enb272 - ln - week 4 - permeability - 6 slides per page -gray

Soil Permeability (Week -4)

ENB272: Geotechnical Engineering 1ENB272: Geotechnical Engineering 1ChamindaChaminda

Chapter 2Chapter 2

Contents• What is permeability• Applications of soil permeability• Factors affect on soil permeability• Ground water flow• Laboratory measurement of

permeability coefficient (k)• Field measurement of k• Permeability of stratified soils

What is permeability(hydraulic conductivity) - k?

A measure of how easily a fluid (e.g. water) canpass through a porous medium (e.g. soil)

Water flow through soil is termed as SEEPAGE

Some applications of soil permeabilityEarth dams•Seepage through dams (or under sheet pile wall)•Stability of dam slopes

Some applications of soil permeabilityPore water pressure and slope stability

Calculation of rate of settlement of clay soildeposits

Typical values of permeabilityThe values of k for different types of soil aretypically within the range shown below

Soil type k (m/sec)

Gravel k > 10-1

Clean Sand 10-2 > k >10-5

Silt 10-5 > k >10-9

Clay k <10-9

Factors affect on soil permeability• Soil type and particle size distribution (fine soils –

low permeability, coarse soil – high permeability)

• Soil structure: moulding water content

Factors affect on soil permeability• Density of soil (high density – low permeability, low

density – high permeability)

• Degree of saturation:

Factors affect on soil permeability• Stress conditions: (higher confining pressure – lower

permeability)

Hydraulic or Hydrostatic (total) headThe total head of water acting at a point in a submerged soilmass is known as hydraulic or hydrostatic head and isexpressed by Bernoulli’s equation

zu

g

vh

ww

++=γγ

2

Total head at a point (h) = Velocity head + Pressure head + Elev ation head

datumchosen a above headElevation

pressure water pore

)kN/m (9.8 water oft Unit weigh

onaccelerati nalGravitatio

waterof velocity seepage

head total

3w

−−

−−−−

z

u

g

v

h

γ

Hydraulic or Hydrostatic (total) headThe seepage velocities in soils are normally so small thatvelocity head can be neglected.

zu

hw

+=γTotal head at a point (h) = Pressure head + Elevation head

•Elevation head (Z) is measured as positive abovedatum•U is pore water pressure at the point due to the watertable above the point

H ZZ)-(H

headelevation head pressure P,at head Total

Z ZP,at headElevation

)( P,at head Pressure

)( P,at pressure water Pore

=+=+=∴

=

−=−=

∴

−=

P

w

w

Pw

wP

h

ZHZHu

ZHu

γγ

γ

γWhat is the total head at P

2 m

5 mX

P

Impermeable stratum1 m

1m

Example: Static water table

1. Calculation of total head at P

Choose datum at the top of the impermeable layer

mhw

wP 51

4 =+=γγ

wPu γ4=

1=PZDatum

2 m

5 mX

P


1m


2. Calculation of total head at X

Choose datum at the top of the impermeable layer

mhw

wX 54 =+=

γγ

wXu γ×=1

4=XZ

The total heads at P and X are identical. Thus this imply that the total head is constant throughout the region below a static water table.

Datum

2 m

5 mX

P


1m

Example: Static water table3. Calculation of total head at P

Choose datum at the water table

044 =−=

w

wPh

γγ

wPu γ4=4−=PZ

Datum

2 m

5 mX

P


1m


4. Calculation of total head at X

Choose datum at the water table

01=−=w

wXh

γγ

wXu γ×=1

1−=XZ

Again, the total head at P and X is identical, but the value is different defending on the datum

Datum

Head

2 m

5 mX

P


1m

• The value of the head depends on the choice of datu m

• Differences in total head are required for flow (no t pressure)

It can be helpful to consider imaginary standpipes (piezometers/manometers) placed in the soil at the points where the head is required

The total head is the elevation of the water level in the standpipe (piezometers/manometers) above the datum

Water flow through soil

∆∆∆∆h

Soil Sample

Darcy found that the flow (volume per unit time -q) was

•proportional to the head difference ∆h (q∝∝∝∝ ∆∆∆∆h )

•proportional to the cross-sectional area A ( q ∝∝∝∝ A )

•inversely proportional to the length of sample l (q ∝∝∝∝ 1/L)

L

Darcy’s law

(2a)Thus

Equation (2a) may be written as

(2b)

where i = ∆h/L the hydraulic gradient

v = q/A the Darcy’s or average velocity

L

hkAq

∆=

where k is the coefficient of permeability or hydraulic conductivity.

kAiq =or kiv =

Darcy’s law (cont..)

where n – the porosity of the soil

kiv =The average velocity, v, calculated from the above equation is for the cross sectional area normal to the direction of flow. However, flow through soil occurs only through the interconnected voids. The velocity through void spaces is called seepage velocity, v’

n

vv ='

Example 1A soil sample 10 cm in diameter is placed in a tube of 1 m long. A constant supply of water is allowed to flow into one end of the soil at A and the outflow at B is collected by a beaker (see the following figure). The average amount of water collected is 1cm3 for every 10 sec. Determine,

(a) Hydraulic gradient(b) Flow rate(c) Average velocity(d) Seepage velocity if e = 0.6 (e) Hydraulic conductivity

Example 1 (cont..): (a) Hydraulic gradient

Step 1: Define the datum position: top of the table

Step 2: Find the total dead at A (hA) and B (hB):

mZu

h

Z

u

w

wA

w

AA

A

wwA

21

1

1

=+=+=

==×=

γγ

γ

γγ

mZu

h

Z

u

wB

w

BB

B

wB

8.08.00

8.0

00

=+=+=

==×=

γγ

γ

Step 3: Find the hydraulic gradient, i

2.11

2.1

0.1

2.18.02

==∆=

==−=−=∆

l

hi

ml

mhhh BA

Example 1 (cont..)Step 4: Determine the flow rate, q

sec/1.010

1 3cmt

Qq ===

Step 5: Determine the average velocity, v

sec/0013.05.78

1.0

5.784

10

42

22

cmv

cmd

A

A

qv

==

=×==

=

ππ

sec10,cm 1Q collected, water of Volume 3 == t

Example 1 (cont..)Step 6: Determine seepage velocity, v’ if e=0.6

38.06.01

6.0

1=

+=

+=

e

en

Step 6: Determine the hydraulic conductivity (coefficient of permeability), k

sec/108.102.1

0013.0

law, sDay' From

4 cmi

vk

kiv

kAiq

−×===

==

n

vv ='

sec/0034.038.0

0013.0' cm

n

vv ===

Determination of coefficient of permeability (k)Coefficient of permeability (or hydraulic conductiv ity) [k] of a soil can be determined one of the following methods

�Laboratory methods:

(a) Constant head permeability test – For coarse gra ined soils(b) falling head permeability test – For fined grain ed soils

�Indirect methods and empirical equations

�In-situ (field) methods: Pumping well test

Determination of coefficient of permeability (k) –Laboratory methods

Constant head permeability test – for coarse soils

Soil specimen at the appropriate density is in a cylinder of cross-sectional area of A

h∆

Prior to running the test, fully saturate the specimen (apply a vacuum to the specimen, use de-aired water)

Vertical flow of water under a constant total head

Once a steady state water flow is achieved, measure the volume of water flowing per unit time (q) and total head difference ∆∆∆∆h

hA

qlk

∆=Then from Darcy’s law


Falling head permeability test – for fine soils

The length of the specimen is l and the cross section area is A. The cross section area of the standpipe is a

Prior to running the test, fully saturate the specimen (apply a vacuum to the specimen, use de-aired water)

The stand pipe is filled with water and water drains into a reservoir of constant level

0h

l1h

Undisturbed specimens are normally tested and containing cylinder may be the sampling tube itself


Falling head permeability test – for fine soils

The fall of water level in the standpipe (relative to the reservoir level) from h0 to h1 is measured during time t1

0h

l1h

1

0

1

log3.2h

h

At

alk =

For more accurate results, a series of tests should be run using different values of h0and h1 and/or standpipes of different diameters ( a)

Example 2In a falling-head permeability test the initial hea d of 1.00 m dropped to 0.35 m in 3 h, the diameter of standpipe being 5 mm. The soil s pecimen was 200 mm long by 100 mm in diameter. Calculate the coefficient of pe rmeability of the soil

L

inlet

measurementsample

inlet

Standpipe

Area - a

Constant level

Porous disks

or filersSample

Area -

A

Reservoir

0h

l1h

1

0

1

log3.2h

h

At

alk =

mA

ma

mmml

t

mh

mh

32

252

1

1

0

1086.74

1.0

1096.14

005.0

2.0200

sec1080036003

35.0

,0.1

−

−

×==

×==

===×=

==

π

π

sec/1085.4log3.2 8

1

0

1

mh

h

At

alk −×==


Triaxial Flexible Wall Permeability Tests

SOIL

INFLOW

OUTFLOW

h in

h iout

Triaxial Flexible Wall Permeability Test

• Advantages– Ability to back pressure

saturate– Reapply isotropic

stresses to simulate field conditions

– Independent measurement of soil sample volume change

• Disadvantages– Requires more

sophisticated equipment– Requires better trained

technicians

SOIL

INFLOW

OUTFLOW

h in

h iout

Determination of coefficient of permeability (k) –Other methods

Indirect method:

Permeability of fine grained soils can be determine d indirectly from the results of consolidation test and will be discussed in ENB371

210

210sec)/( Dmk −=

D10 is effective diameter in mm

Empirical methods (based on research finding):

For sand, Hazen (1892) showed that the approximate value of k is given by

Determination of coefficient of permeability (k) –In-situ methods (field methods)

Pumping well test – suitable for homogeneous coarse soils

1r

2r

2h1h

Pumping well – At least 300 mm in diameter, penetrate to the bottom of the soil stratum under test,

Pumping at constant rate, q , from the well

Steady seepage is established, radially toward the well, resulting in water table being drawn down to form a “cone of depression”

Water levels are observed in a number of boreholes spaced on radial line at various distances from the well


1r

2r

2h1h

Assumption – hydraulic gradient, i, at any distance, r, from the centre of the well is constant with the depth and is equal to the slope of water table

dr

dhir =

At distance r from the well, the area through which seepage takes place, A

rhA π2=From Darcy’s law,

)(

)/log(3.221

22

12

hh

rrqk

−=

π

For unconfined stratum

dr

dhrhkq

kAiq

××=

=

π2


Assumption – hydraulic gradient, i, at any distance, r, from the centre of the well is constant with the depth and is equal to the slope of water table

dr

dhir =

At distance r from the well the area through which seepage takes place, A

rHA π2=

From Darcy’s law,

)(2

)/log(3.2

12

12

hhH

rrqk

−=

π

For confined stratum (confined by two impermeable layers)

1r2r

2h1h

H

dr

dhrHkq

kAiq

××=

=

π2

Example 3If the pumping rate from the well, q is 0.01 m 3/sec, what is the average permeability of soil

)(

)/log(3.221

22

12

hh

rrqk

−=

π

mh

mh

mr

mr

7.11)4.19.1(0.15

5.11)6.19.1(0.15

30

15

2

1

2

1

=+−==+−=

==

sec/000475.0)5.117.11(

)15/30log(01.03.222

mk =−

××=π

Sources of Error in Hydraulic Conductivity Testing in Laboratory

• Use of non-representative samples• Voids formed during sample preparation• Smear Zones: heterogeneous soil sample• Alteration in Clay Chemistry• Air in Sample• Growth of Microorganisms• Menisci Problems in Capillary Tubes• Temperature• Volume Change Due to Stress Change• Flow Direction

Effect of Temperature

• Permeability varies with viscosity of water, which is temperature dependent.

• Lab k normally specified at 20°C• kT = k20/(ηT/η20)• ηT/η20 = temperature correction factor which is

a ratio of viscosity values, can be obtained from a chart or table

ηT/η20

Equivalent hydraulic conductivity in stratified soil (1)

x

z

If the layers are anisotropic, k 1 and k 2 represent the equivalent isotropic coefficients for the layers

Consider two isotropic and homogeneous soil layers of thicknesses H1 and H2 , and respective coefficients of permeability are k1 and k2

1k

1k1H

2H2k

2k

In a stratified soil deposit where the hydraulic co nductivity for a flow in a given direction changes from layer to layer

)( 21

2211

HH

kHkHkx +

+= xk is the equivalent permeability coefficient in horizo ntal direction (Derivation – Appendix 1)

1k

1k1H

2H2k

2k

Equivalent hydraulic conductivity in stratified soi l (2)

+

+=

2

2

1

1

21

k

H

k

H

HHkz

If there are n number of layers

).........(

..........

321

332211

n

nnx HHHH

kHkHkHkHk

++++++=

zk is the equivalent permeability coefficient in verti cal direction (Derivation – Appendix 2)

++

+

+

++++=

n

n

nz

k

H

k

H

k

H

k

H

HHHHk

...............

..................

3

3

2

2

1

1

321

is generally less than - sometimes as much a s 10 times lessxkzk

x

z

A layered soil is shown in the following figure

Example 4

mH 11 =

mH 5.12 =

mH 23 =

sec/101 41 cmk −×=

sec/101.4 53 cmk −×=

sec/102.3 22 cmk −×=

Calculate:• Horizontal equivalent permeability coefficient,

• Vertical equivalent permeability coefficient,

• Ratio of horizontal to vertical equivalent permeabi lity,

zk

xk

z

x

k

k

x

z

• The equivalent permeability coefficient in horizont al direction

)( 321

332211

HHH

kHkHkHkx ++

++=

Example 4 (2)

( )( )25.11

101.4232005.1101 5

++××+×+×=

−

xk

( )( ) sec/1007.1107.1070

5.4

102.4818 255

cmkx−−

−

×=×=×=

mH 11 =

mH 5.12 =

mH 23 =

sec/101 41 cmk −×=

sec/101.4 53 cmk −×=

sec/102.3 22 cmk −×=

x

z

* The equivalent permeability coefficient in vertic al direction

+

+

++=

3

3

2

2

1

1

321

k

H

k

H

k

H

HHHkz

Example 4 (3)

mH 11 =

mH 5.12 =

mH 23 =

sec/101 41 cmk −×=

sec/101.4 53 cmk −×=

sec/102.3 22 cmk −×=

510

1.40.2

32005.1

100.1

0.25.10.1 −×

+

+

++=zk

( ) ( ) ( ) sec/1065.7105883.0

5.410

488.000047.01.0

5.4 555 cmkz−−− ×=×=×

++=

x

z

* The ratio of horizontal to vertical equivalent pe rmeability, R

1401065.7

1007.15

2

=××== −

−

z

x

k

kR

Example 4 (4)

mH 11 =

mH 5.12 =

mH 23 =

sec/101 41 cmk −×=

sec/101.4 53 cmk −×=

sec/102.3 22 cmk −×=

sec/1065.7 5 cmkz−×=

sec/1007.1 2 cmkx−×=

Summary

�Soil permeability�Factor affect on soil permeability�Bernoulli’s equation and Darcy’s law�Laboratory determination of permeability

coefficient ( k)�Field determination of k� k of stratified soils

x

z

Flow parallel to soil layers

1k

1k1H

2H2k

2k

When the flow is parallel to soil layers, the hydra ulic gradient is the same at all points

xxx iii 21 ==Flow through the soil mass as a whole is equal to t he sum of the flow through each of the layers

xxx qqq 21 +=

Appendix -1: Equivalent Hydraulic conductivity in Horizontal direction (1)

Flow parallel to soil layers

xxx qqq 21 +=xxx iii 21 ==

From Darcy’s law, q=Aki

221121

2211

222111

)1()1(1)( kHkHkHH

kAkAkA

ikAikAikA

x

x

xxxx

××+××=××+

+=

+=

)( 21

2211

HH

kHkHkx +

+= xk is the equivalent permeability coefficient in horizontal direction

1k

1k1H

2H2k

2k

Appendix -1: Equivalent Hydraulic conductivity in Horizontal direction (2)

x

z

Flow normal to soil layers – seepage in the vertical direction

1k

1k1H

2H2k

2k

When the flow is normal to soil layers, the vertica l velocity in each layer is the same

zzz vvv 21 ==From Darcy’s law, v=ki

2

22

1

11

21 )( H

hk

H

hk

HH

hk zzz

∆=∆=+∆

∆∆∆∆h is the total head loss, ∆∆∆∆h1 and ∆∆∆∆h2 are the head losses in each of the layers

Appendix -1: Equivalent Hydraulic conductivity in Vertical direction (1)

Flow normal to soil layers – seepage in the vertical direction

For flow normal to soil layers, the head loss in th e soil mass is the sum of the head losses in each layer.

21 hhh ∆+∆=∆

2

22

1

11

21 )( H

hk

H

hk

HH

hkz

∆=∆=+∆

)()( 21

2

221

1

1 HH

hH

k

k

HH

hH

k

kh zz

+∆×+

+∆×=∆

+

+=

2

2

1

1

21

k

H

k

H

HHkz

zk is the equivalent permeability coefficient in vertical direction

Appendix -1: Equivalent Hydraulic conductivity in Vertical direction (1)

enb272 - ln - week 4 - permeability - 6 slides per page -gray

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