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Schwaggs commented that:

"Thus, for every gallon of water that is evaporated, it's the same as running a 0.65 hp chiller for one hour. Assuming that a 0.65 hp chiller would require around 1000 watts of electricity and a typical clip on fan requires less than 15 watts [3] (http://www.epinions.com/Electric_Fans--reviews--clip_on), a fan is about 67 times as efficient."

This conclusion assumes the clip-on fan can evaporate a gallon of water in an hour. Do you know if this is accurate? It also assumes there is no cost for removing the water from the air (which we will pay Also, the evaporation rate is highly dependant on several variables such as tank water temp, ambiant air temp, relative humidity of the room air, water surface area, etc. On a humid day (without house AC on), the clip on fan would be hard pressed to evaporate anything.

I found this article that covers evaporation rates in swimming pools, which I suppose is fairly close to our environment:

http://www.thermexcel.com/english/program/pool.htm

http://www.thermexcel.com/english/images/for_eva3.gif

W = Rate of evaporation at the surface of the water level (kg/h m2)
Pw = Vapor pressure at saturation taken at the temperature of surface of water, in kPa
Pa = Vapor pressure at the dew point according to the temperature of the ambient air of the room, in kPa
V = Air velocity above at the surface of water, in m/s
Y = Latent heat necessary according to the change of state of the water vapor at the temperature of surface of water, in kJ/kgY = 2330 kJ/kg for our case.
For air velocity, I'd guess a fan is about 20 mph. I don't have an anemometer handy. 20mph ~= 9m/s

Which leads to:

W = ((Pw - Pa) * (0.089 + 0.0782 * 8.94)/2330) * 3600
W = (Pw - Pa) * 1.2176776

According to this page (http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/watvap.html), at 80F (a typical tank), Pw would be 25.38 mm Hg or 3.38 kPa.

W = (3.38 - Pa) * 1.2176776

Pa should depend on the ambient air's temperature and humidity. According to this page (http://www.natmus.dk/cons/tp/atmcalc/atmoclc1.htm) (variables changed to match those above), "one calculates Pw, the saturation vapour pressure at the ambient temperature. The actual water vapour pressure, Pa, is:
Pa = Pw * RH% / 100".

Assuming 60% humidity and our 3.38 kPa Pw, this yields Pa = 2.028

Thus:
W = (3.38 - 2.028) * 1.2176776 = 1.65 kg/h m2

There's 3.68 kg per gallon, so that's 2.23 gallons per hour per sq meter.

So, assuming:
- 80 degree F tank temperature
- 20 mph fan speed, equal over the entire tank surface
- 60% ambient humidity
- 48" x 18" tank surface == .55 m2

Then you'd get an evaporation rate of 1.22 gallons per hour. Seems a bit high, but 60% ambient humidity is a little low in a GA summer, and 20mph over the entire tank seems a little high.

I need to put this all in an excel spreadsheet. Can someone check my numbers?

Mojo,

Your math looks right until the last step where you divided 3.68 kg per gallon by 1.65 kg/h m2 instead of the other way around. The answer should be .44837 gallons/hour/sq meter. I have attached a graph of the gallons per hour per square foot at various relative humidity levels if you would like to add it to the WIKI.

Oops- thanks for catching that. I'm borrowing an anemometer from someone today to do some quick tests.

I'd like to come up with a spreadsheet or javascript "app" that we could host on the wiki so that people could plug in their own values.

A Evaporation calculator is a great idea. It figures that a static chart wouldn't be cool enough for a java guru! :D

Lol- it just makes it easier for everyone... :)

I just need to figure out what we want as variables. Probably:

- water surface area
- tank temperature
- ambient humidity
- wind speed

and then have it figure out gallons per hour of evap...

ok WOW. I had no idea this was the case re: fan efficiency! Am I much better off (realizing if it cools too much, that I will have the heater also running) at using fans etc on the canopy and sump versus a chiller?


Thus, for every gallon of water that is evaporated, it's the same as running a 0.65 hp chiller for one hour. Assuming that a 0.65 hp chiller would require around 1000 watts of electricity and a typical clip on fan requires less than 15 watts [3] (http://www.epinions.com/Electric_Fans--reviews--clip_on), a fan is about 67 times as efficient.

It depends. You still have to deal with the humidity and heat buildup in your house, which means you have to either a) vent it or b) run your house A/C at some point.

I did some more number crunching last night - a typical fan that we use runs at about 8-10mph, which changes some things up.

how much does it change things?

Just a quick note to see how much heat gets added by a hot canoppy.... TEMP BEFORE propping canopy open 3.5 inches (max) - 84.2
TEMP AFTER propping open the canopy open (max) - 80.8. Needless to say, a small fan will be going into my canopy shortly to remove the hot air better.