Derivation of the
k -ε model
Start with the RANS equations (RANS ( RANS), ), here for ρ = const Du i i ∂ 2 u i i −1 ∂p = + ν ρ ∂x i i ∂x j j ∂x j j Dt
∂u i u j
−
∂x j j
,
(RANS)
where we defined the substantial derivative over the mean flow as ∂φ ∂φ Dφ := + u j j . ∂t ∂x j j Dt
We don’t know the Reynolds stresses u i u j . This is referred to as an unclosed term. We could derive a transport equation of the form
Du i u j
Dt
= . . . ,
but this will lead to more unclosed terms like the triple correlation u i u j u k , which is still unclosed. This unfortunate situation is called closure problem. To resolve this situation, we need additional equations which give us the Reynolds stress u i u j . The k -ε model solves two additional PDEs to find the stresses, which means it’s a two equation
model.
Firstly, we assume that we can model the Reynolds stress sort of like the stress in a Newtonian fluid. In a Newtonian fluid, we have (with Stokes’ assumption)
dev dev (σ ) = 2µs .
(NF)
In the Boussinesq eddy viscosity ansatz , we have dev(ρCov (u )) )) := −2µT s ⇔ 1 aij := u i u j − u k u k δij := −2ν T T s ij ij . 3
(EV)
As far as I see, the minus −2ν T ( EV)) stems from the fact that strictly speaking, the kinematic T in (EV Reynolds stress tensor is − u i u j , as can be seen in (RANS ( RANS). ). This ν T T is known as the kinematic eddy viscosity . It is not a property of a fluid like a real viscosity, but instead a property of the flow. We then define the turbulent kinetic energy as
k :=
1 u u . 2 i i
(TKE)
In the following subsection, we derive a transport equation for k
Derivation of the transport equation for the turbulent kinetic energy k Start Start with the usual incompr incompressi essible ble Navier-S Navier-Stok tokes es momentum momentum equation equationss ( NS NS)) without body forces ∂u i i ∂u i i ∂ 2 u i i −1 ∂p + u j = + ν . (NS) ∂ t
∂x j j
ρ ∂x i i
∂x j j ∂x j j
Subtract the RANS equation (RANS (RANS)) from it to get a transport equation for the fluctuations
u i ∂u i u j ∂u i ∂u i ∂u i ∂u i ∂ 2 u i −1 ∂p + u j + u j j + u j = + ν + . ∂t ∂x j j ∂x j j ∂x j j ρ ∂x i i ∂x j j ∂x j j ∂x j j
1
(NSFLUCT)
Then, multiply both sides by u i , average and simplify to get a transport equation for k
Dk = P k + T k + Dk − ε Dt P k := −u i u j s ij turb. production
∂ 1 T k := − ∂x j 2
u j u i u i +
1 ρ
p u j turb. transport
(KTSP)
∂ 2 k Dk := ν turb. diffusion ∂x j ∂x j ∂u ∂u ε := ν i i dissipation. ∂x j ∂x j
In the transport equation, P k , T k and ε are still unclosed. Closing turbulent production P k
To close P k , we first employ the eddy viscosity ansatz (EV) to get rid of the Reynolds stress u i u j . P k = −u i u j s i j = 2ν T s i j s i j .
Now, how do we address ν T ? We use dimensional analysis to set it proportional to k α ε β for some α and β . The target variable ν T has dimensions m2 s 1 . k has dimensions m2 s 2 and ε has dimensions m 2 s 3 as can be seen from (KTSP). This yields the following system of equations −
−
−
2
2 −2 −3
Solving it, we determine α = 2, β =
2 α β
=
−
1
.
(NUDIM)
1, so we set
−
ν T :∝
k 2 . ε
(NUPROP)
The proportionality constant c µ of (NUPROP) is a model parameter.
Closing turbulent transport T k Instead of specifying equations for u j u i u i and 1ρ p u j , we directly model T k as gradient diffusion (Fickian)
T k := div
ν T grad(k ) . σk
(TKGRADDIFF)
σk is called the turbulent Prandtl number and is a model constant.
Derivation of the transport equation for the dissipation ε We could also start from the Navier-Stokes equations and then derive an equation for the ∂u ∂u
dissipation ε = ν ∂x j i ∂x j i . Doing so would however lead to a zoo of new unclosed terms which we can’t treat appropriately because we can’t measure them nicely in experiments. So we surrender 2
and just assume that the structure of the transport equation is analogous to the one for k (KTSP) Dε = T ε + P ε − Γε Dt
ν t T ε := div grad(ε) dissipation transport σε P k ε P ε := c ε1 production of dissipation k ε2 Γε := c ε2 destruction of dissipation, k
(EPSTSP)
where σε , c ε1 , c ε2 are model constants.
When to use this model The k -ε works well in free shear flows (high Re , away from walls). Near walls, the k -ω model is better. It fails in strongly curved flows, swirling flows and flows with recirculation zones because of the eddy-viscosity approximation.
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