Box 4 Drawing a Flow Net for an Unconfined System with a Water Table Boundary

Unconfined groundwater systems have a water table boundary which requires special consideration when drawing a flow net because the location of the water table boundary is not known until an acceptable flow net has been drawn.

To illustrate inclusion of a water table in a flow net, consider flow through an earthen dam with a hydraulic conductivity of 0.2 m/d resting on an impermeable base (Figure Box 4-1). The dam extends 55 m in the direction perpendicular to the diagram, and is capped with impermeable material such that water is prevented from infiltrating the surface of the dam. Water enters from the up-gradient reservoir on the left and exits to the downstream reservoir on the right.

Begin the steps for drawing a flow net as described in section 2.3 and shown in Figure Box 4-1. The position of the water table is not known until the flow system is revealed by following the rules for drawing a flow net, so the initial sketch indicates this uncertainty by using a dashed line with “?”.

Figure showing the boundaries of a region for which a flow net is constructed

Figure Box 4-1 – Step 1 – Draw the system to scale, Step 2 – Draw equipotential lines to coincide with head boundaries, Step 3 – Draw flow lines to coincide with no-flow boundaries.

Drawing a flow net is a trial-and-error process because equipotential and flow lines are adjusted until curvilinear squares are formed. The added complication when drawing an unconfined flow net is that the position of the upper boundary (the water table) and length of the seepage face are also adjusted while working to create curvilinear squares.

The next step is to draw flow lines along paths where you envision groundwater will flow, ensuring they are perpendicular to equipotential lines on the boundaries (Figure Box 4-2). There is no recharge from the impermeable upper boundary, so the uppermost flow line forms the water table. The flow line should meet the 25 m equipotential line (constant head boundary) at a right angle. The downstream end of the water table should meet the dam surface at an elevation higher than the surface of the downstream reservoir (forming a seepage face as shown in Figure Box 4-2), and at a slope equal to the slope of the dam face. In the same way that the initial position of the water table is unknown until after the flow net is drawn, the length of the seepage face is unknown until after a valid flow net is drawn. Hydraulic head along a seepage face is equal to the elevation of the ground surface because the gage pressure along the seepage face is zero. Unlike the water table, the location of the seepage face boundary is known because it will be on the downgradient face of the dam, only its length is unknown before sketching the flow net.

Although, for expediency, we have drawn the flow lines in the correct position here, it is likely the first attempt to draw flow lines will require adjustment when drawing a flow net. It is expected that the flow net lines will be erased and redrawn as needed until the criteria for a valid flow net are met.

Figure showing the drawing of flow lines

Figure Box 4-2 – Step 4: Draw flow lines along paths where you envision groundwater flowing, ensuring they are perpendicular to equipotential lines on the boundaries. It is likely the first attempt to draw flow lines will require adjustment. Drawing flow nets is a trial-and-error process.

As one draws the flow net, it is important to remember that the hydraulic head value of an equipotential line is equal to the elevation at which it meets the water table (or the seepage face) because hydraulic head is the sum of pressure head (in terms of a height of a column of water) and elevation. Gage pressure is typically used for quantifying pressure, with atmospheric pressure being equivalent to zero gage pressure. At the water table and along the seepage face, the gage pressure is zero so the hydraulic head is equal to the elevation. For this reason, it is useful to add lines of equal elevation before sketching the equipotential lines (Figure Box 4-3) as a guide for drawing the flow net.

Figure showing addition of lines of equal elevation before sketching equipotential lines

Figure Box 4-3 – Add lines of equal elevation before sketching the equipotential lines, because the value of an equipotential line is equal to elevation where it intersects the water table and seepage face, so the elevation lines provide a guide for placing the equipotential lines.

Once the elevation grid is drawn, proceed to sketch the equipotential lines at right angles to no-flow boundaries and flow lines. Make sure that equipotential lines meet the water table and the seepage face at an elevation that is the same as the hydraulic head of the equipotential line. The equipotential lines and flow lines should intersect to form curvilinear squares. As before, one way to decide if you are creating curvilinear squares is to draw a circle between the intersecting lines. If the circle fits roughly within the shapes, then they are approximately curvilinear squares (Figure Box 4-4).

Figure showing circles fitting in curvilinear squares of a flow net

Figure Box 4-4 – The shapes are curvilinear squares if circles fit approximately within them, but some flow nets may include partial flow tubes as shown here by the narrow flow tube at the bottom of the flow net. The flow net has 18 head drops and 4 flow tubes.

A rough sketch of the equipotential lines provides an estimate of the number of head drops that will create curvilinear squares. In this case, it is determined that 18 head drops create curvilinear squares. So, the contour interval is determined by dividing the total head drop of 10 m by 18 to obtain a value of 0.56 m head drop between each pair of equipotential lines. This establishes the values of the contour lines and knowing that they must meet the water table at the elevation equal to their values further constrains the position of the lines. For example, the first equipotential line to the right of the upper reservoir will have a value of 24.44 m and so the intersection of the equipotential line and the water table should be at that elevation.

Often, both equipotential lines and flow lines need to be erased and redrawn repeatedly before achieving curvilinear squares with equipotential lines meeting the water table at right angles and at an elevation equal to their value. Even after adjustment, a hand drawn flow net is only an approximate solution to the flow equations. For the purpose of this book, a fairly precise flow net is shown as Figure Box 4-5. The flow net does not provide precision to the 3 significant figures shown in the contour labels in the diagram. Three significant figures are shown, not because the system is known to high precision, but to adequately illustrate the difference in head between adjacent contour lines.

Figure showing a flow net for a system with a water table.

Figure Box 4-5 – Draw equipotential lines for the unconfined flow net, ensuring that 1) their value is equal to the elevation of the water table where they meet the water table and to the elevation of the ground surface where they meet the seepage face, 2) that they meet flow lines at right angles, and 3) that the intersecting lines form curvilinear squares. This result was obtained after sketching and erasing until all of the criteria were met.

Given that the dam extends 55 m into the image and the hydraulic conductivity of the earth material was 0.2 m/d, the volumetric flow rate through the dam is:

\displaystyle Q_{total}=KH\frac{n_{f}}{n_{d}}w = \displaystyle \left ( 0.2\ \frac{\textup{m}}{\textup{d}} \right )(10\ \textup{m})\left ( \frac{4}{18} \right )(55\ \textup{m}) = \displaystyle 24.44\ \frac{\textup{m}^3}{\textup{d}} ~ \displaystyle 25\ \frac{\textup{m}^3}{\textup{d}}

which is about 125 oil drums full of water each day, and would take about 100 days to fill an Olympic-size swimming pool. As noted earlier, it is important to recognize that the volumetric flow rate determined from a flow net is an approximate value.

If we had started by sketching equipotential lines with a round number contour interval, for example, a 1-m interval, there would be 10 head drops ([25-15]/1 = 10). In that case, the flow net would have fewer flow tubes and it would not be possible to form curvilinear squares for all tubes because to obtain the ratio of 4/18=0.222 with 10 head drops, 2.22 flow tubes are needed. Consequently, one of the flow tubes needs to be a 0.22 portion of a curvilinear square as shown by the deepest flow tube of Figure Box 4-6. In that case, the volumetric flow rate through the dam is:

\displaystyle Q_{total}=KH\frac{n_{f}}{n_{d}}w = \displaystyle \left ( 0.2\ \frac{\textup{m}}{\textup{d}} \right )(10\ \textup{m})\left ( \frac{2.22}{10} \right )(55\ \textup{m}) = \displaystyle 24.42\ \frac{\textup{m}^3}{\textup{d}} ~ \displaystyle 25\ \frac{\textup{m}^3}{\textup{d}}

Figure showing a flow net with a partial flow tube

Figure Box 4-6 – Some flow nets may include partial flow tubes as shown here by the narrow flow tube at the bottom of the flow net. The flow net has 10 head drops and 2.22 flow tubes.

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Graphical Construction of Groundwater Flow Nets Copyright © 2020 by The Authors. All Rights Reserved.