cooling system formulas

After having two record days of heat in a row, you asked:

Q: HOW IS YOUR COOLING SYSTEM WORKING NOW!!!???

A: Great! The details follow... but... in a nutshell... yeeha! unfriggin' believable!!!

60 gallons of water evaporated so far today.

It feels like each level of the house is about one level lower than without the system running. (i.e. 3rd floor feels like 2nd... 2nd like 1st... 1st like basement)

That's about 10-20 degree drop downstairs... and up here (on the 3rd floor) it is about 20-30 degrees less.

While sidd was in town, he was gracious enough to help my pea brain through the math (see below.)

On a typical hot day, I'm evaporating 40 gallons of water.

Sooo... the math suggests that each gallon of evaporated water is equal to running a typical air conditioner for 1 hour.

Or, you could say I'm saving 40-60 air conditioning hours a day.

Or, that I would have to run 40-60 air conditioners for 1 hour to achieve the same affect.

THE FORMULAS

L = latent heat of evaporation / unit Volume
Q = heat removed
Q = VL
L = 8,094 BTU / gallon

notes:
1 BTU = 1055 Joules
1 BTU heats 1 pound of water by 1 degree F
Lfusion = 80 calories (to freeze water)
Levaporation = 536 calories
1 calorie = 4.2 joules
1 calorie raises the temperature of 1 gm (or cc) of water by 1 degree C
1 US gallon = 3.8 liters = 3800 cc

More Of The Evaporation Math

The Waterfall House

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2003 Philadelphia Spirit Experiment & KingArthur.com

The expression of these ideas by others is strictly for non-commercial use. Written permission is required prior to any commercial application.

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