It's all Plain to Sea

[mrhobbithhnet]

or... The Basics of How the SHCS for a Solar Heated and Cooled Greenhouse would work...

addressing some of your other questions - like how to get to grow food all year for with free solar recycled water, heating and air conditioning.

Assuming you can put up a solar efficient 1000 sq foot long tunnel house, the basic physics works like this:

In order to evaporate water, heat (BTU's) is required to be added to the water. That heat can be in the form of direct radiation (sunshine), conduction (contact with hot things), chemical reaction (photosynthesis in the leaf) fuel burners, electric heat etc. Whatever the source, heat must be added to water for it to go to a vapor state. Even if you physically force water to evaporate (spray a fine mist into the air for example) because water has changed phase, the heat equation must be and so will be met - in this case heat is extracted from the air volume. Because it now contains less heat, it can do more work above ground and it feels much cooler. In the meantime, the excess heat remains underground, partly from conductive transfer and partly from phase change. The phase change portion remains underground as a deposited condensate or recirculates back into the sun space as 'refrigerated' vapor.

In order to condense vapor back to water, heat must be extracted from the water vapor. Virtually the exact amount of heat is required to evaporate a pound of water as is required to condense a pound of water as vapor back to a liquid.

The amount of water (in pounds of water in the form of vapor) that is present in air is directly related to the Temp and RH of that air. Every cu/ft of air has a known weight of water in it if the temp and RH is known. That amount changes at different air pressures, but for most figuring, air pressure within a normal range for folks doesn't change enough to make a whole lot of difference in ad hoc figuring.

Using this chart, you can immediately know how many lbs of water are present in any volume of space in any greenhouse - just measure it's total volume, temp and RH and you have your answer from the chart ... just read the lbs/1kcuft and multiply that figure by totalvolume/1000cuft.

By measuring the input air to a SHCS, you can tell how many lbs of water is in that air as it enters the system. By measuring the output air for temp and RH, you can do the same. When you subtract the two, you have the number of lbs of water per 1000 cuft of air that is being condensed underground.

Because the amount of BTU's needed to condense water is a fixed quantity (approx 970 BTU's/lb) and because you know how many cuft of air you are moving through the SHCS (at least 5 times the volume of the greenhouse every hour and as much as 20x) you have all the data you need to get an exact call on how many BTU's the system is extracting in an hour... in other words, you get the BTU/hr of your 'cooling system'. You can then use that number to compare it to other industry standards for cooling systems. They are rated at BTU's/hr (and in 'tons' of refrigeration - that is a known constant conversion factor of 12,000 BTU/hr. As well, one boiler horsepower is the same as 33,470 BTU/hr.)

The beauty of using the subsoil for a heat sink is that you can, at your convenience, for heating or cooling needs, choose to use those extracted BTU's and the condensed water carrying those BTU's to heat and humidify the same space at night, OR use it to force evaporate water (it takes heat to evaporate water...) by blowing outside air through the system at night and therefore creating a 'refrigerated zone' and a dryer zone for cooling the next day.

Normal evaporative cooling systems take outside air during the day (when it is hottest!!) force evaporate water to take the heat out of the air, move that chilled air through the greenhouse and then blow it all outside. Because you are using hot outside air, some energy is extracted in the evap process of the wet pads, seldom getting even a 30 deg F drop in air temp to cool the greenhouse with. In the process you are saturating the GH with humid air that is also quickly saturated with more exposure to the plants and then exits eventually with some heat, but all of your water as well (water from the evaporative cooling machine AND from the plants). The plants get cooled, but because the air coming in is already saturated, so they can't transpire as well in the process. It's much better to cool them with cool, dry air rather than this higher temperature, higher energy, higher humidity air.

Working with a properly designed SHCS that is sufficiently low in sq/ft costs and operating expense, there would be no need to use high powered venting. Using just good passive venting for a good portion of the year, significant portions of the solar gain and water can be recycled while passive vents are closed and powered vents off. You'd want to design it so that the margins in the extreme ranges are such that you would only need to vent for a shorter than normal portion of the year and evap cool for an even smaller portion of the year. Because the SHCS fan running costs is by design considerably lower than the cost of fans running for venting and lower by even more than the costs for evap fans running, your savings would be in electrical costs of fan operation for venting and evap cooling, potentially a 99% drop in annual running expense. Hitting that sweet spot in the design would be the job of an HVAC engineer qualified to work the equations and pick the market machinery selections to get you there.

Our job (me and all the other interested, curious, inspired SHCS solar greenhouse designers etc) is to simply put up and eventually work directly with some experimental greenhouse test beds to see what variables do what so there can be some solid decision making opportunities for larger bio-regional operations. With those kinds of numbers, and a company set up with the right engineers and experience in underground system installs of ADS tubing, rock beds and fans, you could go out on the market with a design process that guarantees X number of $ savings/earnings per sq/ft of converted greenhouse. Bio-region scale of development is easily a possibility with good design.

Till then, only the smaller systems that are installed in the spirit of hope and promise are being done. All built to spec have demonstrated their worthiness with dramatically better plant performance.... but none have been decked out well enough scientifically to provide anyone with the repeatability needed for riskier commercial sized ventures.

For now, you can use these figures to demonstrate a normal, everyday SHCS Solar Heated Greenhouse performance, probably achievable for at least 6 months of the year:

Input air @ 80 deg F and 80% RH

From this chart, that represents 1.75 lbs of water per 1000 cu ft of air entering the SHCS

Output air @ 65 deg F and 80% RH

From the chart that represents 0.67 lbs of water per 1000 cu ft exiting the SHCS.

Therefore, you are extracting 1.08 lbs of water per 1000 cu ft of air moving through the SHCS. It takes approx 970 BTU's of energy to evaporate or condense a point of water in this range of temperature, so you are extracting 970 x 1.08 = 1026 BTU's per 1000 cuft of air moving through the SHCS.

Cross checking this number with a good conversion chart online at :

http://chuck-wright.com/calculators/watts.html

we find that 1.08 lbs of water evaporated represents 1048 BTU's... about 2% out off line from eyeballing that chart.

For real world figures, we can use the SHCS online calculator to find that if a greenhouse averages 10 feet high and we move that air underground 5 times an hour, we will be moving 50 cuft of air per sq ft underground every hour and extracting (50/1000)x1048BTU=52.4 BTU/hr/sqft of greenhouse or 306.8 watts for every square foot of greenhouse.

Considering that we are designing systems now that can do this in a 2000 square foot greenhouse with just a 300 watt input (two 150 watt fans), you are looking at an efficiency potential of 2000 times for heat sequestering or air cooling alone... one 300 watt fan stores over 300 watts per hour for every one of those 2000 square feet as long as there is that much heat absorbed by the greenhouse. That makes possible a whopping 200,000% return on your electric expense as a baseline to start working with!!! (not to mention the gallons of water that is saved right in the greenhouse where the plants need it to carry the whole process on)

If I wasn't already involved full time developing another ground breaking carbon sequestering phenomenon (using the Strawjet process to build houses almost entirely out of agriculture residue normally burned or rotted in the field), I would be putting in test beds right now...

What's the catch? there is one. So far, it's been the hassle of putting in the static hardware designed well enough to do the numbers. And the cost of that hardware (probably at least $0.50/sqft) But if those expenses could be factored into and compared to a bonafide repeatable, third party verified SHCS... you'd have a winning technology worth getting some serious development costs behind. And another catch? Yes, you can't store heat from the sun if there is none shining and the amount you can store is limited to how much the greenhouse absorbs. Conventional wisdom says that 1000 watts per sq yard is all the power in the sun to begin with (1000/27=37 watts per sq foot). So, even if you can absorb 300 watts per sq ft, there isn't that much available steadily for the whole day. I'd say that working 100%, the best you can feasibly do would be maybe 30 watts/sq ft max. as a total for the whole sun filled day.

For now, I am giving it all away under a Creative Commons Share Alike license... for all the adventurous ones to take it and run with it the rest of the way. Which basically means you can copy all my work and figures for nothing but an attribution as to where your data and technique originated... and that your additional input and improvements are also shared alike. This way, the world gets the answers for free, and we will just have to finance the delivery of those answers by building our own greenhouses that work better in the market place . It's a win-win. It beats oil wars and/or nuclear plants in your back yard...
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[Andy]

John,

Some great references here to calculate the performance of a system, thanks.

Maybe I've misunderstood, but don't you also have to consider the heat energylost from the air itself as it cools during its passage through the tubes? Though smaller than the latent heat transfered in the water phase change it is surely significant (and adds to the total heat transfer, making your dollar figures better still?)


Andy

[Hex]

Alledgedly you gain 1btu from 55 cubic feet of dry air dropping 1 deg F.

1000 cubic feet of dry air dropping by 20F will give you about 363 btu (106w) worth of energy. Its not a lot compared to the phase change portion but its definitely extra

[Andy]

That clarifies it - thanks, Hex.

I wonder how much of the heat trapped underground is 'banked' in the subsoil and how much drains away as hot water? I suspect this might be quite a large proportion, but I suppose this will depend on the exact design and how much the plant roots are pulling straight back up.

[Hex]

Hi Andy, John and all

I`ve been looking at the conduction aspect a bit more lately.

Compared to the phase change component its not great but could be a useful yardstick when judging the minimum system performance.

Dealing with conduction alone is a bit easier to calculate with a few necessary assumptions. I`ve assumed that 55 cubic feet of dry air =1 btu per deg F temperature change and the perforated tubing has R-value of 1.

Using my greenhouse layout as an example,
volume: 900cu.ft and 330ft of 80mm tube.

Total surface area of tube: 272 sq.ft
Conductive heat transfer = 272x 1degF/ R-1 = 272btu/hr/degF
Reduced to a per minute value: 272/60 = 4.53 btu/min/degF

The total tube surface transfers 4.53btu per minute/per degF differential between the air and soil temperature.

BTU versus temperature for 900cu.ft of air: 900/55 = 16.36 btu/degF
(900cu.ft of air must gain or lose 16.36 btu to change temp by 1degF)

Heat transfer capacity via conduction: 16.36/4.53 = 3.611 min
The tube surface area requires this much time to transfer sufficient heat to change the air temp by 1degF (in a perfect world )

Airchanges per hour: 60/3.611 = 16.61x
Fan required: 16.61 x 900/60 = 250cfm

Logically increasing the fan speed wont make a great difference to conductive heat transfer as its limited mostly by the tube surface area and capacity of the air to hold heat.

I used some air/soil differential steps to see what performance may be possible:

10F TD 45.3 btu/min sink or source (2,718btu/hr= 0.8kw/hr)
20F TD 90.6 btu/min sink or source (5,436btu/hr= 1.6kw/hr)
30F TD 135.9 btu/min sink or source (8,154btu/hr= 2.4kw/hr)
40F TD 181.2btu/min sink or source (10,872btu/hr= 3.185kw/hr)
50F TD 226.5 btu/min sink or source (13,590btu/hr= 4kw/hr)

Not sure 50F is very likely though

To round things off, the average solar irradiation for June falling on the floor area of my greenhouse means a peak heat load of about 198 btu/minute.

In theory at least, conduction alone (at 250cfm) may be sufficient provided the air/soil differential is around 44F.

Lots of unseen variables to throw the proverbial spanner in the works but may be enough for roughish estimate.

I tested it with Andy`s system specs and it seems to max out (conductively speaking) with a fan speed of 113cfm, which is 9.8 airchanges an hour.

Useful or not it was a fun exercise

[Hex]

Are you all on a sunny beach somewhere? only drizzle and rain here

I`ve been playing with the figures again and would welcome your opinions.

As far as conductive heat transfer is concerned it doesn`t seem to matter what diameter of tube you use.. the ratio of surface area to air volume remains the same

If you use a 10ft tube length, according to my figures a 3.2 second air dwell time in that tube is sufficient to transfer heat to the soil via conduction alone at a ratio equal to the air/soil differential.

If this is correct then it suggests a larger number of fairly short tubes would be preferable to long ones.
Once the air has lost all its heat it doesn`t want to be loafing along in the tube..ideally it needs to be back into the space as soon as possible which makes room for more warm air to enter the tube

As conduction is the lesser heat transfer mode, logically it will require a somewhat longer dwell time than the phase change needs to reach dewpoint.

In any case, (if the figures are to be believed) it appears the conductive transfer mode plays a significant role in taking the air to dewpoint.

what do you think?

[mrhobbithhnet]

Hex,

Your last comment is exactly the point. You cannot get dewpoint without at first having a fast and efficient conductive exchange. And without dewpoint for a consistent and lengthy period of time (each time it's reached, more conductive loss is necessary to continue a dewpoint condition) The system is a dynamically changing phenomenon and as such is difficult to figure with just math tables and formula. IF you can maintain 10 to 20% of the total volume at dewpoint all of the time, then you would be doing something extraordinary and worth all the effort. Otherwise, not much. Yonaton was finding that conductive exchange was responsible for somewhere in the neighbourhood of 40% of the total exchange for his test system last winter. Personally, I believe better designs could lower that and increase the dewpoint/phase change exchange to more like 80%.

But it takes a test site to manage this topic properly. Your math examples make for some good solid thinking to begin to emerge so that we can assemble good, useful and valid test models. Thanks for that ...
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[Hex]

Hi John
It wont be long before i start digging the trenches hence the recent math frenzy
I have it in my mind that there must be a tube length range where the majority of the conductive losses have taken place and adding more tube length after that may not be as efficient as adding another tube to the plenum.

I may have to bury a tube with sensors every foot to see how fast the air temp drops and whether it levels out. If everything happens in the first 10ft, it should be better to use 3x 10ft tubes than 1 x 30ft tube.

The apparent increase in system performance for 20 airchanges/hr instead of 5 may point towards the same thing. The faster airspeed in a long tube will reduce the dwell time but utilises the tube length better.

I agree its very difficult to figure out mathematically and very easy to fry some braincells in the process

[mrhobbithhnet]

hex,

Correct me if I am wrong but it sounds like u r designing only for conductive ex

That is not good

U must have enuf tubing for phase change ex too

What the extra length should be is the ?

Your #'s for cone ex are good only as a starting point

The magic bullet is knowing dwell time to use

Then it's just length cuz 20x is the fan

What we need is a dwell time factorthat we can be using to simplify design process

The dwell time hmmmmm

Maybe we can start with a %? That is the % of time the air is underground

10%? 20? Your guess is as good as mine

Work the numbers for 20xfan and length/#tubes at different % dwell

U might find a pts answer from that exercise
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[Hex]

[Hex]

Hi John
more brainstorming and conjecture

I think the conduction dwell time may be less important in the case of elevated airflow rate.

Take this scenerio, the entire volume of greenhouse air sent through the tubes at high speed presumably cools it en mass but only by a very small degree per pass.

After a number of passes, all the small incremental drops in temp add up so eventually you should have a huge volume of air somewhere close to or at dewpoint both in the tubes and also in the greenhouse itself?

I guess once the entire air mass gets down to dewpoint the phasechange process could be self perpetuating..at least until the air mass gets hit with a pretty sizable heat influx which would cause it to stall.

It may be that shorter than optimum tube lengths can be offset simply by greater turnover rates?

[mrhobbithhnet]

Hmmmm, that sounds interesting. You are right that if you chill it faster than it gains, then you could be dewpointing all the length of the tubes... but that assumes you have an infinite heat sink. The reality is that as the process progresses the latency in the soil heat sink removes the delta and thus the system bottoms out. For summer time cooling though, you could do some low speed air movement from outside and thru to outside again to re-chill the sink at night. That too would bottom out when your delta approach zero. The best approach to summer operation is highly effective passive venting for most of the summer... and putting the SHCS to sleep.

Passive venting design effectiveness is another hugely interesting challenge I've been continually finding more about. Bi-directional cross wind driven venturi effects built into the upper venting arrangements is the key... and working out the upper vent total area as a larger effective size than the lower vent area (hot air rises to your advantage, but it also expands so needs more room up top to exist after it's heated) Building interior shape also matter, the more 360 dome curves the better the passive flows too.

The classic teepee shape for example, and your dome shape in particular have very effective self supported upward flow patterns. Slight adjustment with subtle fin assemblies surely enhance the flow by encouraging a vortex flow.

If there was a way to store this summer gain for re-use in another season then even the venting issue would pass from interest. How about highly efficient heat pumps condensing into sand storage? I doubt that solar power could run a heat pump, but it could.

And for amplified delta for winter days, how about parabolic troughs with hot liquid piped from focal point and down through the SHCS tubing matrix?

What about re-positional solar hot water panels as summer shade structures inside the house and then heat pump sand storage? ...
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[Hex]

The amount of solar gain we`ve had this summer could probably be stored in an eggcup
I have 360 degree passive intakes at the bottom of the dome from the riser wall, in total about 25 sqft worth not including the door. I think there should be some crossflow if the upper vents were closed.
The height difference of the vents is close to 9ft so it should provide a reasonable flowrate depending on the temperature difference.

Now i`ve found a reasonable price of perf drainage tube (£36 for 50m) an external cooling loop option is economically viable. If its uninsulated, and non hatted it may cool enough at night via diffusion and radiation to make it useful the following day.

If we continue with these short wet summers i think cooling capacity will be less important than heating.

[mrhobbithhnet]

Your focus has to be on heat in the dead of winter. What you have will work just fine... with only slight if any mods to the upper vents to introduce a venturi effect.
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