Calculations on the compressed air to an engine cycle

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sicetaitsimple
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Re: Calculations on the compressed air cycle for a motorization




by sicetaitsimple » 01/06/20, 19:59

Eric Dupont wrote:I simply observe that if we produce photovoltaic electricity at 3 cents and that we associate it with an energy storage system with liquid nitrogen it is interesting to use this storage system to store energy when the photovoltaic does not produce and which is produced by other systems in order to make the storage system profitable more quickly.


More seriously than my previous post ....
It's wrong.
You don't pay for a storage system faster because you use it more often. This requires that the losses inherent in any storage system are low enough so that the "value" of what you remove from storage is greater than the "value" (the price, the cost) of what you have stored. .
And therefore contrary to your initial assertion, the overall performance of the chain is very important.
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Re: Calculations on the compressed air cycle for a motorization




by Eric DUPONT » 02/06/20, 08:24

well you say 60% yield is more than enough, already if we are just talking about electrical production there is no need to store everything since a good part is consumed as you go and therefore the good news is that there will be no need to build new EPRs.
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Re: Calculations on the compressed air cycle for a motorization




by obelix39 » 05/06/20, 08:23

I do not think I am mistaken in saying that all the losses which cause a drop in yield are thermal, whether mechanical friction, heating of electric cables by the passage of current, etc. ... During a production of potential mechanical energy, whether by raising a mass to take advantage of gravity, compressing air to take advantage of the pressing force, or rotating a mass at high speed to take advantage of its inertial energy, etc. ... If we recover all of the dissipated thermal energy, we obtain a yield of 100%, is that correct?
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Re: Calculations on the compressed air cycle for a motorization




by Remundo » 06/06/20, 00:00

obelix39 wrote:I do not think I am mistaken in saying that all the losses which cause a drop in yield are thermal, whether mechanical friction, heating of electric cables by the passage of current, etc. ... During a production of potential mechanical energy, whether by raising a mass to take advantage of gravity, compressing air to take advantage of the pressing force, or rotating a mass at high speed to take advantage of its inertial energy, etc. ... If we recover all of the dissipated thermal energy, we obtain a yield of 100%, is that correct?

it depends on how you count the energy and what you want to do with it. Numerically yes, the energy is conserved, so if you are the losses + the mechanical work recovered, you will find everything that has been supplied to the machine

in thermodynamics, a thermal engine using a hot source is forced to yield heat to the cold source, even assuming that there is no friction.

Try to read this a bit,
https://fr.wikipedia.org/wiki/Machine_thermique
because I have the impression that you do not have the notion of "energy" very clear, all energies are not equal, even if they are 100J each...
100 J mechanical, it's the best, 100 J thermal from a source at 1000 ° C, it can be used, 100 J thermal at 10 ° C, you won't do much in temperate latitudes ...

Good otherwise the pneumatic compressors are not thermal machines, but converters of mechanical energy (rotation) into another mechanical energy (of pressure). If you do something stupid, the performance can be very disappointing, and even if you don't do it, it is not excellent in general.
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Re: Calculations on the compressed air cycle for a motorization




by Eric DUPONT » 06/06/20, 20:14

if the heat of the compression is evacuated as the compression is isothermal and reversible at 100% with the heat present during the relaxation provided that the ambient heat is identical. Of course there is no friction.

But compressed air poses a problem because storage is expensive, large tanks are required, the pressure of which constantly changes, which poses regulatory problems.

this is why the energy will be stored with liquid nitrogen with a larger market than for the battery or hydrogen.
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Re: Calculations on the compressed air cycle for a motorization




by Remundo » 06/06/20, 20:19

Certainly Eric, liquid nitrogen considerably densifies the reserve of compressed air, but poses other problems (cryogenics, adiabatic chambers ...).

I think there is something to dig into this mode of cryogenic storage. Not easy, however, with "small means". Finally in this thread, it's a bit HS, we talked about it in your specific thread.
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Re: Calculations on the compressed air cycle for a motorization




by Eric DUPONT » 06/06/20, 20:28

I am not sure that this is a question of means. THE proof being that I invented something that I patented when I have very little means. Those who have a lot of means are not necessarily capable of inventing and also do not want to develop a few things that are not their own past.
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Re: Calculations on the compressed air cycle for a motorization




by GuyGadebois » 06/06/20, 20:41

Remundo wrote:Certainly Eric, liquid nitrogen considerably densifies the reserve of compressed air, but poses other problems (cryogenics, adiabatic chambers ...).

I think there is something to dig into this mode of cryogenic storage. Not easy, however, with "small means". Finally in this thread, it's a bit HS, we talked about it in your specific thread.

I might say bullshit, but why not run the engine directly with nitrogen? : Cheesy:
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Re: Calculations on the compressed air cycle for a motorization




by sicetaitsimple » 06/06/20, 21:10

Eric Dupont wrote:if the heat of the compression is evacuated as the compression is isothermal and reversible at 100% with the heat present during the relaxation provided that the ambient heat is identical.


It is the 73rd law of thermodynamics .....
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Re: Calculations on the compressed air cycle for a motorization




by Remundo » 07/06/20, 01:28

what Eric says here in somewhat approximate terms is correct, in fact he speaks of an isothermal compression of a perfect gas with an ideal machine, followed by expansion under the same conditions.

For slow machines with large heat exchangers, it is playable, but of course there are still losses.
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