Solar Cover and Reel Recommendations?

[kaybinster]

My chlorine is kept very low, but yes I keep the cover on all the time-- well except when we go swimming (he he he!) I partially agree with you on the fact that you will gain more solar heat with the cover off during the day. However, your net energy flow will not be better by doing that as you will greatly increase the rate of water evaporation with it off and as we all know evaporations will very quickly cool the pool. So, I strongly believe you will better off to leave it on all the time except when swimming.

[matt4x4]

Evaporation - BAD WORD!
I live in the country, no real water sources, trucking it in is expensive - one of the reasons the cover stays on - I keep my water in the pool.
The heat gain without a cover is actually higher even if you take evaporative cooling into account, I think it was proven umpteen times on the forum, but then you have to top it off with cold water, plus you went through all the daily troubles of rolling that thing up!
I'm a firm believer in keeping it simple - the less work my pool is the better, that's why Ben's 3 B's are perfect, I hardly had to do a thing to my pool last year except add bleach and toss teh barracuda in once a week and it was always sparkling!

[duraleigh]

Hmm, well, I always love a good discussion. It would be my uneducated opinion that evaporative cooling would exceed radiant heat gain in MOST cases. I would say at 10% relative humidity and a 30mph wind, I would take evaporative as the winner everytime. 90% relative humidity and no wind would probably tip the scales the other way.

The variables are so great I'm not sure how it could be tested in a real pool. I will soon contact my engineer friend who deals with these issues daily and post back if he offers something interesting.

Dave S.

[kaybinster]

I don't buy it that solar input would ever be greater than losses from evaporation. Here is why. It takes one BTU to raise the temperature of one pound of water one degree F, BUT for each pound of water that evaporates the remaining water will loose roughly 970 BTU's! That is a very big difference. Look at it another way, to raise the temperature of a pool that has 30,000 gallons by one degree requires roughly 250,000 BTU's of energy. That same energy is lost for each 31 gallons of water that evaporate from that 30,000 gallons pool.

[duraleigh]

Hi, Kay,

If I read your post correctly, "that solar input would ever be greater than losses from evaporation." wouldn't that mean that a pools' temperature would never rise during the daytime? Mine sure does.

As to whether the cover adds or subtracts from heat gain is, I think, dependent on all the other variables associated with the cover off (namely wind, RH, color of the pool, etc.)

[kaybinster]

No, did not say that. Just stating that there is a huge loss of heat due to evaporation. I strongly believe that you are better off with the cover on as it reduces evaporative losses. I guess you could try it for a week with the cover on and a week with the cover off and (if conditions are the same) see which works better for you. I have done it both ways and find when I leave the cover on during the summer my heatpump almost never runs, but when I leave it off during the day the heatpump has to run more.

[matt4x4]

HEAT energy is needed for evaporation to occur, this means that you're putting heat into your pool, only, some of that heat is used to evaporate some of the water. The remaining heat will warm the remaining water, meaning it is inputting BTUs , not removing. This is why your pool will warm up during the day and your water level will drop at the same time.
You cannot input BTU and take away BTU at the same time, there are many factors at play here, wind itself plays a big role, so does relative humidity, since the humidity is high where we are, we lose way less water in a day than some person in Arizona on an identical day, yet, both our pools managed to warm up over the course of the day - mine probably warmed up more because i did not lose as much water to evaporation, but both will have warmed to some degree.
You could never get actual numbers using a pool since the environment is not controlled and outside factors are ever changing, however, in general terms, it has been shown numerous times in this forum that a pool with the cover off will warm up more than a pool with the cover on.
Mind you, these tests were likely performed on two days back to back where the weather was similar, yet never identical.
If you use a non scientific logical approach, you can deduct that this is true. Placing a cover over the water does stop evaporation, BUT, the cover reflects much of the heat so you have a big heat loss due to reflection. This is heat that NEVER even gets to your water, so you've just lost efficiency, it also does not allow much heat to penetrate through the cover, essentially the cover does warm up and transfer some of that heat into the water directly below it and also loses some of that heat back into the air above it. The water below the cover does NOT heat the water beneath it very deeply since heat rises, so it just reheats the cover which reheats the water BUT also the air above it.....so you get a 1-2" warm layer under your cover and the remaining 4 feet are unaffected.
If you take the cover off, the heat travels through the water without much loss since water is relatively opaque for the depth of a pool and not much is lost to reflection or absorption, these energy rays (heat) are now heating the floor and some of the walls of the pool - adding the floor area and partial wall area together also shows that there is more surface area now absorbing heat (yes, there will also be some reflection), this heat will transfer to two adjacent surfaces as well, the ground beneath the floor and the water above it. BUT, the heat that transfers to the ground will help warm the water since it will RISE back into the floor of the pool rewarming the ground and WATER above it etc etc...
The water warming at the BOTTOM of the pool will help heat the water above it as the heat rises, the water at the surface of the pool is affected the least since most of your heat transfer happens at the bottom and dissipates as it rises leaving the top surface at a relatively stable temperature while the lower water warms at a much higher rate because you have a lot less wasted energy to the outside environment.

In the end, all I care about is that my pool warms up in the summer, not cools down.

[kaybinster]

Well this is contrary to my experience. Further, I disagree with your statement that heat input is needed for evaporation. If you believe that then explain to me how if you put a glass of water on your kitchen counter the level will drop over time. The water and the kitchen are at the same temperature. Water will evaporate, cooling the water in the glass, which then absorbs heat from the air in the kitchen to maintain an equal temperature with the room. Thus, although trivial in amount, the evaporation process is removing heat from the air in the kitchen. The same happens with the pool. If the evaporation causes the temperature in the pool to drop below the air/ground then it too will absorbe heat, if it is higher it will not.

HEAT energy is needed for evaporation to occur, this means that you're putting heat into your pool, only, some of that heat is used to evaporate some of the water. The remaining heat will warm the remaining water, meaning it is inputting BTUs , not removing. This is why your pool will warm up during the day and your water level will drop at the same time.
You cannot input BTU and take away BTU at the same time, there are many factors at play here, wind itself plays a big role, so does relative humidity, since the humidity is high where we are, we lose way less water in a day than some person in Arizona on an identical day, yet, both our pools managed to warm up over the course of the day - mine probably warmed up more because i did not lose as much water to evaporation, but both will have warmed to some degree.
You could never get actual numbers using a pool since the environment is not controlled and outside factors are ever changing, however, in general terms, it has been shown numerous times in this forum that a pool with the cover off will warm up more than a pool with the cover on.
Mind you, these tests were likely performed on two days back to back where the weather was similar, yet never identical.
If you use a non scientific logical approach, you can deduct that this is true. Placing a cover over the water does stop evaporation, BUT, the cover reflects much of the heat so you have a big heat loss due to reflection. This is heat that NEVER even gets to your water, so you've just lost efficiency, it also does not allow much heat to penetrate through the cover, essentially the cover does warm up and transfer some of that heat into the water directly below it and also loses some of that heat back into the air above it. The water below the cover does NOT heat the water beneath it very deeply since heat rises, so it just reheats the cover which reheats the water BUT also the air above it.....so you get a 1-2" warm layer under your cover and the remaining 4 feet are unaffected.
If you take the cover off, the heat travels through the water without much loss since water is relatively opaque for the depth of a pool and not much is lost to reflection or absorption, these energy rays (heat) are now heating the floor and some of the walls of the pool - adding the floor area and partial wall area together also shows that there is more surface area now absorbing heat (yes, there will also be some reflection), this heat will transfer to two adjacent surfaces as well, the ground beneath the floor and the water above it. BUT, the heat that transfers to the ground will help warm the water since it will RISE back into the floor of the pool rewarming the ground and WATER above it etc etc...
The water warming at the BOTTOM of the pool will help heat the water above it as the heat rises, the water at the surface of the pool is affected the least since most of your heat transfer happens at the bottom and dissipates as it rises leaving the top surface at a relatively stable temperature while the lower water warms at a much higher rate because you have a lot less wasted energy to the outside environment.

In the end, all I care about is that my pool warms up in the summer, not cools down.

[matt4x4]

Sorry, I should not have said heat input, I should have said energy input, energy is needed, this energy is converted to heat which in turn evaporates your water.
Leaving a glass sitting on your counter doesn't mean the world stops turning, there is air circulating (energy), there is light (energy), your house still warms and cools during the day (energy).....
Just like a microwave, you don't just put in something, press a button and voila - magically it's now warm - it must be magic, because the inside of the microwave is still cool, so we didn't add heat, and really it just sat there for 2 minutes and now it's hot, it must be magic.
The microwaves that are generated transfer energy to the particles that make up the food, the particles get excited by this energy and convert it to heat.

Here, this is from the US govenment....

Heat (energy) is necessary for evaporation to occur. Energy is used to break the bonds that hold water molecules together, which is why water easily evaporates at the boiling point (212° F, 100° C) but evaporates much more slowly at the freezing point. Net evaporation occurs when the rate of evaporation exceeds the rate of condensation. A state of saturation exists when these two process rates are equal, at which point the relative humidity of the air is 100 percent. Condensation, the opposite of evaporation, occurs when saturated air is cooled below the dew point (the temperature to which air must be cooled at a constant pressure for it to become fully saturated with water), such as on the outside of a glass of ice water. In fact, the process of evaporation removes heat from the environment, which is why water evaporating from your skin cools you.