How is potential energy lost when a water droplet is dropping down slowly on a wall? When a water droplet is on a vertical wall, it usually drops slowly, which is different from free falling. As the dropping speed is slower than free drawing, so I guess some energy must be lost. 
I guess it is lost as internal energy, but if it is true, how a water droplet gains internal energy when dropping slowly in microscope view?
 A: You are not dealing with dry friction here.  What is involved is viscous flow within the water droplet.  There is a viscous circulation pattern set up within the water droplet, and the flow results in viscous dissipation of the (mechanical) potential energy to internal energy. So, if the water droplet where somehow insulated from its surroundings thermally, its temperature would rise (a tiny amount).  In reality, the thermal energy is transferred to the surroundings.
A: The energy loss is frictive in nature. As the drop slides down the surface, there is a complex interaction at the interface between the droplet and the wall. This will go into changing the temperature of the water droplet, kicking some molecules off, etc...
Perhaps someone with a litter better skills in thermodynamics could help here, but I think you should be able to estimate the temperature change in the water droplet if you know its velocity as a function of time down the wall. The work done by friction should have some relationship to the amount of heat transferred into the droplet. My gut says those two are equal, but I am having trouble justifying that answer. Obviously in vacuum, there should be no heating.
A: Energy lost because it create a layer of water as it slides down, and layer created due to the viscous force between the liquid - liquid layer and liquid - container layer. 
What happens is the fluid experience, s a Shear force as it slides (tangential) , and this Shear force is cause of energy. Loss, as Water flows down drop become small and energy lost will continue till the size of drop become negligible. 
