Fig. 2.9 Section of a flow net.
Referring to Fig. 2.8:
If we assume that the soil is homogeneous and isotropic then k is the same for all elements and it is possible to draw the flow net so that b1 = l1, b2 = l2, b = l. When we have this arrangement, the elements are termed ‘squares’ and the flow net is a square flow net. With this condition:
Since square ABCD has the same flow lines as A1B1C1D1,
Since square ABCD has the same equipotentials as A2B2C2D2,
i.e.
Hence, in a flow net, where all the elements are square, there is the same quantity of unit flow through, and the same head drop across, each element.
No element in a flow net can be truly square, but the vast majority of the elements do approximate to squares in that the four corners of the element are at right angles and the distance between the flow lines, b, equals the distance between the equipotentials, l. Some relaxation is needed when asserting that a certain element is a square and some elements will be more triangular in shape but provided that the flow net is drawn with a sensible number of flow channels (generally five or six), the results obtained will be within the range of accuracy possible. The more flow channels that are drawn, the more the elements will approximate to true squares, but the apparent increase in accuracy is misleading and the extra work involved (if drawing by hand, for example, up to perhaps twelve channels) is not worthwhile.
2.10.2 Calculation of seepage quantities
Let
Nd = number of potential drops
Nf = number of flow channels
h = total head loss
q = total quantity of unit flow
Then
(2.21)
2.10.3 Drawing a flow net
The first step is to draw in pencil the first flow line, upon which the accuracy of the final correctness of the flow net depends. There are various boundary conditions that help to position this first flow line, including:
1 buried surfaces (e.g. the base of the dam, sheet piling), which are flow lines as water cannot penetrate into such surfaces;
2 the junction between a permeable and an impermeable material, which is also a flow line; for flow net purposes, a soil that has a permeability of one‐tenth or less the permeability of the other may be regarded as impermeable;
3 the horizontal ground surfaces on each side of the dam, which are equipotential lines.
The procedure is as follows:
1 draw the first flow line and hence establish the first flow channel;
2 divide the first flow channel into squares checking visually that b = l in each element;
3 project the equipotentials beyond the first flow channel, which gives an indication of the size of the squares in the next flow channel;
4 determine the position of the next flow line (remembering that b = l) draw this line as a smooth curve and complete the squares in the flow channel formed;
5 project the equipotentials and repeat the procedure until the flow net is completed.
Fig. 2.10 Example of flow net construction. (a) Problem. (b) Boundary conditions. (c) Step a. (d) Step c. (e) Step d. (f) Final flow net.
As an example, suppose that it is necessary to draw the flow net for the conditions shown in Fig. 2.10a. The boundary conditions for this problem are shown in Fig. 2.10b, and the sketching procedure for the flow net is illustrated in figures c, d, e and f of Fig. 2.10.
If the flow net is correct, the following conditions will apply:
1 equipotentials will be at right angles to buried surfaces and the surface of the impermeable layer;
2 beneath the dam, the outermost flow line will be parallel to the surface of the impermeable layer.
After completing part of a flow net, it is usually possible to tell whether or not the final diagram will be correct. The curvature of the flow lines and the direction of the equipotentials indicate if there is any distortion, which tends to be magnified as more of the flow net is drawn. This gives a good indication of what was wrong with the first flow line. This line must now be redrawn in its corrected position and the procedure repeated again, amending the first flow line if necessary, until a satisfactory net is obtained.
Generally, the number of flow channels, Nf will not be a whole number, and in these cases, an estimate is made as to where the next flow line would be if the impermeable layer was lower. The width of the lowest channel can then be found (e.g. in Fig. 2.10f, Nf = 3.3).
Note: In flow net problems, we assume that the permeability of the soil is uniform throughout the soil's thickness. This is a considerable assumption and we see therefore that refinement in the construction of a flow net is unnecessary, since the difference between a roughly sketched net and an accurate one is small compared with the actual flow pattern in the soil and the theoretical pattern assumed. This is where numerical models have particular advantage over hand‐drawn flow nets.
Example 2.5 Flow net seepage
Using Fig. 2.9f, determine the loss through seepage under the dam in cubic metres per year if k = 3 × 10−6 m/s and the level of water above the base of the dam is 10 m upstream and 2 m downstream. The length of the dam perpendicular to the plane of seepage is 300 m.
Solution:
From the flow net Nf = 3.3, Nd = 9
Total head loss, h = 10 − 2 = 8 m