Modern Engineering Thermodynamics

19.4 Thermoelectric Coupling 775
b. The absolute Peltier coefficients at each junction are given by Eq. (19.33) as π ch = α ch T and π al = α al T, and we can
compute π ch-al = π ch − π al : At the 0.00°C = 273.15 K junction,
π ch = ð23:0 × 10 −6 V/KÞð273:15 KÞ = 6:28 × 10 −3 V
π al = ð−18:0 × 10 −6 V/KÞð273:15 KÞ = − 4:91 × 10 −3 V
and
π ch-al = ½6:28 − ð−4:91ÞŠ × 10 −3 V = 11:2 × 10 −2 V
At the 100.°C = 373.15 K junction,
π ch = ð23:0 × 10 −6 V/KÞð373:15 KÞ = 8:58 × 10 −3 V
and
π al = ð−18:0 × 10 −6 V/KÞð373:15 KÞ = −6:71 × 10 −3 V
π ch-al = ½8:58 − ð−6:71ÞŠ × 10 −3 V = 15:3 × 10 −3 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
where
ϕ a − ϕ f = ðϕ a − ϕ b Þ + ðϕ b − ϕ c Þ + ðϕ c − ϕ d Þ + ðϕ d − ϕ e Þ + ðϕ e − ϕ f Þ
ϕ a − ϕ b = −
Z Ta
α cu dT =
Z Tb
α cu dT
Tb
Ta
Z Tc
ϕ b − ϕ c = − α al dT
Tb
Z T d
ϕ c − ϕ d = − α ch dT
Tc
Z Te
ϕ d − ϕ e = − α al dT
Te
and
Z T f
ϕ e − ϕ f = − α cu dT
Te
Now, the contribution of the copper lead wires is
Δϕ cu = ϕ a − ϕ b + ϕ e − ϕ f =
Z T b
Z T f
α cu dT +
α cu dT
Ta
Te
and if, as the circuit diagram shows, T a = T f and T b = T e , then the influence of the copper leads is
0 1
Z Tb
Z Ta
Δϕ cu = α cu dT + α cu dT =
Z Tb
B
α cu dT + @ −
Z Tb
C
α cu dTA= 0
Ta
Tb
Ta
Ta
(Continued )