For anybody having a swimming pool...

The standard lines from pool builders is that 90% of the heat loss from a pool is via the surface, only 10% into the ground.
It turns out this originates from some website in Florida, and is probably true in Florida if you have a pool at +30C and totally open, no cover.
The rest of the time it is bollocks and is propagated by pool builders who are too thick and lazy to bother with insulation. The whole UK pool business works on the principle of the rolls royce buyer: if you need to know the mpg you obviously cannot afford one.
So they stick a 40kW boiler in there and waste your money.
I have been operating two outdoor pools over 10 years, and done some very careful measurements. They are quite an eye opener.
Case 1:
12x5m concrete pool, depth from 1m to 2.5m NO insulation anywhere. Cover is 6mm air-bubble PVC (standard £400 sort of thing).
Heat loss is around 10kW for a 20C water-air difference, and about 2/3 of this is into the ground, through the pool sides and bottom and the pipes which are traditionally buried directly in the ground. The air/ground loss split was determined by laying heavy insulation on top of the surface (2 layers of industrial 1/2" thick bubblepack) and seeing how much difference this makes - not a lot.
This pool is expensive to heat... you would spend £5000/year at current gas prices heating it all year round, and that is assuming you have one of those inflatable domes on it for much of the year.
Case 2:
10x4m concrete pool, depth from 1m to 1.7m Insulated with 50mm Sellotex around sides, no insulation underneath (there doesn't appear to be a suitable material which would carry the weight). One idiot pool builder walked off the job because he didn't want to be bothered. Most others wouldn't even quote. Eventually I found one. It's really easy of course. Cover is a 'solar' slatted cover (£13000).
Heat loss is around 2kW for a 20C water-air difference, and about 3/4 of this is into the ground.
The division of this loss is interesting.
There are 4 pipes coming out of the pool, and one going back in. All were insulated with fibreglass wool but only crudely. For half their run they are bare but laying in an air-filled buried duct. So not directly in soil anywhere. But the pipes are exposed in the plant room, where the ambient is just that (due to big vents for a heat pump).
About 1/2 of the loss is via the pipes! IOW, if I stop the filtration pump at night, the total heat lost from the pool is halved. One achieves significant savings by closing all but one of the outlet pipes.
Of course the expensive solar cover makes a hugh contribution; around 15-20kW from the 40m2 area in any sunlight. But one pays for that in the purchase price.
The thing is that insulation costs only a few hundred quid on the purchase price... for that you could insulate ALL the pipework with 100mm Armaflex too.
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While I accept most of what you say, you are not comparing like with like. One pool has more than twice the amount of water in it than the other, so heating costs for the same change of temperature will reflect that. It'll be interesting to see if your pipe theory is correct by experiment - I guess you'll be checking losses with the pump off at night etc over a period of time.
TonyB
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Yes, but not twice the surface area. I was measuring the cost of holding a *constant* temperature, which is not affected by the water volume but merely by heat losses.

The pipe losses are measured, not theoretical. I have been measuring the temperature drop, for a given water-ambient difference, for various combinations of open pipes, with no heating going on.
In fact the pipe losses are greater than it appears because when the pump is running, one is putting about 500W (3/4HP pump, which draws ~ 600W according to a true power wattmeter) into the water from the pump only.
Unfortunately there is nothing I can do to add insulation - except inside the plant room.
I am doing this with a PT100 thermometer, accurate to 0.1C.
I suppose the key point is that insulation is a piece of cake. It takes only minutes extra to stuff a load of Cellotex down the side of the pool before filling the cavity up. And insulating pipes with some pipe insulation is equally quick.
I think the reason the pipe losses are so significant is this:
Say the soil temp is +10C. The pool is +25C. The heat pump produces an increment of about 1.5C (a Calorex 910X, 6kW on a good day...) at the particular flow rate. If the loss along that pipe is just 0.3C (which doesn't sound much, given +25C water) you have wasted 1/5 of the heat pump output, just like that :) When dealing with such small temperature changes, heat losses become critical.
In the trade, the builder chucks in a big boiler which costs a fortune to run, and when this is running it is producing a big temperature increment so that is OK, but one has to run the filtration pump all the time even if the boiler is not on, and that is where the heat loss goes.
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wrote

Rate of heat loss will depend, amongst other things, upon the temperature differential between the water and the air. The higher that differential is, the greater the rate of loss will be. Therefore a large volume of water, which will maintain it's higher temperature for longer, will have a disproportionately higher rate of loss than a smaller volume over the same period of time.
TonyB
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Of course. Newton found this. Conductive heat flow is proportional to temp difference. Convection and radiation are more complicated.
What I reported was a 5x energy saving due to the most basic insulation. Dead easy to achieve.
The difference in the surface area (applicable to conductive losses) between the two pools is only about 1.5x
The funniest thing (for some) was when one pool builder resigned from the job on the day before he was due to start, and after the contract was signed, because he thought the insulation requirement was too difficult for him.
All we need now is a lot of Polish pool builders :)
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actually the above is incorrect!
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wrote

Why? TonyB
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Taking the steady state condition, all that matters is heat loss from the boundary of the object. The object volume, specific heat capacity, etc, don't come into it at all.
The heat loss (assuming the simple case of conduction only) is proportional to the temperature gradient across the boundary.
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TonyB wrote:

Let's see -
volume of a cube = a^3 surface area of a cube = 6a^2
As volume increases, surface area increases at a lower rate. As heat loss can only occur from a surface, the greater the volume, the proportionally less is the increase in surface area and thus the proportionately lower will be the heat loss, compared to smaller volume of water over the seame period of time.
Mice suffer from heat loss - elephants suffer from over-heating..
-- Sue
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dunno about mice, but i keep pet rats, and they suffer over heating easily, because thye have a large body but only a small part of it can transfer excess heat... their tail, they dont pant like dogs, and dont sweat like humans, so their tail is their heat regulator,
How does that work into the equasions :)
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Yes of course that's correct if the two bodies of water are at the same temperature.
My point was that there will be a factor, which may turn out to be insignificant I acknowledge, whereby the larger body of water ( assuming the both start at the same temperature ) will retain it's heat for longer than the smaller body. Therefore after say 4 hours, the smaller body of water will have lost a large amount of heat and it's temperature will be lower than the larger body after the same 4 hours. Because the differential of temperature will be greater in the larger body compared to the smaller, the rate of loss will be corresponding greater from the larger body.
I've just thought of a better way to explain it - which body will reach ambient temperature first? Clearly the smaller body as Sue described, therefore at that point the larger body will still be losing heat and will therefore have a greater heat loss than the smaller. During the cooling cycle, at some point that crossover must occur, and it's due to the larger body retaining more heat than the smaller, therefore the differential to the outside temp becomes greater. TonyB
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TonyB wrote:

Tony, Volume is irrelevant to his problem.
Surface area, temperature differential, insulation, wind chill, and more - they all come in to it. But not volume.
What volume will give you is thermal inertia, which means that the same rate of heat loss (in kW in this case) will result in less drop in temperature - but it will take just as much energy to heat it up again, and just as much energy to keep it warm.
Andy
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Yes, I've acknowledged that it probably is insignificant, but I still think there's a point here that is being missed. Put it this way:
Two identical volumes of water, swimming pools if you like, one has water at 50 C, the other at 15 C. The ambient temperature is 10 C. Which one will lose heat faster? Right, the one at the higher initial temperature.
Now take two different volumes of water at 50 C and the same ambient 10 C. The higher volume will retain a higher temperature for longer than the smaller volume as they both have the same surface area and lose heat at an equal rate. After a while you will have two volumes at different temperatures and we've established above that in that case one will have a higher rate of loss than the other. Therefore volume is important when comparisons are made between different volumes!
TonyB
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TonyB wrote: <snip>

That's a neat trick.
-- Sue
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One's deeper than the other.... but I know what you mean - I should have said "exposed" surface area!
TonyB
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TonyB wrote:

I see what you are saying - but since he is using a pump to *keep* it warm, it doesn't matter to him.
Palindrome,
a cubical pool 2x2x2 has a total area (underground and to air) of 12, and a volume of 8.
A cuboidal pool 1x1x5.5 has the same area, but a volume of only 5.5.
Andy
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Andy Champ wrote:

Of course - different geometrical shapes have different equations for volume and surface areas. However, for any particular simple geometric shape, as volume increases, surface area increases at a lower rate.
Swimming Pools all tend to be near enough the same very approximate depth, independent of volume. Few are spherical. Most have width and length far greater than depth. There are, of course, always exceptions. The Royal Navy has a pool with a depth far greater than width or length, IIUC. Used for submariner escape training.
Hence discussions about how heat loss from cubical swimming pools increases with volume tends to be somewhat theoretical/academic.
--
Sue








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How does a pump keep it warm? He's complaining that he's losing heat so it surely does matter to him.
However, for any particular simple geometric

Succintly put, wish I'd thought of saying it like that :-)

But, however academic, it does happen. I wasn't sure how real or academic it is - but I think we've established that it's not significant - I did say that in my first reply.
I think the Navy's training pool for submariners is an indoor one, so heat losses would be less anyway.
TonyB
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TonyB wrote: <snip>

Nope. It's a bloody great tower, visible for miles. http://www.kentscuba.com/sett.htm
However, unlike it may appear in the picture - it doesn't actually levitate..
70 year's old.
Rather a bargain "experience of a lifetime" at <40GBP.
And yes, as a taxpayer, I pay to heat the water....
-- Sue
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TonyB wrote:

Heat pump. IAC, any pump will warm the fluid slightly if only from friction. He has said he's referring to the steady state.

Look up "Square-Cube law".

Of course real pools aren't cubical. They tend to be about 1m at the shallow end, and 2-3m at the deep end (or part - sometimes the middle) however much the surface area changes. Diving pools excepted of course.
All I was trying to show is that you *can* change the volume without changing the area, and vice versa.
Andy
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