Store thermal energy with palm oil? Natural Phase Change Material (PCM)

[izentrop]

Only risk of fire and complications to manage the overpressure. That's what is written in the document you quoted.

We no longer talk about oil above 400 °, see page 20

the direct generation of steam without intermediate fluid.
This avoids problems related to the use of oils
thermal (high cost of the fluid, pollution problems,
fire hazards, limitation of the temperature of use,
thermal degradation of the fluid over time with
hydrogen production),

http://www.ademe.fr/sites/default/files ... 1-6915.pdf

[Christophe]

So the plants currently using oil with temperatures around 400 ° C see more, are time bombs? ^^

No because there is (logically) no oxygen in the pipes and the circuit ... so no worries as long as there is no escape of course ...

[izentrop]

Precisely, as the flash point of the mineral oil used is 110 °, the slightest leak and it is the fire with pressure of 10 bars. This type of installation is classified Seveso in France.

[curieux80000]

The study is more advanced on rapeseed oil and shows that finally the use is limited to 220 ° C ... it is rather limited, it greatly limits the yield of a solar power plant ^^

[Mystiik]

I allow myself to unearth this thread because it is extremely interesting.
Reading his 37 pages are a treat, for the eyes as much as for the meninges, the views of the possible applications.

After some calculations given to the tastes of the day of Dr. Maloche, I find myself with:

- Energy efficient house: 0.25 W / m².K
- 100m² house, assume 4 walls of 10m by 3m high: 120m² of wall surface
- It is assumed a constant temperature difference of 30 ° in winter between the inside and the outside of the house: -10 ° C ext; 20 ° C int (it will make a slight oversize)
- Have a week of heating in winter thanks to the buffer

We therefore have as loss:
0.25W / m².K x 30 ° x 120m² = 900W

Which gives a need in energy on 7j of:
0.9kJ / sx 7j x 24h / dx 3600s / h = 544 000kJ
divided by the 159 kJ / kg of the buffer = 3.5m3

That's half of what Maloche had found, and for 2x more time

I just changed the heat loss in his calculation, which is more realistic for my taste for new homes (and that can be quite easily completed with straw-clay that said).


I tend to do some experiments so reading this thread made me mouth watering, so I tried to buy between 1 and 5kg of palm oil to test.


Here is what I found:
The current price per kilo of palm kernel oil is about 0.65e / kg according to these sites:
https://ycharts.com/indicators/palm_kernel_oil_price
https://www.indexmundi.com/commodities/ ... kernel-oil

Only impossible to find at this price .. import requires, but from there to go to 5-10e / kg ...

Here is the minimum price I found: 3.6e / kg ... Only the 70e of USPS shipping costs are prohibitive.
https://www.soapgoods.com/palm-kernel-o ... p-245.html

If anyone has an idea where you can buy palm oil at a fair price (or pure lauric acid as a substitute), I'm interested !!
I will continue to look for my side in the meantime !!


In any case I am delighted to have discovered this forum (which seems to have existed for a long time) and who shares totally my willingness to experiment inexpensive and easy to implement, in order to relieve a little our planet.

[izentrop]

Hello,
I read the beginning and nothing understood .... 45 °, heat mass less than 2 times less than water? What differentiates this process from the classic hot water tank?

- Energy efficient house: 0.25 W / m².K

How did you measure or calculate it?

[Mystiik]

45 °, specific heat less than 2 times less than water? What differentiates this process from the classic hot water tank?

We use the energy released by the phase change of materials, read these 2 pages for more details on the method and calculations:
solar-thermal / store-of-the-thermal-energy-with-from-oil-palm-of-t7421.html
Solar-thermal / store-of-the-thermal-energy-with-from-oil-palm-of-t7421-20.html

- Energy efficient house: 0.25 W / m².K

How did you measure or calculate it?

I had lessons from a design office in isolation quite astride the RT2012, they give an optimal thermal resistance R = 3.5 m².K / W

I just took 4 and not 3.5 and then reversed to get my value of 0.25 W / m².K
4 m².K / W = 0.25 W / m².K
3.5 m².K / W = 0.2857 ... W / m².K, it's not practical; p

From memory, it corresponds to a DPE class A, but I do not know how to find it properly, by cons we can do it with the ladle

We have 0.25 W / m².K and we want kWh / m² per year, if we assume an int-ext difference over the year of 20 ° on average, we have:
0.25 W / m².K x 20 ° x 8760h / year = 43.8 kWh / m², so in my opinion (and I could be wrong), we would be in a class A.


I had a little meditation last night on my calculation and 120m² of wall that I assumed, in fact we must take into account the roof surface (and that of the ground?), So we would have 120m² of walls + 100m² of flat roof = 220m² of waste to the outside.

So we would be 1650W loss, say 2000W to predict the leakage of 100m² soil, which will necessarily be less conductive than the roof
We would then need 1 200 000kJ to heat the house during 7j in winter
Divided by 159 kJ / kg of buffer = 7.6m3

We start to join Maloche, we must not forget that in the hypotheses I took -10 ° C ext and 20 ° C int continuously on 7j in winter, it's quite generous for France.

[Mystiik]

So I had another idea, why encapsulate the acid in round beads and put them in water when you can put 100% acid.
The answer is that the acid will gradually crystallize around the exchanger, until completely prevent exchanges with it.

My question is: which layer of crystal is one willing to tolerate? 5mm? 10mm? 50mm?
I intuitively want to say 10mm, the system will have a little inertia due to the solid layer formed, which is appreciated for underfloor heating, moreover, the more the buffer will be "emptied" in energy, the less it will exchange with the outside, even if it remains overall at 42 ° C.

So I propose a solution based lamella:
In the case of our 7.6m3, we have a cube of 2m about.
By taking plates of metal (which does not react with lauric acid) of 2mx2mx1mm, then one can create a system in lasagna:
1 layer: lauric acid 10mm
2 layer: 1mm of resistant metal
3 layer: 2mm of water
4 layer: 1mm of resistant metal
5 layer: lauric acid 10mm

We thus have 20mm of acid for 4mm useful, so we are at 83% acid, exchanger understood, in our system.


By taking:
1 layer: lauric acid 50mm
2 layer: 1mm of resistant metal
3 layer: 2mm of water
4 layer: 1mm of resistant metal
5 layer: lauric acid 50mm

Acid 100mm is obtained for 4mm useful: 96% acid, exchanger included


For a variation of T ° from one kg of 35 to 50 ° C:
211 kJ + 2x (50-35) = 241 kJ for lauric acid

The house needs 1 200 000kJ during 7j in winter
Either divided by 241 kJ / kg buffer = 5m3 !!

We win 30% of volume

[thibr]

but is not more than 30% more expensive?

[Mystiik]

I created a little excel that I join you.

I get, for a house requiring 1 200 000kJ heating, a need for:

EXCEL_2019-01-06_16-08-14.png (7.15 KIO) Accessed 4399 times

I also took into account the density of the buffer, which we currently approximated at "1000kg / m3".


Here is the excel