Storing Cookies (See : http://ec.europa.eu/ipg/basics/legal/cookies/index_en.htm ) help us to bring you our services at overunity.com . If you use this website and our services you declare yourself okay with using cookies .More Infos here:
https://overunity.com/5553/privacy-policy/
If you do not agree with storing cookies, please LEAVE this website now. From the 25th of May 2018, every existing user has to accept the GDPR agreement at first login. If a user is unwilling to accept the GDPR, he should email us and request to erase his account. Many thanks for your understanding

User Menu

Custom Search

Author Topic: Electrical igniter for gas engines A keystone to understanding by Magluvin  (Read 247304 times)

forest

  • Hero Member
  • *****
  • Posts: 4076
dump one capacitor into two empty connected in series, then in parallel
what is the result ?

poynt99

  • TPU-Elite
  • Hero Member
  • *******
  • Posts: 3582
dump one capacitor into two empty connected in series, then in parallel
what is the result ?

Specify the 3 values....then you provide the answer. ;)

.99

woopy

  • Hero Member
  • *****
  • Posts: 608
Thank's p.99

I am happy that we are OK on the first point.

Now i will go on structuring my mind.

When i look at "hyperphysics.com", they explain the stored energy in a cap with a comparison with the  stored energy in an air tank. And i can very very well visualize this. So the voltage in the cap would compared at the pressure in the ai tank.

So 2 caps of same capacity would be compared to 2 air tanks of same capacity.

So the first  air tank(cap) is  at 99 BAR pressure and the receiver cap is empty (or has only the atmosphere pressure that is  about 1  BAR).

There is a tubing with a valve between the 2 tanks,

So when i open the valve on the tubing. depending of the opening  the tanks  will more or less fastly equalise to 50 BAR each.

 So as the  pressure (exactly as the voltage  in energy stored formula for the caps) is SQUARED in the formuly 1/2 * C * V^2, so after the transfer, (of course if we operate carrefully ),  the pressure in the air tank is 25 % in each tank and together it is 50 % lost of energy as perfectly expected.

So we did not loose one molecule of air in the transfer, the quantity of air is the same but the pressure of this air is simply divided by 2 (or almost 2), and almost nothing was lost in the tubing.

 The transfer was almost perfect (almost 100%) but the pressure lost is comparable at the Voltage lost in the direct cap transfer. And the end result is almost 50 % lost in energy .

Does it make sense ?

Thank's

Laurent

poynt99

  • TPU-Elite
  • Hero Member
  • *******
  • Posts: 3582
Assuming that the air tank analogy is a good one, then yes the combined potential energy in the tanks is reduced by 50%. Keep in mind also, that Hyperphysics does not touch on the efficiency or amount of transfer issues.

I would stay away from thinking that the process was 100% as you say. This is what I was trying to say before as it only adds confusion to the discussion, imho.

.99

Magluvin

  • Hero Member
  • *****
  • Posts: 5884
If you had two ideal capacitors (ESR=0 Ohms) of the same value connected together with an ideal wire (Z=0 Ohms) and no diode, then the transfer of energy from C1 to C2 would be instantaneous (with no oscillation), and there would be no energy loss.

So if C1 started with 1.00V and C2 with 0.00V, the end result would be with both caps at 0.707V.

.99

hmmm, well this is interesting.

If in the "normal" world,   100v on one cap and zero on the other, if we make direct transfer, we end up with 50v on each cap. Each cap is 1/4 and if we put them in parallel we have a total of 50% of the total energy left. 50% gone.

Now we go to superconductive world.....

100v in a cap, do a direct transfer, we end up with 70.7v in each cap. We combine them in parallel, we end up with what, 100% energy left?

100v 10uf = 70.7v 20uf   So if we skimmed 20.7v from 70.7 20uf to end up with 50v 20uf, we lost 50% of the energy in the 20uf cap?

I suppose thats correct. Then the 50v 20uf cap, we skim 15v to have 350v 20uf, we lose another 50%.  Then skim 10v to 250v, another 50% loss.

Its the same as my chart through the 4 stages.

Its hard to fathom, but seems correct.

It seems like a far throw for 70v 20uf =100% where 50v 20uf is only 50%. But it is what it is.

So in the BC if we start at 1kv and we do a cutoff of 703v, which gives 703v in each cap, we are just close to ideal world conversion. We have a bit of loss from 707v to 703v.  So with the BC we are very closely approaching ideal world conversions, without superconducting properties.

ps  An ideal wire has no inductance in superconducting world?

Mags

Magluvin

  • Hero Member
  • *****
  • Posts: 5884
Thank's p.99

I am happy that we are OK on the first point.

Now i will go on structuring my mind.

When i look at "hyperphysics.com", they explain the stored energy in a cap with a comparison with the  stored energy in an air tank. And i can very very well visualize this. So the voltage in the cap would compared at the pressure in the ai tank.

So 2 caps of same capacity would be compared to 2 air tanks of same capacity.

So the first  air tank(cap) is  at 99 BAR pressure and the receiver cap is empty (or has only the atmosphere pressure that is  about 1  BAR).

There is a tubing with a valve between the 2 tanks,

So when i open the valve on the tubing. depending of the opening  the tanks  will more or less fastly equalise to 50 BAR each.

 So as the  pressure (exactly as the voltage  in energy stored formula for the caps) is SQUARED in the formuly 1/2 * C * V^2, so after the transfer, (of course if we operate carrefully ),  the pressure in the air tank is 25 % in each tank and together it is 50 % lost of energy as perfectly expected.

So we did not loose one molecule of air in the transfer, the quantity of air is the same but the pressure of this air is simply divided by 2 (or almost 2), and almost nothing was lost in the tubing.

 The transfer was almost perfect (almost 100%) but the pressure lost is comparable at the Voltage lost in the direct cap transfer. And the end result is almost 50 % lost in energy .

Does it make sense ?

Thank's

Laurent

See, this is why I feel that we didnt lose in resistance from heat production. Not 50% in heat losses.  I feel we lost 50% because of the containers vs pressure.

Like woopy says, did we lose 50% in the air lines due to heat losses?

Very good call woopy.  ;]

mags

poynt99

  • TPU-Elite
  • Hero Member
  • *******
  • Posts: 3582
I am not well versed in fluid dynamics so I'll let someone else answer that.

.99

poynt99

  • TPU-Elite
  • Hero Member
  • *******
  • Posts: 3582

ps  An ideal wire has no inductance in superconducting world?

Mags

A real superconducting wire still possess inductance, and it can be wound into an inductor.

The hypothetical situation I posted where the wire was ideal (i.e. Z=0 Ohms), is strictly hypothetical.

.99

Magluvin

  • Hero Member
  • *****
  • Posts: 5884
Assuming that the air tank analogy is a good one, then yes the combined potential energy in the tanks is reduced by 50%. Keep in mind also, that Hyperphysics does not touch on the efficiency or amount of transfer issues.

I would stay away from thinking that the process was 100% as you say. This is what I was trying to say before as it only adds confusion to the discussion, imho.

.99

I can sort of agree on this.  We shouldnt count the air molecules as a count of energy, just the pressure, the closeness of the air molecules or atoms, which ever element in involved is being pressurized.
But the count of air molecules or atoms can be accounted for hear also, as we didnt lose any in the process. The 2 tanks hold them all in total, we just lost pressure.

But if there are no heat losses in the air tank conversion, should there not be 70.7 bar in each tank after direct connection of a 100bar tank, to an empty tank?  If we only have 50bar in each, then where did our 50% go?

Was it lost in moving the air from 1 tank to the other? As we would need outside energy in order to re compress all back to 1 tank.

So I believe it is more of a container vs pressure that gives us these results rather than resistance heat losses.

On the other hand, if we had a way to use a flywheel to use the energy being transferred from tank to tank, instead of direct decompression from tank to tank, we could then get most of the pressure from the first tank to the second one.

I see that with the flywheel, we avoid the 50% loss by not wasting the energy put into the direct transfer that only ends up in 50% of the input. And I think that waste in direct transfer is just a container vs pressure deal, not a loss in resistances.

Mags





Magluvin

  • Hero Member
  • *****
  • Posts: 5884
I am not well versed in fluid dynamics so I'll let someone else answer that.

.99

It can be said that waves of sound can be considered fluid like. But fluids are not usually compressible, where a gas or electrical charge can.

Mags

Magluvin

  • Hero Member
  • *****
  • Posts: 5884
A real superconducting wire still possess inductance, and it can be wound into an inductor.

The hypothetical situation I posted where the wire was ideal (i.e. Z=0 Ohms), is strictly hypothetical.

.99

So if we had a non hypothetical situation, the 2 caps would in fact oscillate forever. And all we would need is a diode for complete and direct transfer. And with no resistance, this conversion would, in theory, happen instantaneously, with the only limit here being the speed of electricity through the superconducting medium.
Is the speed of electric transfer, from say NY to LA in a normal wire, with real resistance involved, move at light speed(or close) ?  Or does that speed change also in superconductors?

Mags

Magluvin

  • Hero Member
  • *****
  • Posts: 5884
Mag, that is an excellent question.  Are you talking speed of electrons, speed of EMF,  (Which is not correctly measured, but detected by electronswe motion.  See the problem...)  or speed of charge propagation, which, for the guys at MIT, is a hotly debated subject.

To me, at least, these questions are at the heart of the matter.....  I can only answer one, and that's the speed of current, or electron motion, which is snail slow.  The others I need help with, as it depends on detection methods.

Here is the best explanation of electron flow I can describe.

Sound waves are like ac current. the sound uttered from your mouth pressurizes air. The actual air particles around the area of your mouth dont make it very far, just the pressure of the vibrating air propagates to push and pull on the air surrounding your head and expands from there.

If we have a pipe from NY to LA, we could hear sounds from one end to the other, but it is only a pressure wave that moves.  Same as in AC lines, the electrons never make it from the power plant to the other end of the wires. They are just like a long row of people just doing the bump( disco dance =] ) at 60hz. Eventually that bump gets to the other end of the line.

At sea level sound waves propagate at near 750mph (this no. has changed over the years) In water, it is different. its faster.

So maybe there is a difference in the speed of electricity in a superconductor vs real world wire.

As in the speed of sound, a different medium will affect it, maybe in different mediums, electric flow is also changeable.

So if the sound wave pipe from NY to LA were to just have DC pressure, that pressure would take a bit of time to reach LA, but this time the particles are on the move, from NY to LA, up to speeds of 750mph.  Im not sure if any faster would cause the pipe to burst.

I tend to think that electrons can move very fast. just because AC only moves them back and forth in a conductor, doesnt mean they are not getting from here to there and back fast. The freq of the AC component would determine how far they go and come back.  =]

But maybe electricity has just one constant. Even through a resistor, the other end will feel the input at the speed of light, just not a lot of current flow.

Im just trying to grasp the superconducting advantages and limits here. =]

Mags




Magluvin

  • Hero Member
  • *****
  • Posts: 5884
Hey Loner

Any links to that Mit debate?

Mags

poynt99

  • TPU-Elite
  • Hero Member
  • *******
  • Posts: 3582
So if we had a non hypothetical situation, the 2 caps would in fact oscillate forever.
No, the caps have an internal resistance, so we are back at square 1.

Quote
Is the speed of electric transfer, from say NY to LA in a normal wire, with real resistance involved, move at light speed(or close) ?  Or does that speed change also in superconductors?

Mags
The velocity of propagation for a pair of wires (twisted for example) varies anywhere from 0.6c to 0.85c, depending on the characteristics of the wire and the wire jacket.

Because the length of the line is an appreciable portion of the signal wave length, here too we need to start thinking about the wires as transmission lines.

Now, the drift velocity in a superconductor? That's a topic for another day.

.99

woopy

  • Hero Member
  • *****
  • Posts: 608
Wow

always more interesting here

I go,  on the air tank theory  and assume and accept that, the 50% energy lost is a real fact and that's it.

But the lost of stored energy is not due to the tubing transfer  (or wire resistance heat lost in the case of cap transfer ), but simply to the DOUBLE VOLUME for the air to expand.  So the air pressue is simply divided by 2 in each tank. Seems very logical.

And always because of the SQUARED calculation of pressure (or voltage ) the result is mathematically and always giving a 50 % in energy lost    always because 1/2 * C * V^2  formula   and that's it.

And now the question is , how and why ,   in the Believe Circuit (BC) of Magvulin    it seems  possible to improve this  result ? I mean we really    in real life    do not lose 50 % of energy

I mean and it is a fact  by using the BC we can reach in real life i insist , a 82  % (with very basic circuitery ) and  more transfer energy ( see all the above post ),

For instance if i put an air turbine in place of the valve in the air tank transfering tubing, and this turbine should activate a flywheel during the transfewr , do you think that the stored energy in the flywheel could be strong enough to go on emptying the source tank to almost 1 BAR and to recompress this air in the receiver tank to almost (let's say 90 BAR )  ?

yep

sliping is needed

good luck at all

Laurent