Turbulence Reynolds Averaged Numerical Simulations …gtryggva/CFD-Course2010/2010-Lecture-18.pdf · A wire hoop ahead of the equator ... Reynolds Averaged Numerical Simulations!

Computational Fluid Dynamics I

Turbulence Modeling—I!

Grétar Tryggvason!Spring 2010!


Why turbulence modeling!Reynolds Averaged Numerical Simulations!Zero and One equation models!Two equations models!Model predictions!Wall bounded turbulence!Second order closure!Direct Numerical Simulations!Large-eddy simulations!Summary!

Outline!


Most engineering problems involve turbulent flows. Such flows involve are highly unsteady and contain a large range of scales. However, in most cases the mean or average motion is well defined. !


Above: A jet in a cross flow!

Left: cross section of a jet!


Flow over a sphere!

The drag depends on the separation point!


Dye in water shows a laminar boundary layer separating ahead of the equator and remains laminar for almost one radius. It then becomes unstable and quickly turns turbulent. ONERA photograph, Werle 1980. From "An Album of Fluid Motion," by Van Dyke, Parabolic Press. !

Instantaneous flow past a sphere at R = 15,000.!


The classical experiment of Prandtl and Wieselsberger is repeated here, using air bubbles in water. A wire hoop ahead of the equator trips the boundary layer. It becomes turbulent, so that it separates farther rearward than if it were laminar (compare with top photograph). The overall drag is thereby dramatically reduced, in a way that occurs naturally on a smooth sphere only at a Reynolds numbers ten times as great. ONERA photograph, Werle 1980.

Instantaneous flow past a sphere at R = 30,000 with a trip wire


Examples of Reynolds numbers:!

Flow around a 3 m long car at 100 km/hr:!

Flow around a 100 m long submarine at 10 km/hr:!

�

Re = LUv

= 3× 27.781.5 ×10−5

= 5.5 ×106

Kinematic viscosity !(~20 °C)!

Water ν = 10-6 m2/s!Air ν = 1.5 ✕10-5 m2/s!

1km/hr = 0.27778 m/s!

�

Re = LUv

= 100 × 2.7810−6

= 2.78 ×108

Water flowing though a 0.01 m diameter pipe with a velocity of 1 m/s!

�

Re = LUv

= 0.01×110−6

=104


It can be shown that for turbulent flow the ratio of the size of the smallest eddy to the length scale of the problem!

�

δL≈O(Re−3 / 4 )

If about 10 grid points are needed for Re=10 (the driven cavity problem) !

Re ! 3d ! ! 2d!103!~ 3003 ! ~ 1002!104! ~ 20003 ! ~ 3002!105!~ 100003 ! ~ 10002!

Largest computations today use about 40003 points!

�

δL≈O(Re−1/ 2)

In 3D! In 2D!


Reynolds Averaged Navier-Stokes (RANS): Only the averaged motion is computed. The effect of fluctuations is modeled!Large Eddy Simulations (LES): Large scale motion is fully resolved but small scale motion is modeled!

Direct Numerical Simulations (DNS): Every length and time scale is fully resolved!


Reynolds Averaged

Navier-Stokes Equations!


To solve for the mean motion, we derive equations for the mean motion by averaging the Navier-Stokes equations. The velocities and other quantities are decomposed into the average and the fluctuation part !

a = A + a'


< a > = A< a' > = 0< a + b > = A+ B< ca > = cA< ∇a > =∇A

a = A + a'Defining an averaging procedure that satisfies the following rules:!

This will hold for spatial averaging, temporal averaging, and ensamble averaging!


There are several ways to define the proper averages !

For homogeneous turbulence we can use the space average !

For steady turbulence flow we can use the time average!

For the general case we use the ensemble average!

< a > = 1Ladx

0

L

∫

�

< a > = 1T

adt0

T

∫

< a > = ar (x,t)ensambles∑ a = A + a'


�

∂∂tu + ∇ ⋅uu = − 1

ρ∇p+ ν∇2u

u =U + u'p = P+ p'

< a >=A< a' >= 0< ca >=cA< ∇a >=∇A

Start with the Navier-Stokes equations!

Decompose the pressure and velocity into mean and fluctuations:!

a = A + a'

Or, in general, for any dependant variable:!


�

∂∂tU + ∇ ⋅UU = - 1ρ ∇P + ν∇2U + ∇⋅ < u'u'>

Applying the averaging to the Navier-Stokes equations results in:!

�

< u'u'>=< u'u'> < u'v'> < u'w'>< u'v'> < v 'v'> < v 'w'>

< u'w'> < v 'w'> < w'w'>

⎡

⎣

⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥

Reynoldʼs stress tensor!


Physical interpretation!

< uv >

Fast moving fluid particle!

Slow moving fluid particle!

Net momentum transfer due to velocity fluctuations!


Closure:!

Since we only have an equation for the mean flow, the Reynolds stresses must be related to the mean flow. !

No rigorous process exists for doing this!!

THE TURBULENCE PROBLEM!


Turbulence Modeling—II!




Outline!


Zero and One equation models!


Introduce the “turbulent eddy viscosity”!

�

νT = l02

t0

�

< u'u'>ij = −νT∂Ui

∂x j

+ ∂Ui

∂x j

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

where!


Zero equation models!

�

νT = l02 dUdy

Prandtlʼ mixing length!

�

l0 = κy

Smagorinsky model!

Baldvin-Lomaz model!

�

νT = l02 2S ijS ij( )1/ 2

�

νT = l02 ω iω i( )1/ 2

�

S ij = 12

∂Ui

∂x j

+∂U j

∂xi

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

�

ω i = ∂Ui

∂x j

−∂U j

∂xi

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟


One equation models!

�

νT = k1/ 2t0

Where k is obtained by an equation describing its temporal-spatial evolution!

However, the problem with zero and one equation models is that t0 and l0 are not universal. Generally, it is found that a two equation model is the minimum needed for a proper description !


Two equation models!


To characterize the turbulence it seems reasonable to start with a measure of the magnitude of the velocity fluctuations. If the turbulence is isotropic, the turbulent kinetic energy can be used:!

�

k = 12

< u'u'> + < v 'v'> + < w'w'>( )The turbulent kinetic energy does, however, not distinguish between large and small eddies.!


To distinguish between large and small eddies we need to introduce a new quantity that describe!

�

ε ≡ ν ∂u'i∂u'i∂x j∂x j

Usually, the turbulent dissipation rate is used!

Smaller eddies dissipate faster!


�

νT = Cµk 2

ε

�

∂∂tU + ∇ ⋅UU = − 1

ρ∇P + ν + νT( )∇2U

Solve for the average velocity!

Where the turbulent kinematic eddy viscosity is given by!


�

∂k∂t

+U j∂k∂x j

= τ ij∂Ui

∂x j

−ε + ∂∂x j

ν ∂k∂x j

− 12ui'ui'u j' − 1

ρp'u j

'⎡

⎣ ⎢

⎤

⎦ ⎥

The exact k-equation is:!

where!

�

τ ij = − ui'u j'

The exact epsilon-equation is considerably more complex and we will not write it down here.!

Both equations contain transport, dissipation and production terms that must be modeled!


∂k∂t

+U ⋅∇k = ∇ ⋅Dk∇k + production − dissipation

∂ε∂t

+U ⋅∇ε = ∇ ⋅Dε∇ε + production − dissipation

The general for for the equations for k and epsilon is:!

These terms must be modeled !Closure involves proposing a form for the missing terms and optimizing free coefficients to fit experimental data!


Here!

The k-epsilon model!

�

νT = C k 2

ε

�

τ ij =< ui'u j' >= 2

3kδij −νT

∂Ui

∂x j

+∂U j

∂xi

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ and!

�

C1 = 0.09; C2 =1.0; C3 = 0.769; C4 =1.44; C5 =1.92

�

DkDt= +∇ ⋅ (ν + C2νT )∇k - τ ij

∂Ui

∂x j

−ε

�

DεDt= ∇ ⋅ (ν + C3νT )∇ε + C4

εkτ ij

∂Ui

∂x j

−C5ε2

kProduction! Dissipation!

Turbulent!transport!


Two major numerical difficulties!

The equations may be stiff in some regions of the flow requiring very small time step. This can be overcome by an implicit scheme.!

In reality k goes to zero at the walls. In simulations this usually takes place so close to the wall that it is not resolved by the grid. To overcome this we usually use a “wall function” or a damping function!


Other two equation turbulence models:!!RNG k-epsilon!!Nonlinear k-epsilon!!k-enstrophy!!k-lo!!k-reciprocal time!!etc!


Turbulent transport of energy and species concentrations is modeled in similar ways.!

For temperature we have:!

�

∂T∂t

+ ∇ ⋅uT = α∇2T

�

u = U + u'T =< T > +T'

�

∂ < T >∂t

+ ∇ ⋅U < T >= α∇2 < T > −∇⋅ <UT >

Gradient Transport Hypothesis:!

�

<UT >≈αT∇ < T >


Model Predictions!


Spreading rates:!

! ! exp ! k-e ! !Cmott!Plane jet !0.10 - 0.11 ! 0.108! !0.102!Round jet !0.085-0.095 ! 0.116! !0.095!Mixing layer !0.13 - 0.17 ! 0.152! !0.154!


From: C.G. Speziale: Analytical Methods for the Development of Reynolds-stress closure in Turbulence. Ann Rev. Fluid Mech. 1991. 23: 107-157!




Results!



Wall bounded turbulence!


Wall bounded turbulence!

Fundamental assumption: determined by local variables only!

Mean flow!

Only the mean shear rate and the properties of the fluid are important!

�

τw = dUdy, ρ, ν


Define a “shear velocity:”!

�

v* = τwρ

�

τw = µ du dy, ρ, ν

kg /ms2[ ], kg /m3[ ], m2 /s[ ]

Normalize the length and velocity near the wall!

�

u+ = uv*

y + = y v*

v

Called “wall variables”!


For parallel flow!

�

v* = τw

ρ

τw = µ du dy

u+ = uv*

y + = y v*

v�

ρ ddy

< u'v '>= − dpdx

+ ddy

µ dudy

�

ρ ddy

< u'v '>= ddy

µ dudy

⎛

⎝ ⎜

⎞

⎠ ⎟ 0

y∫ dy

�

ρ < u'v '>= µ dudy

− τw

Integrate from the wall to y:!

Resulting in:!

0! Near the wall the fluid “knows nothing” about what drives it. Thus we ignore the pressure gradient!


Very close to the wall:!

�

v* = τw

ρ

τw = µ du dy

u+ = uv*

y + = y v*

v

�

< u'v '>≈ 0

so approximately!

Integrating!

�

µ dudy

= τw

�

u(y) = τwµy

Using the nondimensional values!

�

uv*

= τwµ

yv*

=ρ v*( )2

µyv*

= v*yν

or:!

�

u+ = y + Very close to the wall!

�

ρ < u'v '>= µ dudy

− τw


Further away from the wall:!

�

v* = τw

ρ

τw = µ du dy

u+ = uv*

y + = y v*

v�

< u'v '>=νTdudy

so approximately!

Taking:!

Giving:!

�

µ dudy

≈ 0

�

ρ < u'v '>= −τw

�

κy dudy

= −τwρ

�

l0 = κy

yields!

�

u'≈ y dudy

Or simply assume:!

�

< u'v '>= lo2 dudy

dudy

= κy dudy

⎛

⎝ ⎜

⎞

⎠ ⎟ 2

�

νT = lo2 dudywhere! and!


�

v* = τw

ρ

τw = µ du dy

u+ = uv*

y + = y v*

v

Using the nondimensional values!

�

κy dudy

= −τwρ

�

κy + du+

dy + = −τwv

*

ρ=1

integrating!

�

du+ = 1κ∫ dy +

y +∫

�

u+ = 1κln y + + C

We have!

giving!


Thus, the velocity near the wall is!

Velocity versus distance from wall!

�

u+

�

y +

�

u+ = y +

�

u+ = 1κln y + + C

10!

�

κ = 0.4C = 5.5

�

v* = τw

ρ

τw = µ du dy

u+ = uv*

y + = y v*

v

Buffer layer!

Outer layer!

Viscous sub-layer!


For a “practical” engineering problem!

L = 1m; U = 1m/s; ν = 10-6 (water)!

The Reynolds number is therefore:!

For a flat plate, the average drag coefficient is!

�

Re = LUv

=106

�

CD = FD12 ρU

2LW

�

CD = 0.592Re−1/ 5 where!


The average shear stress is therefore!

�

τw = FDLW

= CD12 ρU

2 = 3.74

And we find!

�

v* = 0.06The average thickness of the viscous sub-layer is 10 in units of y+:!

�

CD = 0.0037

Or!


�

y = 10νν *

= 10 ×10−6

0.06=1.667 ×10−4m = 0.1667 mm

Thickness of the viscous sub-layer!

Find the thickness of the boundary layer!

�

δL

= 0.37Re−1/ 5

�

δL

= 0.0233m = 23.3mm

To resolve the viscous sublayer at the same time as the turbulent boundary layer would require a large number of grid points!


To deal with this problem it is common to use “wall functions” where the mean velocity is matched with an analytical approximation to the viscosus sublayer.!

For a reference, see: Patel, Rodi, and Scheuerer, “Turbulence Models for Near-Wall and Low Reynolds Number Flows: A Review. AIAA Journal, 23, page 1308!


Second order closure!


The k-epsilon and other two equation models have several serious limitations, including the inability to predict anisotropic Reynolds stress tensors, relaxation effects, and nonlocal effects due to turbulent diffusion.!

For these problems it is necessary to model the evolution of the full Reynolds stress tensor!


Derive equations for the Reynolds stresses:!

�

∂ui∂t

+ ∇uiu j = - 1ρ ∇p+ ν∇2ui

The Navier-Stokes equations in component form:!

�

ui∂ui∂t

+ ∇uiu j = - 1ρ ∇p+ ν∇2ui⎛ ⎝ ⎜

⎞ ⎠ ⎟

Multiply the equation by the velocity!

and averaging leds to equations for !

�

∂∂t

uiu j


The new equations contain terms like!

which are not known. These terms are therefore modeled!

�

uiuiu j

The Reynolds stress model introduces 6 new equations (instead of 2 for the k-e model. Although the models have considerably more physics build in and allow, for example, anisotrophy in the Reynolds stress tensor, these model have yet to be optimized to the point that they consistently give superior results.!

For practical problems, the k-e model or more recent improvements such as RNG are therefore most commonly used!!


Turbulence Modeling—III!




Outline!


Direct Numerical Simulations!


In direct numerical simulations the full unsteady Navier-Stokes equations are solved on a sufficiently fine grid so that all length and time scales are fully resolved. The “size” of the problem is therefore very limited. The goal of such simulations is to provide both insight and quantitative data for turbulence modeling!


Channel Flow!

Streamwise velocity!

Flow direction!

Periodic streamwise and spanwise boundaries!

Wall!


Streamwise vorticity!


Streamwise vorticity!

Turbulent shear stress!

Turbulent eddies generate a nearly uniform velocity profile!

Channel Flow!


Turbulence are intrinsically linked to vorticity, yet laminar flows can also be vortical so looking at the vorticity is not sufficient to understand what is going on in a turbulent flows. Several attempts have been made to define properties of the turbulent flows that identifies vortices (as opposed to simply vortical flows.!

One of the most successful method is the lambda-2 method of Hussain.!


Visualizing turbulence!

�

∇u =

∂u∂x

∂u∂y

∂u∂z

∂v∂x

∂v∂y

∂v∂z

∂w∂x

∂w∂y

∂w∂z

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ �

S = 12∇u +∇Tu( ) = 12

2∂u∂x

∂u∂y

+ ∂v∂x

∂u∂z

+ ∂w∂x

∂v∂x

+ ∂u∂y

2∂v∂y

∂v∂z

+ ∂w∂y

∂w∂x

+ ∂u∂z

∂w∂y

+ ∂v∂z

2∂w∂z

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥

�

Ω = 12∇u -∇Tu( ) = 12

0 ∂u∂y

− ∂v∂x

∂u∂z

− ∂w∂x

∂v∂x

− ∂u∂y

0 ∂v∂z

− ∂w∂y

∂w∂x

− ∂u∂z

∂w∂y

− ∂v∂z

0

⎡

⎣

⎢ ⎢ ⎢ ⎢ ⎢ ⎢

⎤

⎦

⎥ ⎥ ⎥ ⎥ ⎥ ⎥


It can be shown that the second eigenvalue of !

�

S2 + Ω2

define vortex structures!Referece: J. Jeong and F. Hussain, "On the identification of a vortex," Journal of Fluid Mechanics, Vol. 285, 69-94, 1995.!

Other quantities have also been used, such as the second invariant of the velocity gradient:!

�

Q = ∂ui∂x j

∂u j

∂xi

�

λ2


�

λ2 = −0.3

�

λ2 = −0.2


Large Eddy Simulations!


Unsteady simulations where the large scale motion is resolved but the small scale motion is modeled. Frequently simple models are used for the small scale motion. Most recently some success has been achieved by “intrinsic” large eddy simulations where no modeling is used but monotonicity is enforced by the methods described in the lectures on hyperbolic methods !


In the simplest case, the Smagorinsky eddy viscosity is used in simulation of unsteady flow, thus resulting in a viscosity that depends of the flow. !

�

νT = l02 2S ijS ij( )1/ 2

�

S ij = 12

∂Ui

∂x j

+∂U j

∂xi

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

Since the viscosity increases, the size of the smallest flow scales increases and lower resolution is needed !


Summary!


Turbulence models are used to allow us to simulate only the averaged motion, not the unsteady small scale motion.!Turbulence modeling rest on the assumption that the small scale motion is “universal” and can be described in terms of the large scale motion.!

Although considerable progress has been made, much is still not known and results from calculations using such models have to be interpreted by care!!


For more information:!

D. C. Wilcox, Turbulence Modeling for CFD (2nd ed. 1998). !

The author is one of the inventors of the k-ω model and the book promotes it use. The discussion is, however, general and very accessible, as well as focused on the use of turbulence modeling for practical applications in CFD!

Documents

Turbulence Reynolds Averaged Numerical Simulations …gtryggva/CFD-Course2010/2010-Lecture-18.pdf · A wire hoop ahead of the equator ... Reynolds Averaged Numerical Simulations!