Lecture 3a (Shamma) 1.17.12 Permeability.ppt

8/18/2019 Lecture 3a (Shamma) 1.17.12 Permeability.ppt

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Permeability and Seepage

N. Sivakugan

Flow Nets

Philip B. Bedient

Civil & Environmental Engineering

Rice University


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Permeability and Seepage

Flow Nets


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What is permeability?

A measure of how easily a fluid (e.g., water)can pass through a porous medium (e.g.,soils)

Loose soil

- easy to flow

- high permeability

Dense soil

- difficult to flow

- low permeability

water


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Bernoulli’s Equation

1. Kinetic energy

datum

z

fluid particle

The energy of a fluid particle ismade of:

2. Strain energy

3. Potential energy

- due to velocity

- due to pressure

- due to elevation (z) with respect to a datum


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Total head =

datum

z

fluid particle

Expressing energy in unit of length:

Velocity head

+

Pressure head

+

Elevation head


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Total head =

datum

z

fluid particle

For flow through soils, velocity (and thus

velocity head) is very small. Therefore,

Velocity head

+

Pressure head

+

Elevation head

0

Total head = Pressure head + Elevation head


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Some NotesIf flow is from A to B, total head is higher at

A than at B.

water

AB

Energy is dissipated inovercoming the soil

resistance and hence

is the head loss.


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Some Notes

Pressure head = pore water pressure/w

Elevation head = height above the selected datum

At any point within the flow regime:


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Some Notes

Hydraulic gradient (i) between A and B isthe total head loss per unit length.

water

AB

AB

B A

l

TH TH i

length AB, along the

stream line


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Darcy’s Law

Velocity (v) of flow is proportional to the

hydraulic gradient (i) – Darcy (1856)

v = k i

Permeability

• or hydraulic conductivity

• unit of velocity (cm/s) (ft/S)


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Large Earth Dam

SHELL

FOUNDATION

SHELL

CORE

blanket

filter

cutoff

crest

riprap

free board


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Permeability Values (cm/s)10-

3

10-

6

100

clays gravelssandssilts

CoarseFines

For coarse grain soils, k = f(e or

D10)


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Flow Net Theory

1. Streamlines and Equipotential lines

are .

2. Streamlines are parallel to no flowboundaries.

3. Grids are curvilinear squares, where

diagonals cross at right angles.

4. Each stream tube carries the same

flow.


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Flow Net Theory

C i h ©2001


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Flow Net in Isotropic Soil

Portion of a flow net is shown below

S t r e a m t u b e

C i h ©2001


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The equation for flow nets originates

from Darcyᾼs Law.

Flow Net solution is equivalent to

solving the governing equations of flow

for a uniform isotropic aquifer with well-

defined boundary conditions.

C i ht©2001


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Flow through a channel betweenequipotential lines 1 and 2 perunit width is:

q = K (d m

x 1)( h 1

/dl )

d m

h1

dl

q

h2

q

n

m

C i ht©2001


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Flow through equipotential lines 2 and 3is:

q = K (d m x 1)( h 2 /dl )

The flow net has square grids, so thehead drop is the same in each potentialdrop: h 1 = h 2

If there are n d such drops, then:

h = ( H /n )where H is the total head loss

between the first and last equipotentiallines.

C i ht©2001


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Substitution yields:

ᾶ

q = K (d

m

x dl )(H/n )

This equation is for one flow channel. If

there are m such channels in the net,

then total flow per unit width is:

ᾶ q = ( m /n )K (d m /dl )H

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Since the flow net is drawn withsquares, then d m dl , and:

q = ( m /n )KH

where:ᾶ q = rate of flow or seepage per unit width

ᾶ m = number of flow channels

ᾶ n = number of equipotential drops

ᾶ h = total head loss in flow system

ᾶ K = hydraulic conductivity

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Drawing Method:

1. Draw to a convenient scale the cross

sections of the structure, water

elevations, and aquifer profiles.

2. Establish boundary conditions and draw

one or two flow lines and

equipotential lines near theboundaries.

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Method:

3. Sketch intermediate flow lines andequipotential lines by smooth curves

adhering to right-angle intersections and

square grids. Where flow direction is a

straight line, flow lines are an equal distanceapart and parallel.

4. Continue sketching until a problem

develops. Each problem will indicatechanges to be made in the entire net.

Successive trials will result in a reasonably

consistent flow net.

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Method :

5. In most cases, 5 to 10 flow lines are

usually sufficient. Depending on the

number of flow lines selected, the

number of equipotential lines willautomatically be fixed by geometry

and grid layout.

6. Equivalent to solving the governing

equations of GW flow in 2-dimensions.

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Seepage Under Dams

Flow nets for

seepage through

earthen dams

Seepage underconcrete dams

Uses boundary

conditions (L & R)

Requires

curvilinear square

grids for solution

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Two Layer Flow System withSand Below

Ku / K l = 1 / 50

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Two Layer Flow System withTight Silt Below

Flow nets for seepage from one side of a channel through two

different anisotropic two-layer systems. a) K

u

/ K

l

= 1/50 . b)

K

u

/ K

l

= 50 . Source: Todd Bear, 1961.

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Effects of Boundary Condition

on Shape of Flow Nets

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Radial Flow:

Contour map of the piezometric surface near Savannah,

Georgia, 1957, showing closed contours resulting from

heavy local groundwater pumping after USGS

Water-

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Flow Net in a Corner:

Streamlines

are at right

angles to

equipotential

lines

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Flow Nets: an example

A dam is constructed on a permeable

stratum underlain by an impermeable

rock. A row of sheet pile is installed at

the upstream face. If the permeable soilhas a hydraulic conductivity of 150

ft/day, determine the rate of flow or

seepage under the dam.


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Flow Nets: the solution

Solve for the flow per unit width:

q = ( m /n ) K h

= (5/17)(150)(35)

= 1544 ft 3/day per ft

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Flow Nets: An Example

There is an earthen dam 13 metersacross and 7.5 meters high.TheImpounded water is 6.2 meters deep,

while the tailwater is 2.2 meters deep.The dam is 72 meters long. If thehydraulic conductivity is 6.1 x 10 -4 centimeter per second, what is the

seepage through the dam if n = 21

K = 6.1 x 10 -4cm/sec

= 0.527 m/day

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From the flow net, the total head loss,

H , is 6.2 -2.2 = 4.0 meters. There are 6 flow channels ( m ) and 21

head drops along each flow path ( n ):

Q = ( KmH /n ) x dam length

= (0.527 m/day x 6 x 4m / 21) x (dam

length) = 0.60 m 3/day per m of dam

= 43.4 m 3/day for the entire 72-meter

length of the dam

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py g

http://homepage.eircom.net/~jmcgeever/Soil%20Mechanics/Soil%20Permeability%20-%20Flow%20Nets.htm


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Stresses due to FlowDownward Flow

hw

L

flow

X

soil

z

v = whw +

satz

w hw + w(L-hL)

(z/L)

v' = ' z + wiz

At X,

hL u = w hw

u = w

(hw

+L-

h

… as for static case

= w hw + w(z-iz)

= w

(hw

+z) -w

iz

Reduction due to flow

Increase due to flow

u =

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Stresses due to Flow

flow

Upward Flow

hw

LX

soil

z

v = whw +

satz

w hw + w(L+hL)

(z/L)

v' = ' z - wiz

At X,

hL

u = whw

u = w

(hw+L+hL)

… as for static case

= w hw + w(z+iz)

= w

(hw

+z) +w

iz

Increase due to flow

Reduction due to flow

u =

Copyright©2001

Q i k C di i i G l S il


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Quick Condition in Granular Soils

During upward flow, at X:

v' = ' z - wiz

flow

hw

LX

soil

z

hL

i z w

w

'

Critical hydraulic gradient (ic)

If i > ic, the effective stresses is negative.

i.e., no inter-granular contact & thus failure.

- Quick condition

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S T i l


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Seepage Terminology

concrete dam

impervious strata

soil

Stream line is simply the path of a water molecule.

datum

hL

TH = 0TH = hL

From upstream to downstream, total head steadily decreasesalong the stream line.

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S T i l


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Seepage TerminologyEquipotential line is simply a contour of constant

total head.

concrete dam

impervious strata

soil

datum

hL

TH = 0TH = hL

TH=0.8

hL

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Fl


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Flownet A network of selected stream lines and equipotential

lines.

concrete dam

impervious strata

soil

curvilinearsquare

90º

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Q tit f S (Q)


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Quantity of Seepage (Q)

d

f

L N

N

khQ ….per unit length normal to the plane

# of flow channels

# of equipotential drops

impervious strata

concrete

dam

h

L

head loss from upstream to

downstream

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H d t P i t X


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Heads at a Point X

impervious strata

concrete

dam

datum

X

z

h

LTH = hL TH = 0

Total head = hL - # of drops from upstream x h

h

Elevation head = -z

Pressure head = Total head – Elevation headd

L

N h

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Pi i i G l S il


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Piping in Granular Soils

datumconcrete

dam

impervious strata

soil

h

L

At the downstream, near the dam,

h = total head dropl

l hiexit the exit hydraulic gradient

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Pi i i G l S il


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datumconcrete

dam

impervious strata

soil

h

L

If iexit exceeds the critical hydraulic gradient (ic), firstly

the soil grains at exit get washed away.

no soil; all water

This phenomenon progresses towards the upstream, forming a

free passage of water (“pipe”).

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Pi i i G l S il


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Piping is a very serious problem. It leads to downstream

flooding which can result in loss of lives.

concrete

dam

impervious strata

soil

Therefore, provide adequate safety factor against piping.

exit

c piping

i

i

F

typically 5-6

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Pi i F il


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Piping FailuresBaldwin Hills Dam after it failed by

piping in 1963. The failure occurredwhen a concentrated leak developed

along a crack in the embankment,

eroding the embankment fill and

forming this crevasse. An alarm wasraised about four hours before the

failure and thousands of people were

evacuated from the area below the

dam. The flood that resulted when thedam failed and the reservoir was

released caused several millions of

dollars in damage.

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Pi i F il


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Piping Failures

Fontenelle Dam, USA (1965)

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FiltersUsed for:

facilitating drainage

preventing fines from being washed away

Used in:

earth dams

retaining walls

Filter Materials:

granular soils

geotextiless

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Granular Filter DesignTwo major criteria:

(a) Retention Criteria

(b) Permeability Criteria

- to prevent washing out of fines

- to facilitate drainage and thus avoid build-up of pore pressures

Filter grains must not be too coarse

Filter grains must not be too fine

granular filter

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Granular Filter DesignRetention criteria:

D15, filter < 5

D85, soil

- after Terzaghi & Peck (1967)

Permeability criteria:

D15, filter > 4

D15, soilaverage filter pore size

D15, filter < 20 D15,

soilD50, filter < 25 D50,

soil

- after US Navy (1971)

GSD Curves for the soil and filter must be parallel

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Ss h

t

K 2h

v v

1

x v

2

y v

3

z

Gradient :

“del operator”

Ґi

x Ґ j

y Ґk

z

w Ґi w x

Ґ j w y

Ґk w z

Divergence:

Diffusion Equation:

2 f 2 f

x 2

2 f

y 2

2 f

z 2 Laplacian:

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“del operator” i

j

k


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Ss h

t

K 2h

v v

1

x v

2

y v

3

z

Gradient :

“del operator”

Ґi

x Ґ j

y Ґk

z

w Ґi w x

Ґ j w y

Ґk w z

Divergence:

Diffusion Equation:

2 f 2 f

x 2

2 f

y 2

2 f

z 2 Laplacian:

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Ss h

K 2h“Diffusion Equation”


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2h

x2 + 2h

y2 + 2h

z2 =

SsKht

1r r

r hr

+ 1r 2

2h

2 +

2h

z2 =

SsKh

t

Kr

2hr 2

+Kr

r hr

+ Kz

2hz2

= Ssht

2h

r 2 + 1r

hr

= SsKr

ht

= STht

Cartesian Coordinates

Cylindrical Coordinates

Cylindrical Coordinates,

Radial Symmetry ∂h/∂ = 0

Cylindrical Coordinates,

Purely Radial Flow

∂h/∂ = 0 ∂h/∂z = 0

Ss t

K h Diffusion Equation

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Flow beneath Dam


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Vertical x-section

Flow toward Pumping Well,

next to river= line source

= constant head boundary

Plan view

River Channel

Lecture 3a (Shamma) 1.17.12 Permeability.ppt

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