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Entropy steam table

Steam tables indicate an arbitrary zero internal energy and entropy for water in its liquid state, at the triple point of water. [Pg.7]

From the steam tables, we have initial values for entropy, enthalpy, and specifi volume ... [Pg.123]

Correct the saturation values for the moisture of the steam in the initial state. Sketch the process on a pressure-volume (P-V), Mollier (H-S), or temperature-entropy (T -S) diagram, Fig. 19.2. In state 1, y = moisture content = 15%. Using the appropriate values from the saturation-pressure steam table for 400 psia (2758.0 kPa), correct them for a moisture content of 15% ... [Pg.607]

The final pressure, 1700 psia (11,721.5 kPa), is known, as is the final entropy, 1.3686 Btu/(lb) (°F) [5.7 kJ/(kg)(°C)], with constant-entropy expansion. The T-S diagram (Fig. 19.7) shows that the steam is superheated in the final state. Enter the superheated steam table at 1700 psia (11,721.5 kPa), project across to an entropy of 1.3686, and read the final steam temperature... [Pg.614]

The steam/water enthalpy and entropy values for calculation of AG came from the steam tables and Mollier chart. See also Kotas (1995), p. 239, for moist standard atmosphere analysis. [Pg.156]

The first step is to look up the steam conditions, enthalpies, and entropies in steam tables ... [Pg.346]

There are now several ways to proceed. The most correct is to use the steam tables, and to use either the energy balance or the entropy balance and do the integrals numerically (since the internal energy, enthalpy, entropy, and the changes on vaporization depend on temperature. This is the method we will use first. Then a simpler method will be considered. [Pg.42]

Although the steam tables list steam properties against pressure and temperature, the fact that specific enthalpy and specific entropy are themselves thermodynamic variables means that it is possible to construe the steam tables as providing specific enthalpy as a tabular function of pressure and specific entropy. Alternatively, we may regard the tables as providing specific entropy as a tabular function of pressure and specific enthalpy. These two statements may be summarized mathematically by the two equations ... [Pg.194]

The procedure above has been chosen because of its close relationship to the entropy relations used in steam tables. It may be noted in passing that an alternative procedure exists for applying the gas tables to expansion problems, based on the use of tabulated relative pressures. This procedure may be regarded as marginally more attractive because it replaces the calculation and subsequent addition of R Xn pa/Pa) by a calculation of PrPa Pa, where p, is the tabulated relative pressure. An explanation of such a procedure will normally accompany the gas tables. [Pg.195]

Table 16.3 Specific enthalpy and entropy for superheated steam comparison of approximating function results with steam tables... Table 16.3 Specific enthalpy and entropy for superheated steam comparison of approximating function results with steam tables...
Table 16.2 compares the results of equations (16.76), (16.77) and (16.79) with steam tables. It will be seen that there is a close correspondence. Table 16.3 uses the calculated results of Table 16.1 together with the enthalpy/entropy conversion equations (16.63) and (16.64). The value of specific heat needed for the approximations was found using equation (16.80). [Pg.199]

If the working fluid is steam, calculate the specific entropy at stage inlet from the inlet specific enthalpy and the inlet pressure using either steam tables or equation (16.64). Check whether the steam is initially superheated or saturated by comparing the inlet specific entropy, So, with the saturated specific entropy at inlet pressure, Sglpo). If 5o.i < i,(Po), use equation (16.75) to estimate the initial dryness fraction and Zeuner s equation (15.112) to estimate the isentropic index, y< for the first stage. [Pg.202]

Note that since entropy is a state propeny, once two properties of a one-phase system, such as temperature and pressure, are fixed, the value of the entropy is also fixed. Consequently, the entropy of steam can be found in the steam tables or the Mol Her diagram, and that of methane, nitrogen, and HFC-134a in the appropriate figures in Chapter 3. In the next section we consider entropy changes for an ideal gas. and in Chapter 6 we develop the equations to be used to compute entropy changes for nonideal fluids. [Pg.122]

From the steam tables (using interpolation to obtain the entropy of steam at 214. L C and I bar), we have... [Pg.127]

From these equations, it is evident that to determine the mass distribution between the phases, we need to specify a sufficient number of variables of the individual phases to fix the thermodynamic state of each phase (i.e., the degrees of freedom F) and V — thermodynamic properties of the multiphase system in the form of Eq. 7.6-3. For example, if we know that steam and water are in equilibrium at some temperature T (which fixes the single-degree freedom of this two-phase system), the equation of state or the steam tables can be used to obtain the equilibrium pressure, specific enthalpy, entropy, and volume of each of the phases, but not the mass distribution between the phases. If, in addition, the volume tor enthalpy or entropy, etc.) per unit mass of the two-phase mixture were known, this would be sufficient to determine the distribution of mass between the two phases, and then all the other overall thermodynamic properties. [Pg.316]

Let us recall that the steam tables give the temperature at which water liquid and water vapor are at equilibrium for a given pressure. They also give the specific values for enthalpy, entropy, and volume of both liquid and vapor phases. Do these tables of values constitute a mathematical model ... [Pg.49]

The data are taken from steam tables [14, p. 447]. Traditionally, in the steam tables, the specific quantities are given rather than the molar quantities. Moreover, the entropy is normalized not to absolute zero, but to 0 °C. The full curves are the chemical potentials for the liquid phase at fixed pressure as a function of the temperature. The curves are accessible by measurement only at temperatures below the boiling point. The dashed curves represent the chemical potentials for the gaseous phase at fixed pressure as a function of the temperature. These data are accessible by measurement only at temperatures above the boiling point. Both types of chemical potential are extrapolated in the region that is not accessible to measurement. [Pg.231]

In modern steam tables, the reference state is usually taken to be the saturated liquid at its triple point (T gy= 273.16 K, P f= 0.611 kPa) at this state the internal energy and entropy are set to zero u gjr = = 0. An excerpt from such a table... [Pg.118]

Solution Water at 100 °C, 20 bar is compressed liquid and its entropy is not listed in the steam tables that are provided in the appendk. It may be estimated, however, from the tabulated values at saturation. Since the entropy of liquids is largely independent of pressure and a function of temperature only, we may relate it to entropy on the saturation line in two different paths, one of constant temperature and one of constant pressure (see a very similar calculation of enthalpy in Example, 2.12I. [Pg.140]

Use the steam tables to obtain the entropy of vaporization of water at its normal boiling point. Solution From the steam tables we obtain the entropy of saturated liquid and vapor water at 100 °C Sl = 1.3070 kJ/kg K, Sv = 7-3541 kJ/kg K. [Pg.142]

Finally, as a state function, entropy can be tabulated. The steam tables in the appendix include the specific entropy of water in the saturated region (liquid, vapor) and in the superheated vapor region. If tabulated values are not available, then entropy changes maybe calculated from eg. (4.8). The calculation requires the amount of heat that is exchanged along the path and this is generally obtained by application of the first law. This methodology is applied below to a number of special cases. [Pg.145]

Solution Entropy is a tabulated property in the steam tables. We find ... [Pg.145]

Solution (a) Using the steam tables. From the steam tables we find Sa = 7.8356 kJ/kg K. The reversible adiabatic process is isentropic therefore, Sb = Sa = 7.8356 kJ/kg K. In the final state we know pressure (20 bar) and entropy therefore, the state is fully specified. The temperature and internal energy are obtained by interpolation at Pb = 20 bar, Sb = 7.8356 kJ/kg K and the results are summarized in the table below ... [Pg.147]


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