Unravelling the Mystery of Free Energy

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Discussion Overview

The discussion revolves around the concept of free energy, particularly in the context of water dissociation into hydrogen and hydroxide ions, and the potential for harnessing energy from infrared radiation. Participants explore the thermodynamic implications, energy sources, and the feasibility of various energy conversion methods.

Discussion Character

  • Exploratory
  • Debate/contested
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • Some participants question the concept of free energy in relation to the dissociation of water, suggesting that energy is required to break bonds and that this energy cannot be considered "free."
  • Others propose that the energy for dissociation might come from the heat within the liquid, raising the possibility that removing ions could allow more dissociation and potentially lower the temperature of the liquid.
  • There is a suggestion that infrared to electric converters could provide a source of "free energy" due to the abundance of infrared radiation in the environment.
  • Some participants argue against the feasibility of extracting useful energy from ambient infrared radiation, citing the laws of thermodynamics and the low intensity of such radiation.
  • Concerns are raised about the economic viability and efficiency of harnessing infrared energy compared to solar energy, with some noting that solar panels are more effective due to higher energy density.
  • A few participants reference existing technologies, such as radioisotope thermoelectric generators, as examples of energy conversion methods that utilize thermal gradients.

Areas of Agreement / Disagreement

Participants express a range of views on the feasibility of free energy concepts, particularly regarding water dissociation and infrared energy conversion. There is no consensus, as some argue for the possibility of extracting energy while others firmly state it is not feasible under current thermodynamic principles.

Contextual Notes

Limitations include the dependence on specific definitions of energy sources, the unresolved nature of energy extraction methods, and the varying interpretations of thermodynamic laws as they apply to proposed energy systems.

  • #91


A impressively quick reply!

I have read K. Browne’s paper and in a nut shell he shows that for the two geometries studied, the target had a larger surface area than the sending object. Consequently the targets emitted all that they received from the emitter without raising their temperatures. My design is different in that the target is smaller than the emitter and so will receive a concentration of photons as envisaged by Sudhir Panse. Browne does not cover this eventuality even though he says that other geometries may be found, This is why I am still pursing it. Unless I learn something new that explains why it shouldn’t work, I will continue making the device. I expect to complete the device in the next 2 months when all my questions will be answered. If it doesn't work I shall still be looking for answers LOL

I have often heard the answer that the second law of thermodynamics says it can’t be done. People who say this have been unable to tell me why it won’t work or have some spurious illogical answer. There may well be a reason deep down in the realm of quantum mechanics why it won’t work but I have not heard it yet.

I suggest that the second law of thermodynamics was created from 300 or so years of empirical observation and is not tied to any mathematical proof. This leaves it open to future findings.

Your suggestions are welcome and may save me a great deal of work and expense.

As I said, my math may be wrong but here goes :
I used this site to compute the energy collected at the emitter :
http://hyperphysics.phy-astr.gsu.edu/Hbase/thermo/stefan.html#c3
As I only want the energy emitted, not the net difference between two bodies, I used a T cold of 0K and a T hot of 300K with a collector area of 153.94 sq mm

(oops found my error - I put in 153.94 sq cm not 153.94 sq mm)

Collector collects 0.0707037 watts

I would need 1,000,000 / 153.94 collectors to get 1 sq m = 6,496.12 collectors

# of collectors * watts per collector = 0.0707037 * 6,496.12 = 459.27 watts or 0.62 HP

That certainly looks more reasonable, thanks for the pointer.

At this (now corrected ) rate it looks like the emitted energy is just under half that of the solar energy per unit area. The difference however is that radiant energy runs 24/7 and is not reliant on latitude or cloud cover and so is still a very attractive goal.
 
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  • #92


It is trivially easy to prove that this idea of yours is false - anyone who owns a telescope could point it at a wall and expect to be able to burn a hole in a piece of paper!
 
  • #93


You are correct. The telescope won’t burn a hole in a piece of paper. That is simply because the telescope would not focus ALL the radiant energy from the wall, only those minuscule small number of rays that travel in a parallel manner towards the telescope.

I like your site and am particularly impressed with the Hubble image.
 
Last edited by a moderator:
  • #94


Habanabasa said:
As I only want the energy emitted, not the net difference between two bodies, I used a T cold of 0K and a T hot of 300K with a collector area of 153.94 sq mm

Isn't this a problem? I have to admit I don't follow all of the arguments above, but think about what is keeping your emitter at the temperature of 300K. Isn't it just all of the 'random' radiant energy bouncing around between the emitter and its surroundings?
 
  • #95


Habanabasa said:
You are correct. The telescope won’t burn a hole in a piece of paper. That is simply because the telescope would not focus ALL the radiant energy from the wall, only those minuscule small number of rays that travel in a parallel manner towards the telescope.
Unless the telescope is right up against the wall, the amount of radiation that is not parallel will be miniscule! Besides - even if it misses a quarter or half the radiation, it can still focus light to thousands of times the intensity with which it was emitted: looking at a wall with a telescope as big as mine would be painful to your eyes.
I like your site and am particularly impressed with the Hubble image.
Thanks!
 
  • #96


Anyway, sorry, but this discussion is explicitly against forum guidelines. We deal only in established science here.
 

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