Swimming Pool Loads

Pool Loads

The main concern with pool loads especially outdoor pools, is the surface loss.

Surface Loss = 12 BTU's per square foot per hour per degree rise. The degree rise would be the difference between the desired pool temp and the surrounding air temp.

For instance; a 20' x 40' outdoor pool maintaining 85 deg. water temp on a 65 degree day would be

20' x 40' x 12 BTU's x 20 delta T = 192,000 BTU/hr.

To find out how many hours it would take to heat up, will require the basic formula of the BTU. (the amount of energy required to raise 1 pound of water 1 degree Farenheit)

So lets take that same pool figured with an average depth of 6'. Knowing that 1 cubic foot of water is equal to 7.5 gallons and 1 gallon equals 8.33 pounds the formula is:

20' x 40' x 6' average depth x 7.5 gallons x 8.33 pounds x delta T (20) = 5,997,600 Total BTU requirement.

Divide that by the surface loss per hour and that will give you the amount of hours it will take for 192,000 BTU/hr to heat the pool. 5,997,600 / 192,000 = 31.2 hours.

hope this helps.
kf

And try a \"radiant\" floor!

Truly works! Have done one, placed tubing in floor of pool and used house boiler to heat the pool without the usual heat exchanger and not having to interlock the pool circ pump with the heater. Control with a slab sensor and a tekmar controller. If you have a 90+ efficient boiler then you have a 90+ pool heater too!