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Method for calculating thermocouple cold junction compensation: From millivolts to temperature: Measure the cold junction temperature, convert it to the corresponding millivolt value, add it to the millivolt value of the thermocouple, and thereby determine the temperature； From temperature to millivolts: The actual temperature and the cold junction temperature are measured, converted into millivolt values respectively, and by subtracting one from the other, the millivolt value is obtained, which represents the temperature

This post was last edited by xxkhc on 2019-1-13 at 17:43. This should be an exam question; considering practical production conditions, since the compensation wire has the same thermoelectric properties as the thermocouple, and according to the law of homogeneous conductors for thermocouples, no thermoelectrical potential difference will occur at the 40°C end. Therefore, the actual temperature is: 33.3 + 1.495 = 34.795 mV. Referring to the E-scale table, the corresponding temperature is 472.67°C.

To determine the actual temperature, it is first necessary to know the actual thermoelectromotive force. Assuming the actual temperature is T and the actual thermoelectromotive force is Et, then Et = E(t, t0) + Et40 = 33.3 + 2.419 = 35.719 mV. Since 35.382 mV < 35.719 mV < 36.190 mV, it follows that E(480,0) < Et < E(490,0). According to the given conditions, the thermoelectrical potential can be considered linear within the range of 10°C; therefore, for a temperature increase of 1°C, the potential change is E1 = (36.190 – 35.382)/10 = 0.0808 mV. The temperature T can be calculated as T = 480 + [Et – E(480,0)]/E1 = 480 + (35.719 – 35.382)/0.0808 ≈ 480 + 4.17°C ≈ 484.17°C. Thus, the actual temperature is 484.17°C.

This post was last edited by cqdfwy on 16-1-2019 at 16:42; the calculation method is correct. However, the thermoelectric potential Et (which should be E(t, t0)) = 33.3 mV is measured at both ends of the compensation wire; that is, t0 is 25°C (this can be expressed as: E(t, 25) = 33.3 mV). Therefore, the thermoelectric potential E(t, 0) should be E(t, 25) + E(25, 0) = 33.3 + 1.495 = 34.795 mV. In practice, the temperature value can be obtained by referring to an E-type thermometric scale; however, the question only provides the thermoelectric potential values corresponding to a few temperature points, and it states that \"within the range of 10°C, the thermoelectric potential can be considered linear.\" In other words, it is not possible to use the thermometric scale to obtain the value, and instead an approximate value must be calculated. The thermoelectric potential calculated earlier, E(t,0)=34.795 mv, lies between 34.574 mv and 35.382 mv; that is, E(470,0) < E(t,0) < E(480,0), which means 470°C < t < 480°C℃ ； Furthermore, based on the condition that the thermoelectromotive force can be considered linear within the range of 10°C, the thermoelectromotive force at 1°C is given by E(1,0) = (35.382 – 34.574)/10 = 0.0808 mV. Thus, the approximate temperature value is t = 470 + (34.795 – 34.574)/0.0808 ≈ 470 + 2.735°C ≈ 472.735°C.