Posted By: zenbiscuit on 14 Januarywill do a detailed write up on my website when I get the time

how did it go?

In the theme of annual heat storage has anybody any experience of solar pond technology ?
http://www.green-trust.org/solarpond.htm

Hi Fostertom,
Not in the seasonal category but still making a contribution in winter and an interesting read=
http://www.ece.vill.edu/~nick/Solar_Heat.pdf

Hi Fostertom
Finance, Took early retirement, interested in solar to keep my mind active(actually interested in lots of things), based West Wales

Just in case it's of any interest, you might want to hear about my friend's swimming pool in the South of France.

It's a new build, in ground pool of (typical) reinforced concrete construction lined with mosaic tiles. Size is 10 m x 4 m x 1.8 m deep. Cleaning is through electrolysis / filtration and it uses a great deal of salt. I know it uses a great deal of salt because I personally put 40 kgs of salt into it (@ 8 euros a kg) when the pool was drained of about 0.5 m of water (to check the pipes for leaks) and then topped up again. The pool has a clear polycarbonate cover over it when not in use. This is to comply with French regs re life forms falling in and drowning but also to prevent water evaporation (any condensate falls back into the pool).

What I thought was of interest at the time (not the conundrum why the pool appears to be losing water) was the temperature of the pool which was 34C. Air temperature was 35C. The pool is unheated. I made a point of measuring all this because I am in the throes of wondering about how to optimise solar heating. Clearly I won't be wanting to use the swimming pool as a heat dump as it was like swimming in bath water already.

As an aside, the water didn't feel like the Dead Sea or taste particularly salty. (Salt was pool salt, labelled 'salt' but not regular sodium chloride.)

just to let you know, I'm watching this thread with interest

Reading various inputs re. interseasonal heat storage, I saw no mention of scale. As a general rule, the larger any such system is, the better it will work, and anyone planning a small project using techniques passive or otherwise intended for a larger one will do well to take account of scale effects. The underlying reason is that store capacity tends to relate to the volume or cube of the scale, whereas heat loss relates more to surface area or square of the scale. This would be one reason why the Drake's Landing system is big - 52 houses.

* jon
* CommentTimeNov 30th 2007

quote

"I ran through a few early tests on this using a very simplified model but could not get worthwhile results out of ground storage because of dissipation. Using an insulated tank does seem to get more worthwhile results. however, trying to annualise the heat storage doesn't work well because of the losses.

I think you're right: A dynamic time based model perhaps that can import 3-D models from CAD would be the only way to do this properly: Does anyone know of such a system? "


Just to get a bit of a handle on things, consider a theoretical spherical u/g heat store radius r1 at temperature t1 surrounded completely by soil to a radius of r2 where the temperature (ambient average) is t2. Suppose t1 might be 20degC, and t2 10degC . A reasonably big store might have a radius r1 of 10m. For r2, lets take one figure for the hemisphere below the centre of r2=infinity, and one for the upper hemisphere of r2= 15m to approximate to the relatively short path from the upper half to outside air.
A formula for the steady state heat flow Q w/sec between concentric spheres is, I think:-
Q=4.pi.K.(t1-t2)/(1/r1 -1/r2) where K is the conductivity of the soil. Say K=1 w/mdegC for ease (a value for saturated silty clay is 1.1)

If the above hasn't got you running for the hills, you will be able to put the numbers chosen in, and see if you agree with my answers:-

Q upper half = 1.87kw
Q lower half = 0.62kw
ie total heat loss = 2.5kw all day every day of the year once steady state is achieved which would need to be solar collected or heat pumped, whatever, over and above useful heat returned to the building. Or choose your own figures and see what happens.

That's all for now. I'm just popping out to buy a new anorak.

edit: I have not allowed for the fact that some of the upward heat loss will be to the building, probably, depending on arrangements, and therefore will be useful input or an extra cooling load depending on season.
edit again: Initial maths error now corrected 18.10.08.

Hello Dave :)

Many thanks for your feedback. Somehow I managed to miss all that interesting stuff on the thread you quote! Old age, perhaps.

Remarkably similar, our approaches - but hadn't read far enough up your calc to see that you were assuming no loss through the upper hemisphere. Sorry 'bout that. As you say, this assumption will underestimate the losses, but I guess between us we are bracketing the problem in hopefully some sort of useful way.

My figure for wet clay came from a table and was just as an example. The range you use is a better way of presenting the results, I think. The 20C figure I do agree is no use without a heat pump. 25C or 30C might be more like it otherwise. The main point though perhaps is that we have a very simple formula for all but the most seriously maths-phobic to use to try out whatever figures they like to see what factors are important and to help understand what happens.

Cheers to you

Mike

Hello Tom - thanks for your feedback.

Posted By: fostertommike7
The conceptual error is assuming that the soil in of the lower outer sphere remains at a heat-absorbing 10oC. After a year or two of absorbing heat from the inner sphere, the outer develops a long flat temp gradient, starting at 20oC at the core of the inner sphere and only reaching 10oC (or even higher, as you go deeper) at infinity.

No error. I have used r2 = infinity on the lower hemisphere.

Inner and outer spheres merge seamlessly. In fact the effective inner storage volume actually expands and contracts a bit, on an annual cycle.

I realise I didn't define my r1 adequately - it is intended to be the smallest radius at which the temp remains more or less constant once steady state is achieved, and thus contains all the 'working' ie fluctuating yearly - volume of the store. An inner core rather smaller than r1 will do most of the work.

So after a year or three, further loss from the inner to the outer sphere and beyond drops to negligible, and keeps on getting better, because the temp gradient becomes so flat.

But not completely flat. It is a shallow curve tending to horizontal at infinity, but with a finite gradient at r1 - check the maths - and therefore small but not negligible. It is a bit like sum of the series 1+1/2 +1/4 + 1/8 + ....... which reaches 2 after an infinite no. of terms.

The practical difference is that the upper outer sphere is largely shielded from loss to the cold surface, by a skirt of insulation overlaid with a membrane, extending to approx 6m radius out from the centre point of the building, laid horizontally (to slight outward falls) 600-1000 below the soil surface.

No problem. The r2 figure for the upper hemisphere can be adjusted to mimic the effect of layers having different conductivity.

If one was trying to look at the interseasonal behaviour of the system within the r1 radius, or the build up to steady state conditions over a few years outside r1, then my model is not much help. The purpose of my calculation is not to give particularly accurate values for particular designs, but to illustrate that there will be some ratio of heat lost to heat recoverable and explore how it varies with scale and other parameters eg conductivity.

Want to rework the figures?

No thanks! Having corrected my initial maths error, our calculations agree apart from mine allowing separate calculation for losses outward and upward to ambient at some finite radius. No particular significance should be given to my numerical answers other than seeing that they are not negligible, and by slotting in figures of your choice you can see how the results are affected. That's the point of it all, really.

Posted By: mike7No error. I have used r2 = infinity on the lower hemisphere

No, that's fine - the error is in assuming it stays at 10oC.

Posted By: mike7it is intended to be the smallest radius at which the temp remains more or less constant once steady state is achieved, and thus contains all the 'working' ie fluctuating yearly - volume of the store. An inner core rather smaller than r1 will do most of the work.

That's fine too.

Posted By: mike7But not completely flat. It is a shallow curve tending to horizontal at infinity, but with a finite gradient at r1 - check the maths - and therefore small but not negligible

But that curve gets flatter and flatter year by year. Yes it always has a finite gradient at r1 (which determines the rate of loss from the inner sphere), but a smaller and smaller one.

Posted By: mike7The purpose of my calculation is not to give particularly accurate values for particular designs, but to illustrate that there will be some ratio of heat lost to heat recoverable and explore how it varies with scale and other parameters eg conductivity

That's great - I'd just suggest that the f'rinstance set of parameters, and the result that you leave on record, could be changed to something much closer to reality. As it is, the f'rinstance result is way out, I think - pity if it was taken as suggesting that losses would be forever high, because that's the reverse of what's true. You came up with something mathematically simple, rather than having to model it in Tas or something. Wd it be difficult to extend it to include the raising of the outer sphere's temp over time?

Deleted

OK, I need to look at the calc again - in the morning. But one thing - the final slope of the curve where y declines from 20 to 10 over an x scale of infinity is surely zero, meaning no loss at all? Actually, at a distance of infinity downward, say to the centre of the earth, the 10 figure should be 7300!

Hi Dave

I've attached a table of soil properties lifted from the GS2000 program. Lots of differences. I added the figures for water at the bottom myself - hope they're right, given my habit of not being. If they are right for water, then water is the best as long as you can get it to hold still. What I don't understand is why when water is the best, the more of it there is in soil material, the worse the figures get.
I was reading the other day about schemes where it is usual to do a test borehole to sample the soil first. I'd guess you are right about the water table in E Anglia - here it's about 50m down, after clay and then chalk.

The units of diffusivity also imply that the loss behaviour of a store over time will be proportional to the square of its linear size.
Putting it another way, if you have a satisfactory annual store in operation, and copy it at half scale, it will work just the same, but over a three month cycle, not a year. That would mean smaller ground heat stores can still be viable in the direction I'm looking just now, which is ground storage of ambient air/solar heat gathered during daytime/milder winter days, then heat pumped in colder spells. An advantage of this is that the heat store would be cycling about a temperature much nearer to average ambient, so much lower heat leakage to ground. But then there's the cost to buy and run it...... It all looks uphill work whichever.

Just do it! say you heated the ground under your house up to only 18 C then this heat would tend to keep the house warmer than without it.

A couple of bore holes with pipes connected to solar panels ( ETs ? ) and you are saving on heating from then on.

Re my estimates of heat loss:
I find the calculation for heat loss in two dimensions ie a long cylinder, not a 3d sphere, gives the results Tom was talking of - gradual reduction of gradient over time to zero ultimately. This may be the origin of our difference.

In practice, I think it would be reasonable to look at losses to a steady ambient at some multiple of the store size. Say perhaps five, not more than ten. Losses to surface within this sort of distance rather than those down into a theoretical infinite ground are likely more important in any case.

The cylinder vs sphere thing occurs when you consider a terrace of houses - a heat store extending along under them all would tend to behave like a cylinder in the centre and more like a sphere at the ends. This leads us back to greater size of store leading to higher efficiency. And there are, I believe, several terraces of existing houses about to apply the system to.

One way of reducing losses upward from the store would be to arrange boreholes in a conical or wigwam pattern. The holes would be concentrated in a smaller area near the surface and thus be easier to insulate from ambient, and a smaller volume of soil would be heated in the shallow subsoil area. The first metre or so of all holes could be insulated individually, possibly by an oversize bore diameter backfilled with insulation.

Estimating heat flows in two dimensions is if I remember rightly much easier than three. I've even done some myself using a grid and an iterative method called relaxation. Not so very relaxing if you do it as I did without a computer. A spreadsheet would make it very easy for a simple scheme, but no doubt there are programs out there for this very thing.

I'll see if I can attach some interesting files I found.... hmm, just the one, then.