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Question
Find the neutral temperature and inversion temperature of a copper-iron thermocouple if the reference junction is kept at 0°C. Use the data given in the following table.
| Metal with lead (Pb) |
a `mu V"/"^oC` |
b `muV"/("^oC)` |
| Aluminium | -0.47 | 0.003 |
| Bismuth | -43.7 | -0.47 |
| Copper | 2.76 | 0.012 |
| Gold | 2.90 | 0.0093 |
| Iron | 16.6 | -0.030 |
| Nickel | 19.1 | -0.030 |
| Platinum | -1.79 | -0.035 |
| Silver | 2.50 | 0.012 |
| Steel | 10.8 | -0.016 |
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Solution
Neutral temperature, \[\theta_n = - \frac{a}{b}\]
\[a_{CuFe} = a_{CuPb} - a_{FePb} \]
\[ = 2 . 76 - 16 . 6 = 13 . 84 \mu V^\circ C^{- 1}\]
\[b_{CuFe} = b_{CuPb} - b_{FePb} \]
\[ = 0 . 012 + 0 . 030 = 0 . 042 \mu V^\circ C^{- 2}\]
Thus, the neutral temperature,
\[\theta_n = \frac{- a_{CuFe}}{b_{CuFe}} = \frac{13 . 84}{0 . 042} = 329 . 52^\circ C = 330^\circ C\]
The inversion temperature is double the neutral temperature, i.e. 659 °C.
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