Thermoelectric generator

[highlandsun]

OK, so commercial thermoelectric modules are designed to operate with a delta-T (cold side vs hot side) of 200C. I found a design for a DC-DC converter that could extract ~130mW from a TE module at delta-T of only 20 degrees. What's the point? This is right around the range of heat that a Luxeon 1W emitter produces. (Bare emitter, 15C/W, Star 17C/W).

So if you put a TE module between the Star and its heat sink, you will very likely get that 20C minimum delta-T. With the module recovering 130mW during operation, you could reclaim 13% of the energy spent driving the LED. In other words, you could extend your runtime by 13%. With an overdriven Luxeon producing even more heat, you could recover more energy (the TE efficiency increases as delta-T increases).

The DC-DC converter design is here http://www.ece.uvic.ca/~jbornema/Journals/064a-97ia-jmd.pdf

Thermoelectric module specs are here
http://www.hi-z.com/index.html

 

[Jonathan]

Based upon my experience with this sort of technology, you are better off simply using a larger heat sink.

If you are recovering 130mW of electrical power with a delta-T of 20 degrees, then assuming perfect efficiency you would require a heat flow of about 2W. Given real Peltier junction efficiency, I would expect that this number is more like 10-20W. This means that the Luxeon would not be producing enough heat to maintain the 20 degrees across the Peltier junction.

Additionally, if there is a temperature differential across the junction, then it _must_ be acting as a thermal resistance. Get rid of the Peltier junction, and you drop your Luxeon temperature. Since the Luxeon is more efficient when cooler, you end up feeding it less power to produce the light.

-Jon

 

[INRETECH]

The statement is correct, thermoelectric modules are not very eff - and you still need a large heatsink to get rid of the heat on the other side

A TM doesn't make electricity from heat, it makes electricity from passing the heat thru it

 

[Jonathan]

That isn't quite correct either, but it makes the basic point: for a TE generator to work, heat must pass through it.

All heat engines (including thermoelectric devices) convert heat to work (electricity). However they only work by having heat pass through them and converting some of that heat to work. The heat will only pass through them by flowing from hot to cold, and the maximum fraction of the heat that can be converted depends upon the temperature difference.

This means, for example, that 100 J of heat might be absorbed at 373K (boiling point of water) and 90J of heat rejected at 273K (freezing), with 10J of heat being converted into electricity.

The larger the temperature difference, the larger the fraction of the heat that can be converted to work. In the limit of the cold side being at absolute zero (and a perfectly efficient heat engine) then _all_ of the heat could be converted to work. The _maximum_ fraction of the heat which could be converted to electricity, going from 373K to 273K is about 27%.

-Jon

 

[Albany Tom]

Thermodynamics was the most depressing course I've ever taken. I'll sum up the rules...

You can't win
You can't break even
You always loose

It gets better when the part of the laws being just that, laws, sinks in. Not "sorta", "kinda", or "most of the time".

It's "always", "provable", and "no exceptions".

Hell, even with the IRS *sometimes* you can win.