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Closing on wall functions - part 7: starting from a profile

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Closing on wall functions - part 7: starting from a profile

Posted April 26, 2022 at 13:12 by sbaffini
Updated April 27, 2022 at 19:57 by sbaffini

It might happens that one doesn't have a turbulent viscosity profile but actually has just an equilibrium profile for velocity or temperature. More specifically:

T^+ = Pr \left(\frac{{s_T^{-1}}^+}{y^+}\right)  y^+

with its obvious extension to the velocity case. In order to go back to the framework presented here one should notice that:

\frac{d}{dy^+}\left({s_T^{-1}}^+\right) = \frac{1}{1+\frac{Pr}{Pr_t}\frac{\mu_t}{\mu}}

from which, it follows that:

\frac{Pr}{Pr_t}\frac{\mu_t}{\mu} = \frac{1}{\frac{d}{dy^+}\left({s_T^{-1}}^+\right)}-1

which is all that is needed to compute (numerically if not doable analytically) the remaining integrals for non equilibrium and/or TKE production terms.

For the non equilibrium terms this leads to the following:

{s_T^i}^+ = \int_0^{y+}{\frac{z^{i+1}}{\left[1+\left(\frac{Pr}{Pr_t}\right)\left(\frac{\mu_t}{\mu}\right)\right]}dz} = \int_0^{y+}z^{i+1}\frac{d}{dz}\left({s_T^{-1}}^+\right)dz

Hence, integration by parts finally leads to:

\left(\frac{{s_T^i}^+}{{y^+}^{i+2}}\right) = \left(\frac{{s_T^{-1}}^+}{y^+}\right)-\frac{\left(i+1\right)}{{y^+}^{i+2}} \int_0^{y+}z^is_T^{-1}dz

Formally, this is the generalized trick that I used here to extend the Reichardt wall law to constant only non equilibrium cases (i.e., i=0).
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