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#1 |
New Member
Boqi Ren
Join Date: Aug 2018
Posts: 2
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Hi everyone
I'm writing here to ask some questions about the calculation of wall shear stress on the rough surface. As we know, the definition of wall shear stress is ![]() so for the rough surface, can I calculate the shear stress in the following way? ![]() Where t means the tangential direction of the local surface, and n means the normal direction. And, could you recommend some relative reference to me? THANK YOU VERY MUCH! ![]() Last edited by Sylorn; October 7, 2021 at 04:57. Reason: to correct the formulations |
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#2 |
Senior Member
Filippo Maria Denaro
Join Date: Jul 2010
Posts: 6,813
Rep Power: 73 ![]() ![]() ![]() |
Quote:
Since the surface is rough you should evaluate the normal derivative at x=x(s), y=y(s), not at y=0. The action of the stress can be better see if you write the contribution of the stress tensor on a finite surface: Int[S] n.Td dS Td being the deviatoric part having zero trace. |
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#3 | |
New Member
Boqi Ren
Join Date: Aug 2018
Posts: 2
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Quote:
Yes, as you said, the wall shear stress should be on the wall, so in the second equation, y=0 is not correct. But I still feel confused with ![]() |
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#4 | |
Senior Member
Lucky
Join Date: Apr 2011
Location: Orlando, FL USA
Posts: 5,704
Rep Power: 66 ![]() ![]() ![]() |
Quote:
It's the tensor form of shear stress. du/dy the way Isaac Newton did it is for 1D. |
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Tags |
rough wall, shear stress |
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