Re: For anybody having a swimming pool...
TonyB
"Peter" <nob...@somewhere-in-the-uk.com> wrote in message
news:42b7u3hk7oark5t95...@4ax.com...
> 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.
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
TonyB
>
> "TonyB" <hatt.j....@tesco.net> wrote
>
>>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.
>
> holding a *constant* temperature, which is not affected by the water
> volume but merely by heat losses.
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
TonyB
>
> "TonyB" <hatt.j....@tesco.net> wrote
>
>>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.
>
Why?
TonyB
Palindrome
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
gazz
"Palindrome" <m...@privacy.net> wrote in message
news:UWbFj.678898$X56.2...@fe06.news.easynews.com...
> Mice suffer from heat loss - elephants suffer from over-heating..
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 :)
TonyB
> 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.
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
Andy Champ
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
TonyB
> 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.
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 º C, the other at 15º C. The ambient temperature is 10º C.
Which one will lose heat faster? Right, the one at the higher initial
temperature.
Now take two different volumes of water at 50º C and the same ambient 10º 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
Palindrome
<snip>
> Now take two different volumes of water.....they both have the same
> surface area
That's a neat trick.
--
Sue
TonyB
said "exposed" surface area!
TonyB
Andy Champ
>
> 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 º C, the other at 15º C. The ambient temperature is 10º C.
> Which one will lose heat faster? Right, the one at the higher initial
> temperature.
>
> Now take two different volumes of water at 50º C and the same ambient
> 10º 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
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
Palindrome
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
TonyB
>> I see what you are saying - but since he is using a pump to *keep* it
>> warm, it doesn't matter to him.
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
> shape, as volume increases, surface area increases at a lower rate.
Succintly put, wish I'd thought of saying it like that :-)
>
> Hence discussions about how heat loss from cubical swimming pools
> increases with volume tends to be somewhat theoretical/academic.
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
Palindrome
<snip>
> I think the Navy's training pool for submariners is an indoor one, so
> heat losses would be
> less anyway.
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
Andy Champ
>
> How does a pump keep it warm? He's complaining that he's losing heat
> so it surely does matter to him.
>
friction. He has said he's referring to the steady state.
>
> However, for any particular simple geometric
>> shape, as volume increases, surface area increases at a lower rate.
>
> Succintly put, wish I'd thought of saying it like that :-)
>
Look up "Square-Cube law".
>>
>> Hence discussions about how heat loss from cubical swimming pools
>> increases with volume tends to be somewhat theoretical/academic.
>
> 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.
>
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
pwate...@gmail.com
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