Challenge Questions

I guess that was intended for me as much as for you: but IF the question was worth asking, it's a bit crass to then turn round and say guesswork will do - it's irrelevant to analyse properly. (Given the size of any difference, I would regard it as purely an intellectual exercise, a techie's equivalent of crossword puzzles)

I hope that the thread has secondary value - in persuading people to think seriously about solar heating. (Unfortunately, when it comes to my home our historic monument people say NON, NIET, NEIN, NOWAY, GO'WAY...)

Fyz

P.S. Regarding hardware working first time - is it worth simulating it - or does that take longer and cost more for the sort of stuff you do than building it several times? (And do you have the data to tell the simulator what questions to answer in the first place?)

hmm.

1 2 3 even 4 ...........

?

I know , I know , I am often off topic and repeat myself often but the links might be useful for something. Now I can return here for references. It's my Squirrely nature.

Thanks Kris

!

Hi Kris

I'm astonished at the idea that anyone invented photo-Voltaic solar heating in 1994. SFIK, there were already experimental installations in existence well before that. One problem was the fundamental limit on efficiency of PVs, due at least in part to their being quantised devices and the sun's energy being spread over a wide energy range.

Even more curious is the concept of using PVs with a preheat tank. Unlike thermal panels, the efficiency of heating using electrical power is unaffected by the temperature of the water. So the major function of a preheat tank would be to increase the surface area from which energy might be lost.

The place I can see PVs being thermally more effective (as opposed to easier to plumb or quieter) than heat exchangers is if there is a problem getting the water to the desired temperature using heat exchangers. So it would make sense to use them to get the "last few degrees" in cold and isolated places where other sources of concentrated power are difficult. And if you need PVs for other applications, it could be heat-for-free whenever they are not in use.

Fyz

Hi Fyz,

Sometimes I do rather well at finding too much ! I have to confess that I quickly culled those links after a few seconds glance (did a quick google). I could have looked more and filled a page with links. Now I'm going to have to look at those I posted ! There are so many possibilities for hybrid systems especially, as you say , for wringing the last few drops out.

Kris

Who's Rube Goldberg? (us Europeans don't always get your US references)

<edit>

Found him Sounds like Heath Robinson who predates rube!

It doesn't matter! Typical engineers, too much theory about nothing. Put it simply, and substitute the solar water panels for two solar cells of different proportions. The potential across the panels will be identical, reqardless of which way they are wired in series. Now try two different hoses of different diametres and length. Connect to a tap and the flowrate at the output will be the same if you connect the smaller hose first or the larger hose first.

The two solar water panels obviously will each raise the temperature of the water by a different amount, but the total rise in temperature is the sum of both. We are adding people, not multiplying, dividing or subtracting.

Therefore if panel A increases the water temp by 7 degrees and panel B by 10 degrees then the total temperature rise is 17 degrees regardless.

I see... so....

If the panels reach say 90C and we have a low pump rate.

The first panel can raise the water ten degrees from say 75 to 85.

And the next panel panel will also raise it 10 degrees from 85 to 95 !!!

Nice!

Elegantly put. But with equal and unchanging panel temperatures as assumed in "the solution", the heat output remains independent of panel order.

Just consider the tubes as 22 identical sections, each acquiring heat from a panel at the same temperature. What difference can the order of the sections of tubes make under those conditions?

Fyz

Inappropriate analogies. But even with solar cells, if you shade the second solar cell with the first, you'd see a different result. That would also apply if you heat the second solar cell depending on the input to the first.

However, for Del's simplified case (equal panel temperatures) there should be no difference - worth identifying the mistake there...

Like-for-like calculations of heating effects for the different orders of the panels

Introduction: one assumption made in the official solution to this problem meant that the heat flows for the different ordering of the panels were actually calculated under different environmental conditions. The result is that the results reflected the change in the environment rather than the change in the order that the water flowed through the panels. What follows is my attempt to correct that oversight; no additional simplifying assumptions are needed.

Generalities:
Call the temperature of water entering the first panel T
Call the temperature of the panel surfaces KA and KB respectively
We retain the implicit assumptions that the heat transfer depends linearly on temperature difference, and that the temperature rise is proportional to the heat transferred. This leads to the following expressions:
The temperature rise on passing through panel A = (KA-TinputA)*A and
the temperature rise on passing through panel B = (KB-TinputB)*B
where A and B are constants that depend on the number of zigzags and the speed of the water. Note that no assumptions no assumptions are made concerning the relative values.

To correspond to the explicit simplifying assumption of the official answer, we retain KA=KB=K, and K is independent of the ordering of the panels. The other simplifying assumption that is retained is that there is no heat loss from the pipe passing between the panels.

Then, if panel B is first, we have:
Temperature rise through B = (K-T)*B
Therefore the output temperature of panel B is T+(K-T)*B
According to the second simplifying assumption, this is the input temperature to panel A -
Therefore, the temperature rise through panel A is (K-T-(K-T)*B)*A, and
the output temperature of panel A is T+(K-T)*B+(K-T-(K-T)*B)*A
Simplifying gives the output temperature of panel A as T+(K-T)*(B+A)-(K-T)*B*A

Similarly, if we place panel A first, we get:
Output temperature of panel B is T+(K-T)*(A+B)-(K-T)*A*B

Clearly, these expressions are identical – so under these assumptions the order in which the water flows through the panels makes no difference to the results.

Discussion: the difference between these assumptions and those of the "official" solution is that for these calculations the input temperature to the first panel is independent of the order of the panels. That is to say that when order of the solar panels is changed, the pair is placed in an otherwise identical environment.
In the case of the official solution, the assumption of equal average temperatures for the input panel means that the input temperature to the pair of panels is different for the two analyses – meaning that the heat flows are analysed with different environments for the different ordering of the panels.

Conclusion: under the assumptions made for the official solution, the order of the panels does not affect the efficiency of the arrangement.

Fyz

Excellent!

At lasts the fully worked response.

I can only appologise for my over simplification and the poor assumption in my official answer.

In my defence, I can only say, I was pressed for an answer and would have preffered to post the challenge with an open verdict.

It has been an excellent dabate, so much so I think Chris is going to buy us all a drink?

Anyhow, thanks FYZ for the response and the debate. (I'm sure someone will still want to argue/discuss!)

I am a contrite cat...I shall curl up and lick my wounds! (or my back side!)

I couldn't vouch for Chris , but Kris is far too mean to shout a round ! I simply hide in the margins and occasionaly encourage . Hats off to you both . I've discovered some great stuff by following this particular Question. In the not too distant future I shall be constructing a system , and the facts learnt here will be invaluable to me. Thanks to all I think.

If you ever find yourself in upstate New York, look me up. I'll be happy to buy a round or two, crack open a few bottles...

That would be a lot of fun ! One day I'll hop over the Atlantic . (it could be an expensive round if everyone turns up !)

The person who has never been wrong is the one who has never done anything new or useful. That is what I keep telling myself when I get discouraged knocking my head against the proverbial intellectual brick wall.

There is huge merit in being big enough to recognise when someone else makes progress that is different from yours - full marks for that - as well as for an excellent basis for a thread.

Thanks are due from all, I think.

Fyz

Absolutely! Well done Del.

We need more smilies - hand clapping and beer mug clanking ones would be appropriate here.

Regrettably, the specified smilies are without my computer-graphics capabilities; what do you think is the appropriate solvent for Fyz?

Hmmm.....

Water and sugar based liquids seem to dissolve Fyz well, as do certain water and alcohol based liquids. Perhaps we should experiement? <hic>

Thanks Fyz, I was starting to think I'd have to sit down and do this over the weekend, now I can get on with finishing the EinP episode for this question and the Martian Moons question. Blaine faces the ultimate challenge....

Blaine will have to be careful. He doesn't want to get tied up in the lagging, or get cross-eyed trying to watch the two moons at the same time (I hope I've not stolen any of your thunder).

BTW, independence of order only applies if the panel temperatures are effectively identical. In general, you want to put the heat into the water as late as possible - because the sooner you input it the more of the system is susceptible to heat losses to the environment

Fyz

Never fear - neither of those plot lines are close.

I think that assumption is reasonable for the situation described.

I should have posted this before but the equations verge on the incomprehensible.

As in most real-life situations, nearly everything depends on several other things so the algebra, let alone the arithmetic gets horribly complicated.

I think that we are all agreed that the sun's input to both panels is deemed identical, that B has a higher thermal capacity than A (so A will heat up faster), thermal conductivity is proportional to temperature difference and the transfer fluid heats up in proportion to the heat transferred to it.

For convenience I shall refer to location within the panels as a number between 0 and 22, referring to the number of bends/part-bends that the transfer fluid (water) has travelled from inlet;. also K for coefficient of conductivity and L for thermal capacity of the water in one unit of pipe length. As we have copper pipe the temperature gradient across each panel should be insignificant.

Then the rate of heating at point n is Kx(T(A)-T(n))/L or Kx(T(B)-T(n))/L. T(A) and T(B) are variables but at an instant of time they can be parameters. The difficulty is that T(n) = T(inlet) plus the integral from 0 to n of Kx(T(A)-T(y)))/L dy or Kx(T(B)-T(y))/L dy. When you add the facts that the formulae for both T(A) and T(B) include two integrals (for heat transferred to the water and lost to the environment) and that my computer keyboard does not have an integral sign, you can see why I didn't fancy posting this.

Since A is hotter than B while the system is working, putting the water through A first initially heats it up faster so the average temperature differential is less so the total heat transferred is less.

I cannot derive a numerical answer without horrendous oversimplification, so the following has holes in it as I'm ignoring second order quantities, but it illustrates.

If the water inlet temperature is 20C, B is 36C and A is 44C and K/L is .05, then the crude approximation to the temperature progression is as follows.

B before A is on the left and A before B is on the right

Position Water Panel Position Water Panel

Inlet 20 36 Inlet 20 44

Bend 1 20.8 36 21.2 44

Bend 2 21.56 36 22.34 44

Bend 3 22.282 36 23.423 44

Bend 4 22.968 36 24.452 44

Bend 5 23.620 36 25.429 44

Bend 6 24.239 36 26.358 44

Bend 7 24.827 36 27.240 44

Bend 8 25.385 36 28.078 44

Bend 9 25.916 36 28.874 36

Bend 10 26.420 36 29.230 36

Bend 11 26.899 36 29.569 36

Bend 12 27.354 36 29.890 36

Bend 13 27.787 44 30.196 36

Bend 14 28.597 44 30.486 36

Bend 15 29.367 44 30.762 36

Bend 16 30.099 44 31.024 36

Bend 17 30.794 44 31.272 36

Bend 18 31.454 44 31.509 36

Bend 19 32.082 44 31.733 36

Bend 20 32.678 44 31.947 36

Bend 21 33.244 44 32.149 36

Exit 33.781 Exit 32.342

Obviously the numbers will change if you change any of the parameters, but the inequality remains valid. I've tried it with some other numbers. (Also, since heat lost to the environment is an increasing function of the panels' temperature the more efficient heat transfer system must reduce that as well).

This doesn't invalidate Fyz's derivation of an equality from a set of assumptions, but my assumption of temperature inequality is more realistic

John

The constraints were as close to Del's as possible. The only reason for that was to demonstrate that this specific case did not correspond to the general result that was being claimed.

With regards to your presentation, I could not follow it at all. Nor can I see why you need multiple integrals for each panel. Once you perform the integral you can perform a single integral to cover the entire panel. Note that, within each panel the temperature should correspond to a section of an exponential decay, with the limit temperature for the decay being the panel temperature. That part is linear, and merely corresponds to a somewhat smaller ratio of the heating ratios than 13:9 - but the specific ratio was not relevant to my calculations.

I couldn't see how you arrived at panel temperatures for the different orders - once you allow these to vary with panel order, that is the crucial factor.

If you state your constraints clearly, I don't see any problem coming up with an explicit equation. Whether it will be quite as simple to follow as the version above is another matter.

For what it is worth, the sort of condition under which I have been able to generate higher output temperatures for having panel B first is the dynamic situation when the panels are preheated and the water first starts flowing - and continuing pumping after the heat source disappears compensates for this.

Regards

Fyz

Fyz

With regards to your presentation, I could not follow it at all.

Sorry. I tried to simplify, but clearly not enough. The example is taken at one instant in time, but provided T(A)>T(B) the inequality holds so it must apply throughout the time that the system is running. The times when T(A)<T(B) occur when the system is not operating. The numerical example is simply using stated assumptions and an excel spreadsheet but ignoring second-order effects (i.e. excel won't do integrals that depend on the answer so I treated the heat transfer along each bend as being a function of the starting temperature rather than the average - with a rise of less than 1C, the error is negligible and the difference between the two errors is even more so).

A thorough analysis is made horribly complicated since the heat transfer is proportional to the temperature difference but, except at the instant when the pump is switched on, both temperatures are a function of the heat already transferred to the water and the panel temperatures are also a function of the heat lost to the environment, as well as the time since sunrise, cloud cover, the geographical orientation of the panels etc etc. So T(A) and T(B) are each a function of heat absorbed less transferred or lost and the formula for the value at a given time involves a couple of integrals over time. The heat being transferred to the water is calculable by integrating (T(A)-T(W)) and (T(B)-T(W)) over the length of pipe and the heat already transferred is found by integrating that over time since the pump started.

There must be an explicit equation but when I got to 8 unknown parameters as well as four interdependent variables, and some double integrals, I decided that it wasn't worth it.

continuing pumping after the heat source disappears compensates for this.

If the panels were still hotter than the water after sunset, this might be the case, but in practice this just doesn't happen.

Regards

John

If there is no stored heat after sunset, all that means is that the input power before sunset becomes sufficiently low that the stored heat is recovered before sunset. Put crudely, under these conditions, the heat energy does not simply vanish - eventually it either goes to the environment or to the water. (In practice, it stabilises in a time that is short compared with a day, so transient effect should cancel out)

Returning to what I was trying to ask: as I interpret your text, panel A is at 44-degrees and panel B is at 36 degrees. Are you saying that this is a condition the pump has been running a while are running at (quasi steady state), or is this supposed to be the condition when the pump is first turned on?

Fyz

the stored heat is recovered before sunset.

Not exactly, but at some point in late afternoon the difference between the panel temperature and the water temperature falls below the cut-off point so the pump cuts out and the remaining stored heat is lost to the environment (some after sunset). The latter isn't significant, but the possibility of a cut-off when the water and panels are 40C or 50C is significant because it means that the process is asymmetrical.

The 44C, 36C illustration is supposed to be after the pump has been running for a short while (to avoid any arguments about whether the water can transfer heat to B if A heats it up very fast). Also I don't want to nitpick about the water in the pipes being at virtually the same temperature as the panels when the pump is switched on because that effect will have exponential decay. The actual numbers are not that important, but I picked some that looked plausible, because as long as T(A)>T(B), more heat is transferred to the water by going through B first. (The illustrative numbers can be easily checked).

With an infinite supply of cold water to be heated, you would get the total heat transferred over the whole day being close to equal whichever was first because the higher panel temperature in the afternoon on an A first set-up would lead to more heat transfer offsetting the higher heat transfer in the morning on a B first set-up. However in the case of domestic water heating in England, which is what I was looking at, we have an asymmetric daily cycle and at all times when the system is working, B before A is more efficient.

Regards

John

If the water has been running for a "short" while, I think that you have to allow for the heat that is transferred from the first panel to the second by the initial water, and all sorts of other considerations. In addition, the temperature difference you have specified corresponds to the entire thermal capacity of each panel being due to the water.

However, that is not the main problem. What this does is to use the transient overshoot as the measure of system performance. It is a relevant measure if either the transient makes measurement simpler, or if it cause problems - but not if what you are concerned with is the average over a period, as is needed to impart heat to the water cylinder.

Also, once the system has been running even a short time, the individual panel temperatures are dependent on the order in which they are placed.

Regards

Fyz

The greater differential between heat source and heated medium water produces the greater rate of heat transfer

Therefore the panel with most copper pipe should be utilized first and the panel with least copper pipe used second.

If the panels are in series the flow in each will be the same but the heat transfer to the water will be greater in the first panel as the inlet water and panel temperature will have the greater differential the closer the panel temperature and the water temperature become the less heat transfer takes place.

Totally incomprehensible. Heat the water earlier - more opportunity to lose heat to the environment. That is the reality.

Loss to the enviornment only becomes a factor when the fluid temp.

nears equilibrium with the collector temp.

(Inadequate pump size or saturation of the heat sink water.)

From the published performance (much), earlier this is not the case.

If there is no loss to the environment, all the insolation power is transferred to the water, so the order of the panels is irrelevant. In any event, the large temperature difference between the panel and the water can be indicative of limited conductivity between panel and pipes as much as that all the heat is transferred to the water.

Fyz

If we knew the equilibrium temp. of the panels 'system-off` versus the
equilibrium temp.s of the panels 'system-on` for a given insolation,
(which will change over time anyway), we could evaluate the importance of loss
to the enviornment.

Let's face it already. We don't really have enough data to answer this question
as there are too many variables open.

The knee jerk answer 'A` to 'B` assumes that there is enough gain to make putting
the more efficient panel last effective. This might or might not be the case, or might
change from hour to hour.

Anyway, its been a fun thread.

"I didn't attend the funeral, but I sent a nice letter saying
I approved of it." - Mark Twain

I believe that there is an error in the calculations presented as the answer for this problem. It is in the second heat gained calculation H(BA) and it is that the fluid temperature in the second panel will be the same as if for the first case (T+4). As the area of the 13 coil panel is 13/9 of the first, the entering fluid temperature for the second will be T+4*13/9. Putting this into the H(BA) equation gives:

H(BA) = (k-T)*13*E/9 + {k - (T + 4*13/9)}*E

H(BA) = (E/9)*{13*k - 13*T + 9*k - 9*T - 4*13/9}

H(BA) = (E/9)*{22(k-T) - 52}

Which is the same as for H(AB).

This is the same conclusion that another contributor noted that with a constant heat source (the box temperature in each panel is the same as it is generated by the sun), the order of the coils doesn't matter if the total length is the same.

I think we should both have done this sooner. However, your calculations are rather oversimplified - the rise across the first panel will change by somewhat less than the 13/9 factor you give - because the increased temperature rise means that the average temperature in the pipe comes closer to the temperature of the plate.

Also, the error was not the use of T and T+4, but that T was the average temperature in the first panel, and this will not be the same for both configurations. The calculations (as opposed to the words) were compatible with T being the midrange temperature, rather than the average - and this difference has to be half the temperature difference between input and output - regardless of panel order.

Best regards

Fyz

Hi fellows

I found this a very Interesting read. and just to throw in another curve to all this
I don't believe that I read anywhere in the blog where the expansion of the tubing do to the heating that takes place and what effect that would have on this. Isn't bigger better? More exposure to the rays.

Also the spacing of the tubes apart wouldn't this have a effect on heat retention?

All things aside It was very informative and I believe Black plastic tubing would be a better source for the tubing then the Copper as The copper likes to retain the heat.

Black plastic has a better transfer rate then the copper.

Compared to suitably painted copper, black plastic is more radiative in the IR where you don't want it. It's much less good at transferring heat from tube to water, or from absorption plate to tube.

Episode 6 is up and ready to read!

Two things Del,

First, I still think the heat gradients in the aluminium are the crucial factor: so, how thick is the aluminium so that we can work out what the gradients are.

Second: did you think about using some heat transfer compound (the stuff you put between a processor and heat sink) between the pipe and the panels.

Answers!

a) 3mm

b) Yes (but it's expensive, messy and I'd need a lot!)

How about getting a bit barbaric - bash the tube into a flatter profile (after it's been bent into zig-zags). OK , 'gently squeeze' sounds nicer. Also , there must be something akin to to techy conductive paste that may add a little without much cost ? I don't suggest pancaking the tube , just enough to increase contact area a little without causing flow problems ).

Is hell about to freeze over?

Fyz

ahem ;

(or flatten the base of the pipes).

by getting barbaric or gently squeezing perhaps. Re-using stuff is good , but old rust filled radiator panels ?

My stilettos would skitter about on ice. I assure you Fyz , there is no greater pleasure than applying a hammer to bits of stuff in the shed. I can vent steam this way faster than I can calculate. . Sometimes I multitask and do both.

So what's the Kelvination in aid of?
N.B. Stilettos are fine on ice if you keep them properly sharpened. Keeping the soles (should that be souls?) pointing in the intended direction may be more difficult.

Venting steam: wasn't the idea to modify the shape without creating leaks?

I don't recollect recommending recycling panel radiators - hence my comments about cost-free availability. In any case, I would flush the panels first - they would be no worse than what you already have in central heating systems that commonly share the water with the heating coil in the hot water cylinder. I'd be far more concerned about leaks.

Any thoughts on (virtually) cost-free thermal fillers?

Fyz

availability as scrap material

Kelvinating......

Hi Kris,

I had considered making a wooden former to squeeze the pipe into a triangular cross section along the straight runs.

(I met a bloke ages ago who was making panels for swinmming pool heating from 15mm copper tube and he'd squeezed them all and soldered them together...but he was a 'never finish it' guy).

I think it would still need your 'techy paste' as there would still only be point contact here and there. It's all a cost benefit trade off...with my simple system more wire ties and a bucket full of tecky paste would work wonders.

Any recipes for home made (cheap) techy paste? (not involving cow dung, or cat fur preferably!) Maybe you (or me) should post it as a new thread?

PS. The fingers belong to my son or daughter, but do look v similar to mine (the faces are hereditry from my side.)

hmmm.

I think it must be possible to find a way of nicely squeezing straight sections - to get a flat bottom , but probably ending up with a flat top as incidental. Changes other than to the bottom don't matter too much , and I think a good level of contact could be achieved. A suitable paste may be harder. I shall consider these points on my googling and let you know if anything comes to mind.

ps - I can't post my painted and cloven hooves , I would surely be banned.Open toed sandals are an unknown luxury to me. Hang on , you said fingers . Dash it , I'm still in trouble.

Good ideas - but:

Most heat sink compound isn't all that conductive - it relies on the film being quite thin. So it wouldn't do that much unless you could bend the aluminium around the pipes (or flatten the base of the pipes).

Temperature drop across the Aluminium: I expect that Del would have commented if the Aluminium sheet had bent much around the pipe when he tied it**. I think that would imply that the panel is at least 2-mm thick - which corresponds to a peak temperature rise of less than 3-degrees C at the worst points on panel A (compared with about half that for panel B***). This is a small portion of the overall 30-degree differential. Anyway, the only thing that should matter as regards choosing the order of the panels is the direction of the difference between the conductivities, not its magnitude.

I suspect that pressed-steel radiator technology would be more effective than a typical lash-up - the material may be less conductive, but the spacing is only one cm as compared with about 8-cm for meandered piping - and you don't have the large temperature drop at the interface between the collector and the pipe. But filled weight and availability as scrap material could be more of an issue...

Regards

Fyz

**In moderation, that might help more than the resistance of the sheet hinders....
***That corresponds to a peak difference due to the extra spacing of about 0.8 degrees (the average will be about 0.6 degrees, as the temperature rises more rapidly near the pipes...) Of course, it will be less than that if there is significant loss to the environment - and that would be the reason it needed to be reduced it in the first place.

There's a radiator in the basement of the building I work in (it was replaced because it leaked...)

After all that money they had a leak ? Must be the moles.

This link is intereseting .