We use thermocouples (type B) to measure the temperature of metals (Fe, Ag, Pd, etc) in a furnace under pressure. We weld two thermocouple wires together using an arc welder. This junction then comes into contact with the metal we're measuring. This looks slightly like this: enter image description here

This is done in a temperature gradient. The junction in this case is at 1200 °C, and it has T gradients of ±100 °C over just a few mm.

Because of the high pressure and temperature, we occasionally get contamination of the junction by the metal we're measuring by diffusion and alloying. Here's how it looks:

enter image description here

The junction is a mixture of the two wire metals (because of welding) and you can see some of the yellow diffusing into the junction.

My question is, does it make a difference? When and where does it make a difference? Where is the thermocouple actually measure? In the entire area of the junction, or only where the contact is? Does it matter if it's in contact with a conductive metal?


Here's a backscattered electron image of a sectioned contact between the thermocouple and the metal:

enter image description here

Point 1 is Pt-Rh alloy (type B tc), point 3 is Fe, and point 2 is Pt-Rh-Fe alloy. Point 2 is basically the Pt-Rh thermocouple contaminated by Fe during the high-T measurement. Scale bar is 100 μm.


1 Answer 1


[Note: answer completely updated]

Based on your SEM picture and the further explanation in the comments, it seems that your system looks like this:

Thermocouple bead

You have materials A and B that form the thermocouple pair, a bead of mixture "AB" as a junction. The bead is embedded in and fused with a substrate material "C". A further complication is that there is a substantial temperature gradient in the $z$ direction. The numbers 0--7 are points of interest; 0 and 7 are the reference points.

Resistance network

We can approximate the circuit above as a resistance network:

Thermocouple resistor network

The wiggly lines are resistors. Each numbered point $i$ has a temperature $T_i$. Each connection from one point $i$ to the next point $j$ is made of a uniform material $X$ without junctions. The electromotive force (emf) over each connection is $$\mathcal E_{ij} = S_X (T_j-T_i),$$ Where $S_X$ is the Seebeck coefficient of material $X$; $X$ can be one of $\{A,B,AB,C\}$. The Seebeck coefficient may be temperature-dependent; for the reasoning here, I'll assume that they are constant or that all temperatures are defined as an offset relative to a high reference temperature.

I call them electromotive forces rather than voltages, because the actual voltages are the result of the electromotive forces combined with the voltage drops over the various resistors. We can assume that the measurement device at points 0 and 7 has a very high internal resistance, so there is no current in the legs 01 and 67. But currents may flow elsewhere in the network.

Let's first look at the voltage between points 1 and 6: $$\mathcal E_{16} = S_{AB}(T_6-T_1).$$ This emf competes with the emf over the path 1256, $$\mathcal E_{1256} = \mathcal E_{12}+\mathcal E_{25} + \mathcal E_{56} = S_{AB}[(T_2-T_1) + (T_5-T_2) + (T_6-T_5)] = S_{AB} (T_6-T_1). $$ Now that's convenient: the temperatures $T_2$ and $T_5$ don't matter. Now let's see what the path 123C456 (via the substrate) does: $$\mathcal E_{123C456} = S_{AB} (T_6-T_1) + (S_C-S_{AB})(T_4-T_3).$$ This is not so nice: you get an extra emf proportional to the difference $T_4-T_3$. The potential difference between points 1 and 6 will now be dependent on the exact values of all the resistances and the actual values of the temperatures along the way. This is already a pain to solve for the simplified resistor network; for the actual case of a 3D bead embedded in the surface, you would have to resort to a numerical simulation - and you still have to know all the Seebeck coefficients and specific resistances of the materials.

Engineering solution

In the comments, you stated that you are more interested in an accurate measurement method than in a full physical analysis of a problematic measurement.

In the equations, you see that difficult-to-interpret emfs will occur if $T_1 \neq T_6$ and/or $T_2\neq T_5$ and/or $T_3\neq T_4$. If you make the system symmetric, all temperature pairs will be equal and your thermocouple will effectively only measure the temperature $T_1$, which is equal to $T_6$. To achieve this symmetry, the bead must be really uniform; the material composition around point 2 must be equal to the composition around point 5. The lead wires must leave the surface in a symmetric way. If you don't want the temperature at point 1 and 6 as in the drawing, but rather the temperature at the same $z$ position as the substrate surface, then you should embed the bead deeper into the substrate.

Actually, you could even consider leaving out the whole AB alloy. Instead, braze/weld the "A" and "B" wires directly and separately to the substrate.

  • $\begingroup$ I edited to show a real-life example of how this set up looks after a measurement. If I understand you correctly, the R1 junction is located where the two thermocouple wires first come into contact in the top right corner of the welded bead. Is the R2 junction located in the contact where the welded bead meets the iron? The contamination zone is only few tens of μm wide. Thus, the T gradient over it is negligible. The Fe metal continues to the bottom left for about a cm, and there is a gradient over it. $\endgroup$
    – Gimelist
    Jun 5, 2016 at 22:34
  • $\begingroup$ Is the SEM image a cross section? Is the junction fused with the substrate? $\endgroup$ Jun 6, 2016 at 6:41
  • $\begingroup$ Yes, it's a cross section. If all of the area of the welded bead and the part where it alloyed with the Fe metal in the same temperature, is there still a requirement for solution in 3D? $\endgroup$
    – Gimelist
    Jun 6, 2016 at 8:26
  • $\begingroup$ I'll get back with more detail in 10 hours or so, but do you really want to solve this as a physics problem? It may be better/easier to treat it as an engineering problem: "what is the best way to measure the surface temperature" rather than "how bad is my measurement if I do it this way". $\endgroup$ Jun 6, 2016 at 8:53
  • $\begingroup$ I'm all for the engineering point of view $\endgroup$
    – Gimelist
    Jun 6, 2016 at 8:57

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