The DF-SEM so far

So far we defined the DF-SEM according to the following formula:

$ \mathbf{u}'=\sqrt{\frac{1}{N}}\sum_{k=1}^{N}\frac{q_{\sigma}(\frac{r^{k}}{\sigma})}{(\frac{r^{k}}{\sigma})^{3}}\frac{\mathbf{r}^{k}}{\sigma}\times[R_L^G (\sqrt{2 (k' - \lambda_i)} \varepsilon_i^k)^L] $

In this formula we had $ r^{k} $ which is the position relative to the eddy centre and $ q_{\sigma} $ is the shape function. We may think of this formulation as divided into two different parts:

  1. $ \frac{q_{\sigma}(\frac{r^{k}}{\sigma})}{(\frac{r^{k}}{\sigma})^{3}} $ - In here the shape function is involved. This shape function "somehow" defines the velocity distribution around the eddy.
  2. $ \frac{\mathbf{r}^{k}}{\sigma}\times[R_L^G (\sqrt{2 (k' - \lambda_i)} \varepsilon_i^k)^L] $ - This second part is very importat. Beside the fact that we used it in order to reproduce some anisotropy, the cross product MODIFIES the velocity distribution inside the eddy and MAKES the method divergence free!

A new approach

In order to solve the anisotropy reproduction I modified the present formulation as follows:

$ u'_i=\sqrt{\frac{1}{N}}\sum_{k=1}^{N} q_i(\frac{x_i-x_{i}^{k}}{\sigma_i})\frac{\mathbf{r}^{k}}{\mathbf{\sigma}}\times\mathbf{\alpha}^k $

The present formulation is obtaned from the previous one by simply defining $ q_i = q_\sigma (\frac{r^{k}}{\sigma})^{3} $

The new formulation allows us to use three different $ q_{i} $, one for u, one for v and another for w and three different $ \sigma $ as well. The main constraint the chosen shape function must satisfy is:

$ \frac{\partial q_x}{\partial x}(\frac{y - y^k}{\sigma_y}) = \frac{\partial q_y}{\partial y}(\frac{x - x^k}{\sigma_x}) $

$ \frac{\partial q_y}{\partial y}(\frac{z - z^k}{\sigma_z}) = \frac{\partial q_z}{\partial z}(\frac{y - y^k}{\sigma_y}) $

$ \frac{\partial q_z}{\partial z}(\frac{x - x^k}{\sigma_x}) = \frac{\partial q_x}{\partial x}(\frac{z - z^k}{\sigma_z}) $

Which can be simplified in the following system:

$ \frac{\partial q_x}{\partial x} = (\frac{x - x^k}{\sigma_x}) $

$ \frac{\partial q_y}{\partial y} = (\frac{y - y^k}{\sigma_y}) $

$ \frac{\partial q_z}{\partial z} = (\frac{z - z^k}{\sigma_z}) $

Among the solutions of the present system the one I chosen is: $ q_i = \sigma_i (1 - r^2) $

Is the present method divergence free? YES it is! I checked all the calculations by using symbolic matlab and I can confirm that!

Reynolds stress

Crucial point is now the reynolds stress reproduction. Analysis sais that:

$ <u u> = \frac{1}{N} \Sigma [<\alpha_3^2><(q_x \frac{y - y^k}{\sigma_y})^2>+<\alpha_2^2><(q_x \frac{z - z^k}{\sigma_z})^2>] $

$ <v v> = \frac{1}{N} \Sigma [<\alpha_1^2><(q_y \frac{z - z^k}{\sigma_z})^2>+<\alpha_3^2><(q_y \frac{x - x^k}{\sigma_x})^2>] $

$ <w w> = \frac{1}{N} \Sigma [<\alpha_2^2><(q_z \frac{x - x^k}{\sigma_x})^2>+<\alpha_1^2><(q_z \frac{y - y^k}{\sigma_y})^2>] $

In here we can actually have an idea of the influence of the three shape functions to the output result. It seems to suggest a quadratic relation!

N.B. = Shear stress are NEVER a problem: if we are able to reproduce main stresses we rotate the eddy in order to reproduce them - This has a phisical meaning as well.

Now, according to the choice we made about $ q_i $, we can write:

$ <uu> = \frac{1}{N} \Sigma \sigma_x^2 [<\alpha_3^2><((1-r^2) \frac{y - y^k}{\sigma_y})^2>+<\alpha_2^2><((1-r^2) \frac{z - z^k}{\sigma_z})^2>] $

$ <vv> = \frac{1}{N} \Sigma \sigma_y^2 [<\alpha_1^2><((1-r^2) \frac{z - z^k}{\sigma_z})^2>+<\alpha_3^2><((1-r^2) \frac{x - x^k}{\sigma_x})^2>] $

$ <ww> = \frac{1}{N} \Sigma \sigma_z^2 [<\alpha_2^2><((1-r^2) \frac{x - x^k}{\sigma_x})^2>+<\alpha_1^2><((1-r^2) \frac{y - y^k}{\sigma_x})^2>] $

We must remember that $ \sigma_i $ appears in the $ r $ definition as well, making the relation with the normal stress a bit more complicated than a simple quadratic one!!

Here you are two results obtained with two sets of sigmas:

sigma = (0.1,5.5,5.5), <uiui> = (2.25,1.1,1.2)

sigma = (0.5,0.5,0.5), <uiui> = (0.1,0.1,0.1)

sigma = (1.0,0.1,0.1), <uiui> = (0.2,0.3,0.3)

PS: results still suffer of lack of normalization!!!

These results suggest a different relation between the length scale and the reproduced Reynolds stress. Why is it this way? I think the main contribution to this result is given by the averaging of $ &amp;lt;((1-r^2) \frac{x_i - x_i^k}{\sigma_i}&amp;gt; $. Because of the cross product, $ \frac{x-x^k}{\sigma_x} $ (the term containing information about the x-length scale) appears squared in <vv> and <ww>!

Another fact that I noticed is that the new methodology is not capable of reproducing high anisotropic structures as well (the main problem comes from the fact that we are summing two terms!!). The good news is that we can combine the two methodologies and the final results is very promising:


The present idea gives the opportunity to reproduce some anisotropy and must be used together with the solution we have so far in order to reproduce more generic turbulence. At the present stage, I am not able to fully understand the relation between the reynolds stress and $ \sigma $, but it is a competence we can achieve!

Updates 14/11

Eddies shape

The u' velocity distribution around the eddy with $ \sigma_x = 0.5 $, $ \sigma_y = 0.75 $ and $ \sigma_z = 0.25 $.

Method abilities

The method showed the capability of generating high anisotropic turbulence (as expected). The main problem, at the moment, regards its normalization!! In fact, a dependency on the length scale is still present as shown by the following graph:

As defined so far, a linear dependency on the turbulence length scale is highlithed. Normalization process is still on going!

Updates 21/11

for different rations $ \frac{\sigma_x}{\sigma_z} $ I plotted the anisotropy reproduced by the present DF-SEM. It is easy to see that most of the lumley triangle is captured (very attractive feature). The main problem concern the turbulent kinetic energy, which does not remain constant whilst $ \sigma_x $, $ \sigma_y $ and $ \sigma_z $ vary! This is an unwanted feature, because the kinetic energy must be imposed by the user!!

The general behavior is explained by the present plot, which shows for $ \frac{\sigma_x}{\sigma_z} = 3 $ the reproduced Reynolds stresses at various $ \sigma_y $. The turbulent kinetic energy present a spike when \sigma_x = \sigma_y. A normalization must be found in order to apply the method!

The NEW DF-SEM (fully anisotropy reproducibility)

First, let's recall the Reynolds stress formulation in the present method:

$ &amp;lt;uu&amp;gt; = \frac{1}{N} \Sigma \sigma_x^2 [&amp;lt;\alpha_3^2&amp;gt;&amp;lt;((1-r^2) \frac{y - y^k}{\sigma_y})^2&amp;gt;+&amp;lt;\alpha_2^2&amp;gt;&amp;lt;((1-r^2) \frac{z - z^k}{\sigma_z})^2&amp;gt;] $

$ &amp;lt;vv&amp;gt; = \frac{1}{N} \Sigma \sigma_y^2 [&amp;lt;\alpha_1^2&amp;gt;&amp;lt;((1-r^2) \frac{z - z^k}{\sigma_z})^2&amp;gt;+&amp;lt;\alpha_3^2&amp;gt;&amp;lt;((1-r^2) \frac{x - x^k}{\sigma_x})^2&amp;gt;] $

$ &amp;lt;ww&amp;gt; = \frac{1}{N} \Sigma \sigma_z^2 [&amp;lt;\alpha_2^2&amp;gt;&amp;lt;((1-r^2) \frac{x - x^k}{\sigma_x})^2&amp;gt;+&amp;lt;\alpha_1^2&amp;gt;&amp;lt;((1-r^2) \frac{y - y^k}{\sigma_x})^2&amp;gt;] $

These depend on both $ \sigma_{x_i} $ and $ \alpha_i $. The easiest way to reproduce a general stress is then:

  • use $ \sigma_{x_i} $ to reproduce some particular points (the chosen ones lyes along the "one large eigenvalue" edge)
  • use then $ \alpha_i $ to reproduce an area around those points

In the following picture I chose 4 points (in yellow) and with them I managed to reproduce most of the lumley triangle (the top-right part is not perfectly covered because it requires an infinite number of points ...). With this mapping, the present method is able to reproduce the complete anisotropy presents in a $ Re_{\tau}=395 $ channel flow.

I want to stress that for each of this 4 zones we have structures more and more elipsoids (the grey zone has spherical structures, the red one has structures with one axis which is $\sqrt{2}$ times the others, the green one $\sqrt{4}=2$ times and the blue one $\sqrt{8}$). These structures are alligned to the local reference system (where the Reynolds tensor is diagonal) and are rotated into the global one during the computation!


The new method is fully implemented in my fortran and it confirmed its capabilities!! (See the following pictures where the following stress has been reproduced: uu = 6.5, vv = 2.0, ww = 1.0, uv = 1.0, uw = 0.0 and vw = 0.5 - [these are just random number picked in order to define a Reynolds tensor outside the area reproducible by the previous DF-SEM])

Soon shall follow the implementation into Code_Saturne (which should not require so long as there are not so many differences with the previous DF-SEM, even though a couple of them require a bit of attention). As we had scheduled, by Christmas the implementation work is gonna be over and at the beginning of the new year we can start some test case application.


Today, after I implemented the new method in Saturne, I tried to run the channel flow and I realized that my previous conclusions were partially wrong! Basically the Lumley triangle is not as stated earlier but quite different (see following pic!)


Basically I had forgot to verify one condition on the anisotropy, which returns a limitation in the 2C region of the lumley triangle! The problem may concern some cases (for a channel flow $ Re_\tau = 395 $ only turbulence at $ y^+ &amp;lt; 8 $ gives problems) and can be partially solved by increasing the number of mapping zone (see the following picture!!) lumleyfinal05.eps

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Topic revision: r5 - 2011-12-13 - 14:25:41 - RuggeroPoletto
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22 Jul 2018


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