Note7_SeepageAnalysis

Seepage Analysis

ChihChih--Ping LinPing LinNational National ChiaoChiao TungTung [email protected]@mail.nctu.edu.tw

Soil Mechanics −

Outline

LaplaceLaplace Equation of ContinuityEquation of Continuity11--D ExampleD Example

Flow Nets (2Flow Nets (2--D)D)

Computations using Flow NetsComputations using Flow Nets

Flow Nets in Anisotropic SoilFlow Nets in Anisotropic Soil

Seepage through an Earth DamSeepage through an Earth Dam

33--D FlowD Flow

Laplace’s Equation of Continuity

Laplace Eqn.

Laplace’s Equation of Continuity

[ ] 0dydxvdydzv

dydxdzz

vvdydzdx

x

vv

zx

zz

xx

=+−

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

∂∂

++⎟⎠⎞

⎜⎝⎛

∂∂

+

0z

v

x

v zx =∂∂

+∂∂

x

hkikv xxxx ∂

∂==

z

hkikv zzzz ∂

∂==

0z

hk

x

hk

2

2

z2

2

x =∂∂

+∂∂

0z

h

x

h2

2

2

2

=∂∂

+∂∂If kx=ky

Laplace Eqn.

1-D Example0

2

2

=∂∂z

h zCCh 21 +=

At z = 0, h = h1, we can get C1A = h1.

( ) AAAA Ckzhkvv 22 / −=∂∂−==

AA kvC /22 −=

At z = HA,

zk

vhzh

AA

21)( −=

( ) BBBB Ckzhkvv 22 / −=∂∂−==

BB kvC /22 −=

At z = HA,

At z = HA+ HB, h = 0, and we get

( ) ( ) BBABABB kHHvHHCC /221 +=+−=

zk

v

k

HHvzh

BB

BAB

22 )()( −

+=

Laplace Eqn.

At z = LA, hA(LA) = hB(LA), so we can get

BBAA kHkH

hv

//1

2 +=

⎟⎟⎠

⎞⎜⎜⎝

⎛+

−= zkHkH

khzh

ABBA

B1)( 1

( )⎟⎟⎠

⎞⎜⎜⎝

⎛−+

+= zHH

kHkH

khh BA

ABBA

A1

For 0 ≤ z < LA,

For LA ≤ z< LA + LB,

Laplace Eqn.

Solution to 2-D flow

Analytical solution possible only for simple Analytical solution possible only for simple boundary conditions. But most seepage boundary conditions. But most seepage problems have complex boundary conditions.problems have complex boundary conditions.

Alternative solutionAlternative solutionFlow net (Graphical solution)Flow net (Graphical solution)

Electrical analogy modelsElectrical analogy models

Numerical solutionsNumerical solutions

Flow through a Dam

Drainageblanket

Phreatic line

UnsaturatedSoil

Flow of water

∂∂

∂∂

2

2

2

20

h

x

h

z+ =

z

x

Flow Nets

Graphical representation of solution1. Equipotentials Lines of constant head, h(x,z)

Equipotential (EP)

Flow Nets

Phreatic line

Flow line (FL)

2. Flow lines Paths followed by water particles -tangential to flow

Graphical representation of solution

Equipotential (EP)

Flow Nets

h(x,z) = constant (1a)

∂∂

∂∂

h

xdx

h

zdz+ = 0Thus: (1b)

Equipotenial slopedz

dx

h x

h zEP

⎡⎣⎢

⎤⎦⎥

= −∂ ∂∂ ∂

/

/(1c)

Properties of Equipotentials

Flow line (FL)

Equipotential (EP)

Flow Nets

∆z∆x

Geometry

vzvx

Kinematics

Properties of Flow Lines

From the geometry (2b)

Now from Darcy’s law

Hence (2c)

dx

dz

v

vFL

x

z

⎡⎣⎢

⎤⎦⎥

=

v kh

xx = −∂∂

dx

dz

h x

h zFL

⎡⎣⎢

⎤⎦⎥

=∂ ∂∂ ∂

v kh

zz = −∂∂

Flow line (FL)

Equipotential (EP)

Flow Nets

Orthogonality of flow and equipotential lines

d z

d x

h x

h zE P

⎡⎣⎢

⎤⎦⎥

= −∂ ∂∂ ∂

/

/

dx

dz

h x

h zF L

⎡⎣⎢

⎤⎦⎥

=∂ ∂∂ ∂

On an equipotential

On a flow line

Hencedx

dz

dx

dzF L E P

⎡⎣⎢

⎤⎦⎥

× ⎡⎣⎢

⎤⎦⎥

= − 1 (3)

Flow line (FL)

Equipotential (EP)

Flow Nets

∆Q

X

y

z

t

T

Y

Z

X

FL

FL

vQ

yx=

∆

v kh

z t=

∆

∆∆Q

k h

yx

zt=

∆∆Q

k h

Y X

Z T=

(4a)

(4b)

(4c)

(4d)

From the definition of flow

From Darcy’s law

Combining (4a)&(4b)

Similarly

Geometric properties of flow nets

∆Q

hh+∆h

h+2∆h

EP

Conclusion

yx

zt

YX

ZT= (5)

Flow Nets

vQ

cd=∆

∆∆Q

k h

cd

ab=

v kh

ab=

∆

∆∆Q

k h

CD

AB=

(6a)

(6b)

(6c)

(6d)

From the definition of flow

From Darcy’s law

Similarly

Combining (6a)&(6b)

Conclusion

AB

CD

ab

cd=

∆Q

a

b

c

d

D

B

C

A

h

h h+ ∆

Geometric properties of flow nets

FL

∆Q

EP( h )

EP ( h + ∆h )

Flow Nets

Graphical Construction of Flow Net

1. Equipotential line ⊥ flow line2. Flow element ~ square

Flow Nets

Boundary conditions:

• Submerged soil boundary – Equipotential

b. Impermeable soil boundary - Flow Line

c. Line of constant pore pressure - eg. phreatic surface

Procedure for drawing flow nets

Mark all boundary conditionsMark all boundary conditions

Draw a coarse net which is consistent with the Draw a coarse net which is consistent with the boundary conditions and which has orthogonal boundary conditions and which has orthogonal equipotentialsequipotentials and flow lines. (It is usually easier to and flow lines. (It is usually easier to visualisevisualise the pattern of flow so start by drawing the the pattern of flow so start by drawing the flow lines).flow lines).

Modify the mesh so that it meets the conditions Modify the mesh so that it meets the conditions outlined above and so that rectangles between outlined above and so that rectangles between adjacent flow lines and adjacent flow lines and equipotentialsequipotentials are square.are square.

Refine the flow net by repeating the previous step.Refine the flow net by repeating the previous step.

Flow Nets

Flow Nets

For a single Flow tube of width 1m: ∆Q = k ∆h

For k = 10-5 m/s and a width of 1m ∆Q = 10-5 x 3 m3/sec/m

For 5 such flow tubes Q = 5 x 10-5 x 3 m3/sec/m

For a 25m wide dam Q = 25 x 5 x 10-5 x 3 m3/sec

Calculation of flowPhreatic line

15 m

h = 15m

h =12m h = 9m h = 6mh = 3m

h = 0

fh

NN

HkQ=Note that per metre width

Computations using Flow Nets

hu

zw

w

= +γ

uw w= − −[ ( )]12 5 γ

Calculation of pore pressure

Pore pressure from

At P, using dam base as datum

Phreatic line

P5m

15 m

h = 15m

h = 12m h = 9m h = 6mh = 3m

h = 0


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Figure 7.13 (a) A weir; (b) uplift under a hydraulic structure


Flow Nets in Anisotropic Soil

α =k

kH

V

kh

xk

h

zH V∂∂

∂∂

2

2

2

2 0+ =

Governing Equation

k

k

h

x

h

zH

Vα∂∂

∂∂2

2

2

2

2 0+ =

+zz

xx

== α

Transformation

+∂∂

∂∂

2

2

2

2 0h

x

h

z+ =

Anisotropic Flow Nets

x

z

Impermeable bedrock

L

H1H2

Example1: Flow net for anisotropic soil

The figure shows the dam drawn at its natural scale

Impermeable dam

Soil layerZ


Transformation


Let us assume that the soil has different horizontal and vertical permeabilities such that kH = 4 kV

α = =

= =

=

42

22

x k

kso

x x or xx

z z

V

V

k

kH

V=


z

Impermeable bedrock

L/2

H1H2

x


The figure shows the dam drawn to its transformed scale

Soil layerZ




Equivalent permeability for anisotropic flow

∆∆

Q k th

xH=

∆∆ ∆

Q k th

xk t

h

x

k

keq eqH

V= =

(7a)

(7b)

Equating 7a and 7b gives VHeq kkk =

Considering horizontal flow we have

(a) Natural scale

(b) Transformed scale

xx

Natural scale transformed scale

∆Qt

h h - ∆h h h - ∆h


Example1: Seepage under a dam

h1 = 13.0 mh2 = 2.5 mkV = 10-6 m/skH = 4 x10-6 m/s

k meq = × × = ×− − −( ) ( ) / sec4 10 10 2 106 6 6

∆ h m=−

=( . )

.1 3 2 5

1 40 7 5

∆Q == ´ ´ ´

= ´ = ´

− −

−

( ) ( . ) . / /

. / / / /

2 10 0 75 1 5 10

6 1 5 9 10

6 6 3

3 6 3

m s m

thus

Q m s m m s m


Seepage in Earth Dam

Seepage through a dam on impervious base


1. Obtain α2. Calculate ∆ and then 0.3 ∆3. Calculate d4. With know values of α and d, calculate L by5. With known value of L, calculate q by


Seepage through a dam on impervious base

3-Dimentional Flow

LaplaceLaplace EqnEqn. becomes. becomes

Difficult to solve for most 3Difficult to solve for most 3--dimentional dimentional problems.problems.

Exception: flow to wellsException: flow to wellsPumping testsPumping tests--confined confined acquiferacquifer

Pumping testsPumping tests--unconfined aquiferunconfined aquifer

3-D Flow

02

2

2

2

2

2

=∂∂

+∂∂

+∂∂

z

h

y

h

x

h

Pumping Tests-Confined Aquifer3-D Flow

Pumping Tests-Unconfined Aquifer3-D Flow

Note7_SeepageAnalysis

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