I bought two solar panels last year. each is 2' x 20' black tubing. Incoming
water flows 20' down the 1st 25 tubes and returns in the 2nd 25 tubes then
it goes into the 2nd panel and does the same thing. I installed it last
summer as directed. divert some filtered water just before it returns to the
pool and send it through the heater. this heated water returned and mixed
with the water directly from the filter. It was hard to tell what the heater
was doing. This spring I set up the panels on a platform in a good location,
full sun from 10:00am to 4:00pm. I have a 2nd small old pool pump that I can
use independent of the pool filter,.

Just for the hell of it I want to try and maximize the amount of heat I can
get out of these panels...

One question is the fact that the panels are about 50' from the pool (12' X
24' AG oval with a deep end }.

What if any impact would using a garden hose instead of 1.5" pool hose for
the 50' connection cause?
I'm using the garden hose now to save $$$.

I have valves to control the flow of water through the panels.

Another issue, In the solar panel, there is a " blockage " that diverts some
of the water down the 1st 25 tubes to the end where there is no blockage.
the water then flows freely to the 2nd 25 tubes, 20' down and out to the
next panel. Some water is allowed to flow around the blockage and bypass the
panels altogether. would using a small hose or reducing the water to a
trickel allow to much water to bypass the whole panel by just flowing
through the diverter? At a low level, the water wouldnt be forced through
the 20' tubes. It would just flow through the blockage. ( when I look down
this 1st end of the tube I can see a star shaped plastic that would block
some water flow directly. If I stuck a pocker chip in the tube it would
fully block this bypass and all water would flow through the full length of
the panels...

I'm sure I could blow the panel apart this way by building to much pressure
in the panels...
But if I kept the flow down I would be heating all of the water...

Everything is set up now and as soon as it stops raining I'm going to try it
out.

Any suggestions would be helpfull???

TIA


Steve

Steve wrote:

I bought two solar panels last year. each is 2' x 20' black tubing...

You might collect 2x2x20x250 = 20K Btu/h max in 250 Btu/h-ft^2 full sun.

What if any impact would using a garden hose instead of 1.5" pool hose for
the 50' connection cause?

Not much. If 5 gpm flows through the garden hose, it might be heated
20K/(5x8x60) = 8 F, without much heat loss to the outdoors.

In the solar panel, there is a " blockage " that diverts some of
the water down the 1st 25 tubes to the end where there is no blockage.

That may be there to help make water flow more uniform, so there are no
hot spots with still water inside...

Nick

Steve wrote:

I bought two solar panels last year. each is 2' x 20' black tubing...

You might collect 2x2x20x250 = 20K Btu/h max in 250 Btu/h-ft^2 full sun.

What if any impact would using a garden hose instead of 1.5" pool hose for
the 50' connection cause?

Not much. If 5 gpm flows through the garden hose, it might be heated
20K/(5x8x60) = 8 F, without much heat loss to the outdoors.




Thanks Nick,

June 6th and the waters 65 degrees...

I havent seen anywhere near 8 F rise in temp. maybe 3 F. (I think it was
advertised at 10 F. ??)

I have it set up now so I can control the flow diverted to the panels.
and I can measure my results in GPM and temperature rise.
Would a low flow of water heat more than a high flow?
Would it be better to take the incoming water from the surface where the
waters warmer or use a hose to pull it from the deep where its much colder?


I'll experiment and see what gives the best results..



Steve

Steve wrote:

Would a low flow of water heat more than a high flow?

High makes solar collection more efficient, with
lower temps and less heat loss to the outdoors.

A clear vs blue solar pool cover and a dark bottom might help.

Nick

On Wed, 6 Jun 2007 21:45:20 -0400, "Steve" wrote:

Steve wrote:

I bought two solar panels last year. each is 2' x 20' black tubing...

You might collect 2x2x20x250 = 20K Btu/h max in 250 Btu/h-ft^2 full sun.

What if any impact would using a garden hose instead of 1.5" pool hose for
the 50' connection cause?

Not much. If 5 gpm flows through the garden hose, it might be heated
20K/(5x8x60) = 8 F, without much heat loss to the outdoors.




Thanks Nick,

June 6th and the waters 65 degrees...

I havent seen anywhere near 8 F rise in temp. maybe 3 F. (I think it was
advertised at 10 F. ??)

I have it set up now so I can control the flow diverted to the panels.
and I can measure my results in GPM and temperature rise.
Would a low flow of water heat more than a high flow?
Would it be better to take the incoming water from the surface where the
waters warmer or use a hose to pull it from the deep where its much colder?


I'll experiment and see what gives the best results..



Steve


Sorry BUT I missed the beginning of this thread (change in ISP's)..
so I do not know exactokly what was asked...

BUT...The slower the water travels thru the panel itself the more heat
it will absorb ... and I sure would not be looking for a 3 degree
change between inlet and outlet temps...it is a low process but given
time the entire volume of water in the pool will get warmer ..

Bob G.

Bob G. rg327_remove_comcast.net wrote:

... The slower the water travels thru the panel itself the more heat
it will absorb

Hey, the pool might get really warm with zero flow through the panel :-)

Nick

wrote:

... the faster the water moves the more efficient the transfer process is.
... The trick is optimizing between pumping cost and heat rise.

OK. If it's 60 F outdoors and the pool is 70 F and 250 Btu/h-ft^2 of sun
is hitting 80 ft^2 of unglazed pool heaters 2' above the water surface,
how much water should we pump through the heaters to maximize the COP?

It looks like the answer is zero, with zero heat gain for the pool :-)

If 250x80 = 20K Btu/h = poolgain + airloss, and poolgain = 500gpm(Tf-70)
and airloss = (T-60)80ft^2x2Btu/h-F-ft^2, with final and average heater
temps Tf and T = (70+Tf)/2, Tf = (24K+35Kgpm)/(80+500gpm), which makes
COP = 89.5K/(gpm+6.25gpm^2), with a min COP = 0 at infinite gpm.

20 HEAD=2'pool heater head (feet)
30 FOR GPM=1 TO 5'heater flow
40 HP=HEAD*8.33*GPM/60/550'pump horsepower
50 PE=746*HP*3.412'pump power (Btu/h)
60 TF=(24000+35000!*GPM)/(80+500*GPM)'final heater water temp (F)
70 PS=60*8.33*GPM*(TF-70)'pool solar gain (Btu/h)
80 COP=PS/PE'coefficient of performance
90 PRINT GPM,TF,PS,COP
100 NEXT GPM

1 101.7241 15855.72 12338.92
2 87.03704 17030.23 6626.459
3 81.64557 17461.37 4529.477
4 78.84616 17685.23 3440.661
5 77.13178 17822.32 2773.866

Why pump more than 2 gpm? Going to 4 increases the pool heat gain by 4%
while halving the COP (to a thousand times more than an AC COP of 3 :-)

Nick

wrote:

... If it's 60 F outdoors and the pool is 70 F and 250 Btu/h-ft^2 of sun
is hitting 80 ft^2 of unglazed pool heaters 2' above the water surface,
how much water should we pump through the heaters to maximize the COP?

If it is 60f outside and you have unglazed collectors you are wasting
your money pumping water up there in the first place.

No, in full sun. This is a Small Matter Of Physics.

Nick

wrote in message
...
wrote:

... the faster the water moves the more efficient the transfer process is.
... The trick is optimizing between pumping cost and heat rise.

OK. If it's 60 F outdoors and the pool is 70 F and 250 Btu/h-ft^2 of sun
is hitting 80 ft^2 of unglazed pool heaters 2' above the water surface,
how much water should we pump through the heaters to maximize the COP?

It looks like the answer is zero, with zero heat gain for the pool :-)

If 250x80 = 20K Btu/h = poolgain + airloss, and poolgain = 500gpm(Tf-70)
and airloss = (T-60)80ft^2x2Btu/h-F-ft^2, with final and average heater
temps Tf and T = (70+Tf)/2, Tf = (24K+35Kgpm)/(80+500gpm), which makes
COP = 89.5K/(gpm+6.25gpm^2), with a min COP = 0 at infinite gpm.

20 HEAD=2'pool heater head (feet)
30 FOR GPM=1 TO 5'heater flow
40 HP=HEAD*8.33*GPM/60/550'pump horsepower
50 PE=746*HP*3.412'pump power (Btu/h)
60 TF=(24000+35000!*GPM)/(80+500*GPM)'final heater water temp (F)
70 PS=60*8.33*GPM*(TF-70)'pool solar gain (Btu/h)
80 COP=PS/PE'coefficient of performance
90 PRINT GPM,TF,PS,COP
100 NEXT GPM

1 101.7241 15855.72 12338.92
2 87.03704 17030.23 6626.459
3 81.64557 17461.37 4529.477
4 78.84616 17685.23 3440.661
5 77.13178 17822.32 2773.866

Why pump more than 2 gpm? Going to 4 increases the pool heat gain by 4%
while halving the COP (to a thousand times more than an AC COP of 3 :-)

Nick


Interesting but I dont know what the hell it all means... :-)
Panels are not "up there" they are 2 feet above the ground on a 4' x 20'
plywood stand. No glazing just 2 panels with 50 1/4" black plastic tubes per
panel.
I'm in New England, 80 would be great but I'd be happy to get to the mid
70's. Now that I have the setup with controls I need to get a sunny day
while I'm at home to run some tests with the amount of flow.

You bring up a good point.. I was thinking low flow with higher heat
transfer but I see the point of having higher water turnover with just a few
degrees rise. Its not how hot I can get the water coming out of the
collectors. the point is to increase the whole pools temp.... That may be
better with the higher volume lower temp increase???

Now wheres the sun !!!! Forcast is for clouds and rain all week...


Argggggg


Thanks

Steve

Steve wrote:

... If it's 60 F outdoors and the pool is 70 F and 250 Btu/h-ft^2 of sun
is hitting 80 ft^2 of unglazed pool heaters 2' above the water surface,
how much water should we pump through the heaters to maximize the COP?

It looks like the answer is zero, with zero heat gain for the pool :-)

If 250x80 = 20K Btu/h = poolgain + airloss, and poolgain = 500gpm(Tf-70)
and airloss = (T-60)80ft^2x2Btu/h-F-ft^2, with final and average heater
temps Tf and T = (70+Tf)/2, Tf = (24K+35Kgpm)/(80+500gpm), which makes
COP = 89.5K/(gpm+6.25gpm^2), with a min COP = 0 at infinite gpm.

20 HEAD=2'pool heater head (feet)
30 FOR GPM=1 TO 5'heater flow
40 HP=HEAD*8.33*GPM/60/550'pump horsepower
50 PE=746*HP*3.412'pump power (Btu/h)
60 TF=(24000+35000!*GPM)/(80+500*GPM)'final heater water temp (F)
70 PS=60*8.33*GPM*(TF-70)'pool solar gain (Btu/h)
80 COP=PS/PE'coefficient of performance
90 PRINT GPM,TF,PS,COP
100 NEXT GPM

1 101.7241 15855.72 12338.92
2 87.03704 17030.23 6626.459
3 81.64557 17461.37 4529.477
4 78.84616 17685.23 3440.661
5 77.13178 17822.32 2773.866

Why pump more than 2 gpm? Going to 4 increases the pool heat gain by 4%
while halving the COP (to a thousand times more than an AC COP of 3 :-)

Silly people in Florida heat their pools with outdoor heat pumps :-)
Smarter ones use AC waste heat.

Interesting but I dont know what the hell it all means... :-)

You expect people to spoon-feed you? :-)

Panels are not "up there" they are 2 feet above the ground on a 4' x 20'
plywood stand.

Perhaps you can lower one long edge to tilt the stand towards the south.

No glazing just 2 panels with 50 1/4" black plastic tubes per panel.

Sounds Sol-R-Roll or EZ-Heat. You may find lots of black dust in the water
and a few broken tubes next spring.

You bring up a good point.. I was thinking low flow with higher heat
transfer but I see the point of having higher water turnover with just a few
degrees rise. Its not how hot I can get the water coming out of the
collectors. the point is to increase the whole pools temp.... That may be
better with the higher volume lower temp increase???

Yes, tho returns diminish with more pump power.

How about a clear solar pool cover?

Nick

wrote:

Silly people in Florida heat their pools with outdoor heat pumps :-)

Silly, with a COP = 3 compared to solar pool heating with a COP = 3000.

Smarter ones use AC waste heat.

Why would anyone need to heat a pool if the AC was on?

Another Small Matter of Physics (SMOP.) Consider Jacksonville in June,
with a 79.3 average temp and a 89.3 F daily max and a humidity ratio
wo = 0.0157 and vapor pressure Pa = 29.921/(0.62198/wo+1) = 0.737 "Hg,
vs Pw = e^(17.863-9621/(460+75)) = 0.887 for 75 F pool water at 100% RH.

ASHRAE says a 30'x40' uncovered pool would lose 24hx30x40x100(Pw-Pa)
= 433K Btu/day, like a poorly-insulated McMansion with a 1000 kWh/mo
indoor electric bill and a 1400 Btu/h-F thermal conductance to outdoors.

Nick