Ok. After the 2 posts I made on the cap to cap issue, here is the clincher....


When I was arguing this back when, we agreed that a 10uf cap at 10v has equal energy to a 20uf cap(2 10uf in parallel) with 7.07v

So like I said earlier, if we dump a cap directly into another, 10uf 10v into another 10uf cap, we end up with 5v each. Same comparison to 100psi air tank into an identical tank we have 50psi each.  It was said back then that if the capacitors were void of resistance(and inductance) as in superconductor capacitors, then the full cap at 10v were dumped into the other identical capacitor, we would end up with 7.07v each. I strongly disagree!

Its funny that 10v cap  dumped into an identical cap ends with 5v each, similar to 2 air tanks doing the same starts with 100psi in 1 and 50psi in 2. Seems very logical, right?

Well if we measured the weight of the mass of the air in the tank at 100psi, would that not be the weight of the air in both 50psi tanks total? If we lost energy in the transfer, did we lose any air mass weight in doing so??? So if we were able to count the excess electrons in the 10v cap, and then measure the excess electrons in each cap at 5v, would the total be the same? Do we have the same amount of excess electrons total?? If not, where did we lose some of the excess electrons? In heat? Did they ride off on the heat wave into the sunset?

So to say that if we eliminated the so called resistance loss in the transfer from cap to cap, how could there be more excess electrons added to the system if each cap is 7.07v of excess electrons each??  Where did the extra electrons come from??  I say it isnt the case. I think the superconducting caps would end with 5v each and we just lost energy by releasing the total voltage(pressure) by letting it be expanded into a larger capacitor value of twice the source cap.

Same with the air tanks. If we eliminated the so called heat loss in the transfer, would we end up with 70.7psi in each tank with a direct transfer?? Well where did the extra air mass weight come from to do such??

Now, we didnt do anything, as in usable work during the transfer of each example. So in a way, 'they' can say that resistance within the transfer creates heat(work) and that it is a loss, if heat is not the desired outcome. But to me thats just an odd reasoning to explain why we didnt get 7.07v in each cap. There are only so many excess electrons in the full cap, and that number is equal to the total of both caps excess electrons after full transfer. When I say excess electrons, it is just an example of the charge difference between the + and - plates.. Could be called electron imbalance of + and- plates. Just easier to say excess electrons. You get what Im saying.

The real loss is pressure. Sharing a full tank with an empty tank reduces the original pressure into twice the volume(2 tanks). The amount of energy it took to fill the full tank to 100psi is more than it would take to fill a tank of twice the volume to 50psi.

Now we do a cap to cap transfer using an inductor between them and we cut off the source cap at 7.07v. The inductor is charged at cutoff and with a diode freewheels its stored energy into the second cap that was already charging during the inductor charge up. If the diode were 0v drop, we would end with 7.07v in each cap.....
Now again, if we count the number of total excess electrons, do we have more total excess electrons in the 2 caps than we started with in the source cap?? Yes!  Where did they come from?  They came from the + side of the receiving cap forced into the - side of the receiving cap via the inductor discharge. They were pumped from the + side of the receiving cap to the - side by the collapsing filed of the inductor.

Same with the air tank. If we use the air motor/pump with a flywheel, and we cut off the source tank at 70.7psi, then let the flywheel pump in 'outside air' into the receiving tank, we end up with a total of more total air weight mass in the tanks. More than we started with in the source tank.

So to me saying that we lost half the energy of the source cap via resistance heat losses by letting the 10v charge of one cap simply expand into another identical cap ending up with 5v each doesnt make a lot of sense. It was stupid losses in my book.

Now. The real question is why 'they' explain the loss as a resistance heat loss when it simply doesnt seem to be the case? What is it that may be hidden in that riddle?

Mags

Indeed John,

Lots of changes with those things over the years.

Don't forget that according to the law of conservation of energy there can only ever be unity,, no over and no under.

When Brad opens the system up to the outside environment,, then there is another input potential added.  The gain comes from the unit quantity of air stored within the fixed volume of the tanks.

hint to MH

That is correct.
And the environmental energy drawn into the system was due to the stored energy in tank A-the pressurized tank at the start of the test,and it was drawn into the system without loss,and resulted in a gain of potential energy.
Can i claim the OU prize now?
What if i can raise 10KGs of weight 1 meter, using only 70 joules of energy?--could i claim the OU prize then ,as we would then have close to 100 joules of potential energy stored in that raised 10KG weight.


Brad

author=MileHigh link=topic=8341.msg483071#msg483071 date=1462464689]



MileHigh


Another lie.
I stated-as mag's did,that no potential energy is lost during the tank to tank transfer.
After  !yes-after! that statement,i added the venturi into the equation--after MH--after.

Yes you are.

 
With the capacitors,the missing energy is dissipated as heat--not lost to heat,and as radiation.

As soon as the pressure starts to drop in tank A,it will start to draw in environmental heat energy. As soon as tank B starts to pressurize,it will start to dissipate the same amount of heat energy to the environment.

As the tests were carried out within a 5 second period,and the tanks were insulated against environmental heat gains and losses,the tests can be considered an isolated test from any environmental impacts or energy factors.

And that there is a horses ass understanding of what just took place. The energy from the environment did not just fall into the tank by itself. The energy stored in tank A is what was responsible for the energy increase. The energy in tank A did work against the environmental energy available outside of the DUT,and it did it without loss.

You dont have to try and work it out,as MarkE already done this.
The only reason you say it is a bait and switch bluff,is because you got it wrong,and you done your slim pickings from each post i made,jumbled them around(as you do often),and once again lied about what i stated.

I thought you were here to correct all my spelling mistake's,as it seems to bother you greatly.
Were you able to read the text in that picture i posted?.

No MH,your postings have been quite clear,and you quite clearly stated that half of the potential energy stored in tank A would be lost during the transfer to tank B.
You are wrong--again.

How is your answer to your question coming along?. The one about the ideal inductor and ideal voltage.?.

Brad

"With the capacitors,the missing energy is dissipated as heat--not lost to heat,and as radiation."

I can agree that there is heat created via a cap to cap transfer. But I believe that heat is due to the high current flow, and the resultant equalization of full cap to empty cap IS the reason for the energy loss. We have let the high pressure convert to a low pressure haphazardly. If there were no heat developed in any way, there would still be half the source energy in the total of the 2 equalized caps. If the caps were superconductive, would the energy in the 2 caps equal the energy of full one before the transfer?    Think on it a bit. Not saying Im right and your wrong. Just trying to get my view understood.  Asking questions to see what you think.


"As the tests were carried out within a 5 second period,and the tanks were insulated against environmental heat gains and losses,the tests can be considered an isolated test from any environmental impacts or energy factors."

Did you do a tank to tank test?  If so I must have missed it.   What is the outcome in pressure of the full tank and the results of each tank once equalized?



Mags

I don't understand your argument mags, re. the cap to cap discharge.

An ideal cap discharged into another ideal cap of equal value would yield half the energy in each, i.e. no energy loss. However, there would be an infinite initial current that would tear a hole through space-time.     Unfortunately, we can not have an ideal 'anything', so there is always a finite resistance in the connecting conductors, no matter how small, and it is there where half the energy is lost.

If you insert a high Q inductor between them, you can approach a lossless transfer.

I must be missing something because this has been a long interesting discussion I started reading when the subject was on the JT transistor switching at low mV but the cap to cap and air tank to tank metaphor seem to be more of an impossibly hypothetical situation.

The discharge of electricity as we know still deals with resistance, which is just an extremity and different label for capacitance and conductance in a way. When you discharge a cap into another cap, the losses seem to occur in not only the resistance in the actual conductors flowing the current from cap to cap, but the 'spark', and initial visualization of that resistance being that this current is moving at the speed of electricity. Once that initial connection is made, should that high amount of amperage persists over time, the conductors of course could heat up. Seems like a normal 'ideal'-ized argument.

If an air tank were to discharge into another air tank, we are dealing with an entirely different set of physics altogether aren't we? It's definitely an interesting thought about having some type of turbine driven by the pressurized air, which is not lost in the process but contained in another tank until equalized, requiring no extra energy added to the system to facilitate this process and have the turbine still spin which could be considered taking advantage of passing energy.
However there's still a great deal of energy that was required to pressurize that first tank.
All kinds of losses surely occur during the transfer of that pressure from one stressed metal container to another, along with the switch/valve, just like in electronic circuits.

Motorizing that pressure during it's release seems to only be going in reverse, since you are providing a back pressure that is ultimately working against the energy you transferred motoring the pressure into the tank in the first place.

Anywho to compare that with a capacitor discharge into another capacitor does seem like a very similar kind of situation except now my brain is hurting trying to figure out what would happen if that air could travel at the speed of light initially while trying to equalize. would it 'backspike' at some point? lol

"If an air tank were to discharge into another air tank, we are dealing with an entirely different set of physics altogether aren't we?"

Are we dealing with a different set of physics?  Read my previous post above this one.

Mags

@Mag

you know it's definitely interesting thinking about it, I just don't get how it could mathematically be worked out that there is some gain when placing a charged capacitor in parallel with a discharged capacitor. The air tank analogy is definitely similar, but I don't think we can say that a battery, or capacitor, or any electrical power 'source' is under the same kind of pressure. it's actually a good analogy for voltage I guess, like water pressure in pipes, but when the charged caps plates coalesce with the neutral cap isn't it just a balancing-out transference at electron speed, and not the result of some form of actual pressure like that of air? when it comes to counting electrons, isn't the 'empty' cap more like a container that already has electrons in it just like the empty tank has air in it before being fed pressurized air from another tank? In ideal terms where we have a 100% transference of electrical energy from one place to another then it seems like we answered our own question. If we are talking about a current flow happening between one cap to another, and assuming there is 0 resistance, then if we're going with the flow and also assuming current will actually pass over 0 resistance in the real world then I don't think a 10v cap would have to balance out to 5v to have 'equalized the pressure' so to speak.

Okay, so now let's discuss an ideal situation where we end up with 7.07 volts in each cap.  It's going to be essentially identical to the case with the simplified model of the air tanks and the ideal flywheel pump pumping in extra air molecules.

Assume Cap A has discharged to 7.07 volts.  You know that current is still flowing, and therefore there is energy in the ideal coil connected between the two ideal caps, and there is also some energy in Cap B.  Like I said before, we don't need to know the size of the coil, or the amount of current flow, or the voltage in Cap B.  The most critical thing is that we are at the instant in time were Cap A is at 7.07 volts.

So what we have to do is instantly change the configuration of the circuit at this instant in time to realize the goal of 7.07 volts in each cap.

So the new configuration is this:  Cap A is completely disconnected from the circuit, and the inductor is across Cap B only with an ideal diode.

So what will happen is that the inductor will discharge all of its energy into Cap B and you end up with 7.07 volts in Cap B.   That is the "ideal flywheel air pump."  The ideal inductor supplies the extra electrons.  You have to remember that an ideal inductor can supply an infinite amount of electrons into a zero ohm load.  So in effect the ideal inductor "pumps up" Cap B to 7.07 volts.

Of course, we did an "impossible switching event" where we reconfigured the circuit and put the inductor across Cap B to make this all happen.  Such is the beauty of a thought experiment using ideal components.  The critical point being that you need extra electrons to reach 7.07 volts and the inductor supplies them by sucking them off of one plate and depositing them on the other plate.

It's also worth noting that without this "trick" then you can never get 7.07 volts across each cap.  Even in the ideal case we are discussing, charge has to be conserved.

MileHigh