Chapter (7)Chapter (7)

Instructor : Dr. Jehad HammadInstructor : Dr. Jehad Hammad

2011-20102011-2010

Elemental Cube:Saturation S=100 %Void ratio e= constantLaminar flow

Continuity:

Elemental Cube:Saturation S=100 %Void ratio e= constantLaminar flow

Continuity:0dyqdxqqq y

qyx

qxyx

yx

0dydx y

q

xq yx

outin qq

xq dxq xq

xx

dyq y

qy

y

yq

dx

dy

Laplace’s Equation

• Continuity:

• Darcy’s law:

• Replacing:

• if kx=ky Laplace’s Equation!

(isotropy):

0dydx y

q

xq yx

1dykAikq xh

xxxT

1dxdyk1dydxk0 2T

2

2T

2

y

hy

x

hx

2T

2

2T

2

y

hy

x

hx kk0

2T

2

2T

2

y

h

x

h0

Laplace’s Equation

• Typical cases– 1 Dimensional:

linear variation!!

– 2-Dimensional:

– 3-Dimensional:

2T

2

x

h0

ittancons xhT

xbahT

2T

2

2T

2

y

h

x

h0

2T

2

2T

2

2T

2

z

h

y

h

x

h0

Laplace’s Equation Solutions

• Exact solutions (for simple B.C.’s) • Physical models (scaling problems)• Approximate solutions: method of fragments• Graphical solutions: flow nets• Analogies: heat flow and electrical flow• Numerical solutions: finite differences

Flow Nets

• The procedure consists on drawing a set of perpendicular lines: equipotentials and flow lines.

• These set of lines are the solution to the Laplace’s equation.

• It is an iterative (and tedious!) process.• Identify boundaries:– First and last equipotentials– First and last flow lines

Flow Nets

How to draw flow nets

b= l

a

Seepage Calculation from a Flow Net

Seepage Calculation from a Flow Net

Flow Nets

• gradient:

• flow per channel:

• total flow:

Flow channel

Equipotential lines b= l

a

q

h= equipotential drop

bbh

lh

i eNh

Ab

kAlh

kq eNh

e

ff N

Nba

hkNqq

Flow Nets

Number of flow Channels Nf

Number of potential drops Nd

Seepage Calculation from a Flow Net

Seepage Calculation from a Flow Net

Example

Example

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Example - Uplift Pressure

1 2 3 5 64

Point 1: P1 = [(10 + 3) – 0.85]γw = 12.15 γw kN/m2

Potential Drop = H/Nd = (10 – 1.5)/10 = 0.85 m

Point 2: P2 = [(10 + 3) – 2×0.85]γw = 11.3 γw kN/m2

Example – Flow Calculation

Nf = 3

Nd = 10q = kH Nf/Nd = 10-5 ×8.5 × 3/10 = 2.55 × 10-5 m3/s/m

Uplift Pressure

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Mathematical Solutions

S/T΄

q/kH

Previous ExampleFlow Net Method

q = 3×2.45×10-5 m3/s/m

Mathematical Method

q = 0.5×4.2×10-5×3.5 m3/s/m

Nf Δq

q = 7.35×10-5 m3/s/m

q = 7.35×10-5 m3/s/m

From Figure k H

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