Question 19.3: The chromel-alumel thermocouple circuit shown below has its ...

a. The relative Seebeck coefficient for the chromel-alumel circuit is

α_\text{ch-al} = α_\text{ch} − α_\text{al} = [23.0 −(−18.0)] × 10^{−6}  \text{V/K} = 41.0 × 10^{−6}  \text{V/K}

and for a constant α_\text{ch-al}, Eq. (19.15) can be integrated to give

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)

−ϕ_\text{ch-al} = α_\text{ch-al} (T_H −T_C) = (41.0 × 10^{−6}  \text{V/K} (100.−0  \text{K}) = 4.1 × 10^{−3}  \text{V} = ϕ_\text{al-ch}

(note that ϕ_\text{al-ch} = −ϕ_\text{ch-al}).

b. The absolute Peltier coefficients at each junction are given by Eq. (19.33) as π_\text{ch} = α_\text{ch} T and π_\text{al} = α_\text{al} T, and we can compute π_\text{ch-al} = π_\text{ch} − π_\text{al}. At the 0.00°\text{C} = 273.15  \text{K} junction,

π_\text{ch} = (23.0 × 10^{−6}  \text{V/K}) (273.15  \text{K}) = 6.28 × 10^{−3}  \text{V}

π_\text{al} = (−18.0 × 10^{−6}  \text{V/K}) (273.15  \text{K}) = −4.91 × 10^{−3}  \text{V}

and

π_\text{ch-al} = [6.28 – (−4.91)] × 10^{−3}  \text{V} = 11.2 × 10^{−2}  \text{V}

At the  100.° \text{C} = 373.15  \text{K}  junction

π_\text{ch} = (23.0 × 10^{−6}  \text{V/K}) (373.15  \text{K}) = 8.58 × 10^{−3}  \text{V}

π_\text{al} = (−18.0 × 10^{−6}  \text{V/K}) (373.15  \text{K}) = −6.71 × 10^{−3}  \text{V}

and

π_\text{ch-al} = [8.58 – (−6.71)] × 10^{−3}  \text{V} = 15.3 × 10^{−3}  \text{V}

c. To determine the influence of the copper lead wires, we use Eq. (19.15) for the complete cu-al-ch-al-cu circuit connected to the potentiometer. Using the junction notation shown in Figure 19.8, the potentiometer reading is ϕ_{af} = ϕ_a − ϕ_f. We can move around the circuit to evaluate this reading as

ϕ_a − ϕ_f = (ϕ_a − ϕ_b) + (ϕ_b − ϕ_c) + (ϕ_c − ϕ_d)+ (ϕ_d − ϕ_e) + (ϕ_e− ϕ_f)

where

ϕ_a − ϕ_b = − \int_{T_b}^{T_a} α_\text{cu} dT = \int_{T_a}^{T_b} α_\text{cu} dT

ϕ_b − ϕ_c = − \int_{T_b}^{T_c} α_\text{al} dT

ϕ_c − ϕ_d = − \int_{T_c}^{T_d} α_\text{ch} dT

ϕ_d − ϕ_e = − \int_{T_e}^{T_e} α_\text{al} dT

and

ϕ_e − ϕ_f = − \int_{T_e}^{T_f} α_\text{cu} dT

Now, the contribution of the copper lead wires is

Δϕ_\text{cu} = ϕ_a − ϕ_b + ϕ_e − ϕ_f =\int_{T_a}^{T_b} α_\text{cu} dT + \int_{T_e}^{T_f} α_\text{cu} dT

and if, as the circuit diagram shows, T_a = T_f  and T_b = T_e, then the influence of the copper leads is

Δϕ_\text{cu} = \int_{T_a}^{T_b} α_\text{cu} dT + \int_{T_b}^{T_a} α_\text{cu} dT = \int_{T_a}^{T_b} α_\text{cu} dT + (- \int_{T_a}^{T_b} α_\text{cu} dT ) = 0

Thus, if the copper lead wires have the same end point junction temperatures, then they do not contribute any net Seebeck voltage to the circuit. Adding up all the potential differences around the circuit and using T_a = T_f and T_b = T_e gives

Δϕ_{af} = \int_{T_a}^{T_b} α_\text{cu} dT + \int_{T_b}^{T_a} α_\text{cu} dT + \int_{T_b}^{T_c} α_\text{al} dT + \int_{T_c}^{T_d} α_\text{ch} dT + \int_{T_d}^{T_b} α_\text{al} dT

Now,

\int_{T_b}^{T_c} α_\text{al} dT + \int_{T_d}^{T_b} α_\text{al} dT = \int_{T_d}^{T_c} α_\text{E} dT = – \int_{T_c}^{T_d} α_\text{al} dT

so

ϕ_ {af} = \int_{T_c}^{T_d} (α_\text{ch} − α_\text{E}) dT = \int_{T_c}^{T_d} (α_\text{ch-al}) dT = α_\text{ch-al} (T_d −T_c)

Consequently, the potentiometer measures the chromal-alumel thermoelectric effects only so long as all the lead wires have equal junction temperatures.