For T > 523 K: use ideal gas law for v h = 2971 + 1.67 x (T- 523) + 3.14 x 10-fP - 523*) s = 7.71 + 1.67 x ln(T/523) + 6.28 x 10"* (T- 523)

For T > 623 K: use ideal gas law for v h = 3116 + 1.82 x (T - 623) + 2.63 x 10-" fP - 73*)62 s = 6.75 + 1.82 x ln{T/623) + 5.25 x 10"* (T - 623)

For T > 523 K: use ideal gas law for v h = 2971 + 1.67 x (T- 523) + 3.14 x 10-fP - 523*) s = 7.71 + 1.67 x ln(T/523) + 6.28 x 10"* (T- 523)

For T > 623 K: use ideal gas law for v h = 3116 + 1.82 x (T - 623) + 2.63 x 10-" fP - 73*)62 s = 6.75 + 1.82 x ln{T/623) + 5.25 x 10"* (T - 623)

Note: Hie first row in each table gives saturation values.

PQ, MN, etc. in the T-v diagram of Figure 2.9. In Table 2.3, the first row in each subtable corresponds to saturation conditions. In Table 2.4, the last row of the subtables correspond to saturation.

Formulas replace numerical entries for condition beyond the range of die latter. For example, in the p =19 kPa subtable in Table 2.3, the ideal-gas law very accurately yields the specific volume v for temperatures above 423 K. In this same range, the heat capacity of water vapor is linearly-dependent on temperature. This linear function gives h and s from Equations (2.14) and (3.22):

For the purpose of analytically extrapolating h and s, the reference temperature, along with the property value at this temperature, can be arbitrarily chosen. T^ in these tables has been selected as the temperature that separates the tabular values from the analytical representations of the two properties.

The formulas for h and s in Table 2.4 result from a quadratic variation of CF with temperature.

In using the single-phase tables, the complications associated with the quality of a two-phase mixture are absent. Given any two of the five thermodynamic properties treated in these tables, the others can be found. By the way that the tables are constructed, determination of the state is most easily done if pressure and temperature are given, although even in this case, double interpolation in the table is generally required. If the specified pair of variables is v and h, for example, determining the state of the system requires considerable trial-and-error hunting in either Table 2.3 or Table 2.4. For problems involving single-phase water, the computer-based steam table calculation is a vast improvement over the hand calculation. Problems 2.7,2.8, and 2.12 to 2.15 involve manipulations of the steam tables.

The tabular mode of thermal-property data representation in the single-phase tables for water is quite different from the usual method of presenting EOS information for gases or single condensed phases (solids or liquids). For gases, p-v-T behavior more commonly expressed by an equation of state such as the Van der Waals formula. Because such formulas have a limited number of parameters, their accuracy for steam is not as high as the values listed in Table 2.3. Problem 2.10 shows how to use data from Table 2.3 to determine the Van der Waals coefficients for steam.

For nonvolatile liquids and solids, the p-v-T equation is contained in the coefficient of thermal expansion and the coefficient of compressibility. Isobars in the liquid region of the T-v plot of Figure 2.9 all have about the same slope, which means that the coefficient of thermal expansion of compressed water is approximately constant. Examination of Table 2.4 shows that a varies by about a factor of two over the p and T ranges covered. Similarly, the coefficient of compressibility of water also varies about twofold in the p-T range covered by Table 2.4.

Tables 2.3 and 2.4 do not provide a and p nor the specific heats Cv and CVP. However, these properties can be estimated from the steam tables by numerical differentiation, as shown in the following examples from Table 2.4.

Example: The coefficient of theimal expansion at 5 MPa between 40 and 60°C is obtained by using the formula for V below the 5 MPa subtable:

1 A(vxlQ3) _ 1 1.018"-1.009 vxlO3 AT 1.0 60-40

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Solar Panel Basics

Solar Panel Basics

Global warming is a huge problem which will significantly affect every country in the world. Many people all over the world are trying to do whatever they can to help combat the effects of global warming. One of the ways that people can fight global warming is to reduce their dependence on non-renewable energy sources like oil and petroleum based products.

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