Prepared By: Assist. Prof. Halil Ibrahim FEDAKAR (AGU, Dept. of Civil Eng.)
Reference Book: Coduto, D.P. Geotechnical Engineering: Principles and
Engineering, Prentice Hall, NJ. (Chapter 7: Groundwater-Fundamentals and
One-Dimensional Flow)
1
«All streams flow into the sea, yet the sea is never full. To the place the streams come from, there
they return again.» Ecclesiastes 1:7 (NIV)
«… engineering practice, difficulties with soils are almost exclusively due not to the soils
themselves, but to the water contained in their voids. On a planet without any water there would be
no need for soil mechanics» Terzaghi (1939)
2
▪ The presence of water, or at least the potential for its presence, is a key aspect of
most geotechnical analyses. Therefore, this is a topic worthy of careful study.
▪ Specific-water related geotechnical issues include the following:
➢ The effect of water on the behavior and engineering properties of soil and rock
➢ The potential for water flowing into excavations
➢ The potential for pumping water through wells or other facilities
➢ The effect of water on the stability of excavations and embankments
➢ The resulting uplift forces on buried structures
➢ The potential for seepage-related failures, such as piping
➢ The potential for transport of hazardous chemicals along with the water
3
▪ Geotechnical engineers are mostly interested in
the portions of the hydrologic cycle that occur
underground.
▪ Groundwater (phreatic surface) can be located by
installing observation wells and allowing the
groundwater to seep into them until it reaches
equilibrium.
▪ The water level inside these levels is the
groundwater table, where the pore water pressure
is equal to zero.
▪ The groundwater table location is important, and
it is one of the primary objectives of a site
characterization program.
4
▪ The groundwater table elevation often changes
with time, depending on the season of the year,
recent patterns of rainfall, irrigation practices,
pumping activities, and other factors.
▪ At
some locations, these fluctuations are
relatively small (perhaps less than 1 m), while in
other places the groundwater table has changed
by 20 m or more in only a year or two.
▪ Thus, the groundwater conditions encountered in
an exploratory boring are not necessarily those
we use for design.
▪ Often we need to use the observed conditions as a
basis for estiamting the worst-case conditions that
are likely to occur during the project life.
5
▪ Phreatic zone (below the groundwater table): Positive pore
water pressure as a result of the weight of the overlying
water.
▪ Vadose zone (above the groundwater table): Negative pore
water pressure and capillary action.
▪ Aquifers: Sands and gravels. Good candidates for wells.
▪ Aquicludes: Clays.
▪ Aquitards: Silty sand
▪ Unconfined aquifer: The bottom flow boundary is defined
by an aquiclude, but the upper flow boundary
(groundwater table) is free to reach its own natural level.
▪ Confined aquifer: Both the upper and lower flow
boundaries are defined by aquicludes.
▪ Artesian conditions can occur only in confined aquifers.
6
▪ One-dimensional
flow: The water always
moves parallel to some axis and through a
constat cross-sectional area.
▪ Two-dimensional flow: All
of the velocity
vectors are confined to a single plane, but vary
in direction and magnitude within the plane.
▪ Three-dimensional flow: The velocity vectors
vary in the x, y, and z directions.
7
▪ Steady-state flow: The direction and velocity of fluid flow is constant with time and,
therefore, the flow rate, Q, also remains constant with time.
▪ Unsteady flow (or transient flow): The direction and velocity of fluid flow vary with
time.
8
▪ Laminar flow occurs when the velocity is low.
▪ Turbulent flow occurse when the velocity is high.
▪ For most soils, the velocity is low, so the flow is laminar. This is important because
our analyses are only valid for laminar flow.
▪ There are cases in very coarse soils, such as clean, poorly-graded gravels, in which
velocisities can be much higher and turbulent flow conditions apply.
9
▪ Potential energy: Due to its elevation above the datum
▪ Strain energy: Due to the pressure in the water
▪ Kinetic energy: Due to its velocity
▪ Head: Energy divided by the acceleration of gravity.
▪ Elevation head (hz): The difference in elevation between the
datum and the point (B in the figure)
▪ Pressure head (hP): The difference in elevation between the
point (B) and the water level in a piezometer attached to the
pipe.
▪ Velocity head (hv): The difference in water elevations
between the piezometer and the Pitot tube. (=v2/(2g))
▪ Total head (h) (Bernoulli equation): hz+hP+hv
▪ The Bernoulli equation is a convenient way to compare the
energy at two points.
10
▪ Water always flows from a point of
high total head to a point of low total
head.
▪ The energy of the water flowing from
A to B is lost due to friction, a quantity
known as the head loss, Δh.
▪ Hydraulic gradient (i): The change in
total head per unit length in the
direction of flow. (=-Δh/Δl)
▪ The velocity of water flow in soil is
much lower, so the velocity head is
very small (less than 5 mm). Thus, it
can be neglected for practical
seepage problems. (h= hz+hP)
11
▪ Pore water pressure (u): The pressure in the water within the soil voids.
▪ For points below the groundwater table, u=γwhP
▪ Hydrostatic condition: The pore water pressure is due solely to the force of gravity
acting on the pore water. This is the case so long as the soil is not in the process of
settling or shearing. The associated pore water pressure is the hydrostatic pore water
pressure (uh).
▪ If all of the following conditions have been met, then the pressure head is simply the
difference in elevation between the groundwater table and the point where the
pressure head is to be computed.
➢ The aquifer is unconfined
➢ The groundwater is stationary or flowing in a direction within about 30o of the
horizontal.
▪ Therefore, the pore water pressure: u=uh=γwzw (zw: Depth from the groundwater table to
the point)
12
▪ Example 1: Compute the pore water pressure at Points A and B shown in the
following figure. Point A is located in the upper unconfined aquifer and Point B is
located in the lower confined aquifer.
13
▪ Solution 1:
Since Point A is in an unconfined aquifer, the
pressure head at that location is simply equal to the distance
below the phreatic level surface (zw) and the pore water
pressure (u) is:
u=γwzw
u=(62.4 lb/ft3)(260 ft – 250 ft)
u=624 lb/ft2
▪ Since Point B is located in a confined aquifer, it is necessary to install a
piezometer to determine the pore water pressure. The pressure head (hP)
is equal to difference in elevation between Point B and the elevation to
which the water rises in the piezometer and the pore water pressure is
u=γwhP
u=(62.4 lb/ft3)(268 ft – 240 ft)
u=1747.2 lb/ft2
▪ Note that the water in the piezometer located at Point B rises above the
water table in the unconfined aquifer. This indicates that the lower aquifer
is in artesian condifion.
14
▪ If the soil is in the process of settling or shearing, then excess pore water pressures
will be present and the hydrostatic condition does not exist.
▪ If the water is flowing at a significant angle from the horizontal, or if it is confined,
then it is necessary to perform a two- or three-dimensional analysis.
▪ Above the groundwater table, we normally consider the pore water pressure to be
zero. In reality, surface tension effects between the water and the solid particles
produce a negative pore water pressure above the groundwater table, and this
negative pressure is sometimes called soil suction.
15
▪ Soils below the groundwater table are saturated, and
have a pore water pressure (u=uh=γwzw).
▪ For many engineering problems, this simple model is
sufficient. However, the real behavior of soils is rarely
so simple. One important aspect not addressed by this
model is capillarity.
▪ Capillarity
(or capillary action) is the upward
movement of a liquid into the vadose zone, which is
above the level of zero hydrostatic pressure. This
upward movement occurs in porous media or in very
small tubes.
▪ Capillary action is the result of surface tension at
water-air interfaces. This can be demonstrated by
inserting a small-diameter glass tube into a pan of
water as shown in the figure.
16
▪ The theoretical height of capillary rise, hc, in a glass
tube of diameter d, at a temperature of 20oC is:
hc=0.03/d (hc in m and d in mm)
▪ Capillary rise in soils is more complex because
soils contain an interconnected network of pores of
different sizes.
▪ However, using 0.2D10 as the equivalent d generally
produces satisfactory results for sands and silts.
Thus, the height of capillary rise in these soils is
approximately:
hc=0.15/D10
▪ D10 is grain size corresponding to 10% finer (mm).
17
▪ The theoretical capillary rise in clays can be in excess of 100 m, but in reality it is
much less. The following figure shows the relationship among the vadose, phreatic,
and capillary zones.
18
▪ Geotechnical engineers often need to predict the flow rate, Q, through a soil.
▪ Darcy’s law is expressed as:
Q=kiA
▪ k is hydraulic conductivity (coefficent of permeability), i is hydraulic gradient, and A is area
perpendicular to the flow direction.
▪ Darcy’s law is valid for a wide range of soil types, from clays through coarse sands.
▪ The primary exceptions are clean gravel, where its accuracy is diminished because of the
turbulent flow, possibly in clays with low hydraulic gradients, where the absorption of water to the
clay particles significantly influences flow.
19
▪ Hydraulic conductivity, k, depends on properties of both the soil and the liquid flowing through it,
including the following:
➢ Soil properties
✓ Void size (depends on particle size, gradation, void ratio, and other factors)
✓ Soil structure
✓ Void continuity
✓ Particle shape and surface roughness
➢ Liquid properties
✓ Density
✓ Viscosity
▪ The most common unit of measurement for k is cm/s.
20
▪ The typical values of k for different soil types are given in the following table.
▪ The low k in clays is due to the small particle size (and therefore small void size). It is not due to
water being absorbed by the clay.
21
▪ Confined aquifer flow: In a confined aquifer of constant thickness, the flow is one dimensional so
long as there are no rivers, lakes, or wells that affect the flow of water in the aquifer. In this case,
the flow per unit width of the aquifer is computed using the following equation.
Q=kiA
▪ Example 2: The figure shows a confined aquifer with two piezometers measuring the pressure
head at two different locations 1500 ft apart. The aquifer has a uniform thickness of 10.5 ft and a
hydraulic conductivity of 2.4x10-4 ft/s. Assume the width of the aquifer into the page is 6 miles and
the flow is one dimensional from left to right. Compute the total flow in the aquifer in ft3/s.
22
▪ Solution 2: The flow in a confined aquifer is similar to the flow in a soil filled pipe, except in this
case instead of a round pipe we have a rectangular-shaped aquifer with a cross-sectional area of
10.5 ft by 6 miles. Darcy’s law applies.
▪ Using Darcy’s equation,
Q=kiA
i=Δh/Δl=6 ft/1500 ft=0.004
Q=kiA
Q=(2.4x10-4 ft/s)(0.004)(10.5 ft)(6 mile)(5280 ft/1 mile)=0.319 ft3/s
23
▪ Example 3: A 3.2 m thick silty sand stratum intersects one side of a reservoir as shown in the
following figure. This stratum has a hydraulic conductivity of 4x10-2 cm/s and extends along the
entire 1000 m length fo the reservoir. An observation well has been installed in this stratum as
shown. Compute the seepage loss from the reservoir through this stratum.
24
▪ Solution 3: The observation well indicates a water level above the top of the silty sand strata.
Therefore, it is a confined aquifer.
i=Δh/Δl=(167.3-165)/256=0.009
A=(3.2)(1000)=3200 m2
k=(4x10-2 cm/s)(1 m/100 cm)(3600 s/1 hr)(24 hr/1 day)(30 day/1 mo)=1000 m/mo
Q=(1000)(0.009)(3200) =30,000 m3/mo
25
▪ Apparent velocity (Darcian velocity) (va=ki): It is not the velocity at which the water moves
through the soil. It is an artificial velocity computed when the flow rate is divided by the crosssectional area of the soil. The area includes both soil solids and the voids.
▪ However, the water flows only in the voids. We must account for the difference between the total
cross-sectional area and the cross-sectional of voids to determine how rapidly water actually flows
through the soil. Seepage velocity (vs) is used to describe this velocity.
vs=(ki)/(ne)
▪ Effective porosity (ne) represents the percentage of the cross-sectional area, A, that actually
contributes to the flow.
▪ The seepage velocity is especially important in geoenvironmental engineering problems
because it helps us determine how quickly contamintants travel through the ground.
▪ In sandy soils, ne is equal to porosity ((Vv/V)x100). However, clayey soils contain a static layer of
water around the particles, so the actual flow area is less than the void area.
▪ If no test data is available, the Environmental Protection Agency uses ne=0.10 in clays.
26
▪ When viewed on a microscopic scale, the true velocity at
which an element of water is moving within the soil will be
greater than the seepage velocity because the actual flow
path is circuitous route around the soil particles.
▪ The seepage velocity is based on the equivalent straight line
movement as shown in the figure.
▪ However, the true velocity is primarily of theoretical interest,
and the seepage velocity is more useful for solving practical
contaminant transport problems.
27
▪ Example 4: Assume a chemical solvent, which is denser than water is spilled into the reservoir in
Example 3. How long will it take for the contaminant to reach the observation well assuming only
advection occurs in the aquifer. The silty sand in the aquifer has a void ratio of 0.75.
▪ Advection: The contaminants simply move along with the groundwater flow.
28
▪ Solution 4: If only advection is occurring, then the contaminant will travel at the same velocity as
the groundwater. The seepage velocity is the appropriate velocity to use in this case. Assume the
effective porosity is equal to the porosity for this silty sand.
n=e/(1+e)=0.75/(1+0.75)x100%=43%
vs=ki/ne=(4.2x10-4 cm/s)(0.009)(1 m/100 cm)(86,400 s/1 day)/(0.43)=0.72 m/day
time=(distance)/(vs)=(256 m)/(0.72 m/day)=355 days
29
▪ Constant-head test
➢ A laboratory hydraulic conductivity test that applies a constant head of water
to each end of a soil sample in a permeameter as shown in the figure.
➢ The head on one end is greater than that on the other, so a flow is induced.
➢ Q, i, and A are determined from the test results and then k is computed.
30
▪ Example 5: A constant head test is performed using a permeameter similar to
the one shown in the figure. The graduated cylinder collects 892 ml of water in
112 seconds. The other data from the test are as follows:
➢ Soil specimen diameter=18.0 cm
➢ Elevation of water in upper-most piezometer=181.0 cm
➢ Elevation of water in lowest piezometer=116.6 cm
➢ The piezometer inlets are evenly spaced at 16.7 cm on center
▪ Compute the hydraulic conductivity, k.
31
▪ Solution 5: Q=V/t=(892 ml/112 s)(1 cm3/1 ml)=7.96 cm3/s
A=(πD2)/4=(π182)/4=254 cm2
Note how A includes both the voids and the solids.
i=-(Δh/Δl)=-(116.6-181)/(3x16.7)=1.29
k=Q/(iA)=7.96/(1.29x254)=2x10-2 cm/s
32
▪ Example 6: If the soil specimen in Example 5 is sandy with a void ratio of 0.85,
compute the seepage velocity through the specimen.
33
▪ Solution 6: The soil is sandy, so the effective porosity equals to the porosity, n.
n=e/(1+e)=0.85/(1+0.85)x100%=46%
vs=ki/ne=(2x10-2)(1.29)/(0.46)=5.6x10-2 cm/s
34
▪ Falling-head test
➢ The water in the standpipe is not replenished as it is
in the constant-head reservoir.
➢ Thus, as the test progresses, the water level in the
standpipe falls.
➢ This method is more suitable for soils with very low
hydraulic conductivities, such as clays, where the
flow rate is small and needs to be precisely
measured.
➢ The hydraulic gradient is not constant, which means
that the flow rate also is not constant.
35
▪ Falling-head test
𝑄 = 𝑘𝑖𝐴 = 𝑘
∆ℎ
𝑑(∆ℎ)
𝐴 = −𝑎
𝐿
𝑑𝑡
∆ℎ1
𝑘𝐴 𝑡
𝑑 ∆ℎ
න 𝑑𝑡 = −𝑎 න
𝐿 0
𝑑ℎ
∆ℎ0
𝑘𝐴𝑡
∆ℎ1
= −𝑎 ln(
)
𝐿
∆ℎ0
𝑘=
𝑎𝐿 ∆ℎ0
ln(
)
𝐴𝑡 ∆ℎ1
36
▪ Empirical estimates of hydraulic conductivity
➢ Hazen’s correlation: For loose, clean sands: k=CD102
✓ k in cm/s; C: Hazen’s coefficient 0.8-1.2; D10 in mm
✓ Note: Be sure to use the stated units for k and D10
✓ The accuracy of Hazen’s correlation is limited because it considers only a single measure of
the grain size of a soil and does not account for the distribution of grain sizes found in
natural soils.
✓ Hazen’s work was intended to be used in the design of sand filters for water purification, but
can be used to estimate k in the ground.
✓ However, its applicability is limited to soils with 0.1 mm<D10<3 mm and a coefficient of
uniformity, Cu<5.
37
▪ Empirical estimates of hydraulic conductivity
➢ Kozeny-Carman: 𝑘 =
𝛾
1
𝑒3
(
)( )
𝜇 𝑇 2 𝑆02 1+𝑒
✓ γ is unit weight of pore fluid; μ is viscosity of pore fluid; T is dimensionless factor accounting
for the shape of the pores; S0 is specific surface of the soil particles, area per unit volume.
✓ T2 can be taken to be approximately 5 for most soils. The above equation can be simplified
to
3
1
𝑒
𝑘 = 1.99𝑥10−4 ( 2 )(
)
𝑆0 1 + 𝑒
✓ k in cm/s and S0 in 1/cm
✓ If the soil was made up of uniform spheres then the specific surface would be
𝑎𝑟𝑒𝑎
𝜋𝐷2 6
𝑆0 =
=
=
𝑣𝑜𝑙𝑢𝑚𝑒 𝜋𝐷3 𝐷
6
38
✓ However, real soils are not made of uniform spheres, «6» in the
equation is replaced with a shape factor, SF.
S0=SF/D
✓ All that is necessary is to account for the different sizes of soil
particles that exist in a natural soil. The Kozeny-Carman
equation becomes
✓ 𝑘=
100%
1 2 𝑒3
4
2
1.99𝑥10 (σ[𝑓 /(𝐷0.404 −𝐷0.595 )]) (𝑆 ) (1+𝑒)
𝐹
𝑖
1𝑖
𝑠𝑖
✓ k in cm/s; fi is fraction of soil between two sizes; D1i is the size of
the openings in the larger of the two sieves in cm; Dsi is the size
of the openings in the smaller of the two sieves in cm.
✓ This equation is still limited to sands. It is not accurate for clays
or gravels.
39
▪ Example 7: A poorly-graded, rounded, medium dense sand has a void ratio of 0.49 with the
particle size distribution shown in the following table. Estimate the hydraulic conductivity using
both Hazen’s correlation and the Kozeny-Carman equation.
Sieve Opening (cm)
Sieve Size
Percent Passing
0.475
#4
100
0.2
#10
94
0.085
#20
70
0.0425
#40
55
0.015
#100
10
0.0075
#200
4
40
▪ Solution 7:
𝑘=
100% 2 1 2 0.493
) ( )
= 4.4x10−4 𝑐𝑚/𝑠
3049.42% 6.2
1 + 0.49
1.99𝑥104 (
41
▪ The hydraulic conductivity in some layers is often much greater than in
others, so groundwater flows horizontally much more easily than
vertically.
▪ Such soils are said to be anisotropic with respect to hydraulic
conductivity, so we need to determine two values of k: kx is the
horizontal hydraulic conductivity and kz is the vertical hydraulic
conductivity of the layered soil.
▪ The equivalent horizontal and vertical hydraulic conductivity is
determined for layered soils by considering two constant-head
permeameter tests of a layered soil.
i=i1=i2=i3=Δh/L
Q=Q1=Q2=Q3=kxiA
𝑘𝑥 =
σ 𝑘𝑖 𝐻𝑖
σ 𝐻𝑖
▪ ki is the hydraulic conductivity of horizontal stratum i and Hi is the
thickness of horizontal stratum i.
42
▪ For the case of flow normal to the soil layering, the total head loss
in the system must be equal to the sum of the head loss in each
layer. And the flow in each layer must be the same and equal to the
total flow.
Δ h= Δh1+ Δh2+ Δh3
Q=Q1=Q2=Q3=kziA
σ 𝐻𝑖
𝑘𝑧 =
𝐻
σ( 𝑖 )
𝑘𝑖
▪ Many groundwater problems involve alluvial soils because they
are often near rivers abd often have shallow groundwater tables.
▪ Most alluvial soils have horizontal stratifications, so kx>kz. Varved
clays also have kx>kz.
43
▪ Example 8: A certain varved clay consists of alternating horizontal layers of silt and clay. The silt
layers are 5 mm thick and have k=3x10-4 cm/s; the clay layers are 20 mm thick and have k=6x10-7
cm/s. Compute kx and kz.
44
▪ Solution 8:
𝑐𝑚
3𝑥10−4
σ 𝑘𝑖 𝐻𝑖
𝑠
𝑘𝑥 =
=
σ 𝐻𝑖
𝑘𝑧 =
0.5 𝑐𝑚 + 6𝑥10−7
0.5 𝑐𝑚 + 2 𝑐𝑚
𝑐𝑚
𝑠
2 𝑐𝑚
= 6𝑥10−5 𝑐𝑚/𝑠
σ 𝐻𝑖
0.5 𝑐𝑚 + 2 𝑐𝑚
=
= 7𝑥10−7 𝑐𝑚/𝑠
𝐻
0.5 𝑐𝑚
2 𝑐𝑚
σ( 𝑖 )
+
𝑐𝑚
𝑐𝑚
𝑘𝑖
3𝑥10−4
6𝑥10−7
𝑠
𝑠
45