Yes normal without a solar cover. Go out and buy one of those fish that put
out the invisible solar cover. I have the one with the battery operated
pump, every 1 hour it puts out a little juice that puts a film on the pool
to slow down the evaporation it works great and only cost me $19 and lasts
for 60 days for my size pool (14X25).
check this out
Excessive Splash Out or Evaporation
Lets touch on the splash out and evaporation issue first and see if it
can quickly be eliminated.
If the pool is not being used frequently (then obviously), it is
unlikely that the problem is splash out. On the other hand, if it is
the middle of the Summer, with high temperatures and lots of kids
getting in and out (dripping wet) this could, in fact, be a real cause.
To eliminate evaporation or splash out pool water loss as a causes,
here is a quick test that can be run on your swimming pool:
* Place a 5 gallon bucket on the second step or bench seat and fill
it with water so that it is exactly the same level as the swim pool.
* If evaporation is the cause of your loss, both levels should drop
at the same rate.
* If the bucket drops faster, you probably have a thirsty dog - so
be sure to secure all pets BEFORE the test.
* If you have a pool leak, then the pool will drop faster than the
water level in the bucket.
Once you figure out if you are actually losing water in the pool via a
leak, and that is not just evaporating or splashing out, there are a
few more tests that can done - but now might be a great time to call in
a pool professional to evaluate the situation.
Here in the desert southwest, evaporation from open bodies of water is about
7.5 feet per year, probably more than a foot in a typical hot summer month.
Days with high temperatures and wind experience the greatest loss. The 1/2"
to 1" inch you cite, if it occurs in the summer, seems to be well within the
range you might expect. On the other hand, such losses in January might
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
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.
Whereabouts, more specifically?
What's the pool temp and air temp and RH?
Is it in sun or shade?
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