The thermo e.m.f. of a thermocouple varies wi | vedclass

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(D) The thermo $e.m.f.$ is given by $E = a	heta + b	heta^2$.The neutral temperature $	heta_n$ is the temperature at which the thermo $e.m.f.$ is ma

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No neutral temperature is possible for this thermocouple

Vedclass · Answer

(D) The thermo $e.m.f.$ is given by $E = a	heta + b	heta^2$.The neutral temperature $	heta_n$ is the temperature at which the thermo $e.m.f.$ is maximum,i.e.,$\frac{dE}{d	heta} = 0$.$\frac{dE}{d	heta} = a + 2b	heta = 0$.$	heta_n = -\frac{a}{2b}$.Given that $\frac{a}{b} = 700\,^{\circ}C$,we substitute this into the equation:$	heta_n = -\frac{1}{2} 	imes (700\,^{\circ}C) = -350\,^{\circ}C$.Since the neutral temperature must be greater than the cold junction temperature $(0\,^{\circ}C)$ for a standard thermocouple,a negative value indicates that no neutral temperature is physically possible for this specific thermocouple configuration.

The thermo $e.m.f.$ of a thermocouple varies with the temperature $\theta$ of the hot junction as $E = a\theta + b\theta^2$ in volts,where the ratio $a/b$ is $700\,^{\circ}C$. If the cold junction is kept at $0\,^{\circ}C$,then the neutral temperature is:

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