The thermocouple compensation wire is connected in reverse
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What happens if the thermocouple compensation wire is connected in reverse? I also saw posts on the forum saying that the size changes – it gets bigger, smaller, or stays the same. Which one is correct? Please advise, experts
It becomes smaller; the positive pole of the compensation wire has stronger electron mobility than the negative pole. If they are connected in reverse, it causes some of the electrons in the thermocouple to move in the opposite direction, thereby achieving a new equilibrium in electron movement at a certain fixed temperature.
It can generate additional measurement errors, and the magnitude of these errors is related to the errors at both ends of the compensation wire. Expression: No inversion: e (at the surface) = e (at the hot end) + e (at the cold end) ； Reversed: e (hot side) = e (hot end) - e (cold end). If the temperature difference is zero, then e (cold end) = 0, and the instrument reading will have no additional error. If the temperature of the cold end of the thermocouple is higher than the temperature at the instrument’s input terminal, then e (cold end) > 0; in this case, the instrument reading will be twice lower than the actual value by the amount of the temperature difference. For example, if the actual temperature is 100°C, the temperature of the cold end is 25°C, and the temperature at the instrument’s input terminal is 15°C, then the instrument reading will be around 80°C. If the temperature of the cold end of the thermocouple is lower than the temperature at the instrument’s input terminal, then e (cold end)
If the thermocouple compensation wire is connected in reverse, it depends on the temperatures at the cold end of the thermocouple and at the instrument’s wiring terminals. We assume that the temperature at the measuring end of the thermocouple is t, at the cold end it is t1, and at the instrument’s terminal connections it is t2. If the compensation wire is connected in reverse, the total thermoelectric potential input to the instrument becomes E = E(t, t1) – E(t1, t2) = E(t, 0) – E(t1, 0) – [E(t1, 0) – E(t2, 0)] = E(t, 0) – E(t1, 0) – E(t1, 0) + E(t2, 0). If t1 = t2, then -E(t1, 0) + E(t2, 0) = 0, so the value remains unchanged. If t1 > t2, then -E(t1, 0) + E(t2, 0) < 0, meaning the value decreases. If t1 < t2, then -E(t1, 0) + E(t2, 0) > 0, meaning the value increases. This is my personal understanding; I’m not sure if it’s correct, so it’s open for discussion~~~~