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I know that thermocouples produce electric current when there is a temperature difference between their two ends. This current is directly proportional to the temperature difference. I also understand that thermocouples are not very efficient by themselves, ranging between 3-8% efficiency on average.What I would like to know is that can thermocouple arrays-connected in a combination of series and parallel-produce moderately sufficient power (from 30-50%)?

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The fact that thermocouples produce a (relatively low) EMF is key to understanding how to get power out of them.

You can't do anything about the EMF of a single junction (for a given temperature difference, which is limited by your heat source and material melting points) but you can increase the current by reducing the electrical resistance (i.e. using thicker wire).

However this also reduces the thermal resistance, conducting heat from the hot junction directly to the cold junction - and this is where the efficiency limit comes from. You cannot conduct more current from that junction without conducting more precious heat away from the hot junction.

he crucial parameters in a thermocouple are thus:

  1. The potential (EMF)
  2. The ratio of electrical conductivity to thermal conductivity.

It's one of those annoying facts of life that good electrical conductors are normally also good thermal conductors, so (2) tends to be pretty similar (and lower than you want for this purpose) across materials.

Typical metals yield efficiencies of 3% or below : to get 8% you need fairly exotic materials (bismuth and tellurium compounds I think). I don't see any prospect of improving this without some fundamental breakthrough in solid state physics.

Series or parallel combinations of junctions cannot affect this : all those thermal shorts are always in parallel. You can increase power by using more in series (increasing the voltage) or more in parallel (another way of thickening the wire) but the efficiency won't improve : those thermal shorts are still always in parallel.

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I know that thermocouples produce electric current when there is a temperature difference between their two ends. This current is directly proportional to the temperature difference.

No, they produce a voltage that's proportional to the temperature difference between their two ends. Any current that flows is drawn by a connected load. The largest available electrical power output occurs when the load resistance is equal to the thermocouple internal resistance, this is, at half the open circuit voltage, and half the short circuit current.

Combining thermocouples in series raises the output voltage. This is usually necessary as most power converters will not work with the voltage output of one alone.

Combining thermocouples does not change the overall heat to electrical power efficiency.

There are two main types of thermocouple you can buy. There are temperature measurement thermocouples, which plug into a meter. With their long non-copper leads, these are relatively high resistance and not able to produce much power. Flame sensor thermocouples fit into boilers and gas hobs, and actually power the gas hold-in valve solenoid directly. These are very low resistance, and more suitable for power. But then in normal use, they have a lot of 'free' input power available, and a huge temperature difference to work between.

To get maximum power out of a thermocouple, consider the following thought experiment. Take two long thermocouple wires, and make two junctions. Now cut each wire in two, and make 4 junctions. The output voltage will double for the same resistance, so you can extract more power. However, the length of the heat-leak path through the wires has halved, so you need more thermal power to maintain the temperature difference. If you had the heat available, you could repeat this halving of length and doubling of number of junctions to increase both thermal input and electrical output without limit, but always wit the same heat to electrical efficiency.

If you want to decrease the amount of heat you use for any given electrical output, then you need to increase the thermal resistance of the heat path. Unfortunately, with metals, both electrical and thermal conductivity tend to vary together, so it's not possible to get a good ratio with metals.

Enter semiconductors, and meta-materials. Semiconductors have a much higher voltage output than metals, and a higher electrical to thermal conductivity ratio. That's why they are used in Peltier devices. Much research is going into understanding thermal conductivity, and designing new meta-materials that will have better ratios, and so make more efficient Peltiers.

If you want something more efficient than metal thermocouples, throw them away, and switch to a Peltier.

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  • \$\begingroup\$ Thermocouples,as I understand, are not very efficient. A single thermocouple is about 3-8% efficient. What I'm getting at is if the thermocouples were combined in series, parallel or combination would their power add up, increasing their overall efficiency? \$\endgroup\$
    – user153670
    Sep 15, 2017 at 8:03
  • \$\begingroup\$ @Mats Which bit of combining thermocouples does not change the overall heat to electrical power efficiency don't you get? Whether combined in series, parallel or however, their power would add up, and their heat consumption would add up, resulting in nominally the same efficiency. There may be small differences if one combination had more or less heat leak, or more or less extra wiring resistance, but that's down to how well the combination is put together, not to the combination of the thermocouples per se. \$\endgroup\$
    – Neil_UK
    Sep 15, 2017 at 8:06
  • \$\begingroup\$ Fine,then tell me this. If I wanted to increase my panel's overall power capacity, what would be the best way to go about it? \$\endgroup\$
    – user153670
    Sep 15, 2017 at 8:10
  • \$\begingroup\$ @Mats What is a panel, a panel of what? What is its present power capacity? And don't say 'duh! thermocouples', say how many thermocouples of what technology (material), with what configuration, what temperature difference, what o/c voltage., what s/c current, manufactuer's part numbers if they're purchased. \$\endgroup\$
    – Neil_UK
    Sep 15, 2017 at 8:13
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    \$\begingroup\$ Which bit of be nice don't you get? \$\endgroup\$
    – uhoh
    Sep 15, 2017 at 13:31
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I found out away to increase the efficiency by two ways.

  1. Setting up multiple Peltier's in a matrix series and parallel.
  2. Change the metals by using high temp super conductivity mix in the making of the Peltier's unit.
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    \$\begingroup\$ And what is the thermal conductivity of a superconductor? What overall efficiency have you found for this? \$\endgroup\$
    – user16324
    Feb 3, 2021 at 11:33

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