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William Dhaeseleer- 2011Fluidummechanic 1

Chapter 5Introduction to

Differential Analysis of

Fluid Motion

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Main Topics

Conservation of Mass

Stream Function for Two-Dimensional

Incompressible Flow

Motion of a Fluid Particle (Kinematics)

Momentum Equation

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Main Topics



Incompressible Flow


Momentum Equation

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Basic Law for a Control Volume

Rate or change of mass Net rate of mass flux out

inside the control volume through the control surface0

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Basic Law for a Control Volume

(1)

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Infinitesimal Control Volume;Rectangular Coordinate System

At center O,

density andvelocity

Use Taylor series expansion

and neglect higher order

terms.

V u i v j wk

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Infinitesimal Control Volume;Evaluate density and velocity at faces of cube

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Infinitesimal Control Volume;Evaluate density and velocity at faces of cube

Similarly for y and z faces;

Velocity components assumed to be in

positive coordinate direction;

The area normal is positive on each

face.

Must then evaluate in- andoutflow through 6 faces of

mass flux density

.V V dA

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Infinitesimal Control Volume;Examples of fluxes

( )

2 2

dx u dxLeft x u dydz

x x

1

2

uu dydz u dxdydz

x x

( ) 2 2

dx u dx

Right x u dydzx x

1

2

uu dydz u dxdydz

x x

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Infinitesimal Control Volume;Leads to, for sum of all 6 faces:

u v wu v w dxdydz

x x y y z z

u v wdxdydz

x y z

.

small CV

V dA

(1)

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Recall, Basic Law for a Control Volume

(2)

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Rate of change of massin small Control Volume

(2)

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Rectangular Coordinate System

Continuity Equation

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Rectangular Coordinate System

Del Operator or Nabla

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Vector form (all coordinate systems)

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Rectangular Coordinate System &Vector Form

Incompressible Fluid:

Steady Flow:

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Cylindrical system

r r z z r zV e V e V e V rV V zV 1

r z

r r z

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Cylindrical system; Note on del operator1

V V VV r z with V V

r r z

1 1 zr V VV rVr r r z

rand r

Must distinguish between grad, div and dyadic product

Recall that:

1

... ...r r zVV V V

V rr r zzr r r z

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Result for Cylindrical Coordinate System

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Cylindrical Coordinate System

Incompressible Fluid:

Steady Flow:

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Main Topics



Incompressible Flow


Momentum Equation

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Stream Function for

Two-Dimensional

Incompressible Flow

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Stream Function forTwo-Dimensional Incompressible Flow

Considerincompressible flow in 2D

Concept ofStream Function

Allows to represent the two velocity components

u(x,y,t) and v(x,y,t) by a single function

Stream function defined by

( , , )x y t

u and vy x

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Consider the continuity equation forincompressible 2D flow

Substituting a clever transformation

GivesThis is true for any smooth

function (x,y)

. 0V 0u v

x y

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Why do this? Mathematically

Single variable replaces (u,v). Once is

known, (u,v) can be computed

Physical significance

Curves of constant are streamlines of the flow

Difference in between streamlines is equal tovolume flow rate between streamlines

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Recall from earlier thatalong a streamline

Change in along

streamline is zero

Physical Significance

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Difference in between streamlines is equal

to volume flow rate between streamlines

Note on notation: Recall


. .A A

m V dA and Q V dA

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Consider cut AB

Along AB, x=const, or

Therefore


2 2

1 1

y y

y yA

Q V dA udy dy

y

d dyy

2

12 1Q d

No flow across streamline

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Likewise consider cut BC

Along BC, y=const, or

Again:


1

22 1

Q d

2 2

1 1

x x

x xA

Q V dA vdx dxx

d dxx

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Since flow rate is constant between two streamlines

Velocity will be high when streamlines are close together

and conversely


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Cylindrical Coordinates

Stream Function (r, )

0V or

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Stream Function for SteadyTwo-Dimensional Compressible Flow

For 2D steady compressible flow:

And then:

. 0V so that u and vy x

2 1m

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Main Topics



Incompressible Flow


Momentum Equation

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Motion of a Fluid Particle(Kinematics)

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- Consider finite fluid element, with infinitesimally-small element dmand volume dxdydzat time t

-After time duration dtfinite element has movedand changed shape (severe distortion possible)

- But changes of elementary fluid particle limited tostretching & shrinking and rotation of sides

Because of infinitesimal

particle and time dt, thesides remain straight!

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Elementary particles motion can be decomposed

into fourfundamental components:

translation

rotationlinear deformation

angular deformation

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Important distinction between rotation andangular deformation!

Pure rotation does not involve deformation;

angular deformation does. And fluid deformationgenerates shear stresses and thus viscosity.

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1) Fluid Translation:

Acceleration of a Fluid Particle in a Velocity Field

2) Fluid Rotation

3) Fluid Deformation Angular Deformation Linear Deformation

Organisation of Section:

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2) Fluid Rotation

3) Fluid Deformation Angular Deformation Linear Deformation


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1) Fluid Translation:Acceleration of a Fluid Particlein a Velocity Field

( , , ; )V V x y z t

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Following Newtons law, one might think:

( , , ; )V V x y z t

Va

t

Since is a field

it describes motion of

whole flow and not just ofparticle

V

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Need to keep the field description for fluid propertiesand obtain expression foracceleration of fluidparticle as it moves in a flow field.

Given flow field:

To find acceleration of fluid particle:

( , , ; )V V x y z t

pa

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At time t, particle is at x,y,zand has velocitycorresponding to the velocity of the fieldat that point: ( , , ; )p

tV V x y z t

At timet+dtparticle hasmoved to new position, where

a different velocity applies:

( , , ; )pt dt

V V x dx y dy z dz t dt

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At time t+dt, particle is at x+dx,y+dy,z+dzand hasvelocity

p pV dV

p p p p

V V V V dV dx dy dz dt

x y z t

p p p p

p

dV dx dy dzV V V V dt a

dt x dt y dt z dt t dt

p p pdx dy dzu v w

dt dt dt But

p

p

dV V V V Va u v w

dt x y z t

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To stress the acceleration in a velocity field,special notation:

DV

Dt

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Substantial Derivative or Material Derivative

Physical Interpretation:

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Substantial Derivative or Material Derivative

Physical Interpretation:

Measure fish concentration in a river; since fish aremoving:C=C(x,y,z;t)

- at fixed place on a bridge

- moving in a motor boat with velocity

- moving along in a floating boat, with velocity

C

t

U

VDC

Dt

dCdt

M i f Fl id P i l (Ki i )

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In vector notation:

. .

DC C DV C V

Dt t Dt t

. .dC C d

U C Udt t dt t

M ti f Fl id P ti l (Ki ti )

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Total fluid particle accelerationin vector notation:

.DV V

V VDt t

Applies anywhere in a velocity field (function

of x,y,z and t)Eulerian description method


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Even in steady (non time-dependent) flow,

Particle may undergo convective acceleration,even in a steady velocity field,because of its motion to locations where adifferent velocity applies.

.

p

DVa V V

Dt

M i f Fl id P i l (Ki i )

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Fluid Translation: Particle AccelerationIn Cartesian Coordinates


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Fluid Translation: Particle Acceleration

In Cylindrical Coordinates

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Recall the four components of motion:


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2) Fluid Rotation

3) Fluid Deformation Angular Deformation

Linear Deformation



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2) Fluid Rotation:

Rotation of a Fluid Particle characterized by

x y zi j k

is rotation about xaxis etcx


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Fluid Rotation about z-axis (in x-y plane):

(a): fluid particle at time t sides oa & ob; lengthsx &y

(b): fluid particle at time t+t rotated and deformed


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Fluid Rotation:

(b)= result of (c) + (d)

(c): pure rotation

(d): pure angular deformation


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Fluid Rotation:

Define rotationas average of angular motionsof edges oa& ob

Leads to: Note that is in negative

direction 1

2


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Fluid Rotation:

For small angles:

Also in timet:Pt o moves utand vt;Pts a & b move a bit further:

see figure below.

x and y

u uu y t u t y t

y y

v vv x t v t x t

x x

Figure on left taken from

Munson et al; with symbols Aand B, and instead of.


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Fluid Rotation:

Angular velocity about z axis defined becomes:

0 0 0

1 112 22lim lim lim

zt t t

v ut t

x y x y

t t t

1

2z

v u

x y


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Fluid Rotation:

Similarly for rotation about x & y axes; such that:

1

2z

v u

x y

1

2x

w v

y z

1

2y

u w

z x

12 V


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Fluid Rotation:

Similarly for rotation about x & y axes; such that:

1

2z

v u

x y

1

2x

w v

y z

1

2y

u w

z x

With

sometimes written as:

1 2 3( , , ) ( , , ) ( )iV u v w V V V V V V

1, ,

2

jki

j k

VVi j k cyclic

x x


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Fluid Rotation:

Define vorticity

Define circulation

ds elemental element of arc along curve C

2 V

C

V ds

:C A A

Stokes V ds V dA dA with A area within C


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Fluid RotationIn Cartesian coordinates:

In Cylindrical coordinates:

1 1 1 1 2

z r z rr

V rVV V V Ve e kr z z r r r r


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Fluid Rotation: Physical Interpretation

For rotational flow

For irrotational flow

Rotational flow differs from motion with circular field

lines (vortex flow);

fluid particles can rotate in circular motion, but do

not have to!

0

0


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Fluid Rotation: Physical InterpretationConsider two circular streamline flows: (Ex 5.6)

- solid rotation (rigid body motion; forced vortex):

- irrotational motion (free vortex or line vortex)

0r

V

z const or V r r const

0 constor rV const V r


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z const or V r r const 0constor rV const V

r

:notation u V


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z const or V r r const 0constor rV const V

r

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2) Fluid Rotation


Linear Deformation



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2) Fluid Rotation


Linear Deformation



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Recall:

combined rotation and angular deformation

(a): fluid particle at time t sides oa & ob; lengthsx &y

(b): fluid particle at time t+t rotated and deformed


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Fluid Rotation:

(b)= result of (c) + (d)

(c): pure rotation

(d): pure angular deformation


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Fluid Rotation:

Define rotationas average of angular motionsof edges oa& ob


direction

12


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Fluid Angular Deformation:

Define angular deformationas complement torotation to obtain picture (b)


direction

1

2


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Fluid Angular Deformation: (identical as before)

For small angles:

Also in timet:Pt o moves utand vt;

Pts a & b move a bit further:see figure below.

x and y

v vv x t v t x t

x x

u uu y t u t y t

y y

Figure on left taken from

Munson et al; with symbols Aand B, and instead of.


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Angular Deformation Rate (a.d.r.) in xy plane becomes:

0 0 0

1

22 lim lim limt t t

v ut t

x y x y

t t t

v u

x y


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Similarly for deformation in yz and zx planes; such that:

. . .v u

a d r in xy planex y

. . .w v

a d r in yz planey z

. . .u w

a d r in zx planez x

These expressions will later

be used forgeneralization of

the viscous stresses;

recall from before for

Newtonian fluid:

yx

du

dy

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ot o o a u d a t c e ( e at cs)

Recall the four components of motion:


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( )



2) Fluid Rotation


Linear Deformation



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Linear Fluid Deformation:Angles now remain unchanged;only changes of length

Element can change in x direction only if

Similarly for other directions. Hence:

0ux

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Main Topics

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p



Incompressible Flow


Momentum Equation

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Momentum Equation

Up to the Navier-Stokes Equation

Momentum Equation

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q

Newtons Second LawApply Newtons second law to infinitesimal fluidparticle of mass dm

Recall for system:

But for acceleration fluid particle in a velocity

field; must use the substantial derivativeD/Dt

( )

system

mass systemsystem system

dP dVF with P V dm or dF dm

dt dt

Momentum Equation

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q

Newtons Second LawFor fluid particle in velocity field

.VdF dm V V t

Momentum Equation

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q

Newtons Second LawFor fluid particle in velocity field

Need to find expression for forces on fluidparticle:

S BdF dF dF

Momentum Equation

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q

Forces Acting on a Fluid Particle

Volume element

Stresses at center of small cube in

the x direction:

Net force in x direction is sum of all

(stresses).(dydzordxdzordxdy)shown in figure

, ,xx yx zx

dV dxdydz

Momentum Equation

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2 2xxx xx

S xx xx

dx dxdF dydz dydz

x x

2 2

yx yx

yx yx

dy dydxdz dxdz

y y

2 2

zx zxzx zx

dz dzdxdy dxdy

z z

Momentum Equation

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After simplification:

x

yxxx zxSdF dxdydz

x y z

Momentum Equation

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Forces Acting on a Fluid ParticleAlso for other directions such that together:

x

yxxx zx

SdF dxdydzx y z

y

xy yy zy

SdF dxdydzx y z

z

yzxz zzSdF dxdydz

x y z

Momentum Equation

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Define Stress Tensor

xx xy xz xx xy xz

yx yy yz yx yy yz ii ii

zx zy zz zx zy zz

as

,

ij i j i

i j

e e with e unit vector

Momentum Equation

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Define Divergence of a Tensor:

, , , ,

.ij

ij i j k ij i j k i j

i j k i j i j k k k

e e e e e e e ex x

,

ij ij

j ki j

j k i j ik i

e e

x x

ij

i ix

Vector with components

Momentum Equation

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Comment on Notation(for later convenience)

ij i je e I Unit tensor

ijWith the Kronecker delta

Momentum Equation

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Forces Acting on a Fluid ParticleSurface force in vector form:

j

xj yj zj xj yj zj ij

Si i

dF dxdydz dxdydz dV x y z x y z x

.S

dF dV

Momentum Equation

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xB B xdF dxdydz g or dF dxdydz g

BdF gd V

Momentum Equation

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.S BdF dF dV gdV

Momentum Equation

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Differential Momentum Equation

. . .S BV V

dF dF dm V V dV gdV dV V V

t t

V

g V Vt

Momentum Equation

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William Dhaeseleer- 2011Fluidummechanic

105


.V

V V gt

DVg

Dt

VV V V V V

V

V Vt t t

0 .V cont eq

t

Use

and

V VV gt

V g VV t

Momentum Equation

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106


V g VV t

Last expression can be put in full

conservation form

0

t

AM

With

and

g or gh

I

V VV I t

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107

Momentum Equation

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108

Newtonian Fluid: Navier-Stokes Equations

Generalization of yz du dy

Determination of in terms of

velocity gradients

1 2 3( , , ) ( , , ) ( )

iV u v w V V V V V V

i

j

V

x

with

Momentum Equation

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109


Recall rotation & angular deformation

of fluid particles

Paradox:

Viscous shear stress leads to fluid-

particlerotation, butrotationdoes notexplicitly appear in stress tensor!

Motion of a Fluid Particle (Revisited)

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110



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111


Fluid particles rotate ( finite):

- when subject to a torque by shear stresses

Body forces and normal (pressure) forces may accelerate &deform particles but cannot generate a torque!

Rotation of fluid particles will always occur when shear

stresses, i.e., whenever there is viscous flow!


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112


Fluid particles also rotate ( finite):

- if they were launched with an initial rotation

(e.g., solid rotation)

But,

For solid body rotation,there is no relative movement of fluid layers, hence nofriction, and hence no viscous shear stress!


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113


Recall:1

2z

v u

x y

Whenever

then there is no rotation =0

When solid body rotation

then no deformation

u vy x

u v

y x

. . .v u

a d r in xy planex y

See both formulae and figure!


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William Dhaeseleer- 2011

Fluidummechanic114


In general, usually

then rotation is accompanied by

angular deformation

u v

y x


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Fluidummechanic115


In general, usually

rotation is implicitly present in stress tensor

via the presence ofangular deformation

u v

y x

Viscous stress tensor (BSL)

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Fluidummechanic116

Following Bird, Stewart & Lightfoot (BSL - 2nd Ed p

18-19) and Landau & Lifshitz, Fluid Mechanics (2ndEd p 44-45) one can show that:

2

3

1 2 323

1 2 3

( )

( )

T

BSL

j iBSL ij ij

i j

V V V I

V V V V V

x x x x x

This will now be made plausible.

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Fluidummechanic118

BSL BSLp or p

E.g., Newtons law

according to BSL

according to F&McD

xyx

dV

dy

yx

du

dy


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Fluidummechanic119

Viscous internal friction is due to relativemotion of

various parts of the fluid. It is related to the gradientsof the velocity:

Assuming Newtonian fluids, there is a linearrelationship:

iBSLj

Vf V f

x

kBSL ijklij kll

V

x

Can be simplified: No need for 81 viscosity coefficients


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Fluidummechanic120

A solid rotation does notlead to internal friction of the

viscous type. Consider uniform rotation of fluid characterized by angular

rotation vector so that:

For satisfying this cross product, with a constant vector,

- the partial derivatives are simplyzero;

- the combinarions also vanish.

The viscous stress must be proportional to this sortof symmetrical combinations to guarantee theabsence of friction for uniform rotation!

12

( )V r same as V

V

i iV x

i j j iV x V x for i j


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Fluidummechanic121

Therefore, one can write:

1 2 3

1 2 3

( )j i

BSL ij ij

i j

V V V V V a b

x x x x x

( )TBSL a V V b V I

With aand bconstants


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Fluidummechanic122

It is customary to write:

1 2 3

1 2 3

2( )

3

j iBSL ij ij

i j

V V V V V

x x x x x

( ) 2

22 3

T

BSL

V VV I

Or

For our purposes, can assume =0, so that:


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Fluidummechanic123

( ) 1

22 3

T

BSL

V VV I

Or with 1 1

2 3

T

S V V V

2BSL S I With thus

Viscous stress tensor reduces to:


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124

BSL Derivation so far ignored the pressure contribution

BSLp

2p S

Must be added explicitly. Therefore:

ii iip

And thus:

Momentum Equation

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Fluidummechanic

125

Newtonian Fluid: Summary Stress Tensor

Generalized relationship stresses and velocity gradients;

Generalization of yx du dy

2p S x x y y z ze e e e e e

1 1

3 3xx yy zzp Tr

1 1

.2 3

T

S V V V

Unit tensor

Local thermodynamic pressure

Rate of strain tensor

Momentum Equation

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Fluidummechanic

126

Newtonian Fluid: Summary Stress Tensor

xy yx

v u

x y

yz zy

w v

y z

zx xzu wz x

22

3xx

up V

x

22

3yy

vp V

y

2 23

zzwp Vz

F&McD Eq. (5.25)

Momentum Equation

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Fluidummechanic

127


. . 2DV DV

g g p S Dt Dt

Forconstant and incompressible flow with 0V

. 2DV

g p SDt

2DVg p V

Dt

Navier-Stokes

Momentum Equation

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128


2

x

Du pg u

Dt x

2

y

Dv pg v

Dt y

2

z

Dw pg w

Dt z

Momentum Equation

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129


Momentum Equation

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Special Case: Eulers Equation

0

Momentum transport

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pinterpretation

Recall momentum balance equation

Momentum Equation

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. .V

V V gt

.DV

gDt

VV V V V V

V

V Vt t t

0 .V cont eq

t

Use

and

V VV gt

V g VVt

Momentum Equation

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V g VV t

BSLp With

BSL BSLp VV p VV

Divergence

part

Momentum-flux tensor (BSL)

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BSL (2nd Ed p 36) combine the following:

BSLVV pI

VV

Combined momentum-flux tensor

Molecular momentum-flux tensor

BSL Viscous momentum-flux tensor

p Pressure

Stress/Momentum-flux interpretation(BSL)

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Shear force per unit area identified as follows:

the force in the x direction on a unit area

perpendicular to the y direction

force is exerted by the fluid of lesser y on

the fluid of greater y

(in elasticity theory and in F&McD the convention is reversed!)

(BSL)

x

yx

dV

dy


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Shear stress in the x direction

on the surface of constant y

x

y

yx


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Transport of x-momentum interpretation:

the flux of x-momentum in the positive y

directionflux = flow per unit area

momentum flows downhill;

velocity gradient is driving force for

momentum transport

x

yx

dV

dy


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Viscous flux of x momentum in

the y direction

x

y

yx


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the y direction

x

y

yx yx

Both interpretations combined:


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Compare stress sign conventions:

F&McD BSL


P d i f (BSL Ed2 16 17)

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Pressure and viscous forces (BSL Ed2- pp 16-17):


P d i f (BSL Ed2 16 17)

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pressure force (in x direction) per unit

area of constant x (or perp to x direction)

x x

pe p

viscous force (with x, y, z components) per

unit area of constant x (or perp to x

direction)x

x xx x xy y xz z xx x xy y xz ze e e (Note that our notation BSL notation)

x xe


P d i f (BSL Ed2 16 17)

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Molecular stresses are defined as a combination

of the thermodynamic pressure and the viscous

stresses

ij ij ijp


Press re and isco s forces (BSL Ed2 pp 16 17)

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Molecular stresses can be interpreted in two

ways:

Normal stresses; Shear stresses

( 1, 2, 3)

kk kk

k

p

( )

ij ji

i j


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the y direction

x

y

yx yx

Recall:


Pressure and viscous forces (BSL Ed2 pp 16 17):

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In summary:

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Momentum-flux interpretation (BSL)

Convective momentum transport (BSL Ed2 pp 34 37):

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momentum flux across the

three areas perpendicular to

the respective axes

( ) , ( ) , ( )x y zV V V V V V

Convective momentum transport (BSL Ed2- pp 34-37):

x yV V = convective flux of y-momentum acrossa surface perpendicular to the x direction

(or of constant x)

To be compared with

xy = molecularflux of y-momentum across a

surface perpendicular to the x direction

(or of constant x)

Momentum-flux interpretation(BSL)


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Convective momentum flux tensor:


i i i ii i

i i j j i i j ji j i j

i j i j i j i jij ij

VV V V e V V

V V e V e V

VV e e VV

Momentum-flux interpretation(BSL)


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Convective momentum flux tensor components:


vectorstensor components

Momentum-flux tensor (BSL)

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Thus in summary (recall):

BSLVV pI

VV

Combined momentum-flux tensor

Molecular momentum-flux tensor

BSL Viscous momentum-flux tensor

chapter 5 fm wdh_2011-2

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