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So I've been experimenting creating thermocouples with copper and constantan wire, just by twisting them together at one end and heating the junction with a candle and using a voltmeter to measure the emf at the other end. I thought that changing the gauge of the copper wire might change the emf output, and it appears that the smaller the diameter of the wire the larger the voltage is, but I'm not sure why this is. From research, it looks like the thermoelectric efficiency is determined by the Seebeck coefficient, the thermal conductivity, and the electrical conductivity, as well as the temperature difference.

Since the conductivities are intrinsic properties of the copper and I'm keeping the temperature constant, I'm not sure why the output of the thermocouple is changing. I thought that I might be changing the conductivity in some way by changing the gauge, but is this possible?

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  • \$\begingroup\$ Weld them. Twisting them creates a distributed "thermocouple" region, thus all bets are off. Weld them. In one tiny spot. \$\endgroup\$ Jan 13, 2018 at 4:05

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Thicker wires mean you are heating the 'cold' end more (and, as in the comment) drawing more heat from the hot end. To a rough approximation the no-load voltage is proportional to the temperature difference between the ends.

A long thick wire (long enough the cold end does not heat appreciably) should not be any different from a long thin wire pair if the alloys are similar and negligible current is drawn.

Think of a heat engine, if that makes sense, when you start drawing current the resistance figures into it as well as the heat flow down the wires.

To get a good reading you have to have both ends at controlled temperatures. For a rough measurement (maybe within 2-3 degrees C) you can put one end (the junction) in boiling water and the other (transition junction(s) to copper on both sides) into an ice-water slurry.

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  • \$\begingroup\$ So basically the internal resistance of the wire to the current is causing the cold-end to heat more, effectively reducing the temperature difference? \$\endgroup\$
    – Josh B
    Jan 12, 2018 at 2:28
  • \$\begingroup\$ Right, thermal resistance, which, like electrical resistance, is inversely proportional to cross-sectional area of the wire. \$\endgroup\$ Jan 12, 2018 at 2:30
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    \$\begingroup\$ Thicker wires also mean you're heating the hot end less (same amount of heat energy gives less temperature rise). \$\endgroup\$ Jan 12, 2018 at 2:52
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You may be having types of difficulty.

  1. The voltage may be affected by the quality of the junction. Twisting the wires together is not going to make a good thermocouple junction. They should be welded together at a single point.

  2. When you connect the voltmeter to the thermocouple wires, you create two more thermocouples. There are various ways to compensate for that. One is to make two thermocouples with the thermocouple wire. Connect two copper wires to opposite ends of the constantan wire. Measure the voltage at the ends of the copper wires. Has one of the copper-constantan junctions leave the other at room temperature or put it in ice water. The voltage should be proportional to the temperature difference between the two junctions.

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