Process Engineering Division – Instrumentation and Automation Section – Daily Topics – Topic No. 2020-12-27
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Process Engineering Section – Instrumentation and Automation Section – Daily Topics – Topic No. 2020-12-27: Discussion question: 339. Someone reversed the polarity of the K-compensation wire. When the electric furnace is controlled at 800°C, if the temperature at the thermocouple junction box is 50°C and the temperature at the instrument wiring board is 40°C, what is the potential measured by the instrument under these conditions? (Just write the expression)
Someone reversed the polarity of the K-compensation wire. When the electric furnace is controlled at 800°C, if the temperature at the thermocouple junction box is 50°C and the temperature at the instrument wiring panel is 40°C, determine the potential measured by the instrument under these conditions. Solution: E = E(800, 50) – E(50, 40). Using the table, we find that E = (33.2754 – 2.0231) – (2.0231 – 1.6118) = 30.841 (mv). Answer: The potential measured by the instrument at this time is 30.841 mv.
Solution: E = E(800°C, 50°C) – (50°C, 40°C)
E=Ek(800,50)-Ek(50,40); using the table, E=(33.275-2.023)-(2.023-1.612)=30.841(mv).
E=Ek(800,50)-Ek(50,40); using the table, E=(33.275-2.023)-(2.023-1.612)=30.841(mv).
Solution: E = E(800, 50) – E(50, 40). Using the table, we find that E = (33.2754 – 2.0231) – (2.0231 – 1.6118) = 30.841 (mv).
E=Ek(800,50)-Ek(50,40); using the table, E=(33.275-2.023)-(2.023-1.612)=30.841(mv)
E=Ek(800,50)-Ek(50,40); using the table, E=(33.275-2.023)-(2.023-1.612)=30.841(mv)