14.330 Soil Mechanics Hydraulic Conductivity &...

of 37 Revised 2/2012

14.330 Soil Mechanics Hydraulic Conductivity & Seepage

BERNOULLI’S EQUATION

Zg

vu hw

++=2

2

γWhere:

h = Total Head u = Pressure v = Velocity g = Acceleration due to Gravity γw = Unit Weight of Water



BERNOULLI’S EQUATION IN SOIL

Zg

vu hw

++=2

2

γ

Therefore:

Zu hw

+=γ

v ≈ 0 (i.e. velocity of water in soil is negligible).

v ≈ 0



Figure 5.1. Das FGE (2005).

CHANGE IN HEAD FROM POINTS

A & B (Δh) BA h hh −=Δ

⎟⎟⎠

⎞⎜⎜⎝

⎛+−⎟⎟

⎠

⎞⎜⎜⎝

⎛+=Δ B

w

BA

w

A ZuZu hγγ

BA h hh −=Δ

Lh i Δ

=

BA h hh −=Δ

Δh can be expressed in non-dimensional form

Where: i = Hydraulic Gradient L = Length of Flow between Points A & B




VELOCITY (v) vs. HYDRAULIC GRADIENT (i) General relationship shown in Figure 5.2

Three Zones: 1. Laminar Flow (I) 2. Transistion Flow (II) 3. Turbulent Flow (III) For most soils, flow is

laminar. Therefore:

v ∝ i



DARCY’S LAW (1856)

v = kiWhere: v = Discharge Velocity (i.e. quantity of water in unit time through unit cross-sectional area at right angles to the direction of flow) k = Hydraulic Conductivity (i.e. coefficient of permeability) i = Hydraulic Gradient

* Based on observations of flow of water through clean sands



Kk w

ηγ

=

HYDRAULIC CONDUCTIVITY (k)

Where: η = Viscosity of Water

K = Absolute Permeability (units of L2)

Soil Type k

(cm/sec) k

(ft/min) Clean Gravel 100-1 200-2

Coarse Sand 1-0.01 2-0.02

Fine Sand 0.01-0.001 0.02-0.002

Silty Clay 0.001-0.00001

0.002-0.00002

Clay < 0.000001 <0.000002

Typical Values of k per Soil Type

after Table 5.1. Das FGE (2005)




DISCHARGE & SEEPAGE VELOCITIES

svvAvAq ==Where: q = Flow Rate (quantity of water/unit time) A = Total Cross-sectional Area Av = Area of Voids vs = Seepage Velocity



DISCHARGE & SEEPAGE VELOCITIES

nv

eev

VVVV

vVVVvv

LAAAv

AAAvv

vAAAvq

s

v

s

v

v

svs

v

sv

v

svs

svsv

=⎟⎠

⎞⎜⎝

⎛ +=

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+

=+

=

+=

+=

=+=

11

)(

)()()(




EXAMPLE PROBLEM

CL (Impervious Layer)


SM (k = 0.007 ft/min)

12 ft

175 ft

25 ft

15 ft

12°

GIVEN: REQUIRED: Find Hydraulic Gradient (i) and Flow Rate (q)



EXAMPLE PROBLEM – FIND i, q



SM (k = 0.007 ft/min)

12 ft

175 ft

25 ft

15 ft

Hydraulic Gradient (i):

067.0

12cos17512

=

⎟⎠

⎞⎜⎝

⎛=

Δ=

ftfti

Lhi

12°

ftftq

ftftftq

kiAq

min//109.6

)1)(12)(cos15)(067.0(min

007.0

33−×=

=

=

Rate of Flow per Time (q):

GIVEN: SOLUTION:



LABORATORY TESTING OF HYDRAULIC CONDUCTIVITY Constant Head (ASTM D2434)

Falling Head (no ASTM)

Figure 5.4. Das FGE (2005). Figure 5.5. Das FGE (2005).



LABORATORY TESTING OF HYDRAULIC CONDUCTIVITY

Constant Head (ASTM D2434)

AhtQLk =

tkiAAvtQ )(==Where: Q = Quantity of water collected over time t t = Duration of water collection

tLhkAQ ⎟⎠

⎞⎜⎝

⎛=




LABORATORY TESTING OF HYDRAULIC CONDUCTIVITY Falling Head (No ASTM)

2

110303.2hhLog

AtaLk =

dtdhaA

Lhkq −==

⎟⎠

⎞⎜⎝

⎛−=hdh

AkaLdt

Where: A = Cross-sectional area of Soil a = Cross-sectional area of Standpipe

Figure 5.5. Das FGE (2005). 2

1loghhe

AkaLt =

Integrate from limits 0 to t

Integrate from limits h1 to h2

after rearranging above equation

after integration

or



EMPIRICAL RELATIONSHIPS FOR HYDRAULIC CONDUCTIVITY

210sec)/( cDcmk =

Uniform Sands - Hazen Formula (Hazen, 1930):

Where: c = Constant between 1 to 1.5 D10 = Effective Size (in mm)

eeCk+

=1

3

1

Sands – Kozeny-Carman (Loudon 1952 and Perloff and Baron 1976):

Where: C = Constant (to be determined) e = Void Ratio

85.024.1 kek =

Sands – Casagrande (Unpublished):

Where: e = Void Ratio k0.85 = Hydraulic Conductivity @ e = 0.85

⎟⎟⎠

⎞⎜⎜⎝

⎛

+=

eeCkn

12

Normally Consolidated Clays (Samarasinghe, Huang, and Drnevich, 1982):

Where: C2 = Constant to be determined experimentally n = Constant to be determined experimentally e = Void Ratio



EXAMPLE - ESTIMATION OF HYDRAULIC CONDUCTIVITY (NORMALLY CONSOLIDATED CLAYS)

Void Ratio (e) k

(cm/sec) 1.2 0.6 x 10-7

1.52 1.52 x 10-7

⎥⎦

⎤⎢⎣

⎡

+

⎥⎦

⎤⎢⎣

⎡

+=

21

1

22

1

12

2

1

eeC

eeC

kk

n

n

Example 5.6 Das PGE (2005).

GIVEN: Normally consolidated clay with e and k measurements from 1D Consolidation Test.

SOLUTION: Using (Samarasinghe, Huang, and Drnevich, 1982) Equation:

⎟⎠

⎞⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛=2.252.2

52.12.1

sec/52.1sec/6.0 n

cmcmSubstituting

known quantities

REQUIRED: Find k for same clay with a void ratio of 1.4.



EXAMPLE - ESTIMATION OF HYDRAULIC CONDUCTIVITY (NORMALLY CONSOLIDATED CLAYS) (continued)

sec/101.14.11

4.1sec)/710581.0(

sec/10581.0

2.112.1sec/106.0

1

75.4

4.1

72

5.47

1

121

cmxcmxk

cmxC

cmx

eeCk

e

n

−=

−

−

=⎟⎟⎠

⎞⎜⎜⎝

⎛

+−=

=∴

⎟⎟⎠

⎞⎜⎜⎝

⎛

+=

⎟⎟⎠

⎞⎜⎜⎝

⎛

+=

Example 5.6 Das PGE (2005).

5.42.252.2

52.12.1

sec/52.1sec/6.0

=∴⎟⎠

⎞⎜⎝

⎛⎟⎠

⎞⎜⎝

⎛= ncmcm n



EQUIVALENT HYDRAULIC CONDUCTIVITY IN STRATIFIED SOILS – HORIZONTAL DIRECTION


Considering cross-section of Unit Length 1.

Total flow through cross-section can be written as:

nn HvHvHvqHvq

••++••+••=

••=

1...111

2211

Where: v = Average Discharge Velocity v1 = Discharge Velocity in Layer 1



nnnHH

eqeqH

ikvikvikvikv

===

=

;...;; 222111

)(


Substituting v=ki into q equation and using H to

denote Horizontal Direction

Noting that ieq=i1=i2=…=in

)...(1 2211)( nHnHHeqH HkHkHkH

k +++=

Where kH(eq) = Equivalent Hydraulic Conductivity in Horizontal Direction

EQUIVALENT HYDRAULIC CONDUCTIVITY IN STRATIFIED SOILS – HORIZONTAL DIRECTION



n

n

hhhh

vvvv

====

====

...

...

21

21

nVnVeqV ikikHhk === ...11)(


Total Head Loss = h h = Sum Head Loss in Each Layer

and

Using Darcy’s Law (v=ki) into v equation and using V to denote

Vertical Direction

Where kV(eq) = Equivalent Hydraulic Conductivity in Vertical Direction

EQUIVALENT HYDRAULIC CONDUCTIVITY IN STRATIFIED SOILS – VERTICAL DIRECTION



rhdrdhkq π2⎟⎠

⎞⎜⎝

⎛=

∫∫ ⎟⎟⎠

⎞⎜⎜⎝

⎛=

1

2

1

2

2 h

h

r

rhdh

qk

rdr π

FIELD PERMEABILITY TEST BY PUMPING WELLS UNCONFINED PERMEABLE LAYER UNDERLAIN BY IMPERMEABLE LAYER


)(

log303.2

22

21

2

110

hhrrq

k field−

⎟⎟⎠

⎞⎜⎜⎝

⎛

=πField Measurements Taken:

q, r1, r2, h1, h2

q = Groundwater Flow into Well q also is rate of discharge from

pumping

Equation:

can be re-written as

Solving Equation:



rHdrdhkq π2⎟⎠

⎞⎜⎝

⎛=

∫∫ =1

2

1

2

2hh

r

rdh

qkH

rdr π

FIELD PERMEABILITY TEST BY PUMPING WELLS WELL PENETRATING CONFINED AQUIFER


)(727.2

log

21

2

110

hhHrrq

k field −

⎟⎟⎠

⎞⎜⎜⎝

⎛

=Field Measurements Taken: q, r1, r2, h1, h2

q = Groundwater Flow into Well q also is rate of discharge from

pumping Equation:

can be re-written as

Solving Equation:



after Casagrande and Fadum (1940) and Terzagi et al. (1996).

SOIL PERMEABILITY & DRAINAGE



From FHWA IF-02-034 Evaluation of Soil and Rock Properties.

SOIL PERMEABILITY & DRAINAGE

192

Good drainage

10-8

10-9

10-7

10-6

10-5

10-4

10-3

10-2

10-11.0101

102

10-8

10-9

10-7

10-6

10-5

10-4

10-3

10-2

10-11.0101

102

Poor drainage Practically impervious

Clean gravelClean sands, Clean sand and gravel mixtures

Impervious sections of earth dams and dikes

“Impervious” soils which are modif ied by the effect of vegetation and weathering; f issured, weathered clays; fractured OC clays

Pervious sections of dams and dikes

Very f ine sands, organic and inorganic silts, mixtures of sand, silt, and clay glacial till, stratif ied clay deposits, etc.

“Impervious” soils e.g., homogeneous clays below zone of weathering

Drainage property

Application in earth

dams and dikes

Type of soil

Direct determination

of coefficient of

permeability

Indirect determination

of coefficient of

permeability

*Due to migration of f ines, channels, and air in voids.

Direct testing of soil in its original position (e.g., well points). If properly conducted, reliable; considerable experience required. (Note: Considerable experience

also required in this range.)

Constant Head Permeameter; little experience required.

Constant head test in triaxial cell; reliable w ith experience and no leaks.

Reliable;Little experiencerequired

Falling Head Permeameter;Range of unstable permeability;* much experience necessary to correct interpretation

Fairly reliable; considerable experience necessary (do in triaxial cell)

Computation:From the grain size distribution(e.g., Hazen’s formula). Only applicable to clean, cohesionless sands and gravels

Horizontal Capillarity Test:Very little experience necessary; especially useful for rapid testing of a large number of samples in the f ield w ithout laboratory facilities.

Computations:from consolidation tests; expensive laboratory equipment and considerable experience required.

10-8

10-9

10-7

10-6

10-5

10-4

10-3

10-2

10-11.0101

102

10-8

10-9

10-7

10-6

10-5

10-4

10-3

10-2

10-11.0101

102

COEFFICIENT OF PERMEABILITY

CM/S (LOG SCALE)

Figure 90. Range of hydraulic conductivity values based on soil type.

Another method of estimating hydraulic conductivity empirically through grain size was developed by GeoSyntec (1991) and is presented as Figure 91. As with Hazen’s equation, grain size distribution information is necessary to develop the input parameter (i.e., d15, grain size at 15% passing by weight) of the material. Using this value and the band of values corresponding to gradient and confining stress, a k value can be estimated. This method, as with other empirical methods, can only be used to provide an estimated value within 1 to 2 orders of magnitude of the in-situ condition. The lower the permeability and the higher the variability of grain size in the soil, the higher the error in using empirical relationships based on uniform particle distribution.



LAPLACE’S EQUATION OF CONTINUITY Steady-State Flow

around an impervious Sheet

Pile Wall

Consider water flow at Point A:

vx = Discharge Velocity in x Direction

vz = Discharge Velocity in z Direction

Y Direction Out Of Plane Figure 5.11. Das FGE (2005).

x

z

y



dxdydzzvv

dzdydxxvv

zz

xx

⎟⎠

⎞⎜⎝

⎛∂

∂+

⎟⎠

⎞⎜⎝

⎛∂

∂+

LAPLACE’S EQUATION OF CONTINUITY Consider water flow at Point A (Soil Block at Pt A shown left)

Rate of water flow into soil block in x direction: vxdzdy

Rate of water flow into soil block in x direction: vzdxdy


Rate of water flow out of soil block in x,z directions:



[ ]

0

0

=∂

∂+

∂

∂

=+

−⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛∂

∂++⎟

⎠

⎞⎜⎝

⎛∂

∂+

zv

xv

ordxdyvdzdyv

dxdydzzvvdzdydx

xvv

zx

zx

zz

xx



Total Inflow = Total Outflow



02

2

2

2=

∂

∂+

∂

∂

∴

⎟⎠

⎞⎜⎝

⎛∂

∂−==

⎟⎠

⎞⎜⎝

⎛∂

∂−==

zhk

xhk

zhkikv

xhkikv

zx

zzzz

xxxx



Using Darcy’s Law (v=ki)



FLOW NETS – DEFINITION OF TERMS Flow Net: Graphical Construction used to calculate groundwater flow through soil. Comprised of Flow Lines and Equipotential Lines. Flow Line: A line along which a water particle moves through a permeable soil medium. Flow Channel: Strip between any two adjacent Flow Lines. Equipotential Lines: A line along which the potential head at all points is equal. NOTE: Flow Lines and Equipotential Lines must meet at right angles!



Figure 5.12a. Das FGE (2005).

FLOW NETS FLOW AROUND SHEET PILE WALL



Figure 5.12b. Das FGE (2005).




1. The upstream and downstream surfaces of the permeable layer (i.e. lines ab and de in Figure 12b Das FGE (2005)) are equipotential lines.

2. Because ab and de are equipotential lines, all the flow lines

intersect them at right angles. 3.  The boundary of the imprevious layer (i.e. line fg in Figure

12b Das FGE (2005)) is a flow line, as is the surface of the impervious sheet pile (i.e. line acd in Figure 12b Das FGE (2005)).

4. The equipontential lines intersect acd and fg (Figure 12b Das

FGE (2005)) at right angles.

FLOW NETS – BOUNDARY CONDITIONS




FLOW NETS Flow under a Impermeable Dam



nqqqq Δ==Δ=Δ=Δ ...321


Δq = k h1 − h2l1

#

$%

&

'(l1 = k

h2 − h3l2

#

$%

&

'(l2 = k

h3 − h4l3

#

$%

&

'(l3 = ...

h1 − h2 = h2 − h3 = h3 − h4 = ... =HNd

Rate of Seepage Through Flow Channel (per unit length):

Using Darcy’s Law (q=vA=kiA)

Potential Drop Where: H = Head Difference Nd = Number of Potential Drops

FLOW NETS Seepage Calculations



Figure 5.12b. Das FGE (2005).

d

f

NHN

kq =

If Number of Flow Channels = Nf, then the total flow for all channels per unit length is:

Therefore, flow through one channel is:

dNHkq =Δ




GIVEN:

Flow Net in Figure 5.17. Nf = 3 Nd = 6 kx=kz=5x10-3 cm/sec DETERMINE:

a.  How high water will rise in piezometers at points a, b, c, and d.

b.  Rate of seepage through flow channel II.

c.  Total rate of seepage.


FLOW NETS FLOW AROUND SHEET PILE WALL EXAMPLE



mmm 56.06

)67.15(=

−

SOLUTION: Potential Drop =

dNH

At Pt a: Water in standpipe = (5m – 1x0.56m) = 4.44m At Pt b: Water in standpipe = (5m – 2x0.56m) = 3.88m At Pts c and d: Water in standpipe = (5m – 5x0.56m) = 2.20m






SOLUTION:

dNHkq =Δ

k = 5x10-3 cm/sec = 5x10-5 m/sec Δq = (5x10-5 m/sec)(0.56m) Δq = 2.8x10-5 m3/sec/m

fd

f qNNHNkq Δ==

q = (2.8x10-5 m3/sec/m) * 3 q = 8.4x10-5 m3/sec/m


14.330 Soil Mechanics Hydraulic Conductivity &...

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