0
$\begingroup$

As part of a school project, I would like to calculate whether the efficiency of the Peltier effect is greater, less or equal at higher voltages. In other words, I'm testing if the temperature differential as a function of voltage is linear. For thoroughness, instead of testing the temperature of both sides of the plate directly, I would like to calculate the power output/drain of both sides of the plate. To do this, I need the equation for heat dissipation from an flat, uninsulated surface into air, given the temperature of the plate, the ambient temperature, and the conductivity of the plate material. I will set up a fan to blow air across the plate to ensure that the air around the plate remains at ambient temperature, so convection currents can (I think) be ignored.

NOTE: If you have any suggestions regarding the experiment, I will be happy to hear them, but the question still primarily concerns the equation itself.

$\endgroup$
1
$\begingroup$

There is no good theoretical model for the heat dissipation into air. You could work with really large heat sinks, but even that will be hard to model with any precision.

What you are trying to do is called calorimetry and it is usually done with large thermal baths that are well insulated from the environment. For your purposes I would suggest a large water bath on one side of the Peltier element. If you fill it with water and ice and you keep stirring the temperature will stay constant. The thermal connection between the water and the Peltier should be done with a finned heat sink and the water needs to be agitated. If you use one of the pre-assembled Peltier elements, you don't have to work too hard on this. You probably already know that they look something like this: http://www.mpja.com/images/15312.jpg

Putting one side of this assembly into a water bath is going to make a much more reliable heat transfer than air cooling. You could also use a copper or thick walled stainless steel pot with an even bottom, if you can borrow one from your Mom's kitchen. Bolting the Peltier to the bottom might be a bit harder than with an industrial heat sink that you can drill into or that already comes with the Peltier attached. You can probably come up with other ways to make a good thermal connection to the ice-water. That's part of the fun... there is always more than one solution.

The other side of your element should be thermally connected to a well insulated thermal mass with very good thermal conductivity. Ideally you could use a copper or aluminum block. This thermal mass has to be well insulated with e.g. styrofoam. While you are doing your experiment, you will be "pumping" heat in and out of this second thermal mass while monitoring its temperature with a thermometer or thermo-element, if you have access to electronic measurement equipment. Curiously people are even selling aluminum blocks for this kind of purpose... see e.g. http://www.capitolscientific.com/Benchmark-Scientific-BSW01-Digital-Dry-Bath-Solid-Aluminum-Heating-Block-for-Microscope-Slides-or. OK, that's a bit pricy for what you are trying to do... try to find a chunk of metal with an even surface for less money. Saving cost is also one of the things that the scientists has to learn.

Use a thermal compound when connecting the Peltier to the heat sink and your thermal mass to ensure that the heat flow is as good as possible. The white aluminum oxide paste kind is cheap, just be careful not to get it on your clothes (wear a lab coat as a proper science special effect, anyway!).

Now that you have one side of your Peltier on a constant temperature stabilized by the melting ice you can focus on the temperature change on the other side. This makes your life much easier. You can still model the entire system dynamically with thermal resistances and multiple thermal masses, if you want to, but if you chose a large enough thermal mass for your measurement, so that the thermal mass of the Peltier doesn't matter much, then this won't be necessary. Moreover, your system automatically resembles a near optimal application. In real life the device will probably never perform nearly as well as it will on the water bath, i.e. your measurement will approximate the best case scenario, already.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.