(This subject follows solar-thermal / solar-thermal-with-storage-capsules-kaplan-t7405.html )
It would be possible to store, at a very low cost and small footprint, thermal energy (low T °: 40-45 ° C) thanks to the latent heat of fusion of ... coconut oil or palm oil , a 100% natural and inexpensive product with a melting temperature between 35 and 45 ° C!
See: http://fr.wikipedia.org/wiki/Acide_dod% ... o%C3%AFque
et http://fr.wikipedia.org/wiki/Huile_de_palme
Edit: following many strong protests (and coming from people who have not made the effort to read the subject) about the use of palm oil whose production methods are detrimental to local biodiversity, 2 points to be specified:
a) palm oil is taken as an example: any natural body whose melting point is between 35 and 45 ° C could be suitable! We seek to find a natural body, to avoid using chemical or synthetic MCPs whose econological cost would be much higher!
b) the oil is not consumed! So if 200 kg of oil used in heat buffer during 30 years can save tens of tons of wood or fuel oil, the environmental balance is surely more than positive, even with palm oil ...
So how much of the 1kg calorie of palm oil (which will be assimilated into an 1er time to lauric acid) can it store? And especially how much water L that corresponds?
I remember that water is one of the best heat transfer fluid but are used in thermal buffer requires significant amounts because it is difficult to use changes of state!
So if we could use "oil-doped" water (and not the other way around) it could, perhaps, be interesting! Let's make a 1st estimate! For the lazy, read the conclusion at the bottom.
How many calories are released in 1 Kg of oil that solidifies?
It's all stupid, just read: according to the table of this page, we have:
Melting point: 41-43 ° C, take 42 ° C
Latent heat: 211,6 (kJ / kg)
(compare: for info with water: melting temperature: 330 kJ / kg but especially evaporation: 2200 kJ / kg)
Each kg liquid oil that solidifies will release 211 kJ at a level of 42 ° C, or 42 ° C is a VERY interesting temperature for:
a) the ECS
b) Low temperature heating
It corresponds to how much energy 211 kJ?
The specific heat of the water is about 1 calorie is 4.18 kJ / kg ° C.
- 1 kg of oil at 42 ° C thus makes it possible to heat 1 L with water of 211 / 4.18 = 50 ° C
- On a "working" temperature delta of a thermal buffer of 20 ° C (between 50 and 30 ° C for example), each L of water therefore stores 20 * 4.18 = 83.6 kJ
- So we could say that 1 kg of oil stores as much energy as 211 / 83.6 = 2,5 L water but the advantage of the oil is that the T ° is 42 ° C, a more interesting temperature that 30 ° C!
- at 42 ° C, each kg of oil stored
Good truce blabla, here is a small comparative practical example: a buffer water buffer 2000L and a hybrid water-oil balloon
Buffer of 2000 L of water at 50 ° C.
It is estimated that the T ° of use is exploitable up to 30 ° C.
We have an energy stored on delta 20 ° C of:
E = 2000 * 4.18 * 20 = 167,2 MJ is 46 kWh or enough to take about 23 showers (assuming the last showers will be 30 ° C).
Hybrid buffer: "palm oil - water" with 500L of palm oil and 1500L of water at 50 ° C
The volume occupied in the house is the same as the other balloon.
The oil is immersed in water in containers or not (that's not the problem yet).
500L of liquid oil = 500 * 0.862 = 431 kg.
1500 L water = 1500 kg
What is the stored energy?
- Latent heat of solidification of the oil: 431 * 211 = 90 940 kJ released (and that is what is important) at 42 ° C!
- Massive heat:
Between 50 and 30 ° C for water: 1500 * 4.18 * 20 = 125 400 kJ
Between 50 and 42 ° C for liquid oil: 431 * 2.27 * 8 = 7800 kJ
Between 42 and 30 ° C for solid oil: 431 * 1,76 * 12 = 9100 kJ
In total:
E = 90.9 + 125.4 + 7.8 + 9.1 = 233.2 MJ is 64 kWh is enough to take 32 showers around ... but in fact much more see below!
The gain does not seem so important that ca? Except that a large part of these calories are at a more exploitable temperature ... so there comes the question:
c) What is the energy gain thanks to the oil between 50 and 42 ° C (or rather between 50 and 41.9 ° C)?
- 1er we have E = 2000 * 4.18 * 8 = 67 MJ = 18.6 kWh
- 2ieme case one has E = 1500 * 4.18 * 8 + 431 * 2.27 * 8 + 431 * 211 = 50.2 + 7.8 + 90.9 = 149 MJ = 41.4 kWh
Conclusion: between 50 and 42 ° C the hybrid flask containing 1 / 4 by volume of palm oil to provide 2.2 times more energy than a buffer tank identical to water!
Well I learn about the wholesale price of palm oil to put in our big heat buffer!
Well it remains full of questions to solve, the 1ere being: what is the optimum% of oil according to the desired T ° of use? If it is around 42 ° C the answer is simple: 100% oil!
Question: how is it that no industrial heating or even university can think of it yet?
Warning: there is a strong difference in density between liquid and solid oil which must be taken into account in the case of an "operational" test: 1007 kg / m3 for the solid and 862 kg / m3 for the liquid.
