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Old 08-02-2008, 11:09 PM   #41
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CFB has it right on the device.

For the really techie minded or those with insomnia here is more detail.

You make one electrode from a semi-conductor material and immerse it in an aqueous electrolyte. The other electrode can be similar to those used in a fuel cell and is also immersed in the electrolyte.

You then shine light on the semi-conductor electrode and create conduction band/high energy electrons and/or holes. These then react directly with the water in the electrolyte, electrons being injected from the electrode into the electrolyte at one electrode where oxygen in created and removed from the other electrolyte where hydrogen is created.

Both electrodes have a catalyst layer coated on the surface to decrease the voltage loss needed to make the reactions go. The claim of the MIT group is that they have developed a better catalyst for the oxygen electrode which as you can see is only one subcomponent of the device.

The device is a way to convert solar energy directly into chemical energy in the form of hydrogen but although it sounds like a pretty cool thing I'm pretty sure that it is more efficient to first capture the solar energy with a conventional solar cell and then turn it into H2 in a separate electrolysis plant if you need hydrogen or use it directly as electrical power since that is probably the ultimate use anyway.

MB
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Old 08-02-2008, 11:31 PM   #42
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Or, here's an idea: Instead of shining the light on a semi-conductor electrode, shine it on thousands of acres of farmland. Eat the crops produced. Turn the crop waste into methanol through very well established, mature processes.

Burn the methanol directly in vehicles. If you want, convert the methanol to gasoline through the process develpoed by Mobil many years ago and burn it in existing cars.

Entirely carbon neutral: The plants took the carbon out of the air during photosynthesis last year, now you are burning it and returning it to the atmosphere. That's as "carbon clean" as it is possible to get.

Ooops--sorry. No hydrogen to compress, figure out some way to transport, and needing a tank as big as a phone booth to get a reasonable range. Hydrogen is the ANSWER --I keep forgetting to start from that point.

Back to the regularly scheduled discussion . . .
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Old 08-03-2008, 03:58 AM   #43
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Entirely carbon neutral: The plants took the carbon out of the air during photosynthesis last year, now you are burning it and returning it to the atmosphere. That's as "carbon clean" as it is possible to get.

Ooops--sorry. No hydrogen to compress, figure out some way to transport, and needing a tank as big as a phone booth to get a reasonable range. Hydrogen is the ANSWER --I keep forgetting to start from that point.

Back to the regularly scheduled discussion . . .
There you go, introducing well established and understood methods into a beautiful but un-necessary Hydrogen ANSWER.
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Old 08-03-2008, 09:11 AM   #44
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Hydrogen is the ANSWER --I keep forgetting to start from that point.
Yep, the laws of unintended consequences mates easily with starting at an answer and backing into the question, doesnt it?

I've been thinking about that whole idea of having a zeppelin in my back yard. That might be pretty cool. Drive that around instead of the car. Hang my tv antenna on it, bet it'd get good reception.

Wow, brain flash...we could all LIVE in hydrogen filled zeppelins that would fill the sky, solar panels mounted on top so you dont have that rainy day problem, and we could all migrate to parts of the world that were the sunniest. Sort of like boating but without the dry rot.

No problems with pollution either. Just throw everything over the side. I'm sure the folks that cant afford zeppelins wont mind too much.
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Old 08-03-2008, 11:15 AM   #45
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mb,

You got me thinking. Maybe this process reduces the overvoltage. That's all I can think of.
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Old 08-03-2008, 12:03 PM   #46
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No, the V squared relationship is only for a capacitor, where: E = CV^2 / 2.

For a battery or fuel cell you have to go a step backwards to E = integral VI dt, and then integrate over the voltage-current curve which is different for each different chemistry.
MB,

I'm still confused on this point. If I have a "black box" with a constant output voltage of 1.2V and another with a constant output voltage of 0.8V, the 0.8V device will deliver 44% ( (2/3)^2 ) of the energy of the 1.2V device into the same resistive load over the same period of time, regardless of what's going on inside the "black box". In other words, the integral VI dt, is going on inside the box. Isn't all I care about the terminal voltage of the box (independent of whether the box is a battery, a fuel cell, or something else)?
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Old 08-05-2008, 08:10 AM   #47
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FIRE'd@51, you are correct that .8V applied to a resistive load will produce 44% of the energy that 1.2V will. I'm not a chemist, but I suspect that the voltage involved in electrolysis is not as simple as a resistive load.

Take this example - some cells (batteries) have a voltage of 1.5V, some 3.3V depending on chemistry. All else being equal, the 3.3V cell will produce more energy connected to the same R, but that alone does not make the 3.3V cell more efficient, it's just a different chemistry at a different operating point.

There may be some of this 'operating point' stuff going on with the catalyst, to some degree. Maybe the chemists can confirm/deny that explanation for us.

-ERD50
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