Advanced reynold stresses

 \begin{equation*}  <u'u'> = \cfrac{V_{b}}{N\sigma^{3}}\sum_{k=1}^{N}<[q_{\sigma}(r^{k})]^{2} \cfrac{[(\alpha_{3}^{k})^{2}(y-y^{k})^{2}+(\alpha_{2}^{k})^{2}(z-z^{k})^{2}]}{(r^{k})^{6}}>  \end{equation*}(1)

The <uu> component, whose derivation may be found in this page, has to be semplified. In fact, in such a complex format, it is very difficult to make any consideration. We can multiply the fraction by $ [q_{\sigma}(r^{k})]^{2} $ and divide the summ as follows:

$ &amp;lt;u'u'&amp;gt; = \cfrac{V_{b}}{N\sigma^{3}}\sum_{k=1}^{N} &amp;lt;[q_{\sigma}(r^{k})]^{2} \cfrac{ (\alpha_{3}^{k})^{2}(y-y^{k})^{2}}{(r^{k})^{6} }&amp;gt; +&amp;lt;[q_{\sigma}(r^{k})]^{2}\cfrac{(\alpha_{2}^{k})^{2}(z-z^{k})^{2}}{(r^{k})^{6}}&amp;gt; $

Since $ &amp;lt;\alpha^2&amp;gt; $ is independent from all the other terms we are allowed to average it separately and to substitute its own average:

 \begin{equation*}  &amp;lt;u'u'&amp;gt; = \cfrac{V_{b}}{N\sigma^{3}}\sum_{k=1}^{N} &amp;lt;(\alpha_{3}^{k})^{2}&amp;gt;&amp;lt; \cfrac{[q_{\sigma}(r^{k})]^{2} (y-y^{k})^{2}}{(r^{k})^{6} }&amp;gt; +&amp;lt;(\alpha_{2}^{k})^{2}&amp;gt;&amp;lt;\cfrac{[q_{\sigma}(r^{k})]^{2} (z-z^{k})^{2}}{(r^{k})^{6}}&amp;gt;  \end{equation*}(2)

Since we know that $ &amp;lt;\alpha^2&amp;gt; =1 $ we obtain:

$ &amp;lt;u'u'&amp;gt; = \cfrac{V_{b}}{N\sigma^{3}}\sum_{k=1}^{N} &amp;lt; \cfrac{[q_{\sigma}(r^{k})]^{2} (y-y^{k})^{2}}{(r^{k})^{6} }&amp;gt; +&amp;lt; \cfrac{[q_{\sigma}(r^{k})]^{2} (z-z^{k})^{2} }{(r^{k})^{6}}&amp;gt; $

If we now join again the sum under the average we notice that in the upper part there is almoust the radius formulation. In order to get it we add and subtract $ (x-x^{k})^{2} $:

$ &amp;lt;u'u'&amp;gt; = \cfrac{V_{b}}{N\sigma^{3}}\sum_{k=1}^{N} &amp;lt;[q_{\sigma}(r^{k})]^{2} \cfrac{ (y-y^{k})^{2} + (z-z^{k})^{2} +(x-x^{k})^{2} - (x-x^{k})^{2} }{(r^{k})^{6}}&amp;gt; $

Since, by definition, $ (r^{k})^{2} = (x-x^{k})^{2}+ (y-y^{k})^{2}+(z-z^{k})^{2}$

$ &amp;lt;u'u'&amp;gt; = \cfrac{V_{b}}{N\sigma^{3}}\sum_{k=1}^{N} &amp;lt;[q_{\sigma}(r^{k})]^{2} \cfrac{ (r^{k})^{2} - (x-x^{k})^{2} }{(r^{k})^{6}}&amp;gt; $

 \begin{equation*}  &amp;lt;u'u'&amp;gt; = \cfrac{V_{b}}{N\sigma^{3}}\sum_{k=1}^{N} &amp;lt;[q_{\sigma}(r^{k})]^{2}(\cfrac{(r^{k})^{2} - (x-x^{k})^{2} }{(r^{k})^{6}} )&amp;gt;  \end{equation*}(3)

This is the final stage. The first consideration is that, even if semplified, it is very difficult to understand the way $ q_{\sigma}(r^{k}) $ affects the results. In the old SEM, in fact, the shape function was chosen to satisfy $ \int_{V_b}f^2(\mathbf{x})d\mathbf{x}=1 $ has no contribution to the final Reynold Stress, which is instead controlled

$\begin{cases}  \lim_{r \to 0}\cfrac{q(r)}{r^{3}}=Ar^{k} &amp; A \ne 0,r \ge 0 \\ q(r)\epsilon C^{2}\\ \int_{0}^{\infty}q^{2}(r)dr=1\\ q(r)_{r&amp;gt;1}=0 \\ \end{cases}  $

These conditions, together with the following shape function: $ q(r) = A sin^2(\pi r) r $, help us in defining the A constant, which is: $ A = \cfrac{8 \pi}{\sqrt{8 \pi^2 -15}} $

Using the function created we have:

$ \iiint_{V_b} [q_{\sigma}(r^{k})]^{2} ( 1 - \cfrac{ (x-x^{k})^{2} }{(r^{k})^{6}} ) d\mathbf{x}^k = 3.4360 $ (Matlab result - see the script below!!)


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Topic revision: r3 - 2010-06-03 - 17:50:57 - RuggeroPoletto
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25 Mar 2019


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