Читать книгу Non-equilibrium Thermodynamics of Heterogeneous Systems - Signe Kjelstrup - Страница 32

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 dE_eq = 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 V_w = 10³ kg/m³.

•Solution: The only contribution to the chemical potential gradient is from the pressure (dμ_w^c), so dμ_w,T = V_wdp. The pressure on water at a distance x from the surface is given by the weight of the sand, p = ρgx. This gives dμ_w,T/dx = V_wρg, 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

By using , Eq. (4.15) follows.

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 J_solvent = 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, ∂μ_T/∂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 TcdS_irr/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:

Δ_rG = μ_D − μ_C − μ_B

In the absence of chemical equilibrium, there are three independent chemical potentials. The contribution to σ is

Derive this expression for σ_chem, 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 = σ_vect + σ_scal, we find

In this way, we find the vectorial contributions due to diffusion and the scalar contribution due to the reaction.