3
\$\begingroup\$

Specifically, I've been working with pre-existing temperature cables which are T-type thermocouples which share the same Constantan wire. My question is if one of the thermocouple junctions is heated significantly higher than the others (such as 30 C - 40 C higher), will that hot junction affect the measurement of cooler junctions? Please see the drawing for a quick illustration. If Junction 2 is heated to 60 C and Junctions 1 and 3 are at 25 C, is it correct that the measurements at Junctions 1 / 3 could be incorrectly read as low as 17 C?

Along these lines, I guess another way to frame the question is would the Seebeck Effect take place at Junctions 1 / 3 if the only measurement connection was for Junction 2 (would charge carriers move with respect to temperatures even if left unconnected)? Also, would the moving charge at the heated Junction 2 affect Junctions 1 / 3 and their movement of charges?

We have our own circuit taking measurements, and I have noticed a drop in temperature on the cooler junctions when taking their measurements. I've repeated this result using a purchased thermocouple reading device.

3 thermocouples sharing a common wire.  Measurements are taken one at a time, not in parallel.

Any time or clarification is greatly appreciated.

\$\endgroup\$
8
  • \$\begingroup\$ I think I know what's happening, but I'm struggling to describe the physical model to you. Before I try, are you creating your own combo system by fusing the dissimilar metals yourself at those junctions? (Not buying pre-made thermocouples, but actually using a supply of the right kind of both metals and forming a junction by some kind of weld you make yourself?) Or is this a different setup? \$\endgroup\$
    – jonk
    Commented Oct 24, 2018 at 21:31
  • \$\begingroup\$ @jonk We are purchasing pre-made cables which are like this. My best guess as to why is for lower cost because they will get broken and need replacement somewhat often. I know the company will join these two metals together (from what I can tell they are soldered together). \$\endgroup\$
    – user43655
    Commented Oct 24, 2018 at 23:07
  • \$\begingroup\$ As I understand it, how this works is that the dissimilar metals, bonded, force their Fermi levels together all the time. Physicists use \$\infty\$ as their "zero-reference" for energy. For atoms, this means "vacuum, far away from the nucleus." The work functions and Fermi levels and so on are all referenced against this as the default zero-energy point. Differing metals have different Fermi levels. And these Fermi levels for a given metal are different at any given different temperature. \$\endgroup\$
    – jonk
    Commented Oct 24, 2018 at 23:36
  • \$\begingroup\$ Bonding them together locks the Fermi levels together, so to speak. So their "energy distance" relative to "vacuum, far away from the nucleus" moves about, instead. Since the Fermi levels are locked tight now, it's instead the difference between one metal's energy distance to vacuum, vs the other metal's, that the thermocouple measures. The sea of electrons in the conduction band will distribute immediately throughout all of the metal structures, regardless of kind of metal, such that they maintain a cloud with its electron particles as far apart from each other as can be achieved. \$\endgroup\$
    – jonk
    Commented Oct 24, 2018 at 23:36
  • \$\begingroup\$ This means that there are very long ranging effects, as the electrons redistribute throughout the entire system. I almost cannot see a way that the highest temperature joint would not affect the others. \$\endgroup\$
    – jonk
    Commented Oct 24, 2018 at 23:38

1 Answer 1

1
\$\begingroup\$

if any of the junctions is in a temperature gradient it will influence the reading seen on more distant junctions. Junctions should be evenly heated - this includes any heat conducted in via the constantan wire.

if multiple copper wires are passing current at the same time resistance in the constantan will skew the results. read the junctions one at a time.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.