The criteria for potential flow theory I am learning aerodynamics. In this course a potential flow is denoted that a flow in which the rotation is zero everywhere. But the book told me that we can add vortex into a flow field, and we can also use potential theory to analysis it. I get confused about this. If a vortex is added in a flow field, I don't think there exist any potential. The rotation of a vortex is infinite! Could anyone explain to me, why potential flow theory can be used while there are some vortexes in the flow field?
 A: Potential flow usually means, as you correctly said, that your flow is irrotational, so that your velocity field $\vec{u}(\vec{x},t)$ has the property:
$$
\vec\nabla\times \vec{u} = 0 \;\;\Longrightarrow\;\; \vec{u}=\vec\nabla\phi
$$
However this is not the general case, because you usually have some rotation in your flow which is called vorticity and is defined as:
$$
\vec\omega = \vec\nabla\times \vec{u} 
$$
Now this vector can be whatever it wants, constant and finite for example, there is no reason for it to be infinite. Coming back to potentials, in case a non-null vorticity is present, the velocity field is can still be decomposed into an irrotational part expressed by $\phi$ and a rotational part expressed by the so-called stream function $\vec\psi$ in the following way:
$$
\vec{u}=\vec\nabla\phi + \vec\nabla\times\vec\psi
$$
In this sense people still talk of potential flow even when the flow is not irrotational. 
The reason is that this formalism is very useful in two dimensional flows, where the vorticity and the stream function are vector fields always pointing in the $z$-direction and are linked by the relation:
$$
\nabla^2\psi = -\omega
$$
I hope this clarifies a bit your doubts, if you have some more just ask away!
A: As contradictory as it sounds, you can have a vortex that satisfies $$\vec{\omega} = \vec{\nabla} \times \vec{u} = 0$$  and $$\nabla^2\phi=0$$The potential is given in terms of polar coordinates $r$ and $\theta$ by $$\phi = \frac{\Gamma}{2\pi}\theta$$  which gives $u_{\theta}=\frac{\Gamma}{2\pi r}$ and $u_r = 0$.  Evaluating the curl everywhere but at the origin gives $$\vec{\omega} = \frac{\partial ru_{\theta}}{\partial r}-\frac{\partial u_r}{\partial \theta} = 0 $$
