The noncirculating potential flow around a circular cylinder has zero lift because its velocity and pressure fields are symmetric above and below the cylinder. A circulating cylinder flow is obtained by superposing that solution with an ideal free vortex centered on the cylinder axis.

Take a uniform stream of speed $U$ in the positive $x$ direction around a cylinder of radius $a$. The noncirculating cylinder flow has

$$v_r=U\left(1-\frac{a^2}{r^2}\right)\cos\theta,$$

$$v_\theta=-U\left(1+\frac{a^2}{r^2}\right)\sin\theta.$$

Now add a free vortex with circulation $\Gamma$, using the convention that

$$\boxed{\Gamma>0}$$

means counterclockwise circulation when viewed from the positive $z$ direction. Its tangential velocity is

$$\boxed{v_{\theta,\Gamma}=\frac{\Gamma}{2\pi r}}.$$

Because incompressible irrotational potential-flow solutions can be superposed, the combined field is

$$\boxed{v_r =U\left(1-\frac{a^2}{r^2}\right)\cos\theta},$$

$$\boxed{v_\theta =-U\left(1+\frac{a^2}{r^2}\right)\sin\theta +\frac{\Gamma}{2\pi r}}.$$

The flow remains irrotational everywhere outside the cylinder because the vortex singularity lies inside the excluded solid region.

The cylinder remains impermeable

At

$$r=a,$$

the radial velocity is still

$$\boxed{v_r(a,\theta)=0}.$$

Adding circulation changes only the tangential component, so the same circle remains an exact no-penetration boundary.

The surface tangential velocity is

$$\boxed{ v_\theta(a,\theta) =-2U\sin\theta+rac{\Gamma}{2\pi a} }.$$

This is the key effect of circulation: it adds the same signed tangential contribution all around the cylinder, strengthening the flow on one side while weakening it on the other.

Stagnation points move with circulation

A surface stagnation point requires

$$v_\theta(a,\theta_s)=0.$$

Therefore

$$-2U\sin\theta_s+rac{\Gamma}{2\pi a}=0,$$

so

$$\boxed{ \sin\theta_s=\frac{\Gamma}{4\pi aU} }.$$

When

$$|\Gamma|<4\pi aU,$$

there are two stagnation points on the surface. Increasing the magnitude of circulation moves them away from the front and rear positions of the noncirculating case.

At

$$|\Gamma|=4\pi aU,$$

the two surface stagnation points merge at the top or bottom. For still larger circulation magnitudes, the relevant stagnation points move off the cylinder surface.

Pressure becomes asymmetric

For steady incompressible irrotational flow at constant elevation, one Bernoulli constant applies throughout the connected outer region:

$$p+\frac12\rho v^2 =p_\infty+\frac12\rho U^2.$$

On the cylinder surface the radial velocity is zero, so

$$v^2=v_\theta^2.$$

Define

$$\lambda=\frac{\Gamma}{2\pi aU}.$$

Then

$$\frac{v_\theta}{U}=-2\sin\theta+\lambda,$$

and the surface pressure coefficient is

$$\boxed{ C_p =1-\left(-2\sin\theta+\lambda\right)^2 }.$$

For

$$\Gamma=0,$$

this reduces to the symmetric noncirculating result

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

For nonzero circulation, the cross term in the square is proportional to

$$\Gamma\sin\theta,$$

which changes sign between the upper and lower halves of the cylinder. The pressure distribution is therefore no longer symmetric about the $x$ axis.

Direction of the pressure asymmetry

With the convention above, positive circulation is counterclockwise.

At the top of the cylinder,

$$\theta=\frac\pi2,$$

so the noncirculating flow has negative tangential velocity. Positive $\Gamma$ partially cancels that motion and reduces the speed magnitude there.

At the bottom,

$$\theta=\frac{3\pi}{2},$$

the same positive circulation reinforces the positive tangential velocity and increases the speed magnitude.

Bernoulli therefore predicts relatively higher pressure above and lower pressure below, producing a downward net lift for positive counterclockwise circulation under this sign convention.

Clockwise circulation,

$$\Gamma<0,$$

reverses the asymmetry and produces upward lift.

Worked example

Take

$$a=0.50,\mathrm m,$$

$$U=10,\mathrm{m/s},$$

and clockwise circulation

$$\Gamma=-10,\mathrm{m^2/s}.$$

Then

$$\lambda =\frac{-10}{2\pi(0.50)(10)} \approx-0.318.$$

At the top,

$$\frac{v_{\theta,top}}{U} =-2-0.318=-2.318,$$

so

$$C_{p,top}=1-(2.318)^2\approx-4.37.$$

At the bottom,

$$\frac{v_{\theta,bottom}}{U} =2-0.318=1.682,$$

so

$$C_{p,bottom}=1-(1.682)^2\approx-1.83.$$

The upper surface has the lower pressure, so the net transverse force is upward.

What circulation has changed

The circulation does not create a radial flow through the body and does not make the regular outer flow locally rotational. Instead it changes the global topology and tangential velocity of the potential flow around the excluded cylinder interior.

The resulting pressure asymmetry produces a net force perpendicular to the free stream. Integrating that pressure distribution leads to the Kutta-Joukowski lift relation.