Geotechnical Engineering

DHARM

166 GEOTECHNICAL ENGINEERING

‘flow channels’ of equal size. Since the flow is purely vertical, there cannot be flow from one
channel into another. Since there are four flow channels in, say, the x-direction, and the since
the tube is a square one, there are also four flow channels in the y-direction, i.e., perpendicular
to the page. Thus there will be a total of 16 flow channels. If the flow in one channel is found
the total flow is obtained by multiplying it by 16.

Datum – 200 4 400 800 1200 1600

H=

1600 mm

1600

mm

1200

1000

800

600

400

200

0

Tube of 400 × 400 mmcross section

Pressure head

Datum head

Total head

Head mm

h=

280 mm

p

h=

100 mm

p

h=

140 mm

p

Equipotential

lines

H

0.8 H

0.6 H

0.4 H

0.2 H

Total head

Flow

lines

(a) Flow through tube (b) Heads (c) Flow net

a

b

Soil

K=0.5

mm/s

Soil

K = 0.5

mm/s

Fig. 6.1 One-dimensional flow
In the figure, dashed lines indicate the lines along which the total head is a constant.
These line through points of equal total head are known as ‘equipotential lines’. Just as the
number of flow lines is infinite, the number of equipotential lines is also infinite.
If equipotential lines are drawn at equal intervals, it means that the head loss between
any two consecutive equipotential lines is the same.

A system of flow lines and equipotential lines, as shown in Fig. 6.1 (c), constitutes a ‘flow
net’. In isotropic soil, the flow lines and equipotential lines intersect at right angles, indicating
that the direction of flow is perpendicular to the equipotential lines. An orthogonal net is
formed by the intersecting flow lines and equipotential lines. The simplest of such patterns is
one of the squares. From a flow net three very useful items of information may be obtained:
rate of flow or discharge; head; and hydraulic gradient.

First, let us see how to determine the rate of flow or discharge from the flow net. Con-
sider square a in the flow net–Fig. 6.1(c). The discharge qa through this square is
qa = k · ia · Aa
The head lost in square a is given H/nd , where H is the total head lost and nd is the

number of head drops in the flow net. ia is then equal to

H

nld.

, where l is the vertical dimen-

sion of square a. The cross-sectional area Aa of square a, as seen in plan, is b as shown in the
figure, since a unit dimension perpendicular to the plane of the paper is to be considered for
the sake of convenience.