# Peltier Calculation Conundrum

I have a peltier (TEC) device capable of 400W Qmax, a 160W thermal load that I'd like to keep at ambient, and this water block running at about 1.5 gpm w/ room temp water. I think this system is right on the border of what is possible, but I'm struggling to determine the exact efficiency of my peltier unit.

The datasheet for my peltier indicates a Qmax of 400W and claims that maximum applied power is about 690W -- does this mean that I have a 57% efficiency? To me that seems extraordinarily high for a TEC.

Because I don't believe 57% is right, let's assume a 30% efficiency. In that case, my peltier will need to be operated at around 533W in order to dissipate the 160W load + the load from the peltier. And now, I have 693W total being dumped into my water block.

Am I doing this calculation correctly?

Other notable factors: - This system will be operated in a vacuum environment so convective cooling is a no-go. - There is a thin (~1/8") aluminum plate between the heat load and the peltier.

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• Possible that electronics.stackexchange.com has more folks with practical experience with these devices. – BowlOfRed Aug 9 '18 at 23:00
• What are the temperatures? How tightly do you need temperature control? If your water is at room temperature, and your ambient is just above room temperature do you need a TEC at all? Look at the Q vs I plot at various delta Ts to see how much heat you can pump and at what cost (using the other plot - V vs I) – D Duck Aug 10 '18 at 3:33
• Great suggestion D Duck -- I hadn't really thought about just using the water block without the TEC. The tricky thing is that I really have two scenarios for this test setup: 1.) I want to do what is described above and simply dump heat out of the system to prevent overheating. 2.) I want to use the system as a thermal vacuum plate and cool 'small objects' to within the ballpark of -15 degC. Because of the second case, it makes it easier to leave the TEC in so that I don't have to install/ uninstall it constantly. I'll try running just the water block and see how effective that is. – Austin Prater Aug 10 '18 at 17:08

Your water is at 20 C and the block is at -15 C. Temperature differential is 35 C.

That's a pretty big gradient for the TEC. The plot of Q vs I says that you can shift no more than about 125 W but you'll need 28 A and that will cost you about 675 W. (this is from the plots on the data sheet)

If the source of heat is 25 W or so it will require 10 A and cost you about 100 W (10 V * 10 A). This is where I would use the TEC you specified.

This seems to indicate that you'll need either:

1. To settle for a lower amount of heat shifted or
2. More TECs in series to reduce the temperature gradient of each one or
3. More TECs in parallel to increase the heat pumped or
4. Settle for a smaller temperature differential

or a combination of the above.

PS: You'll need to do option 3 when you apply option 2 because now you need to get rid of the energy used to pump the heat from the colder stage.

• Hey D Duck -- I'm interested in your statement "this is where I would use the TEC you specified". I've been using peltier devices for some time now, however, one of my longtime hunches has been determining which unit to use in which situation. From what I can tell, Qmax is the end-all-be-all property for determining "how powerful" a particular unit is. What makes you say that 25 W heat load is "where you would use" that particular TEC? – Austin Prater Aug 10 '18 at 23:55
• My thinking: What is the temp difference? (usually a laser, keep it near ambient but such that you're always cooling (or heating) so you don't need a bipolar drive) Not too cold, there's problems with condensation and that adds heat load. Your heat sink is at +20 and the cold side is -15 C. Delta T is 35. ~40 so you can look up the plot (Qo vs I) dashed red. I would want to control around a bit of that curve that's still sloped. More current, more heat pumped. If it's horizontal, more current, no more heat shifted. Starting cooling some capacity is required to cool from ambient. 75W tops. – D Duck Aug 11 '18 at 0:09
• Can't +1 because of this dumb "reputation" score thing, but thanks! – Austin Prater Aug 11 '18 at 0:13

TEC efficiency (COP=$Q_C/P_{el}$) depends on many factors, including temperature difference, and can exceed 1 (https://www.meerstetter.ch/compendium/peltier-element-efficiency). So one cannot say that 57% is too high without a more detailed analysis.

• Thanks ahkmeteli. I'll continue running calibration testing to see if I can hone-in on my COP. Cheers – Austin Prater Aug 10 '18 at 17:12