week 09 - ns eqs

8/13/2019 Week 09 - NS Eqs

1/19

11/10/2

Computational Fluid Dynamics

Constitutive Relationship for Viscous Fluids

For Newtonian fluids, such as water, theconstitutive relationship between the stress

tensor and the strain-rate tensor is simple:

Where is a constant called the coefficient of

viscosity.

This constitutive relationship preserves the

defining property of a fluid; that is, that any

small shear stress will produce strain.

10-Nov-13

38

ijij 2


Stress Tensor for Viscous Fluids

Now the stress tensor in terms of strains, and

then in terms of the velocity gradients is:

10-Nov

-13

39

z

v

y

v

z

v

x

v

z

v

y

v

z

v

y

v

x

v

y

v

x

v

z

v

x

v

y

v

x

v

zzyzx

zyyyx

zxyxx

zzzyzx

yzyyyx

xzxyxx

ij

2

1

2

1

2

1

2

1

2

1

2

1

22

z

v

y

v

z

v

x

v

z

v

y

v

z

v

y

v

x

v

y

v

x

v

z

v

x

v

y

v

x

v

zzyzx

zyyyx

zxyxx

2

2

2

8/13/2019 Week 09 - NS Eqs

2/19

11/10/2


Newtons 2ndlaw for Viscous Fluids

We can now substitute this expression for thestress tensor, into the expression for Newtons

2ndLaw (the viscous forces term):

10-Nov-13

40

z

v

y

v

z

v

x

v

z

v

y

v

z

v

y

v

x

v

y

v

x

v

z

v

x

v

y

v

x

v

Pkg

PkgDt

vD

zzyzx

zyyyx

zxyxx

ij

2

2

2


Newtons 2ndlaw for Viscous Fluids

By carrying out the divergence operator on the

stress tensor we get:

10-Nov

-13

41

z

v

y

v

z

v

x

v

z

v

y

v

z

v

y

v

x

v

y

v

x

v

z

v

x

v

y

v

x

v

zzyzx

zyyyx

zxyxx

2

2

2

z

v

zy

v

z

v

yx

v

z

v

x

y

v

z

v

zy

v

yx

v

y

v

x

x

v

z

v

zx

v

y

v

yx

v

x

zzyzx

zyyyx

zxyxx

2

2

2

8/13/2019 Week 09 - NS Eqs

3/19

11/10/2


Newtons 2ndLaw for Viscous Fluids

This can be expanded to produce secondderivatives and cross-derivatives:

10-Nov-13

42

zyv

zxv

yv

xv

zv

zy

v

yx

v

z

v

x

v

y

v

zx

v

yx

v

z

v

y

v

x

v

yxzzz

zxyyy

zyxxx

22

2

2

2

2

2

2

22

2

2

2

2

2

2

22

2

2

2

2

2

2

2

2

2



Finally, we separate the first term in each row

and rearrange as follows:

10-Nov

-13

43

2

222

2

2

2

2

2

2

2

2

22

2

2

2

2

2

2

22

2

2

2

2

2

2

2

2

z

v

zy

v

zx

v

z

v

y

v

x

v

zyv

yv

yxv

zv

yv

xv

zx

v

yx

v

x

v

z

v

y

v

x

v

zyxzzz

zyxyyy

zyxxxx

8/13/2019 Week 09 - NS Eqs

4/19

11/10/2



The last three terms in each row can now beexpressed as the partial derivative of a sum:

The sum in brackets is the divergence of the velocity:

It is zero for incompressible flow.

10-Nov-13

44

z

v

y

v

x

v

zz

v

y

v

x

v

z

v

y

v

x

v

yz

v

y

v

x

v

z

v

y

v

x

v

xz

v

y

v

x

v

zyxzzz

zyxyyy

zyxxxx

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

z

v

y

v

x

vv

zyx



Therefore the last three terms in each row drop

out and we end up with the three components

of the viscous force vector:

10-Nov

-13

45

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

z

v

y

v

x

v

z

v

y

v

x

v

z

v

y

v

x

v

z

v

y

v

x

v

z

v

y

v

x

v

z

v

y

v

x

v

zzz

yyy

xxx

zzz

yyy

xxx

8/13/2019 Week 09 - NS Eqs

5/19

11/10/2



For example, the viscous force per unit volumein thex-direction is:

The term in the parentheses is often

expressed via another vector operator called

the Laplacian defined as:

10-Nov-13

46

2

2

2

2

2

2

z

v

y

v

x

v xxx

2

2

2

2

2

22

z

f

y

f

x

ff



We now have the last piece to the Navier-Stokes

equations.

Substituting the expression for the divergence of the

stress tensor we arrive at the final expression showing

the balance of forces as per Newtons 2ndlaw:

These three equations (one for each direction) are

known as the Navier-Stokes equations.

They apply to incompressible Newtonian fluids that

follow the constitutive relationship:

10-Nov

-13

47

vPkgDt

vD

2

ijij 2

8/13/2019 Week 09 - NS Eqs

6/19

11/10/2



Again, this equation is deceptively concise; itcan be written out as:

10-Nov-13

48

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

z

v

y

v

x

v

z

Pkg

z

vv

y

vv

x

vv

t

v

z

v

y

v

x

v

y

P

z

vv

y

vv

x

vv

t

v

z

v

y

v

x

v

x

P

z

vv

y

vv

x

vv

t

v

zzzz

z

z

y

z

x

z

yyyy

z

y

y

y

x

y

xxxx

z

x

y

x

x

x


Example

Couette Flow

Set up the equations and boundary conditions

to solve for the following problem at steady

state and fully developed:

10-Nov

-13

49

8/13/2019 Week 09 - NS Eqs

7/19

11/10/2


Solution

Only 2D Steady Navier Stokes:

First eliminate all the t and y components.

10-Nov-13

50

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

z

v

y

v

x

v

z

Pkg

z

vv

y

vv

x

vv

t

v

z

v

y

v

x

v

y

P

z

vv

y

vv

x

vv

t

v

z

v

y

v

x

v

x

P

z

vv

y

vv

x

vv

t

v

zzzz

z

z

y

z

x

z

yyyy

z

y

y

y

x

y

xxxx

z

x

y

x

x

x


Next eliminate all the Vz terms due to

impermeability of walls.

Fluid can not pass through walls so there is no

flow in the z direction.

10-Nov

-13

51

2

2

2

2

2

2

2

2

z

v

x

v

z

Pkg

z

vv

x

vv

zv

xv

xP

zvv

xvv

zzzz

zx

xxxz

xx

8/13/2019 Week 09 - NS Eqs

8/19

11/10/2


Solution

If the flow is fully developed the velocity in theflow direction doesnt change.

10-Nov-13

52

z

Pkg

z

v

x

v

x

P

x

vv xxxx

0

2

2

2

2

z

Pkg

z

v

x

P x

2

2


Example

The velocity profile for fully developed laminar

flow between two parallel plates separated by

distance 2b is given by

umaxis at y = 0.

Determine the shear force per unit volume on a

fluid element in the x-direction. Find the maximum value of this quantity for this

flow, when b =1 m, umax=2 m/s and =10-1N s/m2

(SAE-10W).

10-Nov

-13

54

2

2

maxb

y1uu

umaxx

y

8/13/2019 Week 09 - NS Eqs

9/19

11/10/2


Solution

Given:1. Velocity Profile u(y)

At y = 0, u = umax

At y = b, u = 0

Find:

1. Shear Force,

2. The maximum value of the result of the above.

10-Nov-13

55

zy

zxyx

umax


Solution

Governing equation: Naviersequations x-

component, with

Shear force per unit volume along the x-

direction is given by

10-Nov

-13

56

y

u

y

u

x

vyx

0z

u

x

wzx

0 00

2

2

0y

u

zy

u

yzy

zxyx

2max

2

b

yu

y

u

2

max

2

2 2

b

u

y

u

constant22

max

b

u

zy

zxyx

8/13/2019 Week 09 - NS Eqs

10/19

11/10/2


Solution

Maximum value =

N/m3

This value is fixed everywhere in the channel.

10-Nov-13

57

4.)1(

)2()10(2 1


CONSERVATION OF ENERGY

10-Nov

-13

58

8/13/2019 Week 09 - NS Eqs

11/19

11/10/2

Computational Fluid Dynamics 59

Conservation of energy

First law of thermodynamics: rate of change of energy of a fluid particle is equal

to the rate of heat addition plus the rate of workdone.

Rate of increase of energy is DE/Dt.

Energy, E = i + (u2+v2+w2)/2. Here, iis the internal (thermal energy).

(u2+v2+w2)/2 is the kinetic energy.

10-Nov-13



Potential energy (gravitation) is usually treatedseparately and included as a source term.

We will derive the energy equation by settingthe total derivative equal to the change inenergy as a result of work done by viscousstresses and the net heat conduction.

Next we will subtract the kinetic energyequation to arrive at a conservation equationfor the internal energy.

10-Nov

-13

8/13/2019 Week 09 - NS Eqs

12/19

11/10/2


Work done by surface stresses in x-direction:

Conservation of energy10-Nov-13

61

x

z

y

zyx

x

upup

2

1.

zyx

x

upup

2

1.

zyz

z

uu zxzx

2

1.

yxz

z

uu zxzx

2

1.

zxy

y

uu

yx

yx

2

1.

zxy

y

uu

yx

yx

2

1.

zyxx

u

u

xx

xx

2

1

.

zyx

x

uu xxxx

2

1.



The total rate of work done by surface stresses

is calculated as follows:

For work done by x-components of stresses add all

terms in the previous slide.

Do the same for the y- and z-components.

Divide by xyzto get the work done per unitvolume by the surface stresses:

z

w

y

w

x

w

z

v

y

v

x

v

z

u

y

u

x

updiv

zzyzxzzyyy

xyzxyxxx

)()()()()(

)()()()()(

u

10-Nov

-13

8/13/2019 Week 09 - NS Eqs

13/19

11/10/2


Energy flux due to heat conduction

Conservation of energy10-Nov-13

63

x

z

y

yxzz

qq zz )

2

1.(

yxzz

qq zz )

2

1.(

zyxx

qq xx )

2

1.(

zyxx

qq xx )

2

1.(

zxyy

qq

y

y )2

1.(

zxyy

qq

y

y )2

1.(


Conservation of energy Add all terms and divide by xyzgives the net rate of heat transfer to

the fluid particle per unit volume:

Fouriers law of heat conduction relates the heat flux to the localtemperature gradient:

In vector form:

Thus, energy flux due to conduction:

This is the final form used in the energy equation.

qdivz

q

y

q

x

qzyx

z

Tkq

y

Tkq

x

Tkq zyx

Tgradkq

)( Tgradkdivdiv q

10-Nov

-13

8/13/2019 Week 09 - NS Eqs

14/19

11/10/2



Setting the total derivative for the energy in a fluidparticle equal to the previously derived work andenergy flux terms, results in the following energyequation:

Note that we also added a source term SEthat includessources (potential energy, sources due to heatproduction from chemical reactions, etc.).

E

zzyzxzzyyy

xyzxyxxx

STgradkdiv

z

u

y

w

x

w

z

v

y

v

x

v

z

u

y

u

x

updiv

Dt

DE

)(

)()()()()(

)()()()()(

u

10-Nov-13



To derive a conservation equation for the kineticenergy of the fluid multiply the u-momentum equation by u,

the v-momentum equation by v, and

the w-momentum equation by w.

Then add the results together to obtain thefollowing equation for the kinetic energy:

Mzzyzxzzyyyxy

zxyxxx

zyxw

zyxv

zyxupgrad

Dt

wvuD

Su

u

.

.)]([ 222

2

1

10-Nov

-13

8/13/2019 Week 09 - NS Eqs

15/19

11/10/2



Internal energy equation Subtract the kinetic energy equation from the energy

equation.

Define a new source term for the internal energy

Si = SE - u.SM. This results in:

i

zzyzxzzyyy

xyzxyxxx

STgradkdiv

zu

yw

xw

zv

yv

x

v

z

u

y

u

x

udivp

Dt

Di

)(

u

10-Nov-13



Enthalpy equationAn often used alternative form of the energy equation

is the total enthalpy equation. Specific enthalpy h = i + p/.

Total enthalpy h0= h + (u2+v2+w2) = E + p/.

0

0

( )

( ) ( )

( ) ( )( ) ( )

( ) ( ) ( )( ) ( )

yx xyxx zx

yy zy yzxz zz

h

h

div h div k grad T t

u vu u

x y z x

v v ww u

y z x y z

S

u

10-Nov

-13

8/13/2019 Week 09 - NS Eqs

16/19

11/10/2


Equations of state

Fluid motion is described by five partial differential equations for mass,momentum, and energy.

Amongst the unknowns are four thermodynamic variables: ,p, i, andT.

We will assume thermodynamic equilibrium, i.e. that the time it takes

for a fluid particle to adjust to new conditions is short relative to the

timescale of the flow.

We add two equations of state using the two state variables and T:p=p(,T) and i=i(,T).

For a perfect gas, these become: p=RTand i=CvT.

At low speeds (e.g. Ma < 0.3), the fluids can be consideredincompressible. There is no linkage between the energy equation, and

the mass and momentum equation. We then only need to solve for

energy if the problem involves heat transfer.

10-Nov-13


Viscous stresses

A model for the viscous stresses ijis required.

We will express the viscous stresses as functions of the

local deformation rate (strain rate) tensor.

There are two types of deformation:

Linear deformation rates due to velocity gradients.

Elongating stress components (stretching).

Shearing stress components. Volumetric deformation rates due to expansion or compression.

All gases and most fluids are isotropic: viscosity is a scalar.

Some fluids have anisotropic viscous stress properties,

such as certain polymers and dough. We will not discuss

those here.

10-Nov

-13

8/13/2019 Week 09 - NS Eqs

17/19

11/10/2


Viscous dissipation

Substituting the stresses in the internal energy equationand rearranging results as follows:

Here Fis the viscous dissipation term. This term isalways positive and describes the conversion ofmechanical energy to heat.

iSTgradkdivdivpidivt

ienergyInternal F

)()(

)(: uu

2

22

2222

)(3

2

2

udivy

w

z

v

x

w

z

u

x

v

y

u

z

w

y

v

x

u

F

10-Nov-13


SUMMARY

10-Nov

-13

75

8/13/2019 Week 09 - NS Eqs

18/19

11/10/2


Equations in conservation form

0)(: u div

tMass

MxSugraddivx

pudiv

t

umomentumx

)()(

)(:

u

MySvgraddivy

pvdiv

t

vmomentumy

)()(

)(:

u

MzSwgraddivz

pwdiv

t

wmomentumz

)()(

)(:

u

iSTgradkdivdivpidivtienergyInternal F

)()()(: uu

TCiandRTpgasperf ectforge

TiiandTppstateofEquations

v

:..

),(),(:

10-Nov-13


The system of equations is now closed, with seven equations for seven

variables: pressure, three velocity components, enthalpy, temperature, and

density.

There are significant commonalities between the various equations. Using a

general variable , the conservative form of all fluid flow equations can usefully

be written in the following form:

Or, in words:

General transport equations

Sgraddivdivt

u

Rate of increase

of of fluid

elementNet rate of flow

of out of

fluid element

(convection)

Rate of increase

of due to

diffusion

Rate of increase

of due to

sources=+ +

10-Nov

-13

8/13/2019 Week 09 - NS Eqs

19/19

11/10/2


Integral form

The key step of the finite volume method is to integrate the differential equationshown in the previous slide, and then to apply Gauss divergence theorem,which for a vector astates:

This then leads to the following general conservation equation in integral form:

This is the actual form of the conservation equations solved by finite volumebased CFD programs to calculate the flow pattern and associated scalar fields.

ACV

dAdVdiv ana

dVSdAgraddAdVt CVAACV

)()( nun

Rate of

increase

of

Net rate of

decrease of due

to convectionacross boundaries

Net rate of

increase of due

to diffusionacross boundaries

Net rate of

creation

of =+ +

10-Nov-13

week 09 - ns eqs

Documents