Posted by *nicksanspam* on June 8, 2007, 12:36 pm

*>... 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 PEt6*HP*3.412'pump power (Btu/h)

60 TF=(24000+35000!*GPM)/(80+500*GPM)'final heater water temp (F)

70 PS`*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

Posted by *gfretwell* on June 8, 2007, 5:16 pm

On 8 Jun 2007 08:36:36 -0400, nicksanspam@ece.villanova.edu wrote:

*>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?*

If it is 60f outside and you have unglazed collectors you are wasting

your money pumping water up there in the first place.

Posted by *nicksanspam* on June 8, 2007, 5:53 pm

*>>... 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

Posted by *gfretwell* on June 8, 2007, 8:06 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.

Posted by *Steve* on June 9, 2007, 12:30 pm

*>>... 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 PEt6*HP*3.412'pump power (Btu/h)*

*> 60 TF=(24000+35000!*GPM)/(80+500*GPM)'final heater water temp (F)*

*> 70 PS`*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

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