[blockMesh] About Simple Grading in blockMesh

ce73stargazer · January 10, 2016, 21:47

I understand simple grading is the ratio between the size of the end cell and the size of initial cell.
How about the cell size in between, what kind of changes they follow to reach the final cell?
Like constant changes?
For instane, the first cell is size 10 and the last cell is size 5, the middle cells will have a size 10-dx*(N-1) ? where the dx is the constant changes between cells, N is the number of cells in that direction.

This is how i understand it, not sure if thats right.

I am writing a small code to output the cell ID by just inputting the x,y coordinates, however i realised that it only work fine for uniform distributed mesh, not the one with gradings. Hence i came out with this question. Any one have any suggestions? (Or if anyone has faster way to output cell ID (in paraview) and dont mind share with me)

Many Thanks.

Antimony · January 12, 2016, 05:45

Hi,

The expansion is geometric, not arithmetic.

Here is a useful tool that gives you an idea about the expansion from the start cell to the end cell:

https://openfoamwiki.net/index.php/S...ng_calculation

Zetison · May 7, 2020, 06:50

What is the exact formula used here? With

being the number of cells, $r=\frac{\delta_e}{\delta_s}$ being the expansion ratio,

the total length, and based on the distribution of the nodes
$x_i = x_0 + L\left(\frac{i}{n}\right)^g\quad i=0,\dots,n$
the following formula seems reasonable to me
$\delta_s = \frac{L}{n^g}$
$\delta_e = L\left[1-\left(\frac{n-1}{n}\right)^g\right]$
Alternatively, the formulas for a geometric expansion
$x_i = x_0 + L\frac{g^i-1}{g^n-1}\quad i=0,\dots,n$
$\delta_s = L\frac{g-1}{g^n-1}$
$\delta_e = L\left[1-\frac{g^{n-1}-1}{g^n-1}\right]$
but neither matches the program (or OpenFoam?). What is wrong here?

Tobermory · May 8, 2020, 08:47

Apologies, I did not try to debug your maths above, but for simplicity I will give you the correct expressions below in a simpler and more standard terminology.

For a set of

cells, with vertices

, where

varies from 0 to

, then the cell width is the gap between the vertices. If we define a constant stretching rate of $\beta$ then:

$\delta_i = (x_{i+1} - x_i) = \beta \delta_{i-1}$

Armed with this, you can get to the following:

$\delta_i = \beta^i \delta_0$

now $r = \delta_{n-1} / \delta_0$ , the ratio of the last to the first cell width, so:

$r = \beta^{n-1}$

The $\delta_i$ 's form a geometric progression, the sum of which is:

$Sum_i = \delta_0 + \delta_1 + ... + \delta_i = \delta_0 (1 - \beta^i)/(1 - \beta)$

So, the coord of vertex

is:

$x_i = x_0 + Sum_i = x_0 + \delta_0 (1 - \beta^i)/(1 - \beta)$

Finally, the cell centre is defined as

$xc_i = x_i + 0.5 \delta_i$

so you can then calculate the cell coords using

. This is probably what you are outputing from OpenFOAM when you are doing the comparison.

Here's a worked example:
n = 5
beta = 1.3
r = 2.8561

i, x, dx, xc
0 0.000 0.100 0.050
1 0.100 0.130 0.165
2 0.230 0.169 0.315
3 0.399 0.220 0.509
4 0.619 0.286 0.762
5 0.904

Hope this helps - good luck!

(PS - can someone tell me how to format a table in a post, to get the spacing right?)

Zetison · May 9, 2020, 06:37

Thanks for your reply, your formula gets me very close. Given a length

, the width of the start cell $\delta_0$ and the end cell $\delta_{N-1}$ one can compute the number of cells

using the equations

$r = \beta^{n-1}=\frac{\delta_{n-1}}{\delta_0}$

Solving yields
$\beta = \frac{L-\delta_0}{L-\delta_{N-1}}$
such that
$N = 1+\frac{\ln\frac{\delta_{N-1}}{\delta_0}}{\ln\frac{L-\delta_0}{L-\delta_{N-1}}}$
For $L = 1, \delta_0 = 0.05, \delta_{N-1}=0.2$ I then get
$\beta = 1.1875$ and $\lfloor N\rceil = 9$ (rounded from

)

Comparing this to the "Scripts/blockMesh grading calculation"-tool we have

Code:

     
                                        
My computation           1.1875          9.066873951   
Calculator-tool          1.189207115     9.066872597

Ps. I do not think it is possible to create tables here considering the lacking response on this post making table in cfd-post with example

January 10, 2016, 21:47	About Simple Grading in blockMesh	#1
ce73stargazer Member Zhifang Hu Join Date: Sep 2015 Location: Brisbane Posts: 54 Rep Power: 11	I understand simple grading is the ratio between the size of the end cell and the size of initial cell. How about the cell size in between, what kind of changes they follow to reach the final cell? Like constant changes? For instane, the first cell is size 10 and the last cell is size 5, the middle cells will have a size 10-dx*(N-1) ? where the dx is the constant changes between cells, N is the number of cells in that direction. This is how i understand it, not sure if thats right. I am writing a small code to output the cell ID by just inputting the x,y coordinates, however i realised that it only work fine for uniform distributed mesh, not the one with gradings. Hence i came out with this question. Any one have any suggestions? (Or if anyone has faster way to output cell ID (in paraview) and dont mind share with me) Many Thanks. pedroamaral likes this.

January 12, 2016, 05:45		#2
Antimony Senior Member Join Date: Aug 2013 Posts: 407 Rep Power: 16	Hi, The expansion is geometric, not arithmetic. Here is a useful tool that gives you an idea about the expansion from the start cell to the end cell: https://openfoamwiki.net/index.php/S...ng_calculation Neshat, Mondal131211, SHANRU and 2 others like this.

May 7, 2020, 06:50		#3
Zetison New Member Jon Vegard Venås Join Date: Apr 2014 Posts: 16 Rep Power: 12	What is the exact formula used here? With being the number of cells, $r=\frac{\delta_e}{\delta_s}$ being the expansion ratio, the total length, and based on the distribution of the nodes $x_i = x_0 + L\left(\frac{i}{n}\right)^g\quad i=0,\dots,n$ the following formula seems reasonable to me $\delta_s = \frac{L}{n^g}$ $\delta_e = L\left[1-\left(\frac{n-1}{n}\right)^g\right]$ Alternatively, the formulas for a geometric expansion $x_i = x_0 + L\frac{g^i-1}{g^n-1}\quad i=0,\dots,n$ $\delta_s = L\frac{g-1}{g^n-1}$ $\delta_e = L\left[1-\frac{g^{n-1}-1}{g^n-1}\right]$ but neither matches the program (or OpenFoam?). What is wrong here? Last edited by Zetison; May 8, 2020 at 09:46.

May 8, 2020, 08:47		#4
Tobermory Senior Member Join Date: Apr 2020 Location: UK Posts: 746 Rep Power: 14	Apologies, I did not try to debug your maths above, but for simplicity I will give you the correct expressions below in a simpler and more standard terminology. For a set of cells, with vertices , where varies from 0 to , then the cell width is the gap between the vertices. If we define a constant stretching rate of $\beta$ then: $\delta_i = (x_{i+1} - x_i) = \beta \delta_{i-1}$ Armed with this, you can get to the following: $\delta_i = \beta^i \delta_0$ now $r = \delta_{n-1} / \delta_0$ , the ratio of the last to the first cell width, so: $r = \beta^{n-1}$ The $\delta_i$ 's form a geometric progression, the sum of which is: $Sum_i = \delta_0 + \delta_1 + ... + \delta_i = \delta_0 (1 - \beta^i)/(1 - \beta)$ So, the coord of vertex is: $x_i = x_0 + Sum_i = x_0 + \delta_0 (1 - \beta^i)/(1 - \beta)$ Finally, the cell centre is defined as $xc_i = x_i + 0.5 \delta_i$ so you can then calculate the cell coords using . This is probably what you are outputing from OpenFOAM when you are doing the comparison. Here's a worked example: n = 5 beta = 1.3 r = 2.8561 i, x, dx, xc 0 0.000 0.100 0.050 1 0.100 0.130 0.165 2 0.230 0.169 0.315 3 0.399 0.220 0.509 4 0.619 0.286 0.762 5 0.904 Hope this helps - good luck! (PS - can someone tell me how to format a table in a post, to get the spacing right?) ruloz and GregNordin like this.

May 9, 2020, 06:37		#5
Zetison New Member Jon Vegard Venås Join Date: Apr 2014 Posts: 16 Rep Power: 12	Thanks for your reply, your formula gets me very close. Given a length , the width of the start cell $\delta_0$ and the end cell $\delta_{N-1}$ one can compute the number of cells using the equations $r = \beta^{n-1}=\frac{\delta_{n-1}}{\delta_0}$ Solving yields $\beta = \frac{L-\delta_0}{L-\delta_{N-1}}$ such that $N = 1+\frac{\ln\frac{\delta_{N-1}}{\delta_0}}{\ln\frac{L-\delta_0}{L-\delta_{N-1}}}$ For $L = 1, \delta_0 = 0.05, \delta_{N-1}=0.2$ I then get $\beta = 1.1875$ and $\lfloor N\rceil = 9$ (rounded from ) Comparing this to the "Scripts/blockMesh grading calculation"-tool we have Code: $\beta$ My computation 1.1875 9.066873951 Calculator-tool 1.189207115 9.066872597 Ps. I do not think it is possible to create tables here considering the lacking response on this post making table in cfd-post with example ashvinc9 likes this.

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