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December 22, 2006, 01:03 
pressure drop of the periodic boundary!

#1 
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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? 

November 17, 2012, 09:43 

#2 
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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


November 17, 2012, 12:58 

#3  
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Lucky Tran
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Quote:
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. 

November 19, 2012, 03:47 
Periodic B.C

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Quote:
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? 

November 19, 2012, 03:57 

#5  
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Lucky Tran
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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. 

November 19, 2012, 04:16 
Periodic B.C

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Thanks once again 

November 19, 2012, 04:39 

#7  
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Lucky Tran
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February 29, 2016, 11:59 
pressure drop in periodic boundaries

#8  
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Quote:
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? 

March 1, 2016, 17:05 

#9 
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Lucky Tran
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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. 

April 2, 2016, 14:20 

#10  
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Quote:
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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