Course
Year
: S0705 – Soil Mechanic
: 2008
TOPIC 2
STEADY STATE FLOW THROUGH SOIL
CONTENT
•
•
•
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PERMEABILITY / CAPILARITY (SESSION 7 : F2F)
SEEPAGE (SESSION 8 : F2F)
FLOW NET / JARINGAN ALIRAN (SESSION 9 – 10 : F2F)
SESSION 7
PERMEABILITY / CAPILARITY
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INTRODUCTION
•
DEFINITION :
THE VELOCITY OR THE CAPABILITY OF WATER/FLUDI PASS
THROUGH POROUS MEDIA
•
NOTATION :
k
•
UNIT :
m/s, cm/s
•
PURPOSE :
– To evaluate the seepage through dam
– To evaluate the uplift force or seepage force under hydraulic structure for
stability analysis
– To control seepage velocity
– To determine the consolidation time rate
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DETERMINATION OF PERMEABILITY COEFFICIENT
– LABORATORY
• CONSTANT HEAD (TINGGI KONSTAN)
• FALLING HEAD (TINGGI JATUH)
– FIELD
• UNCONFINED AQUIFER (AKIFER BEBAS)
• CONFINED AQUIFER (AKIFER TERKEKANG )
• INCONSTANT WATER HEIGHT (TINGGI AIR TIDAK TETAP)
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CONSTANT HEAD (TINGGI KONSTAN)
– SUITABLE FOR SANDY SOIL, SAND OR GRAVEL WHICH HAVE
BIG VOID RATIO VALUE
– BASIC EQUATION :
 h
Q  A.v.t  A.(k.i).t  A. k. .t
 L
Q.L
k
A.h.t
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FALLING HEAD (TINGGI JATUH)
– MORE ECONOMICAL FOR LONG TERM TEST
– BASIC EQUATION :
v-
dh
dt
qmasuk  -a
dh
dt
qkeluar  A.v  A.k .i  A.k .
qmasuk  qkeluar  -a
k
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a.L h1
ln
A.t h2
h
L
dh
h
 A.k .
dt
L
UNCONFINED AQUIFER (AKIFER BEBAS)
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UNCONFINED AQUIFER (AKIFER BEBAS)

dh
k . . h22  h12
Q  k . .2. .r.h 
r
dr
ln 2
r1
r2
r1
k
 . h22  h12
Q. ln

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

Q
CONFINED AQUIFER (AKIFER TERKEKANG )
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CONFINED AQUIFER (AKIFER TERKEKANG )
r1
2,3.Q. log
r2
k
2..h o .h1  h 2 
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INCONSTANT WATER HEIGHT (TINGGI AIR TIDAK TETAP)
2r
r y
40. .
y t
k

L 
y 
 20  r . 2  L  



y
L
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y
RANGE OF k VALUE
2
k  C.D10
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Cm/s
RANGE OF k VALUE
C is coefficient of soil roughness (published by Hazen), the value
range from 40 to 150 i.e. as follow :
C
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Sand (one or all of the next characteristics)
40-80
Very fine, well graded or contain some fine particle
80-120
Rather rough, poor graded or clean, rough and well graded
120-150
Very rough, very poor graded, contain gravel and clean
EQUIVALENT COEFFICIENT OF PERMEABILITY OF LAYERED SOIL
– Equivalent Coefficient of Vertical Permeability (kv’)
– Equivalent Coefficient of Horizontal Permeability (kh’)
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Equivalent Coefficient of Vertical Permeability (kv’)
Basic Concept
– qin = qout
– v constant
v  kv'.i  k 1 .
h1
h
h
 k 2 . 2  ...  k n . n
H1
H2
Hn
h
H1
h H
h H
H
h
 1 ; 2  2 ; 3  3 ..... n  n
k1
v k2
v k3
v
kn
v
H
h1 h 2 h 3
h
H H
H


 ...  n  1  2  3  ...  n
v
v
v
v
k1 k 2 k 3
kn
H1  H 2  H 3  ...  H n  L
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v  kv'.
h
L
kv' 
H1
k1

H2
k2

L
H3
k3
 ... 
Hn
kn
Equivalent Coefficient of Vertical Permeability (kv’)
q  A.vaverage  L.kh'.i
L.kh'.i  k 1 .H1 .i  k 2 .H2 .i  ...  k n .Hn .i
k 1 .H1  k 2 .H 2  ...  k n .H n
kh' 
L
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EXAMPLE 1
q = 1 ft3/hr
Question :
Find the permeability coefficient of sand in ft/min
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EXAMPLE 1
SECTION 1
q
h1
 k.
A1
L1
h1 
q.L1
A1 .k
h 2 
q.L 2
A 2 .k
SECTION 2
q
h 2
 k.
A2
L2
TOTAL
h t  h1  h 2
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h t 
q.L1
q.L 2

A1 .k
A 2 .k
EXAMPLE 1
1.400 1.600
20 

20.k
10.k
k = 4 ft/hour = 6,67x10-2 ft/min
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EXAMPLE 2
q
Section 1
Questions :
- determine h
- determine q in cc/sec
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Section 2
EXAMPLE 2
Determination of h
Section 2
Section 1
q1  k 1 .i1 .A1
q2  k 2 .i 2 .A2
h5
q2  0.007.
.25
40
50  h
q1  0.02.
.25
40
q1  q 2
0.02.(50  h)  0.007.(h  5)
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h = 38.33 cm
EXAMPLE 2
Determination of water flow rate
q 2  k 2 .i 2 .A2
q1  k 1 .i1 .A1
or
q  0.02.
50  38.33
.25
40
q = 0.15 cc/s
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SESSION 8
SEEPAGE
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INTRODUCTION
• DEFINITION
VOLUME OF WATER/FLUID FLOW IN OR FLOW OUT AT A MEDIA
OR CERTAIN SOIL MASS
• PURPOSE
– TO DETERMINE THE INFLUENCE OF SEEPAGE TO THE STABILITY
OF STRUCTURE OR DAM
– TO ESTIMATE THE FLOW VELOCITY AND FLOW RATE AT
DEWATERING WORK
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WATER FLOW EQUATION
– BASIC OF THEORY
• DARCY LAW
v  k .i
h
i
L
k .i
v' 
n
• BERNOULLI LAW
v12
p1
v 22
p2

 g.z 1 

 g.z 2
2g  w .g
2g  w .g
• CONTINUITY EQUATION
q  v1.A1  v2 .A2  cons tan t
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= constant energy
WATER FLOW EQUATION
Inflow water per time unit:
qinf low  v x dydz  v y dxdz  v z dxdy
Outflow water per time unit:
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v y 

v
v




qoutflow   vx  x dx dydz   v y 
dy dxdz   vz  z dz dxdy
x 
y
z 




WATER FLOW EQUATION
qinflow = qoutflow
v y
 v x
v z



 x
y
z

v y
 v x
v z



 x
y
z


dV

dxdydz


dt


1 Ww
1
e




 w t
1  eo t

CONTINUITY EQUATION
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WATER FLOW EQUATION
STEADY STATE CONDITION:
e
 0
t
v y
 v x
v z



 x
y
z

WATER FLOW VELOCITY :
h
x
h
v y  k y i y  k y
y
h
v z  k z i z  k z
z
v x  k x i x  k x
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

0

WATER FLOW EQUATION
  h     h    h 
 k y
   k z
kx

0
 x  x  y   y   z  z 
HOMOGEN SOIL
k constant at x,y,z direction
 2h
 2h
 2h
kx
 ky
 kz
0
x 2
y 2
z 2
ISOTROPY SOIL
kx = k y = k z = k
 2h  2h  2h
 h


0
x 2
y 2
z 2
2
 2h  2h
 2 0
TWO DIMENSION  h 
2
x
z
2
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LAPLACE
EQUATION
WATER FLOW EQUATION
• SEEPAGE SOLUTION
– CLOSED FORM SOLUTION
– MODEL SOLUTIONS
– APPROXIMATE SOLUTIONS
• NUMERICAL SOLUTIONS
• GRAPHICAL SOLUTIONS  FLOW NET
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SESSION 9-10
FLOW NET / JARINGAN ALIRAN
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DEFINITION
Combination of 2 line group which perpendicular each
other i.e. :
– Flow Line
A bundle of line or points which defined flow direction
– Equipotential Line
The position of point which have same total head
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FLOW NET / JARINGAN ALIRAN
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FLOW NET / JARINGAN ALIRAN
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FLOW NET / JARINGAN ALIRAN
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FLOW NET / JARINGAN ALIRAN
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FLOW NET / JARINGAN ALIRAN
• GUIDANCE OF FLOW NET DRAWING
– The water surface of upper stream and lower stream is equipotential
line
– The interface line of water and soil is equipotential line
– The flow line perpendicular with equipotential line
– The impermeable surface of boundary line is flow line
– The shape area formed by flow line and equipotential line is square
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FLOW NET / JARINGAN ALIRAN
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FLOW NET / JARINGAN ALIRAN
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FLOW NET / JARINGAN ALIRAN
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FLOW NET / JARINGAN ALIRAN
q
a
h+h
b
h
q  A.v  A.k .i  (a.1).k .
h
b
h 
h1  h 2
Nd
H  h1  h 2
a  h1  h 2 

q   q  N f .k . 
b  Nd 
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Nf
a
qk
.H 
Nd
b
a=b
Nf
qk
.H
Nd
EXAMPLE 3
Sheet Piling
4,50 m
Datum
0,5 m
D
B C
A
6,0 m
8,60 m
E
k = 1,5 x 10-6 m/s
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EXAMPLE 3
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EXAMPLE 3
Nd = 12
Nf = 4,3
H = 4,0 m
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EXAMPLE 3
Nf
qk
.H
Nd
q  1.5 x10 6.
4,3
.4.00  2.15 x10 6 m3 / s.m
12
nd
hP 
.H
Nd
10
hP  .4  3.33m
12
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EXAMPLE 4
5m
15 m
k = 2,5 x 10-5 m/s
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12 m
EXAMPLE 4
Nd = 15
Nf = 4,7
H = 4,0 m
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EXAMPLE 4
Nf
qk
.H
Nd
4,7
q  2.5 x10 .
.4.00  3.1x10 5 m3 / s.m
15
5
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SEEPAGE FORCE / GAYA REMBESAN
H
h1
h2
L
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SEEPAGE FORCE / GAYA REMBESAN
 w . h2 . A
L
Soil weight = t.L.A
 w . h1 . A
TOTAL FORCE
 F   t .L.A   w .(h1  h 2 ).A
BODY FORCE (GAYA BADAN)
Body Force( F ) 
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Total Force
volume
SEEPAGE FORCE / GAYA REMBESAN
F
 t .L. A   w .(h1  h2 ). A
L. A
H L
F   t   w
   t   w (1  i )
 L 
bouyancy = t - w
F   bouyancy  i. w
SEEPAGE BODY FORCE (j)= i . w
CRITICAL CONDITION
 bouyant  i. w  0
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ic 
 bouyant
w

Gs  1
1 e
H
h1
h2
L
EXAMPLE 5
Questions :
1. Water Flow Rate
2. Flow Velocity
3. Seepage Velocity
4. Seepage Force at point A
k = 1x10-3 cm/s
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n = 0.67
EXAMPLE 5
• Water Flow Rate
q  k .i .A
i
H
4
 1
L
4
q  1x105.1.A  1x105 A
• Flow Velocity
v  k.i
v  1x10 5 .1  1x10 5 m / s
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EXAMPLE 5
• Seepage Velocity
k .i v
v' 

n n
1x10 5
v' 
 1.5 x10 5 m / s
0.67
• Seepage Force
Fs  i. w
Fs  1.1000  1000 kg / m2
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