Non-equilibrium Thermodynamics of Heterogeneous Systems. Signe Kjelstrup
the equations that describe the time evolution of the system. All the transport coefficients in Eq. (1.2) are then constant. Prigogine showed that stationary states of discontinuous systems, maintained for instance by a temperature difference, have a minimum entropy production [18]. This is sometimes called the Prigogine principle. For stationary states far from global equilibrium, the transport coefficients are no longer constant across the whole system, and the stationary state has no longer a minimum entropy production compared to non-stationary states [120, 121].
In the design of industrial systems, it is fruitful to ask a different question. We can ask for the path that a system must follow to obtain minimum entropy production [122]. This question leads to an optimization problem that is constrained by the performance, for instance by the composition of the chemicals that are produced. The transport coefficients of the system are then considered as given. The nature of systems with minimum total entropy production dSirr/dt has been studied extensively since 1982 by Bejan and coworkers, see [87, 88] (mostly mechanical systems) and by Kjelstrup and coworkers since 1995, see [32] (mostly chemical systems).
Johannessen and Kjelstrup [94] found that much of the path of an optimal chemical reactor was characterized by constant local entropy production, provided that the system had enough degrees of freedom to find this path. This was called the highway hypothesis. In all cases, the entropy production, and not the dissipation function, was used as proper objective function. Bejan [88, p. 227] showed that the state of minimum entropy production was equal to the state of maximum power, contrary to other claims [123], see also [32].
4.3Examples
The following illustrate how single contributions arise in the entropy production. Some exercises are meant to illustrate the theory. The remaining give numerical and physical insight. Transport of heat at low temperatures and chemical reactions give relatively large losses of work.
Exercise 4.3.1.Consider the special case that only component number j is transported. The densities of the other components, the internal energy, the molar volume and the polarization densities are all constant. Show that the entropy production rate is given by
•Solution: In this case, Eq. (4.10) reduces to
The rate of change of the entropy is therefore given by
Now we use Eq. (4.3) for component j and obtain
Comparing this equation with Eq. (4.1), we may identify the entropy flux and the entropy production as
Exercise 4.3.2.Consider the case that only heat is transported. The molar densities, the molar volume and the polarization densities are all constant. Show that the entropy production is given by
•Solution: In this case, Eq. (4.10) reduces to
The rate of change of the entropy is therefore given by
Equation (4.7) reduces to
By substituting this into the equation above, we obtain
By comparing this equation with Eq. (4.1), we can identify the entropy flux and the entropy production as
Exercise 4.3.3.Find the entropy production due to the heat flux through a sidewalk pavement by a hot plate placed d = 8 cm under the top of the pavement. The plate has a temperature of 343 K, and the surface is in contact with melting ice (273 K). The Fourier-type thermal conductivity of the pavement is 0.7 W/m K.
•Solution:Fourier’s law for heat conduction is J′q = −λ(dT/dx). The entropy production per square meter is rather large:
It is typical for heat conduction around room temperature that losses are large.
Exercise 4.3.4.Consider a system with two components (n = 2), having dT = 0, dp = 0, and dEeq = 0. Show, using Gibbs–Duhem’s equation, that one may reduce the description in terms of two components to one with only one component.
•Solution: Gibbs–Duhem equation, Eq. (3.27), gives
c1dμ1 + c2dμ2 = 0
The entropy production of Eq. (4.13) reduces for these conditions to
The equation contains only one independent force. Energy is dissipated as heat by interdiffusion of the two components. We can also write this entropy production by
with v12 ≡ J1/c1 − J2/c2. This is the velocity of component 1 relative to the velocity of component 2. Note that v12 is independent of the frame of reference.
Exercise 4.3.5.We are interested in the filtering of water (w) across a sandy layer of 1 m at 293 K. Evaluate the entropy production for a water flux of 10−6 kg/m2s. The density of the sand at 293 K is ρ = 1940 kg/m3 [124]. The volume of water per unit of mass is Vw = 103 kg/m3.
•Solution: The only contribution to the chemical potential gradient is from the pressure (dμwc), so dμw,T = Vwdp. The pressure on water at a distance x from the surface is given by the weight of the sand, p = ρgx. This gives dμw,T/dx = Vwρg, and
This value is considerable smaller than the value for transport of heat across a pavement (see Exercise 4.3.3).
Exercise 4.3.6.Give the details of the derivation of Eq. (4.15) from Eq. (4.13).
•Solution: We first rewrite the negative force as
By substituting this result into Eq. (4.13), we obtain