How much power can I expect to obtain from 60W thermoelectric cooler if I use it as a generator with a temperature difference of about 50C?
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Answer:
First 60 watts means nothing, it translate to 60 watts maximum transfer of heat when temperature difference is zero!Second the generated voltage is proportional to the Seebeck effect, and the cell is made of many couples in series parallel.For an estimation take a cell made of 126 couples, Seebeck constant 0.05 volts/K (SM), then you need a difference of 100 °K to get a useful voltage from a cell with 3 ohm internal resistance (RM). But let's use 50 °K (DT).Use https://www.ferrotec.com/technology/thermoelectric/thermalRef13/formulas.Maximum Power PM= (SM x DT)^2/(4 x RM)= (0.05 x 50)^2/(4 x 3)= 0.52 watts.Even worse when you observe that max power load means 3 ohm with a voltage of 1.25 volts! at 0.4 amperes.Maybe for experimental reasons you go ahead with 50 °K, then you need to pump some heat (Qc) into the module with hot surface at say 323 °K with thermal conductivity 0.7 watts/°K (Kc) Qc= SM x Th x I- 0.05 x I^2 x RM+ Kc x ÎT, the minus sign for self heatingQc= 0.05x323x0.04-0.05x0.04^2x3+0.7x50= 35.6 watts of heat inAnd heat out on the cold plate of 35.6-0.52 or 35.1 watts at 273 °KTry to scale this to a useful power of say only 10 watts!You are stuck with an efficiency below 1.5%, and in the best case below 3% with ÎT of 100 °K The real problem is not the low power output, is the big waste of heat that must get out on the cold plate.
Miguel Gonzalez at Quora Visit the source
Other answers
You can calculate this for yourself if you know the properties of the junction material in your thermo-electric cooler. See http://en.wikipedia.org/wiki/Thermoelectric_materials
Steve Blumenkranz
Very little. It will never recoup its cost.
George Gonzalez
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