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How does a thermocouple exploit the Seebeck effect?
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[QUOTE="rude man, post: 6011343, member: 350494"] Yes, thermocouples operate on the Seebeck effect. A typical setup is three junctions: T, T[SUB]0[/SUB] and T[SUB]R[/SUB] where T is temp. to be measured; T[SUB]0[/SUB] is a reference temperature, often an ice bath; and T[SUB]R[/SUB] is the temperature of the measuring device, typically a potentiometer at room temperature so no current flows across the T[SUB]R[/SUB] junctions. See attached "Seebeck effect.jpg" taken from [I]Heat and Thermodynamics[/I] by M W Zemansky of CCNY, hopefully visible. Each wire type A, B and C (note the same wire type C connecting T[SUB]0[/SUB] to T[SUB]R[/SUB]) has its own characteristic "entropy transport parameter" S* and depends on the material and temperature of the wire. The emf's developed are across the wires so there are Δemf's developed between junctions a to c, c to e, e to d and d to b. Then the respective transport parameters are integrated over temperature over these four wires; the result after cancellation of the S*[SUB]c[/SUB] parameters is Δemf[SUB]a,b[/SUB] [SIZE=5][SIZE=4] ≈ (T - T[SUB]0[/SUB])(S*[SUB]A[/SUB] - S*[SUB]B[/SUB]) as desired. [/SIZE][/SIZE] [SIZE=5][SIZE=4][/SIZE][/SIZE] [SIZE=5][SIZE=4]Clearly we want to use wires A and B with ΔS* as different from one another as possible. I seem to remember copper and constantan.[/SIZE][/SIZE] [SIZE=5][SIZE=4][/SIZE][/SIZE] [SIZE=5][SIZE=4]There are other thermoelectric effects, e.g. Peltier and Joule. They're different phenomena.[/SIZE][/SIZE] [/QUOTE]
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How does a thermocouple exploit the Seebeck effect?
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