Dimensions for Heat Conducting Material & the Temperature Gradient

1- temperature gradient being measured normal to surface is independent of "length", the gradient is the change in temperature per unit thickness so your question is wrongly formulated you ask NOT for the gradient but for the "temperature difference or drop".

2- Fourier equation is Q= -λ*A* [dθ/dx] where dθ/dx is the gradient. λ is the conductivity of considered material.

3-You have now the equation which allows you to define what you ask for.

4- This equation is valid as it is for plane heat conduction if you want to define insulation for cylindrical surfaces look in a book where it is indicated how to do it.

to nickname: You could also have a temperature gradient from end to end (i.e., over the length of a piece of material).

to the OP: You need to specify more information. I could, with big (i.e., concentrated) enough heat sources or sinks create such a gradient for a variety of lengths. As a probably silly example--get two ovens and run one at 200 degrees and one at 350 degrees.

Take a (long enough) bar of steel and stick one end in one oven and the other end in the other oven. You now have a 150 degree temperature gradient in the bar of steel between one end and the other.

Thank you both for sharing this info. I should have explain my problem more clearly. The situation is this; i have a peltier element which is said to have maximum rated temperature of 150C in either sides when used to generate current using a temperature difference on its both ends. My heat source is a cooker(350C not quite sure 1 which use firewood) so i need to reduce this temperature to 150C using a heat conducting material & this should be in a convenient distance from the cooker. There for i need to calculate proper length etc for a conducting material to apply them in a scientific manner in my project. I think i have made it clear now & sorry for my improper English. Any help is highly appreciated.

That helps, but, I still have questions:

• Are you sure your cooker gets to 350 degrees C? That's about 662 degrees F, which sounds rather high. Maybe not, maybe it doesn't matter.
• What will happen to the Peltier device above 150 degrees C--will it actually fail due to the joint between the two dissimilar metals failing, or will it just fail to produce proportionally more electricity above that temperature?

But, in any case, I think I understand what you want to do. You want to connect something to the cookpot that will project outside the fire and, such that, by radiating (and/or convecting, or conducting) heat away from that something, get the end of it to 150 degrees C so you can attach your Peltier device there without chance of damage.

That should be possible, but I have a feeling it is going to be sort of tricky / fidgety. It will be hard to get an exact calculation, and the temperature at the end will vary depending on a variety of factors like the ambient temperature, wind conditions, impingement of flames on that device, how this bar is attached to the cooker, and so forth.

I'd be inclined to talk to the manufacturer of the Peltier device to find out what will happen when it is exposed to temperatures over 150 degrees C.

What is your objective? Are you trying to generate electricity from waste heat? I would think that you could get more electricity from something like a stirling engine driving a small generator.

Actually that was the idea. Several of these elements can be arranged to get more power right? As this unit is used indoors can we make few assumptions of the room temperature etc & do the calculations? If you can provide the method generally then it can be used even if there are few variations in the above values such as the cooker temperature & device working temperature. It is also important that the other side of the device remain in room temperature 25C.

Peltier device said fail due to the failing in the material. It will damage the device as said. This i bought from ebay & its 400W 12V peltier cooler.

I don't know how to do that calculation. But, note that you'd also have to consider the heat loss through the Peltier device.

Is that 400 watts an electrical rating or a heat rating. I mean, does it take a 125 degree C (150 -25) difference in temperature and produce 400 watts of electricity, or does it take 400 watts of heat energy and produce some smaller amount of electric energy? In either case, that's a significant flow of heat.

I'd be inclined to experiment--build something without the Peltier device and measure the temperature at the end of it. When you have something that seems to work to get close to the right temperature, add the Peltier device and see how much lower the temperature drops.

Place a pot of water on the "cooker". Let the steam vent off to atmosphere. Place one end of your conductor in the boiling water. You now have a constant 100C source. I'd start my testing from there. No regulator needed

"As i'm a undergraduate student who is in the final year of the B.Eng Electronic Engineering degree program conducted by the Sheffield Hallam UK"

Is this question a part of your project?

Nickjayr - It looks to me as though you are ignoring the fundamentals of this problem. You will only get a temperature differential in a heat conducting conduit if you have a flow of heat energy. Let's assume for a moment that there is no leakage from the conduit because it is perfectly insulated on all boundaries except the ends. Then in order for the low temperature end to be at the right temperature it has to conduct the same heat at the desired temperature differential as the peltier device can absorb at its maximum temperature. Then note that the conductor at the hot side of the peltier device will have an ideal surface area. This cross section will likely be a good amount larger than your heat conductor and that an insulated transition will be needed.

This is simple linear mathematics. The heat flow is equal to the temperature differential times the thermal conductance. Get yourself a chart of thermal conductivity of various low cost easy to fabricate materials and do some engineering work. You will probably want to insulate it pretty well to eliminate the unpredictable heat losses to air and mounting points to the system structure.

Don't be afraid to have two different materials in series. One a main conductor from the heater that have a high conductance and an easily replaceable second conductor, perhaps a high temperature plastic shim or a thin sheet of ceramic, mica or insulating paper that lets you fine tune the conductance of heat to the peltier. Then get your thermometers ready and start testing. Don't forget that each joint in the conductor will have its own insulating properties due to a thin film of air. That's what they use heat sink grease for.

Ed Weldon