CFD Online Logo CFD Online URL
www.cfd-online.com
[Sponsors]
Home > Forums > General Forums > Main CFD Forum

VISCOUS DISSIPATION

Register Blogs Members List Search Today's Posts Mark Forums Read

Like Tree20Likes

Reply
 
LinkBack Thread Tools Search this Thread Display Modes
Old   April 24, 2016, 15:26
Default
  #21
Member
 
misagh
Join Date: Apr 2012
Posts: 64
Rep Power: 14
misagh is on a distinguished road
Quote:
Originally Posted by Zanje Sachin Bhaskar View Post
But for experimental which is from 110 degrees to 98 degrees which is realistic at distance of x/d=2.5
How much is the error?
How are you sure about the experimental results, since experimental investigations have some error.
The fact that your numerical results do not exactly fit with the expwrimental results can be due to poor mesh quality or bad boundary condition or use of inappropriate models in the solver.
BlnPhoenixKA likes this.
misagh is offline   Reply With Quote

Old   May 28, 2019, 09:23
Default
  #22
Senior Member
 
A. Min
Join Date: Mar 2015
Posts: 305
Rep Power: 12
alimea is on a distinguished road
Hi all,

I have read many papers in which use two different formulas for Viscous Dissipation:

tau : grad(U)
tau : D

where D = 0.5* ( grad(U) + transpose(grad(U)) )

Which one is correct?
Basically, Is there any difference between Stress Work & Viscous Dissipation?

Thanks
alimea is offline   Reply With Quote

Old   May 28, 2019, 09:29
Default
  #23
Senior Member
 
Filippo Maria Denaro
Join Date: Jul 2010
Posts: 6,764
Rep Power: 71
FMDenaro has a spectacular aura aboutFMDenaro has a spectacular aura aboutFMDenaro has a spectacular aura about
Quote:
Originally Posted by alimea View Post
Hi all,

I have read many papers in which use two different formulas for Viscous Dissipation:

tau : grad(U)
tau : D

where D = 0.5* ( grad(U) + transpose(grad(U)) )

Which one is correct?
Basically, Is there any difference between Stress Work & Viscous Dissipation?

Thanks





the correct definition is mu*S:S wherein S is the simmetric velocity gradient
FMDenaro is offline   Reply With Quote

Old   May 28, 2019, 10:47
Default
  #24
Senior Member
 
A. Min
Join Date: Mar 2015
Posts: 305
Rep Power: 12
alimea is on a distinguished road
Quote:
Originally Posted by FMDenaro View Post
the correct definition is mu*S:S wherein S is the simmetric velocity gradient
Thanks, but this formula is just for Newtonian fluid. I want to use the general form of viscous dissipation to calculate for viscoelastic fluid in which tau is a independent tensor.
alimea is offline   Reply With Quote

Old   May 28, 2019, 11:14
Default
  #25
Senior Member
 
Filippo Maria Denaro
Join Date: Jul 2010
Posts: 6,764
Rep Power: 71
FMDenaro has a spectacular aura aboutFMDenaro has a spectacular aura aboutFMDenaro has a spectacular aura about
Quote:
Originally Posted by alimea View Post
Thanks, but this formula is just for Newtonian fluid. I want to use the general form of viscous dissipation to calculate for viscoelastic fluid in which tau is a independent tensor.



In non-Newtonian fluids the viscosity is a function of the strain
FMDenaro is offline   Reply With Quote

Old   May 28, 2019, 11:25
Default
  #26
Senior Member
 
A. Min
Join Date: Mar 2015
Posts: 305
Rep Power: 12
alimea is on a distinguished road
Quote:
Originally Posted by FMDenaro View Post
In non-Newtonian fluids the viscosity is a function of the strain
Not for all of them!
for viscoelastic fluids, we have a complex relation between tau and strain
alimea is offline   Reply With Quote

Old   May 28, 2019, 11:48
Default
  #27
Senior Member
 
Lucky
Join Date: Apr 2011
Location: Orlando, FL USA
Posts: 5,665
Rep Power: 65
LuckyTran has a spectacular aura aboutLuckyTran has a spectacular aura aboutLuckyTran has a spectacular aura about
Quote:
Originally Posted by alimea View Post
Basically, Is there any difference between Stress Work & Viscous Dissipation?
I apologize for mixing two notation styles but expressing the dissipation term is a little difficult (for me) in a general sense using vector notation. I haven't mastered the two notations yet. =(

The stress work in the energy equation is:

\nabla\cdot (u\cdot\tau)
and this can be expanded to:
\nabla\cdot (u\cdot\tau)=u\cdot(\nabla\cdot\tau)+\tau_{ij}\frac{\partial u_{i}}{\partial x_{j} }
The dissipation function is only the last part
\epsilon=\tau_{ij}\frac{\partial u_{i}}{\partial x_{j} }
The other term corresponds to a reversible transfer of kinetic/potential energy of the flow following the work-energy theorem (you can think of it as u times the momentum equation).


An example of this effect is in a developing pipe flow with uniform velocity inlet, and this is most easily seen in the laminar case. The effect of viscous forces accelerates the flow in the center of the pipe to higher than the inlet (into a parabolic profile). If stress work was purely dissipative, the flow wouldn't accelerate.


Only the last term corresponds to the irreversible conversion into thermal energy. Here it is not shown how the last term is irreversible. To do this proof, you have to show that it is positive semi-definite (and maybe invoke some thermodynamic arguments). Physics aside, that leads back to your actual question...

Quote:
Originally Posted by alimea View Post
Hi all,

I have read many papers in which use two different formulas for Viscous Dissipation:

tau : grad(U)
tau : D

where D = 0.5* ( grad(U) + transpose(grad(U)) )
It looks like tau : grad(U) is more correct, unless I made some mathematical mistakes. At least, you know where it comes from.

However, D is computationally efficient because it is a symmetric tensor (less memory and faster to operate on). In numerical codes, it would make sense to use always compute \tau:D instead of \tau:\nabla U. Tau can be arbitrary without loss of generality so it works for newtonian and non-newtonian fluids.
alimea likes this.
LuckyTran is offline   Reply With Quote

Old   May 28, 2019, 11:54
Default
  #28
Senior Member
 
A. Min
Join Date: Mar 2015
Posts: 305
Rep Power: 12
alimea is on a distinguished road
Quote:
Originally Posted by LuckyTran View Post
The stress work in the energy equation is:

\nabla\cdot (u\cdot\tau)
and this can be expanded to:
\nabla\cdot (u\cdot\tau)=u\cdot(\nabla\cdot\tau)+\tau_{ij}\frac{\partial u_{i}}{\partial x_{j} }
The dissipation function is only the last part
\epsilon=\tau_{ij}\frac{\partial u_{i}}{\partial x_{j} }
The other term corresponds to a reversible transfer of kinetic/potential energy of the flow following the work-energy theorem. Only the last term corresponds to the irreversible conversion into thermal energy.


I apologize for mixing two notation styles but expressing the dissipation term is a little difficult in a general sense using vector notation.
Thanks for your complex answer, Licky.
It seems that is what I want. However, I have 2 other questions:
1. why do some papers use
viscous diss = tau : D
instead of what you mentioned (tau:gradU)?

2. what is exactly the first term? when is it important ? why didn't usually we see it in fluid mechanics papers?

regards,
Ali
alimea is offline   Reply With Quote

Old   May 28, 2019, 12:17
Default
  #29
Senior Member
 
Filippo Maria Denaro
Join Date: Jul 2010
Posts: 6,764
Rep Power: 71
FMDenaro has a spectacular aura aboutFMDenaro has a spectacular aura aboutFMDenaro has a spectacular aura about
Quote:
Originally Posted by alimea View Post
Not for all of them!
for viscoelastic fluids, we have a complex relation between tau and strain



The relation you are referring to is tau = mu(strain)*S. The relation is similar to that of Newtonian fluid but is non linear in the viscosity. The definition of the viscous dissipation (a term you will find in the balance of kinetic energy and, with changed sign, in the balance of internal energy) is straightforward.
FMDenaro is offline   Reply With Quote

Old   May 28, 2019, 14:56
Default
  #30
Senior Member
 
Lucky
Join Date: Apr 2011
Location: Orlando, FL USA
Posts: 5,665
Rep Power: 65
LuckyTran has a spectacular aura aboutLuckyTran has a spectacular aura aboutLuckyTran has a spectacular aura about
Quote:
Originally Posted by alimea View Post
Thanks for your complex answer, Licky.
It seems that is what I want. However, I have 2 other questions:
1. why do some papers use
viscous diss = tau : D
instead of what you mentioned (tau:gradU)?

2. what is exactly the first term? when is it important ? why didn't usually we see it in fluid mechanics papers?

regards,
Ali
1. There are a lot of vector/tensor identities that might be hiding that I don't recognize, so I hesitate to say that tau : D is wrong. However, as I mentioned, D is a symmetric matrix, and numerically this is much more efficient to handle than the full grad(U) tensor.

2. First, you find it in the energy equation. Fluid mechanics papers which only write down the continuity and momentum equations (i.e. the Navier-Stokes equations) will never show you this dissipation term (since dissipation is energy and not momentum). Second, the first term exactly balances the kinetic energy and potential energy so you can easily do some manipulations on your energy equation and it cancels some other terms. Depending on which form of energy equation you write down (and there are dozens of equally valid forms all in popular usage), you may or may not see the term.
alimea likes this.
LuckyTran is offline   Reply With Quote

Old   May 29, 2019, 04:21
Default
  #31
Senior Member
 
A. Min
Join Date: Mar 2015
Posts: 305
Rep Power: 12
alimea is on a distinguished road
Quote:
Originally Posted by LuckyTran View Post
I apologize for mixing two notation styles but expressing the dissipation term is a little difficult (for me) in a general sense using vector notation. I haven't mastered the two notations yet. =(

The stress work in the energy equation is:

\nabla\cdot (u\cdot\tau)
and this can be expanded to:
\nabla\cdot (u\cdot\tau)=u\cdot(\nabla\cdot\tau)+\tau_{ij}\frac{\partial u_{i}}{\partial x_{j} }
The dissipation function is only the last part
\epsilon=\tau_{ij}\frac{\partial u_{i}}{\partial x_{j} }
The other term corresponds to a reversible transfer of kinetic/potential energy of the flow following the work-energy theorem (you can think of it as u times the momentum equation).


An example of this effect is in a developing pipe flow with uniform velocity inlet, and this is most easily seen in the laminar case. The effect of viscous forces accelerates the flow in the center of the pipe to higher than the inlet (into a parabolic profile). If stress work was purely dissipative, the flow wouldn't accelerate.


Only the last term corresponds to the irreversible conversion into thermal energy. Here it is not shown how the last term is irreversible. To do this proof, you have to show that it is positive semi-definite (and maybe invoke some thermodynamic arguments). Physics aside, that leads back to your actual question...



It looks like tau : grad(U) is more correct, unless I made some mathematical mistakes. At least, you know where it comes from.

However, D is computationally efficient because it is a symmetric tensor (less memory and faster to operate on). In numerical codes, it would make sense to use always compute \tau:D instead of \tau:\nabla U. Tau can be arbitrary without loss of generality so it works for newtonian and non-newtonian fluids.
Dear Kuck,
That was really helpful.
Thanks for participation in this discussion.
alimea is offline   Reply With Quote

Reply

Thread Tools Search this Thread
Search this Thread:

Advanced Search
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are Off
Pingbacks are On
Refbacks are On


Similar Threads
Thread Thread Starter Forum Replies Last Post
calculating the viscous stress tensor tau for viscous dissipation heating dgadensg OpenFOAM Programming & Development 5 January 17, 2023 15:20
viscous dissipation modelling mahjoob FLUENT 0 January 31, 2007 18:52
temperature rise due to viscous dissipation Marek Main CFD Forum 6 December 30, 2004 16:24
VISCOUS heat dissipation Joseph CFX 0 October 1, 2004 08:52
Is the viscous dissipation term fully included? yann Phoenics 0 June 5, 2003 10:23


All times are GMT -4. The time now is 06:02.