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pressure drop of the periodic boundary!

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Old   December 22, 2006, 01:03
Default pressure drop of the periodic boundary!
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Hi ! I'v been compute a 2D channel use periodic boundary and got a covergence ,also I found the value of pressure grident changed in "periodic conditions" form.

My question is: whether or not I can treat the pressure gradient as the pressure drop?for example if the pressure gradient is 23Pa/m , as for 0.5m channels then the pressure drop is 0.5*23=11.5Pa? is that correctly?

My second question is:there are two channels which have same profile,one is 2D model the other is 3D model and generally how difference the results of the two situations?
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Old   November 17, 2012, 09:43
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I am also stuck with the same problem, i checked in the user guide and they say the pressure drop is the sum of the linear component (pressure gradient) and a periodic component if you are using the pressure based solver! I can not see how you get the periodic component, also when you get the pressure drop using the derivative dp/dz if z is your streamwise direction the results is different from the pressure gradient! let me know if you get a way out
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Old   November 17, 2012, 12:58
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I am also stuck with the same problem, i checked in the user guide and they say the pressure drop is the sum of the linear component (pressure gradient) and a periodic component if you are using the pressure based solver! I can not see how you get the periodic component, also when you get the pressure drop using the derivative dp/dz if z is your streamwise direction the results is different from the pressure gradient! let me know if you get a way out
Once you choose to do a periodic simulation, the linear component (the pressure gradient) is removed from the governing equations. It is represented by a body force (similar to gravity). The pressure gradient that shows up on the periodic conditions dialog box is the linear pressure gradient.

Because Fluent has removed the linear pressure gradient, if you calculate dp/dx from your your simulation, you will actually get the derivative of the periodic component only and not the overall or actual pressure gradient. To get the overall pressure gradient you need to add back, the linear component.
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Old   November 19, 2012, 03:47
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Originally Posted by LuckyTran View Post
Once you choose to do a periodic simulation, the linear component (the pressure gradient) is removed from the governing equations. It is represented by a body force (similar to gravity). The pressure gradient that shows up on the periodic conditions dialog box is the linear pressure gradient.

Because Fluent has removed the linear pressure gradient, if you calculate dp/dx from your your simulation, you will actually get the derivative of the periodic component only and not the overall or actual pressure gradient. To get the overall pressure gradient you need to add back, the linear component.
Dear Lucky
Thanks for the reply. I will check with that. if i take the derivative at the period boundaries does it represent the actual pressure drop?
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Old   November 19, 2012, 03:57
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Dear Lucky
Thanks for the reply. I will check with that. if i take the derivative at the period boundaries does it represent the actual pressure drop?
No it does not.

Please be careful when saying pressure drop versus pressure gradient and carefully denote whether you mean pressure gradient at a specific location or gradient over the entire periodic length of the domain.

The derivative of pressure at one of the periodic boundaries yields only the local periodic pressure gradient.

The overall pressure drop is also the linear pressure drop since the periodic pressure are boundaries are equal (no contribution to the gradient from the periodic pressure). You can get this overall pressure drop by taking the linear pressure gradient and multiplying by the length of the periodic sector.
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Old   November 19, 2012, 04:16
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No it does not.

Please be careful when saying pressure drop versus pressure gradient and carefully denote whether you mean pressure gradient at a specific location or gradient over the entire periodic length of the domain.

The derivative of pressure at one of the periodic boundaries yields only the local periodic pressure gradient.

The overall pressure drop is also the linear pressure drop since the periodic pressure are boundaries are equal (no contribution to the gradient from the periodic pressure). You can get this overall pressure drop by taking the linear pressure gradient and multiplying by the length of the periodic sector.
Thanks very much Lucky, am interested in the pressure drop across the periodic sector and your explanation making sense. I have checked my solutions and they are reasonable. Are you using workbench? I would like to expose this linear component as my output for parametric analysis but in the periodic panel it is not possible! Is there any other way I can get this linear component in post processing??
Thanks once again
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Old   November 19, 2012, 04:39
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Thanks very much Lucky, am interested in the pressure drop across the periodic sector and your explanation making sense. I have checked my solutions and they are reasonable. Are you using workbench? I would like to expose this linear component as my output for parametric analysis but in the periodic panel it is not possible! Is there any other way I can get this linear component in post processing??
Thanks once again
I don't use workbench, only Fluent directly so I am not sure of workbench specifics. Except manually looking at the periodic conditions, the linear pressure gradient is not obtainable directly, it is not stored or exported as a variable. But you should be able to get it from some integral of wall shear stress (and use area, length relations to put that into pressure gradient) and that is exported. Should be doable.
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Old   February 29, 2016, 11:59
Default pressure drop in periodic boundaries
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Quote:
Originally Posted by LuckyTran View Post
I don't use workbench, only Fluent directly so I am not sure of workbench specifics. Except manually looking at the periodic conditions, the linear pressure gradient is not obtainable directly, it is not stored or exported as a variable. But you should be able to get it from some integral of wall shear stress (and use area, length relations to put that into pressure gradient) and that is exported. Should be doable.

Dear Lucky,

Your conversation previously, helped me somewhat. But I m still not very much satisfied. I m stuck with the problem of heat transfer and friction in a long pipe of 2m in length. From experiment, I modeled the section only where the thermocouples are mounted in the test section and also across which the pressure drop is desired (215mm).
I modeled that section only and took periodic on both sides. Initially I took velocity inlet and pressure outlet as the two ends, but being not satisfied with results I took periodic. Now, I stuck more, since I have not inlets and outlets so where to measure the inlet and outlet temperatures and where to get outlet velocity?
For pressure drop I am gradually observing pressure gradient/m in peridic button am waiting if it is no more varying. I will multiply it with modelled section length to get delp. What is the linear pressure drop and the wall shear stress you talked about?
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Old   March 1, 2016, 17:05
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The wall shear stress is the wall shear stress on the pipe.

The linear pressure drop is the pressure gradient in the periodic button. It is the linear component of the pressure variation.
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Old   April 2, 2016, 15:20
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Quote:
Originally Posted by LuckyTran View Post
The wall shear stress is the wall shear stress on the pipe.

The linear pressure drop is the pressure gradient in the periodic button. It is the linear component of the pressure variation.

The convergence of the mointor of statistical residual in monitors takes very very long to converge. Please tell me how to see convergence in this regard?
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