Text 2344, 94 rader
Skriven 2006-06-11 11:32:00 av Robert E Starr JR (2790.babylon5)
Ärende: Re: Interesting technolog
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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!
                              
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