Thermocouple Sensitivity Calculation

The sensitivity of a thermocouple measures the change in output voltage (potential) per degree Celsius change in temperature. For this Type T thermocouple, we need to find the rate of change of junction potential \( E \) with respect to temperature \( \theta \).

Sensitivity Function Derivation

The junction potential \( E \) (in \( \mu V \)) is given as a function of temperature \( \theta \) (in $^\circ C$):

$ E(\theta) = 38.74\theta + 3.3\times 10^{-2}\theta^2 + 2.07\times 10^{-4}\theta^3 - 2.2\times 10^{-6}\theta^4 + \dots $

Sensitivity \( S \) is the derivative of \( E \) with respect to \( \theta \):

$ S(\theta) = \frac{dE}{d\theta} $

Differentiating the provided polynomial terms:

$ S(\theta) \approx \frac{d}{d\theta} \left( 38.74\theta + 3.3\times 10^{-2}\theta^2 + 2.07\times 10^{-4}\theta^3 - 2.2\times 10^{-6}\theta^4 \right) $

$ S(\theta) \approx 38.74 + (2 \times 3.3\times 10^{-2})\theta + (3 \times 2.07\times 10^{-4})\theta^2 - (4 \times 2.2\times 10^{-6})\theta^3 $

$ S(\theta) \approx 38.74 + 6.6\times 10^{-2}\theta + 6.21\times 10^{-4}\theta^2 - 8.8\times 10^{-6}\theta^3 $

Calculating Sensitivity at 100 $^\circ C$

Substitute \( \theta = 100 \text{ }^\circ C \) into the sensitivity equation:

$ S(100) \approx 38.74 + (6.6\times 10^{-2})(100) + (6.21\times 10^{-4})(100)^2 - (8.8\times 10^{-6})(100)^3 $

Evaluate each term:

• \( 6.6\times 10^{-2} \times 100 = 6.6 \)
• \( 6.21\times 10^{-4} \times 10000 = 6.21 \)
• \( 8.8\times 10^{-6} \times 1000000 = 8.8 \)

Combine the terms:

$ S(100) \approx 38.74 + 6.6 + 6.21 - 8.8 $

$ S(100) \approx 45.34 + 6.21 - 8.8 $

$ S(100) \approx 51.55 - 8.8 $

$ S(100) \approx 42.75 $

The approximate sensitivity at $100 \text{ }^\circ C$ is $42.75 \text{ }\mu V/\text{ }^\circ C$.