[granzer]

Hello, when using a RANS-EVM for turbulence is the does the turbulence KE dissipation act as a source term in the energy equation or do we need to switch on viscous dissipation to take care of this internal energy source?
If turbulence energy dissipation is not acting as an energy source term where does the dissipated energy go?
What is the difference between turbulence dissipation and viscous dissipation?(I suspect viscous dissipation has to do with the RANS averaged/mean velocity gradient causing shear and turbulence dissipation deals with the dissipation due to variations caused by turbulence but not sure)

[FMDenaro]

Originally Posted by granzer

Hello, when using a RANS-EVM for turbulence is the does the turbulence KE dissipation act as a source term in the energy equation or do we need to switch on viscous dissipation to take care of this internal energy source?
If turbulence energy dissipation is not acting as an energy source term where does the dissipated energy go?
What is the difference between turbulence dissipation and viscous dissipation?(I suspect viscous dissipation has to do with the RANS averaged/mean velocity gradient causing shear and turbulence dissipation deals with the dissipation due to variations caused by turbulence but not sure)

Not sure to understand the key of your question. You can first think about the viscous dissipation of kinetic energy without introducing the RANS equation. It is given explicitly in the kinetic energy equation as (mu*D:D). This term is present in the equation of the internal energy (with opposite sign).

Then, if you apply a statistical averaging and define and eddy viscosity you can analyze also the further terms.

[sbaffini]

Originally Posted by granzer

Hello, when using a RANS-EVM for turbulence is the does the turbulence KE dissipation act as a source term in the energy equation or do we need to switch on viscous dissipation to take care of this internal energy source?
If turbulence energy dissipation is not acting as an energy source term where does the dissipated energy go?
What is the difference between turbulence dissipation and viscous dissipation?(I suspect viscous dissipation has to do with the RANS averaged/mean velocity gradient causing shear and turbulence dissipation deals with the dissipation due to variations caused by turbulence but not sure)

Ideally, when you solve for the total energy, that includes k, and the viscous dissipation is not actually a source/sink but a conservative term in it.

When you go to the k equation, it has a source term (production) which is not really the turbulent dissipation but the mean energy lost to turbulent energy.

Finally, k equation also has a sink term (destruction) that is directly proportional to the turbulent dissipation (which has its own equation).

So, is your question: should the turbulent dissipation in k equation actually appear as source for the total energy equation in order to increase the temperature?

The answer is: really not, because, ideally, you already have k tracked in the total energy and there is no actual sink in it toward the k equation. It's just that your k equation has model sources/sinks that rebalance the distribution of the total energy with respect to k.

Remember, total energy is conserved

[sbaffini]

An additional note.

When you solve for the total energy, in practice, you still store the temperature. So, how comes that you are solving for total E? It comes from how you build convective and diffusive terms.

In the latter you just compute kdT/dn + u*tau*n on faces (with tau including the turbulent part). For the convective term, the total energy has H (enthalpy) in it, so it comes down to build H from T and eventually k. For constant Cp you just do:

H = Cp T + 0.5 V^2 + k + any potential term (depending from how you decide to treat them)

In practice k is typically negligible here and most codes don't include it

[andy_]

Originally Posted by granzer

Hello, when using a RANS-EVM for turbulence is the does the turbulence KE dissipation act as a source term in the energy equation or do we need to switch on viscous dissipation to take care of this internal energy source?
If turbulence energy dissipation is not acting as an energy source term where does the dissipated energy go?
What is the difference between turbulence dissipation and viscous dissipation?(I suspect viscous dissipation has to do with the RANS averaged/mean velocity gradient causing shear and turbulence dissipation deals with the dissipation due to variations caused by turbulence but not sure)

The work done by the mean flow against the Reynolds stresses represents the transfer of energy from the mean flow to the turbulent kinetic energy. This primarily occurs over the largest length scales of the turbulent motion. It is a source term of turbulent kinetic energy. With an eddy viscosity model for the Reynolds stresses it is always positive. In reality and with higher order models like Reynolds stress transport models it can be negative.

The work done by the turbulent motion against the viscous stresses represents the transfer of energy from the turbulent motion to heat. This primarily occurs over the smallest scales (largest velocity gradients). It is always a sink term in the turbulent kinetic energy equation.

The action of vortex stretching progressively reduces the scales of the turbulent motion as it is transported and so the largest areas of production and dissipation are not necessarily in the same place.

Viscous dissipation (of turbulent kinetic energy) is usually the dominant mechanism for dissipating the turbulent motion but, as mentioned above, it can also be put back into the mean flow although this is normally a small term (and a zero term with an eddy viscosity model for the Reynolds stresses). So the viscous dissipation component and the full dissipation are usually pretty much the same but context might matter depending on what is being discussed.

[granzer]

Originally Posted by FMDenaro

Not sure to understand the key of your question. You can first think about the viscous dissipation of kinetic energy without introducing the RANS equation. It is given explicitly in the kinetic energy equation as (mu*D:D). This term is present in the equation of the internal energy (with opposite sign).

Then, if you apply a statistical averaging and define and eddy viscosity you can analyze also the further terms.

Originally Posted by sbaffini

Ideally, when you solve for the total energy, that includes k, and the viscous dissipation is not actually a source/sink but a conservative term in it.

When you go to the k equation, it has a source term (production) which is not really the turbulent dissipation but the mean energy lost to turbulent energy.

Finally, k equation also has a sink term (destruction) that is directly proportional to the turbulent dissipation (which has its own equation).

So, is your question: should the turbulent dissipation in k equation actually appear as source for the total energy equation in order to increase the temperature?

The answer is: really not, because, ideally, you already have k tracked in the total energy and there is no actual sink in it toward the k equation. It's just that your k equation has model sources/sinks that rebalance the distribution of the total energy with respect to k.

Remember, total energy is conserved

Originally Posted by andy_

The work done by the mean flow against the Reynolds stresses represents the transfer of energy from the mean flow to the turbulent kinetic energy. This primarily occurs over the largest length scales of the turbulent motion. It is a source term of turbulent kinetic energy. With an eddy viscosity model for the Reynolds stresses it is always positive. In reality and with higher order models like Reynolds stress transport models it can be negative.

The work done by the turbulent motion against the viscous stresses represents the transfer of energy from the turbulent motion to heat. This primarily occurs over the smallest scales (largest velocity gradients). It is always a sink term in the turbulent kinetic energy equation.

The action of vortex stretching progressively reduces the scales of the turbulent motion as it is transported and so the largest areas of production and dissipation are not necessarily in the same place.

Viscous dissipation (of turbulent kinetic energy) is usually the dominant mechanism for dissipating the turbulent motion but, as mentioned above, it can also be put back into the mean flow although this is normally a small term (and a zero term with an eddy viscosity model for the Reynolds stresses). So the vicious dissipation component and the full dissipation are usually pretty much the same but context might matter depending on what is being discussed.

@FMDenaro, @sbaffini, and @andy_ Thank you for taking the time to explain. I got this question when I was trying to understand what happens when "viscous heating" is not included.
If I understand @sbaffini correctly, then if the "viscous dissipation" term is not included in the energy equation then there won't be any temperature change either due to viscous dissipation (since the term is off)or turbulent dissipation (since this term is a part of the viscous dissipation term in the energy eq, and which has been turned off in the energy eq)? If so there won't be any change in 'kinetic energy' in the Total Energy equation due to the viscous dissipation but the TKE does get and produced and dissipated. Where does the dissipated TKE go in this case?

@andy_ When you say "Viscous dissipation (of turbulent kinetic energy) is usually the dominant mechanism for dissipating the turbulent motion but, as mentioned above, it can also be put back into the mean flow although this is normally a small term (and a zero term with an eddy viscosity model for the Reynolds stresses)" you mean TKE can be put back because the "work done" against Reynolds stress BY the mean flow can be negative (in models like RSM) so its actually work done ON the mean flow?

[andy_]

Originally Posted by granzer

I got this question when I was trying to understand what happens when "viscous heating" is not included.
If I understand @sbaffini correctly, then if the "viscous dissipation" term is not included in the energy equation then there won't be any temperature change either due to viscous dissipation (since the term is off)or turbulent dissipation (since this term is a part of the viscous dissipation term in the energy eq, and which has been turned off in the energy eq)? If so there won't be any change in 'kinetic energy' in the Total Energy equation due to the viscous dissipation but the TKE does get and produced and dissipated. Where does the dissipated TKE go in this case?

If you turn off the viscous dissipation then the work done by the mean flow against the Reynolds stresses will accumulate in the turbulent kinetic energy, raising the turbulent viscosity and progressively turning the fluid into treacle which may stop the flow if the BC allow, blow up the solution or transport the treacly fluid out the solution region having created a strange flow pattern.

Originally Posted by granzer

@andy_ When you say "Viscous dissipation (of turbulent kinetic energy) is usually the dominant mechanism for dissipating the turbulent motion but, as mentioned above, it can also be put back into the mean flow although this is normally a small term (and a zero term with an eddy viscosity model for the Reynolds stresses)" you mean TKE can be put back because the "work done" against Reynolds stress BY the mean flow can be negative (in models like RSM) so its actually work done ON the mean flow?

Yes. It can lead to interesting effects with turbulent curved flows.

[granzer]

Originally Posted by andy_

If you turn off the viscous dissipation then the work done by the mean flow against the Reynolds stresses will accumulate in the turbulent kinetic energy, raising the turbulent viscosity and progressively turning the fluid into treacle which may stop the flow if the BC allow, blow up the solution or transport the treacly fluid out the solution region having created a strange flow pattern.

Yes. It can lead to interesting effects with turbulent curved flows.

Why will there be accumulation of TKE if we switch off viscous dissipation term in the energy equation? Won't the TKE be dissipated by the turbulence dissipation term in the TKE equation as mentioned by sbafinni "
Finally, k equation also has a sink term (destruction) that is directly proportional to the turbulent dissipation (which has its own equation).
"?

[granzer]

Originally Posted by andy_

If you turn off the viscous dissipation then the work done by the mean flow against the Reynolds stresses will accumulate in the turbulent kinetic energy, raising the turbulent viscosity and progressively turning the fluid into treacle which may stop the flow if the BC allow, blow up the solution or transport the treacly fluid out the solution region having created a strange flow pattern.

Yes. It can lead to interesting effects with turbulent curved flows.

Why will there be accumulation of TKE if we switch off viscous dissipation term in the energy equation? Won't the TKE be dissipated by the sink term in the TKE equation as mentioned by sbafinni "
Finally, k equation also has a sink term (destruction) that is directly proportional to the turbulent dissipation (which has its own equation).
" ? In fluent while using the the pressure based solver the viscous dissipation is optionally switched on.

[andy_]

Originally Posted by granzer

Why will there be accumulation of TKE if we switch off viscous dissipation term in the energy equation? Won't the TKE be dissipated by the turbulence dissipation term in the TKE equation as mentioned by sbafinni "
Finally, k equation also has a sink term (destruction) that is directly proportional to the turbulent dissipation (which has its own equation).
"?

We would seem to have something crossed but I am not sure what.

If you keep the production term and remove the dissipation term in the transport equation for turbulent kinetic energy then it will get larger rapidly. No idea why you would want to do this but that is what will happen.

Perhaps the confusion is at my end. By "energy equation" are you referring to the transport equation for the kinetic energy in the turbulent motion or some other form of energy? If some other form what is it and does it include or exclude the kinetic energy in the turbulent motion?

[FMDenaro]

Do not forget you are talking about RANS, not LES. The equations are steady.

[LuckyTran]

Keep in mind that the First Law of Thermodynamics is that energy cannot be created or destroyed.

Note that in Fluent, Star-CCM, and OpenFOAM (and just about any CFD solver), when you activate the viscous heating option, it accounts for only the conversion of bulk mechanical energy into internal energy (i.e. temperature rise due to viscosity). That is, tau*u is calculated using the effective viscosity. I'm not aware of any implementation that accounts for the conversion of turbulent kinetic energy into internal energy. All you would have to do is put the turbulent dissipation rate, epsilon, as a "source" term in the energy equation.

Originally Posted by granzer

@FMDenaro, @sbaffini, and @andy_ Thank you for taking the time to explain. I got this question when I was trying to understand what happens when "viscous heating" is not included.
If I understand @sbaffini correctly, then if the "viscous dissipation" term is not included in the energy equation then there won't be any temperature change either due to viscous dissipation (since the term is off)or turbulent dissipation (since this term is a part of the viscous dissipation term in the energy eq, and which has been turned off in the energy eq)? If so there won't be any change in 'kinetic energy' in the Total Energy equation due to the viscous dissipation but the TKE does get and produced and dissipated. Where does the dissipated TKE go in this case?

If you neglect the viscous heating term, then indeed you do not see a temperature rise (i.e. a change in internal energy). The dissipative term is not really a source/sink term, it is a mechanism for converting mechanical energy into internal energy. The total energy is still conserved.

A practical example is: say you have flow in a pipe with insulated walls, the (total) enthalpy of the flow is constant regardless of whether it is laminar or turbulent or whatever the viscosity is. If the flow in the pipe is turbulent, or it has a viscosity, then the amount of internal energy/tke/pressure varies throughout the pipe, but the (total) enthalpy remains constant regardless.

Correctly accounting for all the dissipative effects only matters when you actually care about the correct % distribution of the energy represented by temperature, by bulk kinetic energy, by turbulent kinetic energy, by pressure-work, etc.

Only when the flow in the pipe has a heat flux through the walls (or there is a magical volumetric heat source somehow) does the total energy get affected. I have to say magical volumetric heat source because remember that First Law. Chemical reactions are just mechanisms for converting chemical potential into internal energy. Nuclear reactions are just mechanisms for exchanging mass and internal energy. If I run a current through the flow, I am converting electric potential into heat. Nothing is disappearing or appearing out of nowhere. I can write the energy equation in a way where it looks deceptively like a source term, but none of these are sources of energy if I write down the correct expression for total energy. And if I write down the wrong expression for the total energy, then I have made a clerical error.

[andy_]

Thanks for the post which has cleared up my misreading of the OP as a KE (turbulence model) question and likely answered it as well.

[granzer]

Originally Posted by andy_

We would seem to have something crossed but I am not sure what.

If you keep the production term and remove the dissipation term in the transport equation for turbulent kinetic energy then it will get larger rapidly. No idea why you would want to do this but that is what will happen.

Perhaps the confusion is at my end. By "energy equation" are you referring to the transport equation for the kinetic energy in the turbulent motion or some other form of energy? If some other form what is it and does it include or exclude the kinetic energy in the turbulent motion?

@andy_ I am sorry for using vague terms! By "energy equation" I meant the total energy equation so it does contain kinetic energy. But the exact form of the equation I was thinking of was the Total energy equation with mechanical/kinetic energy ignored as mentioned in the fluent guide for pressure based equation.(With both kinetic energy and pressure work ignored in the enthalpy form of the equation to be exact. I hope that my understanding is correct! https://www.afs.enea.it/project/nept...th/node107.htm )

Originally Posted by LuckyTran

Keep in mind that the First Law of Thermodynamics is that energy cannot be created or destroyed.

Note that in Fluent, Star-CCM, and OpenFOAM (and just about any CFD solver), when you activate the viscous heating option, it accounts for only the conversion of bulk mechanical energy into internal energy (i.e. temperature rise due to viscosity). That is, tau*u is calculated using the effective viscosity. I'm not aware of any implementation that accounts for the conversion of turbulent kinetic energy into internal energy. All you would have to do is put the turbulent dissipation rate, epsilon, as a "source" term in the energy equation.



If you neglect the viscous heating term, then indeed you do not see a temperature rise (i.e. a change in internal energy). The dissipative term is not really a source/sink term, it is a mechanism for converting mechanical energy into internal energy. The total energy is still conserved.

A practical example is: say you have flow in a pipe with insulated walls, the (total) enthalpy of the flow is constant regardless of whether it is laminar or turbulent or whatever the viscosity is. If the flow in the pipe is turbulent, or it has a viscosity, then the amount of internal energy/tke/pressure varies throughout the pipe, but the (total) enthalpy remains constant regardless.

Correctly accounting for all the dissipative effects only matters when you actually care about the correct % distribution of the energy represented by temperature, by bulk kinetic energy, by turbulent kinetic energy, by pressure-work, etc.

Only when the flow in the pipe has a heat flux through the walls (or there is a magical volumetric heat source somehow) does the total energy get affected. I have to say magical volumetric heat source because remember that First Law. Chemical reactions are just mechanisms for converting chemical potential into internal energy. Nuclear reactions are just mechanisms for exchanging mass and internal energy. If I run a current through the flow, I am converting electric potential into heat. Nothing is disappearing or appearing out of nowhere. I can write the energy equation in a way where it looks deceptively like a source term, but none of these are sources of energy if I write down the correct expression for total energy. And if I write down the wrong expression for the total energy, then I have made a clerical error.

Thank you so much, Professor!! I think I have almost got it. I wrongly used the term "source term" when talking about the total energy equation. If in the Total energy equation the Kinetic energy term is ignored, am I correct in calling the Viscous dissipation term as a "source term"?

[LuckyTran]

Originally Posted by granzer

Thank you so much, Professor!! I think I have almost got it. I wrongly used the term "source term" when talking about the total energy equation. If in the Total energy equation the Kinetic energy term is ignored, am I correct in calling the Viscous dissipation term as a "source term"?

Yes! And this is from where it has its meaning as a source term. If you write a transport equation for temperature or internal energy, then there are temperature sources and internal energy sources. But they are not sources of total energy. Likewise, if there are chemical reactions and nuclear reactions, these appear as sources of temperature and and internal energy. But there is no transport of temperature phenomenon (i.e. there's no way to advect temperature) that leads to a generation of temperature. That is because the temperature source has origins in other physical phenomenon (i.e. it is transferred from mechanical energy or some other potential field). Notice I am saying temperature source and not total energy source.

[FMDenaro]

To provide some more details, let us consider the equation of total energy :


d (rho*E) /dt + div [rho*v*(E+p/rho)] = div q + div (v.T)


that is the Eulerian formulation of the law dE/dt = Q -W. Considering that the work is done by a reversible part due to the isotropic component of the stress tensor and an irreversible part due to the deviatoric part (that is the viscous stress), after applying the statistical averaging, you get :


div <q> = div <[rho*v*(E+p/rho)]> - div <(vp)> - div <(v.Tau)>


That explains the physical elements that balances the mean heat flux. The role of the molecular viscosity appears by means of the irreversible work.

You can similarily deduce the kinetic energy equation and check the dissipative term in it. If you introduce an eddy viscosity model, you just introduce a "turbulent dissipation" that mimics the above term.

[granzer]

Originally Posted by LuckyTran

Yes! And this is from where it has its meaning as a source term. If you write a transport equation for temperature or internal energy, then there are temperature sources and internal energy sources. But they are not sources of total energy. Likewise, if there are chemical reactions and nuclear reactions, these appear as sources of temperature and and internal energy. But there is no transport of temperature phenomenon (i.e. there's no way to advect temperature) that leads to a generation of temperature. That is because the temperature source has origins in other physical phenomenon (i.e. it is transferred from mechanical energy or some other potential field). Notice I am saying temperature source and not total energy source.

Last question..and I think the whole thing would be completely clear. If I am using kepsilon turbulence modle with total energy form of the energy eq(without ignoring mechanics energy etc) then there is no actual 'transfer' of energy from the total energy eq to the turbulence kinetic energy eq k. In this way total energy is conserved but the turbulence kinetic energy is created and destroyed seperately(ie no trasfer from/to total energy eq and turbulence kinetic energy eq). We just assum it's (ie the he turbulence kinetic energy)already a part of the total energy in turbulence fluctuation is being modelled by the k and the epsilon eqn.

[LuckyTran]

The production of k and dissipation of k is included in tau*u. It is straightforward to include TKE in the total energy if you wanted to. Actually the transport equation for tke can be derived from the total energy equation. Write down the total energy transport equation (what Filippo just wrote) and then apply the Reynolds decomposition. And then you just group terms based on their interpretations. Just like you can add two balanced equations together (since 0+0=0) you can also pull apart one equation and turn it into two equations the same way if the terms correctly balance–and they do. That is, if my total energy equation is conceptually a+b =0 and I can show that a=0, then I can write a=0 and b=0 where a is my mean total energy equation and b is my transport equation for tke. You're not really assuming anything. You're just drawing parentheses and brackets around (a)+(b)=0

[granzer]

Originally Posted by FMDenaro

To provide some more details, let us consider the equation of total energy :


d (rho*E) /dt + div [rho*v*(E+p/rho)] = div q + div (v.T)


that is the Eulerian formulation of the law dE/dt = Q -W. Considering that the work is done by a reversible part due to the isotropic component of the stress tensor and an irreversible part due to the deviatoric part (that is the viscous stress), after applying the statistical averaging, you get :


div <q> = div <[rho*v*(E+p/rho)]> - div <(vp)> - div <(v.Tau)>


That explains the physical elements that balances the mean heat flux. The role of the molecular viscosity appears by means of the irreversible work.

You can similarily deduce the kinetic energy equation and check the dissipative term in it. If you introduce an eddy viscosity model, you just introduce a "turbulent dissipation" that mimics the above term.

I am sorry Professor I completely missed this post, maybe because it got posted when I was typing a reply! Yes I have deduced the Kinetic energy eq from the NS eq and Reynolds Averaged. Thank you for your guidance.

Originally Posted by LuckyTran

The production of k and dissipation of k is included in tau*u. It is straightforward to include TKE in the total energy if you wanted to. Actually the transport equation for tke can be derived from the total energy equation. Write down the total energy transport equation (what Filippo just wrote) and then apply the Reynolds decomposition. And then you just group terms based on their interpretations. Just like you can add two balanced equations together (since 0+0=0) you can also pull apart one equation and turn it into two equations the same way if the terms correctly balance–and they do. That is, if my total energy equation is conceptually a+b =0 and I can show that a=0, then I can write a=0 and b=0 where a is my mean total energy equation and b is my transport equation for tke. You're not really assuming anything. You're just drawing parentheses and brackets around (a)+(b)=0

Thank you for your insight and guidance! I have a better grasp of the RANS-based models now!. I went through my collected notes and found that I had applied Renolds decomposition and averaging to the total energy eq but left it a that and had gone to TKE eq derivation from the stress eq or NS eqs! Your input about separating the balanced eqns has been just the insight required. I had'nt thought about that!!