Page 484 - Advanced thermodynamics for engineers

476 CHAPTER 20 IRREVERSIBLE THERMODYNAMICS

Substituting for L 22 ,L 12 and L 21 in Eqns (20.29) and (20.30) gives
L 11 dT dε
J Q ¼ lTS (20.37)
T d‘ d‘
and

lTS dT dε

J I ¼ l (20.38)
T d‘ d‘
Now if there is zero current flow (i.e. J I ¼ 0), Fourier’s law of heat conduction may be applied
giving
dQ=dt dT
J Q ¼ ¼ k (20.39)
A d‘
However, if J I ¼ 0 then Eqn (20.38) gives
dε dT
¼ S (20.40)
d‘ d‘
Substituting this value for dε/d[ in Eqn (20.37) gives

kdT L 11 dT 2 dT
J Q ¼ ¼ þ lS T (20.41)
d‘ T d‘ d‘
Hence

2
L 11 ¼ k þ lTS T (20.42)
Substituting the coefficients into Eqns (20.29) and (20.30) gives the following:

dT dε
for heat flow J Q ¼ k þ lTS 2 lTS (20.43)
d‘ d‘
dT dε
for electrical flow J I ¼ lS l (20.44)
d‘ d‘
The entropy flux is obtained from Eqn (20.31).

2
k þ lTS dT dε
J S ¼ lS (20.45)
T d‘ d‘

J Q
. This is
It is possible to define another parameter, the heat of transport Q*, where Q ¼
J I
T
basically the ‘thermal energy’ which is transported by a flow of electrical energy when there is no
temperature gradient, and indicates the magnitude of the off-diagonal terms in the cross-coupling
matrix in Eqn (20.8a).
From Eqns (20.43) and (20.44), when dT ¼ 0

J Q dε dε
¼ lTS l ¼ TS (20.46)
J I d‘ d‘
T