How to estimate wind speed from a pressure difference? Does a formula (or a rule of thumb) exist to estimate the wind speed between a high pressure area and a low pressure area given the pressure difference between the two areas? Only the wind resulting from a pressure difference is of interest, additional influences like Coriolis force or centrifugal force can be neglected.
 A: I have no direct experience with meteorology, but if you want the "rule of thumb", study the Euler equations. Specifically:
$$
\nabla p = - \rho\frac{\mathrm{D}\vec{v}}{\mathrm{D}t}
$$
where D denotes the material derivative. That's the root of all other derivations.
A: A rule of thumb exists if coriollis force is the dominant force balancing the pressure gradient. This is known as the geostrophic balance :
$$ \overrightarrow{V_g} = {\hat{k} \over f} \times \nabla_p \Phi $$
However if only a pressure gradient is being maintained by some source then the velocity will keep increasing as the pressure gradient results in accelerations as the previous user (@Victor noted). However in the real world the balance will eventually be between pressure and some parameterized viscosity : 
$$ \nabla p = \nu_{eddy} \nabla ^2 v $$
Calculating $\nu_{eddy}$ is non trivial and really depends on case to case basis.
A: In meteorology the Coriolis force can very seldom  be ignored.   It gives rise to  the geostrophic wind. This is a flow parallel to the isobars (lines of constant pressure) in which the force due to the pressure gradient is exactly balanced by the Coriolis force. This gives 
$$
(2\Omega \sin \phi) \rho v_y = \frac  {\partial P}{\partial x}\\
(2\Omega \sin \phi) \rho v_x = -\frac  {\partial P}{\partial y}
$$
Here $\phi$ is the latitude, $\Omega$ is the angular velocity of the Earth ($2\pi$ radians in 24 hrs). The geostrophic  approximation is never exact, but is very  good for most meteorological conditions with exception of tornados and hurricanes.
There is wikipedia page:https://en.wikipedia.org/wiki/Geostrophic_wind 
