2 dimensional flow

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EG&G (AMEC2121) 1 PROF. MARTIN FAHEY

TWO-DIMENSIONAL FLOW: FLOW NETS

1. Two-dimensional FlowAll the problems considered so far, flow has been in a straight line – 1-dimensional flow.

In general, flow can change direction in 3-D space. However, some 3-D problems involve flow that is confined to a single plane, and also the flow pattern would look the same on any cross-section taken through the problem. This is what we mean by 2-D flow.

Examples:

Flow underneath a long retaining wall (the flow pattern that occurs on every cross-section through the problem would look the same)

Flow through and under a long dam (a dam across a wide valley)

Flow into a long trench, or flow out of a long canal

Flow towards a vertical cylindrical well occurs radially inwards. This is a class of 3-D problem that is called ‘axisymmetric’ – the flow pattern on every radial section through the well would look the same.

2-D flow is described using a FLOW NET – a diagram showing water flow paths (called STREAMLINES) and lines joining points of equal total head (called EQUIPOTENTIAL LINES).

The example shown below shows a cross-section through a dam, with a vertical cut-off wall in the centre of the dam protruding down into the soil. It shows 13 equipotential lines, but with two additional ones (the ‘upstream’ and ‘downstream’ ground surfaces), it has a total of 15 equipotential lines. These are shown numbered. The other set of lines are the ‘streamlines’, each of which trace the path of water flow – i.e. a drop of water that enters the ground on the left of the dam (the ‘upstream’ side) at a point where a streamline intersects the surface remains on that streamline until it exits at the right of the dam (the ‘downstream’ side).

The stream lines define ‘stream channels’: water that starts off within a stream channel stays within it.

We could draw an infinite number of equipotential lines, and an infinite number of streamlines. However, if certain rules are followed in drawing the equipotential lines and streamlines, the resulting flow net can give a great amount of useful information about the flow patterns in the problem.

School of Civil & Resource Engineering, The University of Western Australia


1.1 Mathematical background

General flow of fluid (not necessarily in soil): At point x,z on the plane, the velocity vector is v(x,z).

The velocity vector v(x,z) can be broken into two orthogonal components vx and vz. The square shown represents a volume of space, dx by dy by 1 unit thickness ‘back into the page’.

Flow into the volume = 1..d1..d xvzv zx

Flow out of the volume =

For steady state flow, there can be no change of volume of water within this volume flow in = flow out

Re-arranging, and cancelling, gives: This is the CONTINUITY equation.

We can define a mathematical function called a POTENTIAL FUNCTION (x,z).

Definition of potential function: . Substituting for vx and vz in the

continuity equation gives:

This is a Laplace Equation for

If flow can be described by a potential function, this implies that flow is IRROTATIONAL, which requires that:

We can now define a STREAM FUNCTION (x,z), such that:

Combining these two equations gives: This is a Laplace Equation for .


Flow in general direction: velocity v(x,z)

z

xx,z

dx

dz

zzv

v zz d.

xxvv x

x d.xv

zv

Flow in general direction: velocity v(x,z)

z

xx,z

dx

dz

zzv

v zz d.

xxvv x

x d.xv

zv


Thus, the full solution to a 2-D flow problem involves a pair of Laplace equations. It can be shown that these equations are ORTHOGONAL:

The total differential of (x,z) is:

By definition, along an equipotential line, the potential function value is constant – i.e. d = 0 – i.e.

, where is the local (at x,z) gradient of the equipotential line.

For the stream function: the total differential of (x,z) is:

Along a streamline, the value of (x,z) is constant by definition – i.e. d = 0 – i.e.:

, where is the local (at x,z) gradient of the streamline.

The product of these two gradients (at x,z) = –1, i.e. the lines cross each other at right angles.

Therefore, the complete mathematical solution to the flow problem in 2-D consistst of a set of stream lines that are orthogonal to a set of equipotential lines.

1.2 Mathematical solution related to soil

Darcy’s Law for water flow in 2-D (where the total head h(x,z) varies on the plane):

and (Simply says that the velocity component in a particular

direction depends on the permeability in that direction and the gradient of total head in that direction).

The continuity equation derived above is: . Substitute Darcy’s Law into this gives:

This looks ALMOST like the Laplace Equations above. If kx = kz (i.e. ISOTROPIC permeability):

which IS the same as the Laplace Equation for potential. Therefore, for ISOTROPIC permeability, lines of equal total head are also equipotential lines.



2. Flow netsA properly constructed flow net consists of a series of equipotential lines (contours of equal total head), with the same change in head between each successive pair, and a series of flow lines (streamlines) defining a set of ‘flow channels’, with equal flow quantity in each flow channel.

2.1 ‘Shape’ of the meshFor an individual mesh ‘element’, average physical width of flow channel is b, average physical length is ℓ (with depth ‘back into page’ = 1 unit).

Darcy’s Law:

But is the same as , by the very meaning of the stream

lines and the stream function:

If is constant between each successive pair of equipotential lines (same head drop between successive pairs of constant head lines), and is constant between each successive pair of stream lines (same quantity of flow Q in each stream channel), then:

Therefore, a properly constructed flow net has mesh elements with a CONSTANT RATIO of width to length. The easiest constant ratio to deal with is 1 – i.e. a properly constructed flow net has mesh elements with b = ℓ. This is a so-called “square” mesh (the “squares” are curvilinear “squares”). One test is to be able to draw a circle with the element, with the circle just touching each side.

2.2 Total quantity of flow from flow net

The flow net is set up with an equal quantity of flow in each flow ‘channel’. Let nf be the total number of flow channels. Then, total flow rate

.

Also, the total head drop in the problem hT is divided into an equal number of head drops nh, so that the head drop between each pair of equipotential lines is


b ℓ

Q

b ℓ

Q

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for a “square” mesh flow net (where b = ℓ for all mesh elements). Note that nf is the number of flow channels (not the number of flow lines), and nh is the number of head “drops” (not the number of equipotential lines).

In the problem shown above, the number of head ‘drops’ is 14, and the number of flow ‘channels’ is 7.

2.3 Rules for Drawing Flow Nets: Stream lines cross equipotential lines at right angles

Stream lines intersect the ‘input’ and ‘output’ boundaries at right angles (these boundaries are equipotential lines)

Equipotential lines intersect the flow boundaries at right angles (these boundaries are stream lines)

Mesh elements are ‘square ‘ (i.e. b = ℓ for all mesh elements – it should be possible to inscribe a circle in each mesh element, just touching all sides of the element)

A flow net drawn following these rules has equal head drop between equipotentials, and equal flow in each stream channel.

We can choose the number of equipotential lines to draw, or we can choose the number of flow lines to draw, but we can’t choose both. Thus, choosing a certain number of equipotential lines (and hence a certain number of head drops, nh), and drawing the flow net correctly, will always give the same ratio of nf/nh (since the flow rate in the problem is fixed, irrespective of how we draw the flow net).

Choose an integer number for nh – but in some cases this will result in a non-integer value for nf (i.e. we can have 5½ or 3⅓ flow channels)

From flow net we can find:o Quantity of flow (using formula above)o Local seepage velocity along a stream line (between two equipotential lines)o Time for seepage to occur from one side to another along a particular stream line. The time

varies from one stream line to another; the fastest flow is along shortest stream line, which may be one of the flow boundaries.

o Uplift pressure on the base of a dam (or whatever the flow is occurring under)o Stability against sliding or overturning.



3. Anisotropic permeability and flow net sketching

We have said earlier that all this theory only works if permeability is isotropic (i.e. if kx = kz).

What if this is not the case – i.e. permeability is anisotropic? Say that the maximum and minimum directions of permeability correspond to x and z directions (a common situation in sedimentary deposits).

In our previous derivation, before we introduced the requirement for isotropic permeability, we got to the stage of showing that:

, which we can re-write as

Using some geometry transformation, we want to get this into the form of a Laplace Equation. To do so, use a distortion of the x-dimensions – choose an X coordinate is some transformed x coordinate.

Choose this new coordinate to be: . This gives

Now we have:

Substituting this into gives Laplace Equation

So, if we distort the x-scale, by multiplying it by , we end up with a transformed section, and the

possibility of using the flow net approach.

Example: Say kx = 9kz. Then . Reduce all x-dimensions by factor of 3:

12m6m

kx = 9kz

4m6m

ke = 3kz=kx /3

Real section Transformed section

12m6m

kx = 9kz

4m6m

ke = 3kz=kx /3


But we need to know what equivalent permeability ke we have to use with this transformed section to do the calculations for flow.



Real Section Transformed Section Thus, for same flow through the two sections, require:

, or

z

x

Q

kxkz

z

Q

ke

xkk

Xx

z .


Head = hh

Head =h

z

x

Q

kxkz

z

Q

ke

xkk

Xx

z .


Head = hh

Head =h

So, in the example above, where kx = 9kz, the equivalent permeability to use with the transformed

section is .

(Would get same result if we looked at horizontal flow).

3.1 Procedure for anisotropic permeability cases:

Re-draw the geometry, with transformed (distorted) horizontal scale

Draw normal ‘square’ flow net on the transformed section, using the normal rules

Compute quantity of flow using the usual formula, but using the equivalent permeability

If required, re-draw geometry to true scale, and map the flow net back to it from the distorted case. The resulting flow net will show ‘non-square’ mesh elements (b≠ℓ) and with intersections between head lines and flow lines being not necessarily right angles. (Note: lines of equal head are now not equipotential lines, as, by mathematical definition, equipotential lines are normal to flow lines).



4. Field tests to determine permeabilityIn the notes on Permeability, it was explained that laboratory measurements of permeability for the purpose of determining the mass permeability of the ground, is very unreliable. Where accurate mass values are required, field tests are essential.

There are many types of field test. One simple test, called a SLUG TEST, that can be carried out while drilling an exploratory borehole, is as follows:

Allow the water level in the borehole to come to equilibrium with the water table.

Suddenly fill the borehole to a level well above the water table.

Monitor the rate at which the water in the borehole falls back to the level of the water table. The rate at which it occurs clearly depends on the permeability of the soil around the borehole.

Interpretation of such a test is fairly straightforward, with the solution depending on whether the borehole is lined with an impermeable casing (and flow is only out of the base of the borehole), or whether the borehole is uncased (over the full depth or over the lower part of the borehole.

A variation of this test is to extract water from the borehole quickly at the start, and monitor the rate at which the water level rises back to that of the water table.

These tests determine the permeability in the soil just around the borehole, but this is still a rather limited scale. Tests at much larger scale are also used, as described below.

4.1 “Steady state” pumping test in an unconfined aquiferSetup is as shown. Scale can be small (r2 < 10 m) or very large (r2 > 500 m), depending on the depth of the aquifer. The pumping and observation wells penetrate to the base of the aquifer.

Pump at a rate that gives significant lowering of the level in the pumping well, and continue pumping to maintain that level until the pumping rate appears to be constant ( ). At this stage, the total head h(r) is a function of radius, and the hydraulic gradient is i = –dh/dr. The heads in the two observation wells are h1 and h2.


h1h2

r1r2

QPumping well Observation wells

Perforated casing

Original water table

Drawn down water table

h1h2

r1r2


Perforated casing

Original water table

Drawn down water table


Assumptions required: The pumping rate is constant Steady state is reached (h1 and h2 don’t change) The flow is horizontal (radially inwards) At radius r, flow occurs radially inwards across a

cylindrical area with radius r, circumferential length 2r, height h(r), and surface area A = 2rh.

Applying Darcy’s law:

4.2 Confined aquiferSituation is very similar to previous case. The difference is that in a confined aquifer, the ‘water table’ does not exist (there is no level in the aquifer where the water is at atmospheric pressure). Standpipes down into the aquifer define a ‘piezometric surface’, which is the equivalent of a water table.

In the pumping test, the pumping must keep the piezometric surface above the top of the aquifer (at least between the two observation wells). In this case, the equivalent cylindrical area across which flow is occurring has a constant height D (the thickness of the aquifer), irrespective of radius – i.e. A = 2rD. Everything else remains the same (i is still – dh/dr)


h1 h2

r1r2


Perforated casing

Permeable aquifer

Impermeable

Impermeable

Piezometricsurfaces

Dh1 h2

r1r2


Perforated casing

Permeable aquifer

Impermeable

Impermeable

Piezometricsurfaces

D

2 dimensional flow

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