Author Topic: Recover energy from temperature (Read 26573 times)

LibreEnergia · « **Reply #15 on:** January 22, 2014, 01:40:05 AM »

Let me see. Last time I looked,

- Water does not spontaneously flow up hill...
- Gas does not spontaneously compress itself...
- Objects do not spontaneously heat up unless they have a lower temperature than the surroundings
- Heat does not flow from cold to hot.
- Objects at the same temperature are in thermal equilibrium with each other.

All sensible outcomes of the 2nd law of thermodynamics...

Chances of this device working, zero.

rc4 · « **Reply #16 on:** January 22, 2014, 08:30:35 AM »

Energy won for each volume is :

PSe - PS(e') = PS(e-e')

@LibreEnergia: so, let me know where I'm wrong, this system is not very complex, so it must be easy for you to find where I'm wrong.

Gas is compress/depress with adiabatic process.

LibreEnergia · « **Reply #17 on:** January 22, 2014, 09:59:17 AM »

Quote from: rc4 on January 22, 2014, 08:30:35 AM

Energy won for each volume is :

PSe - PS(e') = PS(e-e')

@LibreEnergia: so, let me know where I'm wrong, this system is not very complex, so it must be easy for you to find where I'm wrong.

Gas is compress/depress with adiabatic process.

You don't gain energy during the compression. it REQUIRES an external input of work to compress the gas. Assuming a thermodynamically reversible system you could recover all that external work input during the expansion phase, the net result however is no net work is done and no energy can be extracted from the system.

rc4 · « **Reply #18 on:** January 22, 2014, 10:04:27 AM »

Could you explain where I'm wrong in this system ? I would like to learn where I'm wrong.

The energy is more :

PSe - PS(mean e') = PS(e-mean(e'))

Because e decrease more and more when a volume move to the right?

LibreEnergia · « **Reply #19 on:** January 22, 2014, 10:15:10 AM »

Quote from: rc4 on January 22, 2014, 10:04:27 AM

Could you explain where I'm wrong in this system ? I would like to learn where I'm wrong.

The energy is more :

PSe - PS(mean e') = PS(e-mean(e'))

Because e decrease more and more when a volume move to the right?

What causes the volume to move to the right.?

There must be a pressure gradient for that to occur, correct?
The only way to achieve that is to supply work to the system from an external source, (or alternatively heat from a temperature gradient.)

rc4 · « **Reply #20 on:** January 22, 2014, 10:22:51 AM »

Quote

What causes the volume to move to the right.?

Me, but I move less to the right because vertical red surface move to the left in the same time. If vertical red walls don't move to the left in the same time I move to the right, I'm agree I give the same energy I recover. But Here, it's gas that move vertical red walls to the left, not me, the velocity of magenta walls is V but vertical red walls is V' with V' < V. The energy I gave is PSd, if d decrease (because V decrease for vertical red surfaces) I give less energy. For you the energy give to the system is the same if vertical red walls move to the left in the same time ?

LibreEnergia · « **Reply #21 on:** January 22, 2014, 10:38:27 AM »

Quote from: rc4 on January 22, 2014, 10:22:51 AM

Me, but I move less to the right because vertical red surface move to the left in the same time. If vertical red walls don't move to the left in the same time I move to the right, I'm agree I give the same energy I recover. But Here, it's gas that move vertical red walls to the left, not me, the velocity of magenta walls is V but vertical red walls is V' with V' < V. The energy I gave is PSd, if d decrease (because V decrease for vertical red surfaces) I give less energy. For you the energy give to the system is the same if vertical red walls move to the left in the same time ?

The gas doesn't...

1. 'Move itself' or
2. 'Compress itself'.
3. Neither do the walls containing the gas move without work input.

If it did these things then yes, your machine would work.

rc4 · « **Reply #22 on:** January 22, 2014, 10:45:12 AM »

Could you explain, with formula, velocity, distance, etc. ? I prefer to understand with formulas. I Recover PdV, so I need to give PdV, but I give PSe' not PSe, and like e'<e because V'<V I give less energy. Don't forget, outside volume there is P, but inside volume there is 0.1P.

PiCéd · « **Reply #23 on:** January 22, 2014, 01:21:17 PM »

http://en.wikipedia.org/wiki/Enthalpy

rc4 · « **Reply #24 on:** January 22, 2014, 01:37:01 PM »

If I use valves for red surfaces I can compute force on magenta walls. The mean of pressure is 0.45 bar so I need to give PSe but pressure is lower. I don't recover energy from red valve, it's only for compress volume, so valve move "alone", imagine valve alone, independant.

MarkE · « **Reply #25 on:** January 22, 2014, 01:47:47 PM »

Quote from: rc4 on January 22, 2014, 01:37:01 PM

If I use valves for red surfaces I can compute force on magenta walls. The mean of pressure is 0.45 bar so I need to give PSe but pressure is lower.

RC4 perhaps this will help:

Cut your machine in half at the point of greatest restriction and the point of least restriction. Now what you have is a pair of horns. Now pull them apart. Do you agree that if we take either horn that we must apply external work to move air through that horn one way or another? If you agree with that, then why should gluing two horns back to back change the situation?

rc4 · « **Reply #26 on:** January 22, 2014, 02:13:16 PM »

In the last message it's not PSe but PSe' I give.

Yes, I'm agree with you for 2 horns. But here, look at magenta walls they move at V, but I don't need to move e because valves move to the left in the same time. The mean pressure is 0.45 bar like when I recover energy. But for me, dV is lower due to the difference of e -> e'. I would like to understand this point. And if I recover energy from red valve, the force if Poutside - Pinside in the volume, but here it's logical the distance to move is not e but e', no ?

MarkE · « **Reply #27 on:** January 22, 2014, 02:42:24 PM »

Quote from: rc4 on January 22, 2014, 02:13:16 PM

In the last message it's not PSe but PSe' I give.

Yes, I'm agree with you for 2 horns. But here, look at magenta walls they move at V, but I don't need to move e because valves move to the left in the same time. The mean pressure is 0.45 bar like when I recover energy. But for me, dV is lower due to the difference of e -> e'. I would like to understand this point.

rc4, imagine a large cross-section square tube.

Do you agree that the pressure is constant inside and outside the tube?

Now imagine that we have another tube section with a smaller cross section and a flange plate on the left where the plate is at least as big as the cross section of the first tube. Imagine that we lower this tube into place so that the flange plate lines up with zero clearance to the right hand side of the first tube.

Is the pressure now different inside either tube or outside the tubes?

PiCéd · « **Reply #28 on:** January 22, 2014, 03:15:15 PM »

Since energy can not be destroyed in any case I was wondering if "S multiply by T" was especially the heat in loule of a material ...
S = entropy
T = temperature

rc4 · « **Reply #29 on:** January 23, 2014, 07:11:29 PM »

@MarkE: I understood my error. Thanks

I have another idea if you can help to find the error.

The goal is to let gas to compress each volume from 0.5 bar to 1 bar without lost energy, after put this volume inside a container where there is 0 bar (or vapor pressure) without energy, after recover energy from depress gas from 1 to 0.5 bar and after pass this volume to the first container.

For this, I use gears with teeth like drawing show. Gears don't give a torque or a force, they are only there for assume sealing. So the energy is not recover from gears but from volume when they pass from 1 bar to 0.5 bar: PdV.

For move in a volume inside vacuum container, I place 2 volumes of 1 bar inside a tooth of gear 1 or 2, I can pass without energy, no ?

For move out a volume inside 1 bar container, I place 1 volume of 0.5 bar inside a tooth of gear 3 or 4, I can pass without energy, no ?

Gears 1 and 3 (or 2 /4) are fixed together so they turn at the same rotational speed w. But like radius of gear 3 is 2 times higher of gear 1, when I pass a tooth of gear 1, I pass 2 teeth of gear 3. So the number of volume that move in or move out is the same. I keep always the same number.

Now, gears must assumed sealing (no gas escape), it's possible to imagine gears with very few space between teeth in theory. So I can think the problem come from here. It's possible to reduce number of gears. Here I choose diameter of gear 3 = 2 * diameter of gear 1 and this allow to depress gas from 1 to 0.5 but it diameter is 1/3, I can depress to 0.33 bar and recover more energy.

So what's wrong in this case ?

(http://imagizer.imageshack.us/v2/800x600q90/132/dhtx.png)

(http://imagizer.imageshack.us/v2/800x600q90/203/t571.png)

(http://imagizer.imageshack.us/v2/800x600q90/707/vup4.png)

(http://imagizer.imageshack.us/v2/800x600q90/46/cu5u.png)

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Author Topic: Recover energy from temperature (Read 26573 times)

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