If we have a liquid that is is being actively heated and is rapidly evaporating (let's say around 1g/s for Ag at 1500K) inside a vacuum chamber, can convection around the liquid/vapor interface occur and become a source of heat loss? Or would any sufficiently powerful vacuum pump simply remove any vapor?

I imagine this would only occur when the vapor pressure at the liquid vapor interface is sufficiently high, but even then it would probably be a weak source of heat loss compared to the evaporation itself and radiation. But I could be wrong.

  • $\begingroup$ The vapor is leaving the interface at the same temperature as the liquid, and it's enthalpy is equal to that of the liquid, plus the heat of vaporization. Is there another gas in the heat space other than Ag, or is the vapor phase at the equilibrium vapor pressure of the Ag? $\endgroup$ Commented Feb 8, 2022 at 23:00
  • $\begingroup$ There is no other gas, since this is taking place inside a vacuum chamber $\endgroup$
    – tjsmert44
    Commented Feb 9, 2022 at 7:11
  • $\begingroup$ Then, if this is taking place at essentially steady state, then there will be no temperature gradients in the uniform vapor phase, and no convective heat transfer resistance. $\endgroup$ Commented Feb 9, 2022 at 9:52

1 Answer 1


Convection occurs when there is a temperature difference. Is there one in your case?

If not, then the whole thing is at the same temperature and gas expands compared to the liquid, and just moves somewhere else. No convection here, boring case.

If yes, then what stops the gas from condensing on the cold side? As soon as gas will reach the cold side, it will condense on it, and then flow down with the gravity. Be it into the same container where it evaporated from, making something like a convection, or in a separate chamber.

This evaporation-condensation is in itself a pump, that can transfer extreme amount of energy, equalizing the temperature difference, so both cooling and heating is required. An ordinary pump will have to add as much mechanical work, as heating and cooling is added to equalise material movement in this almost convection and gas being removed by the pump.

In practice pumps are very rarely this powerful. And even if pump is rated much higher than the amount of heat supplied, some material will still go to the cold side and condense there.

You can roughly estimate amount of material that went to the cold side to condense and that was pumped away by the pump by comparing amount of power in the heating and the pump.


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.