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A generalized thermal/dynamic wall function: Part 2

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A generalized thermal/dynamic wall function: Part 2

Posted October 17, 2016 at 09:24 by sbaffini
Updated December 21, 2016 at 10:07 by sbaffini

In the first part of this post we left with the problem of computing the following integral:




I added all the explicit functional dependencies here because we know f^+\left(y^+,1\right) (u^+\left(y^+\right) given by Monkewitz et al. and reported in the previous post), and would be great to use that solution directly. Indeed, while the above integral is still directly solvable by partial fractions (check yourself with Wolfram alpha), the solution is not really appealing, because it involves 3 non trivial roots of a cubic equation.

In order to avoid that, we absorb the ratio Pr/Pr_t in the \mu_t/\mu definition, and multiply both numerator and denumerator by \left(Pr/Pr_t\right)^2, obtaining:


Note that, if all the instances of k were multiplied by \left(Pr/Pr_t\right) we would have done, as the trick would have been to replace k with k\left(Pr/Pr_t\right) in the Monkewitz solution (that, for example, would happen for the mixing length formulation, but unfortunately it is not integrable).

In order to arrive at this case, we can assume that a=a\left(Pr/Pr_t\right) and a\left(1\right)=a_0, with a_0 the value obtained by calibration with the velocity profile (11.489 for us, 10.306 for Monkewitz et al.). Then we request for a to be a solution of the following equation:


It turns out that for meaningful values of a_0 and k the solution is:

a=\frac{\left(1+\eta+\eta^2\right)a_0}{\theta \eta}


\eta = \sqrt[3]{1+\theta\left(\psi+\sqrt{\psi^2+\phi}\right)}

\psi=\frac{\theta \phi}{2}

\theta = 3 k a_0 \frac{Pr}{Pr_t}

\phi = 3\left(k a_0-1\right)

So, with the a value computed as above, we have that f^+ is just the Mokewitz et al. solution given in the previous post. Thus, to complete our solution, we just need an expression for g^+. I won't report it here, as i just got it from Wolfram alpha (sometimes you need to add the last 3 right parentheses to the Wolfram formula, still don't know why):

This is for the case F_T^+ constant. The fact that this is extendable to F_T^+ in the form of a polynomial can be rapidly checked by integrating the separate terms multiplied by a certain power of x (i used x in place of y in Wolfram)... but i leave this as exercise for the reader .

In part 3 of this post (the last one) i'll sum up the steps required to compute the whole solution and give a MATLAB/Octave script that compares it with the numerical one.
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