I have to figure out whether sea water can be lifted and distributed over the solar-heated evaporator panels using capillary action. The Hausmann and Slack text that I used in 1942 was on loan to our grandson, so I went to a local public library and searched "capillary action" and "surface tension". Nothing came up so I searched "physics". When only a juvenile book and an encyclopedia showed up I went to the stacks and found some books on college physics. There seems to be a semantic problem that excludes college physics from just-plain-old physics, but I did not pursue that distraction because a book by Serway and Faughn had what I needed. Other books on the shelf and the physics encyclopedia made no mention of the subject. A Van Nostrand Scientific Encyclopedia had some material that seemed incomprehensible and incorrect to me.
When my Hausmann and Slack (written more than 71 years ago) was returned to me, I saw that liquids that adhere to solid surfaces have a contact angle less than 90 degrees. The cohesive force between liquid molecules at the contact line is represented by a vector that bisects the wedge of liquid, i.e. it runs through the nearby center of mass. The adhesive force between the liquid molecules and the solid wall is represented by a vector perpendicular to the wall. The sum of the cohesive and adhesive forces is a resultant vector that is perpendicular to the liquid surface. This resultant vector nails the surface to the wall, but it neither adds to nor subtracts from the surface tension of the liquid at its air interface.
When a glass tube is inserted in a pool of water a concave meniscus is formed and the water turns upward toward its attachment line on the glass. Various sources give the contact angle between 0 and 20 degrees. The surface tension of water at 20C is 0.073 N/m. The lifting component of the surface tension force is 0.073 N/m multiplied by the cosine of the contact angle. We can assume a value of 1.0 for this cosine. If the radius of the tube is R, the total lifting force is (2 pi R)(0.073 N/m). The weight of the slug of water being pulled up is (pi R2)(density)(9.8 m/s2)(height). If the radius is 50 micrometers, the water will rise 29.1 cm above the surrounding pool.
We only need to raise water about 10 cm from the sea surface to the solar-heated evaporator surface, but we have to feed it at a rate which matches the rate at which water is evaporated into the air. The extra 19.1 cm lift translates into pressure to overcome the viscous forces resisting the flow through the skinny tube. This will determine the transport rate through a single tube. The rate at which water is evaporated per square meter divided by the transport rate of one tube will determine the number of tubes required per square meter. I have to check evaporation rate calculations that I did years ago. I have an alternative if the capillary force won't supply enough water.
Comments