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solar pool heating - Page 3

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Posted by N. Thornton on October 12, 2004, 8:39 pm
nicksanspam@ece.villanova.edu wrote in message

That isnt the point, a collector producing a lake of tepid water would
have been a failure. The fact that it would have been more thermally
efficient is not relevant.

The purpose was to impress people with what you can do easily. That
means impressing people who dont already have much knowledge of the
subject. Impressing was achieved by tapping off a collectorful of
steaming scalding hot water every 40 minutes, and using it to wash the

Something that dogs the solar world today is not putting enough
emphasis on the real world considerations. The question for the end
user is not thermal efficiency, but rather output per input, with the
input being measured in dollars, not sunlight. Thermal efficiency is
obviously important, but it is not the goal itself.

Lets illustrate. If for example one could mass produce 10% efficient
solar water heaters at $ per square metre, solar heating use would
increase, they would be useful. If OTOH you perfect 99% efficient
collectors for $000 per sqaure metre, no-one is buying. IOW thermal
efficiency is not the number one issue.

It doesnt interfere with my objecting to your analysis, no, since I
could spot the numbers looked questionable. It does unfortunately stop
me from calculating the relative merits of differing covers.

a name for thermodynamics that conveys that I dont know the subject,
and makes the reader smile. It worked for me.

Well, I didnt make much sense of that. Not enough information there:
what dimension is getting varied in order to achieve this variable
area per amount of tube? Tube spacing? If thats the right guess, when
A=1/12 the tubes are touching, when A=1 theyre spaced a foot apart. If
its not that, god knows, I cant mindread.

It doesnt lose anything at night cos water isnt pumped thru it then.
I dont know what Vt and T are either. Its no good moaning, I did warn
you! :)

Are you seriously suggesting that we should all have no involvement in
or discssion of the things that interest us if they dont happen to be
our area of expertise? What a poor life that would be.

Also I would like to put some solar thermal heating in here, and
believe it or not, despite my limited knowledge, I do know enough to
get it working acceptably. I think, ha.

Thats not where the problem lies. One has to be able to describe the
model before being able to analyse it.

That will only happen when and if I make some kind of sense of this
thermal modelling, as I said b4. At this point I dont know one end of
the beast from the other.

Even when I dont know much I can still give you a hard time :P :)


Posted by nicksanspam on October 13, 2004, 10:00 am

Can you say "swimming pool"?

Pool collector water might enter at 70 and leave at 90.

How annoying. I seem to recall that a) you suggested that someone use
this pipe for heating a swimming pool (see Subject: above) and b) you
claimed it would be more efficient with a polythene cover. I said use
two pieces of polyethylene film and forget the pipe.

Well then, forget the cover. Perhaps you can easily wiggle your ears.

So they will say "Wow" to themselves with one voice and go buy
a practical pool heating system?

Agreed. Joules per day per pound of investment might be nice.

You seem to be playing a different game, with impressive ignorance.

Weather data will always be questionable.

You might learn how to do that before offering technical opinions.

If water stays in the pipe, it stores and loses heat.

I'm suggesting you learn more solar physics.

Full sun (250 Btu/h-ft^2) shines on 1 foot of 1" pipe. The pipe intercepts
21 Btu/h of sun and converts it to heat. It has 80 F moving water inside,
so the pipe is close to 80 F outside. It loses 2+V/2 Btu/h-F-ft^2 of the
heat from its 0.262 ft^2 of surface to V mph air at Ta (F). The rest of
the heat goes into the water. Clear enough?
    ---        Rp                Rp = 1/(0.262(2+V/2))
|--|-->|-------www--- Ta
    ---    |  I -->
 21 Btu/h  | 80 F    You say you know something about electronics?
           -         What's the net current into the battery
           |                if V = 0 and Ta = 70?
                     Hint: what's I?


Posted by N. Thornton on October 13, 2004, 11:00 pm
 nicksanspam@ece.villanova.edu wrote in message

The scenario is clear but the units and how they translate to v,i,r,p
is a mere guess. As I have repeated.

Lets guess then... Btu/h is evidently a measure of power. But how
you'd work out source v and i from that I dont know. Maybe its
modelled as current, might work. Yes that makes sense. V will be temp

when V=0, (windspeed) Rp (thermal R of the pipe walls? Resistance of
connection from pipe to air? yes)
Rp = 1/.262x2 = 1.9 units, what, btu per C? I dont know. guess it must
be. No, R would be delta C / btu/hr.

the outflow: v-ir, i=v/r, so
i out (in btu/hr?) = 80-70 / 1.9 = 5.3. Its i so btu/hr? I think so.

Inflow 21, outflow 5.3, resulting input to battery/water 15.7 btu/hr.

But... its all guesses, making it worthless.

So what would happen if we put polythene on, if I'm nuts enough to
imagine I got the above right.

First we mod the above to take into account wind. At ave 5mph,
Rp = 1/(0.262(2+V/2))
   = .85

so I out will be 10/.85 = 11.8
with I in of 21.
Btu/hr collected with no polythene = 21-11.8 = 9.2 btu/h

Now with polythene our input is .9 x 21 = 18.9.

Rp is as first eg, so Iout is about 5.3 again. Very roughly.
I captured = now 18.9-5.3 = 13.6 btu/hr, which is 48% more than with
no poly cover and ave 5mph winds. It is also less varied by weather,
ie heat output is more reliable.

However... I'm still just guessing.


Posted by nicksanspam on October 14, 2004, 10:57 am

It's Ohm's law, with different units.

Yes, heatflow, like an electrical current.

V (electrical) would be analogous to temperature...

Rp is a thermal resistance in F-h/Btu. "Fhubs" are like Ohms. So we have:

     ---        1.9  
 |--|-->|-------www--- 70
     ---    |  I -->
  21 Btu/h  | 80 F  
            -         What's the net current into the battery
            |                if V = 0 and Ta = 70?
                      Hint: what's I?



Let's put A ft^2 of US R1 polycarbonate over the 1'x1" pipe...

No, because there is no wind (V=0) under the polycarbonate.

That's the direct sun hitting the pipe, out of 0.9x250A = 225A Btu/h
which pass through the polycarb.

Not quite:

                      T (air temp under cover)
     ---        1.9   |  0.5/A
 |--|-->|-------www-X-*---www--- 70 F
     ---    |  I -->  |
18.9 Btu/h  | 80 F    |
           ---        |   How much heat goes into the battery if A = 1?
            -         |
            |         |   Open this circuit at X, and we can replace  
            -         |   the current source below with a voltage
     ---              |   source Vt = 70+(225-18.9-70)0.5 = 138.1 F.

Here's a simplified circuit, obtained by opening the above circut...

                      T (air temp under poly film)
     ---        1.9   |   0.5
     ---    |  I -->            |
18.9 Btu/h  | 80 F              | 138.1 F
           ---                 ---
            -                   -
            |                   |
            -                   -


Posted by N. Thornton on October 14, 2004, 8:43 pm
 nicksanspam@ece.villanova.edu wrote in message

sure, but what ones was a bit mysterious

I'm amazed.

First of all I needed to change what had already been calculated for
the unglazed collector, as real world V is not zero, as was used
above. Only then can we move on to play with a polythened example.

direct sun in 18.9 btu/hr - thats the tough bit

out: R = 2.4, delta v = 58.1F, in rather than out.
so heat inflow from the hot air under the cover = i=v/r = 58.1/2.4 =
24.2 btu/hr
which would make total input to water 43.1 btu/hr

compared to 15.7 before wi no poly.... is that right?


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