Fluid MechanicsHard
Question
Assume that a drop of liquid evaporates by decrease in its surface energy, so that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible? The surface tension is T, density of liquid is ρ and L is its latent heat of vaporization.
Options
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Solution
If a layer of thickness dr is evaporates then change in surface energy
= (change in surface area) T
= (d(4πr 2))T = 8πrdrT
energy required to evaporate layer of thickness
dr = (4pr2dr)ρ .L
The process of evaporation only starts only if change in surface energy is just sufficient to evaporate the water layer
⇒ (4πr2dr)Lρ = (8πrdr) T
⇒
= (change in surface area) T
= (d(4πr 2))T = 8πrdrT
energy required to evaporate layer of thickness
dr = (4pr2dr)ρ .L
The process of evaporation only starts only if change in surface energy is just sufficient to evaporate the water layer
⇒ (4πr2dr)Lρ = (8πrdr) T
⇒
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