The thermo emf of a thermocouple is given by | vedclass

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(D) The thermo emf is given by $E = aT + bT^2$.At the neutral temperature $(T_n)$,the emf is maximum,and at the inversion temperature $(T_i)$,the emf

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$443$

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(D) The thermo emf is given by $E = aT + bT^2$.At the neutral temperature $(T_n)$,the emf is maximum,and at the inversion temperature $(T_i)$,the emf becomes zero.Setting $E = 0$ for the inversion temperature:$0 = aT_i + bT_i^2$$T_i(a + bT_i) = 0$Since $T_i 
eq 0$,we have $a + bT_i = 0$,which gives $T_i = -\frac{a}{b}$.Given $\frac{a}{b} = -200^{\circ}C$,so $T_i = -(-200^{\circ}C) = 200^{\circ}C$.This $T_i$ is the temperature relative to the cold junction being at $0^{\circ}C$.Since the cold junction is kept at $T_c = 30^{\circ}C$,the actual inversion temperature is $T_{inv} = T_i + T_c = 200^{\circ}C + 30^{\circ}C = 230^{\circ}C$.Converting to Kelvin: $T(K) = 273 + 230 = 503 \ K$.Wait,re-evaluating the standard relation: The inversion temperature $T_i$ is defined such that $T_i - T_c = T_c - T_n$,where $T_n$ is the neutral temperature. The equation $E = aT + bT^2$ implies $T_n = -a/2b = 100^{\circ}C$. If $T_c = 30^{\circ}C$,then $T_i = 2T_n - T_c = 2(100) - 30 = 170^{\circ}C$.Converting to Kelvin: $T = 273 + 170 = 443 \ K$.

The thermo emf of a thermocouple is given by $E = aT + bT^2$,where $\frac{a}{b} = -200^{\circ}C$. If the cold junction is kept at $30^{\circ}C$,then the inversion temperature is ($\varepsilon$ in volt,$T$ in centigrade). (in $K$)

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