|
|
re: swimming pool h2o evaporation.
27 jul 2005
wes schott wrote:
>my swimming pool is losing about 1/2" to 1" of water per day.
>
>is this normal?
maybe not. one ashrae swimming pool formula says a square foot of pool
loses 0.1(pw-pa) pounds per hour or water, ie 0.46(pw-pa) inches per day,
where pw and pa are the vapor pressures near the water and in the air
around the water, in inches of mercury column.
pa = 29.921/(0.62198/w) = rhe^(17.863-9621/(460+ta))/100. for instance,
the humidity ratio w = 0.0071 pounds of water per pound of dry air on
an average 88.9 f july day in dagget, ca (a very hot and dry climate),
so pa = 29.921/(0.62198/w) = 0.3376 'hg.
bowen's 1926 equation says the heat gain from warmer air equals the heat
loss by evaporation at the wet bulb temp, independent of wind speed. if we
add heat gain from sun (2540 btu/ft^2/day or 105.8 btu/h-ft^2 in dagget),
(460+88.9)-tw + 105.8 = 100(pw-0.3376), ie 688.5 - tw = 100pw, with tw in
rankine degrees.
pw = e^(17.963-9621/tw), so ln(100pw) = 4.605+17.863-9621/tw = ln(688.5-tw),
ie tw = 9621/(22.47-ln(688.5-tw). plugging in tw = 460+88.9 = 548.9 r on
the right makes tw = 548.8 on the left, then 548.82, so the pool and air
temps would be about the same.
pw = e^(17.863-9621/tw) = 0.717 "hg, so the pool would evaporate
0.46(0.717-0.3376) = 0.174 inches per day, ie 1" every 5.7 days.
sounds like a leak to me.
> located in southern california
whereabouts, more specifically?
what's the pool temp and air temp and rh?
is it in sun or shade?
nick
|
|
