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Unformatted text preview: these approximations into Eq. 12–22, we find
a dP b dT sat h fg R Ph fg RT 2 a dT b T 2 sat or
a dP b P sat For small temperature intervals hfg can be treated as a constant at some average value. Then integrating this equation between two saturation states yields
ln a P2 b P1 sat h fg R a 1 T1 1 b T2 sat
(12–24) This equation is called the Clapeyron–Clausius equation, and it can be used to determine the variation of saturation pressure with temperature. It can also be used in the solid–vapor region by replacing hfg by hig (the enthalpy of sublimation) of the substance. EXAMPLE 12–6 Extrapolating Tabular Data with the Clapeyron Equation
50°F, using the Estimate the saturation pressure of refrigerant-134a at data available in the refrigerant tables. Solution The saturation pressure of refrigerant-134a is to be determined
using other tabulated data. Analysis Table A–11E lists saturation data at temperatures 40°F and above. Therefore, we should either resort to other sources or use extrapolation cen84959_ch12.qxd 4/5/05 3:58 PM Page 661 Chapter 12
to obtain saturation data at lower temperatures. Equation 12–24 provides an intelligent way to extrapolate: | 661 ln a P2 b P1 sat hfg R a 1 T1 1 b T2 sat 40°F and T2 50°F. For refrigerant-134a, R 0.01946 In our case T1 Btu/lbm · R. Also from Table A–11E at 40°F, we read hfg 97.100 Btu/lbm and P1 Psat @ 40°F 7.432 psia. Substituting these values into Eq. 12–24 gives ln a P2 b 7.432 psia P2 0.01946 Btu> lbm 5.56 psia 97.100 Btu> lbm # 1 R 420 R a 1 b 410 R Therefore, according to Eq. 12–24, the saturation pressure of refrigerant-134a at 50°F is 5.56 psia. The actual value, obtained from another source, is 5.506 psia. Thus the value predicted by Eq. 12–24 is in error by about 1 percent, which is quite acceptable for most purposes. (If we had used linear extrapolation instead, we would have obtained 5.134 psia, which is in error by 7 percent.) 12–4 ■ GENERAL RELATIONS FOR du, dh, ds, cv, AND cp The state postulate established that the state of a simple compressible system is completely specified by two independent, intensive properties. Therefore, at least theoretically, we should be able to calculate all the properties of a system at any state once two independent, intensive properties are available. This is certainly good news for properties that cannot be measured directly such as internal energy, enthalpy, and entropy. However, the calculation of these properties from measurable ones depends on the availability of simple and accurate relations between the two groups. In this section we develop general relations for changes in internal energy, enthalpy, and entropy in terms of pressure, specific volume, temperature, and specific heats alone. We also develop some general relations involving specific heats. The relations developed will enable us to determine the changes in these properties. The property values at specified states can be determined only after the selection of a reference state, the choice of which is quite arbitrary. Internal Energy Changes
We choose the internal energy to be a function of T and v; that is, u u(T, v) and take its total differential (Eq. 12–3):
du a 0u b dT 0T...
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This note was uploaded on 03/09/2009 for the course ME 430 taught by Professor Y during the Spring '09 term at CUNY City.
- Spring '09