basic equations in fluid dynamics_rev

Fluid DynamicsFluid DynamicsME 5313 / AE 5313

Basic Equations in Fluid Dynamics

Instructor: Dr. Albert Y. TongDepartment of Mechanical and Aerospace EngineeringDepartment of Mechanical and Aerospace Engineering

The University of Texas at Arlington

Vector Operations in OrthogonalVector Operations in Orthogonal Coordinate System

Vector operations in orthogonal coordinate systems

Scalar function ),,( 321 xxx

Vector function

U it t i

),,( 321 aaaa

Unit vectors in =

Metric scale factors

321 ,, xxx 321 ,, eee

hhhMetric scale factors 321 ,, hhh

Basic Equations in Fluid Dynamics 2


(d l) t

(del) operator:

31 2

1 1 2 2 3 3

ee eh x h x h x

1 1 2 2 3 3h x h x h x



Cartesian Coordinates (x, y, z)

e1= i e2= j e3= k

h1= 1 h2=1 h3= 1

x x x y

x1= x x2= y x3= z

i j kx y z


y


Cylindrical coordinates (r,,z)r ze1= e2= e3=

h 1 h h 1h1= 1 h2= r h3= 1

x = r x3= zx2=

ˆˆ ˆ

x1= r x3= zx2=

r zr r z



Spherical coordinates (r, , )

r e1= e2= e3=

h 1 h r h r sin h1= 1 h2= r h3= r sin

x1= r x2= x3=

ˆ ˆ

1 2 x3=

rr r r sin



Gradient

The gradient of or grad is defined asThe gradient of or grad is defined as

grad 321

he

he

he

g

332211 xhxhxh



In Cartesian Coordinates:

kji

z

ky

jx

i



Consider a simple 1-D case:= T = T (x)

zTk

yTj

xTiT

zyx

dTidx

i



Consider a surface (x, y, z) = C

0

dzz

dyy

dxx

d zyx

along = C

0)( rdd



⇒ rd⇒ rd

Therefore, is normal to the f

Csurface C



DivergenceThe divergence of is defined asa

1 2 3 1

1 2 3 1

1 ( )a h h ah h h x

3 1 2 1 2 3( ) ( )h h a h h a

3 1 2 1 2 32 3

( ) ( )x x



Cartesian Coordinates:

31 2 aa aa

e.g.x y z

ua u v wux y z



Potential flow:

u

)()()()(zzyyxx

2

2

2

2

2

22

(Laplacian Operator)

222 zyx 2



Cylindrical coordinates:

2a1 21 3

a1a (ra ) (ra )r r z



Spherical Coordinates:

212

1 ( sin )a r a

12 ( )

sinr r

2 3( sin ) ( )r a ra



CurlaThe curl of is defined by

1 1 2 2 3 3h e h e h e1

1 2 3 1 2 3

1Curl a ah h h x x x

1 1 2 2 3 3h a h a h a



Cartesian Coordinates:

i j k

ax y z

x y z

x y za a a

y



aa ( )yz

aaa iy z

( ) ( )yx xzaa aaj k

z x x y

(Vorticity)u



Velocity field: ( , , )V u v w

V

Deformation field

V

Volume dilatation

V

Rotation



Incompressible0V

Incompressible (constant density)

0V

0V

Irrotational0V Irrotational


Integral Theorems

Gauss’ Theorem

a ndS adV

S V

a ndS adV It is also known as the divergence theorem.


Integral Theorems

In Cartesian coordinate system (x,y,z)

x x y y z za n a n a n dS S

yx zaa a dV

yx z

V

a a dVx y z

V

x y zn n , n , n

where


Integral Theorems

Stokes’ Theorem

( )a dl a n dS

C S

It relates a line integralto a surface integral


Vector Identities

(i) 0 (i)

(ii)

0

a a a (ii)

(iii)

a a a

a a a

(iii)

(iv)

a a a

a 0

(iv) a 0


Vector Identities

(v) 1

(v)

( i)

a a a a a a2

2(vi) 2a a a

(vii) a b b a a b

(viii) a b b a a b a b b a


Eulerian and Langrangian Coordinates

Eulerian coordinates:Eulerian coordinates:

Open system (control volume)p y ( )

Lagrangian coordinates:Closed system (control mass)Closed system (control mass)


Eulerian Coordinate

Fixed region in space

i.e.x,y,z,t are independent


Lagrangian Coordinate

F iFocus attention on a particular particle as it movesit moves.

i.e. x,y,z,t are no longer independentindependent


Material Derivatives

In Eulerian coordinates:T (temperature) is a function of x,y,z,and ti.e. T = T(x,y,z,t)

D T TD T TD t t

D t t



In Lagrangian coordinates:T = T(x,t)

DT T t T DT T t T xDt t t x t

DT T T u Dt t x



In a 3D case, T = T(x,y,z,t)

DT T t T x T y T z

Dt t t x t y t z tDT T T T T

DT T T T Tu v w

Dt t x y z

y



In vector form:

DT T u TDt t

Dt tu iu jv kw where

i j kx y z

u u v wx y z



In tensor form:

kDT T Tu

kk

uDt t x

uk xkk = 1 u xk 2k = 2 v yk = 3 w z


Reynolds’ Transport Theorem

D V(t) V

D dV u dVDt t

= any fluid properties(mass)

(mass)(momentum)

u

(energy)e



Proof:

D 1(t)dV lim (t t)dV (t)dV t 0

V(t) V(t t) V(t)

(t)dV lim (t t)dV (t)dVDt t

1 By adding and subtracting V(t)

1 t t dVt


( )


Then

D 1(t)dV lim (t t)dV (t t)dVD

t 0

V(t) V(t t) V(t)

( ) ( ) ( )Dt t

1lim (t t)dV (t)dV t 0

V(t) V(t)

(t t)dV (t)dVt



Second limit term =V(t)

dVt

First limit term =t 0

( ) ( )

1lim (t t)dV]t

V(t t) V(t)



S t t

S t t

dS

S t

dS u n

dV u ndS t



1 =

t 0S(t)

1lim (t t)u ndS tt

( )

= =

S(t)

(t)u ndS

V(t)

(t)u dV

S(t) V(t)



Sub. back into the original equation gives

D (t)dV u dVDt t

In tensor notation, it becomes

V(t) V(t)

kk

D (t)dV u dVDt t x

k

V(t) V(t)Dt t x


Conservation of Mass

Physical Law: matter can neither be created nor destroyeddestroyed

take

D take

D dV 0Dt

V



Using R T T

kk

D dV u dV 0Dt t x

k

V VDt t x

ku 0t x

kt x

u v w0

or (C ti it E ti )

0t x y z

or (Continuity Equation)



Special Case: i) uniform constant density

constant

u v w0

t x y z

u v wu v w 0t

t x x y y z z



, , 0

u v w 0 0

x y z



u v w 0x y z

x y z

u 0

or u 0 or



Special Case: ii) incompressible stratified flowalong a streamline but not uniform throughoutc

D 0D 0Dt

The continuity equation can be written as

0 kk

u 0t x



ku => kk

k k

uu 0t x x

k

k

uD 0Dt x

Dwhere 0Dt

k

ku0

k

k

u0

x

or u 0


Conservation of Momentum

Physical Law: The directional rate of change of t t t l fmomentum = net external force.

tF ma

Two types of force

netF ma

(i) Body force, , e.g. gravityf

(ii) Surface force, , e.g. pressure, shear stressP



Rate of Change of MomentumD udVDt

E t l F

V(t)Dt

PdS f dV

External Force

S V

PdS f dV

D udV PdS f dV

V S V

udV PdS f dVDt



Stress in the x1 direction = 11 n1 + 21 n2 + 31 n3



j ij iP n

D => j ij i jV S V

D u dV n dS f dVDt

V S V



Using Reynolds’ Transport Theorem

j j j kD u dV u u u dV

j j j k

kV V

Dt t x

Using Gauss’ Theorem

ij ij

ij ii

S V

n dS dVx


S V


It yields

ijj j k j

k iu u u f dV 0

t x x

Since V is arbitrary, the integrand must vanish

k iV

ijj j k ju u u f

j j k j

k it x x



On the other hand

jj j

uu u

t t t

and

t t t

jkj k j k

uuu u u ux x x

k k kx x x



=> j j ijkj k j

u uuu u ft t x x x

k k it t x x x

u But from continuity equation, k

k

u 0t x



u u j j ijk j

k i

u uu f

t x x

k it x x

D j ijj

i

Duf

Dt x

iDt x


Conservation of Energy

First Law of Thermodynamics for a closed system(1)dE dKE Q W

d E KE) d E KE)Q W

dt

E = internal energyKE = kinetic energyKE = kinetic energy Q = heat transfer to the system


W = work done by the system


L.H.S. :D 1e u u dVDt 2

R.H.S. : (i)

V(t)Dt 2

Q q n ds R.H.S. : (i)

(ii)s

Q q n ds W u p ds u f dV (ii)

s V

W u p ds u f dV Sign convention:Sign convention:W is positive if work is done by the systemQ is positive if heat is transferred into the system


system


D 1 V s V s

D 1e u u dV u PdS u f dV q ndSDt 2

(2)

Using the Reynolds’ Transport Theorem

V s V s

Using the Reynolds Transport Theorem

D 1 1 1dV dV

(3)kk

V V

e u u dV e u u e u u u dVDt 2 t 2 x 2

(3)



Using Gauss’ Theoremj

jV V

qq n dS q dV dV

x

(4)

js V V

and

j js s

u P dS u P dS

s s



But j ij iP n j j

j ij i j ij iu P dS u n dS (u )n dS j ij i j ij i

s s s

( ) j ij

iV

u dVx

(5)



Substituting (3), (4), and (5) into (2) gives:

1 1 dV j j j j k

kV

e u u e u u u dVt 2 x 2

jj ij j j

qu u f dV

x x

j j j j

i jV

x x



since V is arbitrary, the integrand must vanish

j j j j kk

1 1e u u e u u ut 2 x 2 kt 2 x 2

jqu u f

jj ij j j

i ju u f

x x (6)



Mechanical Energy part can be removedL H S f (6)L.H.S. of (6):

11 e 1 1

1st term: j j j j j je u u e u u u ut 2 t t t 2 2 t

2nd term: kj j k k

k k k

u1 ee u u u e ux 2 x x

kj j k j j

u1 1u u u u u2 x x 2


k k2 x x 2


From Continuity Equation:

kk

u 0t x

k

k

ux t

2nd term becomes:k

k j j k j j

k k

e 1 1e u u u u u ut x 2 t x 2



L.H.S. of (6)

k j j k j je e 1 1u u u u u ut x t 2 x 2 k kt x t 2 x 2

u u j jk j k j

k k

u ue eu u u ut x t x

(7)



R.H.S. of (6):

jj ij j j

i j

qu u f

x x

i jx x

ij j jj ij j j

u qu u f

j ij j j

i i ju u f

x x x



After rearranging, (6) becomes: j j ij j j

k j j k j j j ijk k i i j

u u u qe eu u u u u u ft x t x x x x

j

j j ij j j iju u u u

According to momentum equation:

j j ij j j ijj j k j j j j k j

k i k i

u u u uu u u u u f u u f 0

t x x t x x

j jk ij

k i j

u qe eut x x x


j

Remarks

Number of Equation:Continuity 1Momentum 3Momentum 3Energy 1------------------------Total 5


Remarks

Number of unknowns:e 1uj 3uj 3qj 3σ 9σij 9ρ 1------------------Total 17


Total 17

Remarks

Introducing the constitutive equations:1) Fourier’s Law of Heat Conduction2) Newtonian Fluid2) Newtonian Fluid


Remarks

Fourier’s Law of Heat Conduction

q k T

Newtonian Fluid

jk iij ij ij

uu up

e to a u d

ij ij ijk j i

px x x


Deformation of Fluid Element

An infinitesimal element of fluid at time t=0 and time t=t



1) Translation2) Rigid body rotation3) Distortion3) Distortion4) Volumetric dilatation



∂vδx δt δy

∂x δx. δt δy

C Dx

∂vv

∂v∂x δx v δt δα tan 1

δx

δy∂v∂x δxδt

δα~δyδx



v vt x

u

similarlyuy



Rate of rotation (clockwise)

1 1 u v( )2 2 y x

2 2 y x

Rate of shearu v

Rate of shear

y x


Rate of Deformation Tensor

In a 2-D case

or1 1

1 2ij

2 2

u ux x

eu u

u ux yv v

which can be broken down to1 2x x x y

ij

u v u vu 0 00y x y x1 1xe

ijev v u u v2 20 0 ( ) 0y x y y x



where

u 0x

v

u v0y x1

v u2

Volumedilatation Shear

v0y

v u2 0x y

u v0y x1 Rotation

u v2 ( ) 0y x

Rotation



In a 3-D case 31 2 11

2 1 3 11

32 2 1 2ij

uu u uu 00 0x x x xx

uu u u u1e 0 0 02

j

2 1 2 3 2

3 3 31 2

3 1 3 2 3

x 2 x x x xu u uu u0 0 0x x x x x

32 1 1

1 2 1 3

uu u u0

x x x x

32 1 2

1 2 2 3

3 31 2

uu u u1 02 x x x x

u uu u0


1 3 2 30

x x x x

Constitutive Equations

(i) Stress-strain rate relationship for Isotropic Newtonian floNewtonian flo

jk iij ij ij

uu up

h 0 i j

ij ij ijk j i

px x x

where ij = 0 i ≠ jij = 1 i = j

It is called “Kronecker delta”



juu u jk i

ij ij ijk j i

uu upx x x

= dynamic viscosity = second viscosity coefficient = dynamic viscosity, = second viscosity coefficient(empirical parameters)

p = thermodynamic pressure



Sometimes it is written as

whereij ij ijp

jk iij ij

uu ux x x

It is called the viscous stress tensor

j jk j ix x x

It is called the viscous stress tensor



(ii) Fourier’s Law

q k T

iTq k

or ii

q kx

or

k : thermal conductivity


Navier-Stokes Equations

Recall Conservation of Momentum:

j j ijk j

k i

u uu f

t x x

(1)k it x x

Using constitutive relation for ij

ij jk iij ij

uu up

ij ij

i i k j ip

x x x x x



R.H.S. (1st term):

iji

px

Among these onl the one in hich i j is non e o

1 2 31 2 3

j j jp p px x x

Among these, only the one in which i=j is nonzero

pp iji j

px x



R.H.S. (2nd term):

k kij

u u

ij

i k j kx x x x

Therefore Eq (1) becomes:Therefore, Eq. (1) becomes:

(2)(2)

(Momentum equation)


Navier-Stokes Equations(i) Incompressible and constant viscosity

0ku 0

kx

j ji iu uu u

i j i i j i ix x x x x x x

2 2j ji u uu

j ji

j i i i i ix x x x x x

Therefore, Eq. (2) becomes:

(3)2



(ii) Incompressible and inviscid

u u p

Viscous terms vanish, Eq. (2) becomes:

j jk j

k j

u u pu ft x x

(4)

Known as Euler’s Equation


Energy Equations

j jk ij

u qe eut

(1)jk i jt x x x

Applying constitutive relation for Newtonian Fluids

j j jk iij ij ij

u u uu up

ij ij ij

i k j i i

px x x x x

u u u uu u j j j jk iij ij

i k i j i i

u u u uu upx x x x x x


Energy Equations

2j j jk k iu u uu u u

j j jk k iij

i k k j i i

px x x x x x

u j k

iji k

u upx x

2j jk i u uu u

wherej jk i

k j i ix x x x


Governing Equations for Newtonian Fluids

Conservation Equations:

0kut x

kt x


Governing Equations for Newtonian Fluids

Total number of unknowns = 7e u (3) T p, e, ui(3), T, p

Total number of equations = 5

2 more equations are added:equation of state: p = p( T)equation of state: p = p( ,T)

e.g. p = RTcaloric equation of state: e = e (p, T)

caloric equation of state: e e (p, T)e.g. de = Cv dT

=> All 7 unknowns now can be solved.


Flow Kinematics

The kinematic relations for a fluid areconcerned only with the space timegeometry of the motion. They areg y yindependent of the dynamics and thethermodynamics of the continuum, andy ,are based on the continuity equation.


Flow Kinematics

StreamlineStreamlines are lines whose tangents are everywhere parallel to the velocity vector.

For 2-D flows

dy vdx u

u v

dx uu


Flow Kinematics

If the velocity field is known as a function of x and y (and t if the flow is unsteady) this equation can be(and t if the flow is unsteady), this equation can be integrated to gain the equations of the streamlines.

u xi y j

,u x v y

dy v ydy v ydx u x

dy dx 0dy dxy x


Flow Kinematics

By integrating both sides

ln ln lny x c xy c

The particular

xy c

streamline that passes (1,1)

1xy c


Flow Kinematics

PathlinesA pathline is the line traced out by a given particle as it flows from one point to another

Mathematically,

we have

dx 1t 2t 3t 4t( , )i

i idx u x tdt

0t

1t


Flow Kinematics

StreaklinesA streakline consists of all particles in a flow that have previously passed through a common point


Flow Kinematics

Work Example

Consider a 2-D plane flow:

11 ( )

1 1x xv u

t t

1 1t t

2 2 ( )v x v y

3 0 ( 0)v w


Flow Kinematics

For streamlines

/(1 )dy v ydx u x t

/(1 )dx u x t

2 2 2dx v x

1 1 1 /(1 )dx v x t

dy dx (1 )dy dx ty x


Flow Kinematics

ln (1 ) ln lny t x c ( )y

(1 )ln ln ty c x y (1 )ty cx

( 0)y cx t 2 ( 1)y cx t


Flow Kinematics

2y y 2y cxy cxy y

xx


Flow Kinematics

For pathlines

1 11 1

dx xvdt t

22 2

dx v xdt

1dt t

1 1(1 )x a t

dt

2 2tx a e 1 1( ) 2 2


Flow Kinematics

dx x1dt t

d

dy ydt

Combining x1 and x2 gives

1 1 1( ) /2 2

x a ax a e


Circulation and Vorticity

Definitions:

C

u dl

(i)

u (ii)

ki ijk

u i ijk

jx



ijkif any i,j,k are the same

= if i,j,k is an odd permutation0-1

if i,j,k is an even permutation+1



u dl u ndA ndA

C A A

u dl u ndA ndA

0 0

( ) 0u

is divergence free (solenoidal)


Kinematics of Vortex Tubes

A vortex line is a line whose tangents are everywhere ll l t th ti it t A t t b iparallel to the vorticity vector. A vortex tube is a

region whose side walls are made up of the vortex lineslines

vortex line vortex tube



Consider unwrapping a vortex tube

vortex line



ABCDA u dl u dl u dl u dl u dl

ABCDAABCDA A B B C C D D A

u dl u dl u dl u dl u dl

note that u dl u dl

B C D A

dl dl

ABCDAA B C D

u dl u dl

Therefore,



From Stokes’ theorem

( )u dl u nds

ABCDA Area

Area

nds



Since vortex lines are tangential to the vortex tube0

0u dl u dl u dl

0n 0ABCDA and

0ABCDA A B C D

u dl u dl u dl

A B C D

u dl u dl

A B D C

u dl u dl

both in the clockwise direction


A B D C


However,

and1A B

u dl

2D C

u dl

A B D C

u dl u dl

The circulation is constant over any closed contour

1 2 1 2

u dl u dl or

The circulation is constant over any closed contour about a vortex tube



u dl ndA

1 1

1

1 1 1

C A

u dl ndA

A

A1 A2

C1 C2

1 1 1A

Similarly, 2 2 2A 2 2 2

1 1 2 2A A and

Analogous to 1 1 2 2V A V A for a stream tube


Conservative Force Fields

W F dr

C

If W is independent of the path is said to beF

If W is independent of the path, is said to be conservative.

F



If W is path independent, then must be an F dr

exact differential and can be written as a gradient

of a scalar function.

F

22.

12 1

1

F



dr d

i j kx y z

y

dr i dx j dy k dz

dr dx dy dz dx y z

x y z


Kelvin’s Theorem

The vorticity of each fluid particle will bed if th f ll i i tpreserved if the following requirements are

satisfied.

(i) conservative body force field

(ii) inviscid fluid(ii) inviscid fluid(iii) = constant or P = P()


Kelvin’s Theorem

Proof:Equation of motion,

j j jk iu u uu upu f

k jk j j k i j i

u ft x x x x x x x

F i i id fl id d

jDu P G

For inviscid fluid, and are zero

j

j jDt x x


Kelvin’s Theorem

By definition,

j jD D u dxDt Dt Dt Dt

( )j jj j

Du D dxdx u

Dt Dt

j jDt Dt

1jDu P Gd d d 1jj j j

j j

P Gdx dx dxDt x x


Kelvin’s Theorem

( ) ( )j jj j

D dx Dxu u d

j jx xu d u

j ju u dDt Dt

j kk

u d ut x

x jx

0jxt

( )D d

jk j

k

xu u

x

and

( )jj j j

D dxu u du

Dt


Dt

Kelvin’s Theorem

Therefore, 1D P G

1

j j j jj j

D P Gdx dx u duDt x x

1dP 1 ( )2 j j

dP dG d u u

0dG 1 ( ) 02 j jd u u and

D dPDt


Dt

Kelvin’s Theorem

(i) if = constant1D 1 ( ) 0D dP

Dt

(ii) if P = P() (Barotropic)dP = P’() d

( ) ( ) 0D P d f dDt

0DDt


Dt

Bernoulli Equation

Consider (i) inviscid fluid(ii) conservative force field(ii) conservative force field

Equation of Motion

j jk

k j j

u u P Gut x x x

f G

where

k j j

( )u u u P G

( )

t


Bernoulli EquationRecall vector identity,

1

2

a a a a a a

1u u u u u u

2u u u u u u

1 ( )2

u u u ( )

2

1 1u u u u P G

2u u u P G

t


Bernoulli Equation

1 dPP Note:

1 1dl P dl P

dl P dl P

1 dPdP

1 dPd d dl 1 dPd dP dl


Bernoulli Equation

1u dP 1

2u dPu u u Gt

12

u dPu u G ut

2t


Bernoulli Equation

(i) Steady Rotational Flow:

1 ( )2

dPu u u G u u

( ) 0u u

since

1 02

dPu u u G

2


Bernoulli Equation

Recall ( ) ( ) ( )D uDt t

1( )

2dPu u G

Dt t 2 ( ) 0

for steady flow

1 dP

0t

y

12

dPu u G B

B: Bernoulli constant

along a streamline

B: Bernoulli constant

B can vary from streamline to streamline


Bernoulli Equation

(ii) Steady Irrotational Flow

1 02

dPu u G

2 1 dPu u G B

2u u G B

B is constant through out the entire flow field


Bernoulli Equation

(iii) Unsteady Irrotational Flow

1 02

u dPu u Gt

0

0

(irrotational flow)

0u

( ) 0 Recall: ( )

u


Bernoulli Equation

1( ) 0dPu u G ( ) 0

2u u G

t

1 dP 1 ) 02

dPu u Gt

1 ( )2

dPu u G B tt

B(t) is constant with respect to time through out the flow field


time through out the flow field

Vorticity Equation

Consider: constant

f 0

constant

f 0

N-S Equation:

21( )u u u P ut


Vorticity Equation

1( )u u u u u ( )

2u u u u u

1 P1 ( )PP

212

u Pu u u ut

2t


Vorticity Equation

Take the curl on both sides:

12

u u u ut

2p u

2( )

2( )u

t


Vorticity Equation

where

( ) ( ) ( ) ( ) ( )u u u u u

( ) ( ) ( ) ( )u u

2( ) ( )u u

t


Vorticity Equation

2-Dimensional plane flow:

x yu (u ,u ,0)

z(0,0, )

0

and

0 2( )u

D

( )ut

2 2( )z zz z z

Dut Dt

or


basic equations in fluid dynamics_rev

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