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
A.
B.
C.
D.
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
⇒
Create a free account to view solution
View Solution FreeMore Fluid Mechanics Questions
In steel, the Young′s modulus and the strain at the breaking point are 2 × 1011 N/m2 and 0.15 respectively. T...Which is a RNA virus :-...The radius of a soap bubble is r. The surface tension of soap solution is T. Keeping temperature constant, the radius of...A glass capillary tube is of the shape of a truncated cone with an apex angle α so that its two ends have cross sec...A block of steel of size 5 cm × 5 cm × 5 cm is weighed in water. If the relative density of steel is 7. Its ap...