Anisotropic Flow Nets

7/27/2019 Anisotropic Flow Nets

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7. FLOW NETS FOR ANISOTROPIC MATERIALS

7.1 Introduction

Many soils are formed in horizontal layers as a result of sedimentation through water. Because ofseasonal variations such deposits tend to be horizontally layered and this results in different

permeabilities in the horizontal and vertical directions.

7.2 Permeability of Layered Deposits

Consider the horizontally layered deposit, shown in Figure 1, which consists of pairs of layers the firstof which has a permeability of k1 and a thickness of d1 overlaying a second which has permeability k2and thickness d2.

d1

d2

k=k1

k=k2

Fig. 1 Layered soil deposit

First consider horizontal flow in the system and suppose that a head difference ofh exists between the

left and right hand sides as indicated in Fig. 2. It then follows from Darcys law that:

v kh

LQ k

h

Ld

and

v kh

LQ k

h

Ld

1 1 1 1 1

2 2 2 2 2

= =

= =

;

;

d1

d2

v=v1

v=v2

Fig. 2 Horizontal flow in a layered soil deposit

h h= 0 h h h= 0

L

It therefore follows:

1

Fig. 1 Layered Soil

(2a)

Fig. 2 Horizontal flow through layered soil

(1a)

(1b)


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vQ Q

d dk

h

Lwhere

kk d k d

d d

H

H

=+

+=

=+

+

1 2

1 2

1 1 2 2

1 2

Next consider vertical flow through the system, shown in Fig.3. Suppose that the superficial velocity ineach of the layers is v and that the head loss in layer 1 is h1, while the head loss in layer 2 is h2

d1

d2

v

v

Fig. 3Vertical flow in a layered soil deposit

h h= 0

L

h h h h= 0 1 2

h h h= 0 1

In layer 1: v kh

d= 1

1

1

so hv d

k1

1

1

= (3a)

Similarly in layer 2 v kh

d= 2

2

2

and h

v d

k2

2

2

= (3b)

The total head loss across the system will be h=h1+h2 and the hydraulic gradient will be given by:

ih

d

h h

d d

vd

k

vd

k

d d= =

+

+=

+

+

1 2

1 2

1

1

2

2

1 2

(3c)

For vertical flow Darcys Law gives

v kh

dV=

(3d)

and hence

d

k

d

k

d

kV= +

1

1

2

2

(3e)

Example

2

Fig. 3 Vertical flow through layered soil

(2b)


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Suppose that that the layers are of equal thickness d1 = d2 = d0and that k1 = 10-8 m/sec and that

k2 = 10-10 m/sec, then:

/1005.51010

9

108

mdd

ddk

oo

oo

H

=

+

+=

and

sec/1098.1

1010

10

108

mdd

ddk

oo

oo

V

=

+

+=

Showing that, as is generally the case, the vertical permeability is much less than the horizontal.

7.3 Flow nets for soil with anisotropic permeability

Plane flow in an anisotropic material having a horizontal permeability kHand a vertical permeability kvis governed by the equation:

kh

xk

h

zH V

2

2

2

20+ = (4)

The solution of this equation can be reduced to that of flow in an isotropic material by the followingsimple device. Introduce new variables defined as follows:

x x

and

z z

=

=

(5a)

the seepage equation then becomes

k

k

h

x

h

z

H

V

2

2

2

2

20+ = (5b)

Thus by choosing:

= kk

H

V

(5c)

It is found that the equation governing flow in an anisotropic soil reduces to that for an isotropic soil,viz.:

2

2

2

20

h

x

h

z+ = (5d)

and so the flow in anisotropic soil can be analysed using the same methods (including sketching flownets) that are used for analysing isotropic soils.

Example - Seepage in an anisotropic soil

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Suppose we wish to calculate the flow under the dam shown in Figure 4;

x

z

Impermeable bedrock

L

H1H2

Impermeable

dam

Soil layerZ

Fig. 4 Dam on a permeable soil layer over impermeable rock (natural scale)

For the soil shown in Fig. (4) it is found that k kH V= 4 and therefore

=

=

= =

=

42

22

k

k

so

x x or xx

z z

V

V

(6)

In terms of transformed co-ordinates this becomes as shown in Figure 5

z

Impermeable bedrock

L/2

H1H2

x Soil layer

Z

Fig. 5 Dam on a permeable layer over impermeable rock (transformed scale)

The flow net can now be drawn in the transformed co-ordinates and this is shown in Fig.6

4


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Fig. 6 Flow-net transformed coordinates

5m

Impermeable bedrock

Fig. 6 Flow net for the transformed geometry

It is possible to use the flow net in the transformed space to calculate the flow underneath the dam byintroducing an equivalent permeability

k k keq H V= (7)

A rigorous proof of this result will not be given here, but it can be demonstrated to work for purelyhorizontal flow as follows:

xx

Natural scaletransformed

scale

Q

t

h h -h h h -h

Fig. 7 Horizontal flow through anisotropic soil

For the natural scale

Q k th

xH=

(7a)

For the transformed scale

Q k th

xk t

h

x

k

keq eq

H

V

= = (7b)

From Equations 7a and b it can be seen that k k keq H V=

Example

5


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Suppose that in Figure 6 H1 = 13m and H2 = 2.5m, and that kv = 10-6m/sec and kH=4 10

-6m/sec

The equivalent permeability is:

k meq = = ( ) ( ) / sec4 10 10 2 10

6 6 6 (8a)

The total head drop is 10.5 m and there are 14 head drops and thus:

h m= =( . ) .13 2 514

0 75 (8b)

The flow through each flow tube, Q = keqh = (210-6 )(0.75) = 1.5 10-6 m3/s/m

There are 6 flow tubes and so the total flow , Q = 6 1.5 10-6

= 9.010-6 m3/sec/(m width of dam)

For a dam with a width of 50 m Q = 450 10-6 m3/sec = 41.47 m3/day

7.4 Piping

Many dams on soil foundations have failed because of the sudden formation of a piped shapeddischarge channel. As the store water rushes out the channel widens and catastrophic failure results.This results from erosion of fine particles due to water flow. Another situation where flow can causefailure is in producing quicksand conditions. This is also often referred to as piping failure.

In order to analyse this situation consider water flowing upwards through the element shown in Figure8.

Area =A

(z=z2 , h=h2 , u=u2,)

(z=z1 , h=h1 ,u=u1,)

Elevation

Fig. 7 Analysis of Piping

Plan

u2

u1

6

Fig. 8 Analysis of Piping


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Uplift Force A u u

Force due to weight A z zsat

=

=

( )

( )

1 2

2 1

(9a)

The pore pressure can be calculated from the head and so:

u h zand

u h z

w

w

2 2 2

1 1 1

=

=

( )

( )

(9b)

For piping to occur the Uplift must be greater than the self-weight of the soil

A u u A z z

h h z z z z

h h z z z z

h hz z

sat

w w sat

w sat w

sat w

w

( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( )( )

2 1 2 1

1 2 1 2 2 1

1 2 2 1 2 1

1 2

2 1

>

>

>

>

or alternatively

i i

where

i hydraulic gradienth h

z z

and

i critical hydraulic gradient

crit

crit

sat w

w

>

= =

= =

1 2

2 1

Example

Suppose the dam shown in Figure 6 is 39 metres wide (this may be determined from the scale drawing),the water levels are the same as in the previous example (H1 = 13 m, H2 = 2.5 m), and the saturated unitweight of the soil is 18 kN/m3. Piping is most likely to occur at the toe of the dam, the hydraulicgradient there can be obtained from the flow net:

h1 - h2 = h = 0.75 m (calculated from Fig. 6)

z2 - z1 = 1.125 m (scaled from Fig. 6)

thus

i = =0 75

11250 67

.

..

Now

i crit =

=18 9 81

9 810 83

.

.. (10)

7

(9d)

(9c)


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The safety factor against piping failure is thus icrit/i = 0.83/0.67 = 1.25 which is probably not adequategiven the potentially disastrous consequences of a piping failure.

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Anisotropic Flow Nets