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Alletto - 2022 - Comparison of overset mesh with morphing mesh

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Old   February 18, 2022, 17:22
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Alvi Ahmmed
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Thank you very much. I can get the natural frequency(fn) from St=fnD/U equation right? And am I free to choose U and D for a fixed Re=100?
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Old   February 18, 2022, 18:14
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You can choose the parameters as you want as long all dimensional less parameters are the same. The frequency in the st number is the shedding frequency not the natural frequency
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Old   February 19, 2022, 02:54
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How can I find the value of natural frequency (fn) ? From the FFT plot of the lift coff? Regards.
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Old   February 19, 2022, 03:05
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See https://en.m.wikipedia.org/wiki/Natural_frequency

For the natural frequency of a spring mass system. But I think I mentioned this also in the paper. Read it carefully.

What is done in the python scripts is also explained in the paper. A fft of the lift signal is performed and the maximum amplitude is used to calculate the St number. Read the paper a few times if once is not enough.

I usually need three or four times to fully understand a paper
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Old   February 19, 2022, 03:08
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Thanks very much for your guidence
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Old   March 22, 2022, 02:44
Default How to set the density in the VIV case
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Mehdi Badri
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Hi,

According to the definition of non-dimensional mass, m^{\ast}, on page 18 of your paper, the density of the cylinder, \rho, is calculated as \rho= 4*m / ( m^{\ast}  d^{2} H ).
In your paper (and its corresponding git repo), you have used \rho=14817, whereas it seems that if we substitute m=03575, H=0.12, m^{\ast}=10, and d=0.0016, the density should be \rho=46549.

Could you please clarify how you calculated it (and why my understanding is perhaps incorrect)?
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Old   March 22, 2022, 04:16
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Michael Alletto
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Hello.


I just found a typo in how the fluid density is calculated. The non dimensional mass m*
is defined as follows in the paper:


m^* = \frac{m}{m_f} = \frac{4 m}{\rho D^2 H}


Unfortunately I forgot the pi. It should have been


m^* = \frac{m}{m_f} = \frac{4 m}{\pi \rho D^2 H} = 10


With the second formula we obtain the value of the fluid density specified in the paper
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Old   March 22, 2022, 20:58
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Quote:
Originally Posted by mAlletto View Post
Hello.


I just found a typo in how the fluid density is calculated. The non dimensional mass m*
is defined as follows in the paper:


m^* = \frac{m}{m_f} = \frac{4 m}{\rho D^2 H}


Unfortunately I forgot the pi. It should have been


m^* = \frac{m}{m_f} = \frac{4 m}{\pi \rho D^2 H} = 10


With the second formula we obtain the value of the fluid density specified in the paper


That is a nice explanation and helpful note!

Thanks
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Old   March 23, 2022, 00:21
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It was missing in the formula but the calculation of the density is the correct in the tables and paragraphs.

Quote:
Originally Posted by Mehdi80 View Post
That is a nice explanation and helpful note!

Thanks
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Old   April 5, 2022, 12:12
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Wael Elorfi
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Is there a bibtex file available for citing this paper?
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Old   April 5, 2022, 12:34
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I usually copy the bibtex entry from Google scholar
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Old   April 5, 2022, 14:40
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Wael Elorfi
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Got it. Thanks a lot!

@article{alletto2022comparison, title={Comparison of overset mesh with morphing mesh: Flow over a forced oscillating and freely oscillating 2D cylinder}, author={Alletto, Michael}, journal={OpenFOAM{\textregistered} Journal}, volume={2}, pages={13--30}, year={2022} }
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Old   April 11, 2022, 06:15
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Håkan Nilsson
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This paper has now been updated with the ISSN of the journal and its own DOI: https://doi.org/10.51560/ofj.v2.47
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Old   July 7, 2022, 13:42
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Hi,

Can anyone explain the m in this equation? Is it only the mass of the cylinder?
Does this include the added mass as well?


Quote:
Originally Posted by mAlletto View Post
U* = Uoo/(fn D) , fn = ω/(2 π) = (k / m)0.5/(2 π)

So if you want to achieve a given U* for a given diameter of the cylinder with a given mass and inflow velocity you have to set k accordingly.

Hope this helps
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Old   July 8, 2022, 07:45
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It's only the mass of the cylinder
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Old   July 8, 2022, 14:46
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I need some clarification regarding your paper
Q-1:

You have mentioned that
density of the cylinder is set to 14817 kg/m3. It should be called as the density of the fluid not the cylinder?

Q-2:
In your calculations, in the case of Ur 4, you have used the k = m*(2*pi*f)2 that gives k value of 148.28.

I want to ask about the f used in the calculation. We know that the reduced natural frequency of the system is given as Fr = I/Ur and Ur= Unif/f*D , here f is the frequency in the flow medium . It could be air, water or any medium used in the simulation. In the present case, it is the frequency in the fluid having density 14817 kg.m3.

f= 0.0656/(4*0.0016) = 10.25

As you are using cylinder mass m=0.03575 ,then the k should be calculated as
k=(m+m_added_mass)*(2*pi*f)2

m_added_mass= C_EA*(pi*D2*H*rhoInf/4) =0.003575
k=163.108

Please clarify? I just want to learn that how f is related with k when you are using medium other than air. Should it include the added mass or not?

Quote:
Originally Posted by mAlletto View Post
It's only the mass of the cylinder
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Old   July 9, 2022, 04:35
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Q1: It is the density of the cylinder. If you fix the mass and the dimensions of the cylinder the density can be calculated


Q2: The added mass is not considered. It is actually not known in advance since it is solution dependent. What did you mean with added mass exactly. Can you specify it more precisely?



The purpose of setting the constants in the paper in the way I did it, is to match all relevant non-dimensional constants. This are U*, m* and Re. Actually you can choose all combination of dimensional quantities as you want as long as the non-dimensional quantities remain the same. If the non-dimensional quantities are the same you can compare the non-dimensional solutions.



Have a look at the pi-theorem https://en.wikipedia.org/wiki/Buckingham_%CF%80_theorem


Best


Michael
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