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Old Dec 07, 2012, 01:33 PM
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Originally Posted by ShoeDLG View Post

Still not sure I completely understand where you are trying to go with this. Some additional thoughts:

1. Even though the speed of a particle remains constant (equal to V) on the surface of the cylinder (viewed from the reference frame where the remote air is still), the pressure on the surface (the pressure experienced by the particle) does vary with location. You just need to apply the proper form of the Bernoulli equation to relate the pressure to the velocity (in this case the unsteady form of the Bernoulli equation). This is shown in the attached .pdf.

2. There’s nothing imaginary about the “convex” flow near the stagnation point. Just as you can show that crossing “concave” streamlines to get to the top of the cylinder means the pressure is lower than ambient there, you can also show that crossing “convex” streamlines means that the pressure at the stagnation point is higher than ambient there.
My point all along is to show that Bernoulli can not apply to flow where there is no acceleration. (linear) The basic P(s) = P(t) - P(d) does apply.

1. Pressure changes with the changing normal acceleration as the curvature tightens then relaxes in the "e".
I am having trouble making the pdf work. It may just be just me and am needing a diagram!!

2. I was just trying to justify the use of "stagnation point to stagnation point".
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