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Theoretically, I know that steam will start cooling slowly at constant volume to attain thermal equilibrium as per zeroth law of thermodynamics. After some time, steam will condense to water creating vacuum + water mixture because specific volume of steam is very high compared to water. This vacuum will result in catastrophic failure of metal ball. Question here is at what pressure metal will implode? Assume metal as steel and some thickness. You can take any values as per requirement to explain answer.

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    $\begingroup$ is this school work? $\endgroup$
    – jsotola
    Feb 16 at 15:38
  • $\begingroup$ This vacuum will result in catastrophic failure of metal ball. ... that is not a true statement ... your questions also make an assumption that the sphere will implode $\endgroup$
    – jsotola
    Feb 16 at 15:39
  • $\begingroup$ With about 14 psi external pressure, the metal would need to be thin as paper , or less. $\endgroup$
    – blacksmith37
    Feb 16 at 15:42
  • $\begingroup$ @jsotola you can correct me if I am wrong. Question arised during brainstorming session. I came to that implosion conculsion after searching for "soda can implosion experiment". It looked similar to me. But only missing factor was pressure which I haven't found anywhere. Some related things could be "Compressive strength" of metal and "buckling criteria". But I haven't understood that because no detailed explaination given anywhere. I am also interested in PV diagram of this process. $\endgroup$
    – Mr. Mechanical
    Feb 16 at 17:38
  • $\begingroup$ @blacksmith37 not necessary that metal should be thin. If internal pressure become so low, then atmospheric pressure sufficient to break those vessels. So maybe answer could be atmospheric pressure is a limit for implosion but again I am not sure. $\endgroup$
    – Mr. Mechanical
    Feb 16 at 17:41

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