Joule Thief 101

[minnie]

  If you Wiki "Compressed air energy storage" there is quite a bit of useful info.
  One method seemed to have near 100% efficiency theoretical (70%in practice) very
  much inline with Tinman's claim.
                  John.

[Magluvin]

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

[tinman]

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



MileHigh

It's pure sleaze for you to pose a question and then pull a bait and switch where you draw extra air into the destination tank with a venturi without stating that in the first place.  You have no shame and you were too lazy or sleazy to even define the question properly.

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.

And I will concede that you are most likely correct about the standard air tank test and I am wrong.

Yes you are.

 

This is a setup where the environment itself exchanges energy both ways with the two air tanks whereas for two capacitors there is a one-way exchange of energy with the environment, the heat is considered lost.

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

When the pressurized tank is discharged into the empty air tank, energy is lost to heat in the valve, the pressurized tank does work on the unpressurized tank, the pressurized tank draws heat from the environment, and the unpressurized tank starts to put heat into the environment.  So this is a thermodynamic problem and accounting for everything is pretty tricky when this is not your forte (like me).  I was silly by forgetting about the thermodynamic angle and the energy exchange with the environment.

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.

<<< When the same tests were carried out with the venturi in play,the results showed an increase of 16% of stored potential energy in the two tanks. >>>

Perhaps, but now you are playing in the big leagues and you can't cherry pick if you are going to be thorough.  Energy had to be taken out of the external environment to do that, so there is no real energy gain.

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.

I am not even going to try to work out the specifics for the two tanks and will take your word for it.  I would not even consider the venturi example, it's just a stupid bait and switch on your part.

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.

It just happened again, yes.  But you seriously would not want to have an independent audit of errors in this thread made by both of us.  Because of your OCD, your head would explode if all of your errors were pointed out to you.  For Christ's sake, you read your own quote with a jarring spelling mistake and five minutes later you forgot that fact or it didn't even register in your brain that you had read your own quote.  So you end up accusing me of your own spelling mistake - five minutes after you read your own mistake!

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?.

I did not lie and it's not my fault that your brain cannot process information properly and understand what I stated in my posting.

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

[tinman]

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

[MileHigh]

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

That has got to be about the 500th inane and ridiculous reply from you.  Go see your doctor and get Thesaurus injections.

<<< How is your answer to your question coming along? >>>

Fully answered in post #2372, unless you are really the scarecrow.