Entropy production excess

We all widely utilize aspects of the first law of thermodynamics. The first law mainly deals with energy balance regardless of the quality of that part of the energy available to perform work. We define first law efficiency or thermal efficiency as the ratio of the work output to total rate of heat input, and this efficiency may not describe the best performance of a process. On the other hand, the second law brings out the quality of energy, and second law efficiency relates the actual performance to the best possible performance under the same conditions. For a process, reversible work is the maximum useful work output. If the operating conditions cause excessive entropy production, the system will not be capable of delivering the maximum useful output. 

Second-law analysis can determine the level of energy dissipation from the rate of entropy production in the system. The entropy production approach is especially important in terms of process optimality since it allows the entropy production of each process to be determined separately. The map of the volumetric entropy production rate identifies the regions within the system where excessive entropy production occurs due to irreversible processes. Minimizing of excessive irreversibilities allows a thermodynamic optimum to be achieved for a required task. Estimation of the trade-offs between the various contributions to the rate of entropy production may be helpful for attaining thermodynamically optimum design and operation. 

The quantity of excess entropy production is positive by the Cauchy-Schwartz inequality (similar to the inequality in Eq. (4.101)), indicating that P >... 

The term to the right of the equal sign in Eq. (12.32) is the excess entropy production. Equations (12.31) and (12.32) describe the stability of equilibrium and nonequilibrium stationary states. The term 82S is a Lyapunov functional for a stationary state. 

However, nonequilibrium steady states may be unstable even if the system is stable with respect to diffusion. For a nonequilibrium state, the stability condition for a chemical reaction in terms of excess entropy production is... 

The excess entropy production due to a fluctuation in the concentration of X around a steady-state value is... 

So, the contribution to the excess entropy production becomes negative, and the system becomes unstable... 

Equation (12.55) indicates that the quantity d(S2S)/dt has the same form for the perturbations from the equilibrium state as well as the nonequilibrium state. In the vicinity of equilibrium, the quantity 2 Y)8Jj is called the excess entropy production, and it shows the increase in entropy production. The quantities Sf and 8Xf denote the deviations of /, and A, from the values at the nonequilibrium steady state. The increase in entropy production for a perturbation from a nonequilibrium state is... 

The excess entropy production can become negative if kf kh. and hence the stationary state may become unstable. 

The enzyme phosphofructokinase is allosteric, that is, it is made up of equivalent units that possess specific reaction sites for the fixation of the substrate and product. Each unit exists in two conformational states one active with more affinity for the substrate, and one inactive. The reaction products of phosphofructokinase (FDP and ADP) displace the conformational equilibrium in favor of the active form of the enzyme. This may create a destabilizing effect on the excess entropy production. In the glycolytic cycle, the allosteric properties of the phosphofructokinase may lead to oscillations. Consider the following simple model... 

Figure 4. The time evolution of the excess entropy production 82S in the nonlinear region gives a criterion for the stability of the steady state...

In context with the examples for instable networks in Chapter 6 it is elucidating to look for candidates of processes giving negative contributions to the excess entropy production of (7.45) and thus possibly leading the system into an instability. As an example let us consider an autocatalytic reaction vX + Y (v + 1)X as discussed in Section 6.1. With... 

Whereas for 6 > 0 we always have 6F < 0, the expression for 6J may become positive particularly for small X. Thus, the product 6F 6J may give negative contributions to the left-hand side of (7.45). Whether this negative contribution actually leads to an instability clearly depends on the total balance of the excess entropy production. On the other hand, the reader may easily prove that ordinary chemical reactions always have a positive excess entropy production as separate contributions to the left-hand side of (7.45), cf. problem 1. 

Prove that a single chemical reaction of the type + X2 +. .. where none of the reactants simultaneously appears on both sides of the reaction, always has a positive excess entropy production. 

Argue how it could in principle be possible to have a total negative excess entropy production in a network which consists of capacitances and chemical reactions of the type as assumed in problem 1 (Think of closed loops). 

When heat and mass are transported across heterogeneous systems, the interface may pose a barrier to transport. In Figure 14.2 this happens with low butane concentration. Governing equations are needed for the interface, as these equations give boundary conditions for the transport processes in the homogeneous phases on each side. The boundary equations are determined from the excess entropy production for the interface. ... 

For the surface we obtain the excess entropy production from eq 14.34,... 

In Chapter 14 (equation 14.1.16) we obtained the same equation for perturbations from the equilibrium state. Equation (18.3.6) shows that the time derivative of 5 5 has the same form even under nonequilibrium conditions. The difference is that near equilibrium SFkSJk = J2k > 0 this is not necessarily so far from equilibrium. We shall refer to this quantity as excess entropy production, but strictly speaking, it is the increase in entropy production only near the equilibrium state for a perturbation from a nonequilibiium state, the increase in entropy production is equal to 5P = 5pP -I- 5jP. 

Obtain the excess entropy production and analyze the stability of the stationary states for the following reaction schemes ... 

---