MODULE 2

SOIL WATER AND WATER FLOW

 SYLLABUS

Soil water - Various forms – Capillary rise - Effective stress concepts in

soil - Total, neutral and effective stress distribution in soil -
permeability - Darcy's Law-Permeability measurement in the laboratory
- quick sand condition - Seepage - Laplace Equation - Introduction to
flow nets -properties and uses - Simple problems.
 SOIL WATER
• Natural soil deposits consists of water of various forms. This moisture
or water present in the soil is not stationary and is tends to change
based on the climate and the environmental conditions. This change
in the moisture content of the soil will alter properties as well as the
behaviour of the soil.
• Water present in the void spaces of the Soil mass is called as the “Soil
Water”.
• Soil water will be in the form of free water or Gravitational water and
Held water.
 Gravitational Water
• Gravitational water’ is the water in excess of the moisture that can be
retained by the soil. It translocates as a liquid and can be drained by
the gravitational force. Gravitational water can be subdivided into
(a) free water (bulk water) and (b) Capillary water
• Free Water (Bulk Water) : It has the usual properties of liquid water. It
moves at all times under the influence of gravity, or because of a
difference in hydrostatic pressure head. It can be further classifies in
to free surface water and Ground Water

Free Water: Free surface water may be from precipitation, run-off,

floodwater, melting snow, water from certain hydraulic operations.
 Gravitational Water (Cont)
Ground water: It is that water which fills up the voids in the soil up to

the ground water table and translocates through them.

Capillary water: Water which is in a suspended condition, held by the

forces of surface tension within the interstices and pores of capillary
size in the soil, is called ‘capillary water’.
 HELD WATER
• ‘Held water’ is that water which is held in soil pores or void spaces
because of certain forces of attraction. It can be further classified as
(a) Structural water and (b) Absorbed water. Sometimes, even
‘capillary water’ may be said to belong to this category of held water
since the action of capillary forces will be required to come into play
in this case.
• Structural water. Water that is chemically combined as a part of the
crystal structure of the mineral of the soil grains is called ‘Structural
water’.
• Adsorbed Water is the water in which the soil particles freely absorb
moisture from the atmosphere, by the physical forces of attraction,
and is held by the force of adhesion.
CAPILLARY RISE
 Capillary Rise (Cont)
• The pores of the soil mass will be considered as the series of capillary
tubes found above the water table.

• The rise of water in the capillary tube is because of the action of Surface
Tension, which pulls the water up against gravity force.

• The rise of the capillary water above ground water depends of the
diameter of the capillary and the value of the surface tension.

• Figure shows the view of the capillary tube inserted in the water and the
capillary rise phenomena occurs.
 Capillary Rise (Derivation)
• Let Ts cos α be the vertical component that represents the surface
tension force and it depends on the angle of incidence α between the
meniscus and the tube.
• Let “d” be the inner diameter of the tube, and hc be the height of the
capillary rise.
• When the capillary tube is inserted in the water, the rise of the water
will takes place. When the equilibrium is reached, water will stop
moving further. At this equilibrium position, when the height of
capillary rise hc will be equal to the weight of column of water
× hc × γw --------------------------------- EQ A
 Capillary Rise (Derivation)
• The vertical component of the reaction of meniscus against the inside circumference of the
tube, supporting the above weight of the water column is equal to π d Ts cos α -------- EQ B
• Equating two quantities at the equilibrium ( equating A & B)
× hc × γw = π d Ts cos α
• For a perfectly cleaned and wet tube the value of α will be considered as zero , hence the
maximum capillary rise equation is given by
(hc)max = ----------------------- EQ C
For water at 4. c, Ts = 75.6 dynes / cm = 75.6 × 10 -8 kN / cm and γw= 9.81 KN / m3 . If d is
expressed in cm the above equation reduces to

(hc)max = cm.

At 20 . C (h c)max = cm.
Thus its been observed that the capillary rise will decrease with increase in temperature.
Capillary Tension, Capillary potential & capillary
suction

Relationship between R
and d
Capillary Tension, Capillary potential
& capillary suction
• The magnitude of the stress “u” about the radius “R” of the meniscus,
The relationship between diameter d and the Radius R is given by
= R cos α or d = 2R cos α , sub the value of d in EQ C
We get (hc)max = =
Since uc = hc ×
uc is the maximum tension at the level of the meniscus the above
mentioned equation can be rewritten as
uc = =
 Capillary Tension, Capillary
potential & capillary suction
• The tensile stress caused in water is called as the Capillary tension or the
capillary potential.

• Thus the capillary tension or the capillary potential creates a pressure

deficiency or the negative pressure in the soil mass. This generated pressure
deficiency in the soil mass is termed as the soil suction or the suction
pressure.
 Capillary rise (Worked Examples)
• Compute the maximum capillary tension for a tube of 0.05 mm in
diameter.
Solution:
Capillary tension uc = hc × (= 9.81 KN / m3)

The maximum capillary height hc = = = 61.7 cm = 0.617 m

Capillary Tension uc = 0.617 × 9.81 = 6.05 KN / m2

 Capillary rise (Worked Examples)
• Compute the height of capillary rise in a soil whose D10 is 0.1mm, and
void ratio is 0.6.
Solution
Let the average size of the void be d(mm). Volume of each sphere of
the solids may be assumed proportional to D103 . Since the void ratio is
0.6, the volume od void space corresponding to unit volume of solids,
will be proportional to 0.60 D103.
Hence d3= 0.6 D103
d = 0.6(1/3) D10 = 0.845 × 0.1 = 0.0845 mm = 0.00845 cm
hc = cm = = 36.5 cm
 STRESS CONCEPTS IN SOILS
• At any plane in a soil mass, the total pressure or the unit pressure (σ) is the
total load per unit area. This pressure is may be due to (i) Self weight of the
soil (ii) Over burden on the soil.
• The total stress of a soil consist of two distinct components namely Effective
stress and the neutral stress.
• Effective stress ( σ ’ ) is the pressure transmitted from the particle through
their point of contact with the soil mass.
• Neutral stress (u) or the pore water pressure or the pore pressure id defined
as the pressure transmitted through the pore fluid.
• The total stress at any point is defined as the sum of effective stress and
Neutral stress. [ σ = σ ‘ + u ] and [ σ ‘ = σ - u ]
• At any plane pore water pressure or neutral pressure is equal to the height
height of the Water table to the unit weight of water [ u = hw × γw ]
 Total Stress and
Effective stress
under various
conditions

Case 1 : Submerged soil

mass
• Submerged soil mass
The above figure shows the saturated soil mass up to the depth Z. Submerged
water up to depth Z1. Water level is raised from Level AA to Level CC.
Total pressure at Level AA is given by
σ = γsat Z + Z1 γw
u = γw hw = γw ( Z + Z1).
σ ’
= σ - u = γsat Z + Z1 γw - γw ( Z + Z1) = Z γsat + Z1 γw - Z γw - γw Z1
= Z (γsat - γw ) = Z γ’
Total pressure at Level BB is given by
σ = Z1 γw
u = γw hw = γw Z1

σ ’
=σ - u=0
Total Stress and
Effective stress
under various
conditions

Case 2 : Soil mass with

surcharge
• At level AA , the total stress
σ = q + Z γsat + Z1 γt
u = γw hw = Z γw
σ ’
= σ - u = q + Z γsat + Z1 γt - Z γw
= q + Z1 γt + Z γ’

• At Level BB, The total stress

σ = q + Z1 γt
u =0
σ ’
= σ = q + Z1 γt

• At Level CC, The total stress = effective stress = q

Total Stress and
Effective stress
under various
conditions

Case 3 : Saturated soil with

capillary rise
• In this case the soil mass is saturated of about height Z and soil mass is also saturated with the height Z 1,
but this saturation is of by the capillary water.
• At level AA , the total stress
σ = Z γsat + Z1 γsat = (Z+Z1) γsat
u = γw hw = Z γw
σ ’ = σ - u = (Z+Z1) γsat - Z γw = Z γ’ + Z1 γsat
At Level BB, The total stress
σ = Z1 γsat
u = γw hw = - Z1 γw
σ ’
= σ - u = Z1 γsat - ( - Z1 γw )
= Z1 (γsat + γw )
Stress concepts on soils (Problem 1)
• The water table in a certain area is at a depth of 4m, below the
ground surface, to a depth of 12 m, the soil consists of very fine sand
having an average void ratio of 0.7. above the water table the sand
has an average degree of saturation of 50 % . Calculate the effective
stress at 4m, 10 m, 12m from the ground surface. Assume G = 2.65
GL 0m
γt
W.T 4m 12 m

γ sat 8m
• To find γt we must use the empherical relation ships given data are e=0.7, S = 50 %,
G = 2.65
Relationship between Bulk Unit Weight t , In terms of G, e, w and
=[ ] = 9.81 = 17.31 KN / m3

Relationship between Saturated Unit Weight , In terms of G, e, w and

=[ ] = [ ] 9.81 = 19.33 KN / m3.

Effective stress @ 12 m
σ = 8 × 19.33 + 4 × 17.31 = 154.65 + 69.24 = 224 KN / m 2
u = 8 × 9.81 = 78.48 KN / m2
σ’ = σ – u = 224 – 78.48 = 145.52 KN / m 2
Effective stress @ 10 m
σ = 6 × 19.33 + 4 × 17.31 = 115.98 + 69.24 = 185.22 KN / m 2
u = 6 × 9.81 = 58.86 KN / m2
σ’ = σ – u = 185.22 – 58.86 = 126.36 KN / m 2
Effective stress @ 4 m
σ = 4 × 17.31 = 69.24 KN / m 2
u=0
σ’ = σ – u = 69.24 KN / m2
 Stress distribution diagram
• 0m G.L σ(KN/m2) σ’ (KN/m2) u (KN/m2)
• 4m W.T 69.24 69.24

• 10m 185.22 126.36 58.86

• 12m 224 145.42 78.48

Stress concepts on soils (Problem 2)
• The water table in a deposit of sand layer of 8m was located at a
depth of 3m below the surface. And the sand is saturated with the
capillary water above the water table. Calculate the effective stress at
1m, 3m and 8 m respectively from the surface. The saturated unit
weight of the soil is 19.62 KN / m3. Draw the stress distribution
diagrams.
Solution
Effective stress @ 8 m
σ = 5 × 19.62 + 3 × 19.62 = 156.96 KN / m2
u = 5 × 9.81 = 49.05 KN / m2
σ’ = σ – u = 156.96 – 49.05 = 107.91 KN / m2
Effective stress @ 3 m
σ = 3 × 19.62 = 58.86 KN / m2
u=0
σ’ = σ – u = 58.86 KN / m2
Effective stress @ 1 m
σ = 1 × 19.62 = 19.62 KN / m2
u = - hc × γw = - 2 × 9.81 =- 19.62 KN / m2
σ’ = σ – u = 19.62 – (- 19.62) = 39.24 KN / m2
 PERMEABILITY

• Permeability is defined as the property of porous material, which

permits the passage of water or seepage of water through its
interconnecting voids. A material having continuous voids is called
permeable. Gravels are highly permeable whereas the stiff clay is the
least permeable.
 Darcy’s Law
• The law of motion of water through the soil was studied by Darcy
(1856).
• The law states that in a laminar flow conditions in a saturated soil, the
rate of flow of discharge per unit time is proportional to its hydraulic
gradient (K).
v = ki
q = kiA
Where k = coefficient of permeability
i = Hydraulic gradient
 Discharge velocity & Seepage
velocity
• The velocity of flow “v” is the rate of discharge of water per unit time to the
total cross sectional area (A) of the soil.

• This total area of cross section composed of area of solids A s and area of voids

Av .

• The velocity of flow of water in the voids is called seepage velocity (v s) and it is
defined as the rate of discharge of percolating water per unit cross - sectional
area of voids perpendicular to the direction of flow.
From the definition of the seepage velocity and the discharge velocity we have
q = Av = Avvs
vs = v since = = n
vs =
The seepage velocity, vs is also proportional to the hydraulic gradient
vs = kp i (Where kp = coefficient of percolation)
From darcy’s law v = ki
Kpi = ki / n
Kp =
Determination of Co-Efficient of
permeability
The laboratory method of determination of coefficient of permeability
are

• Constant head permeability test

• Falling head permeability test
CONSTANT HEAD TEST
 CONSTANT HEAD TEST
• Water flows from the over head tank consisting of three tubes namely the
inlet tube, the over flow tube and the outlet tube.
• In this type of laboratory setup, the water supply at the inlet is adjusted in
such a way that the difference of head between the inlet and the outlet
remains constant during the test period. After a constant flow rate is
established, water is collected in a graduated flask for a known duration
• In this test the hydraulic gradient I is calculated by the ratio between the
head (h) to the length(L)
• If Q is the total quantity of flow in a time interval t, from darcy’s equation
we can have
q= =kiA
k = = Where A = total C/S area of the sample.
FALLING HEAD TEST
 FALLING HEAD TEST
• Relatively for less permeable soils

• Water flows through the sample from a standpipe attached to the top of the cylinder.

• The head of water (h) changes with time as flow occurs through the soil. At different
times the head of water is recorded.

• Water from a standpipe flows through the soil , the initial head difference h1 at time
t=0 is recorded and water is allowed to flow through the soil specimen such that the
final head difference at time t = t2 is h2.

• Let h be the head at any intermediate time interval t and –dh be the change in head
with respect to time interval dt. ( - sign indicates h decreases with time increases)
• Hence from Darcy’s law
q=- a=KiA
- a = K A or dt = -
Integrating between time limits we get

= - =

(t2 –t1) = loge (Considering t2 – t1 = t, we get)

K = loge = 2.3 log10

 Quick Sand Condition

Quick sand condition is a condition of flow, in which seepage flow tends

to move towards upward direction. Seepage flow causes a floating
condition of a particle in cohesion less soil, such as sand and fine gravel.
Laplace Equation of two
dimensional flow
The quantity of flowing of water in a saturated soil mass, as well as distribution of
pressure can be estimated through the theory of flow of fluids in porous medium.
Assumptions made in Laplace two dimensional Equation
• The saturated porous medium is incompressible. The size of the pore spaces
doesn’t change with time, regardless of water pressure.
• The seepage water flows under the hydraulic gradient which is due only to gravity
head loss.
• Darcy’s law is valid
• There is no change in the degree of saturation.
• The quantity of water flows in will be equal to the quantity of the water flows
out.
• The hydraulic boundary conditions at entry and the exit are known.
• Water is incompressible.
• Consider the element of soil of size Δx, Δy and of unit thickness perpendicular to
the plane. Let vx and vy be the entry velocity of the components in x and y

directions. Then [ vx + Δx] and [ vy + Δy] will be the corresponding velocity

components at the exit of the element.

• As per the assumption 3, the quantity of water entering the element will be equal
to the quantity of water leaving it.

vx (Δy ) + vy (Δx) = [vx + Δx] (Δy) + [ vy + Δy] (Δx)

From which , + =0 (This is continuity Equation)

• According to assumption 2, vx = kx * ix = kx and vy = ky * iy = ky z

• Where h = hydraulic head under which water flows

kx and ky = coefficient of permeability in x and y directions.

Substituting the above equation in the continuity equation we get

+=0

For an isotropic soil kx = ky = k (say) + = 0

Substituting
Φ = kh = velocity potential, we get + = 0
• Calculate the coefficient of permeability of a soil sample 6 cm in height, 50 cm2 in
cross sectional area, if a quantity of water equal to 450 ml passed down in 10
minutes under an effective constant head of 40 cm. on oven drying the test
specimen weights 495 g. taking the specific gravity of solids as 2.65. calculate the
seepage velocity of water during the test.
K = = = 2.25 10-3 cm / sec = 1.944 m / day.
v = = = 1.5 10-2 cm / sec.

Determination of seepage velocity

Vs = , Dry Unit weight ( = = = 1.65 g / cm3

e = - 1 = - 1 = 0.606.
n = e / 1+e = 0.606/1+0.606 = 0.377.
Vs = = = 3.975 10-2 cm / sec
Falling head permeability test (Problem)

In a falling head permeameter test on a silt clay sample, the following results were
obtained. Sample length 12cm. Sample diameter 80 mm, Initial head 120 cm and
final head 40 cm, time for fall in head 6 minutes, stand pipe 4 mm, Find the
coefficient of permeability.
Given L = 12 cm
A= × 82 = 50.265 cm2 , a = × 0.42 = 0.1257 cm2
t = 6 min , 360 sec
h1= 120 cm , h2= 40 cm
K = 2.3 log10 = 2.303 log10 = 9.159 × 10-5 cm / sec
 FLOW NETS
The Laplace equation gives a two
sets of curves known as the
“equipotential lines” and the
“stream lines or flow lines”.
Mutually orthogonal to each
other. The Equipotential lines
represent the contours of equal
head. The direction of seepage is
always perpendicular to the
equipotential lines. The path
along which the individual
particles of water seep through
the soil are called as the stream
lines or the flow lines.
 Properties of Flow Net
• The flow lines and the equipotential lines meet at right angles to each
other.
• The fields are approximately square, so that circles can be drawn
touching all the four sides of the square.
• The quantity of water flowing each flow channel is the same. Similarly
the same potential drop occurs between two successive equipotential
lines.
• Smaller the dimensions on the field, greater will be the hydraulic
gradient and velocity of low through it.
• In a homogenous soil, every transition in the shape of the curves is
smooth, elliptical or parabolic in shape.
 Applications of Flow Net
Flow net can be utilised for the following purposes.

• Determination of Seepage
• Determination of Hydrostatic pressure
• Determination of seepage pressure
• Determination of exit gradient
 Determination of Seepage
• Figure shows the portion of flow net. The portion between any two
successive flow lines is known as Flow Channel. The portion enclosed
between two successive equipotential lines and successive flow lines is
known as FIELD.
Let b and l be the width and length of the field.
∆h= head drop through the field
∆q=Discharge passing through the flow channel
H= Difference between upstream and downstream fields
From Darcy law of flow through the soils.
∆q = k . ( b × l ) (Considering unit thickness)
If Nd = Total number of potential drops in complete flow net, then ∆h =
Hence ∆q = k.
 Determination of Hydrostatic pressure
• The hydrostatic pressure at any point within the soil mass is given by

U = hwγw where u = hydrostatic pressure, hw = piezometric head

The hydrostatic pressure in terms of peizometric head hw is calculated
by the following relation.

hw= h – z, where h = hydraulic potential at the point under

consideration, z = position head of the point above datum, considered
positive upwards.
• Determination of Seepage pressure: The hydraulic potential “h” at
any point located after n potential drops, each of value ∆h is given by
h = H - n ∆h

• Determination of Exit gradient: The exit gradient is the hydraulic

gradient at the downstream end of the flow line where the
percolating water leaves the soil mass and emerges in to the free
water at the downstream. The exit gradient can be calculated from
the following expression, in which ∆h represents the potential drop
and l the average length of the last field in the flow net, at exit end.
Ie=