5
$\begingroup$

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.

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ Reentry heat isn't caused by friction, but by the compression of the air ahead of the spacecraft. $\endgroup$
    – Daniel Griscom
    Oct 4, 2015 at 20:31
  • 1
    $\begingroup$ My apologies, I'd seen it described as friction before. In any case, wouldn't that effect still be present at a high enough air speed on the ground? $\endgroup$
    – codehearted
    Oct 4, 2015 at 20:45
  • 1
    $\begingroup$ Certainly, and there will be a velocity where the conductive cooling of the air is balanced by the compressive heating of the air. I have no idea what that speed would be, though; I just thought your question would get better answers were it phrased realistically. $\endgroup$
    – Daniel Griscom
    Oct 4, 2015 at 23:17
  • 1
    $\begingroup$ May be of interest: Obligatory XKCD on a similar subject. $\endgroup$
    – CoilKid
    Oct 4, 2015 at 23:30
  • $\begingroup$ @DanielGriscom thank you, I updated the question - hopefully appropriately. $\endgroup$
    – codehearted
    Oct 5, 2015 at 1:06

2 Answers 2

Reset to default
3
$\begingroup$

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.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Am I understanding you correctly that if the incoming compressed air is at a higher temperature than the object, it would still cool the object to a lower temperature than it is? I may have misinterpreted a typo in that second paragraph. $\endgroup$
    – codehearted
    Oct 9, 2015 at 21:50
  • 1
    $\begingroup$ @codehearted Typo. If the body is hotter than the compressed air (which is hotter than the free stream air) then it would cool the body. $\endgroup$
    – Rick
    Oct 9, 2015 at 21:55
  • $\begingroup$ Thanks for that. I picked your answer for the equation, though I see now the other answer was also valid. $\endgroup$
    – codehearted
    Oct 12, 2015 at 17:43
2
$\begingroup$

This value will be temperature dependent. As long as the materials survive, the object will tend to that temperature over time.

There is no "wind chill" for an object that is already at ambient temperature and for which evaporation is not relevant. But it will increase the rate of heat transfer when the temperature differs.

So it will exactly cancel at all speeds if you pick the right temperature.

Improve this answer
$\endgroup$
2
  • $\begingroup$ First off, thank you for that lead, I see how the "wind chill" may not be a factor if the object is at ambient temperature, and all the chill does is accelerate the dissipation of heat. However that would not lead to a stable temperature at all wind speeds, once compressive heating overcomes the ability of the conduction to cool it as fast as it heats, the object will gain heat over time. I'm hoping to find that point. If you could give an example of what range of speeds would be thermally self-regulating, assuming a given ambient temp, that would be helpful. $\endgroup$
    – codehearted
    Oct 7, 2015 at 21:35
  • 2
    $\begingroup$ As the temperature increases, the rate of heat transfer away from the warm body increases, while the heat production is basically constant. Assuming no structural failure, the temperature will rise to the point that the heat removal (portions of which are due to the convecting air) equals the heat input. $\endgroup$
    – BowlOfRed
    Oct 7, 2015 at 21:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.