Process Engineering Division – Instrumentation and Automation Section – Daily Topics – Topic No. 2020-12-09
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Process Engineering Section – Instrumentation and Automation Section – Daily Topics – Topic No. 2020-12-09: Discussion question: 321. A platinum-rhodium-platinum thermocouple is used to measure the temperature inside a furnace. In the absence of any other compensation methods, the electromotive force measured is 10.638 mV, and the temperature at the free end of the thermocouple is 50°C. Determine the actual temperature. (5 points are given only to those who provide the correct answer; additional points are awarded for a correct calculation process or detailed explanation.)
From the table at 50℃, ES(50,0) = 0.299 mV. Es(t, 0℃) = 10.638 + 0.299 = 10.937 mV; according to the table, t = 1115.25 ℃
The furnace temperature was measured using a platinum-rhodium-platinum thermocouple. In the absence of any other compensation methods, a thermoelectromotive force of 10.638 mV was measured, with the free end temperature at 50°C. Referring to the table, the thermoelectromotive force E(50,0) equals 0.299 mV. Therefore, the thermoelectromotive force E(t,0) is equal to E(t,50) plus E(50,0), that is, 10.638 mV + 0.299 mV = 10.937 mV. Again, according to the table, a thermoelectromotive force of 10.937 mV corresponds to a temperature of 1118°C. The actual temperature of the medium under test is 1118°C.
According to the table, at 50℃ the value is 0.299 mV; therefore, (t, 0℃) = 10.638 + 0.299 = 10.937 mV. The table also shows that t = 1118℃
t=1115.25 ℃
From the table at 50℃, ES(50,0) = 0.299 mV. Es(t, 0℃) = 10.638 + 0.299 = 10.937 mV; according to the table, t = 1115.25 ℃
t=1115.25 ℃
The furnace temperature was measured using a platinum-rhodium-platinum thermocouple. In the absence of any other compensation methods, a thermoelectromotive force of 10.638 mV was measured, with the free end temperature at 50°C. Referring to the table, the thermoelectromotive force E(50,0) equals 0.299 mV. Therefore, the thermoelectromotive force E(t,0) is equal to E(t,50) plus E(50,0), that is, 10.638 mV + 0.299 mV = 10.937 mV. Again, according to the table, a thermoelectromotive force of 10.937 mV corresponds to a temperature of 1118°C. The actual temperature of the medium under test is 1118°C.
According to the table, the thermoelectric potential E(50,0) = 0.299 mV. The thermoelectric potential E(t,0) is equal to E(t,50) + E(50,0), that is, 10.638 mV + 0.299 mV = 10.937 mV. Thus, the temperature corresponding to the thermoelectric potential E(t,0) = 10.937 mV is 1118°C. The actual temperature of the medium under test is 1118°C.