I have been to a friends pool with these and they seem to work well. It was ~97 outside and the pool was 85.
http://www.mistcooling.com/poolcooler.htm
http://www.mistcooling.com/poolcooler.htm
zea3 said:I've been thinking (always a dangerous thing). I have heard people with solar panels can run them at night to provide some cooling for the pool water. What if you have solar panels set up in a heavily shaded area and run them night or day? Would that create a cooling effect? I have tried the fountain but it doesn't provide much of a drop unless you are standing under it.
Our water gets up to 90 or more in the summer and a chiller is cost prohibitive at this time.
ttu1990 said:I see that I am not alone in the world of HOT pool temps! The other day, my pool hit 100.2 degrees! I have a 20x40 saltwater, pool with an 8.5' deep end, and an automatic pool cover. Thus, it does stay covered all the time unless we are swimming. The cover does hold the heat at night, but kind of shades it during the day. I say that, becuase if I uncover the pool in the morning, and leave it uncovered the entire day, you can watch the pool temp rise almost twice as fast with it uncovered, as it does if it is covered. Anyway, back to the topic at had, "Would this work to cool a pool?"...I have an idea to use my WELL, that I use to water the yard, as a geothermal cooling source, to cool the water in the pool. I was thinking I could run a closed loop of pipe, 100' down into the well, and circulate my pool water into that loop. The water table is at about 40' and my pump head is at 100'deep and is nowhere near the bottom. The water that comes out is very cold during the summer. Running a loop 100' down and 100' back up, with the water table at 40', would give me about 120' of thermal transfer in the cold water of the well. Do you think it would be worth it? The pipe I would use would be the same as the what is in the will now... 100' of 1.25" HDPE. It is pretty thin walled and very flexable, and easy to work with. I have had to replace the well pump twice, so I know what is involved from that end... I was thinking I would probably need another water pump to get that water down the well and back up, mounted near my other pool pump, and plumbed with some bypass/diversion valves. Thoughts?
Now that's great thinking I like it!Just-a-PB said:Have you thought about Geothermal.
Go down 5 to 10' and you have a constant around 55 degrees.
Lay in a bunch of 1.5" PVC, and you can cool in the summer and heat in the winter.
Hey, :idea: the bottom of the pools we build are that deep. What if we put a grid of pipes under the pool for this purpose before we shot it.
By golly I am a genius.![]()
Patent office here I come.
Lershac said:danb said:I wonder how much ice (wet) it would take to make a significant temp change in say a 15000 gal ag pool.
Well you significant heat absorber there would be the transformation energy it takes to actually change the ice from a solid to a liquid.
So, an example...
for a 15000 gal pool, thats 15000gal X 3.7854118 kg/gal = 56781 kg (approx) or 56781000g (approx)
Lets say the starting temperature of the pool is 92deg F
Target temperature is 85 deg f
We will ignore the heat loss/gain through the sides and bottom of the pool for this example, as well as to-from the air.
Change in temperature desired is 7 deg F
Thats 33.33 deg C to 29.44 deg C a change of 3.89 deg C
lets also assume that the ice is at -10 deg C starting.
The target temperature for the ice is also 85 deg F (29.44 deg C)
The specific heat of water is 4.186 joule/gram °C
The heat of transformation for water ice is 333 kJ/kg or 333 J/G
The specific heat of water type ice is 2.1 j/g °C at 1 atmosphere (we are going to assume sea level here as well)
so we are going to wind up with some unknown mass amount at 29.44 °C . that mass will be Z
We need an energy budget, so taking water through a 3.89 °C change will require 56781000g (mass of pool water) X 4.186 J/g deg C X 3.89 = 924595685 joules to be absorbed by the ice on it way to 85 deg F
so we have 3 components to getting the ice to 85 deg F. Heating the ice to 0 deg C, melting the ice, then taking the now cold water to 85 deg F.
heating the ice 2.1 j/g deg C X 10 deg C X Z g = 21Z
melting the ice 333 j/g X Zg = 333Z
heating the cold water 4.186 j/g deg C X 29.44 X Zg = 1232.3584Z
Add all those up =1536.3584Z which has to equal 924595685, so solving for Z is 601810g (rounding)
so thats 601 Kg of ice.
or 1326.7617354 lb of ice.
if you buy it in 20 lb bags, thats 66 bags of ice. :blah:
:bowdown:Just-a-PB said:The real answer is build deep. The deeper it is, the cooler it stays. Mine is 10' and I catch my wife turning on the heatpump in the middle of summer to warm it up. Silly girl, like the 4 ac's on the house arent enough usage.
Well you significant heat absorber there would be the transformation energy it takes to actually change the ice from a solid to a liquid.
So, an example...
for a 15000 gal pool, thats 15000gal X 3.7854118 kg/gal = 56781 kg (approx) or 56781000g (approx)
Lets say the starting temperature of the pool is 92deg F
Target temperature is 85 deg f
We will ignore the heat loss/gain through the sides and bottom of the pool for this example, as well as to-from the air.
Change in temperature desired is 7 deg F
Thats 33.33 deg C to 29.44 deg C a change of 3.89 deg C
lets also assume that the ice is at -10 deg C starting.
The target temperature for the ice is also 85 deg F (29.44 deg C)
The specific heat of water is 4.186 joule/gram °C
The heat of transformation for water ice is 333 kJ/kg or 333 J/G
The specific heat of water type ice is 2.1 j/g °C at 1 atmosphere (we are going to assume sea level here as well)
so we are going to wind up with some unknown mass amount at 29.44 °C . that mass will be Z
We need an energy budget, so taking water through a 3.89 °C change will require 56781000g (mass of pool water) X 4.186 J/g deg C X 3.89 = 924595685 joules to be absorbed by the ice on it way to 85 deg F
so we have 3 components to getting the ice to 85 deg F. Heating the ice to 0 deg C, melting the ice, then taking the now cold water to 85 deg F.
heating the ice 2.1 j/g deg C X 10 deg C X Z g = 21Z
melting the ice 333 j/g X Zg = 333Z
heating the cold water 4.186 j/g deg C X 29.44 X Zg = 1232.3584Z
Add all those up =1536.3584Z which has to equal 924595685, so solving for Z is 601810g (rounding)
so thats 601 Kg of ice.
or 1326.7617354 lb of ice.
if you buy it in 20 lb bags, thats 66 bags of ice. :blah: