3 ms·

The amount of energy in a thermos like that is high enough that I'm not sure that would be safe. Quick approximation, assuming adiabatic expansion: Gamma for

The amount of energy in a thermos like that is high enough that I'm not sure that would be safe. Quick approximation, assuming adiabatic expansion: Gamma for CO2 is ~1.29. Pf*Vf^y = Pi*Vi^y Vf = (Pi / Pf) ^ (1/y) * Vi Vf = (100atm / 1atm) ^ (1/1.29) * 1L Vf = 35.51L W = (PfVf - PiVi) / (y - 1) W = (1atm*35.51L - 100atm*1L ) / (1.29 - 1) W = 22.31 KJ For comparison, a .50 cal round has ~15kJ of muzzle energy. Now, a lot of that energy won't be focused (if nothing else, the final temperature with adiabatic expansion is such that a large chunk should sublimate again), but still.

lifeisstillgood 12y ago
I have to ask - can you explain the equations please?
- TheLoneWolfling 12y ago
  They all come from the equations of adiabatic processes of an ideal gas. Wikipedia explains better than me: https://en.wikipedia.org/wiki/Ideal_gas https://en.wikipedia.org/wiki/Ideal_gas https://en.wikipedia.org/wiki/Adiabatic_process https://en.wikipedia.org/wiki/Adiabatic_process (Especially the section "Derivation of discrete formula" in the latter) It's only a first approximation, but meh.
r00fus 12y ago
The 22 KJ seems widely varying form other estimates on this discussion (6KJ, 15KJ). For example, is it clear that the pressure of 100 atm would actually be attained by a simple thermos and dry ice? What lead you to use that number? Furthermore, the .50cal round is focused on a small area, the impulse and destructive force are multiplied by the shape of the bullet. Just like how shape-charges magnify the explosive force of munitions to bust armor.
- TheLoneWolfling 12y ago
  The 6KJ number is flat-out wrong. You cannot multiply pressure * volume like that. It's not an isobaric expansion. I mentioned this in a reply to his comment. As for the 15JK number, I cannot see any estimates in this thread saying 15JK. Could you link the comment? The 100atm figure is assuming that the linked article's calculation of final pressure is correct, assuming the thermos doesn't burst beforehand. Although I fully agree that a standard thermos is unlikely to achieve that number. And as I said a lot of the energy won't be focused. This is just a first approximation, to indicate that yes, potentially the energy is there.
  - deleted 12y ago
    [deleted]
  - TheLoneWolfling 12y ago
    Edit, because YC locks comments for some absurd reason: Apparently the vapor pressure of CO2 is ~60atm, which bumps the number down to ~12.5KJ. Though I haven't checked that number yet.

lifeisstillgood 12y ago

I have to ask - can you explain the equations please?

TheLoneWolfling 12y ago
They all come from the equations of adiabatic processes of an ideal gas. Wikipedia explains better than me: https://en.wikipedia.org/wiki/Ideal_gas https://en.wikipedia.org/wiki/Ideal_gas https://en.wikipedia.org/wiki/Adiabatic_process https://en.wikipedia.org/wiki/Adiabatic_process (Especially the section "Derivation of discrete formula" in the latter) It's only a first approximation, but meh.

r00fus 12y ago

The 22 KJ seems widely varying form other estimates on this discussion (6KJ, 15KJ). For example, is it clear that the pressure of 100 atm would actually be attained by a simple thermos and dry ice? What lead you to use that number? Furthermore, the .50cal round is focused on a small area, the impulse and destructive force are multiplied by the shape of the bullet. Just like how shape-charges magnify the explosive force of munitions to bust armor.

TheLoneWolfling 12y ago
The 6KJ number is flat-out wrong. You cannot multiply pressure * volume like that. It's not an isobaric expansion. I mentioned this in a reply to his comment. As for the 15JK number, I cannot see any estimates in this thread saying 15JK. Could you link the comment? The 100atm figure is assuming that the linked article's calculation of final pressure is correct, assuming the thermos doesn't burst beforehand. Although I fully agree that a standard thermos is unlikely to achieve that number. And as I said a lot of the energy won't be focused. This is just a first approximation, to indicate that yes, potentially the energy is there.
- deleted 12y ago
  [deleted]
- TheLoneWolfling 12y ago
  Edit, because YC locks comments for some absurd reason: Apparently the vapor pressure of CO2 is ~60atm, which bumps the number down to ~12.5KJ. Though I haven't checked that number yet.