CFD Online Discussion Forums - VISCOUS DISSIPATION

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Quote:

Originally Posted by Zanje Sachin Bhaskar (Post 596483)

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.

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

Quote:

Originally Posted by alimea (Post 734942)

the correct definition is mu*S:S wherein S is the simmetric velocity gradient

Quote:

Originally Posted by FMDenaro (Post 734944)

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.

Quote:

Originally Posted by alimea (Post 734951)

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

Quote:

Originally Posted by FMDenaro (Post 734955)

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

Quote:

Originally Posted by alimea (Post 734942)

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 (Post 734942)

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.

Quote:

Originally Posted by LuckyTran (Post 734960)

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

Quote:

Originally Posted by alimea (Post 734957)

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.

Quote:

Originally Posted by alimea (Post 734962)

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.

Quote:

Originally Posted by LuckyTran (Post 734960)

Dear Kuck,
That was really helpful.
Thanks for participation in this discussion.