Water Evaporation Rate In Kg Per Meter Sq per Hour

This isn't homework, right?

I'll get back to you, really soon.

:-) I'm 58

Kinda old for school. This may help. There may be some really smart guys who can give you the answer. I'm too old to remember those things.

http://en.wikipedia.org/wiki/Clausius-Cl…

Actually, it won't depend on "many" factors (probably just the few already mentioned), but these conditions are more extreme than most psychrometric charts I (and maybe anyone else) have ever seen.

Tornado,

than could you reduce the numbers to the ones you have in the charts and give me a 'modified' answer. It'll still be good for me.

I would expect the evaporation to happen so fast that it would be hard to measure.

First we need a broad phase state diagram for water. Now since 1 atm of pressure is identical to about 101.3 kilo-Pascals that means that 42 atm is 4.26 Mega-Pascals. The super critical fluid, liquid, vapor triple point for water is at 647.096K (about 375°C) and 22.064 MPa. So your water in 'dry' air will only exist as vapor since the pressure is not great enough to make a super critical fluid but there is nearly an order of magnitude more ambient temperature than is needed to create vapor. Now depending on the volume of liquid water exposed to this air and the volume of air available I suppose that liquid water might exist for awhile.

But the OP does not really offer all of the information needed to attempt to calculate the evaporation rate at even normal temperatures and pressure. The most critical things are the volumes of water and air, the surface area between water and air, the circulation in the air, and the relative humidity of the air. Allow me to pose a simple thought experiment at STP (Standard Temperature & Pressure aka room temperatures). If you spill half a glass of water in a dry climate and keep the other half in the glass. In short order the water on the non-porous ground will evaporate while the water in the glass will have only lost a small amount of water.

Thank you guys,

The range I gave is out of the tables, so I am changing it.

What is the pressure is 25.6 atm (2600 kPa) and thetemperature is constant 226 degree C.

The tables don't really have data on the time required. I know that 1 J=1 W and that is the energy needed to rase the temperature of 1 kg of water by 1 degree C for 1 sec. I can't figure it out how to use it though.

Fine, you picked a phase change boundary region between vapour and liquid now so that it you can have water or vapor. You apparently did not understand at all why the spilled water evaporated but the water in a glass stayed mostly liquid. I can only help you so far.

Redfred I hear you,

let's do this. Pick up numbers and conditions that work for you, but give me a result that will have the time for evaporation and mass from 1 sq meter surface.

The numbers I use are from an actual test

It does not look like from actual test. Can you define the application where and how you have encountered water in the 'dry' air at 42 atm pressure and 2900 degree temp? Or else you are simply wasting time of we guys.