The Reynolds Averaged Navier Stokes Equations RANSWe start off by writing down the classical Navier Stokes equations for incompressible, constantviscosity, Newtonian fluids on the form

D

Dt

uvw

= ∂

∂t

uvw

+ (u · ∇)

uvw

= −∇Pρ

+ g + ν∇2

uvw

(1)

where this notation highlights the fact that the operators ∂

∂t, (u ·∇) and (ν∇2) must be applied

to each component of the velocity vector u. To derive the Reynolds Averaged Navier-StokesEquations (RANS) we assume that the dependent variables (velocity and pressure) consist ofa mean part φ and a random fluctuating turbulent part φ′, such that φ = φ + φ′. This iscalled Reynolds decomposition. Before we go on into the algebra, we should note some laws foraveraging.

Laws for averaging

Let the mean of a time signal φ(t) be defined as φ = 1T

∫ T

0φdt, where the integral is taken over

a time period T which is considerably larger than the period of the turbulent fluctuations. Fromthis definition we may define several algebraic rules for manipulation of turbulent signals. Letf = f + f ′ and g = g + g be two turbulent signals:

(1) : f ′ = 0 (2) : f = f (3) : fg = fg

(4) : f ′g = 0 (5) : f + g = f + g (6) : fg = fg

(7) : fg = fg + f ′g′ (8) : ∂f∂s

= ∂f

∂s(9) :

∫fds =

∫fds

Deriving RANS

The process of Reynolds averaging is now to insert u = u + u′ and P = P + P ′ into (1), andthereafter take the average of the whole equation term for term, utilizing the laws above. Theresult is that the transport equations for u involves additional terms called Reynolds stresses,that acts as additional unknowns in the equations, and needs proper modelling to find suitablesolutions. As we will see, these Reynolds stresses emerge from the non linear convective accel-eration term. We will now conduct the Reynolds averaging of the u− momentum equation, andthen generalize the result. Inserting u = u + u′ and p = P + P ′ into the u− momentum, andthen averaging the equation term for term gives:

Time derivative∂(u+ u′)

∂t= ∂u

∂t+ ∂u′

∂t= ∂u

∂t+ ∂u′

∂t= ∂u

∂t+ ∂u′

∂t= ∂u

∂t

1

Pressure gradient

∂(P + P ′)∂x

= ∂P

∂x+ ∂P ′

∂x= ∂P

∂x+ ∂P ′

∂x= ∂P

∂x+ ∂P ′

∂x= ∂P

∂x

Viscous Diffusion

∇2(u+ u′) = ∇2u+∇2u′ = ∇2u+∇2u′ = ∇2u

As seen from the previous calculations, the Reynolds averaging does not produce any additionalterm in the linear terms.

Convective acceleration

Before we calculate the non linear convective acceleration term, we should utilize the followingidentity

(u · ∇)φ = ∇ · (uφ)− φ(∇ · u)

where of course the term φ(∇ · u) = 0, for an incompressible fluid. Reynolds averaging gives:

([u + u′] · ∇)(u+ u′) = ∇ · [(u + u′) (u+ u′] =

∂

∂x((u+ u′)(u+ u′)) + ∂

∂y((v + v′)(u+ u′)) + ∂

∂z((w + w′)(u+ u′)) (2)

Let us work out the first of these terms:

∂

∂x((u+ u′)(u+ u′)) = ∂

∂x(u u+ 2uu′ + u′u′)) = ∂(u u)

∂x+ 2∂uu

′

∂x+ ∂u′u′

∂x

where the terms uu′ = 0 according to the averaging laws, but the terms u′u′ 6= 0, and this is thecause of the Reynolds stresses. We may generalize this results into to yield

∂

∂x((u+ u′)(u+ u′)) = ∂u2

∂x+ ∂u′u′

∂x

∂

∂y((v + v′)(u+ u′)) = ∂(u v)

∂y+ ∂u′v′

∂y

∂

∂z((w + w′)(u+ u′))∂(u w)

∂z+ ∂u′w′

∂z

such that∇ · [(u + u′) (u+ u′] = ∇ · (u u) +∇ · u′u′

We are now in a position to write the full form of the u−momentum RANS equation whichbecomes

∂u

∂t+ (u · ∇)u = −1

ρ

∂P

∂x+ g + ν∇2u+∇ · u′u′

This results may be generalized in 3D to form the complete RANS equations

DuDt

= ∂

∂t

uvw

+ (u · ∇)

uvw

= −∇Pρ

+ g + ν∇2

uvw

+∇ ·

[−u′u′ −u′v′ −u′w′][−v′u′ −v′v′ −v′w′][−w′u′ −w′v′ −w′w′]

(3)

2

The last term of this equations is the divergence of the Reynolds stress tensor. Written in thisform, it illustrates that for each of the momentum equations, the divergence operator should beapplied to the row vector of the corresponding momentum equation. For example if we wereto write out the v−momentum, we must apply the divergence operator to the 2nd row of theReynolds stress tensor.

Continuity

Performing Reynolds averaging of continuity gives us ∇ · (u + u′) = 0→

∇ · u = 0, ∇ · u′ = 0 (4)

So we see that both the mean and fluctuation parts of velocity both satisfy continuity individ-ually.

RANS on tensor notation

The equations in turbulence are notoriously long, tedious and cumbersome to write out in fullvector notation. Therefore it is an absolute must to utilize tensor notation (also called indexnotation). The important thing to remember is to know the Einstein summation conventionthat says that if a term consist of variables with repeated index, the term must be summed overthat index. An example is continuity

∇ · u = ∂ui∂xi

= ∂u

∂x+ ∂v

∂y+ ∂w

∂z

Since the i appears in both ui and xi, we say that that index is repeated, and therefore theterm must by summed over i = 1, 2, 3. If a term has both i and j repeated, the term must besummed over both i and j giving a total of 9 terms if i, j = 1, 2, 3. Using this convention, wemay now write the RANS equations on tensor notation

ρ(∂ui∂t

+ uj∂ui∂xj

) = − ∂P∂xi

+ ρgi + µ∂2ui∂xj2 −

∂

∂xjρu′iu

′j (5)

Note that the Laplacian term ∂2uj∂xi2

= ∂2uj∂xi∂xi

must be summed over i since the i is repeated.Note also that since the Reynolds stress tensor is symmetric, we have a total of 6 additionalunknowns. The question now is how to model the additional unknowns − ∂

∂xj(ρu′iu′j). If we

assume 2D flow where w = ∂

∂z= 0 but w′2 6= 0, we are left with the additional terms −ρu′2,

−ρu′v′ and −ρv′2It is the aim of turbulence models to be able to relate the unknown components of the Reynoldsstress tensor to the mean flow quantities and thus providing closeure to the equations. One lineof modelling known as eddy viscosity models is to relate the Reynolds stress tensor componentsto a the mean rate of strain tensor through a turbulent (eddy) viscosity. This is known as theBoussinesq approximation for turbulence, and takes the form

− u′iu′j = νt

(∂ui∂xj

+ ∂uj∂xi

)− 2

3kδij (6)

where δij is the Kronecker delta function δij = 1 if i = j, and δij = 0 if i 6= j. k is the turbulent

3

kinetic energy defined as k = uiui2 . A transport equation of k can be derived by forming the

dot product of ui with the RANS equations (3), but the result involves many terms.

k - ε turbulence model

Now we need to model the turbulent viscosity νt. The k − ε model, uses

νt = Cµk2

ε(7)

where Cµ is an empirical constant, and ε is the rate of turbulent dissipation. A transport

equations for ε may be derived by applying the operator ν ∂u′i

∂xj

∂

∂xjto the RANS equations

(3), but again the result involves many complicated terms. Instead of incorporating all thecomplexities of the analytically derived k and ε equations, the k− ε equation uses approximatedversions of them involving only the most important terms:

Dk

Dt= ∂

∂xj

(νtσk

∂k

∂xj

)+ νt

∂ui∂xj

(∂ui∂xj

+ ∂uj∂xi

)− ε (8)

Dε

Dt= ∂

∂xj

(νtσε

∂ε

∂xj

)+ C1νt

ε

k

∂ui∂xj

(∂ui∂xj

+ ∂uj∂xi

)− C2

ε2

k(9)

with empirically determined constants

Cmu = 0.09 C1 = 1.44 C2 = 1.92 σK = 1.0 σε = 1.3

These constants should ideally be tuned to the particular flow type at and. Let us now expandthe k equation to get a feeling of their complexity. Expanding the k equation term for termgives:

1. ∂

∂xj

(νtσ

∂k

∂xj

)= ∂

∂x

(νtσ

∂k

∂x

)+ ∂

∂y

(νtσ

∂k

∂y

)+ ∂

∂z

(νtσ

∂k

∂x

)= ∇ ·

(νtσ∇k)

2. ∂ui∂xj

∂ui∂xj

= ∂u

∂xj

∂u

∂xj+ ∂v

∂xj

∂v

∂xj+ ∂w

∂xj

∂w

∂xj= [∇u · ∇u] + [∇v · ∇v] + [∇w · ∇w]

3. ∂ui∂xj

∂uj∂xi

= ∂u

∂xj

∂uj∂x

+ ∂v

∂xj

∂uj∂y

+ ∂w

∂xj

∂uj∂z

= [∇u · ∂u∂x

] + [∇v · ∂u∂y

] + [∇w · ∂u∂z

]

In 3 dimensions (1) consists of 3 terms, (2) consists of 9 terms, and (3) consists of 9 terms, sothe RHS of equations (8) consists of 22 terms including the additional ε term. In 2D this isreduced to a total of 10 terms. We also see that the ε equation contains exactly the same tensorterms.

Boundary treatment

The k− ε equations written above are only valid for fully turbulent high Reynolds number flow,and are thus not valid near the wall. Therefore these equations are not integrated all the wayto the walls. Instead we use one of two options: 1) patch a turbulent wall-law (a logarithmicvelocity profile) onto the solution near the walls, and 2) add damping functions to the equations

4

wall law treatment

let up be a velocity component at a node p closest to the wall. We then impose that the velocityat this node should follow the logarithmic wall law according to

upv∗

= 1κln

(v∗ypν

)+B (10)

Kp = v∗2√Cµ

(11)

εp = v∗3

κyp(12)

where v∗ is the wall friction velocity v∗ =(τwρ

)1/2, κ ≈ 0.40 is the Von Karman constant

Implementation of the k - ε model in openFoam

Initial and Boundary Condition

To set initial and boundary data for turbulence we need to specify the turbulence intensityI which is the ratio of the magnitude of the turbulent fluctuations to the magnitude of thecharacteristic mean velocity.

I =

√13(u′2 + v′2 + w′2)

Uref=

√23k

Uref→ k = 3

2(UrefI)2

I is a dimensionless quantity and its value typically has a value ranging from [0, 20]. I = [0, 1]is considered low intensity, [1, 5] is medium intensity, and [5, 20] is considered high turbulentintensity. Note also that if one has access to fluid dynamical experimental equipments, one maymeasure the turbulent fluctuations quite accurately. For fully developed pipe flow we may usethe relation

I = 0.16Re−18

As boundary condition on ε we may use

ε = 0.164k1.5

0.07L

5