Soil Mechanics UII Soil Water and Flow.pptx

CE 6405
Soil Mechanics
Unit II
SOIL WATER AND WATER FLOW

Syllabus
Soil water – static pressure in water - Effective
stress concepts in soils – capillary stress –
Permeability measurement in the laboratory
and field pumping in pumping out tests –
factors influencing permeability of soils –
Seepage – introduction to flow nets – Simple
problems. (sheet pile and weir).

Types of Soil water
Free water
Held water

Free Water
It moves freely in the pores of the soil under influence
of gravity
It flows from one point to another point when there is a
difference of Head( Elevation)
The rate at which the head reducing along the flow
passage is called Hydraulic gradient
= I = h/L
The flow of free water in soil is just like laminar flow
through pipes

Held water
It is retained in the pores of soil. It can not move
under the influence of gravity
Further Classification of Held Water
Structural Water
Adsorbed water
Capillary water

Types of held water :
• 1. Adsorbed water
– “HYGROSCOPIC WATER”
– It is held by electrochemical forces existing on the soil
surface
– Quantity depends upon the colloidal fractions in the soil
– It is significant in the clay soil and negligible in coarse
grained soils
– Remove by oven drying
– Not available to plants

2. Structural water
• It is chemically combined water in the crystals
of structure of the minerals in the soil
• This water can not be removed without
breaking the structure of the minerals
• A high temperature of 300°C is required for
removing this water

CAPILLARY WATER
This water held in the pore space lines(interstices) of soil
due to capillary forces(Surface Tension)
It exists in soil so long as
there is an air- water interface
As soon as the soil submerged in water,
the capillary water become normal

capillarity
Height water will
rise in
cylinder depends
on
diameter of tube;
due to adhesion
of water and
tube
Plastic Glass

Capillary pressure
• Thin tube in open pan water
0
-10
-20 g/cm3

• the smaller the pore space, the higher
capillary water will rise in profile
• Smaller pore space, tighter water is held to
particle surfaces against gravity (i.e., higher
field capacity)
Pan of water
clay silt sand

Seepage Pressure or Seepage Force
Seepage:
Flow of water through a soil under hydraulic gradient
Seepage pressure
When water flowing through soil pores, a viscous friction exerted
by water on the soil surface
due to that
An energy transfer is effected between soil and water
The force corresponding to this energy transfer is called seepage
force or pressure.
Thus
It is the pressure exerted by water on the soil through which it
percolates.

Determination of seepage pressure
the seepage pressure at any point is
equal to the hydraulic potential multiplied by
unit weight of water.
Ps= h* rw
=(H-n * ∆ h )* rw
the seepage pressure acts in the direction of
flow.

Techniques for Finding Seepage
• Inspection (intuition)
• Graphical Techniques
• Analog Models
• Analytical Mathematical Techniques (Calculus)
• Numerical Mathematical Techniques (Computers)

Assumptions Needed For Flow Net Construction
Aquifer is homogeneous, isotropic
Aquifer is saturated
there is no change in head with time
soil and water are incompressible
flow is laminar, and Darcy’s Law is valid
All boundary conditions are known.

16
Flow Net Theory
1. Streamlines Y and Equip. lines  are .
2. Streamlines Y are parallel to flow
boundaries.
3. Grids are curvilinear squares, where diagonals
cross at right angles.
4. Each stream tube carries the same flow.

17
Flow Net in Isotropic Soil
Portion of a flow net is shown below
F
Y
Stream tube

18
The equation for flow nets originates from
Darcy’s Law.
Flow Net solution is equivalent to
solving the governing equations of flow
for a uniform isotropic aquifer with
well-defined boundary conditions.

19
• Flow through a channel between
equipotential lines 1 and 2 per unit
width is:
∆q = K(dm x 1)(∆h1/dl)
dm
Dh1
dl
F1
F3
Dq
F2
Dh2
Dq
n
m

20
• Flow through equipotential lines 2 and 3 is:
∆q = K(dm x 1)(∆h2/dl)
• The flow net has square grids, so the head
drop is the same in each potential drop:
∆h1 = ∆h2
• If there are nd such drops, then:
∆h = (H/n)
where H is the total head loss between the
first and last equipotential lines.

21
• Substitution yields:
– ∆q = K(dm x 1/dl)(H/n)
• This equation is for one flow channel. If
there are m such channels in the net, then
total flow per unit width is:
– q = (m/n)K(dm/dl)H

22
• Since the flow net is drawn with squares,
then dm  dl, and:
q = (m/n)KH [L2
T-1
]
where:
– q = rate of flow or seepage per unit width
– m= number of flow channels
– n= number of equipotential drops
– h = total head loss in flow system
– K = hydraulic conductivity

23
Drawing Method:
1. Draw to a convenient scale the cross
sections of the structure, water elevations,
and aquifer profiles.
2. Establish boundary conditions and draw
one or two flow lines Y and equipotential
lines F near the boundaries.

24
Method cont…
3. Sketch intermediate flow lines and equipotential
lines by smooth curves adhering to right-angle
intersections and square grids. Where flow
direction is a straight line, flow lines are an
equal distance apart and parallel.
4. Continue sketching until a problem develops.
Each problem will indicate changes to be made
in the entire net. Successive trials will result in a
reasonably consistent flow net.

25
Method cont…
5. In most cases, 5 to 10 flow lines are
usually sufficient. Depending on the no.
of flow lines selected, the number of
equipotential lines will automatically be
fixed by geometry and grid layout.

26
Seepage Under Dams
Flow nets for
seepage through
earthen dams
Seepage under
concrete dams
Uses boundary
conditions (L & R)
Requires
curvilinear square
grids for solution

27
Effects of Boundary Condition on Shape of
Flow Nets

29
Flow Nets: an example
• A dam is constructed on a permeable
stratum underlain by an impermeable rock.
A row of sheet pile is installed at the
upstream face. If the permeable soil has a
hydraulic conductivity of 150 ft/day,
determine the rate of flow or seepage
under the dam.

30
Flow Nets: an example
The flow net is drawn with: m = 5 n = 17

31
Flow Nets: the solution
• Solve for the flow per unit width:
q = (m/n) K h
= (5/17)(150)(35)
= 1544 ft3
/day per ft

32
Flow Nets: An Example
• There is an earthen dam 13 meters across
and 7.5 meters high.The Impounded water
is 6.2 meters deep, while the tailwater is
2.2 meters deep. The dam is 72 meters
long. If the hydraulic conductivity is 6.1 x
10-4
centimeter per second, what is the
seepage through the dam if n = 21
K = 6.1 x 10-4
cm/sec
= 0.527 m/day

33
Flow Nets: the solution
• From the flow net, the total head loss, H, is
6.2 -2.2 = 4.0 meters.
• There are 6 flow channels (m) and 21 head
drops along each flow path (n):
• Q = (KmH/n) x dam length =
(0.527 m/day x 6 x 4m / 21) x (dam length)
• = 0.60 m3
/day per m of dam
• = 43.4 m3
/day for the entire 72-meter
length of the dam

Soil Mechanics UII Soil Water and Flow.pptx

DARCY’S LAW
• The law of flow of water through soil was first studied
by darcy in 1856.
• “for laminar flow through saturated soil mass, the
discharge per unit time is proportional to the hydraulic
gradient.
q=k* I *A
q/A = k*I
v = k* I

q = discharge per unit time
A = total c/s area of soil mass.
I = hydraulic gradient = h/L
k = darcy’s coefficient of permeability
 I = h/L = (h1 – h2)/L
 q = k* (h1-h2)*A/L

Co Efficient of Permeability
• Ability of a soil to transmit water
• Depends on :
– Pore size
• Coarse grained soil has higher cond. than fine-grained
because movement through large pores is faster
– Amount of water in soil
• Cond. decreases as water content increases
– Water moves through largest pores first

Factors affecting permeability of soils
i. Particle size.
ii. Properties of pore fluid.
iii. Void ratio of soil.
iv. Shape of particles.
v. Structure of soil mass.
vi. Degree of saturation.
vii. Adsorbed water.
viii.Impurities in water.

1) Particle size :-
permeability varies approximately as the
square of grain size
K=C*(D10)²
k=coefficient of permeability(cm/sec)
D10=effective diameter(cm)
C= constant = between 100 and 150

2)Properties of pore fluid :
The permeability is directly proportional
to the unit weight of water and inversely
proportional to is viscosity.
It is usual practice to report the coefficient of permeability at
27°C.
3)Void ratio of soil :
The coefficient of Permeability varies as e³/(1+e). For a
given soil , greater the void ratio , the higher is the value
of the coefficient of permeability.

4) Shape of particles :
The permeability of soil depends upon the shape of
particles.
5)Structure of soil mass :
stratified soil deposits have greater permeability
parallel to the plane of stratification then that
perpendicular to this plane.
6) degree of saturation :
If the soil is not fully saturated, it contains air pockets
formed due to entrapped air. The presence of air in soils,
causes blockage of passage and permeability is reduced.

7) Adsorbed water :
the fine grained soils have a layer of
adsorbed water strongly attached to their
surface.
8) Impurities in water :
Any foreign matter in water has a tendency
to plug the flow passage and reduce the
permeability of soils.

DETERMINATION OF COFFICIENT OF
PERMEABILITY
1) Laboratory methods
A-: constant head permeability test
B-: falling head permeability test
2) Field methods
A-: pumping out tests
B-: pumping in tests
3) Indirect methods
A-: computation from the particle size
B-: computation from consolidation test.

A) CONSTANT HEAD PERMEABILITY TEST
 The coefficient of permeability of a coarse-
grained soil can be determined in the
laboratory using a constant-head
permeability test.
 The test includes a cylindrical soil specimen
that is subjected to a constant head as shown
in Figure

Equipments :
Permeability mould, internal diameter = 100mm,
effective height =127.3, capacity = 1000 ml,
Constant head tank.
Graduated cylinder, stop water, thermo meter.
Filter paper, vacuum pump.
Weighing balance, 0.1 gm accuracy.

The length of the soil specimen is L and its cross-
sectional area is A.
The total head loss (h) along the soil specimen is
equal to the constant head, which is the difference
in elevation between the water levels in the upper
and lower reservoirs as shown in the figure.
Using a graduated flask, we can collect a volume of
water (V) in a period of time (t ). From this we can
calculate the flow rate
( q= V/t).
50

B) FALLING HEAD PERMEABILITY TEST
All the equipment required for the constant
head permeability test.
In addition A stand pipe
It is recommended for fine grained soil

Procedure :
 For a falling head test arrangement the specimen(soil sample) a
stand pipe shall be connected through the top inlet..
The time interval required for the water level to fall from a known
initial headh1 to a known final head h2 as measured above the centre of
the outlet shall be recorded
The stand-pipe shall be refilled with water and the test repeated till three
successive observations give nearly same time interval.
RECORD OF OBSERVATION:
The dimensions of specimen, length L and diameter D, are measured.
Area “a” of stand-pipe is recorded.
During the test, observations are made of initial time t1, final time t2
initial head h1 final head h2 in stand-pipe and are recorded.

Where, a= c/s area of stand pipe (cm2)
• A= c/s area of soil sample
• L= length of soil sample (cm)
• t= time interval to fall head from h1 to h2
• h1= initial head (cm)
• h2= final head (cm)

2. Field permeability tests
1) Pumping-out tests:-
A-: for unconfined aquifer
B-: for confined aquifer
2) Pumping –in tests:-
A-: open end tests
B-: single packer tests
C-: double packer tests

1) Pumping-out Tests
 For large engineering projects, it is the usual
practice to measure the permeability of soils of
entire aquifer by pumping out tests.
 This method is very useful for a homogeneous
coarse
 The pumping out tests are very costly.

1. Aquifer
2. Aquiclude
3. Aquitard
4. Aquifuge
5. Unconfined aquifer
6. Confined aquifer
Definitions of Technical Terms related to permeability

Important terminologies related to Aquifer

Aquifuge
An impermeable body of rock which contains
no interconnected pore spaces(interstices)
and therefore
neither absorbs nor transmits water.

Typical View of Perched Aquifer

2) Pumping –in tests
Pumping in tests are conducted to determine
The coefficient of permeability of
an individual stratum thorough which a hole is drilled.
The tests are more economical than pumping out tests
The water pumped in should be clean.
impurities such as silt, clay or any other foreign matter may
cause plugging of the flow passages.

Unconfined flow pumping test
71
 
2
1
2
2
1
2
ln
h
h
r
r
q
k












A soil sample 5 cm in length and 60 cm in cross-
sectional area, water percolates through the
sample in 10 minutes is 480 ml under a constant
head of 40 cm. Weight of oven dried sample is
498 gm and specific gravity of soil = 2.65.
Calculate:
(i) Coefficient of permeability
(ii) Seepage velocity.

If during a permeability test on a soil sample
with falling head permeameter, equal time
intervals are noted for drops of head from
h1 and h2 and again from h1 to h2, find a
relationship between h1, h2 and h3.
Solution: For falling head from h1 and h2

Permeability of Stratified Soils:
When a soil profile consists of a number of strata
having different permeability, the equivalent or
average permeability of the soil is different for Flow
of water is parallel or normal to the plane of
stratification.
For flow parallel to layers
the hydraulic gradient in each layer is the same and
the total flow rate is the sum of flow rates in all the
three layers.

Where Kx = Equivalent or average permeability in direction
parallel to the layers.

For flow normal to the layers
the flow rate must be same in all layers for steady flow, and
as the flow area ‘A’ is constant, the flow velocity across layer is also
the same

Where Kz = equivalent permeability for flow normal to the layers.
So the equivalent permeability for flow parallel to the strata is always
greater than that for flow normal to the strata
i.e., Kx is always greater than Kz.

A sand deposit is made up of three horizontal layers of
equal thickness. The permeability of the top and
bottom layers is 2 x10-4
cm/s and that of middle layer
is 3.2 x 10-2
cm/s. Find the equivalent permeability in
the horizontal and vertical direction and their ratio.

Suction Pressure or Capillary Pressure

Effective Stress Equation.
The principle of effective stress, introduced by Karl Terzaghi,
states that the effective stress σ'
(i.e., the average intergranular stress between solid particles)
may be calculated by a simple subtraction of the pore pressure from
the total stress:
where σ is the total stress and u is the pore pressure.












n
i
i
i
n
n
z h
h
h
h
1
2
2
1
1 ...... 




Many Layers of soil, the vertical stress due to self-weight of soil is
given as following.

3
2
2
2
1
1
1 ' h
h
h
h sat
w
z 






 




Point of Stress under water Table
σ z=γ 1h1+γ 1' h2+γ w
h2+γ sat3h3
Cl ay ( w
aterti ght)
W
ater tabl e
Sand
γ 1' h2
γ w
h2
γ 1h1

Determine the stresses at points A, B, C, and D in the soil profile shown.
Point Total Stress Neutral Stress Effective Stress
A 0 0 0
B 440 0 440
C 1040 312 728
D 1790 686 1104

calculations
Point B
Point C
Point D

Capillary Rice
Above the water table up to the height of capillary rise is the capillary zone and
the water is called capillary water.
The water pressure is negative (less than atmospheric) in the capillary zone.
These observations are illustrated in the figures below.

A soil profile is shown with capillary water and free water. The
problem is to determine the pore water pressures at point A,
B, C, and D.

The stratum’s conditions and the related physical characteristics
parameters of a foundation are shown in Fig below. Calculate the
stress due to self-weight at a,b,c. Draw the stress distribution.
w=15.6%
e=0.57
γs=26.6kN/m3
w=22% wL=32%
wp=23%
γs=27.3kN/m3

a σz=0
b σz(upper)=γ’1h1=9.9×2=19.8kPa
σz(Down)=γ’1h1+ γw(h1+hw)=9.9×2+10×(2+1.2)=51.8kPa
c σz=γ’1h1+ γw(h1+hw)+ γsat2h2
= 9.9×2+10×(2+1.2)+20.8×3=114.2kPa

The stratum’s conditions and the related physical characteristics
parameters of a foundation are shown in Fig below. Calculate the
stress due to self-weight at 10m depth. Draw the stress distribution.
Note: For saturated clay, both cases (watertight and non-watertight)
need to consider.
w=8% e=0.7
γs=26.5kN/m3
e=1.5
γs=27.2kN/m3

Soil Mechanics UII Soil Water and Flow.pptx

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