Advice on DIY poly pipe solar heater... I know I know...

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W

waterman74

0
May 18, 2011
15
Ok ,here goes. I know this has been covered ad nauseam. I can get 1320' of 1" poly pipe for $200. My plan was to connect it direct inline past the filter and flake it out on the roof. The problem I'm thinking will be getting enough water flow through that long distance of a pipe. Also, I'm unsure how to Tee into the existing pool return pipe is such a way as to make sure there is flow and not just a static loop. I was also thinking of making 2 manifolds and cutting the quarter mile of pipe into 4 separate coils of 330'. I just don't want to get too complicated here, as I will be needing a new roof soon and everything will be coming down in couple of years, once the new roof goes up, I'll be looking for a real solar heating system. My other option is to buy some used panels on Craigslist and hope for no leaks... btw, are leaks easy to repair in these things?....FYI the pool is 16x30 and about 11200 gallons... I am located in central Florida.

Thanks!
 
One continuous run of 1300 feet is going to be a lot of head loss for your pump to over come. The parallel loops sound like a better idea if they were tied into 2"pvc headers. With that much pipe volume to work with I would just send all the water up thru the solar heater. You will want to install a bypass line so you can shut the solar off and make sure to set it up so that you can drain it in the winter.
 
For the plumbing you will need a 3way solar valve and two check valves to make sure everything moves in the right direction. There are pics of my solar plumbing in the link in my sig. The left pipe goes up to the roof and the right pipe is coming down. There is also some good info on solardirect.com, which is where I bought my stuff.
 
I was also thinking of making 2 manifolds and cutting the quarter mile of pipe into 4 separate coils of 330'.
Click to expand...
That is still the equivalent of 330' of 2" pipe. 8 loops would be better which is the equivalent of 115' of 3" pipe.
 
If I ran the 1320' continuous I would have a gate valve I could throttle to only send a certain amount through the collector, the rest would bypass the collector and go back to the pool. I would need to play with the valve to make sure I was getting the correct flow. I'm not sure the correct procedure for teeing into my return pipe to the pool. I was thinking for the inlet pipe to the collector I would tee in, maybe with a Y? After that a gate valve to precisely control flow, after the gate valve would be the tee or y return from the solar collector. That way I could close down on the gate vale and increase pressure to send flow up to the solar collector, but I'm unsure if this will work. I would think I could get ok flow out of a 1" pipe, but I'm not sure. 1320' is a long way. I was also looking at square feet of a linear foot of 1" pipe which if I'm reading right on the site I found is .344. So 1320'x.344 = 454 Square feet? Is that right? It doesn't sound correct.
 
Going through a long pipe would not be a good choice from a head loss perspective or an efficiency perspective.

The water gets hotter and hotter, which reduces efficiency. It will either stop gaining heat part way through or get too hot for the pipe.
 
I get 110 sq-ft as well which means to maintain efficiency, you need at least 11 GPM through the 1" pipe. 11 GPM through a 1" pipe results in about 6' of head per 100' of pipe or close to 80' of total head which few if any residential pool pumps can do. It simply won't work very well. 4 GPM would be more reasonable but then the efficiency of the panels would suffer greatly.

The ONLY way you can get this to work in any reasonable fashion is to use at least 4 parallel runs.
 
Circumference = π × diameter
Maximum area exposed to sun is 1/2 circumference.
1.3 x pi ÷2 ÷12 x 1320 =224.6 sqft.
However, that's not all direct exposure. So, you have less effective area.
 

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The pipe circumference does NOT determine the capture area of the sun's energy. The sun's rays are parallel and come in only one direction as a plane wave of flux density so it is the normal incidence of the sun's energy on the surface on an object that determines the effective capture area (i.e the projected area of the pipe in the direction of the sun). This is why the capture energy of flat solar panels varies by the cosine of the incident angle off normal. So the sun's energy that is captured is only 1" in that dimension and the true capture area of all the pipe is only 110 sq-ft.
 
If the sun is shining directly on the pipe, wouldn't it hit all of half of the pipe?

Is the area reduced because the sun's rays are not hitting directly except in one spot and the rest of the rays hitting at less than 90 degrees down to 0 degrees at the sides of the pipe so that the net effective dimension is equal to the outside diameter?
 
JamesW said:
Is the area reduced because the sun's rays are not hitting directly except in one spot and the rest of the rays hitting at less than 90 degrees down to 0 degrees at the sides of the pipe so that the net effective dimension is equal to the outside diameter?
Click to expand...
Yes. If you perform an integral around the surface of the pipe with the cosine of the angle of incidence, the solution is the same as the projected area of the pipe (i.e. 1").

Another way to look at it is that the pipe cannot capture more energy then what is incident upon it. Since the sun's energy comes from only one direction (mostly), the shadow that the pipe casts on a plate behind the pipe is going to be only 1" high so that is all the energy that it can possibly capture.

However, it is actually less than 1" because reflection comes into play. Not all the energy incident on the pipe is actually captured (shiny black poly pipe reflects energy more than typical flat black panels).
 
The angle of the sun makes a big difference in the effective area of a collector.

If a flat collector is 4 x 8, it will have an area of 32 square feet only when it is perpendicular to the rays of the sun.

If it is rotated, the area is reduced and the formula is area x the cosine of the number of degrees the panel is rotated away from maximum exposure.

Angle.......area
0.............. 32.00
15............. 30.91
30...............27.71
45...............22.63
60...............16.00
75.................8.25
90.................0.00

So, when calculating the area of your collector, you have to consider the angle of the sun relative to the collector.

The angle will change throughout the day and the year.
 
Correct, that is what I have been saying.

The heat transfer workbook in my Sig takes all that into account and also determines heat loss of both the panels and pool.

So, when calculating the area of your collector, you have to consider the angle of the sun relative to the collector.
Click to expand...
Forgot to mention that when it comes to solar panels, usually one references the effective area for the solar panels under normal incidence which is basically the maximum projected area of the object. But to calculate the heat transfer, then you need to determine the sun's angle and integrate over the time of exposure because it isn't a constant.
 
You guys just blew my mind... I'm buying a heat pump!!!! No seriously thank you all for the good info. After doing some research, this is not going to work... while looking for solar panels on craigslist, I found a guy who fixes heat pumps, he is going to sell me a refurbished 2015 aquacal installed for 1600. with 1 year warranty, hopefully this won't bite me. Thanks again
 
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