Flow nets

Soil Mechanics Flow Nets page 1
Contents of this chapter :
CHAPITRE 10. FLOW NETS .......................................................................................................1
10.1 INTRODUCTION.......................................................................................................................1
10.2 REPRESENTATION OF SOLUTION .............................................................................................2
10.3 SOME GEOMETRIC PROPERTIES OF FLOW NETS ......................................................................3
10.4 COMMON BOUNDARY CONDITIONS...........................................................................................4
10.4.1 SUBMERGED SOIL BOUNDARY = EQUIPOTENTIAL...................................................................4
10.4.2 IMPERMEABLE BOUNDARY = FLOW LINE................................................................................5
10.4.3 LINE OF CONSTANT PORE WATER PRESSURE .......................................................................6
10.5 PROCEDURE FOR DRAWING FLOW NETS .................................................................................7
10.6 CALCULATION OF QUANTITIES OF INTEREST FROM FLOW NETS ................................................7
10.6.1 CALCULATION OF FLOW .......................................................................................................7
10.6.2 CALCULATION OF PORE WATER PRESSURE ..........................................................................8
10.6.3 EXERCISE ...........................................................................................................................8
10.6.4 EXAMPLE – STRANDED VESSEL RESCUE............................................................................13
10.6.5 EXERCISE .........................................................................................................................14
10.7 FLOW NETS FOR ANISOTROPIC MATERIALS...........................................................................15
10.7.1 INTRODUCTION..................................................................................................................15
10.7.2 FLOW NETS FOR SOIL WITH ANISOTROPIC PERMEABILITY......................................................15
10.8 EFFECT OF WATER FLOW IN A SOIL MASS...............................................................................17
10.8.2 SEEPAGE FORCE : GENERAL FORMULATION.........................................................................19
10.9 HYDRAULIC HEAVE OR PIPING ..............................................................................................20
10.9.1 EXERCISE .........................................................................................................................21
Chapitre 10. Flow Nets1
10.1 Introduction
Let us consider a state of plane seepage as for example in the earth dam shown in Figure 1.
1
Réseaux d'écoulement

Drainage
blanket
Phreatic line
Unsaturated
Soil
Flow of waterz
x
Fig. 1 Flow through an earth dam
For an isotropic material, the head h of the water flowing into the soil satisfies Laplace's equations,
thus analysis involves the solution of:
02
2
2
2
=+
z
h
x
h
∂
∂
∂
∂
subject to certain boundary conditions.
10.2 Representation of Solution
At every, point (x,z) where there is flow there will be a value of head h(x,z). In order to represent
these values we draw contours of equal head (red lines) as shown on Figure 2.
Equipotential
(EP)Flow line
(FL)
Fig.2 Flow lines and equipotentials
These lines are called equipotentials
2
. On an equipotential (EP). by definition:
constant),( =zxh
2
équipotentielles

It is also useful in visualising the flow in a soil to plot the flow lines
3
(blue lines noted FL), these are
lines that are tangential to the flow at a given point and are illustrated in Figure 2.
It can be seen from Fig. (2) that the flow lines and equipotentials are orthogonal.
10.3 Some Geometric Properties of Flow Nets
Let us consider another example of seepage under a sheet pile wall. The flow net is represented
on figure 3.
On Fig. 3, each interval between two equipotentials corresponds to a head loss ∆h equal to 1/Nd
of the total head loss h through the soil :
∆h = h / Nd
Where Nd = Total number of equipotentials.
Consider a pair of flow lines, clearly the flow through this flow tube must be constant and so as the
tube narrows the velocity must increase.
Let us consider the flow through abcd delimited by two flow lines and two equipotentials.
The hydraulic gradient is :
i = ∆h/l1 = h / (Nd. l1)
Where l1 is the distance between the two equipotentials.
3
Lignes de courant
Figure 3
Observation pipes

Let l2 be the distance between the two flow lines.
Applying Darcy's law, the velocity in the tube abcd is:
v = k . i = k . h / (Nd. l1)
The flow passing into abcd, per m width of soil, is :
∆qabcd = Area x velocity = l2.1. k .h / (Nd. l1)
If we draw the flow net taking l2 = l1 (a "squared" mesh net is more convenient to draw : it is
possible to draw an inscribed circle, see Fig.4), ∆qabcd = k .h / Nd (the flow through any
quadrilateral of the flow net is thus the same as the one through abcd ).
The total flow will then be equal to :
Q = Nf . ∆q = k .h . Nf / Nd
Where Nf = Number of flow tubes.
From that equation, one can see that the flow is function of the ratio Nf / Nd and thus if the flow net
is refined by dividing each cell in four smaller cells, the ratio will remain unchanged. That means
that Q is independent of the refinement of the flow net! It is thus easy to determine quickly an
estimate of the flow of water passing under a dam or wall.
To calculate quantities of interest, that is the flow and pore water pressures, a flow net must be
drawn.
The flow net must consist of two families of orthogonal lines that ideally define a square mesh, and
that also satisfy the boundary conditions.
The three most common boundary conditions are discussed below.
10.4 Common boundary conditions
10.4.1 Submerged Soil Boundary = Equipotential
Consider the submerged soil boundary shown in Figure 5
Figure 4 Inscribing Circles in a Flow Net

Water
Datum
H-z
z
H
Soil
Figure 5 Equipotential boundary
The head at the indicated position is calculated as follows:
Hz
zH
h
zHu
z
u
h
w
w
ww
w
w
=+
−
=
−=
+=
γ
γ
γ
γ
)(
so
)(
now
That is, the head is constant for any value of z, which is by definition an equipotential.
Alternatively, this could have been determined by considering imaginary observation pipes placed
at the soil boundary, as for every point the water level in the standpipe would be the same as the
water level.
A consequence of this is that all the flow lines arrive perpendicularly to a submerged soil
boundary.
The upstream face of the dam shown in Figures 1 and 2 is an example of this situation.
10.4.2 Impermeable Boundary = Flow Line
At a boundary between permeable and impermeable material the velocity normal to the boundary
must be zero since otherwise there would be water flowing into or out of the impermeable material,
this is illustrated in Figure 6.

Permeable Soil
Flow Line
vn=0
vt
Im permeable M aterial
Figure 6 Flow line boundary
A consequence of this is that all the equipotentials arrive perpendicularly to an impermeable soil
boundary.
The phreatic surface shown in Figures 1 and 2 is also a flow line marking the boundary of the flow
net. A phreatic surface is also a line of constant (zero) pore water pressure as discussed below.
10.4.3 Line of Constant Pore Water Pressure
Sometimes a portion of saturated soil is in contact with air and so the pore water pressure of the
water just beneath that surface is atmospheric. The phreatic surface shown in Figure 7 below is an
example of such a condition. We can show from the expression for head in terms of pore water
pressure that equipotentials intersecting a line of constant pore water pressure do so at equal
vertical intervals as follows:
zh
u
z
u
h
z
u
h
w
w
w
w
w
∆=∆
=∆
∆+
∆
=∆
+=
soand
0now
thus
γ
γ

Figure 7 Constant pore water pressure boundary
10.5 Procedure for Drawing Flow Nets
1. Mark all boundary conditions. Determine the head at the inlet and outlet of the flow net.
2. Draw a coarse net which is consistent with the boundary conditions
and which has orthogonal equipotential and flow lines. (It is usually
easier to start by drawing the flow lines).
3. Modify the net so that it meets the conditions outlined above and so
that the mesh located between adjacent flow lines and equipotentials are
square (you could draw an inscribed circle).
Refine the flow net by repeating step 3.
10.6 Calculation of Quantities of Interest from Flow Nets
10.6.1 Calculation of flow
We have seen at 10.3 that, when the flow net has been drawn so that the elemental rectangles are
approximately square, the total flow is equal to :
Q = Nf . ∆q = k .h . Nf / Nd
It should be noted in the development of this formula it was assumed that each flow tube was of
unit width and so the flow is given per unit width (into the page).
Let us come back to the earth dam example.

15 m
h = 15m
h = 12m h = 9m h = 6m
h = 3m
h = 0
P
5m
Figure 8 Value of Head on Equipotentials
In this example, we have Nf = the number of flow tubes = 5, and Nd = the number of equipotential
drops = 5.
Suppose that the permeability of the underlying soil is k=10
-5
m/sec (typical of a fine sand or silt)
then the flow per unit width of dam is:
5 3
15 10 m /sec (per m width)Q −
= ×
and if the dam is 25m wide the total flow under the dam:
5 3
25 15 10 m /secQ −
= × ×
10.6.2 Calculation of Pore Water Pressure
The pore water pressure at any point can be found using the expression
z
u
h
w
w
+=
γ
Now referring to Fig. 8 suppose that we wish to calculate the pore water pressure at the point P.
Taking the datum to be at the base of the dam it can be seen that z = - 5m and so:
wwwu γγ 17)]5(12[ =−−=
10.6.3 Exercise
1. A sheet pile wall is driven to a depth of 6m into a permeable sand layer (k = 6.10
-3
mm/s) of
13.5 m thickness lying on an impermeable layer. The water on one side of the wall is at a height of
4.5m, while on the other side, pumps maintain the water level at ground level. To design the
pumping system, draw, by hand, the flow net and estimate the flow under the wall in m³/day.

Figure 9
To check your flow net drawn by hand, you can use FDSOLVER.XLA, a excel macro using the
Finite Difference Method. That macro is available on the course Moodle website, together with an
install note
4
and an help file containing examples. That macro can be also be useful in other
domains like heat diffusion, electric potential, fluid flow, vibrations,…
The steps to follow are explained in the help file. You should "draw" the model like this :
4
An update to the install note has been added by a 2011-1012 student, François Diffels, for Excel 2007 and
2010. Thanks François!
Impermeable soil
13.5m (sand layer)
6 m
4.5 m

The light green cells (left and bottom) are the y and x scales (0.5m step) (not mandatory, but it
helps to draw the model to scale.
The yellow cells are either head specified boundaries (the top lines at either side of the wall), or an
impermeable boundary (the wall itself).
To solve the system and get the head values at any point (cell), just select the area with the white
and yellow cells, then select Tools, FDSOLVER, 2D-LAPLACE, then specify the scale if
necessary, then press "run".
The numerical results appear after a while :

Now, select the result area and choose the good graph options (Excel 2003) :
Then by clicking on the left vertical axis :

And by clicking on the right vertical axis (otherwise the graph appears upside down !):
Finally, one obtains the head graph :
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49
S1
S3
S5
S7
S9
S11
S13
S15
S17
S19
S21
S23
S25
S27
4.00-4.50
3.50-4.00
3.00-3.50
2.50-3.00
2.00-2.50
1.50-2.00
1.00-1.50
0.50-1.00
Of course that macro only plots the head lines, as it is the head that follows the Laplace's
equation. The flow lines must be drawn by hand, trying to draw "square" meshes.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49
S1
S3
S5
S7
S9
S11
S13
S15
S17
S19
S21
S23
S25
S27
4.00-4.50
3.50-4.00
3.00-3.50
2.50-3.00
2.00-2.50
1.50-2.00
1.00-1.50
0.50-1.00
Q = k .h . Nf / Nd = 6. 10
-3
x10
-3
x4,5x 4/9 x 60x60x24= 1.04 m³/day / m of wall
10.6.4 Example – Stranded
5
Vessel Rescue
The figure 10 shows a long vessel, 20 metres wide, stranded on a sand bank. It is proposed to
inject water into a well point, 10 metres down, under the centre of the vessel to assist in towing the
vessel off. The water depth is 1 metre.
Figure 10 Stranded Vessel Example
The sand has a permeability of 3 × 10
-4
m/sec. Assuming that a pressure head of 50 m can be
applied at the well point :
1. draw the flow net by hand (and check eventually with the Excel macro) :
5
échoué
20m
10m
20m (not to scale)

2. calculate:
The pore water pressure distribution across the base of the vessel
The total upthrust due to this increase in pore water pressure
The rate at which water must be pumped into the well point.
10.6.5 Exercise
2. The figure 11 shows a concrete dam lying on a permeable soil layer (k = 12.5 10
-3
mm/s).
a) draw the flow net
b) calculate the total flow, per m width of dam, passing under the dam
c) calculate the upthrust due to the water flow under the dam.
Figure 11

10.7 Flow Nets For Anisotropic Materials
10.7.1 Introduction
Many soils are formed in horizontal layers as a result of sedimentation through water.
Because of seasonal variations such deposits tend to be horizontally layered and this
results in different permeabilities in the horizontal and vertical directions.
10.7.2 Flow nets for soil with anisotropic permeability
It can be demonstrated that in the case of plane flow in an anisotropic material having a
horizontal permeability kH and a vertical permeability kv, the solution can be reduced to
that of flow in an isotropic material by doing a variable change
x
x
and
z z
α
=
=
with
αααα =
k
k
H
V
So the flow in anisotropic soil can be analysed using the same methods (including
sketching flow nets) that are used for analysing isotropic soils.
Example - Seepage in an anisotropic soil
Suppose we wish to calculate the flow under the dam shown in Figure 12;
x
z
Impermeable bedrock
L
H1
H2
Impermeable
dam
Soil layer
Z
Fig. 12 Dam on a permeable soil layer over impermeable rock (natural scale)

For the soil shown in Fig. 12 it is found that k kH V= ×4 and therefore
4
2
2
V
V
k
k
so
x
x
z z
α
×
= =
=
=
In terms of transformed co-ordinates this becomes as shown in Figure 13
z
Impermeable bedrock
L/2
H1
H2
x Soil layer
Z
Fig. 13 Dam on a permeable layer over impermeable rock (transformed scale)
The flow net can now be drawn in the transformed co-ordinates and this is shown in
Fig.14
5m
Impermeable bedrock
Fig. 14 Flow net for the transformed geometry

It is possible to use the flow net in the transformed space to calculate the flow underneath
the dam by introducing an equivalent permeability
k k keq H V=
A rigorous proof of this result will not be given here
Example
Suppose that in Figure 12 H1 = 13m and H2 = 2.5m, and that kv = 10-6
m/sec and kH =4 ×
10
-6
m/sec The equivalent permeability is:
k meq = × × = ×− − −
( ) ( ) / sec4 10 10 2 106 6 6
The total head drop is 10.5 m.
There are 14 head drops and thus Nd=14.
There are 6 flow tubes and thus Nf=6
The flow underneath the dam is, ∆Q = keq ∆h Nf/Nd = (2×10
-6
)×10.5x6/14 = 9.0×10
-6
m
3
/sec(/m width of dam)
For a dam with a width of 50 m, Q = 450 × 10
-6
m
3
/sec = 41.47 m
3
/day
10.8 Effect of water flow in a soil mass
Introduction
Fig. 15 illustrates equilibrium conditions in a column of soil. The left-hand tank contains water and
is connected to the right-hand tank containing soil and water. When the water level is the same in
both tanks. there will be no flow of water through the soil.
Figure 15
At level xx :
σ = H . γw + Z . γsat (total stress)
u = (H+Z) . γw (pore water pressure)
σ’ = Z . (γsat - γw) (contrainte effective = poids de la colonne de sol déjaugé)

Any change from equilibrium conditions will cause water to flow through the soil and this will alter
the effective stress and pore water pressure. As it flows, the water exerts a frictional drag on the
soil particles and the effect of this force is known as the seepage pressure.
(a) Flow downwards through the soil.
If the left-band tank is lowered, and the level of water in the right-hand tank is maintained, water
will flow downwards through the soil (Fig. 16).
Figure 16
At level x-x
σ = H . γw + Z . γsat
u = (H+Z-h) . γw
σ’ = Z . (γsat - γw) + h . γw
Thus the effective pressure is increased by h.γw . This quantity h.γw is the seepage pressure
exerted by the flowing water.
(b) Flow upwards through the soil.
If the left-hand tank is raised, water will flow upwards through the soil (Fig. 17).
Figure 17
At level x-x
σ = H . γw + Z . γsat
u = (H+Z+h) . γw
σ’ = Z . (γsat - γw) - h . γw

Thus the effective pressure is decreased by h.γw, the amount of seepage pressure.
If h is increased, it may happen that the effective stress σ’ vanishes : it is the heave due to
seepage of water in the ground also called piping6
. This kind of phenomenon may be catastrophic
and will be developed in the next paragraph.
Figure 18 Piping
10.8.2 Seepage Force : general formulation
It can be demonstrated that the seepage force acting on a soil volume dV, is a force acting in the
direction of the flow given by the formula:
. .
w
J i dVγ=
r r
Where :
i
r
is the hydraulic gradient in the direction of the flow.
dV is a soil volume subjected to three forces :
Figure 19 Seepage Force
- its own weight W = γsat . dV
- the buoyancy (Archimede's thrust) A = γw .dV
- the seepage force dVidVjJ
w
...
ρρρ
γ==
6
« renard »

10.9 Hydraulic Heave or Piping
Many dams on soil foundations have failed because of the sudden formation of a piped
shaped discharge channel. As the store water rushes out, the channel widens and
catastrophic failure results. This results from erosion of fine particles due to water flow.
Another situation where flow can cause failure is in producing ‘quicksand’ conditions. This
is also often referred to as piping failure (Fig. 20).
Figure 20 Piping Failure : 1) initiation and first deterioration, 2) regressive erosion, 3)
formation of flow channel, 4) liquefaction and collapse
We will now show how to check the danger of piping in an example.

10.9.1 Exercise
Verify the safety against hydraulic heave in the following case.
The flow net, obtained with the Visual AEM freeware, gives the following equipotentials around the
base of the wall. (The datum is at the ground level at the right of the wall.)
GWL
1.0m
Sand γk = 20kN/m
3 3.0m
7.0m
H=2.5m
Water
h z
m m
A 1.6 -3
B 1.5 -2.81
C 1.4 -2.46
D 1.3 -1.99
E 1.2 -1.36
F 1.1 -0.69
1 0

As already mentioned in the chapter 8 about Eurocode 7, when considering a limit state of failure
due to heave by seepage of water in the ground (HYD), it shall be verified, for every relevant soil
column, that :
1. the design value of the destabilising total pore water pressure (udst;d) at the bottom of the
column is less than or equal to the stabilising total vertical stress (σstb;d) at the bottom of the
column : udst;d ≤ σstb;d, or
2. the design value of the seepage force (Sdst;d) in the column is less than or equal to the
submerged weight
7
(G´dst;d ) of the same column Sdst;d ≤ G´stb;d
The partial factor set for this case are:
Parameter Symbol
HYD - Partial
factor set
Permanent action (G) Unfavourable γG, dst 1.35
Favourable γG, stb 0.9
Variable action (Q) Unfavourable γQ, dst 1.5
Favourable - 0.0
z u/γw u=(h-z).γw.1.35 σtot=γsat*z*0.9+γw.1.35 u<σtot ? i=dh/ds Seepage Force per unit
m m kN/m² kN/m² kN /m²
-3 4.6 62.1 67.5 YES 0.526315789 -1.35
-2.81 4.31 58.185 64.08 YES 0.285714286 -1.35
-2.46 3.86 52.11 57.78 YES 0.212765957 -1.35
-1.99 3.29 44.415 49.32 YES 0.158730159 -1.35
-1.36 2.56 34.56 37.98 YES 0.149253731 -1.35
-0.69 1.79 24.165 25.92 YES 0.144927536 -1.35
0 1 13.5 13.5 YES
Sdst;d Total Seepage Force (kN/m²) -8.1
G´stb;d Total weight (kN/m²) -13.5
The wall is thus safe against piping, following Eurocode 7.
7
Poids déjaugé

Flow nets

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