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Due to my previous question, where I asked about flux calculation in lattice Boltzmann (LB) method here, I have more or less same question for deviatoric stress tensor calculation due to pseudo-compressibility of LB method. In fact, strain rate tensor is calculated in LB by using non-equilibrium part of distribution functions ($f_{i}^\text{neq}$) as:

$$\hat{\varepsilon}_{\alpha\beta} = -\frac{1}{2\hat{\tau} \hat{\rho} \hat{c}_{s}^{2}} \sum_{i} f_{i}^\text{neq} c_{i\alpha}c_{i\beta}$$

Where hat quantities are dimensionless quantities, $\hat{\tau}$ is dimensionless relaxation time ($\hat{\tau} > 0.5$), $\hat{\rho}$ is dimensionless instantaneous density of the fluid, $\hat{c}_{s}^{2} = \frac{1}{3}$, and $c_{i\alpha}$ is the $i$th discrete velocity in $\alpha$ direction. Deviatoric stress is defined as:

$$\hat{\sigma}_{\alpha\beta} = 2 \hat{\mu} \hat{\varepsilon}_{\alpha\beta}$$

But, we have for dimensionless viscosity: $\hat{\mu} = \hat{c}_{s}^{2} \left(1-\frac{1}{2\hat{\tau}}\right) \hat{\tau}$

So, finally:

$$\hat{\sigma}_{\alpha\beta} = -\left(1-\frac{1}{2\hat{\tau}}\right) \frac{1}{\hat{\rho}} \sum_{i} f_{i}^\text{neq} c_{i\alpha}c_{i\beta}$$

This is in contrast with what usually LB people use, such as this one from Krüger et. al. as:

$$\hat{\sigma}_{\alpha\beta} = -\left(1-\frac{1}{2\hat{\tau}}\right) \sum_{i} f_{i}^\text{neq} c_{i\alpha} c_{i\beta}$$

I understand at incompressible limit ($Mach \rightarrow 0$), $\hat{\rho}$ should be close to 1, but for my simulations where $Mach \sim 0.06$ and $Re \sim 600$, $\hat{\rho}$ may fluctuate quite a bit. So my question is which of these formulas should be used to calculate deviatoric stress? Should I assume $\hat{\rho} \sim 1$ even with my high $Mach$ number? Any suggestion is truly appreciated.

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  • $\begingroup$ Why don't you try to derive it using a Chapman-Enskog analysis? $\endgroup$ – nluigi Dec 30 '19 at 16:08
  • $\begingroup$ Refer to the following work in Eq. (13): Chemical Engineering Science 64 (2009) 52-58 (Ridha DJEBALI, jbelii_r@hotmail.fr) $\endgroup$ – Refer to the following work in Jan 1 at 9:35
  • $\begingroup$ Can you provide the main point of the article? $\endgroup$ – nicoguaro Jan 1 at 21:12
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Generally, the viscosity we speak about (which is linked to the relax. time) is the kinematic viscosity $\nu$, not $\mu$ as you write. So, by replacing, you will find the right expression.

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    $\begingroup$ Sorry, but no, it's not an answer to my question. The dimensionless dynamics viscosity is defined as: $$\hat{\mu} = \frac{\mu\Delta t}{\rho_{f} \Delta x^{2}}$$ Where $\rho_{f}$ is constant density of fluid. You see that dimensionless dynamic and kinematic viscosities are indeed equal: $$\hat{\mu} = \hat{\nu}$$ So it doesn't matter here. The main idea here is that how close instantaneous and constant fluid densities are or in another how small $Mach$ is. $\endgroup$ – Alone Programmer Dec 28 '19 at 18:33
  • $\begingroup$ @AloneProgrammer - I dont understand your point that dimensionless dynamic and kinematic viscosities are equal. I think you have some confusion about dimensions in LB; you are mixing dimensionless variables with variables in lattice units. You should be consistent in any unit system. Most likely i think RDj is correct and you are simply confusing dynamic and kinematic viscosity. $\endgroup$ – nluigi Dec 30 '19 at 16:13
  • $\begingroup$ Dimensionless dynamic and kinematic viscosities are equal: $$\hat{\mu} = \frac{\mu \Delta t}{\rho_{f} \Delta x^{2}} = \frac{\nu \Delta t} {\Delta x^{2}} = \hat{\nu}$$ where kinematic viscosity is defined as $\nu = \frac{\mu}{\rho_{f}}$. $\endgroup$ – Alone Programmer Dec 30 '19 at 19:10

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