heatlosspeltier
I'm looking at peltier elements to heat my project. I seem to find that most are about 5% efficient, but how is the 95% turned into heat distributed?
Does the cold side get 47.5% and the warm side get 52.5% of the energy, or is it more complex depending on the difference between the temperatures? I'm really surprised this hasn't b een asked before.
A TEC is polarized in the sense that how it is connected matters. If you want to be able to heat and cool an element then a full Hbridge will work. This will allow you to pump current both directions across a TEC's terminals.
If you apply a positive voltage to a TEC in one polarization then side A will get warm and side B will cool. If you then reverse polarity then side A will cool and side B will get warm. So if you just need heat pumped in one direction then a half bridge or even direct connection will be fine.
Edit: Notice if you apply a positive voltage from one electrical connection to the other you are starting at "P". If you reverse connections you are still starting from "P" but now hot and cold will be flipped.
May be possible, but not viable.
Have a look at the Datasheet of a TEC1-12714S from Thermonamic Module. This is a 62x62x4.9mm³ PE, maximum current 14A, maximum voltage 17.2V. It is one of the most powerful devices on the marked.
The first problem a PE has is that it also conducts heat from the warm to the cold side, and this linearly depends on the temperature difference. While the heat pumping power is about constant, this means that the effective heat pumping power has it's maximum when there is no temperature difference. As soon as the difference rises, heat is also conducted backwards, and at some point, this "back-flow" equals the pumping power. So, the first diagram shows a linear dependency between temperature difference DT and effective pumping power Qc:
However, as long as it's really hot inside the car, the full pumping capacity will be available.
Note also that the maximum pumping power is about 140W for 14A and DT=0. If you have a look at the second diagramm, this current occurs at about 12V. So the PE consumes 12V*14A=168W of electrical power, already more than what it pumps. If there is a temperature difference, effective pumping power decreases while power consumption increases due to higher voltage. So, if t's really hot outside (40°C) and you want 20°C in your car, this PE will only have an effective pumping power of 100W.
Also, you have to dissipate pumping power plus consumed power at the outside, so 268W for the last example.
Up to now, we did not say much about how much power you need. A hair dryer typically has about/above 1000W, which heats your bath room only slowly. 100W may me much for a solder iron, but it's quite negligible for heating a room. The sun heats your car about 1000W/m².
I did not find numbers about car ACs, but is well-known that an AC takes some kilowatts from your engine. You may not notice it on most cars, but I do. I've only about 50kW, and on the highway with a slight slope, I sometimes call the AC switch my "inverse turbo boost". For sure, the AC takes about 2-5kW. As the pumping power can be higher than the consumed power (see this discussion, though not answered yet), you can estimate the pumping power to at least 3kW.
3kW, that's 30*100W, so 30 peltier elements each consuming 168W, being 5040W or 420A at 12V. Furthermore, you have to dissipate 30*268W=8040W of heat somehow.
I think, the numbers of my last paragraph show that it would be better to think of a compressor-based AC solution...
Best Answer
A Peltier device is not a marvellous means of heating unless you also value the ability to cool.
A Peltier cooler will transfer ALL electrical heat energy to the hot side PLUS any energy that is "extracted from" the cold side. So, if it is 5% efficient and you input 100 Watts of electrical energy you will get 5 Watts of cooling and 5 + 100 = 105 Watts of heating. While this is more efficient than using a resistive heater it is much more complex and costly per Watt than using resistive heating.