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In Leidenfrost effect, the hot steam suspends water droplets above it and prevent the drop from touching the hot pan. My question is, as the steam is less denser than water, should the steam not travel up and make the water drop to touch the pan?

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As the steam gets created from the evaporated liquid at the bottom of the liquid drop, it expands and provides additional pressure acting on the bottom of the liquid drop. The steam does move upwards, but new steam is continually being created so the pressure is able to hold the drop slightly above the surface.

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  • $\begingroup$ Actually it is viscous fluid dynamic drag that is providing the force to support the liquid drops from falling. If the steam were inviscid, the drops would fall. $\endgroup$ – Chet Miller Jul 18 '16 at 11:48
  • $\begingroup$ @ChesterMiller, I agree that in common conditions tangential forces on the liquid surface play role as well. Still, if evaporation is more intense from the bottom of the liquid drop than it is from the top, the reaction on the drop may well provide the necessary force. I do not see how we could test your hypothetical situation where there is no viscous drag. Do you think the contribution to uplift due to viscous forces is greater than the contribution due to the pressure forces? $\endgroup$ – Ján Lalinský Jul 18 '16 at 16:12
  • $\begingroup$ I have trouble seeing how the temperature gradient across a tiny drop could be so high that evaporation from the bottom is significantly more intense than evaporation from the top. But I guess all this could be modeled mathematically. $\endgroup$ – Chet Miller Jul 18 '16 at 17:13
  • $\begingroup$ @Ján Lalinský, thank you for the reply. In your answer you have mentioned that steam moves up, but new steam is continually created. If that is the case, at that high temperature, steam must immediately rise up and the water must immediately fall on pan and become steam and so on leaving no water back immediately. But it looks the process takes long time. You can see water droplets hovering over the pan for seconds. Is this time not too length for the explanation you have stated? $\endgroup$ – user3219492 Jul 19 '16 at 7:02
  • $\begingroup$ @Chester Miller , are you taking about this formula? F = 0.5*density of liquid*(velocity^2)*drag coefficient*area ? As the density of steam is less, around 0.6Kg/m^3, the drag experienced by water while passing through steam must be less. But I have no idea about how big or small the value of drag coefficient will be. $\endgroup$ – user3219492 Jul 19 '16 at 7:13

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