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December 7, 2019, 19:16 
Neumann boundary condition discretization

#1 
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Karnauhov Valery
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Hello. I apologize in advance for the Google translator.
I need help in understanding how in the FVM the discretization of the Neumann condition occurs at the boundary. I use the book F. Moukalled, L. Mangani, M. Darwish, The Finite Volume Method in Computational Fluid Dynamics. It contains a description of how, for example, to find pressure at the boundary of symmetry (fig.). But I do not quite understand how expressions (15.152) and 15.153) are obtained. Can someone explain to me how they are received? Are there any other books where the discretization of the boundary conditions is also painted? Last edited by kveki; December 8, 2019 at 16:11. 

December 7, 2019, 19:33 

#2 
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I don't know what was your starting idiom but, chapeau to Google translator, never got such level of result.
Unfortunately, I don't have the book now, so can't be of any help until Monday. 

December 8, 2019, 05:54 

#3  
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Filippo Maria Denaro
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I have the book, I just read that page and it seems the Eq.(15.153) is just a linear Taylor expansion along the direction d 

December 8, 2019, 16:25 

#4  
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Karnauhov Valery
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Yes, expression (15.153) is obviously Teylor expansion. Although I don’t understand why in this case the gradient is used on the border, and not in the center of cell C. I still don’t understand why the extrapolation is carried out in the direction d _Cb, and not in the direction n. My main question was how expression (15.152) was obtained. It is a difference of vectors, but I don’t understand how it was formed. From the gradient vector at point C, the directional derivative vector is subtracted, and as a result we get the gradient vector at the border? 

December 8, 2019, 16:35 

#5  
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Karnauhov Valery
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attached a picture. I need to understand how the expression (15.152) is derived 

December 8, 2019, 16:37 

#6  
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Filippo Maria Denaro
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I don't have enough time to read the previous pages, in your case the geometry for the unstructuredbased grid reconstruction depends on the colocation of the variables as well as the construction of the boundary values. However, I would simplify the the issue, the symmetry for a scalar function can be prescribed by setting d phi /dn = n.Grad phi = 0 on the symmetry line. The value of of the function phi on the line of symmetry can be extrapolated by setting the problem of computing the unknown value for a known discrete derivatives, computed from the interior using an asymmetric stencil. 

December 8, 2019, 17:17 

#7 
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Karnauhov Valery
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thanks for the answer.
attached a picture. I need to understand how the expression (15.152) is derived 

December 8, 2019, 17:31 

#8 
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Filippo Maria Denaro
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December 9, 2019, 06:00 

#9 
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Ok, probably the matter in the text is not stated in a sufficiently explicit manner for novices. Yet, the operations are quite clear. The very point here is that you should not really look for derivations, as there is no single correct way of applying bcs, as they all are approximations of the true physics.
For example, a very common approach is to simply set a) p_b = p_c. What they do in the text is to actually extrapolate the pressure gradient from c to b, and then constrain it to be parallel to the wall (eq. 15.152). Then, the pressure at a symmetry boundary (subscript b) is extrapolated from the interior (subscript c) using this corrected gradient (eq. 15.153). You can probably see that, with respect to my simpler formula in (a), eq. 15.153 takes into account the tangential pressure gradient ONLY when the center C is not along the boundary normal passing trough b. In all the other cases they are equal. My personal taste is against using cell gradients for specifying boundary values whenever possible and resort instead to simpler formulas. Because cell gradients are computed using those boundary values in the first place. So, in order to correctly restart a simulation, you would need to save on file also gradients, which are going to waste a lot of space. Of course, there can be tradeoffs. For example, one could use formula (a) for gradients and then use gradients for the true bc (like in eq. 15.153). But I dislike inconsistencies even more. 

December 9, 2019, 15:22 

#10  
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Karnauhov Valery
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I apologize for not immediately noticing and many thanks for the clarification. The expression (15.152) is similar to the expression (15.121) for a similar situation. We have grad pC, which has the normal and tangential components with respect to the symmetry boundary. At the boundary, the normal component of the gradient is zero, then only the tangential component remains, which is calculated based on the elementary difference of the vectors. 

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