Enthalpy equations

A recent article reported equations to help calculate the heat of reaction for proposed organic chemical reactions. In that article, enthalpy equations were given for 700 major organic compounds. 

The process is assumed reversible. This defines entropy as constant and therefore ds = 0, making Tds = 0. The enthalpy equation is simplified to... 

A number of authors have considered channel cross-sections other than rectangular [102-104]. Figure 2.17 shows some examples of cross-sections for which friction factors and Nusselt numbers were computed. In general, an analytical solution of the Navier-Stokes and the enthalpy equations in such channel geometries would be involved owing to the implementation of the wall boundary condition. For this reason, usually numerical methods are employed to study laminar flow and heat transfer in channels with arbitrary cross-sectional geometry. 

The Navier-Stokes equation and the enthalpy equation are coupled in a complex way even in the case of incompressible fluids, since in general the viscosity is a function of temperature. There are, however, many situations in which such interdependencies can be neglected. As an example, the temperature variation in a microfluidic system might be so small that the viscosity can be assumed to be constant. In such cases the velocity field can be determined independently from the temperature field. When inserting the computed velocity field into Eq. (77) and expressing the energy density e by the temperature T, a linear equahon in T is... 

For simplicity, consider an incompressible medium flowing through the multichannel domain depicted in Figure 2.36. In a number of practical applications the heat flux in the x-direction will be very small compared with that in the y-direction. Then the volume-averaged enthalpy equation for the solid walls can be written as... 

Figure 2.36 Multichannel flow domain with typical averaging volume for obtaining volume-averaged enthalpy equations.

Numerical computations of reacting flows are often difficult owing to the different time-scales involved and the highly non-linear dependence of the reaction rate on concentrations and temperature. The solution of the species concentration equations in combination with the momentum and the enthalpy equation generally requires an iterative procedure such as the one outlined in Section 1.3.4. A rough sketch of the numerical structure of a stationary reacting-flow problem is given as... 

The energy equation is solved in the form of a transport equation for static temperature. The temperature equation is obtained from the enthalpy equation, by taking the temperature as a dependent variable. The enthalpy equation is defined as,... 

Observation of the empirically linear equation 2 for the gas-phase enthalpy-of-formation data implies that AH(3) is constant for the series of compounds ML 20a b. AH°(3) can also be expressed in terms of the bond dissociation enthalpies (equation 5) by again using Scheme 1. 

For a continuous-flow reactor, such as a CSTR, the energy balance is an enthalpy (H) balance, if we neglect any differences in kinetic and potential energy of the flowing stream, and any shaft work between inlet and outlet. However, in comparison with a BR, the balance must include the input and output of H by the flowing stream, in addition to any heat transfer to or from the control volume, and generation or loss of enthalpy by reaction within the control volume. Then the energy (enthalpy) equation in words is... 

A similar exercise can be made with other anions and cations, producing a list of relative values of standard enthalpies of formation, anchored on Af77°(H+, ao) = 0. This database is rather useful, because it allows the enthalpies of formation (equation 2.53) and the lattice enthalpies (equation 2.47) of many crystalline ionic salts to be predicted, since their solution enthalpies are usually easy to measure. 

Occasionally these relationships have to be fairly eomplex to deseribe the system accurately. But in many cases simplification can be made without sacrificing much overall aeeuraey. We have already used some simple enthalpy equations in the examples of energy balances. 

Using a lumped model for the reactor metal wall and the simple enthalpy equation h= Cj, T, the energy equations for the reaction liquid and the metal wall... 

One may question the need for a four parameter enthalpy equation, i.e., whether describing an acid or base by two parameters each is redundant. The following simple matrix algebra shows the conditions whereby a four parameter model reverts to a less redundant two parameter equation. Letting A be the transformation matrix, E and C represent the parameters for the four parameter model, and a represent the acid parameters for the two parameter model, the following equation results ... 

We now look at the change associated with two carbons in cyclopropane and cyclobutane being sp2 instead of sp3, i.e. we consider cyclopropene (16) and cyclobutene (14). The difference of their enthalpies of formation is (120.4 2.9) kJ mol"1. Is this still larger change due to destabilization of cyclopropene and/or stabilization of cyclobutene (cf species 15 with n = 3 and 4) One way of appraising this is to look at the olefination enthalpies (equation 37) of cyclopropane, cyclobutane and butane. These three numbers are (223.8 2.6), (128.3 1.6) and (118.5 1.2) kJ mol"1 showing that cyclobutane is comparatively normal (i.e. more like the unstrained, acyclic propane) while cyclopropane is considerably different. 

Recall how to transform this enthalpy equation into an enthalpy difference equation using enthalpy of each component at the reference condition Hir. Thus... 

A full coupling of the solid and fluid phases can be achieved by solving a single enthalpy equation, common for both phases. Such an approach was used in SOFIE [13], but the use of a structured grid system usually prevented the necessary refinement inside the solids. A fully coupled system is being developed in the C-SAFE project at the University of Utah [31],... 

Remark 6.3. The vector function <5(x, 0) can be arbitrarily chosen (as long as the invertibility of T(x, 0) is preserved), which allows us to describe the slow component of the energy dynamics in terms of the enthalpy/temperature of any one of the units. Furthermore, <5(x, 0) may be chosen in such a way that (0d/50)B(x, 0) = 0. In this case, the model (6.18) will be independent of z and the corresponding Q represents a true slow variable in the system (whereas the original state variables evolve both in the fast and in the slow time scales). For example, on choosing <5(x, 0) as the sum of all the unit enthalpies (Equation (6.13)), it can be shown that indeed (88/89)B(x, 0) = 0. Thus, the total enthalpy of the process evolves only over a slow time scale. 

It was pointed out from the beginning that the concept of the solubility parameter was applicable only to amorphous polymers. In order to adapt the method to highly crystalline polymers some way must be found to deal with the heat of fusion (AHm) in the free enthalpy equation ... 

Table 11.1. S02, 02, N2 and S03 enthalpy values (MJ/kg-mole) at 690 and 820 K. They have been calculated with the enthalpy equations in Appendix G.

An efficient method of calculating heatup path points is to put enthalpy equations directly into cells D8 - J8 of Table 11.2. This is detailed in Appendix I. 

The 2nd catalyst bed heatup path is prepared by re-doing Section 14.9 s calculation for many different temperatures in the bed. Only cells HI5 to K15 are changed (most easily with enthalpy equations in cells, Appendix K). The results are tabulated in Table 14.3 and plotted in Fig. 14.3. 

Table 14.3. Heatup path points for Fig. 14.2 s 2nd catalyst bed. The points are shown graphically in Fig. 14.3. They have been calculated using matrix Table 14.2 with enthalpy equations in cells H15-K15, Appendix K. An increase in gas temperature from 700 K to 760 K in the 2nd catalyst bed is seen to be equivalent to an increase in % SO oxidized from 69.2% to 89.7%.

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