Thermodynamics

660 | Thermodynamicswhere, from Table A–11,v fg 1v g v f 2 @ 20°C 0.035969 0.0008161 0.035153 m 3 >kga dPd T b a ¢Psat,20°C ¢T b P sat @ 24°C P sat @ 16°Csat,20°C 24°C 16°C646.18 504.58 kPa8°Csince T(°C) T(K). Substituting, we get1 kJh fg 1293.15 K2 10.035153 m 3 >kg2 117.70 kPa>K2 a1 kPa # b m3 182.40 kJ/kg 17.70 kPa>KThe tabulated value of h fg at 20°C is 182.27 kJ/kg. The small differencebetween the two values is due to the approximation used in determining theslope of the saturation curve at 20°C.The Clapeyron equation can be simplified for liquid–vapor and solid–vaporphase changes by utilizing some approximations. At low pressures v g v f ,and thus v fg v g . By treating the vapor as an ideal gas, we have v g RT/P.Substituting these approximations into Eq. 12–22, we findora dPdT b Ph fgsat RT 2a dP P b sat h fgR a dTT 2 b satFor small temperature intervals h fg can be treated as a constant at some averagevalue. Then integrating this equation between two saturation states yieldsln a P 2b h fgP 1 sat R a 1 1 bT 1 T 2 sat(12–24)This equation is called the Clapeyron–Clausius equation, and it can beused to determine the variation of saturation pressure with temperature. Itcan also be used in the solid–vapor region by replacing h fg by h ig (theenthalpy of sublimation) of the substance.EXAMPLE 12–6Extrapolating Tabular Datawith the Clapeyron EquationEstimate the saturation pressure of refrigerant-134a at 50°F, using thedata available in the refrigerant tables.Solution The saturation pressure of refrigerant-134a is to be determinedusing other tabulated data.Analysis Table A–11E lists saturation data at temperatures 40°F andabove. Therefore, we should either resort to other sources or use extrapolation