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March 3, 2016, 18:25 |
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#21 | |
Senior Member
Filippo Maria Denaro
Join Date: Jul 2010
Posts: 6,768
Rep Power: 71 |
Quote:
Check the spectrum of the total velocity instead of using the fluctuation |
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March 3, 2016, 19:02 |
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#22 |
Senior Member
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Professor;
I exactly the same spectrum, nothing changed... At this point it does not matter the numerical method is purely the initial condition defined as: U(:,1) = 0.2*sin(pi*x( : )/6) + 0.35*sin(pi*x( : )*0.075/6) + & 0.15*sin(pi*x( : )*(6.0/6)) + 0.1*sin(pi*x( : )*(20.0/6)) + & 0.01*exp(cos(pi*x( : )*(100.0/3.1))) Last edited by juliom; March 3, 2016 at 19:05. Reason: typo |
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March 4, 2016, 03:20 |
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#23 | |
Senior Member
Filippo Maria Denaro
Join Date: Jul 2010
Posts: 6,768
Rep Power: 71 |
Quote:
that means that the average velocity is zero...the peaks in the spectrum are due to your initial condition, you could use an initial spectrum E(k) and then construct the velocity field |
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March 4, 2016, 12:17 |
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#24 |
Senior Member
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Thank you very much professor. Your help is very important for me. I tried to follow your recommendation but I did not know how to do it. Thus, I used another approach, that at least will help me to introduce you the issue.
I used the John Hopkins Turbulence Data Base. I extracted the velocity of the a channel flow data base. My domain is 2*pi and I have 62931 points. So I used the same properties to fetch the data from the database. Then I computed the energy spectrum which looks like: https://goo.gl/photos/ajE3JoFUeybfZ2X6A Then I run my code with that initial condition and at time steps = 5000 I computed the energy spectrum again and it looks like: https://goo.gl/photos/iJrqZn1vAvjZ7rEB6 What is very surprising is the tail of the spectrum at the cut-off wavenumber (pi/dx). Professor, I already talked to my advisor and we do not have an answer for this question and unfortunately there are no other people in the department working on Turbulence. My conclusion is that my cut-off frequency (Nyquist) is way up of the Kolmogorov "wave number". In other words my dx is way small to the kolmogoroov scale and there is no more energy to dissipate... This is my conclusion that could be very wrong. If so, please accept my apologize. I really appreciate your thoughts and ideas. Very Respectfully Julio Mendez |
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March 4, 2016, 12:23 |
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#25 | |
Senior Member
Filippo Maria Denaro
Join Date: Jul 2010
Posts: 6,768
Rep Power: 71 |
Quote:
you started from a fully developed spectrum but when you run the non-forced case, your energy will decay... As a consequence, the relevant characteristic scales (Taylor micri-scale, Kolmogorov) will increase the lenght so decreasing their wavelenght. Your computed spectrum shows this issue, the tail you see is nothing else that zero energy content at double precision. Your Nyquist frequency is fixed and dictated by the extension of frequencies in the initial condition, but during the time the Kolmogorov wavelenght will move to left as it increases. In other words, without forcing, you Re number will diminuish in time due to the dissipation of kinetic energy. In my JCP paper, you can see a figure with the DNS spectra for several times. |
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March 4, 2016, 18:01 |
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#26 |
Senior Member
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Thank you very much professor. I read the papers JCP and I found the same trend, however, the plot cuts the horizontal axis. Did you reduced the range from the "Y" axis professor. ? I am having very problems defining the initial conditions. I tried to follow your papers but I was not ale to understand.
Thank you very much professor, for your kindness relying to my questions! |
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