Text 2415, 112 rader
Skriven 2006-06-12 13:39:00 av Robert E Starr JR (2861.babylon5)
Ärende: Re: Interesting technolog
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"Matt Ion" <soundy@moltenimage.com> wrote in message
news:SBVig.9923$Mn5.2506@pd7tw3no...
> Carl wrote:
> > Since we've been having a discussion on GW and various energy sources, I
> > just thought I'd pass along a few technologies that I ran across that I
> > found interesting.
> >
> > I've seen several write-ups on this one:
> >
> > 1) Capacitors to Replace Batteries?
> >
> > "MIT's Joel Schindall plans to use old technology in a new way with
> > nanotubes. 'We made the connection that perhaps we could take an old
> > product, a capacitor, and use a new technology, nanotechnology, to make
that
> > old product in a new way.' Capacitors contain energy as an electric
field of
> > charged particles created by two metal electrodes, and capacitors charge
> > faster and last longer than normal batteries, but the problem is that
> > storage capacity is proportional to the surface area of the battery's
> > electrodes. MIT researchers solved this by covering the electrodes with
> > millions of nanotubes. 'It's better for the environment, because it
allows
> > the user to not worry about replacing his battery,' he says. 'It can be
> > discharged and charged hundreds of thousands of times, essentially
lasting
> > longer than the life of the equipment with which it is associated.'"
> >
> > Apparently the capacitors they've created can be recharged in seconds.
The
> > technology is expected to hit the market within 5 years.
>
> That's an interesting way of increasing capacitance via increasing
> surface area (as noted, that IS the limitation to storage capacity vs.
> size).  The one thing I don't see them getting around is the charge
> time, which is DIRECTLY propotional to capacitance - time to fully
> charge a capacitor is considered to be 5RC, or five times the resistance
> times the capacitance of the circuit*.  The higher the capacitance, the
> longer it will take to charge, and the only way to reduce charge time is
> to lower the resistance feeding the cap... which will necessarily
> increase the current draw for a very brief period.
>
> * "RC", or resistance times capacitance, defines the time it takes a cap
> to charge to, if memory servers, 63.2% of max capacity, FROM WHERE IT'S
> CHARGE IS CURRENTLY.  So if you start at 0 volts and apply 10 volts to a
> 1 Farad capacitor through a 100 ohm resistor (using nice round numbes
> for easy calculation), you'll see a charge of 6.32 volts after 100
> seconds (1 Farad times 100 ohms)... after another 100 seconds it will
> increase another 2.32576 volts (63.2% of the 3.68 volt differential), to
> a total of 8.64576 volts... after another 100 seconds, you'd be at
> 9.50163968, and so on... technically, the capacitor will never reach
> 100% of the supply voltage, as it just keeps increasing TOWARD that
> level in infinitely smaller steps, but in general electronics, 5 time
> periods is considered to be close enough to be considered "full charge"
> (in this case, we'd be at 9.93251005202432 volts at 500 seconds).
>
> Now here's the rub: 1 Farad is not going to power anything for very
> long, and 10 volts is not very useful except in small electronics. As
> the voltage need goes up, you need a thicker and/or stronger dielectric
> (the non-conductive material between the plates), and thus the size of
> the cap increases (a modern 1 Farad electrolytic capacitor designed to
> run at 12 VDC (such as those designed for high-powered car audio
> systems) is about the size of two smaller coffee cans, stacked).
>
> To get anything really useful - say, to run an electric car - you'd need
> to get into the hundreds or thousands of farads... and thus your charge
> time increases by a factor of hundreds or thousands, unless you can
> lower the resistance... which the increases the initial current demand
> (100 Farads through 100 ohms would require 50,000 seconds to reach full
> charge).  In order to maximize efficiency, you'd want to increase your
> voltage, so the current demands of your load are lower (a load, such as
> a motor, that draws one amp at 10 volts, would only require .1 amp to do
> the same work at 100 volts).
>
> To feed a 100 Farad cap through a 100 ohm resistor would take 50,000
> seconds, or a little shy of an hour and a half, to charge fully, and
> would initially draw one amp of current with a 100 volt supply (current
> drops as the charge nears full).  If you want to lower the resistance to
> 10 ohms, you'll reduce the charge time to only 17 minutes, but your
> initial current draw is then 10 amps... and that has to come from
somewhere.
>
> To be succinct: you need REALLY high-capacity, high-voltage capacitors
> to be useful for anything beyond your radio or pocket shaver; the more
> capacity they have, the longer they take to charge, OR the more current
> supply they require to charge.  You could conceivably strike a balance
> with an automatically-adjustable resistor that starts out at a higher
> resistance to limit current demand, and drops resistance as current
> demand drops to improve charge time, but that will only buy you SO much
> time.
>
> They can pack'em into smaller spaces, which is great, but ye canna
> change the laws of physics!

So, you can't use 'em to power your Chevrolet. Lots of other devices devices
to power, you have noted a couple yourself. It is this kind of technology
which can drive the development of devices which require lower power to
operate and thus use the new power tchnology.

Also, it sounds like it would power the hell out of my X-Mod!

--DwightG
    
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