Joule Thief 101

[Magluvin]

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

[Magneticitist]

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

[poynt99]

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.

[Magluvin]

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.

Hey Poynt

Its just something that clicks in the back of my mind a lot. The back of my mind may not be correct, but the clicks keep coming.

Say if we have a capacitor of particular value capacitance, and we have no charge. Zero.  Now if we charge that cap to 10.0000v, cant we actually say there is a particular number of electrons taken from that positive and the same amount added to the negative in order for there to be a 10.0000v potential in that cap? For example, using imaginary numbers that for the cap to have 10.0000v there needs to be 1 million electrons taken from the positive and 1 million added to the negative.  Wouldnt those numbers stand every time we charge that cap to 10.0000v?

Soo, if we did that cap to cap transfer to 5.0000v each, wouldnt there be only 5 million count missing from each positive plates and 5 million count plus electrons on the negative plates, all vs caps with no charge?

So with that said, if we can associate and determine the voltage charge in the cap if we're able to count the electron differential between the positive and negative plates, then we will always have half the differential in each cap as we did in the 10.0000v source cap to begin with doing a direct cap to cap transfer.   So how, how is it possible to end up with 7.07v in each cap by direct transfer in theoretical superconducting capacitors, connections and switch, etc?  Wouldnt there need to be some electrons added to the circuit in order for that to happen? 7.07 million differential???   

Isnt it odd when you think of it that way?  Thats why I like the air tank analogy because the psi (at particular temp) can be determined by how many air atoms, to say, are pumped into the tank in the same way we can look at caps.  So if the air tanks were to do a direct tank to tank transfer till equalized, how could we start with 100psi in the source tank and end up with 70.7psi in each when done? Eliminate losses, and how did we get the extra air atoms/molecules that we didnt start with?

May sound nutty to others, but to me it sounds correct. Maybe you guys can help me understand this. But I think its going to be a tough cookie to bake.

I hope that what Im saying here makes sense. I reread it a couple times.  Its as good as I can put it to ya.

Mags

[Magluvin]

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