DHARMSEEPAGE AND FLOW NETS 167∴ qa = k. H
nld.. b
Since a square net is chosen, b = l.∴ qa = kH. l
nd.Since all flow through the flow channel containing square a must pass through square
a, the flow through this square represents the flow for the entire flow channel.
In order to obtain the flow per unit of length L perpendicular to the paper, qa should be
multiplied by the number of flow channels, say, nf :
∴ q/L = qa. nf = kH.n
nf
d∴ q/L = k. H.n
nf
d= kH. s ...(Eq. 6.1)the ratio s =
n
nf
dis a characteristic of the flow net and is independent of the permeability k andthe total loss of head H. It is called the ‘shape factor’ of the flow net. It should be noted that nf
and nd need not necessarily be integers; these may be fractional, in which case the net may
involve a few rectangles instead of squares.
The value of s in this case iss =n
nf
d= 4/10 = 0.4and, q/L = k.H.s = 0.5 × 1600 × 0.4 mm^3 /s/mm = 320 mm^3 /s/mm
Then, q = (q/L) × (400) = 320 × 400 = 128000
= 1.28 × 10^5 mm^3 /s.
The value of total seepage is, of course, the same as that obtained by the initial compu-
tation using Darcy’s law directly.
Next, let us see how to use the flow net to determine the head at any point. Since thereare ten equal head drops,^1
10
H is lost from one equipotential to the next. Since it is the totalhead that controls the flow, it should be noted that equipotential are drawn through points of
equal total head. The pressure head may be readily determined since the total head and eleva-
tion head are known.
For example, at elevation 1000 mm,
The total head, h = (8/10) × H = (8/10) × 1600 mm = 1280 mm
Elevation head, he = 1000 mm
∴ Pressure head, hp = (1280 – 1000) mm = 280 mm.
Since this is also the piezometric head, water will rise to a height of 280 mm in a
piezometer installed at this elevation, as shown by the side of the flow net. The pore pressure
at this elevation is 280 × 9.81 × 10–6 N/mm^2 or 2.75 × 10–3 N/mm^2. Similarly, the pressure