modelling multiphase flows
TRANSCRIPT
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CE/ENVE 320 Vadose Zone Hydrology/Soil Physics
Spring 2004 Copyright Markus Tuller and Dani Or 2002-2004
Water Flow in Saturated Soils
Darcys Law
P1
P2
Hillel, pp. 173 - 177
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Flow occurs from locations with high potential energy to locations of
lower potential energy in pursuit of equilibrium state.
The driving force for flow is called potential (energy) gradient, the
difference in potentials between two points in a system separated by
a certain distance.
Non Equilibrium and Flow
Potential Gradient i High potential energy
Low potential energy
1
2L1=1-2LLi
21 ipotential gradient.. potential energyL...distance betweenthe locations [L]
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In general, gradients can develop due to differences in:
- Pressure- Position in a gravity field- Chemical concentration- Temperature- Position in an electrical field
leading to spontaneous flow of mass or energy.
We will focus on flow due to differences in hydraulic potential in this
section (neglecting solute potential).
Hydraulic Potential
h =z + m + p
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Definition of Liquid Viscosity
Newtons Law of Viscosity
Early concepts in fluid dynamics are based on perfect fluids that
are assumed to be frictionless and incompressible. In a perfect fluid
contacting layers can exhibit no tangential forces (shearing
stresses) only normal forces (pressures).
Perfect fluids do not exist. In the flow of real fluids adjacent layers
do transmit tangential stresses (drag), and the existence of
intermolecular attraction causes fluid molecules in contact with
solid surfaces to adhere to it rather than to slip over it.
The flow of a real fluid is associated with the property ofviscosity.
Before we discuss flow in soils it is advantageous to introduce
some basic concepts related to flow in general.
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Liquid viscosity
The nature of viscosity can be visualized considering fluid motion
between two parallel plates; one at rest, the other one moving at
constant velocity.
Under laminar flow conditions water molecules are moving in
adjacent parallel layers. The layers transmit tangential stresses
(drag) due to attraction between fluid molecules.
Motion of fluid between parallel plates
The existence of intermolecular attraction causes fluid molecules
to adhere on the solid walls.
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Newtons law of viscosity
The velocity distribution in the liquid is linear.
Maintaining the relative motion of the plates at constant velocity
requires the application of a constant tangential force to overcomethe frictional resistance in the fluid.
This resistance per unit area of the plate is proportional to the
velocity of the upper plate and inversely proportional to the distance
between the plates. The shearing stress at any point is proportionalto the velocity gradient.The viscosity is the proportionality factor between and thevelocity gradient
d
d
A
F
shearing stress (force F acting on an area A) [M L-1 t-2]dv/dy velocity gradient perpendicular to the stressed area (shear rate) [t-1] viscosity coefficient of the liquid in [Pa s] [M L-1 t-1]
Viscosity is the property of the fluid to resist the rate of shearing
and can be visualized as an internal friction.
Newtons Law of Viscosity
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Fluid flow in cylindrical tubes
yyPLy
2
2 2
Fluid flows through a cylindrical tube having a diameter of 2R and length
L. We assume that the flow is laminar and caused by a pressure gradient
P=P2-P1.
2yPFp
yFf 2
Pressure Force:
Frictional Resistance Force:We equate the pressure
and frictional resistance
forces and solve for
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Flow through cylindrical tubes
Now we can introduce Newton's law of viscosity.
Substituting the integration constant back into our previous result yields
the expression for the velocity profile as a function of distance from the
tube axis
yLPdv
LPy
dydv
22
2
y
L2
P)y(vdyy
L2
Pdv
2
4
R
L
PC0C
2
R
L2
P22
2222
yR
L4
P
4
R
L
P
2
y
L2
P)y(v
The resulting ODE can be solved by integration.
Since we know that the velocity at y=R is equal to zero we can solve for
the integration constant
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Poiseuilles law for flow in cylindrical tubes
L4
RP
v
2
max
We know that the velocity is maximum at the center of the tube where
y=0, and can calculate vmax .
If we divide this expression by the tube cross section we receive the
average flow velocity as:
To calculate the Discharge Rate (volume of water flowing through the
tube per unit time) we have to integrate the velocity profile over the
cross-sectional tube area. This can be done very simple by calculating
the volume of a paraboloid of revolution.
This relationship is known as Poiseuilles law. It shows that the volume
of flow is proportional to the pressure trop per unit distance and to the
fourth power of the tube radius.
L8
RQ
L4
RP
2
R
t
VQ
422
L
P
8
Rv
2
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Example: laminar flow in tubes
What is the average (laminar) flow velocity of water at 20oC
through a 50m long tube having a diameter of d=0.1m under a
pressure difference of 100 Pa ?
Viscosity of water at 20 oC: = 0.001 Pa s
sm625.0v
sPa
m
50
100
001.08
05.005.0v
2
L
P
8
Rv
2
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Water Flow in Soils
Images of porous media pore space reveal that pores do not
resemble uniform and smooth circular tubes that form the basis for
Poiseuilles law.
Flow in porous media is generally described by macroscopic or
averaging terms that replace microscopic description of individual
flow pathways.
The first one able to quantitatively describe saturated flow through
porous media was HENRY DARCY a French engineer.
P1
P2
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Flow of Water in Saturated Soil (Darcys Law)
Historical Background
Henry Darcy, a French engineer, was commissioned by
the city of Dijon to find a solution for cleaning the city'swater supply that was contaminated by the waste of
mustard industry.
Darcy, in search of suitable filtering media, conducted
experiments with sand-packed filters.
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Water flux density (flux) JwVolume of water flowing through a unit
cross section per unit time.
Saturated hydraulic conductivity KsProportionality coefficient between water
flux density and hydraulic gradient.
Flow of Water in Saturated Soil (Darcys Law)
Historical Background
Darcy, in search of suitable filtering media, conducted
experiments with sand-packed filters
The pioneering work of Darcy published in 1856, provided thefundamental law for fluid flow in porous media.
Darcys Lawz
KtA
V
A
QJ hsw
JW water flux density [L/t]Q discharge rate [L3/t]V volume of water [L3]A cross-sectional area [L2]Ks saturated hydraulic conductivity [L/t]h/z hydraulic gradient [L/L]
Henry Darcy, a French engineer, was commissioned by the city
of Dijon to find a solution for cleaning the city's water supplythat was contaminated by the waste of mustard industry.
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Flow of Water in Saturated Soil (Darcys Law)
Darcy, in search of suitable filtering
media, conducted experiments with
sand-packed filters
The pioneering work of Darcy published
in 1856, provided the fundamental law
for fluid flow in porous media.
Darcys Law
zK
tA
V
A
QJ hsw
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Coordinates and conventions
For the application of Darcys law it is convenient to introduce a
sign convention for flux and heads when expressed in energy
per unit weight [L].
Upward flux is given a positive sign
The differences H and z, should be taken at the same order(if taken H=H1-H2 then z=z1-z2)
The negative sign in Darcys law ensures the algebraicconsistency of the equation.
+
-
+-
H1, z1 H2, z2
1 2
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Potentials and Heads
Potential Energy of Soil Water
As previously mentioned the potential energy of soil water can be
expressed in terms of chemical potential (energy/mass), soil waterpotential (energy/volume), or soil water head H (energy/ weight).
g acceleration of gravityw density of water
gw
For many hydrological applications it is advantageous to expresspotential as energy on weight basis (length).
This results in a simple notation for expressing heads as H=h+z
H the hydraulic headh pressure (positive) or matric (negative) headz gravitational head
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Darcys Law - Vertical Flow
ExampleA constant 20 mm of water is ponded on the
surface of a 50 mm long saturated vertical
sand column. What is the water flux from the
bottom of the column if the saturated
hydraulic conductivity is 50 mm/day?
(1) Define a convenient reference level
and designate it as z=0.
(2) Calculate the difference in hydraulic
head across the soil length
Solution
m70mm50mm20zhH ininin
m0mm0mm0zhH outoutout
m70mm0mm70HHH outin
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Darcys results
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Darcys Law - Vertical Flow
(3) Calculate the hydraulic gradient i:
.1mm50mm70
zHi
da/mm704.150iKJ Sw
with units of hydraulic head it is a
dimensionless quantity
Flux is downward
Note the energy loss in the soil!
(4) Calculate the flux.
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Darcys Law - Horizontal Flow
Example
The sand column from case a is now placed horizontally with 90 mm
of water ponded on the left side and 20 mm on the right side. Find:(1) the water flux density, and (2) the volume of water collected at the
outlet during 12 hr if the cross-sectional area of the column was
1000 mm2.
(1) Set the reference
level z=0 to coincide
with the axes of the
column
(2) Mark the columninlet by x=0.
Sign convention
Solution
+
-
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Darcys Law - Horizontal Flow
SolutionContinued
(3) Calculate the difference in hydraulic head across the soil length
m90mm0mm90zhH ininin m20mm0mm20zhH outoutout
m70mm20mm90HHH outin
(4) Calculate the hydraulic gradient i:
(5) Calculate the flux.
with units of hydraulic head it is adimensionless quantity.1mm50
mm70
z
Hi
h/mm917.2day/mm704.150iKJ sw
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Darcys Law - Horizontal Flow
SolutionContinued
(6) Calculate the cumulative volume of flow
ww
m35004917.2121000V
tAJVtA
VJ
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Saturated Flow - Potential Diagram
A constant water pressure of 20
kPa was maintained at the bottom
of a 0.5 m vertical saturated soilcolumn, and the water height at
the columns top was also kept
constant at 20 mm. Given the soils
saturated hydraulic conductivity
Ks = 5 mm/hr, find:
2 0 3 9 m m
2 0 m m
5 0 0 m m
(1) The direction of flow; draw a
system sketch and a potential
diagram
(2) The water flux density Jw
(3) The height of ponded water
on top of the column that causes
a cessation of flow.
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Saturated Flow - Potential Diagram
][kgsm
smkg
sm
kgm
N
s
m
m
kgPah 22
2
22
2
23
First we convert the pressure at the bottom of the column from
potential (kPa) to head (m):
[0.281.91000
200
gh
w
Then we assume the bottom of the column as reference level and
calculate the head at the top and the bottom as:
2 0 3 9 m m
2 0 m m
5 0 0 m m
[55020TOPzTOPpTOPh
[2020BOTzBOTpBOTh
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Saturated Flow - Potential Diagram
With known Ks we now calculate the flux density Jw as:
Positive Jw means flow from bottom to top.
2 0 3 9 m m
2 0 m m
5 0 0 m m
[1.10500
205205
zzK
BOTTOP
BOThTOPh
sw
The flow ceases when the hydraulic head at the top equals the
head at the bottom:
[1520TOPpBOThTOPh
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Saturated Flow - Potential Diagram
2 0 3 9 m m
2 0 m m
5 0 0 m m
50
52 203
52
z[
]
[m]
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CE/ENVE 320 Vadose Zone Hydrology/Soil Physics
Spring 2004 Copyright Markus Tuller and Dani Or 2002-2004
Saturated Hydraulic Conductivity and Flow
Through Layered Soils
Hillel, pp. 185-190 & 193-195
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Measurement of Saturated Hydraulic Conductivity
Constant Head Method
A constant pressure head (50 mm) is
maintained on the top of a saturated
soil column of known cross-sectional
area (1000 mm2) and length (50 mm).
The outflow on the bottom is collected
over a certain period of time (5 hr) and
the outflow volume is determined
(25000 mm3).
With known and determined quantities
we can calculate the saturated
hydraulic conductivity.
Saturated hydraulic conductivity is an important medium property
used in many model calculations for flow and transport in soils.
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Measurement of Saturated Hydraulic Conductivity
First we rearrange Darcys law to receive an explicit expression for
saturated hydraulic conductivity Ksat:
JKz
HKJ wssw
hmm551000
25000
tA
VJw
hmm5.2100
505Ks
Negative sign because of
downward flow
Note that the negative sign
of Darcys law ensures positive Ks(There is no physical meaning to a
negative hydraulic conductivity)
We calculate the flux density from our measurements and the column
dimensions:
We apply Darcys law to calculate Ksat:
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Laboratory Setup
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Falling Head Method
An alternative method for measurement of saturated hydraulic
conductivity does not require the maintenance of a of a constant
head nor any outflow measurement is called Falling Head Method.
Only initial and final depths of water
expressed as pressure head in length units
need to be recorded as a function of time.
The rate of decrease of depth of ponding isequal to the flux density.
Darcys Law
Measurement of Saturated Hydraulic Conductivity
t
d
A
a
tA
VJw
z
KJ sw
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We can equate Darcys law with the expression for flux density
derived from the rate of decrease of ponding and integrate the
resulting expression to derive a relationship for Ksat:
Measurement of Saturated Hydraulic Conductivity
2t
01t
s
2h
1h
s
L
KdhLh
1aL)t(h
L
K
dt
dh
A
a
2
1
2s2
s
2
1
H
HlnA
a
t
LKt
L
K
Lh
LhlnA
a
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Typical Values of Ks in Soils
Saturated hydraulic conductivity Ks
Textural class UNSODA Database[cm/d]
NRCS Soil Survey Database[cm/d]
Sand 506 713
Loamy sand 227 350
Sandy loam 42 106
Loam 39 25
Silt 56 6
Silt loam 31 11
Sandy clay loam 10 31
Clay loam 2 6
Silty clay loam 7 2
Silty Clay 8 0.5
Clay 26 5
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Limitations of Darcys Law
Reynolds Number
vdRe
d effective pore diameterv mean flow velocity liquid density liquid viscosity
inertial forcesviscous forces
At high flow velocities inertial forces are no
longer negligible TURBULENT FLOW
In very fine textured media (clays)
adsorptive surface forces affect flow. The
flux density at low gradients is smaller than
predicted according to Darcys law
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Saturated Steady Flow Through Layered Soil
Under steady-state flow conditions the
flux through both layers is equal.
We solve for H2 and obtain two
equations:
Solving for flux density and introducing a effective saturated
hydraulic conductivity yields:
2
2
2s
1
21
1sw
HK
L
HHKJ
3
2s
2
w2
1s
1w12
HK
LJH
K
LJHH
1s
1
2s
2
31w
21
13effs
K
L
K
L
HHJ
LL
HHK
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Saturated Steady Flow Through Layered Soil
This solution can be generalized to a soil
profile having multiple layers.
The effective hydraulic conductivity for asoil profile consisting of n layers, each
with distinct hydraulic conductivity Ks
and thickness L is obtained by setting
Jw=Ks-eff(Hn-H1)/LiThis solution is valid for flow perpendicular to the layering (harmonicmean).
For flow parallel to the layering we use an arithmetic mean weighed by
layer thickness
n
1i si
i
n
1i
i
)N(effs
K
L
L
K
n
1i
i
n
1i
ii
)P(effs
L
L
K
Ks-eff (N)
Ks-eff (P)
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A 1 m long glass tube having a radius of 1
mm was inserted into a 1 m long saturated
cylindrical soil column with a diameter of100 mm, and Ks of 0.01 mm/min. The water
head at the top of the column and at the
tubes inlet was 0.25 m while the outlets
where at atmospheric pressure.
Effective Saturated Hydraulic Conductivity - Example
2 5
1 0 0 0
250
1000
(1) What would be the total flux through
the column tube system and what
percentage is contributed by the tube?
(2) What is the effective Ks of the column-
tube system
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To apply Poiseuilles law for average flow velocity within the
tube we first have to convert hydraulic head h(m) to hydraulic
potential h (Pa) using the following relationship:
Effective Saturated Hydraulic Conductivity - Example
2 5
1 0 0 0
250
1000
gwh
[125.181.91000h
Note that the hydraulic head is the sum of
water head and tube length.
With known hydraulic potential we now can
solve Poiseuilles law:
LP
8rv 2
[91]sm[533.1
1
1226
001.08
001.0v2
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The tubes flux Qtube is simply the product of average velocity
and the tubes cross-sectional area:
Effective Saturated Hydraulic Conductivity - Example
2 5
1 0 0 0
250
1000
The flux of the soil column QC is given as:
m[2819162Q 32tube
wc JQ
zh(KJ sw
m[01.01000
125001.0Jw
m[98)150(0125.0Q 322C
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Then we calculate the total flux as the sum of column and tube flux:
Effective Saturated Hydraulic Conductivity - Example
The contribution of the tube to the total flux is calculated as:
m[289828783Q 3T
[.910287
287100Q
Q[%]T
tube
The systems (tube & column) flux density JTis simply the ratio of
the total flux QT and the total cross-sectional area AT. The
systems effective saturated conductivity is calculated as:
z(JK TeffS
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Total flux density
Effective Saturated Hydraulic Conductivity - Example
Effective saturated conductivity
m[7.36502879J 2T
m[3.29125
1007.36KeffS