The cold junction of a thermocouple is at 0^{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The thermo e.m.f $E$ is given by $E = 16T - 0.04T^2$.At the temperature of inversion $(T_i)$,the thermo e.m.f $E$ becomes zero.Setting $E = 0$:$0

Vedclass · Accepted Answer

$400^{\circ} C ; 200^{\circ} C$

Vedclass · Answer

(B) The thermo e.m.f $E$ is given by $E = 16T - 0.04T^2$.At the temperature of inversion $(T_i)$,the thermo e.m.f $E$ becomes zero.Setting $E = 0$:$0 = 16T_i - 0.04T_i^2$$16T_i = 0.04T_i^2$$T_i = \frac{16}{0.04} = 400^{\circ} C$.The neutral temperature $(T_n)$ is the average of the cold junction temperature $(T_c)$ and the inversion temperature $(T_i)$.Given $T_c = 0^{\circ} C$:$T_n = \frac{T_i + T_c}{2} = \frac{400 + 0}{2} = 200^{\circ} C$.Thus,the temperature of inversion is $400^{\circ} C$ and the neutral temperature is $200^{\circ} C$.

The cold junction of a thermocouple is at $0^{\circ} C$. The thermo e.m.f produced in the thermocouple is given by the equation $E = 16T - 0.04T^2$,where $T$ is the temperature of the hot junction. The temperature of inversion and neutral temperature of the thermocouple are:

Similar Questions

Cultivation of $Bt$ Cotton has been much in the news. The prefix "$Bt$" means

$A$ closed compartment containing gas is moving with some acceleration in the horizontal direction. Neglect the effect of gravity. Then the pressure in the compartment is

If $\omega$ is a complex cube root of unity,then $225+(3 \omega+8 \omega^2)^2+(3 \omega^2+8 \omega)^2$ is equal to:

Identify the plant shown in the given figure.

Vedclass Test Series

Exam Paper Generator

Online Exam Module