The case of the metal bar is revised from this
post. and this
one. The main error made in these previous posts is that the the heat flux
q was chosen proportional to ∂
xT instead of ∂
x(1/T) according to the phenomenological law. Thus, the profile corresponding to the minimum entropy principle did not correspond to the profile of a steady state (∂
tT=0,
q is constant uniform along x).
By using the correct phenomenological law, the profile of minimum entropy production is also the profile of a steady state; but it is not the traditional linear profile in T, it is a profile such as (1/T) is linear. The linear profile we learnt is only an approximation of the real steady state profile.
First, we need the temperature equation:
where ρ is the density and c
v is the specific heat at constant volume. The phenomenological law gives:
In that case, the steady state ∂
tT=0,
q uniform in x gives a linear profile in 1/T.
The equation corresponding to the
internal entropy production ∂
tS of the system is:
We want to minimize the internal entropy production inside the volume. Using the calculus of variation and the phenomenological law Eq.(2), we need to resolve:
Writing F(1/T,∂
x(1/T)) =
q², we obtain:
The last term is zero because 1/T is fixed at the boundary conditions. We thus obtain the Euler-Lagrange equation:
The term on the lhs gives:
after using Eq.(2). Thus
But
using Eq.(2), so that the solution corresponds to
q constant in x that is it is uniform all along the bar. We have seen that such state corresponds also to the steady state. Thus
the steady state corresponds to a state of minimum entropy production. Calculation of the time rate of change of the entropy production &part
ttS would show that &part
ttS ≤ 0 so that the entropy production is always decreasing and indeed at steady state, the entropy production has reached a minimum.
The case presented above works also for a motionless one-component fluid.
See de Groot and Mazur,
Non-equilibrium thermodynamics, p.47-48.
