Potential flow around a circular cylinder

Uniform irrotational flow of speed $U$ past a circular cylinder of radius $a$, with zero circulation, has velocity potentialOn the cylinder, the tangential speed is $-2U\sin \theta$, so the Bernoulli equation gives

Adding circulation $\kappa$ gives the possibly multivalued velocity potentialThe velocity remains single-valued, and on the cylinder its tangential component is $-2U\sin \theta +\kappa /(2\pi a)$.

For complex potential of a two-dimensional potential flow $w=U(z+a^2/z)-i\Gamma \log z/(2\pi )$, zeros of the complex velocity solve a quadratic. If $0<\gamma <1$, the displayed two roots are distinct and both have modulus $a$; positive circulation places them above the centreline. Their polar angles satisfy $\sin \theta =\gamma$. At the threshold $\gamma =1$ they merge at the top of the cylinder. This follows also from the surface tangential velocity $-2U\sin \theta +\Gamma /(2\pi a)$. The shifted stream function $U(r-a^2/r)\sin \theta -\Gamma \log (r/a)/(2\pi )$ is constant on the cylinder.

For steady two-dimensional inviscid flow of density $\rho$, far-field velocity $\mathbf{U}$, and circulation $\kappa$, the force per unit span is perpendicular to the flow. With counterclockwise circulation taken positive and $\mathbf{U}=U{\mathbf{e}}_x$, it is $-\rho \kappa U{\mathbf{e}}_y$.

For uniform flow $\mathbf{U}$ plus a line vortex of counterclockwise circulation $\Gamma$, the outward advective momentum flux around a large circle is $\rho \mathbf{\Gamma }\times \mathbf{U}/2$. The Bernoulli equation supplies an equal pressure contribution. Their sum is the force on the fluid; the force on an obstacle per unit span is its negative, $\rho \mathbf{U}\times \mathbf{\Gamma }$. Retaining only the advective term loses half the lift and obscures its sign.