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Author Topic: Recover energy from temperature  (Read 26329 times)

rc4

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Recover energy from temperature
« on: January 20, 2014, 11:46:59 PM »
Magenta walls recover energy. If e is very thin, the energy needed to compress volumes is very small and come from pressure too. I need only to give Y*e and if e is near 0, I need to give 0. The efficiency  is 0.5 due to Y forces.  I imagine the velocity of compression of volumes very high compare to the speed of walls. Y forces compress volume too, it's not only X forces (it's very important, in the contrary this idea not works). Do you understand the idea ?

MarkE

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Re: Recover energy from temperature
« Reply #1 on: January 20, 2014, 11:49:48 PM »
You will have to drive this with an external temperature gradient or it will equalize itself and stop.

rc4

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Re: Recover energy from temperature
« Reply #2 on: January 21, 2014, 12:11:22 AM »
External temperature heating the gaz, because it lost temperature in one cycle

MarkE

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Re: Recover energy from temperature
« Reply #3 on: January 21, 2014, 12:17:49 AM »
RC4 it is not enough to just have some ambient temperature above absolute zero.  You need both a high temperature ("hot") reservoir and a low temperature ("cold") reservoir.  Then you can harvest some of the power in the thermal flux between the two.  Once you add the required "cold" reservoir, then your idea will essentially be a thermoacoustic engine.

rc4

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Re: Recover energy from temperature
« Reply #4 on: January 21, 2014, 12:20:36 AM »
This is idea is to recover energy from only one temperature not hot and cold. Look at left volumes when they are compress from X and Y forces. If Y forces don't compress I need to move for each volume the thickness of each volume but like Y decrease in the same time the tickness, I need to  move lower distance. Do you understand  ?
I added an image for show the reduction of thicness of each volume, more and more.

MarkE

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Re: Recover energy from temperature
« Reply #5 on: January 21, 2014, 04:30:04 AM »
This is idea is to recover energy from only one temperature not hot and cold. Look at left volumes when they are compress from X and Y forces. If Y forces don't compress I need to move for each volume the thickness of each volume but like Y decrease in the same time the tickness, I need to  move lower distance. Do you understand  ?
I added an image for show the reduction of thicness of each volume, more and more.
RC4 the second law of thermodynamics is a tyrannical monster that prevents such ideas from working.  Countless people have tried to cheat the second law and it has bested all of them.

rc4

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Re: Recover energy from temperature
« Reply #6 on: January 21, 2014, 08:02:16 AM »
Here, this is pressure (temperature) that compress volumes at left, what is the distance volumes moves ? It's not S*e*P, with P pressure, S surface of all volumes and t the thickness of a volume at start. It's Se'P with e' < e. Each time a volume move at right it lost thickness, e4<e3<e2<e1<e. What energy I need ? It's ( (S1-S)e+(S2-S1)e1+(S3-S2)e2+(S4-S3)e3+(S5-S4)e4 )* P. Sure surface is the same but thickness decrease with time. Do you understand this idea ?

Where pressure is constant at 0.1 bar don't help to recover energy, it's possible to have just usefull things like second image shown.


MarkE

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Re: Recover energy from temperature
« Reply #7 on: January 21, 2014, 09:01:39 AM »
rc4 what I see is two exponential horns wrapped around a closed path and connected back to back.  In the best possible case it can take a very small amount of work to drive fluid (gas) around the loop.  It will still take work, and you will have to drive that with an external energy source. 

rc4

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Re: Recover energy from temperature
« Reply #8 on: January 21, 2014, 10:02:38 AM »
MarkE, forget the last image (I deleted it), think with linear volumes please.  Do you agree with: ( (S1-S)e+(S2-S1)e1+(S3-S2)e2+(S4-S3)e3+(S5-S4)e4 )* P energy needed for compress volumes ?

Look at image below, you can see red lines are the true volume I move inside P.

It's possible to think with linear velocity for understand. If velocity is V for all walls, red surface don't move at V, they move at V' < V, because they are compressed with Y forces. In a time t, if velocity decrease, the distance decrease so I need to give less energy.
« Last Edit: January 21, 2014, 01:28:50 PM by rc4 »

MarkE

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Re: Recover energy from temperature
« Reply #9 on: January 21, 2014, 02:06:14 PM »
rc4 even under the best of circumstances you have to do work to compress.  You will get less than that work back when you expand.  In order to execute a full circuit, you need energy from the outside.  Let's suppose that you use the body of the machine itself as your low temperature reservoir.   Just sitting in the environment brings the machine up to the ambient temperature and you lose your gradient.

rc4

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Re: Recover energy from temperature
« Reply #10 on: January 21, 2014, 04:06:45 PM »
Yes, I need energy to compress but less I recover. And Energy come from temperature, but it's only a unique temperature not a difference in this system. We can discuss about the energy in/out. In theory I can recover energy from PdV but I lost energy give by the formula. Are you agree with the work to give (formula) ? Take the last image to think of the energy. Red surfaces don't move at the same velocity than walls, and the energy I give it's the energy for red walls. If velocity is lower, distance to move is lower and energy too.

MarkE

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Re: Recover energy from temperature
« Reply #11 on: January 21, 2014, 04:14:52 PM »
Yes, I need energy to compress but less I recover. And Energy come from temperature, but it's only a unique temperature not a difference in this system. We can discuss about the energy in/out. In theory I can recover energy from PdV but I lost energy give by the formula. Are you agree with the work to give (formula) ?
rc4 you are at odds with the second law of thermodynamics.  The second law requires that you have a temperature difference in order to do work.  If you have a single heat reservoir then you do not have a temperature difference and by the second law you cannot extract work.  This means two general possibilities exist:  1. You have found an exception to the second law, or 2. You are mistaken.

rc4

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Re: Recover energy from temperature
« Reply #12 on: January 21, 2014, 04:24:45 PM »
Maybe it's 1/ ;)

Please, look at work to give, if you compress walls only with X, I'm agree I need to give the same energy I gave (in theory). But if I use Y forces, the energy come from temperature too for compress, and this allow the system to move at lower distance to the right. Have you understand this principle ?

When gas compress volume, velocity decrease, like all walls must turn a round in one cycle, this would say at one moment some walls must increase its velocity, it's possible inside walls where there is 1 bar and when volume increase (recover energy area).
« Last Edit: January 21, 2014, 06:57:41 PM by rc4 »

PiCéd

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Re: Recover energy from temperature
« Reply #13 on: January 21, 2014, 10:25:24 PM »
Mh, pressure is a no conservative force, here we have a system who not use pressure in permanence
E in joule with pression (P) in pascal is:
E=(0.5?)xPx(the 3 dimensions)²
P=N (in Newton)/m²
If a force of 1 Newton is permanently exerced to an object of 2 cm x 2cm x 2cm, an energy of:
E=(0.5?)x2500x(0.02x0.02x0.02)²=160 or 80 nanojoules
An energy of 80 nanojoules is used in each second if it is in permanence.
If the volume is 1.5 m x 0.04 m x 0.1 m and if the force (1N) is exerced on the surface of 1.5 x 0.04 it is:
E=(0.5?)x16.666666x(1.5x0.04x0.1)²=0.3 or 0.6 millijoules
if the same force is exerced on the surface of 0.04x0.1, it is:
E=(0.5?)x250x(1.5x0.04x0.1)²=4.5 or 9 millijoules
This is some exemples, we can make pressure in permanence without using a generator or something else but it seem to be hard to harverst this energy with something viable.

PiCéd

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Re: Recover energy from temperature
« Reply #14 on: January 21, 2014, 10:32:20 PM »
Ooops, I think I made an error, the volume is not in ²... :o

----------------------------------------
Ah yes, that's it:
E=(0.5?)xPxV
V= volume
E=(0.5?)x2500x0.02x0.02x0.02=10 or 20 millijoules
E=(0.5?)x16.666666x1.5x0.04x0.1=50 or 100 millijoules
E=(0.5?)x250x1.5x0.04x0.1=0.75 or 1.5 joules