14.330 Hydraulic Conductivity - UML

Slide 1 of 37Revised 02/2013

14.330 SOIL MECHANICSHydraulic Conductivity

BERNOULLI’S EQUATION

Zg

vu hw

2

2

Where:

h = Total Headu = Pressurev = Velocityg = Acceleration due to Gravityw = Unit Weight of Water



Zg

vu hw

2

2

Therefore:

Zu hw

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

v ≈ 0

BERNOULLI’S EQUATION IN SOIL



Figure 5.1. Das FGE (2005).

CHANGE IN HEADFROM POINTS A

& B (H)BA h hh

B

w

BA

w

A ZuZu h

BA h hh

Lh i

BA h hh

h can be expressed in non-dimensional form

Where:i = Hydraulic GradientL = 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. Transition Flow (II)3. Turbulent Flow (III)

For most soils, flow is laminar. Therefore:

v i



DARCY’S LAW (1856)

Where:v = Discharge Velocity (i.e. quantity of water in

unit time through unit cross-sectional areaat right angles to the direction of flow)

k = Hydraulic Conductivity (i.e. coefficient ofpermeability)

i = Hydraulic Gradient* Based on observations of flow of water through clean sands



Kk w

HYDRAULIC CONDUCTIVITY (k)

Where: = Viscosity of Water

K = AbsolutePermeability(units of L2)

Soil Typek

(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 AND SEEPAGE VELOCITIES

svvAvAq Where:

q = Flow Rate(quantity of water/unit time)

A = Total Cross-sectional AreaAv = Area of Voidsvs = Seepage Velocity



nv

eev

VV

VV

vV

VVvv

LAAAv

AAAvv

vAAAvq

s

v

s

v

v

svs

v

sv

v

svs

svsv

11

)(

)()()(


DISCHARGE AND SEEPAGE VELOCITIES



EXAMPLE PROBLEM: FIND i AND q

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)





SM (k = 0.007 ft/min)

12 ft

175 ft

25 ft

15 ft

Hydraulic Gradient (i):

067.0

12cos17512

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:

EXAMPLE PROBLEM: FIND i AND q



HYDRAULIC CONDUCTIVITY: LABORATORY TESTINGConstant Head(ASTM D2434)

Falling Head(no ASTM)

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



Constant Head(ASTM D2434)

AhtQLk

tkiAAvtQ )(Where:

Q = Quantity of water collectedover time t

t = Duration of water collection

tLhkAQ


HYDRAULIC CONDUCTIVITY: LABORATORY TESTING



Falling Head(No ASTM)

2

110303.2

hhLog

AtaLk

dtdhaA

Lhkq

hdh

AkaLdt

Where:A = Cross-sectional area of Soila = 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

HYDRAULIC CONDUCTIVITY: LABORATORY TESTING



210sec)/( cDcmk

Uniform Sands - Hazen Formula(Hazen, 1930):

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

eeCk

1

3

1

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

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

85.024.1 kek

Sands – Casagrande(Unpublished):

Where:e = Void Ratiok0.85 = Hydraulic Conductivity @ e = 0.85

e

eCkn

12

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

Where:C2 = Constant to be determined experimentallyn = Constant to be determined experimentallye = Void Ratio

HYDRAULIC CONDUCTIVITY: EMPIRICAL RELATIONSHIPS



Void Ratio (e)k

(cm/sec)1.2 0.6 x 10-71.52 1.52 x 10-7

k1

k2

C2

e1n

1 e1

C2e2

n

1 e2

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

knownquantities

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

HYDRAULIC CONDUCTIVITY: EMPIRICAL RELATIONSHIPSEXAMPLE – NORMALLY CONSOLIDATED CLAYS



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

n

cmcm n

HYDRAULIC CONDUCTIVITY: EMPIRICAL RELATIONSHIPSEXAMPLE – NORMALLY CONSOLIDATED CLAYS



STRATIFIED SOILS: EQUIVALENT HYDRAULIC CONDUCTIVITYHORIZONTAL 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 Velocityv1 = Discharge Velocity in Layer 1



nnnHH

eqeqH

ikvikvikv

ikv

;...;; 222111

)(


Substituting v=ki into q equation and using H to

denote Horizontal Direction

Noting that ieq=i1=i2=…=in

)...(12211)( nHnHHeqH HkHkHk

Hk

Where: kH(eq) = Equivalent Hydraulic Conductivityin Horizontal Direction

STRATIFIED SOILS: EQUIVALENT HYDRAULIC CONDUCTIVITYHORIZONTAL DIRECTION



n

n

hhhh

vvvv

...

...

21

21

nVnVeqV ikikHhk ...11)(


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

and

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

Vertical Direction

Where: kV(eq) = Equivalent Hydraulic Conductivityin Vertical Direction

STRATIFIED SOILS: EQUIVALENT HYDRAULIC CONDUCTIVITYVERTICAL DIRECTION



rhdrdhkq 2

1

2

1

2

2 h

h

r

rhdh

qk

rdr

FIELD PERMEABILITY TESTING BY PUMPING WELLSUNCONFINED PERMEABLE LAYER UNDERLAIN BY IMPERMEABLE LAYER


)(

log303.2

22

21

2

110

hhrrq

k field

Field Measurements Taken:

q, r1, r2, h1, h2

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

pumpingEquation:

can be re-written as

Solving Equation:



rHdrdhkq 2

1

2

1

2

2h

h

r

rdh

qkH

rdr


)(727.2

log

21

2

110

hhHrrq

k field

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

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

pumpingEquation:

can be re-written as

Solving Equation:

FIELD PERMEABILITY TESTING BY PUMPING WELLSWELL PENETRATING CONFINED AQUIFER



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

SOILPERMEABILITY

ANDDRAINAGE



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

Good drainage

10-8 10-910-710-610-510-410-310-210-11.0101102 10-8 10-910-710-610-510-410-310-210-11.0101102

Poor drainage Practically impervious

Clean gravel Clean 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)

Computat ion: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.

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

10-8 10-910-710-610-510-410-310-210-11.0101102 10-8 10-910-710-610-510-410-310-210-11.0101102

COEFFICIENT OF PERMEABILITYCM/S (LOG SCALE)

SOIL PERMEABILITY AND DRAINAGE



LAPLACE'S EQUATION OF CONTINUITYSteady-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 PlaneFigure 5.11. Das FGE (2005).

x

z

y



dxdydzzvv

dzdydxxv

v

zz

xx

Consider water flow at Point A(Soil Block at Pt A shown left)

Rate of water flow into soil block in x direction:

vxdzdyRate of water flow into soil block in z direction:

vzdxdy


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

LAPLACE'S EQUATION OF CONTINUITY



0

0

zv

xv

ordxdyvdzdyv

dxdydzzvvdzdydx

xv

v

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 TERMSFlow 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 NETSFLOW AROUND

SHEET PILE WALL



Figure 5.12b. Das FGE (2005).

FLOW NETSFLOW AROUND

SHEET PILE WALL



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 NETSFLOW UNDER ANIMPERMEABLE

DAM



nqqqq ...321


q k h1 h2

l1

l1 k h2 h3

l2

l2 k h3 h4

l3

l3 ...

h1 h2 h2 h3 h3 h4 ... HNd

Rate of Seepage ThroughFlow Channel (per unit length):

Using Darcy’s Law (q=vA=kiA)

Potential DropWhere:

H = Head DifferenceNd = Number of Potential Drops

FLOW NETS: DEFINITION OF TERMS



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

FLOW NETSFLOW AROUND SHEET PILE WALL EXAMPLE



GIVEN:

Flow Net in Figure 5.17.Nf = 3Nd = 6kx=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.





mmm 56.06

)67.15(

SOLUTION:

Potential Drop = dN

H

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

k = 5x10-3 cm/seck = 5x10-5 m/sec

q = (5x10-5 m/sec)(0.56m)q = 2.8x10-5 m3/sec/m

fd

f qNN

HNkq

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


Documents

14.330 Hydraulic Conductivity - UML