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I'm new to this site so I apologize if I'm asking a previously answered question, but I couldn't find anything anywhere. I was trying to derive the pressure coefficient over a cylinder without using inviscid potential flow derivations. So for the usual potential flow equations, we obtain, for flow on the surface of the cylinder,

at $r = R, V_{\theta} = -U(1+R^2/r^2)\sin \theta = -2U\sin \theta$

where U is the free flow velocity. After plugging into Bernoulli's equation and rearranging,

$C_p = 1 - 4\sin^2\theta$

However, let's assume we don't want to use potential flow derivations, but just resolve the free flow vector into $V_r$ and $V_{\theta}$. In that case, we get

$V_r = U\cos \theta$

$V_{\theta} = -U\sin \theta$

where does the discrepancy come from? Why does the potential flow solution give double that of a resolved vector? Where does the extra $R^2/r^2$ term come into play when we try to resolve it directly?

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  • $\begingroup$ The discrepancy comes from the fact that in your second case there is no cylinder, just the free flow. Clearly, a flow with no cylinder is different from the flow around a cylinder... $\endgroup$
    – Pirx
    Nov 21, 2016 at 22:18

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It seems that when you try to just resolve the freestream velocity onto the cylinder you neglect the fact that from potential theory flow over a cylinder includes a doublet. (Uniform flow + doublet = flow over a cylinder).

What those resolved components basically do is resolve a freestream uniform velocity in cylindrical coordinates at some distance $r = R$ from an arbitrary point.

You'll need to add the contributions of the doublet at a distance $r$ from the center. That is where is $R^2/r$ comes into play.

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