CHAPTER
CHAPTER 2
2
FLOWNET
FLOWNET (DIMENSIO
(DIMENSIONAL
NAL FLOW
FLOW OF
OF WATER
WATER
THROUGH SOILS)
THROUGH SOILS)
In this chapter, you will study the basic principles of two-dimensional flow of water In this chapter, you will study the basic principles of two-dimensional flow of water through soils. The emphasis will be on gaining an understanding of the forces from groundwater through soils. The emphasis will be on gaining an understanding of the forces from groundwater flow that provoke failures.
flow that provoke failures.
You will learn methods to calculate flowrate, uplift/net pressure distribution, uplift You will learn methods to calculate flowrate, uplift/net pressure distribution, uplift forces, and seepage stresses for a few simple geotechnical s
forces, and seepage stresses for a few simple geotechnical s ystems.ystems.
When you complete this chapter, you should be able to: When you complete this chapter, you should be able to:
Calculate flowrate under and within earth structures.Calculate flowrate under and within earth structures.
Calculate seepage stresses, uplift/net pressure distribution, uplift forces, hydraulicCalculate seepage stresses, uplift/net pressure distribution, uplift forces, hydraulic
gradients, and the critical hydraulic gradient. gradients, and the critical hydraulic gradient.
Determine the stability of simple geotechnical systems subjected to two-dimensional flowDetermine the stability of simple geotechnical systems subjected to two-dimensional flow
of water. of water.
Importance Importance
Many failures in geotechnical engineering result from instability of soil masses due to Many failures in geotechnical engineering result from instability of soil masses due to groundwater flow. Lives are lost, infrastructures are damaged or destroyed, and major economic groundwater flow. Lives are lost, infrastructures are damaged or destroyed, and major economic losses occur. The topics covered in this chapter will help you to avoid pitfalls in the analysis and losses occur. The topics covered in this chapter will help you to avoid pitfalls in the analysis and design of geotechnical systems where groundwater flow
design of geotechnical systems where groundwater flow can lead to instabilitycan lead to instability
Definitions of Key Terms; Definitions of Key Terms; Flownet
Flownet a combination of a number of flow lines a combination of a number of flow lines and equipotential lines.and equipotential lines.
Equipotential line,N
Equipotential line,Nee is a line along which the is a line along which the potential head at all points is equal. potential head at all points is equal. Thus,Thus,
if piezometers are placed at different points along an
if piezometers are placed at different points along an equipotentialequipotential line, the water level will rise to the same elevation
line, the water level will rise to the same elevation in all of themin all of them
Flow line,N
Flow line,Nf f is a line along which a water particle will travel from upstream tois a line along which a water particle will travel from upstream to
the downstream side in the permeable soil medium. the downstream side in the permeable soil medium.
Head loss,H
Criteria for Sketching Flownets
A flownet is a graphical representation of a flow field that satisfies Laplace’s equation and comprises a family of flow lines and equipotential lines. A flownet must meet the following criteria:
1.The boundary conditions must be satisfied.
2.Flow lines must intersect equipotential lines at right angles.
3.The area between flow lines and equipotential lines must be curvilinear squares. A curvilinear square has the property that an inscribed circle can be drawn to touch each side of the square and continuous bisection results, in the limit, in a point.
4.The quantity of flow through each flow channel is constant.
5.The head loss between each consecutive equipotential line is constant.
6.A flow line cannot intersect another flow line.
7.An equipotential line cannot intersect another equipotential line.
An infinite number of flow lines and equipotential lines can be drawn to satisfy Laplace’s equation. However, only a few are required to obtain an accurate solution.
Flownet for Isotropic Soils
1. Draw the structure and soil mass to a suitable scale.
2. Identify impermeable and permeable boundaries. The soil – impermeable boundary interfaces are flow lines because water can flow along these interfaces. The soil – permeable boundary interfaces are equipotential lines because the total head is constant along these interfaces. 3. Sketch a series of flow lines (four or five) and then sketch an appropriate number of
equipotential lines such that the area between a pair of flow lines and a pair of equipotential lines (cell) is approximately a curvilinear square. You would have to adjust the flow lines and equipotential lines to make curvilinear squares. You should check that the average width and the average length of a cell are approximately equal by drawing an inscribed circle. You should also sketch the entire flownet before making adjustments.
The flownet in confined areas between parallel boundaries usually consists of flow lines and equipotential lines that are elliptical in shape and symmetrical (Figure 2.1). Try to avoid making sharp transitions between straight and curved sections of flow and equipotential lines. Transitions should be gradual and smooth. For some problems, portions of the flownet are enlarged and are not curvilinear squares, and they do not satisfy Laplace’s equation. For example, the portion of the flownet below the bottom of the sheet pile in Figure 2.1 does not consist of curvilinear squares. For an accurate flownet, you should check these portions to ensure that repeated bisection results in a point.
Figure 2.1 Flownet for a sheet pile.
The flownet in confined areas between parallel boundaries usually consists of flow lines and equipotential lines that are elliptical in shape and symmetrical (Figure 2.1). Try to avoid making sharp transitions between straight and curved sections of flow and equipotential lines. Transitions should be gradual and smooth. For some problems, portions of the flownet are enlarged and are not curvilinear squares, and they do not satisfy Laplace’s equation. For
example, the portion of the flownet below the bottom of the sheet pile in Figure 2.1 does not consist of curvilinear squares. For an accurate flownet, you should check these portions to ensure that repeated bisection results in a point.
A few examples of flownets are shown in Figures 2.1 to 2.3. Figure 2.1 shows a flownet for a sheet pile wall, Figure 2.2 shows a flownet beneath a dam, and Figure 2.3 shows a flownet in the backfill of a retaining wall. In the case of the retaining wall, the vertical drainage blanket of coarse-grained soil is used to transport excess porewater pressure from the backfill to prevent the imposition of a hydrostatic force on the wall. The interface boundary, AB (Figure 2.3), is neither an equipotential line nor a flow line. The total head along the boundary AB is equal to the elevation head.
Figure 2.2 Flownet under a dam with a cutoff curtain (sheetpile) on the upstream end.
INTERPRETATION OF FLOWNET i) Flow Rate
Let the total head loss across the flow domain be ∆ H , that is, the difference between upstream and downstream water level elevation. Then the head loss (∆h) between each consecutive pair of equipotential lines is
Where, Nd is the number of equipotential drops, that is, the number of equipotential lines minus
one. In Figure 2.1, ∆H = H = 8 m and Nd = 18. From Darcy’s law, the flow through each flow
channel for an isotropic soil is
Where; b and L are defined as shown in Figure 2.1. By construction, b/ L ≈ 1, and therefore the total flow is
∑
Where; Nf is the number of flow channels (number of flow lines minus one). In Figure 2.1, Nf =
9. The ratio Nf /Nd is called the shape factor. Finer discretization of the flownet by drawing more
flow lines and equipotential lines does not significantly change the shape factor. Both Nf and Nd
can be fractional. In the case of anisotropic soils, the quantity of flow is
√
or
ii) Hydraulic Gradient, i
You can find the hydraulic gradient over each square by dividing the head loss by the length, L; that is,
You should notice from Figure 2.1 that L is not constant. Therefore, the hydraulic gradient is not constant. The maximum hydraulic gradient occurs where L is a minimum; that is,
⁄
Where; Lmin is the minimum length of the cells within the flow domain. Usually, Lmin occurs at
exit points or around corners (e.g., point A in Figure 2.1), and it is at these points that we usually get the maximum hydraulic gradient.
Critical Hydraulic Gradient, i cr
We can determine the hydraulic gradient that brings a soil mass (essentially, coarse-grained soils) to static liquefaction.
(
)
Where; icr is called the critical hydraulic gradient, Gs is specific gravity, and e is the void ratio.
Since Gs is constant, the critical hydraulic gradient is solely a function of the void ratio of the
soil. In designing structures that are subjected to steady-state seepage, it is absolutely essential to ensure that the critical hydraulic gradient cannot develop.
iii) Porewater Pressure Distribution,U
The porewater pressure at any point j within the flow domain (flownet) is calculated as follows: 1. From Bernoulli’s eqn:
Rearranging :
Where;h – head at any point in the flownet. h represents water level in tube if placed at that point. Example; h p - h located at point P.
z – elevation head relative to datum. Datum usually taken as downstream water level. z can be (-)
2. Select a datum. Let us choose the downstream water level as the datum. 3. There are two method to determine the total hydraulic head loss,h;
i) total hydraulic head loss at point p, h p =
=
(counted potential dropfrom upstream)
ii) total hydraulic head loss at point p, h p =
(counted potential drop from downstream)Where;
H – head difference (driving seepage)
Ne – total no. of equipotential drops n p – no of equipotential
drops at any points (eg: at point P)
Where;
H – head difference (driving seepage)
Ne – total no. of equipotential drops n p – no of equipotential
drops at any points (eg: at point P)
iv) Uplift Pressure Distribution
Consider the dam section shown in Figure below.
Given H = 10m
Pressure head at point D =[
]+ 3.34 = 11.67 mPressure head at point E = [
]+ 3.34 = 10.84 mPressure head at point F = [
]+ (3.34 - 1.67) = 8.75 mPressure head at point G = [
]+ (3.34 - 1.67) = 4.56 mPressure head at point H = [
]+ 3.34 = 5.84 mPressure head at point G = [
v) Uplift Pressure on Dam
(Refer to above Dam)
Uplift Pressure on Dam =
γ
w(area of pressure head diagram)x1m=
= 1714.9 kN/m
Uplift/Net Pressure Distribution on Various Dam
