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Overunity Machines Forum



Silly question about capacitors

Started by dieter, February 14, 2014, 10:48:52 AM

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0 Members and 3 Guests are viewing this topic.

Farmhand

I fail to see how anyone could not agree that if you lower 5000 liters (kg's) of water closer to the ground you would not be lowering the (GPE) of it. Energy must be dissipated, but it
could be used partly for other purposes, exactly like the principal of the Hydro electric power generator, just one example.

As with the low loss series inductor the storing of some of the energy in the magnetic field could increase the efficiency of the transfer. Like storing momentum. Or as you like to say MileHigh "like a flywheel".  The thing is that when I transfer water and equalize rain water tanks so I can catch more water, I know I am reducing the "head" of the water in the fuller tank, but it doesn't matter because I only want the water. And the rain will refill the tanks and recharge the "head" of water for free.  ;D So it makes sense to "just do it" when the time is right, there must be a reason to do a thing or it is 100 % waste.

Cheers

Trust me, I don't pump water up high so I can let if fall under the force of gravity to use it.  ;) I catch it up high if I want to use it by gravity force.

..

MarkE

Quote from: Farmhand on February 15, 2014, 12:51:27 AM
I fail to see how anyone could not agree that if you lower 5000 liters (kg's) of water closer to the ground you would not be lowering the (GPE) of it. Energy must be dissipated, but it
could be used partly for other purposes, exactly like the principal of the Hydro electric power generator, just one example.

As with the low loss series inductor the storing of some of the energy in the magnetic field could increase the efficiency of the transfer. Like storing momentum. Or as you like to say MileHigh "like a flywheel".  The thing is that when I transfer water and equalize rain water tanks so I can catch more water, I know I am reducing the "head" of the water in the fuller tank, but it doesn't matter because I only want the water. And the rain will refill the tanks and recharge the "head" of water for free.  ;D So it makes sense to "just do it" when the time is right, there must be a reason to do a thing or it is 100 % waste.

Cheers

Trust me, I don't pump water up high so I can let if fall under the force of gravity to use it.  ;) I catch it up high if I want to use it by gravity force.

..
Farmhand, if one wanted to extract the energy that otherwise goes to heat then one would need to devise a machine that has a highly variable impedance.  Throughout the transfer the machine needs to represent an impedance that is proportional to the difference in head between the two tanks.  At the start of the transfer, that is not too difficult as the difference is quite substantial.  Near the end of the transfer, the machine needs to present an impedance that is almost zero.  One might imagine a wide water wheel with a tapered diameter that slides along its axis, positioned by a lever connected to floats in each of the two tanks.

Farmhand

Quote from: MarkE on February 15, 2014, 01:12:41 AM
Farmhand, if one wanted to extract the energy that otherwise goes to heat then one would need to devise a machine that has a highly variable impedance.  Throughout the transfer the machine needs to represent an impedance that is proportional to the difference in head between the two tanks.  At the start of the transfer, that is not too difficult as the difference is quite substantial.  Near the end of the transfer, the machine needs to present an impedance that is almost zero.  One might imagine a wide water wheel with a tapered diameter that slides along its axis, positioned by a lever connected to floats in each of the two tanks.

I'm not wanting to design a machine. I'm just saying that for people to say half the energy is lost as heat and that is that is not telling the entire story I don't think.

Going back to the electrical model, how much energy is lost in the circuit I posted with a load when the switching is done so as to get good use of the inductor ? As compared to not using a load and an inductor ?

One of my points is that if there is a good reason to do it then the loss is regrettable but acceptable. If there is no good reason then it is just an experiment to show the loss of a pointless procedure. Like i said I only want the water but if I wanted to I could lower the water and cause it to do work on the way down but not much that would be worth the effort.

I made no claims, I'm simply trying to explain it in a way that people with a lesser training can understand better. Like the bricks analogy. Why make one big stack if you want two small ones anyway ?

..

A good example of Power is not Energy and Energy is not Charge ect. ect..


dieter

Wow, the lots of replies indicates that there is still a little bit of a paradoxon going on.


Personally I find it best to be compared to water. The deeper you get, the higher the pressure, in an exponentional amount, probably just like a cap: ^2.


Take a full waterbottle, make a pinhole near the bottom and see how the pressure is constantly reduced. Even atlhough there is more pressure-energy stored in the 8 to 16v, it can only flow to cap B until both have the same pressure.
Still having problems with the heat dissipation. Is gravity heat dissipation? Considering the need of flow from + to - (or in reverse, depending on popular naming) is just like gravity, I am not sure if you can explain gravity by heat dissipation. What is gravity anyway, as it is opposite to conservative models of centrifugal force.


As we see, we can utilize the energy flow with a load, but this will slow down the pressure equalisation. Now, does that mean we can lower the energy requirement to fill cap A to 16v when we allow more time to be "consumed"?

MarkE

Quote from: Farmhand on February 15, 2014, 02:12:29 AM
I'm not wanting to design a machine. I'm just saying that for people to say half the energy is lost as heat and that is that is not telling the entire story I don't think.

Going back to the electrical model, how much energy is lost in the circuit I posted with a load when the switching is done so as to get good use of the inductor ? As compared to not using a load and an inductor ?

One of my points is that if there is a good reason to do it then the loss is regrettable but acceptable. If there is no good reason then it is just an experiment to show the loss of a pointless procedure. Like i said I only want the water but if I wanted to I could lower the water and cause it to do work on the way down but not much that would be worth the effort.

I made no claims, I'm simply trying to explain it in a way that people with a lesser training can understand better. Like the bricks analogy. Why make one big stack if you want two small ones anyway ?

..

A good example of Power is not Energy and Energy is not Charge ect. ect..
Farmhand you can and should answer your first question by replacing the load resistor with a capacitor of like size as the input capacitor and operating the circuit until the input capacitor depletes to 50% of its starting voltage.  For your machine more than 50% will be lost, because you will have the switching losses on top of the transfer losses.

If your intent is to retain the 50% energy, then you need a third energy store to hold that energy. 

This particular problem is one of energy, so we have to stick with accounting for energy:  beginning, middle, and end.