected to water pressure on
the upstream face and uplift pressure on the base must be evaluated for stability
against sliding. Secondly, since uplift pressure acts within the footprint of a
structure, it also can contribute to overturning.

In the masonry dam and flow net in Fig. 14.17, the total head loss h is 3 m.
There are 14 equipotential spaces in the net, so the head loss represented
by each space is 3  14 ¼ 0.214 m of water. The second equipotential
line contacts the basal upstream corner so the total head at this point is
3 – 2(0.214) ¼ 2.57 m, which equals pressure head plus elevation head. If the
tailwater elevation is taken as the datum, the uplift pressure at this point is
2.57 þ 1.5 ¼ 4.07 m of water.

A similar calculation for the basal downstream corner gives a total head of
1.5 þ 2(0.214) ¼ 1.93 m of water. As the equipotential lines are evenly spaced,

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356 Geotechnical Engineering

Figure 14.20
Uplift pressures at
the base of a dam
determined by
intersections with
equipotential
lines.

the average uplift pressure across the base of the dam is 2.25 m of water with the
center of pressure at the middle of the dam. The distribution of pressure across the
base of the structure is shown in Fig. 14.20.

Sliding friction is reduced because of a reduction in effective stress contact pressure

from the weight of the dam. Effective stress is an important principle in
geotechnical engineering: total stress equals the weight of the dam divided by the
contact area, and effective stress is the same total stress minus the uplift pressure.
It is the effective stress that determines sliding friction. The part of the stress carried
by water sometimes is called the ‘‘neutral stress’’ since it acts in all directions.

14.10 SEEPAGE FORCE AND QUICKSAND

14.10.1 Overview
Water flowing in a stream or river has the power to move solid material ranging
from sand-size to boulders and pickup trucks, depending on the velocity of the
stream, which in turn depends on the hydraulic gradient. Similarly, water flowing
through soil exerts a force on the soil mass that acts in the direction of flow. As
this also is a viscous force acting on all soil grains, it is proportional to the fluid
velocity and therefore to the hydraulic gradient.

Seepage force is a distributed force acting throughout an affected soil mass, so

strictly speaking it is not a force or a pressure; it is a force per unit volume, with
dimensions FL
3.

Perhaps the most sudden and dramatic display of seepage force for one not
accustomed to surprises is quicksand, where vertical seepage forces are sufficient
to lift and separate soil grains so that there is no friction at grain contacts.
Contrary to popular conceptions, stepping on quicksand is not like stepping on
ball bearings, because assorted ball bearings interlock and have friction. Stepping

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Seepage 357

on quicksand is like stepping onto dense liquid and offers virtually no resistance
to sinking, until opposed by a buoyant force as a result of partial submergence.

Quicksand is a dense liquid, so any object with a lower density will float, and
according to the principle of Archimedes the weight of a floating object equals the
weight of the fluid it displaces. Without saturation and an upward flow of water
there is no quicksand. There is no quicksand without water, and the density of
quicksand is such that it is not possible to sink below the armpits unless one
happens to be wearing heavy jewelry.

14.10.2 Calculating Seepage Force

A laboratory experiment to demonstrate seepage force and the development of
quicksand is shown in Fig. 14.21. The net upward force on the bottom of the soil
column is the pressure times the area, or

F ¼ A
w h ð14:44Þ

where A is the cross-sectional area, h is the head loss in the soil, and
w is the unit
weight of water. The volume of the soil is AL where L is the length of the soil
column, so the seepage force per unit volume is

F

w h


A
S¼ ¼
AL

A
L

Since h/L ¼ i,

S ¼ i
w ð14:45Þ

Figure 14.21
Seepage force and
quicksand.

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358 Geotechnical Engineering

where S is the seepage force per unit volume, i is the hydraulic gradient, and
w is
the unit weight of water. This is the basic equation for seepage force that acts
in the direction of seepage. Therefore if seepage is horizontal or at some
angle other than vertical, the seepage force can be combined vectorially
with the weight of the soil in order to obtain a resultant body force acting
on the soil mass. For example, horizontal seepage into an excavation adds a
horizontal component to the weight of the soil, which affects stability of the
excavation walls.

14.10.3 Quicksand
Opposing the seepage force in Fig. 14.21 is the submerged weight of the soil
column. The submerged weight of solids in a unit volume is Vs(G – 1)
w:
W
¼ Vs ðG
1Þ
w ð14:46Þ
AL
where Vs is the volume of solids in a unit volume of soil and G is the specific
gravity of the solids. From the definition of void ratio for a saturated soil,
e ¼ Vw/Vs, and appropriate substitutions it can be shown that Vs ¼ 1/(e þ 1).
Therefore
W G
1
¼
w ð14:47Þ
AL e þ 1
A quicksand condition develops when the seepage force S equals the weight per
unit volume, or
G
1
S ¼ i c
w ¼
w
eþ1
where ic is the critical hydraulic gradient. Then
G
1
ic ¼ ð14:48Þ
eþ1
This is the basic equation for quicksand. As an approximation, G for quartz is
2.65, so if e is 0.65,
ic  1:0
Therefore as a general guide quicksand can be expected to develop if the hydraulic
gradient in an upward direction exceeds 1.0. However, as ic is sensitive to void
ratio, a more accurate determination can be made from Fig. 14.22, which is based
on G ¼ 2.70.

It can be seen that the critical gradient does not depend on size of the soil particles
except insofar as they influence permeability and the quantity of flow necessary to
achieve ic. That is, if k in the expression Q ¼ kiA is large, the flow quantity Q must
have a proportionate increase to create the same hydraulic gradient. Quick
conditions therefore are much more likely to develop in sand than in gravel,
although a quick gravel theoretically is possible. A quick condition with relatively

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Seepage 359

Figure 14.22
Effect of soil
density on the
hydraulic gradient
required to
generate a quick
condition.

little seepage can develop in cohesionless silt, and sometimes occurs where a slip
zone subjected to upward seepage is exposed in the toe area of a landslide.

14.10.4 Occurrences of Quicksand

Quicksand commonly occurs in the bed of a lake or river where water seeps
upward as a result of artesian pressure. The points of emergence are underwater
springs. Quicksand conditions therefore may be seasonal when artesian pressures
are highest, and absent during drier seasons.

Excavations that are pumped dry in preparation of placing foundations for

buildings or bridges can be an invitation for quicksand or ‘‘boiling’’ of soil as
water seeps upward into the excavation. Sand boils also can occur on the
downstream side of an earth dam as the reservoir is filled and develops high
seepage pressures. Sand boils are common on the landward side of temporary
levees built during flood stages of major rivers.

Unpredicted quick conditions can result if an undiscovered aquifer, buried tile

line, gravel layer, or fracture zone in rock conveys water without diminishing the
head. This is illustrated in Fig. 14.23.

14.10.5 Liquefaction
Temporary quicksand conditions can be induced by ground vibrations,
particularly from earthquakes, if they densify saturated, loose sand so that
stress is transferred to the water and the effective stress is reduced or becomes
zero. Then, as the sand settles, the water they displace rises. A vibration-induced
quick condition is called liquefaction. Evidence for liquefaction includes conical
‘‘sand volcanoes’’ where sand is carried up to the ground surface by the ejected

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360 Geotechnical Engineering

Figure 14.23
A fugitive tile line
or stringer of
gravel can cause
quick conditions.
Back-pressure is
applied by
encircling the
seepage area with
sandbags.

water. The occurrence of sand volcanoes buried in ancient sediment layers is used
to date the recurrence interval of earthquakes.

14.10.6 Managing Quicksand

A simple and effective way to stop a quick condition is to reduce the hydraulic
gradient by building a wall of sand bags to pond water, as shown in Fig. 14.23.
A more permanent measure that is more suitable for foundation excavations is
to use interceptor wells to reduce or reverse the gradient.

Another method is to cover a potential quicksand area with a layer of heavy,

coarse-grained material that adds additional weight that must be lifted, but which
also is sufficiently permeable that it allows seepage water to escape without
appreciable resistance. Such a blanket of material is called a protective filter.
A layer of filter fabric or intermediate particle size material can be used to prevent
soil material from being carried upward into the filter. The design of filters is
discussed later in Section 14.11.3.

14.11 FLOW NETS IN EARTH DAMS

14.11.1 Dam Arguments

Earth dams are the largest man-made structures and provoke some of the biggest
arguments. On the positive side, dams are cost-effective and save energy for
production of electrical power, and reservoirs created by dams are essential for
survival of places like Las Vegas. (As grandmother would say, more’s the pity.)
Dams allow irrigation that turns deserts into gardens, and reservoirs are used for
recreation and for flood control. Dams can allow river navigation that is the most
economical means for shipping large quantities of grain, coal, etc.

Negative aspects are that dams drown valuable floodplain soils, displace towns
and people, corrupt nature, and have a limited useful life because of sedimentation
in the reservoir. Sediment trapped behind a dam is not available for delta-building

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Seepage 361

at the mouth of the river, which starts a domino effect, the next domino being the
reduction in sediment to maintain beaches, and encroaching shorelines as a result
of wave erosion.

Of these problems, sedimentation may be the most serious as it will gradually

diminish and eventually will destroy the usefulness of the dam. A bypass canal can
be used to divert sediment-laden flood water, but this may only borrow time, and
the amount of sediment that eventually may be deposited often will be hundreds
of times larger than the volume of the dam itself. Optional procedures include
lowering the reservoir level to let the river entrench and carve out a new
floodplain, or fluidizing and pumping sediment around the dam (Jenkins et al.,
1992). An option that does not yet appear to have been tried is to let sediment-
laden water form a density current along the bottom of a reservoir to an outlet at
the base of the dam.

Engineers should not be saddled with the decision of whether or not to build a
dam, but can provide an objective assessment of the data. Engineers should
develop an awareness of both the positive and negative aspects of dams, and if a
decision is made to proceed, it is the engineer’s obligation to ensure that the dam
and reservoir are safe, functional, and long-lasting. Large dam construction is
particularly active in developing countries of the world.

14.11.2 Flow-Net Boundary Lines in Earth Dams

The lower flow boundary in an earth dam normally is an impermeable layer that
may or may not be at the base elevation of the dam. The upper flow boundary was
investigated by A. Casagrande using dyes introduced at the upstream face of
model dams. He found that the upper flow line can be approximated with a
parabola that is modified at the upstream side to make a 908 angle with the dam
face, which is an equipotential line. Casagrande’s method, while empirical, has
been compared with an analytical method and shown to be adequate for general
use (Jepson, 1968).

Figure 14.24 shows a cross-section of a trapezoidal dam of homogeneous soil.

The parabola begins at the water surface of the reservoir at point A, which is a

Figure 14.24
Casagrande
method for drawing
an upper boundary
flow line through
an earth dam.

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 Seepage

362 Geotechnical Engineering

horizontal distance 0.3 times the horizontal width of the upstream face that is
under water.

Next, the focal point of the parabola, F, is assigned at the boundary of the
drainage layer. The directrix is found by swinging an arc centered at A from F to a
horizontal line and dropping a vertical line to G. One point on the parabola is
halfway between F and G. Another point, H, is directly over F such that F, G,
and H are corners of a square. These points should be sufficient to sketch in the
parabola that then is modified with a smooth curve at B.

Without a drainage gallery the focal point of the parabola would move to F0 , and
extending a parabola to this point indicates that seepage would emerge on the
upstream face, which should be avoided.

14.11.3 Drain and Filters

A layer of a more permeable granular soil often is incorporated into an earth dam
as an underdrain, EF in Fig. 14.24. Another alternative is a toe drain that is
triangular in cross-section, Fig. 14.27. In both instances soil is separated from the
drain by a suitable filter to prevent soil fines from penetrating into voids in the
coarser material. As design involves particle sizes in one medium and void sizes in
the other, it is empirical based on laboratory tests. Representative criteria are as
follows:

d15
55 ð14:49aÞ
D85

d15
between 5 and 40 ð14:49bÞ
D15

D85
2 ð14:49cÞ
hole diameter in pipe drain D50

where d and D refer to particle sizes in the finer and coarser materials,
respectively, and the subscripts denote percentages finer than the respective sizes.
These requirements apply to both boundaries of a filter layer, so if a single layer
does not satisfy all criteria two or more layers can be used. The thickness of each
layer depends on the hand that must be resisted; a rule-of-thumb being that it
should be at least 5 percent of the pressure head.

Another option is the use of a synthetic filter fabric, which is discussed by Koerner
(1990). Whereas filter layers of soils have progressively larger void spaces and
permeability, the capability of a geotextile to transmit water varies, and must be
high enough not to impede drainage. The design therefore involves two criteria,
one for flow of water and the other for retention of the soil.
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 Seepage

Seepage 363

Figure 14.25
Diffraction of flow
lines resulting from
a change in k.

Figure 14.26
Some methods for
reducing underdam
seepage.

A geotextile having a sufficient thickness and transmissivity (sometimes called

permitivity) can substitute for the drain itself. In this case it must resist crushing
under pressure from the soil. Some applications include drains along the backsides
of retaining walls under pavements or earth embankments.

14.11.4 Diffraction of Flow Lines

Earth dams often incorporate a clay core to reduce seepage, and in addition
employ a granular drainage area in the toe. Seepage crossing a boundary between
two permeabilities is diffracted, much in the same way that light is diffracted when
crossing a boundary between two refractive indices, and the same equation
applies. An analogy is soldiers marching in a parade and entering an area of
mucky soil at an angle: as the marchers reach the muck they slow down, which
causes the direction of the column to change. If the velocity decreases, the angle
increases, and vice versa. The equation is
tan 1 k1
¼ ð14:50Þ
tan 2 k2
The derivation for this equation can be found in a physics textbook. It means
that flow lines entering the clay core of an earth dam will be diffracted
downward, lowering the phreatic line, and upon exiting the clay core or entering a
toe drain will be diffracted upward. The nature of the diffraction is indicated in
Fig. 14.25.

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 Seepage

364 Geotechnical Engineering

Figure 14.27
Monitoring
potentials within
an earth dam in
order to detect
leakage.

14.11.5 Controlling Underdam Seepage

A typical alluvial section on which a dam may be founded consists of clay
overlying sand over gravel. Upstream from the dam the river most likely will be
running in sand that connects to the gravel and can become a major path for
leakage. Several methods may be used to reduce such leakage. As shown in
Fig. 14.26(a), an area upstream, in particular the river channel, may be covered
with a compacted clay layer called an ‘‘upstream blanket.’’ In Fig. 14.26(b)
relatively impermeable soil in the dam is extended down into the foundation soils,
which is called a ‘‘key trench.’’ A trench that extends all of the way through the
permeable layers is a ‘‘cutoff trench.’’ A third possibility is a ‘‘curtain wall’’ such
as shown in Fig. 14.26(c), which can be made with sheet pile or with a
concrete-filled trench called a ‘‘slurry trench.’’

14.11.6 Piping
Bad things still can happen after a dam is constructed and in service, and walk-
over inspections should be routinely performed. For example, floodplain soils
supporting earth dams normally are soft and slowly compressible, and over a
period of time a dam may settle unevenly and develop internal tension cracks
running transverse across the axis of the dam. Such cracks form a direct route
through the dam and if undetected and not corrected by grouting can be a prelude
to a major leak and failure.

Any leakage observed through an earth dam must be carefully examined to see
if the water is clouded by soil particles, which would indicate internal erosion
or ‘‘piping.’’ If piping is observed, remedial treatments should immediately be
initiated.

14.11.7 Pinhole Test

Some soils are more susceptible to piping than other soils. Most vulnerable
are silts that are easily erodible, and soils containing clays that tend to
disperse as water reduces the content of flocculating ions. Normally these

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Seepage 365

are sodium clays, and are referred to as ‘‘dispersive clays.’’ Such clays are
regarded as unsatisfactory for use in earth dams. However, their detection
can be complicated, and for this purpose a ‘‘pinhole test’’ was devised
by Sherard et al. (1976). Details of the test are given in ASTM Designation
D-4647.

The pinhole test essentially consists of drilling a 1 mm diameter hole through a

compacted soil specimen and passing water through the hole. A constant head of
50 mm (2 in.) is used initially and the test continued for 5 minutes, but these values
may be modified for additional testing. Erosion is determined from cloudiness of
the water, increasing flow rate as the pinhole enlarges, and final dimensions of the
pinhole. If after 5 minutes the hole has enlarged 50 percent, the soil is considered
dispersive. The test is performed with distilled water, which will tend to increase
dispersion.

14.11.8 Measuring Seepage Conditions within an Earth Dam

Cracks that do not go all of the way through or under a dam still can be
dangerous, but are difficult to detect except as they affect the equipotential lines in
the flow net. Anomalous behavior therefore can be detected by measuring the
pore water pressure at various points within a dam. This is done with tubes
called piezometers that have a porous tip that allows water to enter up into the
tube. The height of the water in the tube is a measure of the total head at that
point in the dam, and can be compared with that predicted from the flow net.
An example is shown in Fig. 14.27. The level of water in a piezometer head is
not at the phreatic surface, but is as determined by the contact with an
equipotential line.

Let us now assume that the head measured by the piezometer is at the h1 level of
the water surface behind the dam instead of at h. That would mean that there is a
direct connection between the location of the piezometer tip and the reservoir
behind the dam, which should be thoroughly investigated as it eventually could
develop into seepage and possible quick conditions in the face of the dam, and
breaching of the dam. Piezometers are a valuable tool for assessing seepage
conditions in and under a dam, and often are permanent installations.
Some major dam failures might have been prevented had the seepage conditions
been monitored with piezometers and corrective measures taken before it was
too late.

Problems
14.1. The change in elevation of a stream is 947 mm/m (5 ft/mile). What is the
hydraulic gradient?
14.2. State the Darcy Law for flow of water through soil. Define the coefficient
of hydraulic conductivity. Is it the same as the coefficient of permeability?
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366 Geotechnical Engineering

14.3. Explain the term ‘‘velocity of approach,’’ as used in connection with the
Darcy Law.
14.4. Some soils are said to be nonisotropic with respect to flow of gravitational
water. What does this mean? What are some examples?
14.5. What is the difference between a constant-head and a falling-head
permeameter? Explain why one type may be preferred for particular soils.
14.6. A constant-head permeameter is set up with a soil specimen 203 mm (8 in.)
long and 76 mm (3.0 in.) in diameter. The vertical distance between the
headwater and tailwater surfaces is 305 mm (12 in.). In a test run, 163.9 kg
(361.5 lb) of water passes through the sample in a period of 18 hr 20 min.
What is the coefficient of hydraulic conductivity of this soil?
14.7. A falling-head permeameter is set up with a soil sample 152 mm (6 in.) long
and 51 mm (2 in.) in diameter. The area of the standpipe is 213 mm2
(0.33 in.2). In a test run, the water in the standpipe falls from an ele-
vation of 1.22 m (48 in.) above the tail water to 0.538 m (21.2 in.) in 7 hr
36 min 32 s. What is the coefficient of hydraulic conductivity of this soil?
14.8. The conductivity coefficient of a soil determined in the laboratory at a
temperature of 24.48C (768F) is 3.5 mm/min. What is the coefficient of the
same soil when the permeating water is at 4.48C (408F)?
14.9. A test well is installed in permeable sand to a depth of 9.1 m (30 ft) below
the groundwater table, and observation wells are put down at distances of
(a) 15.2 m (50 ft) and (b) 30.5 m (100 ft) from the test well. When pumping
reaches a steady state at 965 1/min (255 gal/min), the water table at
(a) is lowered 1.158 m (3.8 ft) and at (b) 0.945 m (3.1 ft). Compute the
coefficient of permeability of the sand.
14.10. A test well 762 mm (30 in.) in diameter is drilled through a water-bearing
sand down to an impervious layer that is 7.32 m (24 ft) below the
normal water table. Steady pumping at the rate of 757 l/min (200 gal/min)
causes the water in the test well to be lowered 2.84 m (9 ft 4 in.). What is the
approximate value of the coefficient of hydraulic conductivity of the sand?
14.11. A 114 mm (4.5 in.) diameter casing is set at 6.7 m (22 ft) depth in a hole with
the water table at l.04 m (3.4 ft) depth. Water added at the rate of 2.08 l/min
(0.55 gal/min) maintains the water level at 51 mm (2 in.) depth. Find k.
14.12. A test head with packers spaced 1.52 m (5 ft) apart is lowered to a
depth of 13.7 m (45 ft) in a 102 mm (4 in.) diameter hole. After
sealing, a pumping pressure of 240 kPa (35 lb/in.2) resulted in a pumping
rate of 2.78 l/s (44 gal/min), increasing to 4 l/s (63 gal/min) after 12 min.
Find the initial and final k and give possible explanations for the
difference.
14.13. A hole 152 mm (6 in.) in diameter is bored in the soil to a depth of 1.90 m
(75 in.) below the water table. After the hole has been pumped out several
times and allowed to refill, it is pumped out again and the rate of rise of
water at an elevation 1.02 m (40 in.) below the water table is observed to be
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Seepage 367

Figure 14.28
Earth dam for
flow-net
construction in
Problem 14.21.

19.0 mm (0.75 in.) in 1 min. What is the coefficient of hydraulic

conductivity of the soil?
14.14. What is the basis for the flow net?
14.15. Define flow line; equipotentiaI line; flow path; equipotential space; three
kinds of head.
14.16. State five fundamental properties of a flow net.
14.17. Identify the boundary flow lines and equipotential lines in Fig. 14.15.
14.18. Assume that the coefficient of hydraulic conductivity of the soil in
Fig. 14.15 is 0.0042 mm/min. Determine the seepage per 30 m (100 ft) of
length of the sheet-pile cofferdam.
14.19. The pervious foundation soil beneath the masonry dam in Fig. 14.17 has a
coefficient of hydraulic conductivity of 2.96 mm/min (0.14 ft/day).
Determine the seepage per 30 m (100 ft) length of the dam.
14.20. The pervious foundation soil in Fig. 14.19 has a coefficient of hydraulic
conductivity of 1.22 mm/min in the vertical direction and 6.1 mm/min (0.02
ft/min) in the horizontal direction. Compute the seepage per linear meter
(foot) of dam.
14.21. Construct a flow net and estimate the seepage per linear meter (foot) of the
dam in Fig. 14.28.
14.22. What is the critical hydraulic gradient of a soil having a specific gravity of
2.70 and a void ratio of 0.85?
14.23. Determine the hydraulic gradient of seepage water at point d in Fig. 14.15.
If the specific gravity of the soil inside the cofferdam is 2.62 and the void
ratio is 1.14, will sand boils develop? What is the factor of safety against
boiling at this point?
14.24. What is the hydraulic gradient at point e in Fig. 14.15?
14.25. Calculate the effective stress on a horizontal plane 4.9 m (16 ft) below the
surface of a soil mass in which the water table is 1.5 m (5 ft) below the
surface. Assume that the dry density of the soil is 14.1 kN/m2 (90 lb/ft2),
the moisture content above the water table is 15%, and the specific gravity
of the soil is 2.70.
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368 Geotechnical Engineering

Figure 14.29
How not to draw a
flow net.

14.26. Permeability tests on samples from several horizontal soil layers under a
proposed dam give the following:
Material k (cm/s) % of total section

Clay 8  10
5 45
Silt 1  10
4 35
Sand 2  10
2 18
Gravel 10
1(est.) 2

(a) Calculate the horizontal and vertical coefficients of permeability.

(b) Suggest methods to reduce the horizontal permeability.
14.27. Identify errors in the flow net in Fig. 14.29.

References and Further Reading

Boast, C. W., and Kirkham, D. (1971). ‘‘Field Measurement of Soil Permeability Using
Auger Holes.’’ Proc. Soil Sci. Soc. Amer. 35, 365.
Casagrande, A. (1937). ‘‘Seepage Through Dams.’’ J. New Engl. Waterworks Assoc.
Reprinted in Contributions to Soil Mechanics 1925–1940, pp. 295–336. Boston Soc. Of
Civil Engrs., Boston, Mass.
Cedergren, H. R. (1989). Seepage, Drainage, and Flow Nets, 3rd ed. John Wiley & Sons,
New York.
Harr, M. E. (1962). Groundwater and Seepage. McGraw-Hill, New York.
Jenkins, S. A., Wasyl, J., and Skelly, D. W. (1992). ‘‘Tackling Trapped Sediments.’’
Civil Engineering (Feb.), pp. 61–63.
Jepson, R. W. (1968). ‘‘Seepage Through Dams in the Complex Potential Plane.’’ ASCE J.
Irrigation and Drainage Div. 94(IR1), 23–39.
Jury, W. A., and Horton, R. (2004). Soil Physics, 6th ed. John Wiley & Sons,
Hoboken, N. J.
Kirkham, D., and Powers, W. L. (1971). Advanced Soil Physics. Wiley Interscience,
New York.
Koerner, R. M. (1990). Designing with Geosynthetics 2nd ed. Prentice-Hall, Englewood N.J.
Leonards, G. A., ed. (1987). Dam Failures. Elsevier, Amsterdam. Reprinted from
Engineering Geology 24,1–4.
Sherard, J. L., Dunnigan, L. P., Becker, R. S., and Steele, E. F. (1976). ‘‘Pinhole Test for
Identifying Dispersive Soils.’’ J. Geotech. Eng. Div. ASCE 102(GT1), 69–85.

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