CEE 262A

HYDRODYNAMICS

Lecture 5

Conservation Laws Part I

1

Conservation laws

VA

The ultimate goal is to use Newton's 2nd law (F=ma) to relateforces on fluid parcels to their acceleration. A natural way to do this is to compute the equations of motion following a volume of fluid; We first begin with the governing equations for a solid, non-deformable object with volumeV, surface area A, and density

Mass of object:

Momentum of object: V

dVuP~~

V

dVm

Newton's 2nd law for the object is given by:

bodysurface

V

FFdVudt

dtP

dt

dF

~~~~~)(

Surface forcese.g. Friction

Body forcese.g. Gravity

Example: Block of mass m pushed with force F along surface with friction coefficient b

0

3

~

3

~

1

~

1

~

~

3

~

3

~

1

~

1

~~

~~~

mgNdt

dwm

Fbudt

dum

amgaNaFabudt

udm

dVagaNaFabudVudt

d

FFdVudt

d

Body

VSurface

V

bodysurface

V

F

m

N

g

u

),(),(),()(

)(

taFdt

datbF

dt

dbdx

t

FdxtxF

dt

d b

a

tbx

tax

Which in 3D becomes:

~ ~ ~( ) ( ) ( )

( , )A

V t V t A t

d FF x t dV dV F u n dA

dt t

velocity of boundary

The problem for fluids is that the volume is generally not fixed in time, and so mass and momentum may leave the volume unless a material volume is employed.

The key to understanding what to do is Leibniz's Theorem:

Consider a volume V(t) for which the bounding surface moves (but not necessarily at the fluid velocity)

u

n

dA

( )A t( )V t

Fixed volume–VF : Flow of fluid through system boundary (control surface) is non zero, but velocity of boundary is zero. For this case we get

Material Volume–VM : Consists of same fluid particles and thus the bounding surface moves with the fluid velocity. Thus, the second term from the Leibnitz rule is now non-zero, so

Using Gauss' theorem:

~( , )

F FV V

d FF x t dV dV

dt t

~ ~ ~( , )

M MV V A

D FF x t dV dV F u n dA

Dt t

~ ~ ~

MA V

F u n dA F u dV

~ ~

( , )M MV V

D FF x t dV F u dV

Dt t

This is Reynolds transport theorem, where D/Dt is the same as d/dt but implies a material volume.

~~

Note that the Reynolds transport theorem is often written in themore general form which does not assume that the control volumeis bounded by a material surface. Instead, the control volume isassumed to move at some velocity and that of the fluid is

defined as relative to the control volume, such that

In this case, V is not necessarily a material surface. If ur=0, thenub=u and we revert to the form on the previous page.

A

rA

bVV

dAnuFdAnuFdVt

FdVtxF

Dt

D~~~~~

),(

bu~

bruuu~~~

~

n

dA

( )A t( )V t

The general form of all conservation laws that we will use is:

Rate of change of F = effects of volume sources + effects of surface sources

M MV V A

DF dV D dV C n dA

Dt

Quantity (F) Volume sources (D) Surface sources (C)

Momentum1 Gravity Stresses (pressure/viscous)

Heat Dissipation Diffusion/radiation

Salt (scalar) None Diffusion

Algae (e.g.) Growth Diffusion

1 Note: momentum is a vector quantity

Conservation of Mass

~

0M MV V

DdV u dV

Dt t

Mass is conserved (non-relativistic fluid mechanics)

For any arbitrary material volume

Since integral is zero for any volume, the integrand must be zero

~

0ut

Process: We have taken an integral conservation law and used it to produce a differential balance for mass at any point

0

~

ii

i

i

ii

i

ii

i

xu

x

u

t

xu

x

uu

xu

However,

andi

i

Du

Dt t x

~

1 Du

Dt

Thus if the density of fluid particles changes, the velocity field must be divergent. Conversely, if fluid densities remain constant,

~0u

Let where is an intensive property (amount/mass)F f f

dVx

fuu

xf

t

f

tf

dVfux

ft

dVfDt

D

M

MM

V iii

i

V

iiV

)(

)()(

dVDt

Df

dVx

fu

t

fu

xt

M

M

V

V iii

i

0)(But

dVDt

DfdVf

Dt

D

MM VV

Any other fluid property (scalar, vector,.. also drop triple integral)

Why is this important/useful?

Because Newton’s 2nd law:

dVDt

uDdVu

Dt

D

MM VV ~

~

dVFdVuDt

D

MM VV

~~

Dt

uDF ~

~

But from above:

Rate of Change of Momentum = Net Applied Force

Net Applied Force = Mass AccelerationIndependent of volume type!

Some Observations

1. Incompressible

~

1u

Dt

D

01

Dt

D

[ No volumetric dilatation, fluid particle density conserved]

0~

i

i

x

uu

Differential form of “Continuity”

2. Slightly Compressible

• Typically found in stratified conditions where

0( , , , ) ( ) '( , , , )x y z t z x y z t

• Boussinesq Approximation

- Vertical scale of mean motion << scale height

- or0

'1

Allows us to treat fluid as if it were slightly incompressible

Note: Sound and shock waves are not included !

Reference density (1000 kg/m3 for water)

Background variation (typ. 1-10 kg/m3 for water)

Perturbation density due to motion (typ.

0.1-10 kg/m3 for water)

Informal “Proof”

2dPc dP c d

d

If a fluid is slightly compressible then a small disturbance caused by a change in pressure, , will cause a change in density . This disturbance will propagate at celerity, c.d

dP

• If pressure in fluid is “hydrostatic”

2

dP d gg

dz dz c

Now

and

d d dz

dt dz dt

wdt

dz [ Streamline curvature small]

2c

gw

dt

d

2

1

c

gw

dt

d

Typically: g ≈10 m/s2 ; c ≈ 1500 m/s ; w ~ 0.1m/s

0~ u

Conservation of Momentum – Navier-Stokes

We have:

~ ~

~ ~

Du uF u u

Dt t

Two kinds of forces:

• Body forces

• Surface forces

Two kinds of acceleration:

• Unsteady

• Advective (convective/nonlinear)

Two kinds of surface forces:

• Those due to pressure

• Those due to viscous stressesDivergence of Stress Tensor

Plan for derivation of the Navier Stokes equation

1. Determine fluid accelerations from velocities etc. (done)

2. Decide on forces (done)

3. Determine how surface forces work : stress tensor

4. Split stress tensor into pressure part and viscous part

5. Convert surface forces to volume effect (Gauss' theorem)

6. Use integral theorem to get pointwise variable p.d.e.

7. Use constitutive relation to connect viscous stress tensor to strain rate tensor

8. Compute divergence of viscous stress tensor (incompressible fluid)

9. Result = Incompressible Navier Stokes equation

Forces acting on a fluid

a) Body Forces: - distributed throughout the mass of the fluid and are expressed either per unit mass or per unit volume

- can be conservative & non-conservative

~g

Force potential

Examples:

(1) force due to gravity (acts only in negative z direction)

(2) force due to magnetic fields

gz

We only care about gravity

b) Surface forces: - are those that are exerted on an area element by the surroundings through direct contact

- expressed per unit of area

- normal and tangential components

dA

ndF

sdF

~dF

c) Interfacial forces:

- act at fluid interfaces, esp. phase discontinuities (air/water)

- do not appear directly in equations of motion (appear as boundary conditions only)

- e.g. surface tension – surfactants important

- very important for multiphase flows (bubbles, droplets,. free surfaces!)

Very important deviation from text!!!

CEE262a (and most others)

Deviatoric (viscous) stress

Full stress

Kundu and Cohen

Full stress

Deviatoric (viscous) stress

Stress at a point(From K&C – remember difference in nomenclature,i.e. ij ←ij)

What is the force vector I need to apply at a face defined by theunit normal vector to equal that of the internal stresses?

~n

Consider a small (differential) 2-D element

1

2

11

12

1112

21

21

22

22

11

12

2122

n

1n

2n

1dx

2dx ds

dF

cut away

1 11 2 21 1 F dx dx force component in x1 direction

1 2 11 11 21

11 1 21 2

11 1 21 2~

cos cos

dF dx dxf

ds ds ds

n n n[ has magnitude of 1]

Defining the stress tensor to be ij

11 12 13

21 22 23

31 32 33

And in general

jjjj nfandnf 2211

jjii nf

d

But [see Kundu p90]ijji

or

or (3D)

ii ij j

total

CS

dFf n

ds

dFf n

ds

dFf n F n dA

dA

“ Surface force per unit area”(note this is a 2D area)

Total, or net, force due to surface stresses

Conservation of momentum

2),( 1

1

112111

dx

xxx

2),( 1

1

112111

dx

xxx

2),( 2

2

212121

dx

xxx

2),( 2

2

212121

dx

xxx

1x

2x

3x

2),( 3

3

312131

dx

xxx

2),( 3

3

312131

dx

xxx

Dimensions:dx1 . dx2 . dx3

Sum of surface forces in x1 direction:

11 1 11 111 11 2 3

1 1

21 2 21 221 21 1 3

2 2

31 3 31 331 31 1 2

3 3

3111 211 2 3

1 2 3

2 2

2 2

2 2

ji

j

dx dxdx dx

x x

dx dxdx dx

x x

dx dxdx dx

x x

dx dx dxx x x

dVx

Defining i component of surface force per unit volume

to be iVF

~For body forces we use gravity = ig g

~gF

Dt

DuV

i

In general :

iV ij

j

Fx

j

iji

i

xg

Dt

Du

“Cauchy’s equation of motion”

Force = divergence of stress tensor

3

Note that usually

-g g e

Important Note: This can also be derived from the Integral From of Newton’s 2nd Law for a Material Volume VM

A jijV iV i dAdVgdVu

Dt

DMM

MM V

i

V i dVDt

DudVu

Dt

D But

and [Gauss' Theorem]M

ijij jA V

j

dA dVx

0M

ijiiV

j

Dug dV

Dt x

ijii

j

Dug

Dt x

Constitutive relation for a Newtonian fluid“Equation that linearly relates the stress to the rate of

strain in a Newtonian Fluid Medium”

(i) Static Fluid: - By definition cannot support a shear stress

- still feels thermodynamic pressure

(in compression)

ij ijp

(ii) Moving Fluid: - develops additional components of stress

(due to viscosity)ij ij ijp Hypothesis

Note difference from Kundu !

Deviatoric stress tensor [Viscous stress tensor]ij

If medium is isotropic and stress tensor is symmetric

only 2 non-zero elements of

ij ij mm ij2 e e

Assume mnijmnij eK

ijmnK 4th order tensor (81 components!) that depend on thermodynamic state of medium

K

which gives

or

ijijij eup 2)ˆ3

2(

See derivation of in Kundu, p 100

Special cases

(i) Incompressible 0ˆ u

ijijij ep 2

(ii) Static 0 ije

ijij p

In summary

Cauchy's equation

Constitutive relation fora compressible, Newtonian fluid.

ijijij eupii 2)ˆ3

2()(

( ) ijii

j

Dui g

Dt x

Navier-Stokes equation

2

2

ii ij ij

j

i iji j

Dug p e

Dt x

pg e

x x

The general form of the Navier-Stokes equation is given by substitutionof the constitutive equation for a Newtonian fluid into the Cauchy equation of motion:

ijijkki

ii eep

xg

Dt

Du 23

2

Incompressible form (ekk=0):

22

2

2

12

2

ijii

i j

jii

i j j i

jii

i j j i j

i ii

eDu pg

Dt x x

uupg

x x x x

uupg

x x x x x

pg u

x

),,,( tzyxfAssuming

23

2

22

2

21

222

x

u

x

u

x

u

xx

uu iii

jj

ii

where

~

2

~

~ ugpDt

uD

If “Inviscid” 0

~

~ gpDt

uD Euler Equation

Or in vector notation

Inertia Pressure gradient

Gravity (buoyancy)

Divergence of viscous stress

(friction)