At relatively slow wind speeds such as 15mph, wind chill drains heat from an object as it flows past, and this conductive cooling effect seems to increase as the wind speed increases. However, at very high wind speeds such as 15,000mph (during atmospheric re-entry from space, for example), friction / compressive heating with the atmosphere has a well-documented heating effect – rather than transferring heat from the object as with wind chill, it adds enough heat to bathe it in flame and/or burn it entirely.
So how fast must one be going for these two effects to cancel out, assuming STP? If it is a smooth gradient of temperature there must be a zero in there somewhere. If not, there must be a dramatic jump in externally induced temperature change that could perhaps be used to generate energy. Please feel free to make any other assumptions necessary to answer (such as the shape of the object) as long as they are stated. References to further info would be appreciated as well.
Best Answer
As BowlOfRed points out the incoming air has a certain temperature and the convective flow will tend to bring the temperature of the object towards the temperature of the incoming air.
In the case where the incoming air was going very fast it gets heated through adiabatic compression before it reaches the object, so the incoming air is at higher temperature than the free stream temperature, but it will still cool the object if it is at an even higher temperature.
Similarly, if you take something out of the freezer, a gentle breeze will actually have a warming effect on the object bringing it closer to the temperature of the incoming air.
If you want to know the relationship between the equilibrium temperature and the free stream temperature vs. velocity that's given by the isentropic flow equation:
$$T=T_0\,\left(1+\frac{\gamma-1}2 M^2 \right)$$
Where $T$ is the surface temperature (in Kelvin or Rankine), $T_0$ is the free stream temperature (in Kelvin or Rankine), $\gamma$ is the ratio of specific heats (1.4 for air), and $M$ is the Mach number.