Seebeck Effect applied to Fire?

GoldenTurtle

Hello Physics Forums,

I'm attempting a portable energy generator that will apply the Seebeck Effect to a campfire, allowing the camper to enjoy his new source of electricity.

Right now I have a basic concept, basically a raised platform on which you stack the logs and then light them on fire.

The platform will be where the Seebeck effect is applied, allowing me to generate electricity from the energy difference between the fire and the bottom of the raised platform.

Any electricity generated will be stored in a battery, and can then be used via outlet.

So I have the following questions:

Is the bottom of the fire the best place?
Is it safe to burn fire directly on top of the Seebeck platform?
If not, then should there be a grill, conductive plastic, or something to separate them?
What materials should I use to build the Seebeck platform?
Are there any guides online as to how to build one at home?
Does it matter what king/brand of wiring I use?
Is there a battery you reccommend?
Is there a preferred method of hooking up a battery to an outlet?

If you have any other advice, please let me know.

Mike_In_Plano

I'd suggest calculating some specifics to get some idea of the viability. Suppose we use 24 gauge, type J thermocouple wire. This is fairly common material...

Next, let's suppose we give ourselves about 2 feet between the heart of the fire (800 C?) and a cooler region (75C?)

Assumming we get an average of 50uV/C, we would have an EMF of 36.3mv
Of course, this is an open circuit condition. When delivering power, the max power point will be when the voltage drops to the halfway point (maximum current x voltage). Thus, we would expect 18.1mv per thermocouple.

Omega specifies a resistance of .878 ohms per foot, or in this case, 1.76 ohms. The supplied voltage divided by the resistance gives the supplied current, 10.3ma.

18.1mv x 10.3ma = .186mw per thermocouple.

It would take roughly 5400 thermouples to produce 1 watt.

MATLABdude

Welcome to PhysicsForums!

Unfortunately, your device probably won't produce a whole lot of usable electricity:
http://en.wikipedia.org/wiki/Thermocouple
http://en.wikipedia.org/wiki/Thermopile

If you needed emergency outdoors power, you'd probably be better off with something hand-cranked or solar-powered (I've seen survival radios and flashlights that were powered in that fashion).

GoldenTurtle

Quote by Mike_In_Plano

I'd suggest calculating some specifics to get some idea of the viability. Suppose we use 24 gauge, type J thermocouple wire. This is fairly common material...

Next, let's suppose we give ourselves about 2 feet between the heart of the fire (800 C?) and a cooler region (75C?)

Assumming we get an average of 50uV/C, we would have an EMF of 36.3mv
Of course, this is an open circuit condition. When delivering power, the max power point will be when the voltage drops to the halfway point (maximum current x voltage). Thus, we would expect 18.1mv per thermocouple.

Omega specifies a resistance of .878 ohms per foot, or in this case, 1.76 ohms. The supplied voltage divided by the resistance gives the supplied current, 10.3ma.

18.1mv x 10.3ma = .186mw per thermocouple.

It would take roughly 5400 thermouples to produce 1 watt.

I appreciate your mathematics very much, they've saved me a lot of time.

However, is it really necessary to have a 2 foot distance?
Since it is intended to operate at night, when the ground is cold, would there be another way to decrease the temperature of the bottom side?
- it is against the cold ground
- against a bed of cold gravel/stones

Your response has also raised several more questions:

Would a decrease in distance lead to more electricity generated?
Does humidity play a role in the resistance of the air?
Would it be a good idea to cover over the entire thermocouple?
What metal would be best for the heated side of the thermocouple?
What metal would be best for the cool side of the thermocouple?

Quote by MATLABdude

Welcome to PhysicsForums!

Unfortunately, your device probably won't produce a whole lot of usable electricity:
http://en.wikipedia.org/wiki/Thermocouple
http://en.wikipedia.org/wiki/Thermopile

If you needed emergency outdoors power, you'd probably be better off with something hand-cranked or solar-powered (I've seen survival radios and flashlights that were powered in that fashion).

This is intended to generate electricity mostly at night time.
I intend to somehow wire the battery to receive electricity from solar panels during the day time, and from the fire in the night time.

Speaking of which, I've never tried wiring a battery like that before, is it possible?

russ_watters

You guys are missing the concept. It's been done and it works. Here's a commercial product designed for truck exhausts: http://hi-z.com/hz2.php

And according to the wiki:

In the early 1990s, Hi-Z Inc designed an ATEG which could produce 1 kW from a diesel truck exhaust system. The company in the following years introduced other designs for diesel trucks as well as military vehicles [1].

In the late 1990s, Nissan Motors published the results of its ATEG which utilized SiGe thermoelectric materials. Nissan ATEG produced 35.6 W in testing conditions similar to running conditions of a 3.0 L gasoline engine in hill-climb mode at 60.0 km/h.

Clarkson University in collaboration with Hi-Z has designed an ATEG for a GM Sierra pick-up truck. The program was funded by the American DOE and NYSERDA. The published literature of this ATEG explained its ability to produce 255 W at a vehicle speed of 112 mph. [4][5]. In 2006, scientists in BSST and BMW of North America announced their intention to launch the first commercial ATEG in 2013.[6]

http://en.wikipedia.org/wiki/Automot...tric_generator

Now will it work well or be economically viable? That's a completely different ball of wax.

I can't help much with building one yourself, but you need to make sure you have a hot side and a cold side. Building a fire on top is fine, but you need to be able to keep the other side cool with a heat exchanger perhaps with a fan.

Since it is intended to operate at night, when the ground is cold, would there be another way to decrease the temperature of the bottom side?
- it is against the cold ground
- against a bed of cold gravel/stones

You'll quickly heat up the ground and lose effectiveness. You need flow of water or air over the back of the device to keep it cool.

MATLABdude

Russ, that's crazy, and I stand corrected. I remember hearing a whole lot of hoopla (in PopSci/Mech and the likes) in the 90s when I was in high school about scavenging "waste" heat using the Seebeck, but then I stopped hearing about it and assumed it just fell off the face of the Earth. Ditto when I quickly Googled for it last night before answering.

225W from a standard car is pretty darn respectable--that's order of magnitude for an alternator!

GoldenTurtle

A flow of water...

What if I built a thin pan, directly underneath the bottom plate (cold side), and have a tiny slit be visible from all sides?

It would have a drainage plug on the bottom, and there'd be a small tube where you could pour water into it.

The water would be cold when drawn from a water source nearby, and it would remain cold with the wind blowing through the slits.

Do you think that would work to cool the bottom side properly and easily?

Also, what material exactly do I make these hot and cold plates from?
They need to be conductive, yet very resistant to heat.

Stan52

GoldenTurtle, you are focused on the mechanical design. You need to get the idea of thermoelectric generation before trying to actually design something. The nice folks have basically told you that you would have a hard time powering a single LED off a campfire. The current state of the art with the best (read: expensive) thermoelectric materials known still have very poor efficiency. You would be better off putting a fan over the fire to let the rising heat spin a generator. Not too sure of that last statement. Someday somebody will make this viable, but it's not ready yet.
That's not to say I know how it works either. Maybe somebody can help me with this: How does the heat have to flow? Which of the following is correct?

heat flow in →░▓░▓░▓░▓→ heat flow out
..............NPNPNPNP (I am using dots to align this because spaces don't work)
................→ → electron flow
Here the heat "carries along" the electrons. Every alternate junction acts like a rectifier, so that electrons that hop across due to the heat can't hop back. Or the flow of heat energy pushes the electrons along. Its directionality gives them directionality. The other junctions are "shorted" somehow so the electrons can flow "up" to the N-type material to the next rectifier. Or it's the heat itself that pumps them up "backwards" to the diode, and the very low forward threshold that keeps them from falling back. That doesn't make sense. Except for a heat-carrying effect, as many would fall back as forward, and the output would be miniscule.

heat flow in →░▓░▓░▓░▓← heat flow in
.............→↑↓↑↓↑↓↑↓↑→ electron flow
heat flow out ←▓░▓░▓░▓░→ heat flow out
Imagine the hot layers connected to the cold layers, like color to like color.
Here the heat flows into one set of junctions and out a different set. The top left material is the positive electrode and the bottom right is the negative. The hot N-P junctions act like rectifiers and the hot P-N's shown are actually open (imagine a tiny space there) so the electrons take a zigzag path from hot to cold, cold to hot, back and forth, while the heat flows one way, of course. On the cold side the electrons have to jump up P to N, (the cold N-P's are also open) which raises the question, how do they do that without losing all the useful energy? Merely keeping those junctions cold isn't going to help. They would have to be shorted so the electrons can efficiently get back into the heat. But if they are shorted, why do they have to be cold? Surely it isn't the cold side that has the rectifying junctions, while the heat acts to "short" or "pump" the hot junctions backward? That would explain the horrible efficiency. (In that case the electrons would flow right to left in the diagram above.)

Anybody? No math, please!

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MATLABdude Blog Entries: 1 Recognitions: Science Advisor	Welcome to PhysicsForums! Unfortunately, your device probably won't produce a whole lot of usable electricity: http://en.wikipedia.org/wiki/Thermocouple http://en.wikipedia.org/wiki/Thermopile If you needed emergency outdoors power, you'd probably be better off with something hand-cranked or solar-powered (I've seen survival radios and flashlights that were powered in that fashion).