# Pressure coefficient over a cylinder without using potential flow equations

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?

• 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...
– Pirx
Nov 21 '16 at 22:18

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.