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k-epsilon implementation? (including volScalarField G)

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Old   August 21, 2016, 11:24
Default k-epsilon implementation? (including volScalarField G)
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Dear All.
I'm using: engineFoam (xiFoam), openfoam 2.2.1.

I'm trying to understand k-epsilon turbulence model in openFoam but I can't get through those k and epsilon equations.
Let's focus on k equation:
Code:
    tmp<fvScalarMatrix> kEqn
    (
        fvm::ddt(rho_, k_)
      + fvm::div(phi_, k_)
      - fvm::laplacian(DkEff(), k_)
     ==
        G
      - fvm::SuSp((2.0/3.0)*rho_*divU, k_)
      - fvm::Sp(rho_*epsilon_/k_, k_)
    );
I'm not sure if kEpsilon implements this equation but it looks similar:
\frac{\partial\rho k}{\partial t}+\frac{\partial}{\partial x_i}(\rho V_i k) = \frac{\partial}{\partial x_j}[(\mu+\frac{\mu_{turb}}{\sigma_k})\frac{\partial k}{\partial x_j}]+\mu_{turb} (\sqrt{2S_{ij}S_{ij}})^2-\rho\epsilon

If that's true, then:
fvm::ddt(rho_, k_) = \frac{\partial\rho k}{\partial t}
fvm::div(phi_, k_) = \frac{\partial}{\partial x_i}(\rho V_i k)
fvm::laplacian(DkEff(), k_) = \frac{\partial}{\partial x_j}[(\mu+\frac{\mu_{turb}}{\sigma_k})\frac{\partial k}{\partial x_j}]
fvm::Sp(rho_*epsilon_/k_, k_) = \rho\epsilon


Probably:
- fvm::SuSp((2.0/3.0)*rho_*divU, k_) = -\frac{2}{3}*\rho*k*\frac{\partial V_i}{\partial x_i}
and maybe this term is to remove trace but as long as I cant understand G, I can't go any further.

Definition of G is
Code:
 volScalarField G(GName(), mut_*(tgradU() && dev(twoSymm(tgradU()))));
and I don't know how to understand this.
I can't figure out how:
G - fvm::SuSp((2.0/3.0)*rho_*divU, k_) is equal to \mu_{turb} (\sqrt{2S_{ij}S_{ij}})^2

If someone can correct me or lead me to correct answer it would be great.
Greetings to all.
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Old   August 23, 2016, 14:42
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Have a look here (page 123, at the top)
http://www.tfd.chalmers.se/~lada/pos...-modelling.pdf


About G: You need volScalarField to enter k equation.

Regards,
Darko
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Old   August 24, 2016, 15:17
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Hello. Thank You for this pdf but I can't find full solution (exact implementation formula).

According to my further research k-epsilon turbulent model is based on this report:
"k-epsilon equation for compressible reciprocating engine flows", Journal of Energy, Vol. 7, No. 4 (1983), pp. 345-353.

Unfortunately it is not free.
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Old   August 25, 2016, 04:34
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Hello.

Here is your paper.

https://www.dropbox.com/s/iwnvw5om0b...48086.pdf?dl=0

About your first post, in exact k equation (have look here)

https://en.wikipedia.org/wiki/Turbulence_kinetic_energy

find production term. Insert Bussinesk assumption
\overline{u_i^{\prime}u_j^{\prime}} = \frac{2}{3}\,k\delta_{ij} + ...
in that production term and you will get terms that you need. In the case of incompressible flow,

-\frac{2}{3}\,\rho\, k\,\frac{\partial V_i}{\partial x_i}

is equal to zero.

Best regards,
Darko
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Old   August 25, 2016, 13:21
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Thank You once again, dear Darko. That 2nd report is very useful for me (eg. C_3 constant).

I'm really sorry but it still isn't clear for me.

I have done what you said.
The production term:

P = -\rho \overline{u'_i u'_j}\frac{\partial \overline{u_i}}{\partial x_j}

After applying Boussinesq Eddy Viscosity:
P = - \frac{2}{3} \rho k \delta_{i,j}\frac{\partial \overline{u_i}}{\partial x_j}+\mu_{turb}(\frac{\partial \overline{u_i}}{\partial x_j}+\frac{\partial \overline{u_j}}{\partial x_i})
\frac{\partial \overline{u_i}}{\partial x_j}

Removing delta Kronecker
P = - \frac{2}{3} \rho k \frac{\partial  \overline{u_i}}{\partial x_i}+\mu_{turb}(\frac{\partial  \overline{u_i}}{\partial x_j}+\frac{\partial \overline{u_j}}{\partial  x_i})
\frac{\partial \overline{u_i}}{\partial x_j}


First term in openfoam:
- fvm::SuSp((2.0/3.0)*rho_*divU, k_) = - \frac{2}{3} \rho k \frac{\partial  \overline{u_i}}{\partial x_i}

But I still don't understand how

G = mut_*(tgradU() && dev(twoSymm(tgradU()))) = \mu_{turb}(\frac{\partial  \overline{u_i}}{\partial  x_j}+\frac{\partial \overline{u_j}}{\partial  x_i})
\frac{\partial \overline{u_i}}{\partial x_j}

tgradU is temporary field of gradient U

tgradU = \frac{\partial \overline{u_i}}{\partial x_j}
twoSymm(tgradU) returns 2*symmetric part of a tensor so:

twoSymm(tgradU) = 2* \frac{1}{2}(tgradU+tgradU^T) =
\frac{\partial  \overline{u_i}}{\partial  x_j}+\frac{\partial \overline{u_j}}{\partial  x_i} (Am I right here? Im pretty sure but...)


dev(twoSymm(tgradU())) - this one returns only deviatoric component of tensor tgradU

dev(twoSymm(tgradU())) = \frac{\partial  \overline{u_i}}{\partial  x_j}+\frac{\partial \overline{u_j}}{\partial  x_i} - \frac{1}{3}(\frac{\partial  \overline{u_k}}{\partial  x_k}+\frac{\partial \overline{u_k}}{\partial  x_k})\delta_{i,j} \mathbb{I}
dev(twoSymm(tgradU())) = \frac{\partial   \overline{u_i}}{\partial  x_j}+\frac{\partial \overline{u_j}}{\partial   x_i} - \frac{2}{3}\frac{\partial \overline{u_k}}{\partial  x_k}\delta_{i,j}  \mathbb{I} Is that right?

And at last:
tgradU() && dev(twoSymm(tgradU())) means "double inner product" and it hard to understand.

Definition of double inner product for two 2nd rank tensors:
A:B = A_{i,j}B_{k,l}\delta_{j,k} \delta_{i,l} = A_{i,j}B_{j,i}

tgradU() && dev(twoSymm(tgradU())) = [\frac{\partial \overline{u_i}}{\partial x_j}] : [\frac{\partial    \overline{u_l}}{\partial  x_k}+\frac{\partial \overline{u_k}}{\partial    x_l} - \frac{2}{3}\frac{\partial \overline{u_m}}{\partial   x_m}\delta_{k,l}  \mathbb{I}]


And I have no idea what to do now.
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Last edited by sheaker; August 25, 2016 at 17:18.
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Old   August 26, 2016, 08:12
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Book at this link has very good introduction part about tensors

https://www.dropbox.com/s/72rz5760b3...ou%29.pdf?dl=0

Sooner or later, you will have to learn it, so it is better to start now.
Pay attention to eq. 1.76, it will make things easier for you many times.

If we use (1.148) from above book, it follows:

tgradU() && dev(twoSymm(tgradU())) =
\mu_t\,{\left({\mathbf{D} + \mathbf{\Omega}}\right)}:\left[2\mathbf{D} - \frac{2}{3}(tr \mathbf{D})\mathbf{I})\right] = \mu_t\left(2\mathbf{D}:\mathbf{D} - \frac{2}{3} (tr{\mathbf{D})^2}\right]

Last term is equal to zero when flow is incompressible. It has been utilized that product of symmetric and asymmetric tensor is zero.

Regards,
Darko
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Old   August 26, 2016, 17:26
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Dear Darko. Thank You for a lot of patience to me.

I forgot about splitting gradU into symmetric and antisymmetric parts. I understand that:
(tgradU() && dev(twoSymm(tgradU())) ={\left({\mathbf{D} + \mathbf{\Omega}}\right)}:\left[2\mathbf{D} - \frac{2}{3}(tr \mathbf{D})\mathbf{I}\right] = \left[2\mathbf{D}:\mathbf{D} - \frac{2}{3} (tr{\mathbf{D})^2}\right]

According to previous hints:
\left[{2\mathbf{D}:\mathbf{D} - \frac{2}{3} \left({tr{\mathbf{D}}}\right)^2}\right] should be equal to \left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}\right)\frac{\partial u_i}{\partial x_j}

And I'm not sure about that.
The first one is a scalar and it looks correct but I can't transform it to that second form.

Darko, wish You best!

Last edited by sheaker; August 27, 2016 at 04:32.
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Old   August 27, 2016, 05:25
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I wasn't clear enough.

\mathbf{D} is rate of strain tensor and from the above mentioned link from Chalmers, page 123, you can see that
{2\mathbf{D}}:\mathbf{D} =2 D_{ij}D_{ij} =
\left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}\right)
\frac{\partial u_i}{\partial x_j}

Term - \frac{2}{3} \left(tr{\mathbf{D}}\right)^2 is not in your starting equation.

When I think better from the very beginning we used assumption of incompressible flow through Bussinesq assumption - in the case of compressible flow, Bussinesq assumption is
\mathbf{T} = 2\mu_t\left(\mathbf{D} - \frac{1}{3}(tr{\mathbf{D}}) \mathbf{I}\right) - \frac{2}{3} \rho k \mathbf{I}

If you again insert Bussinesq assumption into production term of TKE, this forgotten term explains existence of - \frac{2}{3} \left(tr{\mathbf{D}}\right)^2 in Openfoam implementation of k-epsilon.

This agrees well with k equation of
http://turbmodels.larc.nasa.gov/ke-chien.html

Of course, in incompressible flow trace of rate of strain tensor is zero.

Regards,
Darko
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Old   August 27, 2016, 09:07
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Thank You for Your help. I think I get it now. So the therm - \frac{2}{3} \left(tr{\mathbf{D}}\right)^2 = - \frac{2}{3} \left(\frac{\partial u_i}{\partial x_i}\right)^2 SHOULD exist in my TKE equation?

\frac{\partial\rho k}{\partial t}+\frac{\partial}{\partial x_i}(\rho u_i k) = \frac{\partial}{\partial x_j}\left[ \left( \mu+\frac{\mu_{turb}}{\sigma_k} \right) \frac{\partial k}{\partial x_j} \right]+\mu_{turb} \left[ \left( \frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i} \right) \frac{\partial u_i}{\partial x_j} - \frac{2}{3} \left(\frac{\partial u_i}{\partial x_i}\right)^2 \right] - \frac{2}{3}\rho k\frac{\partial u_i}{\partial x_i}-\rho\epsilon

And G(openFoam) =
\mu_{turb} \left[ \left(  \frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}  \right) \frac{\partial u_i}{\partial x_j} - \frac{2}{3}  \left(\frac{\partial u_i}{\partial x_i}\right)^2 \right]
Production P = \mu_{turb} \left[ \left(  \frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}  \right) \frac{\partial u_i}{\partial x_j} - \frac{2}{3}  \left(\frac{\partial u_i}{\partial x_i}\right)^2 \right] -  \frac{2}{3}\rho k\frac{\partial u_i}{\partial x_i}

I think that is correct now.


Kindest regards,
Oskar
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Old   August 27, 2016, 11:46
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Couple hours ago I verified this on link in my above post (but now, at the moment of this writing, above link doesn't work).

If you check eq. (24) and eq. (37) from paper "k-e Equation for Compressible Reciprocating Engine Flows", 1983, I believe you will see that is it.

Regards,
Darko
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Old   May 22, 2018, 09:51
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Hello all,
Thank you for this very useful discussion.
I'm using multiphaseEulerFoam (OFv4) to perform simulation of two-phase (gas dispersed in liquid) dispersed flow. To model turbulence, I used the k-epsilon model.
The code has a little bit changed in version 4 (https://github.com/OpenFOAM/OpenFOAM...lon/kEpsilon.C) compared to what you shared in last posts.
In OFv4, we have source terms "kSource()" and "epsilonSource()". My understanding is that they are used for bubble induced turbulence in the case of two-phase dispersed flow.
Do you know how they are implemented and what is their exact definition?
Thank you.
mhrz
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Old   March 22, 2019, 07:02
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I am also interested in kSource()and epsilonSource(). What is its function and meaning?
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Old   July 1, 2019, 18:29
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They are used to modify the turbulence model, e.g. for buoyancy effects. See https://github.com/OpenFOAM/OpenFOAM...uoyantKEpsilon


This turbulence model inherits from the basic compressible k-epsilon turbulence model and overwrites those methods to implement the effect of buoyancy on turbulence.



Quote:
Originally Posted by calf.Z View Post
I am also interested in kSource()and epsilonSource(). What is its function and meaning?
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Old   May 20, 2021, 09:51
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Quote:
Originally Posted by dradenkovic View Post
Hello.

Here is your paper.

https://www.dropbox.com/s/iwnvw5om0b...48086.pdf?dl=0

About your first post, in exact k equation (have look here)

https://en.wikipedia.org/wiki/Turbulence_kinetic_energy

find production term. Insert Bussinesk assumption
\overline{u_i^{\prime}u_j^{\prime}} = \frac{2}{3}\,k\delta_{ij} + ...
in that production term and you will get terms that you need. In the case of incompressible flow,

-\frac{2}{3}\,\rho\, k\,\frac{\partial V_i}{\partial x_i}

is equal to zero.

Best regards,
Darko
Dear Sir, could you please upload the paper again? The link is showing that the paper is deleted.
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Old   May 20, 2021, 10:03
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Quote:
Originally Posted by sheaker View Post
Thank You for Your help. I think I get it now. So the therm - \frac{2}{3} \left(tr{\mathbf{D}}\right)^2 = - \frac{2}{3} \left(\frac{\partial u_i}{\partial x_i}\right)^2 SHOULD exist in my TKE equation?

\frac{\partial\rho k}{\partial t}+\frac{\partial}{\partial x_i}(\rho u_i k) = \frac{\partial}{\partial x_j}\left[ \left( \mu+\frac{\mu_{turb}}{\sigma_k} \right) \frac{\partial k}{\partial x_j} \right]+\mu_{turb} \left[ \left( \frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i} \right) \frac{\partial u_i}{\partial x_j} - \frac{2}{3} \left(\frac{\partial u_i}{\partial x_i}\right)^2 \right] - \frac{2}{3}\rho k\frac{\partial u_i}{\partial x_i}-\rho\epsilon

And G(openFoam) =
\mu_{turb} \left[ \left(  \frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}  \right) \frac{\partial u_i}{\partial x_j} - \frac{2}{3}  \left(\frac{\partial u_i}{\partial x_i}\right)^2 \right]
Production P = \mu_{turb} \left[ \left(  \frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}  \right) \frac{\partial u_i}{\partial x_j} - \frac{2}{3}  \left(\frac{\partial u_i}{\partial x_i}\right)^2 \right] -  \frac{2}{3}\rho k\frac{\partial u_i}{\partial x_i}

I think that is correct now.


Kindest regards,
Oskar
Dear sir, Could you please share the paper and the book shared by Darko Radenkovi? The link is broken.
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Old   April 6, 2022, 05:54
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Dear Sheaker


Do you have any Idea about G?


Here we have
tmp<fvScalarMatrix> kEqn
(
fvm::ddt(alpha, rho, k_)
+ fvm::div(alphaRhoPhi, k_)
- fvm::laplacian(alpha*rho*DkEff(), k_)
==
alpha*rho*G
- fvm::SuSp((2.0/3.0)*alpha*rho*divU, k_)
- fvm::Sp(Ce(D, KK)*alpha*rho*sqrt(k_)/this->delta(), k_)
+ kSource()
+ fvModels.source(alpha, rho, k_)
);



I have understood the term - fvm::SuSp((2.0/3.0)*alpha*rho*divU, k_)
but still this one is unclear:
alpha*rho*G
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Old   April 6, 2022, 05:56
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Quote:
Originally Posted by sheaker View Post
Thank You for Your help. I think I get it now. So the therm - \frac{2}{3} \left(tr{\mathbf{D}}\right)^2 = - \frac{2}{3} \left(\frac{\partial u_i}{\partial x_i}\right)^2 SHOULD exist in my TKE equation?

\frac{\partial\rho k}{\partial t}+\frac{\partial}{\partial x_i}(\rho u_i k) = \frac{\partial}{\partial x_j}\left[ \left( \mu+\frac{\mu_{turb}}{\sigma_k} \right) \frac{\partial k}{\partial x_j} \right]+\mu_{turb} \left[ \left( \frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i} \right) \frac{\partial u_i}{\partial x_j} - \frac{2}{3} \left(\frac{\partial u_i}{\partial x_i}\right)^2 \right] - \frac{2}{3}\rho k\frac{\partial u_i}{\partial x_i}-\rho\epsilon

And G(openFoam) =
\mu_{turb} \left[ \left(  \frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}  \right) \frac{\partial u_i}{\partial x_j} - \frac{2}{3}  \left(\frac{\partial u_i}{\partial x_i}\right)^2 \right]
Production P = \mu_{turb} \left[ \left(  \frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}  \right) \frac{\partial u_i}{\partial x_j} - \frac{2}{3}  \left(\frac{\partial u_i}{\partial x_i}\right)^2 \right] -  \frac{2}{3}\rho k\frac{\partial u_i}{\partial x_i}

I think that is correct now.


Kindest regards,
Oskar
Dear Sheaker


Do you have any Idea about G?


Here we have
tmp<fvScalarMatrix> kEqn
(
fvm::ddt(alpha, rho, k_)
+ fvm::div(alphaRhoPhi, k_)
- fvm::laplacian(alpha*rho*DkEff(), k_)
==
alpha*rho*G
- fvm::SuSp((2.0/3.0)*alpha*rho*divU, k_)
- fvm::Sp(Ce(D, KK)*alpha*rho*sqrt(k_)/this->delta(), k_)
+ kSource()
+ fvModels.source(alpha, rho, k_)
);



I have understood the term - fvm::SuSp((2.0/3.0)*alpha*rho*divU, k_)
but still this one is unclear:
alpha*rho*G
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Old   May 31, 2022, 10:27
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G refers to the production of the TKE. This is written out as:


Code:
nut*(tgradU() && dev(twoSymm(tgradU())))

which is essentially a double inner over gradU(j,i) and 2*S(i,j) - 2/3 S(k,k) \delta(i,j). The first term is the velocity gradient and the second term represents the model for the Reynolds stress less the nut, which is pre-multiplied.
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Old   June 6, 2023, 19:41
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Dear Sourav Hossain

The book is "Papanastasiou, T., Georgiou, G., & Alexandrou, A. N. (2021). Viscous fluid flow. CRC press."
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