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Discussion board help and admin topics => Half Baked Ideas => Topic started by: rc4 on January 20, 2014, 11:46:59 PM
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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 ?
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You will have to drive this with an external temperature gradient or it will equalize itself and stop.
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External temperature heating the gaz, because it lost temperature in one cycle
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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).
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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.
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Ooops, I think I made an error, the volume is not in ²... :o
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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
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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.
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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.
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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.
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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?
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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.)
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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 ?
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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.
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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.
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http://en.wikipedia.org/wiki/Enthalpy
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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.
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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?
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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 ?
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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?
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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
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@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) (https://imageshack.com/i/3odhtxp)
(http://imagizer.imageshack.us/v2/800x600q90/203/t571.png) (https://imageshack.com/i/5nt571p)
(http://imagizer.imageshack.us/v2/800x600q90/707/vup4.png) (https://imageshack.com/i/jnvup4p)
(http://imagizer.imageshack.us/v2/800x600q90/46/cu5u.png) (https://imageshack.com/i/1acu5up)
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rc4 this looks like a gear pump. Are you thinking that somehow this pump will run itself to increase the pressure difference? The external forces will operate so as to attempt to equalize the pressures.
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Look at images please, it's not a gear pump, gear don't press gas. I use gears only for prevent the gas at 1 bar to move inside vacuum. Gas press volume, and I recover energy inside vacuum recipient. If you see a torque somewhere I will be ok, but here I don't find a torque, so where is lost energy ? The only energy I can see it's lost is the small space between teeth when gears are in action. It can be very low:
http://commons.wikimedia.org/wiki/File:Involute_wheel.gif
You can imagine big radius for wheel, like that the volume lost for each volume is very low. There are gaskets inside teeth of gears.
g5 image: W = win energy, L = lost energy. If I consider gear/gear move in => lost energy, the sum is 0 and is I consider gear/gear move out => lost energy, the sum is 0 too. Look at direction of gears.
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Look at images please, it's not a gear pump, gear don't press gas. I use gears only for prevent the gas at 1 bar to move inside vacuum. Gas press volume, and I recover energy inside vacuum recipient. If you see a torque somewhere I will be ok, but here I don't find a torque, so where is lost energy ? The only energy I can see it's lost is the small space between teeth when gears are in action. It can be very low:
http://commons.wikimedia.org/wiki/File:Involute_wheel.gif
You can imagine big radius for wheel, like that the volume lost for each volume is very low. There are gaskets inside teeth of gears.
g5 image: W = win energy, L = lost energy. If I consider gear/gear move in => lost energy, the sum is 0 and is I consider gear/gear move out => lost energy, the sum is 0 too. Look at direction of gears.
If your mating gear tooth faces are not gas tight then it just leaks. If they are gas tight then it is an air pump. In either case as air pressure changes there will be loss to heating / cooling. This machine like the last one is subject to the Second Law of Thermodynamics like all others.
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There is no gas inside between tooth, it's not an air pump. But, if energy is not lost here, where is it ? I can pass volume with 1 bar (look at 4 images please) inside vacuum container without need energy and I can pass volume of 0.5 bar inside container of 1 bar without energy. Could you help me to find the error ?
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rc4 if the gears rotate then volume between any pair of teeth on one gear and the meshing tooth of the other gear varies as the two gears rotate relative to one another.
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Yes, I understood the torque is not the same, one part act on small radius for one gear and one part act on big radius for another gear.
But here, in the image, if I cancel pressure when the tooth is inside pressure = 0, all the time gas act for 4 surfaces so no torque, and for cancel pressure of this small volume of gas I lost less energy no ? Could you explain please ?
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Yes, I understood the torque is not the same, one part act on small radius for one gear and one part act on big radius for another gear.
But here, in the image, if I cancel pressure when the tooth is inside pressure = 0, all the time gas act for 4 surfaces so no torque, and for cancel pressure of this small volume of gas I lost less energy no ? Could you explain please ?
Your illustration shows operation of a gear pump. If the lower gear rotates CCW gas is being pumped to the lower left. If the lower gear rotates CW, gas is being pumped towards the upper right.
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No, it's not a pump, look at images, where tooth are one inside other, I don't pump gas, I use gears like gasket. Each tooth has a valve, this valve recover PdV for each step. The only place I can loose energy is at the middle, but if the volume between teeth (at the middle) is near 0 (image big radius for gears), I don't lost PdV.
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I understood, it's because gaskets are usefull for one gear, this not works for second. So I lost that I won : 2 PdV
Now if I put the gasket in the center of blue arrow:
http://commons.wikimedia.org/wiki/File:Involute_wheel.gif
I compute torque, it must be the same energy than 2 PdV.
R1 = small radius of gear
R2 = big radius of gear
P = 100 000 Pa
I found the result in the second image, so 1/6 of PdV, where is the error please ?
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Another idea:
Take a screw (worm drive) with N starts (increase lead). The thread is used for move out vacuum objects and recover energy from temperature. The torque I need to give of the screw is T/N with N the number of starts and T the torque I need with one start. In one turn I can move out the same volume of vacuum object than with one start (The slope change when N increase) but here the torque is lower and divided by N. I drawn an image with 3 starts for example. Imagine walls of threads like very (very) thin for put N very high.
The part of thread inside vacuum: add wall for full the thread to obtain a cylinder (a perfect cylinder). Like the screw turn, all object with vacuum in it move up. I need to give a torque for turn the screw, but I recover N times I need to give.
Second image: multiple starts, but imagine thin of thread very thin (where the is solid on the thread)
Third image: I need to give a torque of the slope of one start only, but the lead so the volume of vacuum move out is multiplied by 2. The energy needed fot the torque is C*Θ, Θ = 2pi for one turn. C is the torque for one slope, but the lead is multiplied by 2 so the volume of vacuum object is pi(R2²-R1²)*lead.
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The thickness of worm drive thread is equal to thickness of gear. When I turn one turn I move up a tooth to one pitch, but the thickness of thread increase the distance. I move to 2 teeth. A gear give PV for tooth/notooth (a pitch), so here the sum of energy is 0, but if thickness of thread is near 0, in one turn of worm drive I move up one tooth, the next turn too, but gear give/recover energy only for one tooth, after it's a blank.
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P = 100 000 Pa
Ok 100 000 Pa, but if the volume is too litle 0.000001 m³ or lesser it will be almost nothing. How many anions or cations are produced with this?
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Another idea:
All under pressure P.
Roll up a full solid on a screw. The screw turn only. S1>S2.
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If I take a cycle like that ?
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Surface S1 = Surface S2 but direction are not the same, so the torque are not the same too, no ?
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Increase area where there is 0.5 bar without energy.
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with this ?