Thermodynamics

694 | Thermodynamicsorwhich yieldsa mf i 1dh i T m ds i v i dP m 2 0dh i T m ds i v i dP m(13–26)This is an important result because Eq. 13–26 is the starting equation in thedevelopment of the generalized relations and charts for enthalpy and entropy.It suggests that the generalized property relations and charts for real gasesdeveloped in Chapter 12 can also be used for the components of real-gasmixtures. But the reduced temperature T R and reduced pressure P R for eachcomponent should be evaluated by using the mixture temperature T m andmixture pressure P m . This is because Eq. 13–26 involves the mixture pressureP m , not the component pressure P i .The approach described above is somewhat analogous to Amagat’s law ofadditive volumes (evaluating mixture properties at the mixture pressure andtemperature), which holds exactly for ideal-gas mixtures and approximatelyfor real-gas mixtures. Therefore, the mixture properties determined with thisapproach are not exact, but they are sufficiently accurate.What if the mixture volume and temperature are specified instead of themixture pressure and temperature? Well, there is no need to panic. Just evaluatethe mixture pressure, using Dalton’s law of additive pressures, and thenuse this value (which is only approximate) as the mixture pressure.Another way of evaluating the properties of a real-gas mixture is to treatthe mixture as a pseudopure substance having pseudocritical properties,determined in terms of the critical properties of the component gases byusing Kay’s rule. The approach is quite simple, and the accuracy is usuallyacceptable.T 1 = 220 KP 1 = 10 MPaAIR79% N 221% O 2HeatFIGURE 13–17Schematic for Example 13–5.T 2 = 160 KP 2 = 10 MPaEXAMPLE 13–5Cooling of a Nonideal Gas MixtureAir is a mixture of N 2 , O 2 , and small amounts of other gases, and it can beapproximated as 79 percent N 2 and 21 percent O 2 on mole basis. During asteady-flow process, air is cooled from 220 to 160 K at a constant pressureof 10 MPa (Fig. 13–17). Determine the heat transfer during this processper kmol of air, using (a) the ideal-gas approximation, (b) Kay’s rule, and(c) Amagat’s law.Solution Air at a low temperature and high pressure is cooled at constantpressure. The heat transfer is to be determined using three differentapproaches.Assumptions 1 This is a steady-flow process since there is no change withtime at any point and thus m CV 0 and E CV 0. 2 The kinetic andpotential energy changes are negligible.Analysis We take the cooling section as the system. This is a control volumesince mass crosses the system boundary during the process. We note thatheat is transferred out of the system.The critical properties are T cr 126.2 K and P cr 3.39 MPa for N 2 andT cr 154.8 K and P cr 5.08 MPa for O 2 . Both gases remain above their