Question 19.2: The open circuit voltage of an iron-copper thermocouple is a...

a. From Eq. (19.15), we have that

σ_\text{AB} = −\lim_{ΔT→0} (\frac{ϕ_\text{A} −ϕ_\text{B}}{ΔT}) \mid_{I=0} = −\frac{dϕ_\text{AB}}{dT} \mid_{I=0}                (19.15)

α_\text{fe-cu} = −\frac{dϕ_\text{fe-cu}}{dT} \mid_{I=0}

Thus

α_\text{fe-cu} = −(−13.4 + 0.028  T + 0.00039  T^2) × 10^{−6}  \text{V/K}

where T is in °\text{C} not \text{K}. At T = 100.°\text{C}, this becomes

α_\text{fe-cu} (100.°\text{C}) = 6.70 × 10^{−6}  \text{V/K}

b. From Eq. (19.33) we have^{11}

π = \frac{\dot{Q}_P}{I} = αT            (19.33)

π_ \text{fe-cu} = T_ \text{αfe-cu} = (100. + 273.15  \text{K}) (6.70 × 10^{−6}  \text{V/K}) = 2.50 × 10^{−3}  \text{V}

^{11}Note that, even though most of the temperature to voltage correlations published for thermocouples are written in terms of relative temperature units (°C or °F), the multiplying temperature factor T in the Peltier and Kelvin coefficient equations is always in absolute units (K or R)