Non-equilibrium Thermodynamics of Heterogeneous Systems. Signe Kjelstrup
using
Exercise 4.3.7.Derive the entropy production for an isothermal twocomponent system that does not transport charge. The solvent is the frame of reference for the fluxes.
•Solution: In an isothermal system ∂(1/T)/∂x = 0. Furthermore, there is no charge transport so that j = 0. Finally, the solvent is the frame of reference, so Jsolvent = 0. There remains only one force–flux pair, namely for transport of solute. Using Eq. (4.15), we then find
for the entropy production. Using that σ ≥ 0, it follows that the solute will move from a higher to a lower value of its chemical potential.
Exercise 4.3.8.What is the entropy production for systems that are described by Eqs.(2.1), (2.2) and (2.3)?
•Solution: Substitution of these equations into Eq. (4.15), setting the reaction rate and the displacement current zero, yields
Typical values in an electrolyte are: λ = 2 J/msK, T = 300 K, dT/dx = 100 K/m, D = 10−9 m2/s, ∂μT/∂c = RT/c, c = 100 kmol/m3, dc/dx = 10−5 mol/m4, κ = 400 Si/m, dϕ/dx = 10−2 V/m. The resulting entropy productions are: σT = 0.2 J/Ksm3, σμ = 10−13 J/Ksm3 and σϕ = 10−4 J/Ksm3. Heat conduction therefore clearly gives the largest contribution to the entropy production in electrolytes.
Exercise 4.3.9.What is the first law efficiency ηI for a Carnot machine? Compare this efficiency with the expression for the entropy production of a system that transports heat from a hot reservoir to the surroundings.
•Solution: The Carnot machine transforms heat into work in a reversible way. The efficiency is defined as the work output divided by the heat input [100, 125]. This efficiency is (Th − Tc)/Th, where Th and Tc are the temperatures of the hot and the cold reservoir, respectively. If we do not use the heat to produce work, but simply bring the hot and cold reservoirs in thermal contact with one another one gets a heat flow from the hot to the cold reservoir. The entropy production for the whole (onedimensional) path, with a cross section Ω, is
As there is no other transport of thermal energy, the heat flux is constant and equal to dQ/dt for a unit area. This results in
The work lost per unit of time TcdSirr/dt is identical to the work that can be obtained by a Carnot cycle ΩηI(dQ/dt) per unit of time. This can be used as a derivation of the first law efficiency of the Carnot machine. This machine is reversible and has as a consequence no lost work. This implies that the work done by the Carnot machine must be equal to ΩηI(dQ/dt).
Exercise 4.3.10. Consider the reaction:
B + C ⇄ D
The driving force of the reaction is the reaction Gibbs energy:
ΔrG = μD − μC − μB
In the absence of chemical equilibrium, there are three independent chemical potentials. The contribution to σ is
Derive this expression for σchem, assuming that the reaction takes place in a reactor in which the internal energy is independent of the time.
•Solution: The n components of Eq. (4.3) are B, C and D. The balance equations for mass have a source term from the reaction rate:
Using that the internal energy is independent of the time, Eq. (4.10) reduces to
The rate of change of entropy is therefore given by
We substitute the balance equations in this expression, and obtain
By comparing this equation with Eq. (4.1), we may identify the entropy flux as
and the entropy production as
By writing this entropy production as a sum of a scalar and a vectorial part σ = σvect + σscal, we find
In this way, we find the vectorial contributions due to diffusion and the scalar contribution due to the reaction.
4.4Frames of reference for fluxes in homogeneous systems
In this section, we define the various frames of reference that are used in homogeneous systems. We consider first the transformations between different frames of reference for molar fluxes. We subsequently give the definition of the heat fluxes in the different frames of reference.
4.4.1Definitions of frames of reference
The laboratory frame of reference or the wall frame of reference. The experimental apparatus is at rest in this frame of reference. We define the velocities of the various components in this frame of reference by
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(4.18) |
where cA is the concentration of A in mol/m3.
The flux of A (in mol/m2 s) in a frame of reference which is moving with a velocity vref relative to the laboratory is given by
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(4.19) |
The solvent frame of reference. This frame of the reference is typical when there is an excess of one component, the solvent. For transport of A relative to the solvent, one has
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(4.20) |
The frame of reference