Charging a capacitor without loss

[exnihiloest]

I went through this a while back.

If we had 2 air tanks, say 1 full, 1 empty, connected together via hose to each tanks valve, valves closed, then we open the valves till the pressure levels out in the 2 tanks from the full one.

Now, if we were to measure the amount of energy that 1 full tank could provide to say an air driven motor from full till empty, and compared that to measuring the output of the air motor running on the 2 half full tanks, measuring the total for 1 tank and then the other and adding, do we get the same amount of energy from our air motor using the full tank till empty, and from the total of the 2 half empty tanks?  

Did we lose half of the energy stored in the full tank, when we connected the empty tank to the full tank and leveled their pressures?  
And when we level out the 2 tanks, is there a loss in the system because of heat produced by the transfer in the system?  

The parallel between the two capacitors paradox and the two tanks problem is interesting and there are real similarities nevertheless we are not exactly in the same conditions.

Firstly you must give the way according which you fill one tank with the other: is it adiabatic or not? You may have noticed that I used the term "adiabatically" to qualify the charge of the capacitor from a current source. This meant that we "filled" the capacitor step by step, electron by electron, without creating a wide potential difference. In this case we avoid heat and entropy increase, and this is the interest of the above mentioned method. For pressure, it's the same. If your filling is not adiabatic, as the charge of a capacitor from a voltage source like from another capacitor, you will produce heat and so, loss energy.

The second point is that you suppose that your experiment is made in an environment which is neither at the absolute zero nor at a zero pressure. You must firstly obtain the pressure difference with a work exerted from a start point including the ambient temperature and pressure. The initial energy is this work. At the end of your experiment, the two tanks would be at the same pressure, but the possibility to extract energy from this state depends on the difference from the ambient pressure unlike the case of the capacitors that can be fully discharged in any case whatever the ambient potential because they are dipoles while your tank is like a monopole.

...
If we had 2 large caps, 1 full and the other empty, and we put a light bulb in line, we would get light and heat, and the caps will still be half the original voltage, but we made use of the transfer. Or a motor. Or a JT.

That's correct. When we charge a capacitor from a voltage source we could use half the energy that otherwise would be wasted. It's an other option than the goal of the proposed method, which is to move without loss the energy from a capacitor to another one.

...
Now, it is said, that if it were all supper conducting, that the caps will be more than half of the original voltage. But I dont know.

The caps wouldn't be more than half of the original voltage (see the arXiv papers, especially this one involving thermodynamics).  
If you led the experiment with surperconductors, there would be an infinite current, incompatible with an experimental setup. So what happens in reality?
In fact in this case, if the conductors don't fuse and the strong magnetic field doesn't collapse the superconducting state, then you have to take into account the inductance of the circuit, because it becomes no more negligible in comparison with the circuit resistance. In the real life, all circuits have an inductance. The time constant due to the inductance will limit the current, working exactly as the circuit with the coil, the coil being replace by the natural inductance of the circuit.

[TinMan]

I have done this.  It is possible to transfer energy without loss.  For example, 1/2CV^2.  If you have 10V 1mF cap, you can actually convert it into .32V 1F cap. 

Now I have read Faraday law of electrolysis.  It basically say that the amount of hydrogen produce is proportional to the electrical quantity Q(charge).  Does this means .32V 1F cap would produce more hydrogen than 10V 1mF cap? 

Ok, so you started off with 50m joules of energy-and you ended up with 51.2m joules of energy.
And this was transfering from one cap to another?
So not only didnt you have any loss-you had a gain in energy at the end?

Was the system a closed loop,or was there an out side energy source aswell?

May we have the schematic to the circuit that achieved this?

[GibbsHelmholtz]

hi Tinman,

You can see the extra 1.2 mJ of energy as experimental error or uncertainty. 

[ion]

Ok, so you started off with 50m joules of energy-and you ended up with 51.2m joules of energy.
And this was transfering from one cap to another?
So not only didnt you have any loss-you had a gain in energy at the end?

Was the system a closed loop,or was there an out side energy source aswell?

May we have the schematic to the circuit that achieved this?

Looks to me like there is not enough precision in the answer of 0.32V, it was probably rounded up error, should have been 0.3162V.

Small errors become large errors when the square function is invoked. Why precise measurement is required.

A low cost DMM will not give the degree of precision nor accuracy required.

[GibbsHelmholtz]

I've just made a setup to test Faraday electrolysis. 

First test:
     
   - A 4700uF is charged to about 30 volts and discharge into water.  Gas level is observed.

Second test:

   - Four 4700uF in series is charged to 30 volts each (30x4= 120V) and discharge into water.  Same gas level observed.

In conclusion, Faraday law of electrolysis is correct that gas is proportional to the charge Q.  It is also interesting to note that even the energy ratio is 1:4, the gas ratio is still 1:1.  According to this, hydrogen can be generated in unlimited quantity. 

[Magluvin]

Mags what can I say, you have said it all

Mike

Thanks. I say things that I regret later when it comes to these things, but I stick to this one like glue for some time now.

But what does it mean, if it means anything? If Im right, did 'they' get it wrong? Or is it that things are not what they seem on purpose?


Today I read an article about how the milk industry wants to be able to use aspartame and such in milk products, WITH OUT HAVING IT ON THE LABEL!!  Why?  Does milk and butter need sugar? Let alone an artificial sweetener?  Does gmo milk not taste so good without an added sweetener? The Yogurt companies surly want it omitted from their labels, as the people flock to more  brands that just use sugar.

Is aspartame cheaper than sugar? Is it safer? If its safe, then why all that magic acts to hide it? Why use it at all? Subsidies?

If this is what is happening to milk, then why is it so hard to believe that electronics theory may have a few secrets that 'they' dont want everyone to know about?


Dats what me thinks. 

Mags

[muDped]

Today I read an article about how the milk industry wants to be able to use aspartame and such in milk products, WITH OUT HAVING IT ON THE LABEL!!  Why?  Does milk and butter need sugar? Let alone an artificial sweetener?  Does gmo milk not taste so good without an added sweetener? The Yogurt companies surly want it omitted from their labels, as the people flock to more  brands that just use sugar.

Is aspartame cheaper than sugar? Is it safer? If its safe, then why all that magic acts to hide it? Why use it at all? Subsidies?

It could be all of the above, except the "safer"
possibility.  Milk already contains proprietary
(non-revealed) ingredients which are needed to
extend its shelf life, to keep it from souring
too early.

I suspect the aspartame would tend to mask the
souring in its early stages.  Milk loses its sweetness
as it begins to sour - before it begins to smell.

One thing we know for a certainty - it has nothing
to do with benefiting the health of the consumers.

If this is what is happening to milk, then why is it so hard to believe that electronics theory may have a few secrets that 'they' dont want everyone to know about?

Dats what me thinks. 

Mags

Aye, there are quite a few "secrets" having to do
with electronics and electronic devices which "they'd"
rather keep hidden from the people.

[GibbsHelmholtz]

Additional experiment shown that hydrogen production is independent of the circuit resistance.  Adding resistance to the circuit only slow down the discharge rate and not limit the amount of charge passing through water.  This is in line with hydrogen production is independent of energy.  I would say that it depends on the amount of charge passing through the water.  If resistance is what taking away the circuit's energy, then hydrogen is the product of reactive power. 

[GibbsHelmholtz]

Couple more test:

-   When four 4700uF in parallel (30V) and discharge to water, the gas level is much higher than in series (120).  Again, the same energy level in both cases but different amount of gas. 

-   Additional water cell in series seems to double the production.  I would say the more cells in series, the more gas produced for the same amount of discharge.

Next step is to find out how much hydrogen needed to set unity.

[GibbsHelmholtz]

Some numbers scavenged from the net.

240 kJ/mol of hydrogen 
Faraday constant = 96000 C/mol

There are two grams in a mole of hydrogen so we probably need 2 moles of electrons.  That equates to about 192,000 Coulombs.  That would gives 1.25 Joules/Coulombs.

1.25 = .5V
V = 2.5 volts

Double check:
If we have 1 Farad 2.5V, energy = 3.125 J
Q = 2.5, E = 2.5 x 1.25 = 3.125 J

If we have 5F 2.5V, energy = 15.625 J
Q = 12.5, E = 12.5 x 1.25 = 15.625 J

So there seems to be a threshold to each cell's voltage.