modelling multiphase flows

7/29/2019 Modelling multiphase flows

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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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Copyright Markus Tuller and Dani Or2002-2004

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


7/41Copyright Markus Tuller and Dani Or2002-2004

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



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



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



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



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



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.



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.


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.




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



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



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



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



Darcys results



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.



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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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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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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][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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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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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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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


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:


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:


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:


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:


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 conductivity

m[7.36502879J 2T

m[3.29125

1007.36KeffS

modelling multiphase flows

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