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Thermocouple is connected to Controller by Thermocouple Wire (so called Extension wire)
This wire can be 100 to 200 meters long.
This wire is special one, not normal as electrical wire.
 For example, Thermocouple is connected to controller in a long distance by wire. How difference voltage at Vae by appearance of junction at point “b” and “d”?
Assume: Point “a”, “e” has same temperature T3
Point “b”, “d” has same temperature T2
Point “c” has temperature T1
Wire ab, bc, cd, de has Seebeck Coefficient S1, S2, S3, S4 respectively

Vae = Vab + Vbc + Vcd + Vde
Vae = S1(T2 – T3) + S2(T1 – T2) + S3(T2 – T1) + S4(T3 – T2)
Vae = T1(S2 – S3) + T2(S1 – S2 + S3 – S4) + T3(-S1 + S4)                    (1)
If wire ab has same Seebeck Coefficient with Thermocouple lead connected (bc), then S1 = S2
Also, wire ed same with dc, then S3 = S4
Equation (1) can be expressed as:
Vae = T1S23 + 0 – T3S23 = S23(T1 – T3)
Where S23 is Seebeck Coefficient of Thermocouple
T1 is temperature at measuring point, T3 is temperature at Controller module.
Temperature at junction b and d didn’t impact to measuring! But temperature at b and d must be same.
Wire ab and ed must have same Seebeck Coefficient with Thermocouple lead where it connects to. But this same character is in range of environment temperature; Thermocouple lead can’t act as extension wire because it is too expensive.

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