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Conservation of energy question (input vs output)

06/16/2007 12:41 PM

Hi, I apologize if my question seems silly, unfortunately I don't know anyone who can help me figure this out and cannot find my way around on energy resources on the web (I didn't studied engineering). So my question is kinda simple maybe:

If we build a 3mt vertical tube (or line) with coils and drop a magnet, it will generate some energy we can store someway. Of course not enough to lift the magnet back to the top in any way. It will reach a speed (x) and the energy will be (V).

BUT, if we repeat the same thing on a big big spiral, making the magnet go for like 15 meters all around, ending right below the starting point at 3 meters vertically, it will generate (V2) volts and reach (x2) speed. I know the friction means lower speed at the end. Could this mean more energy than the previous case?

My guess is, the speed x2 will be less than x (first case), this will affect the amount of Volts generated, perhaps it will excite more coils but at that speed rate will mean less volts, being V2 less than V. Off course I don't know how to make the numbers.

I made some silly test with a multimeter and a little motor, and half rpms doesn't mean half volts generated, so there is a clear clue of the speed degrading the system on case 2.

Last thing: In this example I try to eliminate everything that means more weight (difficulty on bringing the magnet up again), so we have lots of coils on the system but the only thing that moves is the magnet. It will mean coils on both sides, up and down. Sphere magnets will mean less friction.

The key factor is, powerful magnets doesn't mean "heavier" magnets

Please don't lock me into the "free energy" thing as is a quick way to declare that somebody is nuts.

thanks

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#1

Re: Conservation of energy question (input vs output)

06/16/2007 1:39 PM

The amount of energy that the magnet has is kinetic. It assumed this energy when you lifted to the top of the tube. You now are transferring that energy. The amount will be less as it spirals down. Due to losses in the form of heat. Friction at the surface of what ever you make the spiral out of. The most the spiral can due will increase the amount of time the same amount of energy would be transfered. So the transfer would be at lower level. Voltage would be less sustained for a longer time.

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#2

Re: Conservation of energy question (input vs output)

06/16/2007 3:38 PM

They both drop the same distance...one goes quick the other goes slow.

What one gains in speed it loses in distance and vice versa.

There may well be differences in the efficiency, but that would all depend on the physical arrangement and this is a theoretical question.

The magnet starts with potential energy due to it's raised position....It's all about how efficiently you can recover this energy. It becomes kinetic energy once it is moving.

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#3

Re: Conservation of energy question (input vs output)

06/16/2007 8:22 PM

"If we build a 3mt vertical tube (or line) with coils and drop a magnet, it will generate some energy we can store someway. Of course not enough to lift the magnet back to the top in any way. It will reach a speed (x) and the energy will be (V)."

Drop a steel ball bearing and you can easily figure it out for your self. It is all about potential energy or energy of position.

Since the 'coils' and how they are configured is not specified we can not comment on how they will affect or be affected by the fall of the magnet or a ball bearing substitute.

HOMEWORK PROBLEM ??? for Physics or Electrical class???

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#4

Re: Conservation of energy question (input vs output)

06/17/2007 12:24 AM

I am probably confused: The elevated magnet has multiple types of potential energy. 1 is based purely on it's mass and height. A stone of the same mass and height would have the same potential energy as the magnet.

Another potential of the magnet is the ability of the magnetic lines of flux to induce a voltage (potential energy) in a conductor. The amount of voltage induced is based on numerous factors including the density of the flux lines, the speed of cutting, and the angle of cutting. This electrical potential energy is completely different and has no reason to be equal to the first type. If you want more voltage, then decrease the air gap between the magnetic pole and the conductors, or pack more conductors in less space.

However it is interesting to realize that 2 magnets containing the same magnetism but different weights or mass should have the ability to generate the same electrical energy during free fall.

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#5
In reply to #4

Re: Conservation of energy question (input vs output)

06/17/2007 1:24 AM

Hi Mevel123, you wrote: "... it is interesting to realize that 2 magnets containing the same magnetism but different weights or mass should have the ability to generate the same electrical energy during free fall."

I don't think so. A magnet moving against a coil that is delivering electrical energy encounters magnetic resistance against the movement. This will slow down a lighter (free falling) magnet more than a more massive one.

Another way to look at it is that the larger mass that is elevated has more potential energy than the smaller one. Under otherwise identical conditions, the energy transfer will be proportional to the amount of potential energy.

-J

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#6
In reply to #5

Re: Conservation of energy question (input vs output)

06/17/2007 12:56 PM

Jorrie, thanks for your input. When I typed my last paragraph on my previous post, I thought to myself..."Sounds like free energy-lump me into the crazy group" But then I convinced my self that my statement was accurate as long as I kept the energy in a potential context. I thought that there would be no magnetic resistance (counter emf) as long as no current flows through the conductors; but equal voltages would still be measured regardless of the mass of the falling magnets. Obviously without current no kinetic energy is being transferred from the falling magnet, making the anomaly unusual, but not actually useful.

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#7

Re: Conservation of energy question (input vs output)

06/17/2007 1:52 PM

"Hi, I apologize if my question seems silly,....."

It is not a silly question but your lack of it being well defined in detail prohibits a definative answer. Try to restate you question with details of the setup.

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#8
In reply to #7

Re: Conservation of energy question (input vs output)

06/18/2007 10:54 AM

Wow, thank you all for the great answers, you really enlight me here.

As for not giving the exact configuration and thus being unable to provide a exact answer, sorry. I'm still figuring this out and would like to try it. By not having any engineering background it is hard to find the answers or basis as I walk around the idea. To my surprise I find myself with no help of friends as most studied things to win the class only... so they don't really have the knowledge to solve somethings (sorry if it sounds rude). So thats why I came here :) And I'm also reading some theories and laws around the web to learn.

I would like to have a configuration right now to provide the basis but I still don't build anything. I ordered a multiple set of NIB magnets to carry on on a series of experimental ideas I have. I live in Central America so most of the useful "toys" you find at radioshack, are not there at the stores.

Thank you very much for your time, I really appreciate it.

As for the questions I found on the thread:

HOMEWORK PROBLEM ??? for Physics or Electrical class???

Not at all, I have been enthusiastic about all this since I was a child and now having the web and some money to buy "toys", I find more interesting to experiment than to watch repeated seasons of tv series :)

I will keep reading. On an electronic forum I was told this idea relates to magnetic flux compression and microwave and pulse fundamentals and somehow it is interesting on theory but in practice doesn't resemble the formula too much. To me this sound like Chinese :) so I have more to read and learn now...

Thank you all for your answers!

------------------

On another hand, I'm totally curious about pulse motors, I saw the magnetic gate - callaway v track and a circular version (which doesn't work on a closed system, neither of them) I'll keep learning. Thanks

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Anonymous Poster
#9

Re: Conservation of energy question (input vs output)

06/19/2007 8:01 AM

Entropy..... You will never get back all that you put in. It is impossible.

To break it down, the energy the magnet has before it drops is Potential energy (mass x gravity x height) which was put in when you first lifted the magnet to the top. It is then converted to kinetic energy ( 1/2 x mass x velocity squared) as the magnet drops. The interaction between the magnet field and the coils then converts the kinetic energy to electrical energy. There are losses built into the system (friction, conversion efficiency, etc.) which result in energy being dissipated as heat or sound which can never be recovered.

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Anonymous Poster (2); hroldan (1); Jorrie (1); Mevel123 (2); Stirling Stan (2); user-deleted-1105 (1)

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