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testing pool solar panels Steve 06-04-2007
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Posted by on June 8, 2007, 4:06 pm
On 8 Jun 2007 13:53:54 -0400, nicksanspam@ece.villanova.edu 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
The physics I am thinking of is the heat loss to the air.
Outdoor pool?
You might actually see some small net gain in water temperature but
you won't get it warm enough to swim in and that is the point isn't
it?
I suppose if you are one of those "polar bear club" folks you can but
most folks I know won't get in water that is much below 80f.
I have about 70% solar to pool surface in South Florida and I have a
hard time maintaining the ambient air temp with the pool uncoverd,
covered will get me about 10 degrees above ambient air but that drops
like a stone as soon as you take the cover off.

PexSupply Save 10 468x60
Posted by Steve on June 9, 2007, 8:30 am

>
>>... 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






Posted by on June 9, 2007, 10:05 am

>> ... 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


Posted by on June 9, 2007, 12:47 pm
On 9 Jun 2007 10:05:56 -0400, nicksanspam@ece.villanova.edu wrote:

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


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


Posted by on June 9, 2007, 2:17 pm

>>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


Page 3 of 3       << first < 1 2 3
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