Influence of water in soil
Prepared by:
Dr. Hetty
Muhammad Azril
Fauziah Kassim
Norafida
Copyright Dr. Hetty et. al. 2014
Important things
How water flow in soil?
How fast water flow in soil?
Does water effect soil condition?
Can water permit in geotechnical
structures (Dam, Retaining Wall and etc.)?
Why do we need to know water flow?
Copyright Dr. Hetty et. al. 2014
Introduction
From an engineering mechanics point of view, groundwater in soil may be one of two types, occurring in two distinct zones separated by the water table or phreatic surface.
Two zones of water in soil are:-
a) Phreatic water or gravitational water which;
-subject to gravitational forces
-saturates the pore spaces in the soil below the water table
-has an internal pore pressure greater than atmospheric pressure
-tends to flow laterally
(b) Vadose water, which may be:
-transient percolating water, moving downwards to join the phreatic water below the water table
-capillary water held above the water table by surface tension forces (with internal pore pressure less than atmospheric)
Copyright Dr. Hetty et. al. 2014
Water zones in soil
Fig. 1: Water zones in soil
Source: Whitlow, 2001Copyright Dr. Hetty et. al. 2014
Water Flows through Soil (1 D)
As water flows through pipes it will also
flow to any mediums that permit
Water could flow through soil medium
that can be explained by the Bernoulli’s
equations
Copyright Dr. Hetty et. al. 2014
Bernoulli Equation
ht= hz + u + v2 = hz + hp
w 2g
where hz = position or elevation head
u = pressure head to pore pressure
w
v2 = velocity head
2g
Water will flow through soil if there exists a difference
of head within 2 points Copyright Dr. Hetty et. al. 2014
Darcy Law for saturated flow
In saturated conditions, one-dimensional flow is governed by Darcy’s Law, which states that the flow velocity is proportional to the hydraulic gradient:
v α i or v = ki
where v = flow velocity
k = the flow constant or coefficient of permeability
i = the hydraulic gradient = ∆ h
∆ L
∆h = difference in total head over a flow path length ( ∆L)
Copyright Dr. Hetty et. al. 2014
Coefficient of Permeability
The permeability of a soil is a measure of its capacity
to allow the flow of water through the pore spaces
between solid particles.
The degree of permeability is determined by applying a
hydraulic pressure gradient, i in a sample and
measuring the consequent rate of flow of water.
The coefficient of permeability is expressed as velocity,
(mm/s2)
Copyright Dr. Hetty et. al. 2014
Coefficient of Permeability
The value of k is used as a measure of the resistance to flow by the soil, and it is affected by several factors:
a) The porosity of the soil
b) The particle size distribution
c) The shape and the orientation of soil particles
d) The degree of saturation/presence of air
e) The type of cation and thickness of adsorbed layers associated with clay minerals (if present)
f) The viscosity of the soil water, which varies with temperature
Copyright Dr. Hetty et. al. 2014
Permeability
“the larger a soil’s void space, the greater
will be its permeability. Conversely, the
smaller the void space, the lesser will be
it’s permeability”
Therefore, coarse soil > fine soil on
permeability value.
Copyright Dr. Hetty et. al. 2014
Coefficient of Permeability
k value (m/s) Type of soil Note
102 Clean gravels Very good drainage
101
1
10-1
10-2 Clean sands
Gravel-sand mixtures10-3 Good drainage
10-4
10-5 Very fine sands
Silts and silty sands
Poor Drainage
10-6
10-7 Clay silts (>20% clay)
10-8 Practically impervious
10-9
Source: Roy Whitlow (2001) Copyright Dr. Hetty et. al. 2014
Question
Could you estimate the coefficient of
permeability if you are given a sandy soil
sample?
The answer
It ranges from 10-4 to 10-5 m/s
Copyright Dr. Hetty et. al. 2014
Determination of k Laboratory
- Falling head
- Constant Head
In situ ( On field Test)
- Steady state Pumping - confined aquifer/unconfined aquifer
-Borehole test- Rising head, Variable head, Constant Head
Copyright Dr. Hetty et. al. 2014
Lab vs. In situ (Advantages)
Lab
1. Reliability is
questionable
Insitu
1. Reliable
Lab vs. In situ (Disadvantages)
• Lab
1. Simple
2. Cheap (Cost)
• Insitu
1. Complex
2. Expensive
Copyright Dr. Hetty et. al. 2014
Laboratory Constant Head
Usually use for coarse-grained soils
The equation use to calculate is k = Q L
Aht
Falling Head
Usually use for cohesive soils
The equation use to calculate is k = 2.303 aL log10 h1
At h2Copyright Dr. Hetty et. al. 2014
Constant Head TestThe total volume of water collected may be expressed as
Q = Avt = A(ki)t = Ak(h/L)t
where Q = volume of water collected
A = area of cross section of the soil specimen
t = duration of water collection
h = head loss
or the equation can be substituted into
Q = A(kh)t
L
L = length of specimen
or
k = QL
AhtCopyright Dr. Hetty et. al. 2014
Constant Head Apparatus
Fig. 2: Constant Head apparatus (M. Das, 2000)
Copyright Dr. Hetty et. al. 2014
Example 1
A test from Constant Head Test give this value :-
L = 18 mm ; A = area of specimen = 3.5 mm2
h = constant head difference = 28 mm
Water collected in a period of 3 minutes = 21.58 mm3
Solution :-
•k = QL
Aht
Given ; Q= 21.58 mm3, L = 18 mm, A = 3.5 mm2, h = 28 mm and t = 3 min
K = (21.58)(18) = 0.022mm/s
(3.5)(28)(3)(60)Copyright Dr. Hetty et. al. 2014
Example 2Determine the time required from the coefficient of
permeability,k =0.0986 cm/s according to following
data:
Quantity of water discharge during test = 250 cm3
Length of specimen = 11.43 cm
Head (difference between manometer level) = 5.5 cm
Diameter of specimen = 10.16 cm
Copyright Dr. Hetty et. al. 2014
Solution :-
k = QL
Aht
Given ; k =0.0986 cm/s Q= 250 cm3, L = 11.43 mm, d
= 10.16 cm, h = 5.5 cm
A = ∏ x 10.162/4 = 81.07 cm2
t = (250)(11.43) = 65 s
(81.07)(5.5)(0.0986)
Copyright Dr. Hetty et. al. 2014
Example 3During a test using a constant-head permeameter, the following
data were collected. Determine the average value of k
Diameter of sample = 100 mm
Distance between manometer tapping points = 150 mm
Quality collected in 2 min
(1x103mm3)
541 503 509 474
Difference in manometer
level (mm)
76 72 68 65
Copyright Dr. Hetty et. al. 2014
Solution
Cross-sectional area of sample = 1002 x ∏/4 = 7854 mm2
Flow time = t = 2 x 60 = 120 s
k = (Q)(150) = 0.159 Q/h mm/s
(7854)(h)(120)
Average k = (k1+k2+k3+k4) / 4 = 1.15 mm/s
Copyright Dr. Hetty et. al. 2014
Falling Head Test
The specimen is first saturated with water.
Water is then allowed to move through the soil specimen under a falling-head condition while the time required for a certain quantity of water to pass through the soil specimen is determine and recorded.
Water is insert through the burette ( (a) cross section area of burette) for several time (t)
Copyright Dr. Hetty et. al. 2014
Falling Head
The rate of flow of the water through the specimen at any time t can be given by
qin = -a(dh) qout = k(h/L)A
(dt)
where q = flow rate L = length specimen
a = cross-sectional area of the standpipe
A = cross-sectional area of the soil specimen
Equating qin and qout gives
-a(dh/dt) = kA(h/L) k = (aL/At) In (h1/h2)
or
k = 2.303 aL log10 h1 At h2
h1 = hydraulic head of
beginning
h2 = hydraulic head of endCopyright Dr. Hetty et. al. 2014
Falling Head Test Set Up
Fig. 4: Schematic of falling permeability setup ( Liu and Evett,
2002)
Copyright Dr. Hetty et. al. 2014
Example 1
In a laboratory, a falling head permeability test was
conducted on a silty soil. The following data were
obtained:
Length of specimen = 15.80 cm
Diameter of specimen = 10.16 cm
Cross section area of burrette = 1.83 cm2
Hydraulic head at beginning h1 = 120.0 cm
Hydraulic head at end h2 = 110.0 cm
Time requirement = 20.0 min
Determine the k value for the soil sample? Copyright Dr. Hetty et. al. 2014
Solution :
k = 2.303 aL log10 h1
At h2
A = ∏ (10.16 cm)2/4 = 81.07 cm2
k = 2.303 (1.83)(15.80) log10 (120)
(81.07)(1200) (110)
k = 2.58 x 10-5 cm/s
Copyright Dr. Hetty et. al. 2014
Example 2
A falling head permeameter mould of 75 mm
diameter and 150 mm height is filled with
soil height.the internal diameter of burette
tube is 5 mm. Determine the time ( in
seconds) required for the hydraulic head to
fall from 1300 mm to 50 mm if the
coefficient of permeability of soil in the
mould is 0.0215 mm/s?
Copyright Dr. Hetty et. al. 2014
Given ; k = 0.0215 mm/s
D = 75 mm, L = 150 mm
Internal diameter of tube, d = 5 mm
h1 = 1300 mm , h2 = 50 mm
Solution :
Specimen cross-section area, A = ∏(752)/4
= 4418.44 mm2
Tube cross-sectional area, a = ∏ (52)/4
= 19.64 mm2
0.0215 = 2.303 (19.64)(150) log10 (1300)
(4418.44)(t) (50)
t = 2.303 (19.64)(150) log10 (1300)
(4418.44)(0.0215) (50)
t = 101 s
Copyright Dr. Hetty et. al. 2014
Example 3Sampel tanah liat berdiameter 100mm sepanjang 150mm diuji
kebolehtelapanya di makmal , set data ujikaji kebolehtelapan
turus menurun adalah seperti jadual berikut:-
Diameter
paip
(mm)
Aras dalam paip(mm)
Bacaan awal Bacaan akhir
Perbezaan masa
(s)
5.00 1200
800
800
400
82
149
9.00 1200
900
700
900
700
400
177
169
368
Dapatkan purata bagi nilai k tanah ini.
Copyright Dr. Hetty et. al. 2014
Given;
Diameter sample = 100 mm, L = 150 mm
Diameter pipe = 5 mm dan 9 mm
Solution:
A = ∏ (1002)/4 = 7855 mm2
a1 = ∏ (52)/4 = 19.64 mm2; a2 = ∏ (92)/4 = 63.63
For a1
k = 2.303 (19.64)(150) log10 h1
(7855)t h2
= 0.864 log10h1
t h2Copyright Dr. Hetty et. al. 2014
For a2
k = 2.303 (63.63)(150) log10 h1
(7855)t h2
= 2.798 log10h1
t h2
a h1 h2 t k =
1 1200 800 82 0.011176
800 400 149 0.006469
2 1200 900 177 0.016476
900 700 169 0.017191
700 400 368 0.008313
kpurata = 0.011925 mm/s
Copyright Dr. Hetty et. al. 2014
In Situ Permeability Test
Permeability determined in a laboratory may not be truly
indicative of the in situ permeability.
Thus, field tests are generally more reliable because the
test are performed on the undisturbed soil.
Other reason are :-
Soil stratification.
Overburden stress.
Location of the groundwater table.
Several field methods for evaluating permeability such as
pumping, borehole and tracer tests.
Copyright Dr. Hetty et. al. 2014
Pumping Test
Pumping tests involve the measurement of a pumped quantity from a well together with observations in other wells of the resulting drawdown of the groundwater level.
A steady state is achieved when at a constant pumping rate, the levels in the observation wells also remain constant.
The pumping rate and the levels in two or more observation wells are then noted.
The analysis of the results depends on whether the aquifer is confined or unconfined.
Copyright Dr. Hetty et. al. 2014
Confined Aquifer
Through an impermeable layer and then a
permeable layer (an aquifer) to another
impermeable layer.
If water is pumped from the well at a
constant discharge (q), flow will enter the
well only from the aquifer and the
piezometric surface will be drown toward
the well as shown on figure.
Copyright Dr. Hetty et. al. 2014
Confined Aquifer (Cont.)
An equilibrium condition will be reached at some time after pumping begins.
The piezometric surface can be located by auxiliary observation wells located at distances r1 and r2 from the pumping well.
The piezometric surface is located at distance h1 above the top of the aquifer at point r1 from the pumping well and at distance h2 at point r2.
Copyright Dr. Hetty et. al. 2014
Unconfined Aquifer
The piezometric surface lies within the
aquifer.
The analysis of this type of well is the
same as that for the confined aquifer.
Copyright Dr. Hetty et. al. 2014
Confined Aquifer
k = q ln (r2/r1)
2D(h2-h1)‘
Unconfined Aquifer
k = q ln (r2/r1)
(h22-h1
2)
Copyright Dr. Hetty et. al. 2014
Steady state pumping test
(unconfined aquifer)
Source: Whitlow (2001)Copyright Dr. Hetty et. al. 2014
Example 1
A pumping test was performed in a well penetrating a confined aquifer to evaluate the coefficient of permeability of the soil in the aquifer. When equilibrium flow was reached, the following data were obtained:
q = 750 l/min
water level (h1 & h2) = 5m & 6 m
Distance from the well (r1 & r2) = 20m & 60m
Thickness of aquifer = 6m
Copyright Dr. Hetty et. al. 2014
Solution :
k = q ln (r2/r1)
2D(h2-h1)
q = 750 l/min = 0.0125 m3/s
D = 6m; h1 = 5m; h2 = 6m; r1 = 20m; r2 = 60m
k = (0.0125m3/s) ln (60m/20m) = 0.0137
2(6m)(6m-5m) 37.7
= 0.00036m/s
Copyright Dr. Hetty et. al. 2014
Example 2
Some conditions as in Example 1, except that the well is
located in an unconfined aquifer.
Copyright Dr. Hetty et. al. 2014
Solution:
k = q ln (r2/r1)
(h22-h1
2)
q = 750 l/min = 0.0125 m3/s
h1 = 5m; h2 = 6m; r1 = 20m; r2 = 60m
k = (0.0125) ln (60/20) = 0.0137
(62-52) 34.54
= 0.00397 m/s
Copyright Dr. Hetty et. al. 2014
Horizontal and Vertical flow in stratified soils
Fig.6 a: Horizontal Flow b: Vertical flow
Source: Whitlow (2001)
Copyright Dr. Hetty et. al. 2014
Horizontal flow
Horizontal flow, occur when the water
flow are tangential to strata
The head lost between the entry and exit
faces will be the same for each layer:
# h1 = h2 = h3 = h
Hence the hydraulic gradients are the
same:
# i1 = i2 = i3 = i
Copyright Dr. Hetty et. al. 2014
Horizontol flow
Starting with Darcy law (q=Aki) the flows in the layer will be:
∆q1 = A1k1i1 ∆q2= A2k2i2 ∆q3 = A3k3i3
which Ā= B (D1 +D2+ D3) and total flow qH = ∆q1+ ∆q2+ ∆q3 = Ā kHi
kH = average horizontal coefficient of permeability
= D1k1+ D2k2+ D3k3
D1+D2+D3
Copyright Dr. Hetty et. al. 2014
Vertical flow Vertical flow is a flow which normal to strata. The rate
of flow will be the same through each layer.
#∆q1 = ∆q2= ∆q3 = ∆qv
The head lost in each layer will be h1,h2 and h3 giving
hydraulic gradients or
#i1=h1/D1 i2 = h2/D2 i3=h3/D3
Copyright Dr. Hetty et. al. 2014
Vertical flow
From Darcy law (q=Aki)
The total flow, qv = kvĀi but the total head lost ∑h =
h1+h2+h3 and L = D1+D2+D3
Therefore kv = D1+ D2+ D3
D1 + D2 +D3
k1 k2 k3
Copyright Dr. Hetty et. al. 2014
Question
A stratified soil consists of sand and silts.
The sand are generally 150mm thickness and
have a permeability k= 6.5 x 10 -1 mm/s, the
silts layer are 1.8m and the k is 2.5 x 10-4
mm/s. Assuming flow conditions are
isotropic determine horizontal and vertical
flow?
Copyright Dr. Hetty et. al. 2014
Capillary water in Vadose zone
Capillary water is held above the water table by surface tension which the attractive force exerted at the interface or surface between materials in different physical states, i.e. liquid/gas, solid/liquid
For example, a water/ air surface exhibits an apparently elastic molecular skin due to the sub-surface water molecules (which are more dense than air) exerting a greater attraction than air molecules. Similarly, water is attracted towards a solid interface because of the greater density and therefore attraction of the solid
Copyright Dr. Hetty et. al. 2014
Capillary water
Fig.7 Capillary water in capillary tube
Source: Whitlow (2001)
Copyright Dr. Hetty et. al. 2014
Capillary rise
Capillary rise, hc = 4T cos / w d, As an approximation for soils put T = 0.000074 kN/m w = 9.81 kN/m3 = 0 and d ed10 (where d10 = effective size)
Giving hc 4 x 0.000074 x 106 = 30
ed10 x 9.81 ed10
This estimate may improved to allow for the effect of grading and shape characteristics, such as irregularity and flakiness (Terzaghi and Peck)
hc = C
ed10
where C=a value between 10 and 40 mm2
Copyright Dr. Hetty et. al. 2014
Total stress, pore pressure and effective stress
Stress or intensity of loading is the load per unit area. The fundamental definition of stress is the ratio of the force ∆P acting on a plane.
Total stress () is the stress carried by the soil particles and the liquids and gases in the void.
= ∑ (h x )
When an external stress is applied to a soil mass that is saturated with pore water the immediate effect is an increase in the pore pressure. This produces a tendency for the pore water to flow away through adjoining voids, with the result that the pore decreases and the applied stress is transferred to the granular fabric of the soil. At a given time after application, therefore, the applied total stress will be balanced by two internal stress components.
Copyright Dr. Hetty et. al. 2014
Effective stress Effective stress (’): This is the stress transmitted through
the soil fabric via inter-granular contacts. It is this stress component that is effective in controlling both volume change deformation and the shear strength of the soil since both normal stress and shear stress is transmitted across grain to grain contacts. Terzaghi (1943) showed that, for a saturated soil, effective stress may be defined quantitatively as the difference between the total stress and the pore pressure:
’ = - u
As a conclusion effective stress () is the stress carried by the soil particles and pore pressure (u) is the pressure of water held in soil pores.
Note: ’ = sat -w where,
’ is submerge unit weight,
sat is saturated unit weight
w is water unit weight
Copyright Dr. Hetty et. al. 2014
Pore pressure
Pore pressure (u): This is the pressure induced in the fluid (either water, or vapor and water) filling the pores. Pore fluid is able to transmit normal stress, but not shear stress, and is therefore ineffective in providing shear resistance. For this reason, the pore pressure is sometimes referred to as neutral pressure.
Pore pressure (u) = wh
While pore pressure for capillary zone ur = -Sr wh/ 100
Copyright Dr. Hetty et. al. 2014
b = 18kN/m3 Sand
b =19.0 kN/m3 Silty Clay
sat = 20 kN/m3 Clay
A
B
C
D
Ground surface
1m
1m
2 m
Copyright Dr. Hetty et. al. 2014
d = 18kN/m3 Dry Sand
b =19.0 kN/m3 S= 50% Silty Clay
sat = 22 kN/m3 Clay
A
B
C
D
Ground Surface
1m
0.5m
2 m
Copyright Dr. Hetty et. al. 2014
Exam question
Soil profile shown in following figure. You are
required to determine and plot stresses
distribution at point A,B,C dan D (Total stress,
pore water pressure and effective stress)
A-B : 3 m , B-C: 2m C-D: 4m
Determine actual maximum capillary rise
according Terzaghi dan Peck if C is 40mm2 .
Copyright Dr. Hetty et. al. 2014
Gs= 2.7, e = 0.6 Dry sand d10 = 0.2mm
d = 14.03 kN/m3 Silty Clay w = 20% d10 =
0.05mm (capillary zone) e = 0.85 Gs = 2.65
sat = 22 kN/m3 Clay
w = 55%
A
B
C
D
Ground surface
Copyright Dr. Hetty et. al. 2014
Exam question
Determine stresses distribution σ, and σ’ on
each soil profile level if the GWL rise up to
ground surface. γb dan γsat of each soil layer as
following:-
A-B: 3 m γb = 16 kN/m3 dan γsat = 18kN/m3
B-C: 2m γb = 17kN/m3 dan γ sat = 19kN/m3
C-D: 4m γb = 18kN/m3 dan γsat = 20kN/m3
Copyright Dr. Hetty et. al. 2014
Flow net
1. Streamlines Y and Equip. lines are .
2. Streamlines Y are parallel to no flow
boundaries.
3. Grids are curvilinear squares, where
diagonals cross at right angles.
4. Each stream tube carries the same flow.
Copyright Dr. Hetty et. al. 2014
Flow Net in Isotropic Soil
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.
Copyright Dr. Hetty et. al. 2014
Method of Drawing
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.
Copyright Dr. Hetty et. al. 2014
Method of Drawing
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.
Copyright Dr. Hetty et. al. 2014
Method of Drawing
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.
6. Equivalent to solving the governing
equations of GW flow in 2-dimensions.
Copyright Dr. Hetty et. al. 2014