As this result is sufficiently complicated to derive, it may be easier to solve the problem by trial and error using a spreadsheet.
We want a refrigerator capable of removing 10 W from a cold box at —5C, rejecting the heat to the environment at 30 C.
Owing to the temperature drops across the heat exchangers, the cold junction must be at -15 C and the hot one at 40 C.
The thermocouple materials have the following characteristics:
a = 0.0006 V/K, AA = 0.015Wcm"1 K"1, pA = 0.002 Q cm, AB = 0.010Wcm"1 K"1, and pB = 0.003 Q cm.
The temperatures are
For optimum geometry,
AR = 3 ^ [a/0.015 x 0.002 + a/0.010 x 0.00^ =120 x 10^6 V2/K.
Applying Equations 5.95, 5.96, and 5.97,
B =\J4 x 120 x 10-6 + 2 x 0.00062 x 571 = 0.02985 V K-1/2
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