Chapter 3: Turbulence and its modeling - Eastern ...me.emu.edu.tr/sezai/ME555/Chapter 3.pdf1 Chapter...

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Chapter 3:Turbulence and its modeling

Ibrahim SezaiDepartment of Mechanical Engineering

Eastern Mediterranean University

Fall 2009-2010

I. Sezai – Eastern Mediterranean UniversityME555 : Computational Fluid Dynamics 2

IntroductionFor Re > Recritical => Flow becomes Turbulent.Chaotic and random state of motion develops.Velocity and pressure change continuously with timeThe velocity fluctuations give rise to additional stresses on the fluid

=> these are called Reynolds stressesWe will try to model these extra stress terms

2


3.1 What is turbulence?

( ) ( )u t U u t′= +


In turbulent flows there are rotational flow structures called turbulent eddies.

Have a wide range of length scales.

Notice the eddies with large, intermediate and small size

3


In turbulent flow:A streak of dye which is introduced at a point will rapidly

break up and dispersed effective mixing

Give rise to high values of diffusion coefficient for;- Mass- Momentum- and heat

All fluctuating components contain energy across a wide range of frequencies or wave numbers

2 fwave number f frequencyUπ

= =


Fig.3.3 Energy Spectrum of turbulence behind a grid

4


Large Eddies:- have large eddy Reynolds number,- are dominated by inertia effects - viscous effects are negligible- are effectively inviscidSmall Eddies:- motion is dictated by viscosity- Re ≈ 1- length scales : 0.1 – 0.01 mm- frequencies : ≈ 10 kHzEnergy associated with eddy motions is dissipated and converted into thermal internal energy

increases energy losses.- Larges eddies anisotropic- Smallest eddies isotropic

vlυ


3.2 Transition from laminar to turbulent flowCan be explained by considering the stability of laminar flows to small disturbances

Hydrodynamic instabilityHydrodynamic stability of laminar flows;

a) Inviscid instability: flows with velocity profile having a point of inflection.

1. jet flows2. mixing layers and wakes3. boundary layers with adverse pressure gradients

b) Viscous instability: flows with laminar profile having no point of inflection

- occurs near solid walls

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Fig. 3.4 Velocity profiles susceptible to

a) Inviscid instability and

b) Viscous instability


Transition to TurbulenceJet flow: (example of a flow with point of inflection in velocity

profile)

Fig. 3.5. transition in a jet flow

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Boundary layers on a flat plate (Example with no inflection point in the velocity profile)

The unstable two-dimensional disturbances are called Tolmien-Schlichting (T-S) waves

Fig. 3.6 plan view sketch of transition processes in boundary layer flow over a flat plate


Fig. 3.7 merging of turbulent spots and transition to turbulence in a natural flat plate boundary layer

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Common features in the transition process:(i) The amplification of initially small disturbances(ii) The development of areas with concentrated rotational

structures(iii) The formation of intense small scale motions(iv) The growth and merging of these areas of small scale motions

into fully turbulent flowsTransition to turbulence is strongly affected by:- Pressure gradient- Disturbance levels- Wall roughness- Heat transferThe transition region often comprises only a very small fraction of

the flow domainCommercial CFD packages often ignore transition entirely (classify the flow as only laminar or turbulent)


3.3Effect of turbulence on time-averaged Navier-Stokes eqns

In turbulent flow there are eddying motions of a wide range of length scalesA domain of 0.1x 0.1m contains smallest eddies of 10-100 μm sizeWe need mesh pointsThe frequency of fastest events ≈ 10 kHz Δt≈100μs neededDNS of turbulent pipe flow of Re = 105 requires a computer which is 10 million times faster than CRAY supercomputerEngineers need only time-averaged properties of the flowLet’s see how turbulent fluctuations effect the mean flow properties

9 1210 10−

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Reynolds EquationsFirst we define the mean Φ of a flow property φ as follows:

The property φ can be thought of as the sum of a steady mean component Φ and fluctuation component φ´

The time average of the fluctuations φ´ is , by definition, zero:

Information regarding the fluctuation part of the flow can, for example, be obtained from the root-mean-square (rms) of the fluctuations:

0

1 ( )t

t dtt

ϕΔ

Φ =Δ ∫ (3.1)

( ) ( )t tϕ ϕ′= Φ +

0

1 ( ) 0t

t dtt

ϕ ϕΔ

′ ′= ≡Δ ∫ (3.2)

( ) ( )1/ 2

2 2

0

1 t

rms dtt

ϕ ϕ ϕΔ⎡ ⎤

′ ′= = ⎢ ⎥Δ⎣ ⎦∫ (3.3)


The rms values of velocity components can be measured (e.g. by hot-wire anemometer)

The kinetic energy k (per unit mass) associated with the turbulence is defined as

The turbulence intensity Ti is linked to the kinetic energy and a reference mean flow velocity Uref as follows:

Rules which govern the time averages of fluctuation properties and :

.

(3.4)2 2 21 ( )2

k u v w′ ′ ′= + +

1/ 2(2 / 3 )i

ref

kTU

= (3.5)

0; ; ;

; ; ; 0

ds dss sϕϕ ψ ϕ

ϕ ψ ϕψ ϕ ψ ϕ ϕ

∂ ∂Φ′ ′= = Φ = Φ = = Φ∂ ∂

′ ′ ′+ = Φ +Ψ = ΦΨ + Ψ = ΦΨ Ψ =

∫ ∫(3.6)

ϕ ϕ′= Φ + ψ ψ ′= Ψ +

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Since div and grad are both differentiations the above rules can be extended to a fluctuating vector quantity a = A + a´ and its combinations with a fluctuating scalar :

To illustrate the influence of turbulent fluctuations on mean flow we consider

Substitute

( ) ( ) ( ) ( ); ;div div div div div div

div grad div grad

ϕ ϕ ϕ

ϕ

′ ′= = = ΦΑ +

= Φ

a A a a a (3.7)

( )

( )

( )

01

1

1

divu pdiv u v div grad ut xv pdiv v v div grad vt yw pdiv w v div grad wt z

ρ

ρ

ρ

=∂ ∂

+ = − +∂ ∂∂ ∂

+ = − +∂ ∂∂ ∂

+ = − +∂ ∂

u

u

u

u

(3.9a)

(3.8)

(3.9b)

(3.9c)

; ; ; ;u U u v V v w W w p P p′ ′ ′ ′ ′= + = + = + = + = +u U u

ϕ ϕ′= Φ +


Consider continuity equation: note that this yields the continuity equation

A similar process is now carried out on the x-momentum equation (3.9a). The time averages of the individual terms in this equation can be written as follows:

Substitution of these results gives the time-average x-momentum equation

( ) ( ); ( )

1 1 ;

u U div u div U div ut t

p P vdiv grad u v div grad Ux xρ ρ

∂ ∂ ′ ′= = +∂ ∂

∂ ∂− = − =

∂ ∂

u U u

0div =U (3.10)

div div=u U

1( ) ( )

( ) ( ) ( ) ( ) ( )

U Pdiv U div u v div grad Ut xI II III IV V

ρ∂ ∂′ ′+ + = − +∂ ∂

U u (3.11a)

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Repetition of this process on equations (3.9b) and (3.9c) yields the time-average y-momentum and z-momentum equations

The process of time averaging has introduced new terms (III).Their role is additional turbulent stresses on the mean velocitycomponents U, V and W.

1( ) ( )

( ) ( ) ( ) ( ) ( )1( ) ( )

( ) ( ) ( ) ( ) ( )

V Pdiv V div v v div grad Vt yI II III IV VW Pdiv W div w v div grad Wt zI II III IV V

ρ

ρ

∂ ∂′ ′+ + = − +∂ ∂

∂ ∂′ ′+ + = − +∂ ∂

U u

U u

(3.11b)

(3.11c)


Placing these terms on the rhs:

21( )U P u u v u wdiv U v div grad Ut x x y zρ

⎡ ⎤′ ′ ′ ′ ′∂ ∂ ∂ ∂ ∂+ = − + + − − −⎢ ⎥

∂ ∂ ∂ ∂ ∂⎢ ⎥⎣ ⎦U

(3.12a)21( )V P u v v v wdiv V v div grad V

t y x y zρ

⎡ ⎤′ ′ ′ ′ ′∂ ∂ ∂ ∂ ∂+ = − + + − − −⎢ ⎥

∂ ∂ ∂ ∂ ∂⎢ ⎥⎣ ⎦U

(3.12b)21( )W P u w v w wdiv W v div grad W

t z x y zρ

⎡ ⎤′ ′ ′ ′ ′∂ ∂ ∂ ∂ ∂+ = − + + − − −⎢ ⎥

∂ ∂ ∂ ∂ ∂⎢ ⎥⎣ ⎦U

(3.12c)

Reynolds equations

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The extra stress terms result from six additional stresses, three normal stresses and three shear stresses:

-These extra turbulent stresses are termed the Reynolds stresses.-The normal stresses are always non-zero-The shear stresses are also non-zeroIf, for instant, u´ and v´ were statistically independent fluctuations the time

average of their product would be zeroSimilar extra turbulent transport terms arise when we derive a transport

equation for an arbitrary scalar quantity. The time average transport equation for scalar φ is

2 2 2xx yy zz

xy yx xz zx yz zy

u v w

u v u w v w

τ ρ τ ρ τ ρ

τ τ ρ τ τ ρ τ τ ρ

′ ′ ′= − = − = −

′ ′ ′ ′ ′ ′= = − = = − = = −(3.13)

*( ) ( ) u v wdiv div grad St x y z

ϕ ϕ ϕΦ Φ

⎡ ⎤′ ′ ′ ′ ′ ′∂Φ ∂ ∂ ∂+ Φ = Γ Φ + − − − +⎢ ⎥∂ ∂ ∂ ∂⎣ ⎦

U (3.14)

2 2 2,u v and wρ ρ ρ′ ′ ′− − −,u v u w and v wρ ρ ρ′ ′ ′ ′ ′ ′− − −

u v′ ′


In compressible flow density fluctuations are usually negligibleThe density-weighted averaged (or Favre-averaged) form of the compressible turbulent flow are:

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Closure problem – the need for turbulence modeling

In the continuity and Navier – Stokes equations (3.8) and (3.9 a-c) there are 6 additional unknowns

the Reynolds stressesSimilarly in scalar transport eqn (3.14) there are 3

extra unknowns:

Main task of turbulence modeling:is to develop computational procedures to predict the 9 extra terms. (Reynolds stresses and scalar transport terms).

,u v and wϕ ϕ ϕ′ ′ ′ ′ ′ ′


3.4. Characteristics of simple turbulent flowsIn turbulent thin shear layers,

The following 2D incompressible turbulent flows with constant P will be considered:Free turbulent flows- Mixing layers- Jet- Wake

Boundary layer near solid walls- Flat plate boundary layer- Pipe flow

Data for U, will be reviewed.These can be measured by 1) Hot wire anemometry 2) Laser doppler

anemometers

1

x y

Lδ

∂ ∂∂ ∂

2 2 2, ,u v w and u vρ ρ ρ ρ′ ′ ′ ′ ′− − − −

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3.4.1 Free turbulent flows

b= cross sectional layer width, y=distance in cross-stream directionx=distance downstream the source

maxmin

max max min max min

for jets for mixing layers for wakes

U UU UU y y yg f hU b U U b U U b

−−⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠ ⎝ ⎠


Eddies of a wide range of length scales are visibleFluid from surroundings is entrained into the jet.

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If x is large enough functions f, g and h are independent of x. such flows are called self – preserving.

The turbulence structure also reaches a self-preserving state albeit after a greater distance from the flow source than the mean velocity. Then

Uref= Umax – Umin for a mixing layer and wakesUref= Umax for jets

Form of functions f, g, h and fi varies from flow to flowSee Fig.3.10

2 2 2

1 2 3 42 2 2 2ref ref ref ref

u y v y w y u v yf f f fU b U b U b U b′ ′ ′ ′ ′⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠


Fig. 3.10 mean velocity distributions and turbulence properties for

(a) two- dimensional mixing layer, (b) planar turbulent jet and (c) wake behind a solid strip

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In Fig.3.10:are maximum when is maximum

u´ gives the largest of the normal stresses (v´and w´) .

Fluctuating velocities are not equal anisotropic structure of turbulenceAs mean velocity gradients tend to zero turbulence quantities tend to zero turbulence can’t be sustained in absence of shearThe mean velocity gradient is also zero at the centerline of jets and wakes. No turbulence there.The value of is zero at the centerline of a jet and wake since shear stress must change sign here.

uy∂∂

2 2 2,u v and w and u v′ ′ ′ ′ ′−

2

max0.15 0.40u

u′= −

u v′ ′−


3.4.2 Flat plate boundary and pipe flow

y: distance away from the wallNear the wall (y small) is small viscous forces dominateAway from the wall (y large) is large inertia forces dominate.Near the wall U only depends on y, ρ, μ and τ (wall shear stress), so

Dimensional analysis shows that

Formula (3.18) is called the law of the wall

Re inertia forcesviscous forces

=

Re Uyv

=

Re y

Re y

( , , , )wU f y ρ μ τ=

( )u yUu f f yu

τ

τ

ρμ

+ +⎛ ⎞= = =⎜ ⎟

⎝ ⎠(3.18)

( )1/ 2wu friction velocityτ τ ρ=

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Far away from the wall:

Dimensional analysis yields

The most useful form emerges if we view the wall shear stress as the cause of a velocity deficit Umax – U which decreases the closer we get to the edge of the boundary layer or the pipe centerline. Thus

This formula is called the velocity – defect law.

( , , , )wU g y independent ofδ ρ τ μ=

U yu guτ δ

+ ⎛ ⎞= = ⎜ ⎟⎝ ⎠

maxU U yguτ δ− ⎛ ⎞= ⎜ ⎟

⎝ ⎠(3.19)


Linear sub layer – the fluid layer in contact with smooth wall

Very near the wall there is no turbulent (Reynolds) shear stresses flow is dominated by viscous shear

For shear stress is approximately constant,

Integrating and using U = 0 at y = 0,

After some simple algebra and making use of the definitions of u+ and y+ this leads to

Because of the linear relationship between velocity and distancefrom the wall the fluid layer adjacent to the wall is often known as the linear sub-layer

( 5)y+ <

( ) wUyy

τ μ τ∂= ≅

∂

wyU τμ

=

u y+ += (3.20)

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Log-law layer – the turbulent region close to smooth wall

Outside the viscous sublayer (30<y+<500) a region exists where viscous and turbulent effects are both important.

The shear stress τ varies slowly with distance; it is assumed to be constant and equal to τw

Assuming a length scale of turbulence lm= κy where lm is mixing length, the following relationship can be derived:

κ=0.4, B=5.5, E=9.8; (for smooth walls)B decreases with roughnessκ and B are universal constants valid for all turbulent flows past smooth

walls at high Reynolds numbers.(30<y+<500) log – law layer

1 1ln ln( )u y B Eyκ κ

+ + += + = (3.21)


Outer layer – the inertia dominated region far from the wall

Experimental measurement show that the log-law is valid in the region 0.02 < y ⁄ δ < 0.2.

For larger values of y the velocity-defect law (3.19) provides the correct form.

In the overlap region the log-law and velocity-defect law have to become equal.

Tennekes and Lumley (1972) show that a matched overlap is obtained by assuming the following logarithmic form:

where A is constantSee Fig.3.11

max 1 lnU U y A Law of the Wakeuτ κ δ− ⎛ ⎞= +⎜ ⎟

⎝ ⎠(3.22)

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- The inner region: 10 to 20% of the total thickness of the wall layer; the shear stress is (almost) constant and equal to the wall shear stress τw . Within this region there are three zones

- the linear sub-layer: viscous stresses dominate the flow adjacent to the surface

- the buffer layer: viscous and turbulent stresses are of similar magnitude- the log-law layer: turbulent (Reynolds) stresses dominate.

- The outer region or law-of-the-wake layer: inertia dominated core flow far from wall; free from direct viscous effects.


For y ⁄ δ > 0.8 fluctuating velocities become almost equalisotropic turbulence structure here. (far away the wall)For y ⁄ δ < 0.2 large mean velocity gradientshigh values of .(high turbulence production) .Turbulence is anisotropic near the wall.

2 2 2,u v and w and u v′ ′ ′ ′ ′−

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3.5 Turbulence models

We need expressions for :- Reynolds stresses:- Scalar transport terms :

2 2 2' , , , , ,u u v u w v v w w′ ′ ′ ′ ′ ′ ′ ′,u v and wϕ ϕ ϕ′ ′ ′ ′ ′ ′


Classical models: use the Reynolds eqn’s.(all commercial CFD codes)Large eddy simulation: the flow eqn’s are solved for the

- mean flow- and largest eddiesbut the effect of the smaller eddies are modeled- are at the research state- calculations are too costly for engineering use.

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The mixing length and k-ε models are the most widely used and validated.They are based on the presumption that“there exists an analogy between the action of viscous stressesand Reynolds stresses on the mean flow”Viscous stresses are proportional to the rate of deformation. For incompressible flow:

Notation:i = 1 or j = 1 x-directioni = 2 or j = 2 y-directioni = 3 or j = 3 z-direction

jiij ij

j i

uuex x

τ μ μ⎛ ⎞∂∂

= = +⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠(2.31)


For example

Turbulent stresses are found to increase as the mean rate of deformation increases.It was proposed by Boussinesq in 1877 thatReynolds stress could be linked to mean rate of deformation

This is similar to viscous stress equation

Except that μ is replaced by μt,

μt = turbulent (eddy) viscosity

1 212

2 1xy

u u u vx x y x

τ τ μ μ⎛ ⎞ ⎛ ⎞∂ ∂ ∂ ∂

= = + = +⎜ ⎟ ⎜ ⎟∂ ∂ ∂ ∂⎝ ⎠⎝ ⎠

jiij i j t

j i

UUu ux x

τ ρ μ⎛ ⎞∂∂′ ′= − = +⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠

(3.23)

jiij ij

j i

UUex x

τ μ μ⎛ ⎞∂∂

= = +⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠

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Similarly; Kinematic turbulent or eddy viscosity.Eqn.(3.23) shows that

Turbulent momentum transport is assumed to be proportional to mean gradients of velocity

By analogy turbulent transport of a scalar is taken to be proportional to the gradient of the mean value of the transported quantity. In suffix notation we get

Where Γt is the turbulent diffusivity.We introduce a turbulence Prandtl / Schmidt number as

ttv μ

ρ=

jiij i j t

j i

UUu ux x

τ ρ μ⎛ ⎞∂∂′ ′= − = +⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠

(3.23)

i ti

ux

ρ ϕ ∂Φ′ ′− = Γ∂

(3.24)

tt

t

μσ =Γ

(3.25)


Experiments in many flows have shown that

Most CFD procedures use Mixing length models:

Attempts to describe the turbulent stresses by means of simple algebraic formulae for μt as a function of positionThe k-ε models:

Two transport eqn’s (PDE’s) are solved:1. For the turbulent kinetic energy, k2. For the rate of dissipation of turbulent kinetic energy, ε.Both models assume that μt is isotropic(i.e. the ratio between Reynolds stresses and mean rate of deformation is

the same in all directions)

1tσ ≈1tσ ≈

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This assumption fails in many categories of flow.It is necessary to derive and solve transport eqn’s for the 6 Reynolds stresses themselves. ( ).

These eqn’s contain:1. Diffusion 2. Pressure strain 3. Dissipationterms whose individual effects are unknown and cannot be measured

Reynolds stress equation models:Solves the 6 transport equations (PDE’s) for the Reynolds stress terms

together with k and ε eqn’s., by making assumptions about- diffusion- pressure strain- and dissipation terms.

2 2 2' , , , , ,u u v u w v v w w′ ′ ′ ′ ′ ′ ′ ′


Algebraic stress models:

Approximate the Reynolds stresses in terms of algebraic equations instead of PDE type transport equations.Is an economical form of Reynolds stress model.Is able to introduce anisotropic turbulence effects into CFD simulations

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3.5.1 Mixing length model

where C is a dimensionless constant of proportionality.Turbulent viscosity is given byThis works well in simple 2-D turbulent flows where the only significant

Reynolds stress is and only significant velocity gradient is

For such flows

Where c is dimensionless constant

xy yx u vτ τ ρ ′ ′= = −

tv Cϑ= (3.26)

t Cμ ρϑ=

Uy

∂∂

Ucy

ϑ ∂=

∂ (3.27)

2

: turbulent velocity scale

: length scale of largest eddies

tm m ms s

ν ϑ→ = i

ϑ


Combining (3.26) and (3.27) and absorbing C and c into a new length scale

This is Prandtl’s mixing length model.The Reynolds stress is

(3.28)2t m

Uvy

∂=

∂

m

2xy yx m

U Uu vy y

τ τ ρ ρ ∂ ∂′ ′= = − =∂ ∂

(3.29)

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The transport of scalar quantity φ is modeled as

Where Γt = μt ⁄ σt and vt is found from (3.28).

σt = 0.9 for near wall flows

σt = 0.5 for jets and mixing layers Rodi(1980)

σt = 0.7 in axisymmetric jetsIn table 3.3:y : distance from the wallκ : 0.41 von Karman’s constant

(3.30)tv

yρ ϕ ∂Φ′ ′− = Γ

∂


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3.5.2. The k – ε model If convection and diffusion of turbulence properties

are not negligible (as in the case of recirculatingflows), then the mixing length model is not applicable.

The k-ε model focuses on the mechanisms that affect the turbulent kinetic energy.Some preliminary definitions:

( )

( )

2 2 2

2 2 2

1 mean kinetic energy2

1 turbulent kineticenergy2

( ) instantaneous kinetic energy

K U V W

k u v w

k t K k

= + +

′ ′ ′= + +

= +

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To facilitate the subsequent calculations it is common to write the components of the rate of deformation eij and the stresses τij in tensor (matrix) form:

- Using gives

xx xy xz xx xy xz

ij yx yy yz ij yx yy yz

zx zy zz zx zy zz

e e ee e e e and

e e e

τ τ ττ τ τ τ

τ τ τ

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟

= =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠( )ij ij ije t E e′= +

( ) ; ( ) ;

( ) ;

1 1( ) ( )2 2

1( ) ( )2

xx xx xx yy yy yy

zz zz zz

xy xy xy yx yx yx

xz xz xz zx zx zx

U u V ve t E e e t E ex x y y

W we t E ez z

U V u ve t E e e t E ey x y x

U We t E e e t E ez x

′ ′∂ ∂ ∂ ∂′ ′= + = + = + = +∂ ∂ ∂ ∂

′∂ ∂′= + = +∂ ∂

′ ′⎡ ⎤ ⎡ ⎤∂ ∂ ∂ ∂′ ′= + = = + = + + +⎢ ⎥ ⎢ ⎥∂ ∂ ∂ ∂⎣ ⎦ ⎣ ⎦∂ ∂⎡ ⎤′ ′= + = = + = + +⎢ ⎥∂ ∂⎣ ⎦

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1 1( ) ( )2 2yz yz yz zy zy zy

u wz x

V W v we t E e e t E ez y z y

′ ′∂ ∂⎡ ⎤+⎢ ⎥∂ ∂⎣ ⎦′ ′⎡ ⎤ ⎡ ⎤∂ ∂ ∂ ∂′ ′= + = = + = + + +⎢ ⎥ ⎢ ⎥∂ ∂ ∂ ∂⎣ ⎦ ⎣ ⎦


The product of a vector a and a tensor bij is a vector c

The scalar product of two tensors aij and bij is evaluated as follows

Suffix notation:1 x – direction2 y – direction3 z – direction

( )11 12 13

1 2 3 21 22 23

31 32 33

1 11 2 21 3 31 1

1 12 2 22 3 32 2

1 13 2 23 3 33 3

ij i ij

T T

j

b b bb a b a a a b b b

b b b

a b a b a b ca b a b a b c ca b a b a b c

⎛ ⎞⎜ ⎟≡ = ⎜ ⎟⎜ ⎟⎝ ⎠

+ +⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟= + + = = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠

a

c

11 11 12 12 13 13 21 21 22 22 23 23

31 31 32 32 33 33

ij ija b a b a b a b a b a b a b

a b a b a b

⋅ = + + + + +

+ + +

27


Governing equation for mean flow kinetic energy K

Reynolds equations: (3.12a), (3.12b), (3.12c)

Multiply x – component Reynolds eqn. (3.12a) by UMultiply y – component Reynolds eqn. (3.12b) by VMultiply z – component Reynolds eqn. (3.12c) by W

Then add the result together.After some algebra the time-average eqn. for mean

kinetic energy of the flow is


Or in words, for the mean kinetic energy K, we have

( )( ) ( ) 2 2

( ) ( ) ( ) ( ) ( ) ( ) ( )

ij i j ij ij i j ijK div KU div P E u u E E u u EtI II III IV V VI VII

ρ ρ μ ρ μ ρ∂ ′ ′ ′ ′+ = − + − − ⋅ + ⋅∂

U U U

(3.31)

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Governing equation for turbulent kinetic energy k1- Multiply x, y and z momentum eqns (3.9a-c) by u´, v´ and w´,respectively2- Multiply x, y and z Reynolds eqns (3.12a-c) by u´, v´ and w´, respectively3- subtract the two of the resulting equationsRearrange:

yields the equation for turbulent kinetic energy k:

( ) 1( ) 2 22

( ) ( ) ( ) ( ) ( ) ( ) ( )

ij i i j ij i j ijij

k div kU div p e u u u e e u u EtI II III IV V VI VII

ρ ρ μ ρ μ ρ∂ ⎛ ⎞′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′+ = − + − ⋅ − ⋅ + ⋅⎜ ⎟∂ ⎝ ⎠u u

(3.32)


The viscous term (VI)

Gives a negative contribution to (3.32) due to the appearance of the sum of squared fluctuating deformation rate e´ij

The dissipation of turbulent kinetic energy is caused by work done by the smallest eddies against viscous stresses.

The rate of dissipation per unit mass (m2/s3) is denoted by

The k – ε model equationsIt is possible to develop similar transport eqns. for all other turbulence quantitiesThe exact ε – equation, however, contains many unknown and unmeasurabletermsThe standard k – ε model(Launder and Spalding,1974) has two model eqns(a) one for k and (b) one for ε

( )2 2 2 2 2 211 22 33 12 13 232 2 2 2 2ij ije e e e e e e eμ μ′ ′ ′ ′ ′ ′ ′ ′− ⋅ = − + + + + +

2 ij ijve eε ′ ′= ⋅ (3.33)

29


We use k and ε to define velocity scale and length scale representative of the large scale turbulence as follows:

Applying the same approach as in the mixing length model we specify the eddy viscosity as follows

Where Cμ is a dimensionless constant

ϑ

3/ 21/ 2 kkϑ

ε= =

2

tkC Cμμ ρϑ ρε

= = (3.34)


The standard model uses the following transport equations used for k and ε

In words the equations are

The equations contain five adjustable constants . The standard k – ε model employs values for the constants that are arrived at by comprehensive data fitting for a wide range of turbulent flows:

( ) ( ) 2tt ij ij

k

k div k div grad k E Et

μρ ρ μ ρεσ⎛ ⎞∂

+ = + ⋅ −⎜ ⎟∂ ⎝ ⎠U (3.35)

2

1 2( ) ( ) 2t

t ij ijdiv div grad C E E Ct k kε ε

ε

μρε ε ερε ε μ ρσ⎛ ⎞∂

+ = + ⋅ −⎜ ⎟∂ ⎝ ⎠U (3.36)

1 20.09; 1.00; 1.30; 1.44; 1.92kC C Cμ ε ε εσ σ= = = = =(3.37)

1 2, , ,kC C and Cμ ε ε εσ σ

30


To compute the Reynolds stresses with the k – ε model (3.34-3.36) an extended Boussinesq relationship is used:

δij = 1 if i = jδij = 0 if i ≠ jConsider the first part of eqn(3.38):If we sum all the normal stresses by letting i=j=1,2,3

( )2 2 2 2 2 20 .1 .1 .1 23 3 3

2 2

u v w k k k k

k k

ρ ρ ρ ρ ρ

ρ ρ

′ ′ ′− + + = − − − = −

− = −

2 223 3

jii j t ij t ij ij

j i

UUu u k E kx x

ρ μ ρ δ μ ρ δ⎛ ⎞∂∂′ ′− = + − = −⎜ ⎟⎜ ⎟∂ ∂⎝ ⎠

(3.38)

Extra term compared with eqn(3.23) of mixing length model

2 2 2 0t ii t tU V WE divUx y z

μ μ μ⎛ ⎞∂ ∂ ∂

= + + = =⎜ ⎟∂ ∂ ∂⎝ ⎠

( )2 2 2u v wρ ′ ′ ′− + +

=2k

Turbulence kinetic energy per unit volume


So an equal 1/3 is allocated to each normal stress component in eqn(3.38) to have –2ρk when summed.

This implies an isotropic assumption for the normal Reynolds stresses.Boundary Conditions:

The following boundary conditions are needed

For inlet, if no k and ε is available crude approximations can be obtained from the

- Turbulence intensity, Ti

- Characteristic length, L (equivalent pipe radius) which are:

31


The formulae are closely related to the mixing length formulae in section 3.5.1 and the universal distributions near a solid wall given below.

Near the wall, at high Reynolds numbers, the k – ε model equations (3.34-3.36) are not used. Instead, universal behaviour of near wall flows is used.

In this region measurements of the budgets indicates that:the rate of turbulence production = rate of dissipation

Using this assumption and , the following wall functions can be derived:

1 ln 30 500u y for yk

+ + += < < (3.21)

( )2 31 ln ; ;P

u uUu Ey ku yC

τ τ

τ μ

εκ κ

+ += = = = (3.39)

( )3/ 22 3/ 42 ; ; 0.07

3 ref ikk U T C Lμε= = =

2 /t C kμμ ρ ε=


κ = 0.41 (Von Karman’s constant)E = 9.8 (wall roughness parameter) for smooth wallsFor heat transfer we use the universal near wall temperature distribution

valid at high Reynolds numbers (Launder and Spalding, 1974):

Finally P is the “pee-function”, a correction function dependent on the ratio of laminar to turbulent Prandtl numbers (Launder and Spalding, 1974)

( ) ,,

,

,

,

with temperature at near wall point turbulent Prandtl number

wall temperature / Prandtl numberwall heat flux thermal conducti

T lw PT t

w T t

p P T t

w T l P T

w T

T T C uT u P

q

T y

T Cq

τ σρσ

σ

σ

σ μ

+ +⎡ ⎤⎛ ⎞−

≡ − = +⎢ ⎥⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦= =

= = Γ =

= Γ =

p

vityC fluid specific heat and constant pressure=

(3.40)

32


At low Reynolds numbers the log-law is not valid so the above-mentioned boundary conditions cannot be used

Modifications to the k – ε model is used in such cases.The equations of the low Reynolds number k – ε model, which replace (3.34-

3.36), are given below:

( ) ( ) 2tt ij ij

k

k div k div grad k E Et

μρ ρ μ μ ρεσ

⎛ ⎞⎛ ⎞∂+ = + + ⋅ −⎜ ⎟⎜ ⎟⎜ ⎟∂ ⎝ ⎠⎝ ⎠

U

2

tkC fμ μμ ρε

=

2

1 1 2 2

( ) ( )

2

t

t ij ij

div div gradt

C f E E C fk k

ε

ε ε

μρε ρε μ εσ

ε εμ ρ

⎛ ⎞⎛ ⎞∂+ = +⎜ ⎟⎜ ⎟⎜ ⎟∂ ⎝ ⎠⎝ ⎠

+ ⋅ −

U

(3.41)

(3.43)

(3.42)


Modifications:A viscous contribution is includedCμ, C1ε and C2ε are multiplied by wall damping functions fμ, f1 , f2

which are functions of turbulence Reynolds number or similar functions

As an example we quote the Lam and Bremhorst (1981) wall-damping functions which are particularly successful:

In function the parameter is defined by . Lam and Bremhorst use as a boundary condition.0yε∂ ∂ =

2

3

21 2

20.51 exp( 0.0165Re ) 1 ;Re

0.051 ; 1 exp( Re )

yt

t

f

f ff

μ

μ

⎛ ⎞⎡ ⎤= − − +⎜ ⎟⎣ ⎦

⎝ ⎠

⎛ ⎞= + = − −⎜ ⎟⎜ ⎟⎝ ⎠

(3.44)

1/ 2 /k y vfμ Re y

33


Assessment of performance

Table 3.5 k-ε model assessment


3.5.3 Reynolds stress equation modelsThe most complex classical turbulence model is Reynolds stress equation model (RSM),

Also called :- second order- or second – moment closure model

Drawbacks of k – ε model emerge when it is attempted to predict

- flows with complex strain field- flows with significant body forcesThe exact Reynolds stress transport eqn can account for the

directional effects of the Reynolds stress field.Let Reynolds stressij

ij i jR u uτρ

′ ′= − =

34


The exact solution for the transport of takes the following form

Equation (3.45) describes six partial differential equations: one for the transport of each of the six independent Reynolds stresses( )Two new term appear compared with ke eqn (3.32)

1. . : pressure strain correction term2. . : rotation term

ijij ij ij ij ij

DRP D

Dtε= + − +Π +Ω (3.45)

2 2 21 2 3 1 2 1 3 2 3 2 1 1 2 3 1 1 3 3 2 2 3, , , , , since ,u u u u u u u and u u u u u u u u u u and u u u u′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ ′= = =

ijΠijΩ

ijR


CFD computations with Reynolds stress transport equations retain the production term in its exact form

The diffusion term Dij can be modelled by the assumption that the rate of transport of Reynolds stresses by diffusion is proportional to gradient of Reynolds stress.

The dissipations rate εij is modeled by assuming isotropy of the small dissipative eddies. It is set so that it effects the normal Reynolds stresses (i = j) only and in equal measure. This can be achieved by

Where ε is the dissipation rate of turbulent kinetic energy defined by (3.33). The Kronecker delta, δij is given by δij = 1 if i = j and δij = 0 if i ≠ j

j iij im jm

m m

U UP R Rx x

∂⎛ ⎞∂= − +⎜ ⎟∂ ∂⎝ ⎠ (3.46)

( )2

with ; 0.09 1.0

ijt tij ij

m k m k

t k

Rv vD div grad Rx x

kv C C andμ μ

σ σ

σε

∂⎛ ⎞ ⎛ ⎞∂= =⎜ ⎟ ⎜ ⎟∂ ∂⎝ ⎠ ⎝ ⎠

= = =(3.47)

23ij ijε εδ= (3.48)

35


For Πij term:The pressure – strain interactions are the most difficult to modelTheir effect on the Reynolds stresses is caused by two distinct physical processes:

1. Pressure fluctuations due to two eddies interacting with each other

2. Pressure fluctuations due to the interactions of an eddy with a region of flow of different mean velocityIts effect is to make Reynolds stresses more isotropic and to reduce the Reynolds shear stressesMeasurements indicate that;The wall effect increases the isotropy of normal Reynolds stresses by damping out fluctuations in the directions normal tothe wall and decreases the magnitude of the Reynolds shear stresses.


A comprehensive model that accounts for all these effects is given in Launder et al (1975). They also give the following simpler form favoured by some commercial available CFD codes:

The rotational term

ωk : rotation vector

eijk = +1 if i, j and k are different and in cyclic order

eijk = –1 if i, j and k are different and in anticyclic order

eijk = 0 if any two indices are the sameTurbulent kinetic energy k is.

1 2

1 2

2 23 3

with 1.8; 0.6

ij ij ij ij ijC R k C P Pk

C C

ε δ δ⎛ ⎞ ⎛ ⎞Π = − − − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= =

( )2ij k jm ikm im jkmR e R eωΩ = − +

( ) ( )2 2 211 22 33 1 2 3

1 12 2

k R R R u u u′ ′ ′= + + = + +

(3.49)

(3.50)

36


The six equations for Reynolds stress transport are solved along with a model equation for the scalar dissipation rate ε. Again a more exact form is found in Launder et al (1975), but the equation from the standard k – εmodel is used in commercial CFD for the sake of simplicity

The usual boundary conditions for elliptic flows are required for the solution of the Reynolds stress transport equations:

(3.51)2

1 1 2

1 2

2

where 1.44 and 1.92

tt ij ij

vD div grad C f v E E CDt k k

C C

ε εε

ε ε

ε ε εεσ⎛ ⎞

= + ⋅ −⎜ ⎟⎝ ⎠= =


In the absence of any information inlet distributions are calculated from

For computations at high Reynolds numbers wall-function-type boundary conditions can be used which are very similar to those of the k-ε model .

Near wall Reynolds stress values are computed from formulae such as where cij are obtained from measurements.

Low Reynolds number modifications to the model can be incorporated to add the effects of molecular viscosity to the diffusion terms and to account for anisotropy in the dissipation rate in the Rij-equations.

Wall damping functions to adjust the constants of the ε-equation and a modified dissipation rate variable give more realistic modelling near solid walls.

( )3 / 22 3 / 4 2

1

2 22 3

2 ; ; 0 .0 7 ; ;3

1 ; 0 ( )2

re f i

i j

kk U T C L u k

u u k u u i j

με ′= = = =

′ ′ ′ ′= = = ≠

ij i j ijR u u c k′ ′= =

1/ 2 2( 2 ( / ) )v k yε ε≡ − ∂ ∂

37


Similar models exist for the 3 scalar transport terms of eqn (3.17) in the form of PDE’s.

In commercial CFD codes a turbulent diffusion coefficient

is added to the laminar diffusion coefficient, where

for all scalars.

/t t ϕμ σΓ =

0.7ϕσ =

iuϕ′ ′


Table 3.6 Reynolds stress equation model assessment.

38


3.5.4 Algebraic stress equation models (ASM)- ASM is an economical way of accounting for the

anisotropy of Reynolds stresses.- Avoid solving the Reynolds stress transport eqns- Instead, use algebraic eqn to model Reynolds stresses.The Simplest method is:

To neglect the convection and diffusion terms altogether

A more generally applicable method is:To assume that the sum of the convection and diffusion

terms of the Reynolds stresses is proportional to the sum of the convection and diffusion terms of turbulent kinetic energy


Introducing approximation (3.52) into the Reynolds stress transport equation (3.45) with production term (3.46), modeled dissipation rate term (3.50) and pressure – strain correction term (3.49) on the right hand side yields after some arrangement the following algebraic stress model:

- appear on both sides (on rhs within Pij ) - eqn(3.53) is a set of 6 simultaneous algebraic eqns. for 6 unknowns,- solved iteratively if k and ε are known.- The standard k-ε model eqns has to be solved also (3.34 - 3.37)

i ju u′ ′

[ ]

( )

transport of ( . . )i j i jij

i ji j ij

Du u u u DkD k i e div termsDt k Dt

u uu u E

kε

′ ′ ′ ′ ⎛ ⎞− ≈ ⋅ −⎜ ⎟⎝ ⎠

′ ′′ ′= ⋅ − ⋅ −

(3.52)

1

2 23 31

Dij i j ij ij ij

C kR u u k P PPCδ δ

εε

⎛ ⎞⎜ ⎟⎛ ⎞′ ′= = + −⎜ ⎟⎜ ⎟

⎝ ⎠⎜ ⎟− +⎜ ⎟⎝ ⎠

(3.53)

ijP

i ju u′ ′

39


For swirling flows:10.55 and 2.2DC C= = (3.54)

Table 3.7 Algebraic stress model assessment


3.5.5 Some recent advancesResearch work has followed two distinct lines:1. The development and optimization of turbulence models for

limited categories of flows.2. The search for a comprehensive and completely general purpose

turbulence modelThe large majority of models use line 1 for specific problems

The nonlinear k-ε models (Speziale, 1987, 1991)

The value of adjustable constant CD was found by calibration with experimental data.

2

32 0 0

2

0

2 23

1 14 . .3 3

. ( ) . .

1.68

ij i j ij ij

D im mj mn mn ij ij mm ij

ij jiij ij ki ki

k k

D

ku u k C E

kC C E E E E E E

E UUwhere E grad E E Et x x

and C

μ

μ

τ ρ ρ δ ρε

δ δε

′ ′= − = − +

⎛ ⎞− − + =⎜ ⎟⎝ ⎠

∂ ∂⎛ ⎞∂= + − +⎜ ⎟∂ ∂ ∂⎝ ⎠=

u

(3.55)

40


The nonlinear k - ε model accounts for anisotropy.Accounts for the secondary flow in non-circular duct flows.

The RNG k - ε model (Renormalization Group)The RNG procedure:

systematically removes the small scales of motion from the governing equations by expressing their effects in terms of large scale motions and a modified viscosity. (Yakhot et al 1992):

( )( ) ( ) 2k eff t ij ijk div k div grad k E E

tρ ρ α μ μ ρε∂

+ = + ⋅ −∂

U

( )2

*1 1 2 2

2

1 2

( ) ( ) 2

;

0.0845; 1.39; 1.42; 1.68

eff t ij ij

eff t t

k

div div grad C f E E C ft k k

kwhere C

with C C C

ε ε ε

μ

μ ε ε ε

ρε ε ερε α μ ε μ ρ

μ μ μ μ ρε

α α

∂+ = + ⋅ −

∂

= + =

= = = = =

U

(3.56)

(3.57)


Only the constant β is adjustable; the above value is calculated from near wall turbulent data. All other constants are explicitly computed as part of the RNG process.This model can be applied with

The isotropic Reynolds stress formula (3.38) (k-ε model)The nonlinear form (3.57) (nonlinear k-ε model )

( )( ) ( )

2*

1 1 2 2

1/ 20*1 1 03

( ) ( ) 2

1 /and ; 2 ; 4.377

10.012

eff t ij ij

ij ij

div div grad C f E E C ft k k

kC C E E

ε ε ε

ε ε

ρε ε ερε α μ ε μ ρ

η η ηη η

βη εβ

∂+ = + ⋅ −

∂−

= − = ⋅ =+

=

U

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