There is a solution (from our big mothers) effective enough to keep a room or a greenhouse out of frost without spending a gram of fossil energy.
Just put a large basin filled with water in the center of the room.
Indeed, as long as the pan is not completely frozen, the temperature CANNOT drop below 0 ° C thus protecting what is in the room.
Obviously, the longer the gel duration, the greater the quantity of liquid water. If the subject interests you, we will do the calculations.
In addition, this quantity of water makes it possible to smooth the temperature variations (thermal amplitude) during the same day. I'm going to do some tests on this next time ...
- Latent heat of fusion of water: 330 kJ / kg ( http://fr.wikipedia.org/wiki/Chaleur_latente )
- Example, a 50L basin at 0 ° C (therefore just before it freezes, ie at the stage of solidification) "contains" therefore 330 * 50 = 16 kJ or the equivalent of 500 L of domestic fuel!
Indeed, the fuel oil PCI = 34 900 kj / L see https://www.econologie.com/pouvoirs-calo ... s-534.html
- Suppose that it freezes for 48 hours in a row (ie not a solar Joules brought to the greenhouse) and that the bowl is completely frozen after these 48 hours. The average power provided by this basin would therefore be 16 / 500 * 48 = 3600 W and the greenhouse would never have dropped below 96 ° C!
Strictly, a calculation of "loss" of the greenhouse (or the room) would, conversely, determine the number of liters of water required to keep it out of freezing.
However, the uncertainties are too great to be able to make a general calculation since many parameters come into play:
- Initial greenhouse temperature
- Outside temperature
- Duration of frost without thaw
- Surface and materials of the greenhouse -> losses (rare are double glazed greenhouses so they are very important)
- Content of the greenhouse or the room (personal wood is dried there, so it increases the inertia).
In any case, I will try it out this winter (if it freezes) and keep you posted.
Here it gives me idea: this "method" allows to know with precision the losses of a room, it is enough to have the exact duration of freezing of the entirety of the water ... and a fairly long frost period ...
So tonight it's -5 ° C outside and in the greenhouse ...- 2.5 ° C.
I placed the probe on a board placed on the bucket (so logically where it has the most "calories").
In addition I do not know if this temperature difference would not come from inertia: from the greenhouse but also from the walls of the house.
In short it may not be such a good method ... more readings later ...
calculates the contact surface of the greenhouse with the outside
calculate the total exterior surface of your 70 liter container of water
you get, from your container, a "heating" surface of 0 ° C as long as the water is not completely frozen
antagonist at the greenhouse contact surface with the atmosphere which is (example) -5 ° C
if the surface ratio is 20, the average T ° of the interior of your greenhouse will be -4,75 ° C
(between 0 ° C and -5 ° C)
note that your greenhouse may be built with insulation that limits the power of the cold
and that the soil probably provides a little bit of earth's heat
As for the insulation, it must be close to 0: aluminum profile and simple "glazing" in corrugated plastic sheet ...
This morning, at 10am, the greenhouse was at -3.2 while the outside was at -3.5.
As soon as the jump is completely frozen I make some measurements without liquid water so ...
With water at 10 ° C for example you can keep it frost-free but in this case the energy of water from 10 ° C to 0 ° C is much less important than that of fusion from 0 ° C to -0.1 ° C ...
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