adding viscous dissipation to openfoam solver

alimea · January 19, 2018, 09:31

Hi
I want to add viscous dissipation to energy eqn. but my fluid is not Newtonian and has a special stress tensor. So the work of stress tensor field is "tau && grad(U)"

1- is it a correct relation for work of stress tensor field? or I should write:
tau && [grad(U) + transpose of (grad(U))]

2- is it correct to write:

solve
(
fvm::div(phi, T)
- fvm::laplacian(alphaEff, T)
- tau && grad(U)
);

thank you

Tobi · January 22, 2018, 18:01

Hi,

I would expect to have the shear heating given as:

$\boldsymbol \tau \boldsymbol \colon (\nabla \otimes \textbf{U})$

However, I am not sure about Non-Newtonian fluids. The only change should be in the viscosity (which does not follow a linear profile). The stress tensor itself should be similar to those of Newtonian fluids. I might be wrong in that case and have to proof it somewhere in literature. The equation of the shear-heating is correct.

alimea · January 23, 2018, 02:34

Quote:

Originally Posted by Tobi

Hi,

I would expect to have the shear heating given as:

$\boldsymbol \tau \boldsymbol \colon (\nabla \otimes \textbf{U})$

However, I am not sure about Non-Newtonian fluids. The only change should be in the viscosity (which does not follow a linear profile). The stress tensor itself should be similar to those of Newtonian fluids. I might be wrong in that case and have to proof it somewhere in literature. The equation of the shear-heating is correct.

Thank you, I found it. Yes, that is correct. but we should add transpose of grad(U) to your formula:
tau && [grad(U) + transpose of (grad(U))]
this relation is valid for viscoelastic fluid, just we should use viscoelastic stress tensor.

Tobi · January 23, 2018, 03:21

Thanks for the reply. Can you give me a reference for proof? Thus, I can add this information to my book.

elones · January 23, 2018, 10:48

Hi,
this is incorrect.
If Tau is Cauchy stress tensor then (does not matter what kind, i.e., viscoelastic fluid etc.)
$\boldsymbol{\tau} : \boldsymbol{\nabla} \boldsymbol{v} = \boldsymbol{\tau} : \frac{1}{2} \left( \boldsymbol{\nabla} \boldsymbol{v} + (\boldsymbol{\nabla} \boldsymbol{v})^T \right),$
but note that by "additing" transpose of gradient of the velocity to the formula, you really gives less information in the formula as
$\frac{1}{2} \left( \boldsymbol{\nabla} \boldsymbol{v} + (\boldsymbol{\nabla} \boldsymbol{v})^T \right),$
is symmetric part of the gradient of the velocity. This is indeed used in many proofs.

Best Ondra

Tobi · January 23, 2018, 13:00

I don't get your point. Who is wrong and what is correct now?
In general the Cauchy stress tensor is defined as $\boldsymbol \sigma$ while the shear tensor is $\boldsymbol \tau$ such as:

$\boldsymbol \sigma = \boldsymbol \tau - p\textbf{I}$

while $\boldsymbol \tau$ is the deviatoric part of the Cauchy stress tensor and the negative pressure represents the hydrostatic part such as:

$\boldsymbol \sigma= \boldsymbol \sigma^\mathrm{dev} + \boldsymbol \sigma^\mathrm{hyd}$

Based on Bird et al. the viscose heating is related to:

$\boldsymbol \tau \boldsymbol \colon (\nabla \otimes \textbf{U})$

but it is mentioned that you have to have an symmetric shear tensor ( $\tau_{xy} = \tau_{yx}$ ). Thus, your relation is only valid for symmetric ones; if this was the thing you wanted to point out, I just clearyfied the things. If there is some mistake, please don't hesitate to correct me.

only valid for symmetric shear rate tensor $\textbf{D}$

while $\textbf{D}$ is given as:

$\textbf{D} = \frac{1}{2} [\nabla \otimes \textbf{U} + (\nabla \otimes \textbf{U})^\mathrm{T}]$

elones · January 23, 2018, 13:24

Quote:

Originally Posted by Tobi

I don't get your point. Who is wrong and what is correct now?
In general the Cauchy stress tensor is defined as $\boldsymbol \sigma$ while the shear tensor is $\boldsymbol \tau$ such as:

$\boldsymbol \sigma = \boldsymbol \tau - p\textbf{I}$

while $\boldsymbol \tau$ is the deviatoric part of the Cauchy stress tensor and the negative pressure represents the hydrostatic part such as:

$\boldsymbol \sigma= \boldsymbol \sigma^\mathrm{dev} + \boldsymbol \sigma^\mathrm{hyd}$
you mean the gradient of the velocity. Note that

Based on Bird et al. the viscose heating is related to:

$\boldsymbol \tau \boldsymbol \colon (\nabla \otimes \textbf{U})$

but it is mentioned that you have to have an symmetric shear tensor ( $\tau_{xy} = \tau_{yx}$ ). Thus, your relation is only valid for symmetric ones; if this was the thing you wanted to point out, I just clearyfied the things. If there is some mistake, please don't hesitate to correct me.

only valid for symmetric shear rate tensor $\textbf{D}$

while $\textbf{D}$ is given as:

$\textbf{D} = \frac{1}{2} [\nabla \otimes \textbf{U} + (\nabla \otimes \textbf{U})^\mathrm{T}]$

Sorry, I forget to cite alimea post. My post was reaction mainly to point-out that it does not matter if the Newtonian fluid is under consideration.

I defined Cauchy stress as Tau not Sigma, but it really does not matter in our discussion. Cauchy stress is always symmetric and the D is defined as symmetric part of the gradient of the velocity. So the formula holds always within classical mechanics of continuum.

Tobi · January 23, 2018, 14:55

I agree, the nomenclature does not play a big deal here. However, do you have any example in which the Cauchy stress tensor, or the general formulation - given in Gurtin et al. - The Mechanics and Thermodynamics of Continua - is not symmetric?

PS: Did you comment in the quotes? I did not mean the gradient. What did you mean there?

elones · January 23, 2018, 15:09

Quote:

Originally Posted by Tobi

I agree, the nomenclature does not play a big deal here. However, do you have any example in which the Cauchy stress tensor, or the general formulation - given in Gurtin et al. - The Mechanics and Thermodynamics of Continua - is not symmetric?

No, as long as conservation of angular momentum holds the Cauchy stress tensor is symmetric.

Hgholami · March 22, 2019, 00:54

Dear alimea
Do you add viscous dissipation to energy equation? I have the same problem, if favor me, I pleasure. thanks

alimea · March 22, 2019, 13:48

Quote:

Originally Posted by Hgholami

Dear alimea
Do you add viscous dissipation to energy equation? I have the same problem, if favor me, I pleasure. thanks

Hi
what's your problem?

Hgholami · March 22, 2019, 14:17

Dear alimea
I want to add a viscous heating term to energy equation as below

alimea · March 23, 2019, 17:21

Quote:

Originally Posted by Hgholami

Dear alimea
I want to add a viscous heating term to energy equation as below

ok. what's the problem?

Hgholami · March 24, 2019, 00:05

I want to add last term to energy equation, So I uses

Quote:

tau=rho_*fluid.nu()*(twoSymm(fvc::grad(U_)))
I22=(tau && fvc::grad(U_)); //for I2/2

Is it true?

alimea · March 24, 2019, 01:09

Quote:

Originally Posted by Hgholami

I want to add last term to energy equation, So I uses Is it true?

Yes. But don't forget that there is a viscosity in your first formula. in fact:

stressWork = (1.0/(rho*Cp))*( tau && fvc::grad(U) );

Regards

Hgholami · March 24, 2019, 01:19

Quote:

Originally Posted by alimea

Yes. But don't forget that there is a viscosity in your first formula. in fact:

stressWork = (1.0/(rho*Cp))*( tau && fvc::grad(U) );

Regards

In first Formula, fluid.nu() is dynamic viscosity that multiply with density to give kinematic viscosity?

aerothermal · November 17, 2024, 17:26

Quote:

Originally Posted by alimea

Hi
I want to add viscous dissipation to energy eqn. but my fluid is not Newtonian and has a special stress tensor. So the work of stress tensor field is "tau && grad(U)"

1- is it a correct relation for work of stress tensor field? or I should write:
tau && [grad(U) + transpose of (grad(U))]

2- is it correct to write:

solve
(
fvm::div(phi, T)
- fvm::laplacian(alphaEff, T)
- tau && grad(U)
);

thank you

Code:

  // Step 1: Calculate deviatoric stress tensor
  volSymmTensorField devRhoReff = turbulence->devRhoReff();

  // Step 2: Compute the velocity gradient tensor
  volTensorField gradU = fvc::grad(U);

  // Step 3: Compute the viscous dissipation term
  volScalarField viscousDissipation("viscousDissipation", devRhoReff && gradU);

  // Optional: Log the min and max of viscous dissipation for debugging
  Info << "Viscous Dissipation (min, max): "
    << min(viscousDissipation).value() << ", "
    << max(viscousDissipation).value() << endl;

  // Step 4: Construct the enthalpy equation
  fvScalarMatrix hEqn
  (
    fvm::ddt(rho, h)
  + fvm::div(phi, h)
  - fvm::laplacian(turbulence->alphaEff(), h)
  ==
    DpDt + viscousDissipation
  );

based on

https://develop.openfoam.com/Develop...sDissipation.C

January 19, 2018, 09:31	adding viscous dissipation to openfoam solver	#1
alimea Senior Member A. Min Join Date: Mar 2015 Posts: 309 Rep Power: 13	Hi I want to add viscous dissipation to energy eqn. but my fluid is not Newtonian and has a special stress tensor. So the work of stress tensor field is "tau && grad(U)" 1- is it a correct relation for work of stress tensor field? or I should write: tau && [grad(U) + transpose of (grad(U))] 2- is it correct to write: solve ( fvm::div(phi, T) - fvm::laplacian(alphaEff, T) - tau && grad(U) ); thank you

January 22, 2018, 18:01		#2
Tobi Super Moderator Tobias Holzmann Join Date: Oct 2010 Location: Bad Wörishofen Posts: 2,713 Blog Entries: 6 Rep Power: 52	Hi, I would expect to have the shear heating given as: $\boldsymbol \tau \boldsymbol \colon (\nabla \otimes \textbf{U})$ However, I am not sure about Non-Newtonian fluids. The only change should be in the viscosity (which does not follow a linear profile). The stress tensor itself should be similar to those of Newtonian fluids. I might be wrong in that case and have to proof it somewhere in literature. The equation of the shear-heating is correct. alimea likes this. __________________ Keep foaming, Tobias Holzmann

January 23, 2018, 03:21		#4
Tobi Super Moderator Tobias Holzmann Join Date: Oct 2010 Location: Bad Wörishofen Posts: 2,713 Blog Entries: 6 Rep Power: 52	Thanks for the reply. Can you give me a reference for proof? Thus, I can add this information to my book. __________________ Keep foaming, Tobias Holzmann

January 23, 2018, 10:48		#5
elones New Member Join Date: Mar 2014 Location: Czech Republic Posts: 29 Rep Power: 15	Hi, this is incorrect. If Tau is Cauchy stress tensor then (does not matter what kind, i.e., viscoelastic fluid etc.) $\boldsymbol{\tau} : \boldsymbol{\nabla} \boldsymbol{v} = \boldsymbol{\tau} : \frac{1}{2} \left( \boldsymbol{\nabla} \boldsymbol{v} + (\boldsymbol{\nabla} \boldsymbol{v})^T \right),$ but note that by "additing" transpose of gradient of the velocity to the formula, you really gives less information in the formula as $\frac{1}{2} \left( \boldsymbol{\nabla} \boldsymbol{v} + (\boldsymbol{\nabla} \boldsymbol{v})^T \right),$ is symmetric part of the gradient of the velocity. This is indeed used in many proofs. Best Ondra alimea likes this.

January 23, 2018, 13:00		#6
Tobi Super Moderator Tobias Holzmann Join Date: Oct 2010 Location: Bad Wörishofen Posts: 2,713 Blog Entries: 6 Rep Power: 52	I don't get your point. Who is wrong and what is correct now? In general the Cauchy stress tensor is defined as $\boldsymbol \sigma$ while the shear tensor is $\boldsymbol \tau$ such as: $\boldsymbol \sigma = \boldsymbol \tau - p\textbf{I}$ while $\boldsymbol \tau$ is the deviatoric part of the Cauchy stress tensor and the negative pressure represents the hydrostatic part such as: $\boldsymbol \sigma= \boldsymbol \sigma^\mathrm{dev} + \boldsymbol \sigma^\mathrm{hyd}$ Based on Bird et al. the viscose heating is related to: $\boldsymbol \tau \boldsymbol \colon (\nabla \otimes \textbf{U})$ but it is mentioned that you have to have an symmetric shear tensor ( $\tau_{xy} = \tau_{yx}$ ). Thus, your relation is only valid for symmetric ones; if this was the thing you wanted to point out, I just clearyfied the things. If there is some mistake, please don't hesitate to correct me. only valid for symmetric shear rate tensor $\textbf{D}$ while $\textbf{D}$ is given as: $\textbf{D} = \frac{1}{2} [\nabla \otimes \textbf{U} + (\nabla \otimes \textbf{U})^\mathrm{T}]$ __________________ Keep foaming, Tobias Holzmann

January 23, 2018, 14:55		#8
Tobi Super Moderator Tobias Holzmann Join Date: Oct 2010 Location: Bad Wörishofen Posts: 2,713 Blog Entries: 6 Rep Power: 52	I agree, the nomenclature does not play a big deal here. However, do you have any example in which the Cauchy stress tensor, or the general formulation - given in Gurtin et al. - The Mechanics and Thermodynamics of Continua - is not symmetric? PS: Did you comment in the quotes? I did not mean the gradient. What did you mean there? __________________ Keep foaming, Tobias Holzmann

March 22, 2019, 00:54		#10
Hgholami Senior Member Hojatollah Gholami Join Date: Jan 2019 Posts: 171 Rep Power: 8	Dear alimea Do you add viscous dissipation to energy equation? I have the same problem, if favor me, I pleasure. thanks

March 22, 2019, 14:17		#12
Hgholami Senior Member Hojatollah Gholami Join Date: Jan 2019 Posts: 171 Rep Power: 8	Dear alimea I want to add a viscous heating term to energy equation as below

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