ENV-2E1Y - Fluvial Geomorphology
2004 - 2005
Slope Stability and Related Topics
Flownet of seepage of water through soil around an obstruction
Section 2
Seepage, Flow of Water, Pore Water Pressures
ENV-2E1Y Fluvial Geomorphology 2004 – 2005
N. K. Tovey
Section 2
Slope Stability and Related Topics
2. Seepage of Water in Soils/Permeability
2.1 Introduction
Both the first an second are clearly important (the first arises
directly from the depth below the water table, whereas there is
ample evidence (in the form of springs to indicate water flow
through soils).
NOTE: FIRST A HEALTH WARNING!!!! There are
some sections of this handout which are in shaded boxes.
These are not essential parts, but complement the main
course. Thus for some of you who are doing ENV-2B31
(Mathematics I) you may object the simplistic approach
sometimes used in the handout. There is a more rigorous
approach for you in these boxes. In other cases, the boxes
show additional information which may be derived which may
be of use in other courses (e.g. Hydrogeology). Please consult
the note before each box.
What about the velocity head?
Typical velocities even in coarse sands will rarely exceed 10
mm s-1, and the magnitude of the last term in this extreme
case will be (remember to convert to metres!!!!):( 0. 01)2
 0. 000005 m
2 x 10
For another set of notes for this section see the University
of West of England WEB Site on the topic.
in terms of total head or 0.00005
kPa in pressure terms.
This is exceedingly small, and in most soils, the velocity will
be many orders of magnitude less than this so in future we can
conveniently neglect the velocity head term.
http://fbe.uwe.ac.uk/public/geocal/SoilMech/water/water.htm
A knowledge of the factors affecting the flow of water through
soils is important as the permeability of a soil affects the way
in which a soil consolidates which in turn affects its
mechanical properties. Equally water pressures may build up
within the soil and as seen the demonstrations can greatly
affect the ability of a soil to resist shearing. There are three
component parts to the water pressure:-
2.2 Hydraulic Gradient
Associated with water pressure is the hydraulic gradient
which is a measure how fast the water pressure is changing.
This in turn affects how fast water will flow and what the
immediate pressure within the soil will be.
i) that pressure arising from a static head
Now consider flow between two points A and B (see Fig. 2.1).
[ at any point at a height Z above the measuring
datum, this pressure will be w Z where w is
the unit weight of water.]
The hydraulic gradient is defined as the rate of change of head
of water with distance (in the direction of measurement)
ii) excess pore water pressure (i.e. a pressure head
differential which actually causes water flow [ this
has the symbol
u]
 w v2
2g
iii) a velocity head and equals
The total pore water pressure (often abbreviated to pwp)
wZ  u 
=
 w v2
2g
.......2.1
For those who have done the Hydrology or Oceanography
options you may already be familiar with Bernoulli's equation
of fluid flow i.e.
H
Z
total = position
head
head

+
u
w
pressure
head
v2
2g

+
Fig. 2.1 Flow of water in a channel of simple horizontal cross
section
Since the flow is horizontal, there is no effect from the static
head of water.
........2.2
velocity
head
The head of water at A is h1
and at B h2
i.e. equation 2.1 is Bernoulli's equation expressed in terms of
pressure rather than head of water.
[To convert from
equation 2.1 to 2.2 all we need do is divide by w.
The excess pressure of the water at A (as the velocity is small)
How significant are these three terms in water flow in soils?
and at B
is
14
w h1
w h2
ENV-2E1Y Fluvial Geomorphology 2004 – 2005
N. K. Tovey
The gradient of the head drop (or pressure drop) is known as
the hydraulic gradient (i). In this simple situation, the
hydraulic gradient is:-
Note the -ve sign as h decreases as we go in direction of
flow of water,
i.e. from B to A.
h1  h 2

i
NOTE: The hydraulic gradient as defined above is
dimensionless (i.e. has no units). In some other
disciplines, it is defined in terms of pressure (rather
than head) and thus is no longer dimensionless. In
this case,
i
= w (Z + h)
the total pwp at A is
= uw
while the total pwp at B is
= uw + duw
= w (Z + dZ + h + dh)
i.e. the difference in head is
 w (h1  h2 )

duw =
w (dZ +
dh)
w dZ is the pressure arising from the head difference
In this case, there are units associated with the
hydraulic gradient (i.e. kN m-3). To keep things
simple we shall use the first definition in this course.
wdh
is the pressure arising from the excess
pressure in the standpipes which will cause flow of water
from B to A.
The above is a simple description of what how to measure the
hydraulic gradient. Strictly speaking we should be talking in
terms of the differential coefficient:-
i 
Section 2
In the alternative definition of hydraulic gradient used in
some disciplines,
dh
ds
i 
du
ds
and the units are kN.m-3
where s refers to a general direction of measurement.
2.3 The Permeameter
For most of you it is sufficient to accept the above definition,
but if you want the full derivation it is given in the following
box (see WARNING given in introduction about these boxes).
The permeability of a soil dictates how quickly water will flow
within the soil, and more importantly how quickly excess
water pressure will dissipate. The permeability of a soil may
be measured with a Permeameter. For silts and sands it is
common to use a constant head Permeameter (Fig. 2.3). For
clays the permeability is so low and a falling head
Permeameter is used.
2.3.1 Constant Head Permeameter (Fig. 2.3)
The apparatus consists of a vertical cylindrical tube in which
is placed the sample. Below and above the sample are porous
stones. Water is fed from a supply to a constant head
reservoir and to the base of the sample. After passing through
the sample the water flows into a measuring cylinder which is
used to estimate the flow rate. In the more accurate
permeameters there are two pressure at a fixed distance apart
in the cylinder wall. Fine bore capillary tubes are inserted into
sample and connected to manometers to measure water
pressure at the two points, and thus the hydraulic gradient
may be determined.
Fig. 2.2 Illustration of two types of water pressure.
Let A and B be two points separated by a small distance
ds, and let the excess p.w.p. be du = w dh
then the hydraulic gradient (denoted by i) is given by:-
i
In the limit as ds
i 
dh
ds
--->
or
The experiment starts with the constant head reservoir at a
given level. Water is allowed to pass through the sample for a
few minutes (depending on the nature of the material under
test) until a steady state is reached. The level of water in the
pressure tappings is measured and water is allowed to flow
into the measuring cylinder for a known period of time.
h
s
0
i 
1 du
 w ds
……….(2.3)
The flow rate Q may be estimated (volume of water collected
in cylinder divided by time). We also measure the total
internal cross-sectional area of the cylinder (At) .
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ENV-2E1Y Fluvial Geomorphology 2004 – 2005
N. K. Tovey
Section 2
of water, the density of water, and a shape factor (i.e. relating
to the shape of the voids). Since this is an introductory course,
and the use of non-dimensional parameters for hydraulic
gradient leads to a simpler set of units, this will be used in this
course and is consistent with those used in the textbooks on
the reading list..
2.5 Experimental Results from Permeameter
The results of a replication of Darcy's experiment are shown
as line A. For many soils, the linearity of the line is good,
confirming Darcy's Law, but in peats, the data is often very
non-linear.
Fig. 2.3 Constant Head Permeameter
The velocity va of the water as it flows in the part of the
cylinder above the sample may be obtained from:-
va 
Q
At
........................................(2.4)
va is also known as the apparent velocity (i.e. the
velocity the water would have when passing through the soil if
the solids occupied zero volume. It is less than the actual
velocity.
Fig.
2.4
Experimental Permeability Results on Leighton
Buzzard Sand (after Schofield and Wroth, 1968).
2.4 Darcy's Law
As the head is increased, the upward pressure gradient across
the sample increases, and eventually at point C the upward
force equals the downwards force from self-weight and
LIQUEFACTION occurs (i.e. we have created a quicksand).
The soil "boils" at this point. If we now reduce the pressure
gradient by lowering the constant head reservoir, we will find
that the sample will settle again, but will occupy a greater
volume than previously.
In 1856 Darcy, using equipment similar to that described
above and continued the experiment by raising the level of the
constant head reservoir while keeping the height of the sample
constant. Two points were noted. First, the flow rate
increased, and secondly the hydraulic gradient also increased.
More readings were taken with further increases in the height
of the constant head reservoir.
Darcy found that the apparent velocity
to the hydraulic gradient
i.e.
We may now repeat the experiment, but this time, although
still displaying a linear trend, the points lie on a line with a
higher gradient. As might be expected, the rate of flow of
water in the loose sample is greater than for the dense sample
and hence the coefficient of permeability (k) has increased.
The results shown in Fig. 2.4 were obtained for the same
Leighton Buzzard Sand as used in most of the demonstrations
described in section 1.
(v) was proportional
(i)
va  k i   k
dh
ds
.
.........(2.5)
k is known as the coefficient of permeability
NOTE: k has the units of velocity (i.e. m.s-1) because the
hydraulic gradient in non-dimensional. In some branches of
Science, the hydraulic gradient is measured in terms of the
pressure (rather than head). In such cases, permeability is
measured in units of m3.s-1.kN-1. The two sets of readings
differ only by a factor of the unit weight of water. There is
further confusion in that the term hydraulic conductivity is
sometimes used instead of permeability.
The cylinder containing the sample has a uniform crosssection and thus there is a uniform hydraulic gradient within
the soil sample and equals up the cylinder and it is equal to
i
h
s
where dh is the difference in the level of the pressure
tappings,
and ds is the distance between the tappings.
This latter term, used in Hydrogeology, differs from
permeability in that it also attempts to allow for the viscosity
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ENV-2E1Y Fluvial Geomorphology 2004 – 2005
N. K. Tovey
The volume of the reduction in water in the capillary tube in a
time t will equal the flow of water through the sample in the
same period.
The initial voids ratio for the (initially medium dense) sand
may be calculated as follows:Let m be the total mass of sand
and
Section 2
a
 its length
then volume occupied = A 
and volume of sand grains
h
t
dh A
k  dt
h
L0
h
1
i .e.  a 
m
= Gs w
where Gs is the specific gravity of the solid particles.
dh khA

dt L
1
2
h
aL
k  2 .3
log10 0 ...............(2.7 )
At 1
h1
Hence
A  m G 
AGs  w
s w
e

1
m
m
Gs  w
...........(2.6)
Thus to measure the permeability using a falling head
Permeameter, the sample is enclosed in the sample tube and
the capillary tube is filled so that the water level is nearly at
the top. A valve remains closed preventing the water flowing
through the sample until the experiment is ready to begin. The
height of the water in the capillary tube is measured and a stop
clock is started as the valve is opened. The time for the level
in the capillary tube to fall to a second level is noted. The
measurements needed are thus:- two heights on the capillary
tube, the diameter of the capillary tube, the length and
diameter of the sample.
2.6 Falling Head Permeameter (Fig. 2.5).
When the permeability of fine grained silts/clays is to be
determined it is found that the flow rate is so low that it cannot
be measured accurately. A falling head Permeameter is then
employed.
2.7 Formation of a Quicksand - Piping
Let a be cross-section of capillary tube
A be cross-section of sample tube
L be length of sample tube
ho be initial height of water at t = 0
h be height at time t
and h1 be height at time t1
Referring back to graph in section 2.5. When point C was
reached the sand appeared to boil as in a quicksand and piping
occurred.
The seepage force is that force exerted by the water seeping
through the soil.
In the situation above, at point C, the seepage force equals the
submerged weight of the sand.
Let critical value of dh at which the quicksand occurs be
dhcrit. the corresponding critical hydraulic gradient (icrit)
will be given by:va 
the upwards force
Q
At
= Atw dhcrit =
whilst the downwards force
At du
= At'dz
thus the critical hydraulic gradient occurs when
i crit  
Gs  1
h crit
A '
 t

z
At  w
1e
.........(2.8)
Fig. 2.5 Falling Head Permeameter
During piping the volume occupied by the sand increases and
consequently the voids ratio. The second line (loose in Fig.
2.4) was obtained by repeating the experiment at the new
voids ratio.
From Darcy
Q = kiA
17
ENV-2E1Y Fluvial Geomorphology 2004 – 2005
N. K. Tovey
At a condition of piping, the sand becomes completely
buoyant and it is as though the effect of gravity had been
reduced to zero.
1) estimations of pump capacities in water resources, and
2) stability calculations on slopes.
The flow of water through soils is directly analogous to two
other processes namely HEAT FLOW and the flow of
electricity. Those of you who have done the Energy
Conservation (ENV-2D02) course will already be familiar
with the equations defining heat flow through a component of
a building (such as the walls). Those of you who did Physics
at school or who have done Geophysics will be familiar with
electrical conductivity.
If flow of water had been downwards, the effect of gravity
would be increased. This effect can be used in model analysis
to study slope stability using models of a slope..
2.8 Typical Values of k
gravels
k > 10 mms -1
sands
10 mms-1 > k > 10-2 mms-1
silts
10-2 mms-1 > k > 10-5 mms-1
Section 2
These analogies are helpful, as we can model both heat flow
and water flow using an electrical analogue model which is
easier to change than is either a brick wall or a layer of
sediment.
k < 10 -5 mms-1
clays
In HEAT FLOW
kA
( 1   2 )

where Q is the heat flow rate
Q
2.9 Actual Seepage Velocity
The actual velocity of seepage through the pores must be
greater than the apparent velocity as calculated by equation
2.4.
1 is the int ernal temperature
 2 is the external temperature
A is the cross  sec tion area
Q
va 
At
 is the thickness of wall
and k is the thermal conductivity
........................................(2.4
repeated)
In the FLOW of ELECTRICITY
If vs is the actual velocity, and
section of the voids
Av
is the actual cross
kA
( E1  E2 )

where I is the heat flow
I
then Q = vsAv = vaAt - i.e. the water flowing through
the soil pores must equal the apparent flow mentioned earlier.
This is the continuity equation..
i.e.
..........(2.9)
E1 is the inlet potential
E2 is the outlet potential
A is the cross  sec tion area
A
V
v
1e
vs  va t  va t  a  va
Av
Vv
n
e
 is the thickness of wall
and k is the electrical conductivity
In the FLOW of WATER in SOILS
where Vt and Vv are the volume of the total sample and
that of the voids respectively.
for dense Leighton Buzzard Sand
kA
( h1  h2 )

where Q is the flow rate
Q
vs is approximately 2.6va
h1 is the int let head
NOTE: In the symbols, upper case V is used for volumes,
but lower case v is used for velocities.
h2 is the outlet head
A is the cross  sec tion area
 is the thickness of wall
2.10 Flow of Water in Soils
and k is the permeability
In the permeameter the shape of the sample is cylindrical and
analysis of the flow of water through the soil is simple. In
nature, however, the boundary conditions are never as simple
and ways must be found to enable estimates to be made of
seepage.
The solution of the problem through a cylinder is
straightforward and can become quite complex in more
general situations. The flow of water (or flow of heat, or flow
of electricity) is governed by Laplace's equations and suitable
solutions to these equations. There are five ways in which
solutions may be achieved:-
This is important for two reasons:
18
N. K. Tovey
ENV-2E1Y Fluvial Geomorphology 2004 – 2005
Section 2
This is the two-dimensional version of Laplace's equation,
and solution to seepage problems from which we can
obtain the water pressure at any point require solution of
this fundamental equation within appropriate boundary
conditions. In this course we shall be using graphical
solutions,
but if you are doing ENV-2B31
(Mathematics), you may like to attempt a mathematical
solution of the problem to simple boundary value
conditions. The example shown in section 2.14 may also
be solved provided that you transform the co-ordinates
using conformal transformation beforehand.
1) Mathematical solutions
a) exact solutions for certain simple
situations
b) solutions by successive approximate e.g. relaxation methods
2) Graphical solutions
3) Solutions using the electrical analogue
4) Solutions using models
Mathematical solutions are beyond the scope of this course
(although anyone doing the Maths options may wish to follow
section 2.11, and then attempt some solutions as applications
for the Maths Course).
There must be continuity: i.e. the water flowing in across
the bottom and left hand border must equal the water
flowing out across the other two borders.
The Relaxation Method requires the development of computer
software. An example is already in use in dynamic heat flow
computations in the Energy Conservation Course as a self
contained computer package, and it is hope to adapt this for
use in water flow in soils for this course in the next few years.
v x z  v z x  ( v x 
i. e.
Graphical Solutions are the method we shall adopt in this
course
v x
v z
. x )z  ( v z 
. z )x
x
z
v x
v z

0
x
z
. . . . . . . . . .( 2. 10 )
for convenience we will make the substitution
Electrical Analogue method are nice to use in practicals, and
were in fact used in the past in this course when more time
was devoted to the topic.

i
k
x
x
i. e.
Solutions using models. These are expensive to construct, but
nevertheless can give useful information particularly in
complex situations when it becomes difficult to define the
conditions easily in mathematical terms. A model of water
flow around an obstruction was constructed many years ago
and is now used in the Hydrogeology option.
thus
vx  
and

x
 = ki

i
k
z
z
and v z  

z
Substituting for vx in Equation (2.10) we get
 2  2

 0 ............(2.11)
x 2  z 2
2.11 Flow Equations - 2-D case
This section contains the Mathematical derivation of the basic
equations governing the flow of water in two-dimensions.
This is strictly for those who are mathematically inclined - for
those who are not - skip the following highlighted box.
Note the analogy with flow of electricity:
 2V
 2V

x 2
z 2

0
where V is the electrical potential or voltage
or heat flow
 2
 2

x 2
z 2

0
where  = temperature
By analogy  is termed the potential in the flow
of water.
Fig. 2.5 Schematic of two-dimensional flow of water
As solutions to the above equations already
exist for heat/electricty flow, we may use these
soltuions to solve flow of water problems.
19
2.12 Graphical Solutions - Flow Nets (Fig.
N. K. Tovey
ENV-2E1Y Fluvial Geomorphology 2004 – 2005
Section 2
2.12 Graphical Solutions - Flow Nets (Fig. 2.6).
In this course we shall only concern ourselves with the
graphical solution to Laplace's Equation. While this may at
first sight seem rather crude, it is nonetheless very effective
and can be used for all shapes of flow channel provided that a
little care is used when drawing. We do this by drawing
flownets - i.e. we draw lines parallel to the line of flow of
water (flowlines), and lines at right angles which are lines of
equal pressure or equipotentials. Experience certainly helps,
and you may wish to practice by drawing arbitrary shapes to
define boundaries and attempt to fill in the appropriate flow
net.
In the permeameter we had a situation where flowlines are
vertical and lines of equal pressure (equi-potentials) are
horizontal.
Fig. 2.7 Asymmetric Flownet
Note: equipotentials are still orthogonal to flow
lines and it can be shown that all regions bounded by
two adjacent flowlines and two adjacent equipotentials
are curvilinear rectangles of similar proportion (or in
most cases squares - See figure 2.8). The following is
merely a proof of the this, you may skip the following
box unless you are particularly interested.
Consider a single flow channel initially of width a1 and a
spacing between equipotentials of b1. The potential drop
between all equipotentials is Dh. At a second point in
the flow channel where its width is now a2 the distance
between the equipotentials is b2. By continuity we must
have the same amount of water flowing in both parts of
the flow channel.
Fig. 2.6 Flowlines and Equipotentials in a simple situation
where all flow lines are parallel
Two points to note:1) flowlines and equipotentials are at right angles to
one another.
2) the cylinder walls are also flowlines.
The distances  between the equipotentials are equal and thus
the head drops between the equipotentials are also equal.
Fig. 2.8
Each flow channel is defined by two adjacent flow lines define
the region in which water moves, i.e. water will not move from
one channel to another (apart from a minor effect from
diffusion).
Let a1 be the width of channel when velocity is v1
and a2 be the width of channel when velocity is v2
then q = v1a1 = v2a2 = v3a3
What happens if cylinder is not of constant cross-section? (see
Fig. 2.7)
but v1 = k i1 = 
By symmetry, flow lines must diverge, but since the package
of water A must still remain within the flowlines c - c, d - d
its velocity will change and become less. If the dashed lines
represent equipotentials, then they must become progressively
further apart as one proceeds towards the right.
thus
20
q   k h
k h
k h
and v2 = k i2 = 
b1
b2
a1
a
  kh 2
b1
b2
ENV-2E1Y Fluvial Geomorphology 2004 – 2005
N. K. Tovey
Hence for equal pressure drops we must have
a1
a
 2
b1
b2
Section 2
nf q f 
the total seepage =
i.e. the ratio is constant.
n f kH
nd
..............
=============
(2.12)
Thus we must draw rectangles of constant proportion.
However this is difficult when we have curvilinear figures
unless a = b (i.e. we have squares - a special case of the above
requirement).
In other words if we require to determine the total volume of
water seeping we need only draw a flow net and count the
number of pressure drops and flowlines.
General Flownets solution
To work out the pressure at any point, which is what we really
want, the procedure is equally simple and will be illustrated
with reference to a particular example (Section 2.13).
- (See figure 2.9)
Solutions are relatively straightforward. We need:1) draw the appropriate flownet
2) count the number of pressure drops in the flow net (over
the relevant distance)
3) count the number of flow lines
4) do a simple calculation as given by equation 2.12 below.
Let total pressure drop between AB and CD be
there be nd pressure drops and nf flow lines.
2.13 Seepage around an obstruction - (See figure
2.10)
The example shown in Fig. 2.10 has a vertical obstruction
which is preventing the normal flow of water. It also
represents the model rig in the Soil Mechanics Lab which was
formerly used as a practical in the fore-runner to this course,
but recently has been used in the Hydrology courses.
H and let
Assuming uniform permeability k and h to be pressure drop
across one square of side.
h
H


nd
On one side of the obstruction, the water level is maintained
at a high level while the water level on the other side is at
approximately the level of the soil. Water will flow in around
the obstruction and upwards on the downward side of it. Thus
there will be an upward flow of water immediately
downstream of the obstruction, and if the upward force of the
water exceeds the downward submerged weight of the column
of soil, then piping will occur (i.e. a quicksand will form).
where qf is the flow
per unit cross-section
By Darcy' s Law v  ki 
and q f  kia 
kH
nd
kHa
nd
but a  
and
a x 1 is the crosssection between flow
lines.
We first draw the flownet and note that piping will occur if
seepage force at A >= buoyant weight of column of sand.
Hence the total flow in a single flow channel is given by:-
qf 
If the difference in the height of the water on the two sides of
the obstruction is H, then the pressure drop between two
equipotential lines will be wH / nd.
kH
nd
and if the number of flowlines is
nf
this space is left blank for notes:
21
N. K. Tovey
ENV-2E1Y Fluvial Geomorphology 2004 – 2005
Section 2
Fig. 2.9 Generalised Flownet
22
N. K. Tovey
ENV-2E1Y Fluvial Geomorphology 2004 – 2005
Section 2
Fig. 2.10 Flow of water around an obstruction
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ENV-2E1Y Fluvial Geomorphology 2004 – 2005
N. K. Tovey
NOTE: we must include w as we are now dealing with
if we know the voids ratio, then from as this may indicate
whether fine material is washed out from the soil because of
high actual seepage velocities.
pressure, and not merely head of water.
If the number of equipotential drops from the base of the
obstruction to the surface of the soil on the downstream side is
Nab.
Then the upward seepage force
vs  va
From equation (2.9)
N ab w H
nd
=
1e
e
At any point within our flow net we can estimate the hydraulic
gradient by dividing the head drop between two equipotentials
by the distance between them
' 
and the downward force of the soil =
where  is
the depth of penetration of the obstruction into the soil. A
i.e.
H
nd
Once again the following is not essential to the course, but
may be considered if you wish.
quicksand will occur if
N ab w H
 '
nd
hence
Section 2
Alternatively we may measure the actual seepage
velocity on a model determine permeability by
rearranging equation 2.16.
Nab H '

..................(2.14)
nd
w
i.e.
k
As an example in the use of a factor of safety, we may predict
such a factor (Fs) as follows
v snd e
.
H 1e
and by Darcy
actual downward force of the soil
-----------------------------------------------force required to just resist seepage force
Factor of safety =
Fs 
 '  nd
.
 w H N ab
va 
. . . . . . . . . . . . . . . . . . . . . . . . .( 2. 15 )
kH
nd
i. e. v s 
1  e kH
.
e
n d  …(2.16)
2.14 Flow nets (Summary)
Rules for drawing flow nets:-
In the above example,
and noting that very approximately
Fs 
i.e. the distance

Nab 3.5
nd = 10 and
1) All impervious boundaries are flow lines.
2) All permeable boundaries are equipotentials
3) Phreatic surface - pressure is atmospheric, i.e. excess
pressure is zero.
A further requirement is that the change in head
between adjacent equipotentials equals the vertical
distance between the points on the phreatic surface.
4) All equipotentials are at right angles to flow lines
5) All parts of the flow net must have the same geometric
proportions (e.g. square or similarly shaped rectangles).
6) Good approximations can be obtained with 4 - 6 flow
channels. More accurate results are possible with higher
numbers of flow channels, but the time taken goes up in
proportion to the number of
channels. The extra
precision is usually not worth the extra effort.
' = w
10 
3. 5 H
must exceed 0.35 times the difference in
head of water.
The following box is not essential to the course, however it is
complementary to the course and to aspects of the Hydrology
courses.
We are sometimes interested in the actual seepage velocity vs
If we wished to work out the quantity of water flowing
around the obstruction then from equation 2.12, and noting
that nd = 10 and the number of flow channels is
approximately 4.25.
Q  n f q f  kH
nf
nd
Note:
· a difficulty in flownet analysis is the determination of the
top flow line in flow down a slope or through an earth
dam.
· It is difficult to give hard and fast rules - best try and see.
However, a good approximation is a parabola.
· In regions of sharply varying flow channels, there are
more advanced methods which allow the subdivision of
squares into quarter size squares just at the key points,
but these techniques are beyond the scope of this course.
 0. 425 kH
If, for example, the permeability
field situation, H = 10 m
( k ) = 1 mms-1 and if a
Q = 0.425 x 10-3 x 10 = 4.25 litres.s-1.
===============
24
N. K. Tovey
ENV-2E1Y Fluvial Geomorphology 2004 – 2005
The next two sections (i.e. 2.15 and 2.16) of this handout
are not essential for the course, but you should appreciate
in general terms what is going on. You will not be
expected to do numeric problems based on the following.
Section 2
2.16 Flow through soil having differing
permeabilities in x and z directions
The above discussion has centred on the assumption that
permeability is uniform in all directions. Not
infrequently there is a difference, through lamination,
between the horizontal and vertical permeabilites.
2.15 Flow between regions of different permeability
Flowlines and equipotentials in Fig. 2.11 in Region A form
squares of side a. In region b, these become rectangles of
size b x  . Let permeability be ka and kb then by
continuity flow in channel must be same in both A and B.
Hence:-
Let these permeabilities be kx and kz respectively.
We may drawn an equivalent flow net if we first adjust
the scale of the original diagram so that we multiply all
horizontal distances by a factor of
kz
kx
(See figure 2.12)
If we do this, we can then still use square (or rather
curvilnear square) flow nets on the transformed diagramd
diagram provided that we recognise that the effective
permeability is also modified as follows:The equivalent permeability is then k = (kxkz)
Fig. 2.11 Water flow across a boundary between regions of
different permeability
q
a
b
k a h  k b h
a

and AB 
so
The geometric modification is a form of conformal
transformation (which those doing ENV-2B31 may have
come across). Indeed the problem of the seepage around
an obstruction may also be solved by transforming the
actual scale into a square net using an appropriate
conformal transformation, although in this case it does
require elliptical functions in the transformation.
ka
b

kb

a
b
a


and AC 

sin  a
sin  b
cos  a
cos  b
ka
tan  b
b


kb

tan  a
Thus in passing between two zones flowlines are refracted
such that the ratio of the tangents of angles of incidence are
proportional to the permeabilities. Note: when drawing flow
nets at boundaries, squares in one zone will become
rectangles in the second zone.
For details see pages 59-64 of Critical State Soil
Mechanics by Schofield and Wroth (1968) - publisher
McGraw Hill - It is in the Library. and I also have a copy
for those interested
so
Fig. 2.11 Example of transformation to allow for anisotropic conditions
This pressure arises from both the static and excess heads.
For an obstruction which has a flat base (which is horizontal,
the static head will be constant along its length and equal to
the depth of the base below the water level on the down stream
end. The excess head will vary along the underside of the
obstruction.
2.17 Uplift on Obstructions
When water seeps under a large obstruction such as a boulder
which is partly buried in the soil, the obstruction experiences
an uplift from the total water pressure exerted on the base.
25
N. K. Tovey
ENV-2E1Y Fluvial Geomorphology 2004 – 2005
Section 2
Fig. 2.13 Uplift on a rectangular boulder by water passing underneath
If the total uplift force exceeds the downward force from the
self weight of the obstruction, then the object will be displaced
downstream. To assess the likelihood of this happening, a
flownet is drawn for the water seeping through the soil. A
graph is plotted with the x-axis as the distance along the base
from the upstream face, and the water pressure as gauged
from the equipotentials (+ the static head) as the y-axis. The
total force is then evaluated by working out the area under the
curve. If this exceeds the with of the obstruction then it will
be displaced
Because the base of the obstruction is 2m below the surface
the uplift force from the static head is 2w multiplied by
width (i.e. 6w kN per metre length). From the excess head
(see graph), the upward force is the area under the curve
multiplied by w. In this example (since the line is nearly
linear), the upward force = 6w kN per metre length, i.e. in
this case it equals the static head uplift. The total uplift will
then be the sum of these two components, i.e. 12w kN m-1.
The pressure distribution shows the excess head an has been
drawn from position of intersection of equipotentials along the
base of the obstruction. The equipotentials are nearly equidistant from each other in this region and the distribution is
nearly linear. By counting equipotentials, the head at the
upstream head is 0.75 of total head, and at the down stream
end it is 0.25 of the total head.
This uplift will considerably reduce the ability of the
obstruction to resist movement through the pressure of water
(and has significance both as potential boulder blockages in a
river and also as man-made drop structure built in river
engineering works to dissipate energy (see RDH's part of the
Course).
There may also be erosion from the possible
quicksand which may form at the down stream end of the
obstruction.
space for further notes.
26