Modern Engineering Thermodynamics

768 CHAPTER 19: Introduction to Coupled Phenomena
V
+
Voltage
−
φ A
Voltage
φ A > φ B
V =φ A − φ
Current I
B and α AB > 0
φ B
Metal A
Metal B
FIGURE 19.1
A schematic of the Seebeck effect.
Metal A
Metal B
Junction
Junction
Heat
Heat
This early discovery gave rise to the relative Seebeck coefficient, α AB ,
defined as (see Figure 19.2).
ϕ
σ AB = − lim A − ϕ I B
= − dϕ AB
ΔT!0 ΔT =0
dT (19.15)
I =0
where ϕ AB is the potential difference (known as the Seebeck voltage),
ϕ A − ϕ B .Thus,α AB is simply the negative value of the slope of the open
circuit voltage-temperature relationship for the pair of conductors. The
sign convention usually adopted for α AB is as follows. If the current flow
in conductor A is from the cooler junction (T C ) to the hotter junction
(T H ), then α AB is positive (see Figure 19.2). Both the terminals across
which ϕ AB is measured must be at the same temperature, as shown in
Figure 19.1b, or else additional Seebeck voltages are generated at these
terminals.
A positive value for α AB corresponds to current flowing from T C to T H in
conductor A, andfromT H to T C in conductor B. Also, it is possible to
assign “absolute” Seebeck coefficients to the pure conductors, α A and α B ,
defined from
FIGURE 19.2
α AB = α A − α B
The sign convention for the Seebeck coefficient, α AB .
since a superconducting material below its transition temperature (i.e.,
where it becomes superconducting) has no measurable thermoelectric
effect (α S = 0). We can determine the absolute Seebeck coefficient of any material, say A, by joining it to a superconductor
and measuring the resultant open circuit voltage-temperature slope. 3 Then,
− dϕ AB
dT = α AS = α A − α S = α A − 0 = α A
I =0
and once we know the absolute Seebeck coefficient for any one material, that material can be used as a reference
material (α R ) to determine the absolute Seebeck coefficients of other materials as
α B = α BR + α R
Finally, we can easily determine a relationship between relative Seebeck coefficients as
α AB = α A − α B = ðα A − α C Þ − ðα B − α C Þ = α AC − α BC
19.4.2 The Peltier Effect 4
Passing a current I through an isothermal bimetallic (or semiconductor) closed circuit causes heat absorption at
one junction and heat release at the other junction (see Figure 19.3).
Conductor A
Peltier
heat
absorption
(cooling)
T C
Current I
T H
Peltier
heat
release
(heating)
Conductor B
FIGURE 19.3
A schematic of the Peltier effect.
3 This technique works only at very low temperatures (below 18 K). Other techniques are used at higher temperatures.
4 Discovered in 1834 by the French watchmaker turned physicist Jean Charles Athanase Peltier (1785–1845).