re: looking for swimming pool heating advice: propane
1 jul 1999
randal bertuccelli wrote:
>...right now i know it takes a gallon of propane an hour for each
a gallon of propane provides about 90k btu of energy, no matter how fast
we burn it (power = energy/time.) burning a gallon per hour makes about
90k btu per hour or about 26 kw of heating power, about the same as
a 5 kw generator.
>...i'm looking at usage for *maintaining* heat. i plan to use it in
>conjunction with a solar bubble-wrap type cover...
we lose heat with the cover on (without much evaporative heat loss)
and with the cover off (when swimming, with lots of evaporative loss.)
>the other two variables mentioned in my first post but not here are:
>24,000 gallon pool and winter temps from 40 to 65 degrees f.
let's assume you have a 4' tall x 32' diameter above-ground pool near
philadelphia with an r1 solar cover and 4" of foamboard (r20) wrapped
around the outside and 2" (r10) on the ground underneath. in january, the
average outdoor temperature is about 30 f, and an average 490 btu/ft^2
per day of sun falls on a horizontal surface, vs 1000 on a south wall.
the average windspeed is 10.3 mph (which varies a lot by location), and
the average humidity ratio w = 0.0025 pounds of water per pound of dry air,
according to nrel.
the pool has about 800 ft^2 of top surface, and 400 ft^2 of sidewall.
the cover's thermal resistance is about 800ft^2/r1 = 800 btu/h-f, and
the sidewall has 40 times less: 400ft^2/r20 = 20. the ground underneath
might have an effective r10 thermal resistance to a 50 f winter temperature,
so the pool would lose about (80f-50f)800ft^2/r20 = 1200 btu/h to the ground.
keeping the covered pool at 80 f requires about 1200+(80f-30f)820btu/h-f
42k btu per hour or = 1013k btu/day. if the cover transmits 80% of the sun,
it provides an average of 0.8x490x800 = 313k, leaving 699k, about $6/day,
if propane costs $1/gallon.
swimming in fairly still air with the cover off (only andrew mckegney's
friends swim under the covers :-) means losing about 100a(pp-pa) btu/h of
heat by evaporation, according to equation 2 on page 4.7 of the 1991 ashrae
handbook of hvac applications, where a is the pool surface in ft^2, and pp
and pa are the water vapor pressures in inches of mercury near the surface
and in the surrounding air. pp = exp(17.863-9621/(80+460)) = 1.047 "hg, an
estimate from the clausius-clapeyron equation. pa = 29.921/(1+0.62198/w)
= 0.120 "hg, so the pool loses about 100x800(1.047-0.120) = 74k btu/h of heat
(and about 9 gallons of water per hour) by evaporation, with the cover off.
this adds another 2 bucks or so to the daily propane bill. the pool's surface
receives an average of 64 gallons per day of rain in pennsylvania.
now suppose we add a serious movable cover, say another 4" of foamboard,
raised up for 6 hours during the day and in place overnight. suppose it's
reflective underneath, and hinged on the north edge, and it bounces a total
of 600k btu/day of sun into the pool.
then 600k btu = 6h(t-30)800ft^2/r1 for the cover, during the day
+ 18h(t-30)800ft^2/r21 for the cover, at night
+ 24h(1200+(t-30)400ft^2/r20) for the bottom and sidewall
+ 2hx74k btu for swimming,
and t = 101 f, with no need for propane.