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Why is Pressure Recovery less than 1 for a smooth wall subsonic compressible flow |
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#1 |
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MWRS
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Wall roughness is set to zero. Then why is total pressure in outlet lower than inlet for a compressible flow?
I have simulated a simple diverging pipe. What exactly is decreasing the total pressure for a smooth pipe? |
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#2 |
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Lucky
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Do you still have a wall? I.e. no slip?
Smooth is a hydraulic term meaning that the wall roughness is small relative to a specific length scale. A smooth wall does not mean there is no friction (it just means there is no roughness). There will be a total pressure drop. Even if you had slip walls, you still have a viscous flow (navier-stokes) and there would still be a small drop in total pressure. |
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#3 |
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yes i have no slip
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#4 |
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so how would there be a small drop due to viscous flow in a no slip condition?
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#5 |
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Lucky
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The fluid has a velocity going into the diffuser. At walls, because of the no-slip condition (for both smooth and rough walls) the fluid adopts the velocity of the wall (presumably 0 m/s). Hence the fluid loses some momentum and this manifests as a total pressure drop.
Even if you have slip walls where there is no frictional force induced by the walls. The area is diverging and because of this you have velocity gradients which gives rise to viscous forces. In a newtonian fluid, wherever you have velocity gradients, you have viscous forces. |
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#6 |
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Many thanks !
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