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bumper July 30, 2019 07:29
Number of equation by using eddy viscosity turbulence models
 
1 Attachment(s)
Hello,

i'm trying to understand the theory behind the eddy viscosity turbulence modelling.

First i'm doing the RANS approach on the conservation equations and i get the Reynolds Stress tensor like in the picture (Reynold_Stress_Tensor) below. Each line characterize the additonal terms that i get in the three momentum equation. By choosing a eddy viscosity turbulence model, i use the approach of Boussinesq for modelling the expressions of the Reynolds Stress Tensor. There is an unknown variable, the turbulent viscosity. The definition of the turbulent viscosity depends on the choosen turbulence model. Let's choose the k-epsilon model for example.

https://www.cfd-online.com/Wiki/Stan...-epsilon_model

In the equations of k and epsilon, you can find the indices i and j which go from 1 to 3 each (direction x,y,z). If i=1, j could be 1,2,3, so i have 3 equations in total for k in x-direction ? All in all i have 9 equations for k and 9 equations for epsilon? Than i have lots of opportunities to calculate the turbulent viscosity and the entries of the Reynolds Stress Tensor.

Or do i have just 9 additonal equations (one for each entrie of the Reynolds Stress Tensor). For example i want to calculate the first entry. Than, i and j are 1 and it "goes down" to the equation of the turbulence model?

So the main question is, how many equations do i get in total. I hope that i describe my problem in a good way. It would be great if someone could explain it to me.

Thanks a lot,

bumper

LuckyTran July 30, 2019 09:50
You get 3 scalar momentum equations because it's a 3-dim vector. Really you could equivalently say you have 1 momentum equation.

You get 1 equation for k and 1 equation for epsilon because they're both scalars. The fact that you see an i,j, or k in the k and epsilon equations doesn't really mean you get 3 new equations each. If you write out the entire equations, you'll find there is just 1 equation each. You have to follow the summation rules for the notation being used.

Even though the Stress Tensor has 9 entries in the matrix, you still get 3 momentum equations. You have to pay attention to the products of vectors and tensors.


What you need to get the turbulent viscosity is k and epsilon. Once the turbulent viscosity is known, you get all the components of the reynolds stress tensor from the boussinesq hypothesis. You get k and epsilon by solving their transport equations, respectively.

bumper July 30, 2019 11:13
Okay, i have three momentum equations, that's clear. 3 entries (one line) of the Reynolds Stress Tensor belong to one momentum equation.

So the i,j or k in the equation for the energy or dissipation doesn"t mean, that i get a 3dimensional equation. There is only one equation for the energy k and one for the dissipation rate.
Which summation rules do you mean? I see just a partial derivation. That's the reason why i was expacting 3 equation each.

Can you show me how the entire equation looks like?

Thanks a lot for your help.

IbizaFR July 30, 2019 11:16
Quote:
Originally Posted by LuckyTran (Post 740493)
You get 3 scalar momentum equations because it's a 3-dim vector. Really you could equivalently say you have 1 momentum equation.

You get 1 equation for k and 1 equation for epsilon because they're both scalars. The fact that you see an i,j, or k in the k and epsilon equations doesn't really mean you get 3 new equations each. If you write out the entire equations, you'll find there is just 1 equation each. You have to follow the summation rules for the notation being used.

Even though the Stress Tensor has 9 entries in the matrix, you still get 3 momentum equations. You have to pay attention to the products of vectors and tensors.


What you need to get the turbulent viscosity is k and epsilon. Once the turbulent viscosity is known, you get all the components of the reynolds stress tensor from the boussinesq hypothesis. You get k and epsilon by solving their transport equations, respectively.
Thank you.

Can you suggest a book with numerical schemes solving k and epsilon equations? Many of CFD books describe for example SIMPLE, SIMPLER, PISO etc algorithms but do not mention numerical methods for solving for example k-epsilon equations.

LuckyTran July 31, 2019 12:20
Quote:
Originally Posted by bumper (Post 740506)
Okay, i have three momentum equations, that's clear. 3 entries (one line) of the Reynolds Stress Tensor belong to one momentum equation.
No that's not how it works. It's not one line of the stress tensor going into each equation, you'll find terms from other lines. 3 versus 1 is not really important. The momentum equation spits out a vector equation. It's 1 equation if you consider the velocity as a 3-d vector, and 3 equations if you consider the velocity in its component form. It's purely a matter of perspective and taste. It's still the same momentum equation.

As an example, the momentum equations look like this when you expand all the summations. Yes this is the navier-stokes & not the RANS. The stress tensor and reynolds stress tensor are similar anyway and the difference is mainly in the effective viscosity. Despite there being 9 combinations of indices, you get only an x,y,z equations. There's no magic here, you just have to follow the rules for the notation being used.

Quote:
Originally Posted by IbizaFR (Post 740507)
Thank you.

Can you suggest a book with numerical schemes solving k and epsilon equations? Many of CFD books describe for example SIMPLE, SIMPLER, PISO etc algorithms but do not mention numerical methods for solving for example k-epsilon equations.
SIMPLE/SIMPLER/PISO is an approach for solving a different problem, the pressure-velocity coupling problem. In CFD the momentum equation gets a lot of attention because there's 1 momentum equation but two variables (pressure and velocity). This does not occur in standard advection-diffusion equations.

Other equations like energy, k, epsilon, etc. don't have this issue. All you need is some type of linear solver (i.e. solve Ax=b). You just need a way to solve this linear system. The dumb way is to invert A. Gauss-Seidel algorithm is another archetype.

Even when you execute the SIMPLE algorithm, you will use similar solvers to solve the individual pieces of the SIMPLE algorithm because the first step in SIMPLE is to solve the momentum equation with an assumed pressure field.

FMDenaro July 31, 2019 12:40
I think that a brief observation is required in case of incompressible flows.

The eddy viscosity model is written for the deviatoric part of the tensor, the isotropic part (the trace) being incorporated into the "pressure".
From that you can write the three scalar equations

bumper August 1, 2019 04:45
1 Attachment(s)
Quote:
No that's not how it works. It's not one line of the stress tensor going into each equation, you'll find terms from other lines. 3 versus 1 is not really important. The momentum equation spits out a vector equation. It's 1 equation if you consider the velocity as a 3-d vector, and 3 equations if you consider the velocity in its component form. It's purely a matter of perspective and taste. It's still the same momentum equation.

As an example, the momentum equations look like this when you expand all the summations. Yes this is the navier-stokes & not the RANS. The stress tensor and reynolds stress tensor are similar anyway and the difference is mainly in the effective viscosity. Despite there being 9 combinations of indices, you get only an x,y,z equations. There's no magic here, you just have to follow the rules for the notation being used.
Okay, let's say we have an incompressible fluid and we ignore the gravitation, the last two summands of each of your equation is not needed anymore. Let"s talk about the component form that we are on the same level, for example just for the x-directon. If i do the RANS approach, i get the quation for x as you can see on the attachment. So if you look on the Reynolds Stress Tensor (first line), i find each entrie of the line in the RANS momentum eqaution for the x-direction. Why is it not one line of the Reynolds Stress Tensor going into each equation? Where do i have my thinking error?

Thanks for your help.

FMDenaro August 1, 2019 05:03
I am not sure to understand the key of these doubts...In the eddy viscosity model, you substitute the full unknown tensor with the modelled tensor. One does not pretend to substitute each entry of the Reynolds tensor with a modelled entry.


At the end you have a term like this: Div(2*mu_total*S) being S the symmetric part of the velocity gradient with zero trace. Now, to get the contribution to one of the scalar momentum equations, just project the term along a direction ei

bumper August 1, 2019 06:16
Quote:
Originally Posted by FMDenaro (Post 740709)
I am not sure to understand the key of these doubts...In the eddy viscosity model, you substitute the full unknown tensor with the modelled tensor. One does not pretend to substitute each entry of the Reynolds tensor with a modelled entry.

At the end you have a term like this: Div(2*mu_total*S) being S the dymmetric part of the velocity gradient with zero trace. Now, to get the contribution to one of the scalar momentum equations, just project the term along a direction ei
I want to know if the additonal expressions in the momentum equations that i get with the RANS-Approach are summarized in the Reynolds Stress Tensor.

Expressions in equation for x mometum: First line
Expressions in equation for y mometum: Second line
And so on...

FMDenaro August 1, 2019 08:07
Quote:
Originally Posted by bumper (Post 740713)
I want to know if the additonal expressions in the momentum equations that i get with the RANS-Approach are summarized in the Reynolds Stress Tensor.

Expressions in equation for x mometum: First line
Expressions in equation for y mometum: Second line
And so on...



So, you are not interested to the eddy viscosity modelling but only on the unresolved Reynolds stress? In such case just consider that this tensor is nothing but as same as you have in the classical convective term Div (vv).

Along the x-direction you write i.Div (v'v'), along y j.Div (v'v') and so on (I did not write the averaging).


for example, the x component:


d/dx(u'u')+d/dy(v'u')+d/dz(w'u')

bumper August 1, 2019 08:25
Sure, i"m interested to the eddy viscosity modelling but i want to understand the Reynolds Stress tensor and the equations behind.

If the tensor is nothing but as same as you have in the classical convective term Div, i'm right with the lines of the Reynolds stress tensor and the equations of momentum, right?

FMDenaro August 1, 2019 08:28
Quote:
Originally Posted by bumper (Post 740734)
Sure, i"m interested to the eddy viscosity modelling but i want to understand the Reynolds Stress tensor and the equations behind.

If the tensor is nothing but as same as you have in the classical convective term Div, i'm right with the lines of the Reynolds stress tensor and the equations of momentum, right?



yes, the expression I wrote above is the same in the figure you posted

bumper August 1, 2019 08:35
To sum it up and to check if i unstand it in the right way, the Reynolds Stress Tensor is the summary of additonal entries which i get through the Reynolds-approach in the equations of momentum right?

FMDenaro August 1, 2019 08:55
Quote:
Originally Posted by bumper (Post 740740)
To sum it up and to check if i unstand it in the right way, the Reynolds Stress Tensor is the summary of additonal entries which i get through the Reynolds-approach in the equations of momentum right?

yes, is exactly as same expression as when you write the convective terms, that is d/dxi(u'i u'j).

bumper August 1, 2019 08:57
Perfect, than thanks a lot for your help

IbizaFR August 1, 2019 09:22
Quote:
Originally Posted by LuckyTran (Post 740634)
No that's not how it works. It's not one line of the stress tensor going into each equation, you'll find terms from other lines. 3 versus 1 is not really important. The momentum equation spits out a vector equation. It's 1 equation if you consider the velocity as a 3-d vector, and 3 equations if you consider the velocity in its component form. It's purely a matter of perspective and taste. It's still the same momentum equation.

As an example, the momentum equations look like this when you expand all the summations. Yes this is the navier-stokes & not the RANS. The stress tensor and reynolds stress tensor are similar anyway and the difference is mainly in the effective viscosity. Despite there being 9 combinations of indices, you get only an x,y,z equations. There's no magic here, you just have to follow the rules for the notation being used.



SIMPLE/SIMPLER/PISO is an approach for solving a different problem, the pressure-velocity coupling problem. In CFD the momentum equation gets a lot of attention because there's 1 momentum equation but two variables (pressure and velocity). This does not occur in standard advection-diffusion equations.

Other equations like energy, k, epsilon, etc. don't have this issue. All you need is some type of linear solver (i.e. solve Ax=b). You just need a way to solve this linear system. The dumb way is to invert A. Gauss-Seidel algorithm is another archetype.

Even when you execute the SIMPLE algorithm, you will use similar solvers to solve the individual pieces of the SIMPLE algorithm because the first step in SIMPLE is to solve the momentum equation with an assumed pressure field.

Thank you.
So lets take the k equation in k-epsilon model. Can we discretize this equation with standard schemes applied for convective-diffusive terms like
ADI and solve it for example with Gauss-Seidel or Jacobi?

LuckyTran August 2, 2019 11:02
Quote:
Originally Posted by IbizaFR (Post 740749)
Thank you.
So lets take the k equation in k-epsilon model. Can we discretize this equation with standard schemes applied for convective-diffusive terms like
ADI and solve it for example with Gauss-Seidel or Jacobi?

Yes. Usually in CFD the advection terms are the trouble maker in terms of accuracy and stability, but specifically the advection term in the momentum equation (it's non-linear) is where all the attention goes. The advection terms in k and epsilon equations are linear with respect to k and epsilon and much easier to solve and give much fewer headaches. That's why we don't really talk about them much. If you can solve the much tougher momentum equation, the rest are easier.

FMDenaro August 2, 2019 11:40
Quote:
Originally Posted by LuckyTran (Post 740905)
Yes. Usually in CFD the advection terms are the trouble maker in terms of accuracy and stability, but specifically the advection term in the momentum equation (it's non-linear) is where all the attention goes. The advection terms in k and epsilon equations are linear with respect to k and epsilon and much easier to solve and give much fewer headaches. That's why we don't really talk about them much. If you can solve the much tougher momentum equation, the rest are easier.



Well, actually there is a problem in the transport of k as it is assumed to be a non-negative variable and a monotonicity-preserving scheme should be properly used.

IbizaFR August 2, 2019 15:54
Quote:
Originally Posted by LuckyTran (Post 740905)
Yes. Usually in CFD the advection terms are the trouble maker in terms of accuracy and stability, but specifically the advection term in the momentum equation (it's non-linear) is where all the attention goes. The advection terms in k and epsilon equations are linear with respect to k and epsilon and much easier to solve and give much fewer headaches. That's why we don't really talk about them much. If you can solve the much tougher momentum equation, the rest are easier.
Can you suggest a scheme to solve it?
Thank you.

bumper August 3, 2019 03:29
I have an additional question about the turbulence modelling. For the calculation of the turbulent viscosity, you're going to solve differential equations for k and epsilon.



I found additional euqations for k and epsilon.

k = \frac{1}{2} \overline{u_i' u_i'} = \frac{1}{2} \left(\overline{u_x'^{2}} + \overline{u_y'^{2}} + \overline{u_z'^{2}} \right) = \frac{3}{2} \overline{u'^{2}}


\varepsilon=\frac{k^{3/2}}{l}


During the simulation, i think the differential euqations will be used for k and epsilon. But when and for what usage are the named equations for k and epsilon?


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