Thermodynamics

cen84959_ch03.qxd 4/1/05 12:31 PM Page 134134 | ThermodynamicsGiven: P and Tv~=v f @Tu =~ u f @Th =~h f @TFIGURE 3–42A compressed liquid may beapproximated as a saturated liquid atthe given temperature.T, °CT = 80°CP= 5 MPaIn the absence of compressed liquid data, a general approximation is to treatcompressed liquid as saturated liquid at the given temperature (Fig. 3–42).This is because the compressed liquid properties depend on temperaturemuch more strongly than they do on pressure. Thus,y ≅ y f @ T(3–8)for compressed liquids, where y is v, u, or h. Of these three properties, theproperty whose value is most sensitive to variations in the pressure is theenthalpy h. Although the above approximation results in negligible error inv and u, the error in h may reach undesirable levels. However, the error in hat low to moderate pressures and temperatures can be reduced significantlyby evaluating it fromh ≅ h f @ T v f @ T (P P sat @T ) (3–9)instead of taking it to be just h f . Note, however, that the approximation inEq. 3–9 does not yield any significant improvement at moderate to hightemperatures and pressures, and it may even backfire and result in greatererror due to overcorrection at very high temperatures and pressures (seeKostic, Ref. 4).In general, a compressed liquid is characterized byHigher pressures (P P sat at a given T)Lower tempreatures (T T sat at a given P)Lower specific volumes (v v f at a given P or T)Lower internal energies (u u f at a given P or T)Lower enthalpies (h h f at a given P or T)But unlike superheated vapor, the compressed liquid properties are notmuch different from the corresponding saturated liquid values.EXAMPLE 3–8Approximating Compressed Liquidas Saturated Liquid5 MPaDetermine the internal energy of compressed liquid water at 80°C and 5MPa, using (a) data from the compressed liquid table and (b) saturated liquiddata. What is the error involved in the second case?80u ≅ u f @ 80°CFIGURE 3–43Schematic and T-u diagram forExample 3–8.uSolution The exact and approximate values of the internal energy of liquidwater are to be determined.Analysis At 80°C, the saturation pressure of water is 47.416 kPa, and since5 MPa P sat , we obviously have compressed liquid, as shown in Fig. 3–43.(a) From the compressed liquid table (Table A–7)(b) From the saturation table (Table A–4), we readThe error involved iswhich is less than 1 percent.P 5 MPaf u 333.82 kJ>kgT 80°Cu u f @ 80°C 334.97 kJ>kg334.97 333.82333.82 100 0.34%