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The temperature of inversion of a thermocouple is $620^\circ C$ and the neutral temperature is $300^\circ C$. What is the temperature of cold junction :
A
$40^\circ C$
B
$20^\circ C$
C
$320^\circ C$
D
$-20^\circ C$
Detailed Solution
The thermo emf varies with the temperature $\theta$ of the hot junction as $e = \alpha(\theta - \theta_C) + \beta(\theta^2 - \theta_C^2)$, where $\theta_C$ is the temperature of the cold junction.
At the inversion temperature $e = 0$: dividing by $(\theta - \theta_C)$ gives $\alpha + \beta(\theta_i + \theta_C) = 0 \Rightarrow \theta_i = -\theta_C - \dfrac{\alpha}{\beta}$
At the neutral temperature $\dfrac{de}{d\theta} = 0$: $\alpha + 2\beta\theta_n = 0 \Rightarrow \theta_n = -\dfrac{\alpha}{2\beta}$
$\Rightarrow \theta_i = 2\theta_n - \theta_C \Rightarrow \theta_n = \dfrac{\theta_i + \theta_C}{2}$
$300 = \dfrac{620 + \theta_C}{2}$
$620 + \theta_C = 600$
$\theta_C = -20^\circ C$
At the inversion temperature $e = 0$: dividing by $(\theta - \theta_C)$ gives $\alpha + \beta(\theta_i + \theta_C) = 0 \Rightarrow \theta_i = -\theta_C - \dfrac{\alpha}{\beta}$
At the neutral temperature $\dfrac{de}{d\theta} = 0$: $\alpha + 2\beta\theta_n = 0 \Rightarrow \theta_n = -\dfrac{\alpha}{2\beta}$
$\Rightarrow \theta_i = 2\theta_n - \theta_C \Rightarrow \theta_n = \dfrac{\theta_i + \theta_C}{2}$
$300 = \dfrac{620 + \theta_C}{2}$
$620 + \theta_C = 600$
$\theta_C = -20^\circ C$