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Shuttle re-entry and friction

  1. Oct 25, 2012 #1
    When the space shuttle reenters the atmosphere, what is the mechanism by which the friction heats it up? Is it air molecules losing their kinetic energy as heat when it collides with the shuttle?

    If this is the case, will all falling objects heat up if it manages to reach a certain velocity?
     
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  3. Oct 25, 2012 #2

    Drakkith

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  4. Oct 25, 2012 #3

    boneh3ad

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    The Wikipedia article is pretty vague though.

    For starers, when you move at very high speeds, you create shockwaves, which cause enormous jumps in pressure, temperature and density over a short distance that get even larger with high speed.

    For example, across a normal shock at Mach 25, the temperature downstream of the shock will be almost 125 times higher than upstream of the shock! The ambient temperature at, say, 120,000 feet, is about -26°F, or 241 K. That means that the temperature behind the shock is 29,515 K or 52,668°F! That is over 5 times the temperature of the surface of the sun! Now, here the ideal gas law doesn't apply and you have to take into account chemistry and ionization, but the concept still holds.

    Of course, the space shuttle quickly passes through these kinds of temperature regimes since it is moving so fast. Once you get down to Mach 10 or so, which is still quite fast, the temperature increase across the shock "only" increases by a factor of 20. Of course that is still 4820 K, or just slightly less than the surface of the sun at the free-stream temperature mentioned before. This is one reason it is so important to decelerate quickly and therefore why they design re-entry vehicles to have very high drag.

    So that explains the most notable way the temperature increases. The rest is just convection. It is the opposite of how you blow on your coffee to help it cool faster. Instead, the air flow tends to try and bring the vehicle into thermal equilibrium with itself, so it gets really hot, really fast. Even Fourier's law of heat conduction would predict extraordinarily high heat transfer rates with a temperature difference like that, but factor in convection and radiation and it gets worse. That is why we need exotic heat shields.
     
  5. Oct 26, 2012 #4

    CWatters

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    Perhaps more obvious to think of it as the Shuttle loosing it's KE when colliding with the air molecules.
     
  6. Oct 26, 2012 #5

    CWatters

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    Yes heat would be produced by friction with the air and at any velocity. However the end result might not be a hotter object. Some things (like a falling cup of coffee) will obviously get cooler if subjected to forced air cooling and that effect would dominate any heating effect caused by air friction.

    As I understand it when a rain drop falls it does warm up due to air friction and that causes some of the water to evaporate carrying off some of the heat and warming the air. The remaining rain drop may not warm up much if at all. It's temperature will depend more on the temperature of the air it falls through.

    The Apollo capsule also had an ablative heat shield. As it got hot the surface was vapourised and some of the heat was carried away with the vapourised material.
     
  7. Oct 26, 2012 #6
    Meteorites, shooting stars.
     
  8. Oct 26, 2012 #7

    D H

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    That's not a good explanation. By the time the airflow hits the Shuttle body, it has zero velocity with respect to the Shuttle. By analogy, look at a blade on a house fan. It's oftentimes dusty, sometimes very dusty. The reason is that there is a stagnation layer near the fan blades. The same is true for a reentering body such as the Shuttle.

    You have to look at what happens to the air before it hits the body. For subsonic flow, the air is merely compressed. This heats the air a bit, but just a bit. Things get much more interesting for supersonic flow. Now a shockwave sets up. This makes for very drastic warming; see boneh3ad's post above.
     
  9. Oct 26, 2012 #8

    boneh3ad

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    Whoa, so you suggest that if you dropped a cup of coffee from orbit it would cool off? This simply is not true. If they temperature surrounding it was truly colder than the coffee, then yes, but as described in my previous post, your coffee would have be be several thousand Kelvins before that was the case when moving at re-entry speeds.

    Actually, ablative heat shields work because the chemical reactions involved in ablation are highly endothermic and require a lot of energy. A lot more energy, in fact, than the thermal mass of the heat shield would otherwise be able to handle. While some energy may be carried away by the ablation products, the much bigger energy sink is the chemical reaction itself.
     
  10. Oct 28, 2012 #9

    CWatters

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    No. That's not what I said...

    "Some things (like a falling cup of coffee) will obviously get cooler if subjected to forced air cooling and that effect would dominate any heating effect caused by air friction. "

    The point I was trying to make is simply that hot objects don't allways get hotter when dropped.
     
  11. Oct 28, 2012 #10
    I think the cup would actually cool off, altough the coffee would disappear rather quickly. Coffee will boil at a very low temperature at these low pressures. If you point the opening in the cup forward, this might not happen however, because the pressure will be much higher.
     
  12. Oct 28, 2012 #11

    boneh3ad

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    So how, pray tell, do you expect an object re-entering the atmosphere where the ambient temperature around it is hotter than the photosphere of the Sun to cool off as it travels? Convection works by making an object closer in temperature to its surroundings.
     
  13. Oct 28, 2012 #12
    No matter what the temperature around the object is, it's still very low density and it won't radiate nearly as much as you'd think.
    0.2 kg of coffee moving at 7 km/s has (1/2) (0.2) (7000)^2 = 4.9 MJ of kinetic energy, while you need 0.45 MJ to evaporate 0.2 kg of water.

    At least half of the 4.9 MJ should be radiated not in the direction of the cup, and there's also heat lost to the atmosphere

    While the remaining heat seems enough to evaporate the coffee and probably burn up the cup afterwards, I don't think it will immediately happen if you let the cup go at 120 km alitude and a 1.5 degree angle below the horizontal.
     
  14. Oct 28, 2012 #13

    boneh3ad

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    I still don't think you understand how this all works. Sure the density will change the boiling point and the rate of heat transfer but it doesn't change the fact that the surrounding atmosphere is nearly 30,000 K and, because of the second law of thermodynamics, heat will flow into that cup unless it is hotter. Of course, if the coffee evaporates, it may briefly draw heat away from the cup itself, but any evaporation is going to be nearly instantaneous and overall there is going to be a much hotter cup for the brief period of time before it vaporizes.

    The heat transfer in this situation will come from two primary sources: convection dominated by the combination of the huge temperature gradients and air flow and radiation, since the air around an object moving that fast becomes ionized and so hot that radiative heat transfer is non-negligible. Despite all this, the second law of thermodynamics would say that the cup must get hotter in such an environment except for perhaps the tiny speck of time where the coffee is actually evaporating.

    Regarding that speck of time, the coffee would have to evaporate fast enough to draw heat away from the mug at a rate faster than it is being added to the mug from outside, and considering the heat transfer rate to the exposed surface of said coffee would be much greater than through the porcelain mug, a much larger portion of the heat added to the coffee will come from there rather than the mug. Consider me doubtful that this would cool the mug any appreciable amount or for any longer than this tiny blip over the course of an otherwise intense heating.
     
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