Permeability and Seepage -2 - Nptel

Prof. B V S Viswanadham, Department of Civil Engineering, IIT Bombay

13


Permeability and Seepage -2


Conditions favourable for the formation quick sandQuick sand is not a type of sand but a flow conditionoccurring within a cohesion-less soil when its effective stressis reduced to zero due to upward flow of water.

Quick sand occurs in nature when water is being forcedupward under pressurized conditions.

In this case, the pressure of the escaping water exceedsthe weight of the soil and the sand grains are forced apart.The result is that the soil has no capability to support a load.

Why does quick condition or boiling occurs mostly in fine sands or silts?


Some practical examples of quick conditionsExcavations in granular materials behindcofferdams alongside rivers

Any place where artesian pressures exist (i.e. wherehead of water is greater than the usual static waterpressure).

-- When a pervious underground structure iscontinuous and connected to a place where head ishigher.

Behind river embankments to protect floods


Contrary to popular belief, it is not possible to drownin quick sand, because the density of quick sand ismuch greater than that of water.

waterquicksand ρρ >>

Consequently, it is literally impossible for a person tobe sucked into quicksand and disappear. So, aperson walking into quicksand would sink to aboutwaist depth and then float.

Some practical examples of quick conditions


• Example• The sand layer of the

soil profile shown in figure is under artesian pressure. A trench is to be excavated in the clay up to a depth of 4m. Determine the depth of water h to avoid boiling.

PSand

Rock

3m

4m5m

Water level

h

stiff clay γ = 18kN/m3

6m

Boiling condition


• Solution:• At P: σ = hγw + (6-4)γclay• = hγw + 2γclay

• At P: u = 5γw

• At P: σ’ = hγw + 2γclay - 5γw

• σ’ = 0 (boiling condition)• hγw + 2γclay - 5γw = 0• h = (5γw - 2γclay )/γw

• = (5 x 9.81 – 2x18)/9.81• h = 1.33m

PSand

Rock

3m

4m5m

Water level

h = 1.33 m

stiff clay γ = 18kN/m3

6m

Boiling condition


Seepage forcesWater flowing past asoil particle exerts a dragforce on the particle inthe direction of flow.

The drag force iscaused by pressuregradient and by viscousdrag.

h

L

B

Wd

(Fs)c


B

Wd

(Fs)cW

FBD of soil

Direction of flow

FBD of grain

At critical condition; h = hc

( ) wcws ALhAL

eeG γγ +=

++

1

+−

==e

GLhi sc

c 11

Seepage forces


Considering FBD of grain in the direction of flow at i = ic( )

( )

( )

( ) ( )

( ) wws

cs

wVws

cs

wsws

cs

dcs

csd

ALnALe

GF

VV

VVALe

GF

VALe

GF

BWFBFW

γγ

γγ

γγ

)1(1

1

1

−−

+=

−−

+=

−

+=

−=

+=

ViF

Ve

GF

wccs

ws

cs

γ

γ

=

+−

=

)(1

1)(

Seepage pressure

ps = icγw

If h < hc then seepage force Js is iγwV


∆l

∆lDirection of flow

∆h

Seepage forceArea per unit width of section = ∆l x 1

Volume affected by seepage force = ∆l2 x 1

Force applied to sand particles = γwh1 -γwh2 (∆l x 1)

h1 h2

h1 > h2


Seepage force

Force applied to sand particles = γw (∆h) (∆l x 1)

= γw (∆h/∆l) (∆l2 x 1)

J = i γw V

Seepage pressure ps = Seepage force per

unit volume

= i γw


Critical hydraulic gradient – Quick conditionThe quick condition occurs at a critical upwardhydraulic gradient ic, when the seepage force justbalances the buoyant weight of an element of soil.(Shear stresses on the sides of the element are

neglected.)

−

=υ

1sc

Gi

The critical hydraulic gradient is typically around 1.0 for manysoils. Fluidized beds in chemical engineering systems rely ondeliberate generation of quick conditions to ensure that thechemical process can occur most efficiently.


Example problem

Determine and plot totalstress, PWP and effectivestress diagrams if: (i) h = 1 m;(ii) h = 4 m; and (iii) h = 2 m.

a

b

c

d

2 m

h

1 m

1 m

Datum

γsat =20 kN/m32hi =

Case – I; i = 0.5;

Case – II; i = 2;

Case – III; i = 1


Case - I i = 0.5

Point PH [m] EH [m] TH [m]

a 0 4 4

b 2 2 4

d 5 0 5

c 3.5 1 4.5


a

b

c

d

2 m

h = 1 m

1 m

1 m

Datum

γsat =20 kN/m320

40

60

20

35

50 10

5

σ σ´u

Stress units are in kN/m2

Case - I i = 0.5


Case - II i = 2


a 0 4 4

b 2 2 4

d 8 0 8

c 5 1 6


a

b

c

d

2 m

h = 4 m

1 m

1 m

Datum

γsat =20 kN/m320

40

60

20

50

80

σ

-10

σ´u

-20

Quick sand condition wouldhave already occurred…

Stress units are in kN/m2Case - II


Case - III i = 1


a 0 4 4

b 2 2 4

d 6 0 6

c 4 1 5


Just subjected to Quick sand condition…

a

b

c

d

2 m

h = 2 m

1 m

1 m

Datum

γsat =20 kN/m320

40

60

20

40

60

σ

0

σ´u

0

0

Stress units are in kN/m2

0

Case - III


A large open excavation was made in a stratum of clay with asaturated unit weight of 17.6 kN/m3. When the depth of theexcavation reached 7.5 m, the bottom rose, gradually crackedand was flooded from below by a mixture of sand and water.Subsequent borings showed that the clay was underlain by abed of sand with its surface at a depth of 11 m.

Compute the elevation for which the water would have risenfrom the sand surface into a drill hole before the excavation wasstarted.

Example problem


γsat = 17.6 kN/m3

11m

Solution


h

For σ´= 0 at A: σ´= 17.6 x (11 – 7.5) - 10 x h = 0

h = 6.16 m

Solution

Elevation for which thewater would have risenfrom the sand surfaceinto a drill hole beforethe excavation wasstarted.

A

7.5 m


3 m

6 m

4 m

Coarse sand; Gs = 2.65; e = 0.833

Silty clay; Gs = 2.70; w = 45.2 %

3 mB

A

C

Determine and plot the total vertical stress, pore waterand effective vertical stress distribution at levels A, Band C.

Example problem


Example problemAn artesian pressureexists in the lowersand layer.

Determine σ, u, andσ ′ at A, B, and C.

(TH)A = 2 + 3 = 5m

(TH)B = 7 + 0 = 7m

1 m

2 m

γclay = 17 kN/m3

A

C

B

2 m

1 m

γd = 18.5 kN/m3

γsat = 19 kN/m3

DATUM

1 m

2 m

1.33 m


At point A:σ = 18.5 x 1 + 19 x 2 = 56.5 kN/m2

u = 10 x 2 = 20 kN/m2

σ′ = 56.5 – 20 = 36.5 kN/m2

At point B:

σ = 18.5 x 1 + 19 x 2 + 17 x 3 = 107.5 kN/m2

u = 10 x 7 = 70 kN/m2

σ′ = 107.5 – 70 = 37.5 kN/m2

Solution


At point c:

σ = 18.5 x 1 + 19 x 2 + 17 x 2 = 90.5 kN/m2

u = 10 x (PH)c = 10 x 5.33 = 53.3 kN/m2

σ′ = 107.5 – 53.3 = 38.2 kN/m2

Solution

(TH)C = (TH)B - i Z

= 7 – (2/3) x 1 = 6.33 m

(PH)C = (TH)C – Zc = 6.33 – 1 = 5.33 m


1 m

2 mA

C

B

2 m

1 mDATUM

1 m

90.5

56.5

53.3

20 36.5

107.5 70

38.2

37.5


Measurement of soil permeabilitiesThe rate of flow of water q (volume/time) throughcross-sectional area A is found to be proportional tohydraulic gradient i according to Darcy’s law:

kiAqv ==

Lhi =

where v is flow velocity and k is coefficient of permeability with dimensions of velocity (length/time).

The coefficient of permeability of a soil is a measure of theconductance (i.e. the reciprocal of the resistance) that itprovides to the flow of water through its pores.


The value of the coefficient of permeability k dependson:

(i) Average size of the pores and is related to theparticle sizes and their packing,

(ii) Particle shape, and(iii) Soil structure.

Measurement of soil permeabilities

The ratio of permeabilities of typical sands/gravels to those oftypical clays is of the order of 106. A small proportion of finematerial in a coarse-grained soil can lead to a significantreduction in permeability.

510

10

dd

edd

=

=


Measurement of soil permeabilities

Laboratory methods:

The constant head test (used for highly permeable soils)The falling head test

(used for relatively impermeable soils)Indirect methods: computation from grain size distribution and During Oedometer test

A number of tests can be used to measure or estimate the permeability of soils

Field methods:1. Pumping tests and 2. Borehole tests


Laboratory measurement of permeabilityIn determining permeability of coarse grainedmaterial, large quantities of flow occur in short periodsof time and small quantities of flow occur over longperiods of time for fine grained soils.

Two aspects that need careful attention for all types of soils are:(i) To ensure that flow occurs only through the soil and not at theinterface between soil and the mould in which the soil iscontained.

(ii) The soil sample is fully saturated before recordingobservations


hL

Constant water level

Water supply

Determination of coefficient of permeability in the laboratory

Constant Head test

AhtQLk

tLhAkQ

Lhi

AvtQ

=

=

=

=A = area of C/S of the specimen


Main features of constant head test

It is suitable for soils having a coefficient ofpermeability in the range of 10-2 m/s to 10-5 m/s,which applies to clean sand and sand-gravelmixtures with less than 10 % fines.

It can be suitable for soils when used in theircompletely disturbed or remolded states such as fordrainage materials and filters to confirm that theirperformance will be adequate.


dh

L

h1h2h

Falling head test or Variable head test

Cross-sectional area of stand pipe = a

Cross-sectional area of specimen = A

At time to head h1

At time t1 head h2

Let h be the head of water at any time t. Let in time dtthe head drop by amount dh


Quantity of water flowing through the sample in time dt from Darcy’s law:

)(

)(

dtALhk

dtkiAdQ

=

= Quantity of discharge can also be expressed as: )(dhadQ −=

=

=− ∫∫ =

2

110log303.2

1

0

2

1

hh

AtaLk

dtLkAdha

t

tt

h

h

Falling head test

As the time increases head decreases!!


Main features of constant head and falling head testsIn the constant head test, permeability is computed on thebasis of fluid that passes through the soil sample.

While in the falling head test, k is computed on the basis offluid flowing into the sample.With the constant head test, time is required to accumulatethe fluid volume necessary to perform computation. Extremecare would be required to prevent leaks in the apparatus andevaporation of discharge water.

With the falling head test, the duration of the test is shortenedand care is required to prevent evaporation of water in the inlettube.


3 m

6 m

4 m

Coarse sand; Gs = 2.65; e = 0.833

Silty clay; Gs = 2.70; w = 45.2 %

3 mB

A

C

Determine and plot the total vertical stress, pore waterand effective vertical stress distribution at levels A, Band C.

Problem

Permeability and Seepage -2 - Nptel

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