4
$\begingroup$

A comment on recent popular question brought me to Stirring Tea and this entry

To boil a cup of water, you'd have to drop it from higher than the top of the atmosphere.

If that is true, in theory a nice coating of ice could make a viable heat shield. The biggest problem that comes to mind is that water ice is relatively fragile.

Assuming I could keep the ice from breaking away on re-entry could I use it for a heat shield on Earth re-entry?

$\endgroup$
2
  • 2
    $\begingroup$ The heat from reentry ought to be related to speed, surface area, and mass of the object. I'd assume the cup of water being dropped is a lot smaller than a space capsule and going a lot slower than orbital velocity. $\endgroup$ Nov 10, 2015 at 20:04
  • 1
    $\begingroup$ Funny thing about water and other volatile ices is that they're quite explosive in such conditions, as also demonstrated by bolides and superbolides. $\endgroup$
    – TildalWave
    Nov 11, 2015 at 11:03

1 Answer 1

7
$\begingroup$

Nice idea, but no, it doesn't work.

It's true that a cube of ice in 100km height has a potential energy of just about a third of what is needed to melt and boil it completely. First point, as you already mentioned: How to keep the liquid water in place?

But, talking about reentry usually applies coming from an orbit and hence having a speed of at least 7.8 km/s. This kinetic energy is about 20 times higher than the potential energy in 100km height. Hence, you need to get rid of about 6 times the energy the ice needs to boil.

And, even worse: So far we just talked about the ice itself. A heat shield is put in place to protect something - and we need to get rid of both potential and kinetic energy of this something as well. If we assume a payload the same size as your block of ice - and that's still no good ratio at all - then we have to spend about 13 times more energy than the ice can take while evaporating.

On the other hand, when bringing some material back from an Asteroid it might be a viable option to bury it deep inside a huge ball of ice - but you better don't hit the wrong spot on earth as something similar already messed up our beloved dinosaurs life some time ago.

When talking about heat shielding, one also has to keep in mind that its purpose is not to absorb all the kinetic energy at reentry. It merely prevents that heat enters the payload taking advantage of its low heat conductivity (a feature water does not have) and it optimizes heat transfer back to the air rushing by. The ablative features of most materials are mostly helping to keep the material and its properties intact, though getting thinner with time.

$\endgroup$
5
  • $\begingroup$ Wolfram Alpha tells me 30 MJ/kg kinetic energy at 7.8km/s, 0.4MJ/kg to raise water temperature 100 K, so ~75 times too much energy. $\endgroup$ Nov 10, 2015 at 21:06
  • 3
    $\begingroup$ Raising water temperature is not taking much energy, evaporating takes another 2.2 MJ/kg and melting 0.3 MJ/kg. If you boil a pot of water, it takes e.g. 10 minutes to reach 100 degrees centigrade, but another hour until the pot is empty! $\endgroup$
    – asdfex
    Nov 10, 2015 at 21:08
  • $\begingroup$ Ah, okay. But the water, unlike the phenolic, is going to flow away easily before boiling, yeah? $\endgroup$ Nov 10, 2015 at 21:15
  • 2
    $\begingroup$ A blunt-body heat shield does not need to dissipate anywhere near all of the entry energy by itself. The beauty of blunt bodies is that the shock wave in front of the blunt body keeps the vast majority of the heat load off the body and sends it around. (Look for the classic Allen and Eggers paper.) So your calculation is way over conservative. $\endgroup$
    – Mark Adler
    Nov 10, 2015 at 21:29
  • 1
    $\begingroup$ @MarkAdler You're absolutely right when it comes to real heat shields. But as the question specifically mentioned the high heat capacity as central argument and hence my answer explains why heat capacity doesn't help. $\endgroup$
    – asdfex
    Nov 10, 2015 at 21:52

Your Answer

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

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