Energy balances enthalpies

B. Explore. Since there are only two unknowns in the mass balances, B and D, we can solve for these variables immediately. Either solve Eqs. f3-ll and (3z2) simultaneously, or use Eqs. (3 ) and (3 4). For the energy balances, enthalpies must be determined. These can be read from the enthalpy-composition diagram fFigure 2-41. Then Q, can be determined from the balance around the... 

Then a system of seven simultaneous equations may be derived from three atom balances (oxygen, hydrogen, and carbon), an energy balance (enthalpy of reactants must equal enthalpy of products), a mole fraction constraint, and the two remaining equilibrium relations. 

The analysis of the heat exchanger network first identifies sources of heat (termed hot streams) and sinks (termed cold streams) from the material and energy balance. Consider first a very simple problem with just one hot stream (heat source) and one cold stream (heat sink). The initial temperature (termed supply temperature), final temperature (termed target temperature), and enthalpy change of both streams are given in Table 6.1. 

It should be stressed that although these symmetry considerations may allow one to anticipate barriers on reaction potential energy surfaces, they have nothing to do with the thermodynamic energy differences of such reactions. Symmetry says whether there will be symmetry-imposed barriers above and beyond any thermodynamic energy differences. The enthalpies of formation of reactants and products contain the information about the reaction s overall energy balance. 

The essential differences between sequential-modular and equation-oriented simulators are ia the stmcture of the computer programs (5) and ia the computer time that is required ia getting the solution to a problem. In sequential-modular simulators, at the top level, the executive program accepts iaput data, determines the dow-sheet topology, and derives and controls the calculation sequence for the unit operations ia the dow sheet. The executive then passes control to the unit operations level for the execution of each module. Here, specialized procedures for the unit operations Hbrary calculate mass and energy balances for a particular unit. FiaaHy, the executive and the unit operations level make frequent calls to the physical properties Hbrary level for the routine tasks, enthalpy calculations, and calculations of phase equiHbria and other stream properties. The bottom layer is usually transparent to the user, although it may take 60 to 80% of the calculation efforts. 

The specific enthalpies ia equation 9 can be determined as described earUer, provided the temperatures of the product streams are known. Evaporative cooling crystallizers operate at reduced pressure and may be considered adiabatic (Q = 0). As with of many problems involving equiUbrium relationships and mass and energy balances, trial-and-error computations are often iavolved ia solving equations 7 through 9. 

An equation representing an energy balance on a combustion chamber of two surface zones, a heat sink Ai at temperature T, and a refractory surface A assumed radiatively adiabatic at Tr, inmost simply solved if the total enthalpy input H is expressed as rhCJYTv rh is the mass rate of fuel plus air and Tp is a pseudoadiabatic flame temperature based on a mean specific heat from base temperature up to the gas exit temperature Te rather than up to Tp/The heat transfer rate out of the gas is then H— — T ) or rhCp(T f — Te). The... 

Compute a new set of values of the T) tear variables by solving simultaneously the set of N energy-balance equations (13-72), which are nonlinear in the temperatures that determine the enthalpy values. When linearized by a Newton iterative procedure, a tridiagonal-matrix equation that is solved by the Thomas gorithm is obtained. If we set gj equal to Eq. (13-72), i.e., its residual, the hnearized equations to be solved simultaneously are... 

Energy balances differ from mass balances in that the total mass is known but the total energy of a component is difficult to express. Consequently, the heat energy of a material is usually expressed relative to its standard state at a given temperature. For example, the heat content, or enthalpy, of steam is expressed relative to liquid water at 273 K (0°C) at a pressure equal to its own vapor pressure. 

Equation 6-10 is the macroscopic energy balance equation, in which potential and kinetic energy terms are neglected. From tliermodynamics, the enthalpy per unit mass is expressed as... 

Thus the total mass flows tn= m, + m,) differ in different cases. Water vapor flow th, is obtained by multiplying the dry air mass flow by the corresponding humidity x (Eq. 4.93). As a basic quantity in humid air mass and energy balance calculations, we use dry air mass flow m and the effect of humidity on the energy balance is noted in the enthalpy h, (Eq. 4.87). 

Alternatively, we can write the energy balance with the help of the enthalpy flows as... 

Many cliciiiical reactions evolve or absorb heat. When applying energy balances (consenatioit law for energy) in tccluiical calculations the heat (enthalpy) of reaction is often indicated in mole units so that tliey can be directly applied to demonstrate its chemical change. To simplify the presentation that follows, examine the equation ... 

Methods have been given for the calculation of the pressure drop for the flow of an incompressible fluid and for a compressible fluid which behaves as an ideal gas. If the fluid is compressible and deviations from the ideal gas law are appreciable, one of the approximate equations of state, such as van der Waals equation, may be used in place of the law PV = nRT to give the relation between temperature, pressure, and volume. Alternatively, if the enthalpy of the gas is known over a range of temperature and pressure, the energy balance, equation 2.56, which involves a term representing the change in the enthalpy, may be employed ... 

Enthalpy-pressure diagram for water-steam 813 Enthalpy term, energy balance equation 74 Entropy 3,28... 

A homogeneous flow basis must be used when thermodynamic equilibrium is assumed. For furtl er simplification it is assumed there will be no reaction occurring in the pipeline. The vapor and liquid contents of the reactor are assumed to be a homogeneous mass as they enter the vent line. The model assumes adiabatic conditions in the vent line and maintains constant stagnation enthalpy for the energy balance. 

The time derivative is zero at steady state, but it is included so that the method of false transients can be used. The computational procedure in Section 4.3.2 applies directly when the energy balance is given by Equation (5.28). The same basic procedure can be used for Equation (5.25). The enthalpy rather than the temperature is marched ahead as the dependent variable, and then Tout is calculated from Hout after each time step. 

The kinetic equilibrium constant is estimated from the thermodynamic equilibrium constant using Equation (7.36). The reaction rate is calculated and compositions are marched ahead by one time step. The energy balance is then used to march enthalpy ahead by one step. The energy balance in Chapter 5 used a mass basis for heat capacities and enthalpies. A molar basis is more suitable for the current problem. The molar counterpart of Equation (5.18) is... 

A dynamic differential equation energy balance was written taking into account enthalpy accumulation, inflow, outflow, heats of reaction, and removal through the cooling jacket. This balance can be used to calculate the reactor temperature in a nonisothermal operation. 

A dynamic model should be consistent with the steady-state model. Thus, Eqs (1) and (4) should be extended to dynamic form. For the better convergence and computational efficiency, some assumption can be introduced the total amounts of mass and enthalpy at each plate are maintained constant. Then, the internal flow can be determined by total mass balance and total energy balance and the number of differential equations is reduced. Therefore, the dynamic model can be established by replacing component material balance in Eq. (1) with the following equation. 

The vapour enthalpies are calculated from the molar heat capacity functions for the vapour components and the latent heats of vaporisation at standard temperature. The vapour overflow, V, is then obtained from the energy balance as... 

Enthalpy-concentration diagrams greatly facilitate the calculation of energy balances involving concentration and phase changes this is illustrated in Example 3.6. 

Himmelbau (1995) or any of the general texts on material and energy balances listed at the end of Chapter 2. The Ponchon-Savarit graphical method used in the design of distillation columns, described in Volume 2, Chapter 11, is a further example of the application of the lever rule, and the use of enthalpy-concentration diagrams. 

In many flash processes the feed stream is at a higher pressure than the flash pressure and the heat for vaporisation is provided by the enthalpy of the feed. In this situation the flash temperature will not be known and must be found by trial and error. A temperature must be found at which both the material and energy balances are satisfied. 

If a material balance is to be solved, then the convergence variables can be taken to be the component molar flowrates. When a material and energy balance is to be solved, the additional convergence variables are usually taken to be pressure and enthalpy. 

This is consistent with the starting assumption that correlations for power greater than 2 MW were to be used. The outlet steam enthalpy is given by an energy balance from Equation 23.22 ... 

An energy balance around the deaerator assuming the enthalpy of the vented steam to be saturated at the deaerator pressure gives ... 

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