bhaazee Posted May 28, 2012 Share Posted May 28, 2012 Recently, I came the topic Thermoelectric Modules, Coolers. I have a basic doubt. Assuming, I am supplying 100w of electric power to a thermocouple and heat starts generating on one side and cooling occurs on the other side. I would like to know what percentage of the electric power is converted into heat? Does heating : cooling ratio equals 50 : 50? Regards Link to comment Share on other sites More sharing options...
Externet Posted May 28, 2012 Share Posted May 28, 2012 The coefficient of performance for Peltier modules is around 1%. Not even near 50:50. Link to comment Share on other sites More sharing options...
John Cuthber Posted May 28, 2012 Share Posted May 28, 2012 Are you sure? "Technically, the word efficiency relates to the ratio of the amount of work one gets out of a machine to the amount of power input. In heat pumping applications, this term is rarely used because it is possible to remove more heat than the amount of power input it takes to move that heat. For thermoelectric modules, it is standard to use the term "coefficient of performance" rather than "efficiency." The coefficient of performance (COP) is the amount of heat pumped divided by the amount of supplied electrical power. The COP depends on the heat load, input power, and the required temperature differential. Typically, the COP is between 0.3 and 0.7 for single-stage applications. However, COPs greater than 1.0 can be achieved especially when the module is pumping against a positive temperature difference (that is, when the module is removing heat from an object that is warmer than the ambient)." from http://www.tetech.com/FAQ-Technical-Information.html#2 (other suppliers are available, but this was the first one Google found for me) Link to comment Share on other sites More sharing options...
swansont Posted May 28, 2012 Share Posted May 28, 2012 The coefficient of performance for Peltier modules is around 1%. Not even near 50:50. Are you thinking of the Seebeck effect performance? That's temperature difference generating electricity, and it's generally few percent. But the numbers John Cuthber posted are right in my experience. I've used modules that drew 5A at 12V and would chill, moving a few tens of Watts. Link to comment Share on other sites More sharing options...
Externet Posted May 28, 2012 Share Posted May 28, 2012 From what I remember, Peltier modules COP is about 1%; Seebeck modules efficiency is about 5% Link to comment Share on other sites More sharing options...
John Cuthber Posted May 28, 2012 Share Posted May 28, 2012 If peltier coolers really dumped 100 times more heat than they pumped, who would buy them? Link to comment Share on other sites More sharing options...
InigoMontoya Posted May 28, 2012 Share Posted May 28, 2012 If peltier coolers really dumped 100 times more heat than they pumped, who would buy them? We use 'em at my place of employment. The application that they generally find themselves in.... - Temporary. IE, the test rig is only going to be in operation for a few days before being torn down. - Extreme noise/vibration. As in, your ear drums would literally rupture if you were in the room. - Hot. Ambient temperatures in the region of 100-120 F. - Lots o' power available. Need 100 amp 480 VAC service? No problem. And through all that you need to keep some electronics cool. Peltier coolers are cheap and easy to throw into a box. They handle harsh environments. We don't care too much about efficiency. What's not to like? 2 Link to comment Share on other sites More sharing options...
swansont Posted May 29, 2012 Share Posted May 29, 2012 http://customthermoelectric.com/tecs_imax.html?gclid=CO7j0NygpbACFUSK4AodNEK6ag All the ones I checked have a Qmax of around 0.5*IV Link to comment Share on other sites More sharing options...
the asinine cretin Posted May 29, 2012 Share Posted May 29, 2012 We use 'em at my place of employment. The application that they generally find themselves in.... - Temporary. IE, the test rig is only going to be in operation for a few days before being torn down. - Extreme noise/vibration. As in, your ear drums would literally rupture if you were in the room. - Hot. Ambient temperatures in the region of 100-120 F. - Lots o' power available. Need 100 amp 480 VAC service? No problem. And through all that you need to keep some electronics cool. Peltier coolers are cheap and easy to throw into a box. They handle harsh environments. We don't care too much about efficiency. What's not to like? Your job sounds pretty sweet. +1. Link to comment Share on other sites More sharing options...
bhaazee Posted May 29, 2012 Author Share Posted May 29, 2012 Made a simple calculation on the ratio of heat removed to heat emitted (values taken from products in ebay) 29% of power supplied produces cooling and the rest is emitted as heat Link to comment Share on other sites More sharing options...
InigoMontoya Posted May 30, 2012 Share Posted May 30, 2012 Your job sounds pretty sweet. +1. I'm pretty partial to it. Me ...and the facility I run. Link to comment Share on other sites More sharing options...
the asinine cretin Posted May 30, 2012 Share Posted May 30, 2012 I'm pretty partial to it. Me ...and the facility I run. Um... Damn. Rockets. . . *picks up jaw off of floor* Link to comment Share on other sites More sharing options...
Enthalpy Posted May 31, 2012 Share Posted May 31, 2012 I confirm the very few % efficiency, like 2%. 5% is already better than what I used long ago. The theoretical limit is that electrical conductors that link the contacts of different materials need a section to reduce ohmic losses, but then they carry heat as well. Metals have the least bad ratio of heat-vs-electricity, but semiconductor offer a larger thermoelectric voltage and a less bad. Superconductors aren't used there (even at suitable temperatures) despite low heat conductivity; I ignore why. Link to comment Share on other sites More sharing options...
John Cuthber Posted May 31, 2012 Share Posted May 31, 2012 And I'm going to repeat my original question, but in a slightly different form. Say I have a processor chip in my PC dissipating 50W (which isn't a lot according to this) http://en.wikipedia.org/wiki/List_of_CPU_power_dissipation Now I want to cool it with a peltier unit. If the efficiency is really only 5% then I need to use 20 times more power to cool it than it dissipates. So I would need a kilowatt of peltier power to keep the processor cool. But that peltier unit is driven from the PC's power supply unit. That power supply isn't rated for anything like 1KW. So the 5% figure is impossible. As I asked before who would use them? Link to comment Share on other sites More sharing options...
swansont Posted May 31, 2012 Share Posted May 31, 2012 I confirm the very few % efficiency, like 2%. 5% is already better than what I used long ago. Confirm? I gave a link that showed devices with ~0.5 COP numbers. How can you "confirm" 2%? Link to comment Share on other sites More sharing options...
Enthalpy Posted June 4, 2012 Share Posted June 4, 2012 The thermal power given is for zero temperature difference, for which one wouldn't need any module. ...Say I have a processor chip in my PC dissipating 50W... I want to cool it with a peltier unit. If the efficiency is really only 5%... I would need a kilowatt of peltier power to keep the processor cool. That's why Peltier modules are bad candidates to cool Cpu. People who want a Cpu cooler than the heat dissipator use a standard cooling cycle like in a fridge, or have a stock of liquid nitrogen. Peltier modules are seducing where little power must be removed and vibration free operation is better, for instance at the sensor of a camera. Link to comment Share on other sites More sharing options...
swansont Posted June 4, 2012 Share Posted June 4, 2012 The thermal power given is for zero temperature difference, for which one wouldn't need any module. And if you're correct, this immediately drops to 0.02 as soon as a temperature difference exists. Link to comment Share on other sites More sharing options...
John Cuthber Posted June 4, 2012 Share Posted June 4, 2012 Enthalpy, You are missing the point. "People who want a Cpu cooler than the heat dissipator " are very rare. Not many people want to chill the CPU below ambient temperature. People actually use Peltier coolers because they work quite well. If they were as poor as has been suggested then to run a decent spec CPU chip you would need to feed more power into the cooler than I use to run my entire central heating system. Why do you think people use Peltier coolers on CPUs? Link to comment Share on other sites More sharing options...
Enthalpy Posted June 5, 2012 Share Posted June 5, 2012 And if you're correct, this immediately drops to 0.02 as soon as a temperature difference exists. At least as soon as you want an interesting temperature difference, say to cool an infra-red imager or a sensitive visible imager, you need several cooling steps. Three steps of 27% efficiency bring you to 2%, and the imager isn't even very cold. Why do you think people use Peltier coolers on CPUs? Looks like they don't. http://www.heatsink-...com/peltier.htm introduction with figures, consistent with our discussion here http://www.tomshardw...cooling-peltier disappointed experimenter, consumes 100W for a Pentium III http://www.tomshardw...ate-suggestions efficiency problem is known http://www.tomshardw...peltier-cooling efficiency problem is known there as well I've checked in case some advance has been made, but Peltiers are still as limited as for 20 years. Pity. Link to comment Share on other sites More sharing options...
John Cuthber Posted June 6, 2012 Share Posted June 6, 2012 (edited) It looks like they do. http://www.ebay.co.uk/sch/i.html?_nkw=peltier&_sacat=58058&_odkw=computing&_osacat=58058 And trying to pretend that they don't just makes you look silly. Incidentally, even you have just tacitly accepted a 27% efficiency. Here are some data (unless you are going to pretend that a major company is telling fibs.) http://docs-europe.electrocomponents.com/webdocs/002a/0900766b8002a346.pdf Let's start with the first one in the table. It pumps 1 Watt. To do so it takes 2.1 A at 0.85V. That's 1.78Watts Why are you seeking to tell us that 1 W is "very few % " of 1.78W? It looks like roughly half to me. Feel free to look at the other actual facts presented and see if any of them supports your claim. Edited June 6, 2012 by John Cuthber Link to comment Share on other sites More sharing options...
Enthalpy Posted June 6, 2012 Share Posted June 6, 2012 JC, I don't recognize you usual style here. As I already explained, this thermal power is given for zero temperature difference, as well in the Pdf you linked. Selling Peltiers for computers on eBay does not mean they're useful. I provided links to user experience telling they're not and why. As I stated, in normal sensible uses - which do not include Cpu cooling - you'd have 3 stages or more, dropping the efficiency to very few %. But if you want to waste money and time to buy and try a Peltier on a Cpu, do so and get disappointed as others did. Link to comment Share on other sites More sharing options...
John Cuthber Posted June 6, 2012 Share Posted June 6, 2012 (edited) Glad you recognise the style. It's called showing evidence, and I plan to carry on with it. It's true that the figures quoted are for delta T of zero, but these things still work when delta T is quite large. Pointing out that if something is inefficient then using lots of them is less efficient isn't really helpful. People have been buying and using them for years. The market isn't that generous: if they didn't work, people wouldn't use them. Anyway, as I said, I usually try to provide data so here's some more. http://docs-europe.electrocomponents.com/webdocs/007b/0900766b8007ba20.pdf page 8, coefficient of performance still exceeds 10% for a delta T of 60C for a single stage. For a smaller delta T the efficiency improves. With a small enough temperature difference, say 20C (which is still just about useful) the COP is a little over 100%. Edited June 6, 2012 by John Cuthber Link to comment Share on other sites More sharing options...
Enthalpy Posted June 7, 2012 Share Posted June 7, 2012 And 10% is exactly what one doesn't want when cooling a Cpu. Link to comment Share on other sites More sharing options...
John Cuthber Posted June 7, 2012 Share Posted June 7, 2012 And 10% is exactly what one doesn't want when cooling a Cpu. Wait a minute. You said the answer was 2%. Even with Delta T of 20C you get an answer above 100% and yet you claimed that you have "confirmed" that it was 2%. Are you actually accepting that 2% or 5% was the wrong answer? How many pages of data showing COP greater than 50% do I need to cite before you agree that 50% is bigger than 5%? Link to comment Share on other sites More sharing options...
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