Author Topic: Death of the Lead-acid battery  (Read 850 times)

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wdyasq

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Death of the Lead-acid battery
« on: September 09, 2007, 03:19:25 PM »
"An Austin-based startup called EEStor promised "technologies for replacement of electrochemical batteries," meaning a motorist could plug in a car for five minutes and drive 500 miles roundtrip between Dallas and Houston without gasoline."


OK, here's the link: http://ap.google.com/article/ALeqM5injP_H-HkCkxCFHZ0ryEBSrvGgWQ


I heard this in a news report while driving. I thought, 'WOW - how much power is that!'. So, I put the famous SWAG calculations to it (Strictly a Wild Axx Guess) while I was driving.


I guessed the trip would take 6 hours. And, I guessed 15HP or 10kW power to propel a vehicle down the road at cruising speeds. This totals 600kWh. Now, this much power will  need to be put in the vehicle in 5 minutes, one twelve th of an hour. So we multiply by 12 to get the transfer rate 12*6 = 72, add the zeros =7200kAh charge rate. How large of a wall-wart am I going to need?


Household current can be 240V so, I can use this formula, P=VI to figure the amperage needed. Lets see, I have P and V, I is needed; P/V = I. So, 7200/240 = 30kA. Remember, we are working in "k" or one thousand as our original guess of power was '10kW'. So, we need only 30,000 Amps for 5 minutes at 240V.


Well, the capacity of a 0000 cable is 302A, according to my chart. It will need 10 of these cables going in, and ten coming out. Man, what an extension cord! Of course, when I plug it in, the small 200A service on my shop would glow and pop like a flash-bulb.


For some reason, I don't think these folks are telling the truth....


Ron

« Last Edit: September 09, 2007, 03:19:25 PM by (unknown) »
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wdyasq

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Re: Death of the Lead-acid battery
« Reply #1 on: September 09, 2007, 09:26:37 AM »
OOPs ... I busted the math ... added a zero up front then took it off at the end ...


60kWh - 720kAh - 3,000A charge rate and still 20 X 0000 cable .....


More coffee ...


Ron

« Last Edit: September 09, 2007, 09:26:37 AM by wdyasq »
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stephent

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Re: Death of the Lead-acid battery
« Reply #2 on: September 09, 2007, 09:45:36 AM »
Interesting calculations Ron...

Wouldn't take that long a cable though, just use a direct plug in at the end of a garage parking space (drive car VERY careful!)

Use a 2 times size cap of the one in the  vehicle and pulse charge it (from grid/RE  source) for a few weeks--drive car/vehicle into garage--position VERY careful and RAM home the connection in low gear (30K amp receptacle is tuff to get plugged in ya know).

Then stepping outside the garage and into neighbors yard next 2 doors over--use remote control that throws the vacuum break(s) switches to get the juice into vehicle cap....the 2x size "charger" cap is so volts/charge equalizes at just vehicle cap capability in both after successful charge.....(left nitpicking math whizzes room on 2x charger cap size on purpose :)

i.e.--successful meaning you still have a vehicle or charger/house still there.

Entirely possible--see?

This is all advanced thinking and TIC of course... :)
« Last Edit: September 09, 2007, 09:45:36 AM by stephent »

finnsawyer

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Re: Death of the Lead-acid battery
« Reply #3 on: September 09, 2007, 09:48:59 AM »
I didn't bother to check your math.  The "five minute charge" is enough for me.  Considering that the energy stored in a capacitor is given by:


  E = 0.5xCxV^2,


how much capacitance C would be needed at say 200 volts?

« Last Edit: September 09, 2007, 09:48:59 AM by finnsawyer »

finnsawyer

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Re: Death of the Lead-acid battery
« Reply #4 on: September 09, 2007, 09:53:30 AM »
O.K., a little follow up here.  It says high voltage.  The question is how high, 100,000 volts?
« Last Edit: September 09, 2007, 09:53:30 AM by finnsawyer »

DamonHD

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Re: Death of the Lead-acid battery
« Reply #5 on: September 09, 2007, 10:27:38 AM »
My understanding was something of the order of 600V.


Rgds


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« Last Edit: September 09, 2007, 10:27:38 AM by DamonHD »
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kell

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Re: Death of the Lead-acid battery
« Reply #6 on: September 09, 2007, 10:41:31 AM »
6 hours is 72 times 5 minutes.

So the 10kW translates into 720kW charging rate.

at 240 volts that's 3000 amps.

Makes you realize the huge amounts of energy

we consume when we drive.

I mean, that Dallas-to-Houston commuter driving down the highway

eats up as much sheer power as

a number of large Houston homes in high summer with their

air conditioning.

Back on topic:

I'm not familiar with any supercaps that hold

a couple of hundred million Joules.
« Last Edit: September 09, 2007, 10:41:31 AM by kell »

feral air

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Re: Death of the Lead-acid battery
« Reply #7 on: September 09, 2007, 11:03:35 AM »
5 minutes I doubt. 5 hours I can believe. It looks to me like the 5 minute claim was made by the author of the article, not by the company doing the R&D.


That said; Batteries, super caps or peanut butter, I don't care how the charge is stored...


I just want an electric that can do 100 miles on a charge, with swappable/stackable storage. That way you could get more batteries (or whatever) and only plug them in when you really do need to go 300 miles. My normal commute is 30miles round trip...why carry the extra weight?


I think they're too focused on "300+ miles". Give me 100 miles and I'm happy....make it stackable to satisfy the people that "have" to be able to drive 300 miles or more.


...my 2¢.

« Last Edit: September 09, 2007, 11:03:35 AM by feral air »

joestue

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Re: Death of the Lead-acid battery
« Reply #8 on: September 09, 2007, 11:14:08 AM »
I believe they are talking about a 3000 volt, 31 farad, 336 pound, ceramic capacitor.

The idea was good, take known super capacitor dielectrics that have that magnitude of capacity, and then claim that they could get the breakdown voltage to 1700 volts.


Of course it didn't work

« Last Edit: September 09, 2007, 11:14:08 AM by joestue »
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Nando

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Re: Death of the Lead-acid battery
« Reply #9 on: September 09, 2007, 12:18:20 PM »
Few days back the same type of thread appeared in the "12 volts group".


I wrote the following regarding such capacitor battery bank:

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


For those interested in the capacitor battery banks.


To store energy, one needs to determine the ultimate KWH and what the power source is and its limitations.


Taking as example what the patent says (if I remember right) 52 KWH hours of storage.


The bank has to accumulate around 3500 volts and one needs to remember the basic capacitor charging formula:


Current = 1 / 2 C *dV / dT and the energy available is 1 / 2 C *V^2


Also we need to make a correction for time


52 kwh = 52000 wh *3600 seconds = 187200000 watt seconds = 1 / 2 C * V^2

C= 187200000 * 2 / 3500^2 = 30.56 Farads.


Now we need to define the charging time, let's assume 5 minutes  = 300 seconds


Current = 1 / 2 * 30.56 * 3500 volts / 300 seconds = 89.1 amps


We need a 89.1 amps current source capable of having a range of at least 3500 volts to charge the capacitor from Zero (0) Volts to 3500 Volts.


One can see that the power source needs to be quite strong and expensive.


Home charging with 230 volts with 30 amps limitations, this 6.9 KW or 6.9 KWH limitations, so the charging time will be the 52 KWH(battery) / 6.9 KWH (house)=  7.536 hours


This time is minimum but we need additional time to take care of the equipment efficiencies and if we assume 80% efficient system then the hours become 7.536 / 0.8 = 9.42 hours


Here you have the generalized analysis of the charging time for such theoretical bank.


Nando

« Last Edit: September 09, 2007, 12:18:20 PM by Nando »

joestue

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Re: Death of the Lead-acid battery
« Reply #10 on: September 09, 2007, 01:19:07 PM »
Well 75 % of the energy storage is between 50 and 100 % of full charge


You would need a 100 amp current source from 1000 volts to 3500 volts, the application for the vehicle requires the same challenges.

ZVR switching is easily capable of providing the voltages and currents needed.

« Last Edit: September 09, 2007, 01:19:07 PM by joestue »
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Nando

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Re: Death of the Lead-acid battery
« Reply #11 on: September 09, 2007, 02:03:07 PM »
To "easy" in your reply and it seems that you do not have done high power and high voltage power conversion.


It is not that easy, you PLUG = ZVR but you do not demonstrate that it is a workable principle for such system.


And what do you mean by ZVR ???


The bank seems to be set to around 3500 volts and for charging without high voltage stresses at high frequency, the design has to be done by examining the bank, which may be divided for easy handling and installation, as well as, vibration protection in blocks, let's say 600 volts, ( and 50 to 75 lbs each), therefore 6 units placed in series, which implies to have 6 units with 6 times the total series capacitance ( in this case of 50 farads total value @ 3500 volts), so each block may be 300 Farads @ 600 volts to give the 50 Farads at 3500 volts.


The charger in this case may be 6 of them in series ( parallel ), each charging 1/6 of the bank, this way a maximum of 600 volts with the calculated current, utilizing constant current power sources, and Proper inter-segment isolation.


Nando

« Last Edit: September 09, 2007, 02:03:07 PM by Nando »

joestue

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Re: Death of the Lead-acid battery
« Reply #12 on: September 09, 2007, 08:04:26 PM »
Zero Voltage Resonant switching. ZVR

So what if you combine 6, 500 volt converters is series?

Industry sums 500 switchers in series to get HVDC transmission.


another to charge it up is a step up transformer @ 3500 volts rectified and a buck regulator. That would give 90% efficiency.


The electrical side of this was the easy part, and why the original claim drew so much attention. the real challenges in the physical issue of having a capacitor very capable of shorting out and either exploding or melting into slag.

« Last Edit: September 09, 2007, 08:04:26 PM by joestue »
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BigBreaker

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Re: Death of the Lead-acid battery
« Reply #13 on: September 10, 2007, 08:23:53 AM »
Actually its more like 4kV.  They take a high C capacitor topology and use a special dielectric that can tolerate huge voltages.  The 1 farad caps from maxwell can only do 1-2 volts.  EEStor is like 1 farad but 4kV.


If they really have these caps in any kind of scale the internal combustion engine is dead.  Seriously - no one will want a gas powered car.

« Last Edit: September 10, 2007, 08:23:53 AM by BigBreaker »

finnsawyer

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Re: Death of the Lead-acid battery
« Reply #14 on: September 10, 2007, 08:36:46 AM »
Which begs the question of why not put the six blocks in parallel to 1800 farads charged to 600 volts?  The energy stored will be the same and 600 volts will be a better match to an electric motor.  This whole thing shows how little real information these news releases have.
« Last Edit: September 10, 2007, 08:36:46 AM by finnsawyer »

finnsawyer

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Re: Death of the Lead-acid battery
« Reply #15 on: September 10, 2007, 08:47:08 AM »
The 10KW for 6 hours comes to 60 KWH, not 600.  At an electrical cost of $0.1 per KWH this would come to $6.00 worth of electricity.  Compare that to the cost of gas for the same trip.
« Last Edit: September 10, 2007, 08:47:08 AM by finnsawyer »

DamonHD

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Re: Death of the Lead-acid battery
« Reply #16 on: September 10, 2007, 10:10:41 AM »
1F @ 4kV is a HUGE amount of energy!


evil cackle

« Last Edit: September 10, 2007, 10:10:41 AM by DamonHD »
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Nando

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Re: Death of the Lead-acid battery
« Reply #17 on: September 10, 2007, 11:39:15 AM »
Not that easy.


to go for 1800 Farads, the cost of the bank may be too high, quite cheaper with high voltage converted down as needed. ( 50 F @ 3500 Volts, the weight may be around 330 pounds) so for 1800 F the weight would be 330 *1800/50 = 11,880 pounds minimum, but with the manufacture geared for a automobile the weight may increase to around 400 pounds for the 50 F @ 3500 Volts bank).


If they go for 1800 F @ 600 volts, the system may not be a good match for the motors since the capacitor bank voltage varies with use and the electronics may be close to the 3500 volts in equivalent energy supplied to the motors.


As we can see weight versus voltage, the high voltage wins handily with extreme weight reduction.


Another member "INSERTED" ZVR, a DC/DC converter system around 15 or so years old that is mostly geared to low power devices, this due to the resonant principles which may present a much higher voltage than the battery voltage by several fold -- this depending on the Q of the system.


Another comment from the same member indicating the use of 500 converters for the transmission of power via extremely high voltage ( 500,000 to 1,500,000 volts) - this technology may have some use for this conversion but not the way that is presently used in GRID power transmission.


Nando

« Last Edit: September 10, 2007, 11:39:15 AM by Nando »

joestue

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Re: Death of the Lead-acid battery
« Reply #18 on: September 10, 2007, 03:10:03 PM »
ZVR has nothing to do with the Q of the system, it is a relatively low power network surrounding the active switching elements to reduce losses, exclusively used in almost all high power switching, cost effective above 5KW.


The original claim, that using barium titanate coated with aluminum oxide and glass could withstand 3500 volts (2 in series at ~1700), and have charge capacities comparable to 5 volt caps of the same farads, was completely and utterly bull$#|+.

no capacitor was ever built, it was a theoretical idea.


Back to the challenge of getting 1K-3.5K @ 30 amps....

It is expensive at the moment, but after mass production, the cost will equalize between all voltage ranges

« Last Edit: September 10, 2007, 03:10:03 PM by joestue »
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finnsawyer

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Re: Death of the Lead-acid battery
« Reply #19 on: September 11, 2007, 09:08:14 AM »
Supposedly they would be putting two 1700 volt capacitors in series.  So, why not put them in parallel to start.  What is so special about 3500 volts anyway?  Beyond that, if they make all the substrates thinner they can get a larger capacitance at lower voltage for the same weight.  They would still have the same electric field in the dielectric.  The energy is stored in the dielectric, and basically you need to keep the amount of that constant.  So, you would still end up with the same weight for the same energy storage capacity as long as you can also reduce the weight of the metal parts by the proportional amount.  So, the real issue is, "What can they actually accomplish?"  If the ceramic dielectrics are that good, then there may be some reason for hope.      
« Last Edit: September 11, 2007, 09:08:14 AM by finnsawyer »

joestue

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Re: Death of the Lead-acid battery
« Reply #20 on: September 11, 2007, 02:32:26 PM »
What was so special: a capacitor with an energy density twice that of the lithium ion battery, 10 times that of the lead-acid.


  1. F @ 3500Vdc = 189MegaJoules = 52 KWH, divide by 336 pounds =
  2. WH/pound or 345WH/KG.


-This has been beat into the ground on several forums, if you want to know why this capacitor cannot exist- Google it.

« Last Edit: September 11, 2007, 02:32:26 PM by joestue »
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BigBreaker

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Re: Death of the Lead-acid battery
« Reply #21 on: September 11, 2007, 03:30:08 PM »
That's why people are paying attention to EEStor.  Their claims are so world changing that they have something of a credibility gap.  Extraordinary claims require extraordinary proof.  So far EEStor has some proving to do.
« Last Edit: September 11, 2007, 03:30:08 PM by BigBreaker »

finnsawyer

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Re: Death of the Lead-acid battery
« Reply #22 on: September 12, 2007, 08:36:15 AM »
Of more interest would be why anyone thought the technology might work in the first place?  Just because they couldn't make the given capacitor doesn't mean they couldn't make a better capacitor.  A 500 pound capacitor that could provide power for a vehicle for two hours might have its use as well.  I presume someone must have noticed that these ceramic substrates have a higher dielectric constant.  So, what is it?  For reference, water has a dielectric constant of 80.
« Last Edit: September 12, 2007, 08:36:15 AM by finnsawyer »

ghurd

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Re: Death of the Lead-acid battery
« Reply #23 on: September 12, 2007, 10:04:17 AM »
Maybe none of it matters now they figured out how to burn salt water.

G-
« Last Edit: September 12, 2007, 10:04:17 AM by ghurd »
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Nando

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Re: Death of the Lead-acid battery
« Reply #24 on: September 12, 2007, 08:42:08 PM »
RESONANT circuits call for oscillatory circuits that may be half sine wave or even full sine waves.


In either case, the resonant needs to generate the sine wave that HAS TO BE much greater than the basic voltage, that in this case is 3500 volts peak, which may force a half sine wave of at least 50 % higher than the basic voltage.


You can see a patent for such ZVR , Down load the patent 5057986.pdf from my files.


May be you could give us a more defined feedback.


and too many high power and high voltage devices.


Nando

« Last Edit: September 12, 2007, 08:42:08 PM by Nando »

joestue

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Re: Death of the Lead-acid battery
« Reply #25 on: September 12, 2007, 10:33:33 PM »
my bad.


What I'm saying is that using a capacitor @3500V to drive a car is no less efficient than current 250 volt battery systems.


Whether you use IGBTs, GTOs, Thyristors, RCTs, MCTs or asymmetrical SCRs it doesn't matter. it's relatively the same at all voltage ranges, none of this has an order of magnitude greater performance. Keep in mind there is a lot of new technology being factored into the mix, 50Amp, 1200 volt silicon carbide diodes are now available.


This paper describes a 2.5KV, 30 amp inverter:

http://www.ece.rutgers.edu/~jzhao/papers/MSF-2004-MPS-Switching.pdf

« Last Edit: September 12, 2007, 10:33:33 PM by joestue »
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Nando

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Re: Death of the Lead-acid battery
« Reply #26 on: September 13, 2007, 09:56:59 AM »
More or less efficiency was not considered.


What is being considered is the storage capacity of a battery and/or a large capacitor to supply enough energy to "move" the car for about 300 miles.


The conversion of the available power to the "energy mover = motor" is what is considered in regards of difficulties, easy to convert and/or the problems in regards to the conversion of a high voltage to the "energy mover" requirements.


The efficiencies of one type to the other may considered, like about 33% for the gas engine to the "motor + conversion" that could reach around 80 %.


One needs to examine the overall loop efficiency, from generating the energy ( well etc or hydro or nuclear or even coal or gas generators) to the ultimate use of such energy "material " and its costs.


Also the storage capacity versus volume and weight of such storage.


The Hybrid system, large Battery + small gas engine is one step to improve the efficiency of the system compared to the gas engine alone which could reach around 40 to 50 % or maybe a bit more.


A single large capacitor energy storage from the generating the energy to the energy going into the wheels to move the vehicle could reach something like an overall 60 to 65 + % efficiency.


I could extend myself here for a long rant, better stop.


Nando

« Last Edit: September 13, 2007, 09:56:59 AM by Nando »

spinningmagnets

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Re: Death of the Lead-acid battery
« Reply #27 on: September 15, 2007, 10:40:19 PM »
Its my understanding (notice how I'm vague, so my "sources" can't be questioned?) that the regenerative braking effect only adds a small amount of efficiency to the energy use profile of an electric or hybrid vehicle. I recently found a reference that suggested that chemical batteries are slow on the uptake and that is why much of the available regen braking is lost. Also accelerating from a standing start draws a significant spike in amps, and then settles down quickly to a small speed maintenance draw from the battery.


I've read recently an article that suggested Toyota and Honda are working on incorporating a supercapacitor into their energy management systems. Its hoped the super caps will power initial acceleration, and will dramatically improve regenerative braking.


They were not very forthcoming with details, but they indicated NiMH or Li-ion chemical batteries still proved to be the best off the shelf solution for most of the energy storage, rather than capacitors only, and I wondered why.


I have a stupid capacitor question, and please, feel free to point and laugh.


Benjamin Franklin experimented with leyden jars, which are a simple glass jar capacitor. Apparently one the size of a large pumpkin can knock a man off his feet. Franklin put in a lightning rod on his roof and ran the cable through his shop on its way down to (literally) ground. Even when there wasn't lightning, dry electricity would occasionally flow through the cable and temporarily magnetize it, drawing a small metal bell and alerting him. When the cable was live, Franklin would charge up his set of leyden jars for later experimentation by touching them to the bare cable (instead of hand-cranking his wool/glass static generator).


If I made a giant leyden jar out of a man made pond lined with a dielectric liner, and then charged it with a lightning rod,...


can a high voltage capacitor be used to charge up a battery pack?


also, can large simple home-made capacitors be used as a dump load, and later used to add to the charge of a battery pack when there's no wind? (batteries are hard to make, maybe capacitors are easier to make?)


"Some mornings, it's just not worth chewing through the leather straps." -Emo Philips

« Last Edit: September 15, 2007, 10:40:19 PM by spinningmagnets »

finnsawyer

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Re: Death of the Lead-acid battery
« Reply #28 on: September 16, 2007, 08:53:27 AM »
When you're drawing a few hundred amps out of a battery, the battery's internal resistance becomes an important loss item, so a capacitor having less internal resistance at high current draw would help the equation.


One thing we haven't discussed is a capacitor's self discharge rate.  This might not be a problem for a hybrid vehicle, but could become significant for a vehicle powered only by a super capacitor.  You put $6.00 worth of energy into the capacitor and then it sits for a week, and all leaks away.  Well, you'd probably have it on charge each day, which would actually maximize the leakage loss over time.  This brings me to the pond idea.  Water is a good dielectric, but it also can be a good conductor.  It is likely your pond capacitor would have a high self discharge rate.


If you connect a high voltage capacitor directly to a battery, the battery voltage must rise to equal the capacitor voltage.  This would result in a large current flow into the battery due to the difference between the capacitor voltage and the battery voltage.  There may also be an effect due to the inability of the chemical processes in the battery to proceed fast enough.  This should show up as an added non linear resistance.  If the battery doesn't blow up the current will quickly drop in value.  One obvious thing you could do is place an inductor between the capacitor and battery to limit the current, but all inductors have resistance too.

« Last Edit: September 16, 2007, 08:53:27 AM by finnsawyer »

spinningmagnets

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Re: Death of the Lead-acid battery
« Reply #29 on: September 16, 2007, 02:04:24 PM »
I am an electronics newbie, so I appreciate your patience. Against the helpful advice of a customer service rep on the phone from Pakistan, I once tried to fix a recalcitrant computer with a sledgehammer, and apparently that makes them perform even worse. (also, never urinate on a computer until you unplug it first...don't ask how I know).


I am collecting information on using wind and water to make high-voltage AC so the alternator can be in the best collection site, and the house can be in the best home site, often far apart. Its my understanding that when the high voltage reaches the garage, it should be run through a step-down transformer to lower voltage, then rectified to a charging DC to the battery pack.


Franklin noted that even when there was no lightning, his cable conducted electricity from the air to the ground quite often. I plan on having a pond, even if its not a giant capacitor. But if a few small changes make it useable as a capacitor, the problem as I see it is, the voltage will be much higher than 120/240, so there would be no off-the-shelf management components available, plus, the voltage would not be a constant, but highly variable. I can buy a 110/12 volt transformer, but a "black box" that takes variable extra-high voltage and gives me a stable 12 volts, well, I don't even know what questions to ask! but theres gotta be a way.


I know that a cars tiny tail-light bulb can be hooked up directly to a 12-volt battery, but if I use cheap thin jumper cables with no resistance (but much thicker than the bulb filament), they will get hot enough to melt.


A Leyden jar is certainly inefficient for a given volume compared to a modern capacitor, but they are very easy and cheap to make. If one the size of a hot tub could kill a man, then it goes without saying that it would need a non-conductive berm and a significant fence and cover with warning signs. However, if its possible to occasionally harvest even a tiny fraction of the "free" electricity around us (even with a high self-discharge rate) I am very intrigued, even though I am "electrically challenged".


"I told my psychiatrist that everyone hates me. He said I was being ridiculous... everyone hasn't met me yet." -Rodney Dangerfield

« Last Edit: September 16, 2007, 02:04:24 PM by spinningmagnets »

finnsawyer

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Re: Death of the Lead-acid battery
« Reply #30 on: September 17, 2007, 08:44:16 AM »
For the situation you describe I usually advise people to investigate the alternator design that I propose in my diary, which would give more voltage at a higher frequency for the same investment in magnets, but in your case I won't.  You should be aware that the output voltage will likely only vary by a factor of two, from say 100 to 200 volts between cut-in and furling.  This will also be true of the lower voltage after going through the transformer.


Why don't you make a test capacitor using pond water.  Get a wide mouth (large pickle bottle?) glass jar, fill it with pond water and add two conducting concentric pipes of different diameters.  Then charge it up and see what you get.


I used to light up a fluorescent light from a short wave antenna during snow storms, but the current was too low to be of any use.  Any time you have air movement you could get charge build up on a wire and run it into a capacitor, but this runs into the issue of the rate of self discharge for the capacitor.  You certainly can experiment.  Have fun.  If you wish to continue this discussion, I suggest you start a new posting.

« Last Edit: September 17, 2007, 08:44:16 AM by finnsawyer »