The cold junction of a thermocouple is mainta | vedclass

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Question

(C) The thermo $e.m.f.$ $(E)$ in a thermocouple is given by the relation $E = \alpha 	heta + \frac{1}{2} \beta 	heta^2$,where $	heta$ is the temper

Vedclass · Accepted Answer

$270$

Vedclass · Answer

(C) The thermo $e.m.f.$ $(E)$ in a thermocouple is given by the relation $E = \alpha 	heta + \frac{1}{2} \beta 	heta^2$,where $	heta$ is the temperature difference between the junctions.For a thermocouple,the neutral temperature $	heta_n$ is the temperature of the hot junction at which the thermo $e.m.f.$ becomes zero,given the cold junction is at temperature $	heta_c$.The relationship between the neutral temperature $	heta_n$,the cold junction temperature $	heta_c$,and the inversion temperature $	heta_i$ (where $e.m.f.$ is zero) is given by $	heta_n = \frac{	heta_c + 	heta_i}{2}$.Given,cold junction temperature $	heta_c = 10\,^{\circ}C$ and inversion temperature $	heta_i = 530\,^{\circ}C$.Therefore,$	heta_n = \frac{10 + 530}{2} = \frac{540}{2} = 270\,^{\circ}C$.

The cold junction of a thermocouple is maintained at $10\,^{\circ}C$. No thermo $e.m.f.$ is developed when the hot junction is maintained at $530\,^{\circ}C$. The neutral temperature is ................ $^{\circ}C$.

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