Thermodynamics

cen84959_ch03.qxd 4/1/05 12:31 PM Page 143more reduced property called the pseudo-reduced specific volume v R asv R v PP R =actualPcr(3–21)Z =…RT cr >P crChapter 3 | 143When P and v, or T and v, are given instead of P and T, the generalizedcompressibility chart can still be used to determine the third property, but itwould involve tedious trial and error. Therefore, it is necessary to define oneNote that v R is defined differently from P R and T R . It is related to T cr and P crinstead of v cr . Lines of constant v R are also added to the compressibilitycharts, and this enables one to determine T or P without having to resort totime-consuming iterations (Fig. 3–54).vv R =RTcr /P cr(Fig. A–15)EXAMPLE 3–12Using Generalized Charts to Determine PressureDetermine the pressure of water vapor at 600°F and 0.51431 ft 3 /lbm, using(a) the steam tables, (b) the ideal-gas equation, and (c) the generalized compressibilitychart.Solution The pressure of water vapor is to be determined in three differentways.Analysis A sketch of the system is given in Fig. 3–55. The gas constant,the critical pressure, and the critical temperature of steam are determinedfrom Table A–1E to be(a) The pressure at the specified state is determined from Table A–6E to beThis is the experimentally determined value, and thus it is the mostaccurate.(b) The pressure of steam under the ideal-gas assumption is determinedfrom the ideal-gas relation to beTherefore, treating the steam as an ideal gas would result in an error of(1228 1000)/1000 0.228, or 22.8 percent in this case.(c) To determine the correction factor Z from the compressibility chart (Fig.A–15), we first need to calculate the pseudo-reduced specific volume andthe reduced temperature:v R v actualRT cr >P crR 0.5956 psia # ft 3 >lbm # RP cr 3200 psiaT cr 1164.8 Rv 0.51431 ft 3 >lbmfP 1000 psiaT 600°FP RTv 10.5956 psia # ft 3 >lbm # R211060 R2 1228 psia0.51431 ft 3 >lbm10.51431 ft 3 >lbm213200 psia210.5956 psia # ft 3 >lbm # R211164.8 R2 2.372T R T T cr 1060 R1164.8 R 0.91 ∂P R 0.33FIGURE 3–54The compressibility factor can also bedetermined from a knowledge of P Rand v R .H 2 OT = 600°Fv = 0.51431 ft 3 /lbmP = ?FIGURE 3–55Schematic for Example 3–12.