electricity storage: graphene ultracapacitor-600Wh / kg!

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danielj
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Graphene super-capacitors




by danielj » 25/11/14, 11:42

Hello, hello everyone. Efficiency of 50% charge of a capacitor,
I answer and demonstrate that the yield is indeed 50%. I take again the graph of 'dirk pitt'

dirk pitt wrote:then ONCE FOR EVERYONE and this is the last one because I feel it is starting to roll. YES the 1 / 2CV² formula is valid all the time. it is like 1 / 2mV² for kinetic energy or what do I still know but NO the energy stored in a constant current charged capacitor is not half that injected. it is the same as that injected minus the low joule losses of the circuit, in particular ESR of the capacitor.
we can estimate ESR elsewhere from the deltaV at the start of charging and beginning of charging.

here is a typical curve of charge then charge at constant current. apart from the small vertical offsets, the area under the voltage curve is almost the same at load and unload, right? However, since I is constant in charge and discharge, the power is therefore also almost the same in charge and discharge AND as the charge and discharge times are the same, the energy recovered is very much the same as that injected. at least not half. QED.

Image


and I say, follow carefully:

I take the graphic provided by 'dirk pitt'.

-1) To have a current (even constant), you need a non-zero voltage. If zero voltage, zero current, ok ...

-2) On the graph, we see the constant current supplied by an electronic circuit, this circuit must be supplied by a voltage energy supplier U (take 1 volt for simplicity). It is he who ultimately provides the energy to the capacitor, ok ...

-3) during the charging time (1 second to simplify) of the 1 amp constant current capacitor (C) (to simplify the calculations), the voltage rises linearly across its terminals. Okay.

-4) The power given by the energy supplier is U (ct) x I (ct) or P. The energy (watt hour) is therefore P x ​​T (charging time) = W = 1v * 1a * 1s = 1 watt..ok.

-5) The power absorbed by the capacitor during this time is u (function of t) x I (ct). The power absorbed by the capacitor is also a variable as a function of time t. It is also a line which is superimposed on the line (inclined) of the voltage u across the terminals of C. I took 1amp, and Umax1volt).

-6) The energy stored by C, is: 1 Amp x average voltage of ux T (charging time). The average voltage of u is easy to calculate; it varies linearly from zero to 1 (v) or 0,5 volts!

-7) The stored energy is therefore 1amp x 1 sec x 0.5 volt or 0.5 Watt.
It is also the area of ​​the surface of the right triangle under the line u (ft) and the abscissa) so half the rectangle under the line U (ct) x I (ct) ...

-8) to compare with the 1 Watt given by the energy supplier in §4 ...

The yield is therefore 0.5, or 50% ...!

I can not break down and simplify more to explain!
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Re: Graphene super-capacitors




by Gaston » 25/11/14, 11:55

DanielJ wrote:-2) On the graph, we see the constant current supplied by an electronic circuit, this circuit must be supplied by a voltage energy supplier U (take 1 volt for simplicity). It is he who ultimately provides the energy to the capacitor, ok ...
It is on this point that the error of reasoning is situated.
This circuit must be supplied by an energy supplier, of course, but its voltage does not have to be constant over time (for a constant current charge, it is even impossible).

DanielJ wrote:-b] The yield is therefore 0.5, or 50% ...! [/ b]
As long as we keep the constant voltage generator hypothesis, we arrive at this result : Mrgreen:
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Graphene super-capacitors




by danielj » 25/11/14, 12:31

Except that my friend, an electronics engineer, confirmed the 50% yield. using the digital simulator (SPICE, known worldwide), he simulated the circuit and measured the energies well, in accordance with theory, whatever the means of charge!

And then if you do not supply the current generator with a fixed voltage, what do you supply it with?

I am not trying to convince the faith of some, I am a scientist!
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Re: Graphene super-capacitors




by Gaston » 25/11/14, 14:27

DanielJ wrote: And then if you do not supply the current generator with a fixed voltage, what do you supply it with?
In this case, the entire internal diagram of the current generator should be taken into account.
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by Gaston » 25/11/14, 14:44

I too am a bit of a scientist.

So let's try to detail a little:

In a circuit comprising a voltage generator E, a resistor R and a capacitor C, we can write:

E = RI (t) + Uc (t) where I (t) represents the current in the circuit at time t and Uc (t) represents the voltage across the capacitor at the same time.

Furthermore, Uc (t) and I (t) are linked by the relation I (t) = C (dUc / dt) which expresses the charge of the capacitor.

By replacing I (t) with this expression in the first, it comes:
E = RC (dUc / dt) + Uc (t)

This differential equation, added to the initial condition Uc (0) = 0 has for solution Uc (t) = E (1-exp (-t / RC)).

So I (t) =
C (dUc / dt) =
CE d (-exp (-t / RC)) / dt

Since the derivative of exp (ax) is equal to a * exp (ax), the derivative of -exp (-t / RC) is 1 / RC * exp (-t / RC).

So I (t) = CE * 1 / RC exp (-t / RC) = (E / R) exp (-t / RC)

The energy supplied by the generator is therefore the integral of EI (t) dt = integral of E * (E / R) exp (-t / RC) dt.
As E and R are constant, we can get them out of the integral: there remains E² / R integral of exp (-t / RC) dt.

As the integral of exp (-ax) dx is 1 / a, the integral of exp (-t / RC) dt is RC.
So the energy supplied by the generator is worth E² / R * RC = CE².

The energy received by the capacitor is equal to
integral of Uc (t) I (t) dt =
integral of E (1-exp (-t / RC)) * (E / R) exp (-t / RC) dt =
E² / R [integral of exp (-t / RC) dt - integral (exp (-2t / rc) dt =
E² / R [RC - RC / 2] =
CE² / 2

The energy stored in the capacitor is therefore half the energy supplied by the generator.
In that case the yield is 50%.


Now let's examine what happens if we charge the capacitor twice.
The first time with a voltage generator V = E / 2.
The above calculation shows that the capacitor stores energy CV² / 2 = CE² / 8 and that the generator supplied energy CV² = CE² / 4.

Then connect this already charged capacitor at voltage E / 2 across a voltage generator E.
Taking the time of this connection as the new time origin, the equation of the beginning RC (dUc / dt) + Uc (t) = E is still valid, but the initial condition is now Uc (0) = E / 2 .
It then comes Uc (t) = E (1-exp (-t / RC) / 2) and therefore I (t) = (E / 2R) exp (-t / RC)

The energy supplied by the generator is therefore equal to:
integral of E * I (t) dt =
integral of E * (E / 2R) exp (-t / RC) dt =
E² / 2R integral of exp (-t / RC) dt =
E² / 2R * RC =
CE² / 2.

The generator supplied less energy because the capacitor was already partially charged.

In total, the capacitor has accumulated an energy of CE² / 2 twice.
The first generator supplied an energy of CE² / 4 and the second CE² / 2 for a total of 3CE² / 4.

The yield is therefore (CE / 2) / (3CE² / 4) = 2/3

Conclusion: the yield depends on the generator (s) used.
The value of 50% is only valid if a single fixed voltage generator is used and the other elements are comparable to pure resistance.
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by Gaston » 25/11/14, 16:24

For charging by a generator with constant current, we can even do without integral calculus:

The relation I = C dUc / dt gives us, since I is constant, Uc = It / C.

Charging stops when a voltage E is reached across the capacitor, so at time tc = CE / I

L'energy dissipated in the resistance during the charge is equal to RI²tc = RICE.

L'energy stored in the capacitor is CE² / 2 (it doesn't change).

L'total energy supplied by the generator therefore worth RICE CE² + / 2 (energy conservation).

And the yield is (CE² / 2) / (RICE + CE² / 2), i.e. after simplification by CE: E / (E + 2Ri)

Unlike the constant voltage load, this time the efficiency depends on R, and the lower the RI, the closer the efficiency is to 1.
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Re: Graphene super-capacitors




by chatelot16 » 25/11/14, 19:19

DanielJ wrote:Except that my friend, an electronics engineer, confirmed the 50% yield. using the digital simulator (SPICE, known worldwide), he simulated the circuit and measured the energies well, in accordance with theory, whatever the means of charge!

And then if you do not supply the current generator with a fixed voltage, what do you supply it with?

I am not trying to convince the faith of some, I am a scientist!
8)


well let's see if it's simulated with spice it's true :P

a constant current generator is a constant current generator ... no matter where the energy comes from

a constant current generator is not necessarily a linear current regulator which only limits the current coming from a constant voltage source with a transistor which dissipates power as stupidly as a resistance

a constant current generator can be a dynamo with the right regulation, which is supplied with mechanical energy, and which has a good efficiency despite the voltage variation

with cutting in an inductor we do everything we want with a maximum theoretical yield of 100% and a real yield as good as we want according to the quality of material that we put in it
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Re: Graphene super-capacitors




by Obamot » 25/11/14, 22:48

Two words on the methodology, if I may, something bothers me:

DanielJ wrote:Except that my friend, an electronics engineer confirmed the 50% yield.

In which case? And in the other? And with what variable / s of adjustment? And which circuit exactly? etc ...

If you are "scientific", you will agree that posing a problem in a certain way (rather than another) in order to propose to achieve the result that one hoped for in "asking to confirm it", is perhaps not the best way to check if other opposable hypotheses would be" good "(or not) it involves a big risk. Because indeed, it seems to me that under these conditions: the one who makes the simulation will inevitably fall on the result "hoped".

And even more precise, I would say it would be better to ask: "what value will we find with this or that scheme VS another type "? Without specifying what is expected ... Already so as possibly not to influence the experimenter on performance "expected"! And even more ethical still: some say"an anonymous request to an independent laboratory»Who does not know you (or to make sure that he cannot know from whom the request comes directly), to be sure of an impartial result, etc ...

Thus the result of the simulation could / should have given "values", and no "Confirmation / invalidation of an expected result". It smells of a lack of rigor ... scientific, in the way of presenting things to somewhere. So we wonder: where is the protocol / experimental plan in your explanation?

In the other case, we had to test the two hypotheses (to verify and / or invalidate / confirm them, one and / or the other), which does not appear either in the way whose things you are presenting. Since the result would have been: "Such / s value / s in one case VS such / s value / s in the other ..." And at the end of the day, you also had to give the result in the proposal that was opposed to you (this would have been a loyal and ethical approach amha ... and if so much is impartial.)

Ditto for all other parameters. So the value of that: no comment, but be careful to keep the best possible rigor in your approaches!

I'm not trying to convince either, but "the faith"can also possibly arise from pre-existing conviction / s or whatnot, based on acquired knowledge which would be no less so, but which for various reasons would have caused the paradigm to have changed since, precisely according to more up-to-date theoretical approaches ... (pure speculation on my part, just an idea like that ...) Yet I say that, I say nothing, I AM NOT scientific! So maybe I am wrong? even if it did, it wouldn't be a big deal if I had humbly learned something more today!

In any case, I am delighted to know shortly the epilogue of your discussions on the real coefs.

It gets full-bodied : Cheesy:

Sincerely,

8)
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Graphene super-capacitors




by danielj » 26/11/14, 15:25

Hello, hello everyone!
I see that 'gaston' the matheux 'obamo' the electro; 'chatelot' the tecno-philo; 'surivai' and 'dirk pitt' etc ... are good contacts!
'obamo' in his last post asked me to make a general demonstration, downright a course, there I can not, I do not feel capable! In addition, I do not have the ease of gaston in math!

And on all I did not plan to spend as much time blogging on this thread. I landed on this thread out of curiosity and when I read that the capacitors were trimmed of all qualities, I reacted by pointing out that the charge of the capacitors was 'normally' affected by a theoretical efficiency of 50%. .. Not everyone knew it seems to me ...
Note: the charge through a choke brings the same concern, and even more because there is certainly a risk of causing the current to oscillate so as to transmit nothing more!

That being said, If we have a "native" power supply, variable in voltage And delivering a constant current, can the theoretical efficiency improve?

I think so ! , because we get rid of exponential growth, but I can not demonstrate it here, I had not thought of it ...

The use of a dynamo controlled in voltage AND imposed current is a good example to achieve this; still it is necessary that the mechanical power of the drive motor can be properly adjusted. But the accumulation of yields, of the engine then of the generator risks compromising the desired goal! And above all the complexity, the weight, the cost risks putting this model in oblivion!

Capacitors were supercapacitors are effective to store small amounts of energy and to release it in a very short time, therefore with a very important instantaneous power
Voila!
600wh per kg at most, this is to be compared with a liter of rapeseed oil contained in a bottle (at 10 cents) which already provides 10 wh!

The regenerative energy when braking a car is done very well with current batteries, and this is sufficient (you should not brake too hard, otherwise you will lock the wheels).

Now, on the demo of Gaston from 25-11, I suspect a bug in the discharge of the 1st condo in the second, it has been shown (4 to 5 days ago) that in the case of discharging from one condo in another 50% of the energy is wasted, it's a textbook case!

Well I did not try to find where the error is (or not) because I am spending time right now on the construction of a hot water boiler using a "poly-fuel vaporization pot" (in particular of vegetable oil). See the thread in this site ...

I am above all a technician "jack of all trades" who DIY (in the literal sense of the word and without pejorative notions) and who does a lot of things ...

Well, to be honest I'm running out of time to run after all the "pots I have on the stove"

I let you continue your research and find an efficient and practical system for building, charging the capacitor ... I am interested!
We can then measure. ;)
Well I spent another hour with you ... a + :D
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Re: Graphene super-capacitors




by Obamot » 26/11/14, 16:15

Thank you for these details and for pointing this out Danielj. Rhôoo, the soufflé is falling! Already? Pity! : Cheesy:

... well, well: and a big tip of the hat and a thousand thanks to Dirk Pitt et Gaston for their time spent at "explain"them too, it would not eat much to tell them warmly: well, here it is: I did it! Ditto for Chatelot and his "practical good sense" (And between us I don't think there is anything wrong with the remarks of Dirk non-plus, which develops in particular charge / discharge control units for LiFePO4 batteries among others, and this for vehicles actually in circulation, which I have already had the opportunity to check what he is talking about at the Geneva Motor Show or one of his partner's vehicles was on display, we had discussed it with Bernardd and can be trusted: as to Gaston, hats off, without forgetting other "sizes" of the forum as Remundo, Etc.).

Score 2 at zero bullet in the center, I leave it to whoever wants to conclude: as it is, we can therefore say that ultracapacitors are not a big waste of energy (it depends ...) nor Has Been as we understand: so much the better!
Last edited by Obamot the 26 / 11 / 14, 16: 18, 1 edited once.
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