Modern Engineering Thermodynamics

19.4 Thermoelectric Coupling 773
The Peltier and Kelvin coefficients are not independent from the Seebeck coefficient, so they need not be
introduced into the phenomenological coefficients. For example, the Peltier heat _Q P is the heat transfer rate that
occurs when there is no temperature difference, dT/dx = 0 = X Q . Therefore, from Eq. (19.12),
or
Also, from Eq. (19.13),
J Q
XQ
=0 = _ Q P
A = _q P = L QEX E
_Q P = AðαT 2 k e Þ − 1 T
dϕ
dx
dϕ
= −αATk e
dx
or
J E
XQ
=0 = I/A = L EEX E
I = ATk e − 1 T
dϕ
dx
Then, the Peltier coefficient π given in Eq. (19.16) becomes
π =
_ Q P
I
dϕ
= −Ak e
dx
= αT (19.33)
Thus, the Seebeck and Peltier coefficients for a pure conductor are directly related by Eq. (19.33).
EXAMPLE 19.1
A 0.0100 m diameter copper bus bar is maintained at a constant temperature of 20.0°C and is subjected to a voltage gradient
of 1.00 V/m. Determine the Peltier heat flow, _Q P . The Seebeck coefficient for the copper is α cu = 3.50 × 10 −6 V/K, and its
resistivity is ρ e = 5.00 × 10 −9 ohm meters.
Solution
The Peltier heat flow is given by Eq. (19.33) as
_Q P = πI = ðαTÞI
where
I = AJ E = ρ A
− dϕ
e dx
Here, −dϕ/dx = voltage gradient = 1.00 V/m, so
Then
I = ðπ/4Þð0:0100 mÞ2 ð1:00 V/mÞ
5:00 × 10 −9 Ω gm
= 1:57 × 10 4 V/Ω = 1:57 × 10 4 A
_Q P = ð3:50 × 10 −6 V/KÞð20, 0 + 273:16 KÞð1:57 × 10 4 AÞ = 16:2V.A ¼ 16:2W
EXAMPLE 19.2
The open circuit voltage of an iron-copper thermocouple is approximately given by
ϕ fe-cu = ð−13:4 T + 0:014 T 2 + 0:00013 T 3 Þ × 10 −6 V
where T is in °C, not K. At 100.°C, determine
a. The relative Seebeck coefficient α fe-cu .
b. The relative Peltier coefficient π fe-cu .
(Continued )