Some Pool Heater Math ... Is this correct?

[TriPool]

So planning on getting a 400,000 BTU NG heater for a new build 15,000 - 15,500 gallon pool and was trying to do some cost estimates of running the heater. First my assumptions:

• from some research it appears that 1 BTU will heat 1 pound of water 1 degree.
• There are 8.34 pounds of water per gallon.
• 400,000 BTU heater puts out 400,000 BTUs per hour
• 1 CCF is roughly 102,500 BTU

So for this pool and heater I can expect to raise the pool 400,000 / (15,500 * 8.34 ) = 3.09 degrees per hour of run time.

My current cost for NG is $.19 per Ccf. so roughly 4 Ccf per hour to get 400,000 BTU or about 80 cents per hour of run time. Or if I want to think about it in cost per degree roughly 80 cents / 3.09 degrees = 25.85 cents per degree of temperature increase.

Is my math correct? It seems like it will just cost me 26 cents and 20 minutes to raise my pool 1 degree.

 

[jblizzle]

Pretty close ... the heater will actually put less BTUs into the water and there are going to be losses to evaporation and the surroundings. So might be closer to use 80% of the heater output ... or 320k BTU. So maybe 2.5 degrees an hour. Even this might be optimistic.

 

[JasonLion]

Most heaters are right around 80% efficient.

Heat loss that needs to be made up by the heater varies depending on the difference between the unheated temperature and the actual temperature, as well as if you are using a solar cover or not. As the temperature difference goes up the heat loss that needs to be made up goes up.

 

[Flippy]

I did the math a few years back and I was at 4 degrees per hour @ 4 dollars per hour. It really depends on if the sun is shinning and if it is windy.

 

[TriPool]

Thanks,

So 4 degrees an hour for an 11K pool if I use the calcs for a 15.5 K pool at 80% efficiency I get 3.09 * 80% efficiency * 15.5/11 = 3.48 which doesn't seem too far off.

So bottom line it looks like 400K BTU will use roughly 4 Ccf per hour and I can probably expect somewhere in the range of 2.5 degrees of increased temperature per hour depending on the efficiency of the system ( 2.5 at 80% )

Thanks everyone!