and
∂
∂
∂
∂
∂
∂
∂
∂
∂
∂
y y y
a a a
n x
x
n z
z n
v v v
= +
= - ssin ( sin )- + scos2 = s Thus
∂
∂
∂
∂
y f
n = s or approximately
D D
D D
y f
n = s (2.18)
Seepage Dh h
Nd (2.20)
and
q N q= fD
Hence, from Equation 2.19 q khN
= Nf
d
(2.21) Equation 2.21 gives the total volume of water flowing per unit time (per unit dimension in the y-direction) and is a function of the ratio Nf/Nd. Between two adjacent equipotentials, the hydrau- lic gradient is given by
i h
= Ds
D (2.22)
Example of a flow net
As an illustration, the flow net for the problem detailed in Figure 2.8(a) will be considered. The figure shows a line of sheet piling driven 6.00 m into a stratum of soil 8.60 m thick, underlain by an impermeable stratum. On one side of the piling the depth of water is 4.50 m; on the other side the depth of water (reduced by pumping) is 0.50 m. The soil has a permeability of 1.5 × 10–5 m/s.
The first step is to consider the boundary conditions of the flow region (Figure 2.8(b)). At every point on the boundary AB the total head is constant, so AB is an equipotential; simi- larly, CD is an equipotential. The datum to which total head is referred may be chosen to be at any level, but in seepage problems it is convenient to select the downstream water level as datum. Then, the total head on equipotential CD is zero after Equation 2.1 (pressure head 0.50 m; elevation head –0.50 m) and the total head on equipotential AB is 4.00 m (pressure head 4.50 m; elevation head –0.50 m). From point B, water must flow down the upstream face BE of the piling, round the tip E and up the downstream face EC. Water from point F must flow along the impermeable surface FG. Thus, BEC and FG are flow lines. The other flow lines must lie between the extremes of BEC and FG, and the other equipotentials must lie between AB and CD. As the flow region is symmetric on either side of the sheet piling, when flowline BEC reaches point E, halfway between AB and CD (i.e. at the toe of the sheet piling), the total head must be half way between the values along AB and CD. This principle also applies to flowline FG, such that a third vertical equipotential can be drawn from point E, as shown in Figure 2.8(b).
The number of equipotential drops should then be selected. Any number may be selected;
however, it is convenient to use a value of Nd, which divides precisely into the total change in head through the flow region. In this example, Nd = 8 is chosen so that each equipotential will represent a drop in head of 0.5 m. The choice of Nd has a direct influence on the value of Nf. As Nd is increased, the equipotentials get closer together such that, to get a ‘square’ flow net, the flow channels will also have to be closer together (i.e. more flow lines will need to be plotted).
This will lead to a finer net with greater detail in the distribution of seepage pressures; however, the total flow quantity will be unchanged. Figure 2.8(c) shows the flow net for Nd = 8 and Nf = 3.
These parameters for this particular example give a ‘square’ flow net and a whole number of flow channels. This should be formed by trial and error: a first attempt should be made and the
positions of the flow lines and equipotentials (and even Nf and Nd) should then be adjusted as necessary until a satisfactory flow net is achieved. A satisfactory flow net should satisfy the fol- lowing conditions:
●
● All intersections between flow lines and equipotentials should be at 90°.
●
● The curvilinear squares must be square – in Figure 2.8(c), the ‘square-ness’ of the flow net has been checked by inscribing a circle within each square. The flow net is acceptable if the circle just touches the edges of the curvilinear square (i.e. there are no rectangular elements).
Due to the symmetry within the flow region, the equipotentials and flow lines may be drawn in half of the problem and then reflected about the line of symmetry (i.e. the sheet piling in this case).
In constructing a flow net, it is a mistake to draw too many flow lines; typically, three to five flow Figure 2.8 Flow net construction: (a) section, (b) boundary conditions, (c) final flow net including a check of the ‘square-ness’ of the curvilinear squares, and (d) hydraulic gradients inferred from flow net.
Seepage channels are sufficient, depending on the geometry of the problem and the value of Nd which is most convenient.
In the flow net in Figure 2.8(c), the number of flow channels is three and the number of equi- potential drops is eight; thus the ratio Nf/Nd is 0.375. The loss in total head between any two adjacent equipotentials is
Dh h
= N = =
d
4 00 m
8 0 5
. .
The total volume of water flowing under the piling per unit time per unit length of piling is given by
q khN
= N = ´ ´ ´
= ´
-
- f d
m /s
1 5 10 4 00 0 375 2 25 10
5
5 3
. . .
.
A piezometer tube is shown at a point P on the equipotential with total head h = 1.00 m, i.e. the water level in the tube is 1.00 m above the datum. The point P is at a distance zP = 6 m below the datum, i.e. the elevation head is –zP. The pore water pressure at P can then be calculated from Bernoulli’s theorem:
u h z
h z
P w P P
w P P
kPa
= éë - -
( )
ùû=
(
+)
= ´ +
( )
= g g
9 81 1 6 68 7
. .
The hydraulic gradient across any square in the flow net involves measuring the average dimen- sion of the square (Equation 2.22). The highest hydraulic gradient (and hence the highest seepage velocity) occurs across the smallest square, and vice versa. The dimension Δs has been estimated by measuring the diameter of the circles in Figure 2.8(c). The hydraulic gradients across each square are shown using a quiver or vector plot in Figure 2.8(d) in which the length of the arrows is proportional to the magnitude of the hydraulic gradient.
Example 2.1
A river bed consists of a layer of sand 8.25 m thick overlying impermeable rock; the depth of water is 2.50 m. A long cofferdam 5.50 m wide is formed by driving two lines of sheet piling to a depth of 6.00 m below the level of the river bed, and excavation to a depth of 2.00 m below bed level is carried out within the cofferdam. The water level within the cof- ferdam is kept at excavation level by pumping. If the flow of water into the cofferdam is 0.25 m3/h per unit length, what is the coefficient of permeability of the sand? What is the hydraulic gradient immediately below the excavated surface?
Solution
The section and boundary conditions appear in Figure 2.9(a) and the flow net is shown in Figure 2.9(b). In the flow net there are six flow channels (three on either side of the plane of symmetry) and ten equipotential drops. The total head loss is 4.50 m. The coefficient of permeability is given by
k q
h N N
=
= × × = × −
( / ) .
. ( / ) .
f d
0 25 m/s
4 50 6 10 602 2 6 105 Figure 2.9 Example 2.1.
Seepage
The distance (∆s) between the last two equipotentials is measured as 0.9 m. The required hydraulic gradient is given by
i h
= s
= ´ =
D D
4 50 10 0 9. 0 50
. .
Example 2.2
The section through a dam spillway is shown in Figure 2.10. Determine the quantity of seepage under the dam and plot the distributions of uplift pressure on the base of the dam, and the net distribution of water pressure on the cut-off wall at the upstream end of the spillway. The coefficient of permeability of the foundation soil is 2.5 × 10–5 m/s.
Figure 2.10 Example 2.2.
Solution
The flow net is shown in Figure 2.10. The downstream water level (ground surface) is selected as datum. Between the upstream and downstream equipotentials the total head loss is 5.00 m. In the flow net there are three flow channels and ten equipotential drops.
The seepage is given by q khN
= N = ´ ´ ´
= ´
-
- f d
m /s
2 5 10 5 00 3 10 3 75 10
5
5 3
. .
.
This inflow rate is per metre length of the cofferdam. The pore water pressures acting on the base of the spillway are calculated at the points of intersection of the equipotentials with the base of the spillway. The total head at each point is obtained from the flow net, and the elevation head from the section. The calculations are shown in Table 2.2, and the pressure diagram is plotted in Figure 2.10.
The water pressures acting on the cut-off wall are calculated on both the back (hb) and front (hf) of the wall at the points of intersection of the equipotentials with the wall. The net pressure acting on the back face of the wall is therefore
u u u h z h z
net b f b
w f
w
= − = −
− −
g g
The calculations are shown in Table 2.3, and the pressure diagram is plotted in Figure 2.10.
The levels (z) of points 5–8 in Table 2.3 were found by scaling from the diagram.
Table 2.2 Example 2.2
Point h (m) z (m) h–z (m) u = γw(h–z) (kPa)
1 0.50 −0.80 1.30 12.8
2 1.00 −0.80 1.80 17.7
3 1.50 −1.40 2.90 28.4
4 2.00 −1.40 3.40 33.4
5 2.30 −1.40 3.70 36.3
Table 2.3 Example 2.2 (contd.)
Level z(m) hb (m) ub/γw (m) hf (m) uf/γw (m) ub – uf (kPa)
5 −1.40 5.00 6.40 2.28 3.68 26.7
6 −3.07 4.50 7.57 2.37 5.44 20.9
7 −5.20 4.00 9.20 2.50 7.70 14.7
8 −6.00 3.50 9.50 3.00 9.00 4.9
Seepage