hydraulic conductivity

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L.12-13 Hydraulic Conductivity

CIVE 431 SOIL MECHANICS & LAB

FALL 2014-15

Why the Interest in Water Flow

To recover as a resource (wells)To evaluate movement of pollutantsTo evaluate water loses from

reservoirs

To evaluate water pressures and their effect on stresses in the soil

To evaluate migration of soil particles - Piping

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S7-6

TYPES OF DAMS

CONCRETE ARCH

CONCRETE GRAVITY

CONCRETE BUTTRESS

EARTH EMBANKMENT

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S7-7

Embankment Dam

S7-8

Qaraoun DamRock-fill with Concrete Liner

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Bernoullis Equation

1. Kinetic energy fluid particle

The energy of a fluid particle consists of:

2. Strain energy

- due to velocity

- due to fluid pressuredatum

z

3. Potential energy- due to elevation (z) with respect to a datum

Bernoullis Equation

Total head =

Expressing energy in unit of length (head):

Velocity head +

Pressure head +

Elevation head

fluid particle

datum

z

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Bernoullis Equation

Total head =

For flow through soils, velocity (and thus velocity head) is very small. Therefore,

Velocity head +

Pressure head +

Elevation head

fluid particle

datum

z

0

Total head = Pressure head + Elevation head

Definition of Head at a Point

P

z(P)

Datum

Note

z is measured vertically upfrom the datum

Total Head at Point (P)

Elevation Head at (P)

Pressure Head at (P)

(P)w

(P)(P) z

uh

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Example - Static Water Table1. Calculation of head at P

Choose datum at the top of the Rock Layer

2 m

5 mX

P

Rock

1 m

1m

Location of Water TablewwwP hu 4)(

m1)( Pz

)()(

)( Pw

PP z

uh

w4 mw

= + =1 5)(Ph

Example - Static Water Table2. Calculation of head at X

Choose datum at the top of the Rock Layer

2 m

5 mX

P

Rock

1 m

1m

Location of Water TablewwwX hu 1)(

m4)( Xz

)()(

)( Xw

XX z

uh

1mw

w 4 5)(Xh

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Example - Static Water Table

2 m

5 mX

P

Rock

1 m

1m

Location of Water Table

Total Head at point X is equal to the Total Head at P

Same Energy at the two points No flow between the two points

Remark

If flow is from A to B, total head is higher at A than at B.

Energy is dissipated in overcoming the soil resistance and hence is the head loss.

water

AB

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Example - Change the Datum1. Calculation of head at P

Choose datum at the Water Table

2 m

5 mX

P

Rock

1 m

1m

Location of Water TablewwwP hu 4)(

m4)( Pz

)()(

)( Pw

PP z

uh

mww

4 4 0)(Ph

Example - Change the Datum1. Calculation of head at X

Choose datum at the Water Table

2 m

5 mX

P

Rock

1 m

1m

Location of Water TablewwwX hu 1)(

m1)( Xz

)()(

)( Xw

XX z

uh

)(Xh1

mww

1 0

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Example - Change the Datum

2 m

5 mX

P

Rock

1 m

1m

Location of Water Table

Again the Total Head at point X is equal to the Total Head at P, although the position of datum is different

The location of the datum affects the Value of the Total Head but not the value of the pressure head.

Darcys Experiment

h

Soil Sample

Stand Pipe

Porous Stones Length of

Soil Sample

Head Difference between 2 ends of sample

A = Cross sectional area of soil sample

2h

1h

L

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Darcys Experiment

Soil Sample

AL

1h

h2h

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

proportional to the head difference h

proportional to the cross-sectional area A

inversely proportional to the length of sample L

Darcys Law

Flow Rate (Volume/time)

Gross cross-sectional area of flow (Length2)

Length of Soil Specimen

Head Difference between two points

Constant of ProportionalityHydraulic Conductivity or Permeability(Length/time)

ALhkq

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Darcys Law

Darcys Law can be written as:

Where i = h/L is the hydraulic gradient

Where v = Darcy Velocity (Length/Time)v = q/A = k i

OR

q = k i A

ALhkq

Darcy Velocity, v can be related to the actual seepage velocity, vs by: vs= v/n (where n is porosity)

Hydraulic Gradient

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

water

AB

length AB, along the stream line

AB

BA

lhhi

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Measuring k, Constant Head

Manometers

L

inlet

H

constant headdevice

device for flowmeasurement

porous disk

Constant Head Permeameter

sample

Constant Head Permeameter The volume discharge Q during a suitable

time interval T is collected. The difference in head H over a length L is

measured by means of manometers. Knowing the cross-sectional area A, Darcys

law gives

LHkA

TQ

HATQLk

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Measuring k, Falling HeadAnalysis

H2

H1H

L

Standpipeof area a

Sample of area

A

Consider a time interval t

The flow in the standpipe =

The flow in the sample =

a Ht

k AHL

a dHdt

k AHL

Measuring k, Falling HeadAnalysis

H2

H1H

L

Standpipeof area a

Sample of area

A

Starting from:

The equation has the solution:

Initially H = H1 at Time = T1H = H2 at Time = T2

a dHdt

k AHL

a n H kALt cons t ( ) tan

k aLAn H Ht t

( / )1 2

2 1

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Typical Permeability Values

10-1 10-2 10-3 10-4 10-5 10-6 10-7 10-8 10-9 10-1010-1110-12

Gravels Sands Silts Homogeneous ClaysFissured & Weathered Clay

Typical Permeability Ranges (meters/second)

Soils exhibit a wide range of permeability and while particle size may vary by about 3-4 orders of magnitude permeability may vary by about 10 orders of magnitude.

Hydraulic Conductivity, k

Pore Size Distribution Fluid ViscosityGrain Size DistributionVoid Ratio Specific Area of Solids (the higher the specific area,

the lower the value of k)

Structure of Clay ParticlesDegree of Saturation (k decreases significantly as

degree of saturation decreases)

The hydraulic conductivity depends on several factors:

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Hydraulic Conductivity, k

Hazens Formula (Fairly Uniform Sand):Empirical Relationships for k:

210)/( cDscmk

Grain Size in mmConstant varying from 1 to 1.5

Kozeny-Carmen Formula:

e

eSk

scmkoo 111)/(

3

2

Viscosity of Permeant Kozeny-Carman

Coefficient = 5.0

Specific Surface Area per unit volume of particles (1/cm)

Example 1 - Perpendicular Flow

(1) Find the Flow Rate inside thelayered system?

(2) How much head loss occurs in each layer (h1=? h2=?) The flow rate, Q is constant

in all layers

The total head loss H is divided into two head losses, h1 and h2 such that:

Clay: K2 = 10-8 cm/s

Sand: K1 = 10-2 cm/s

H=4mQ (flow rate)

H = h1 + h2

2m

2m

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hSand: K1 = 10-2 cm/s

H=4mQ (flow rate)Q1= k1i1A1 = k1(h1/L1)A1

Q2= k2i2A2 = k2(h2/L2)A2Q= kequivalantiA = keq(/L)A

H = h1 + h2ButQL/KeqA = Q1L1/K1A1 + Q2L2/K2A2

Given that:A = A1 = A2Q = Q1 = Q2

L/Keq= L1/K1 + L2/K2

2

2

1

1

KL

KL

LKeq

2m

2m




H=4mQ (flow rate)Generalization:

i

i

ieq

KLL

K

For this problem:

sec/10999.1

102

102

22 8

82

cmKK eqeq

2m

2m

Least permeable layer controls the hydraulic conductivity of an equivalent layered system if flow is perpendicular to the soil layers

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H=4mQ (flow rate)

Find Q ??

2m

2m

iAkQ eqAssume 1mx1m

28 100100400400/10999.1 cm

cmcmscmQ

scmQ /10999.1 34

yearlitersQ /307.6




H=4mQ (flow rate)

Find Head loss in each layer?

2m

2m

Layer 1:

11

111 AL

hkQ 10010010 20010999.1 2

4

11

111

AKLQh

cmh 000399.01 Layer 2:

22

222 AL

hkQ 10010010 20010999.1 8

4

22

222

AKLQh

cmh 999.3992

Almost all of the head loss occurred in the clay layer (very low permeability layer)

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Example 2 - Parallel Flow

(1) Find the Total Flow Rate inside the layered system?

(2) Find Flow rate in each layer?

The head loss now is the same in all layers

The total flow now is divided between layer 1 and layer 2 such that:



H=4m

Q (flow rate)

QTotal= Q1 + Q2Lo = 100 cm




H=4m

Q (flow rate)

Lo = 100 cm

Q1= k1i1A1 = k1(h1/Lo)L1Q2= k2i2A2 = k2(h2/Lo)L2Q= kequivalantiA = keq(/Lo)L

Q = Q1 + Q2But

keq(/Lo)L = k1(h1/Lo)L1 + k2(h2/Lo)L2Given that:h1 = h2 = hL

LKLKKeq 2211

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Generalization:

For this problem:

Most permeable layer controls the hydraulic conductivity of an equivalent layered system if flow is horizontal through horizontal layers



H=4m

Q (flow rate)

Lo = 100 cm

i

iieq L

LKK

sec/10500006.022

210210 282 cmKK eqeq


Find Q ??Clay: K2 = 10-8 cm/s


H=4m

Q (flow rate)

Lo = 100 cm

iAkQ eq400x100cm

22 100400100400/10500006.0 cm

cmcmscmQ

scmQ /00004.800 3yearlitersMillionQ /22.25

Layer 1 and Layer 2: scmcmcmcmscmAikQ /800100200

100400/101 3221111

scmQ /00004.0 32

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Example 3

Clay: K2=8x10-7 ft/min

hSilt: K1 = 1x10-5 ft/min

740 ft

770 ft

790 ft

810 ft

760 ft

(1) Calculate the Flow rate q in ft3/min through the soil system if the area is 1 ft2:

c

c

s

seq

KL

KL

LK

min/1026.1

10830

10120

50 6

75

ftKeq

min/10265.11505010265.1 366 ftiAkQ eq

Example 3

Clay: K2=8x10-7 ft/min

hSilt: K1 = 1x10-5 ft/min

740 ft

770 ft

790 ft

810 ft

760 ft

(2) Calculate the head loss in the silt layer and in the clay layer. What is the fraction of each head loss compared to the total head loss. Is it reasonable?

ALhkQsilt

siltsilt

ftAK LQh siltsiltsilt 53.211012010265.1

5

6

ftAK LQh clayclayclay 46.4711083010265.1

7

6

Fraction hclay =94.92% and Fraction hsilt = 5.06%

hydraulic conductivity

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