Thermodynamics

cen84959_ch03.qxd 4/1/05 12:31 PM Page 131Chapter 3 | 131EXAMPLE 3–4Pressure and Volume of a Saturated MixtureP or TA rigid tank contains 10 kg of water at 90°C. If 8 kg of the water is in theliquid form and the rest is in the vapor form, determine (a) the pressure inthe tank and (b) the volume of the tank.Solution A rigid tank contains saturated mixture. The pressure and the volumeof the tank are to be determined.Analysis (a) The state of the saturated liquid–vapor mixture is shown inFig. 3–38. Since the two phases coexist in equilibrium, we have a saturatedmixture, and the pressure must be the saturation pressure at thegiven temperature:(b) At 90°C, we have v f 0.001036 m 3 /kg and v g 2.3593 m 3 /kg (TableA–4). One way of finding the volume of the tank is to determine the volumeoccupied by each phase and then add them:Another way is to first determine the quality x, then the average specific volumev, and finally the total volume:andP P sat @ 90°C 70.183 kPa 1Table A–42V V f V g m f v f m g v g 18 kg210.001036 m 3 >kg2 12 kg2 12.3593 m 3 >kg2 4.73 m 3x m gm t 2 kg10 kg 0.2v v f xv fg 0.001036 m 3 >kg 10.22312.3593 0.0010362 m 3 >kg4 0.473 m 3 >kgV mv 110 kg210.473 m 3 >kg2 4.73 m 3Discussion The first method appears to be easier in this case since themasses of each phase are given. In most cases, however, the masses of eachphase are not available, and the second method becomes more convenient.EXAMPLE 3–5Properties of Saturated Liquid–Vapor MixtureAn 80-L vessel contains 4 kg of refrigerant-134a at a pressure of 160 kPa.Determine (a) the temperature, (b) the quality, (c) the enthalpy of the refrigerant,and (d) the volume occupied by the vapor phase.Sat. liquidv gSat. liquidv fv f v f < v < v g v g vFIGURE 3–37The v value of a saturatedliquid–vapor mixture lies between thev f and v g values at the specified T or P.T,°C90Tm g= 90°C= 2 kgm f = 8 kgP= 70v f = 0.001036 v g = 2.3593FIGURE 3–38Schematic and T-v diagram forExample 3–4..183 kPav, m 3 /kgSolution A vessel is filled with refrigerant-134a. Some properties of therefrigerant are to be determined.Analysis (a) The state of the saturated liquid–vapor mixture is shown inFig. 3–39. At this point we do not know whether the refrigerant is in thecompressed liquid, superheated vapor, or saturated mixture region. This can