Microsoft Word - Cengel and Boles TOC _2-03-05_.doc

When Pand v, or Tand v, are given instead of Pand T, the generalized
compressibility chart can still be used to determine the third property, but it
would involve tedious trial and error. Therefore, it is necessary to define one
more reduced property called the pseudo-reduced specific volumevRas

(3–21)

Note that vRis defined differently from PRand TR. It is related to Tcrand Pcr
instead of vcr. Lines of constant vRare also added to the compressibility
charts, and this enables one to determine Tor Pwithout having to resort to
time-consuming iterations (Fig. 3–54).

vR

vactual

RTcr>Pcr

Chapter 3 | 143

(Fig. A–15)

P P

R =Pcr

vR =

RTcr/Pcr

Z =...

v

FIGURE 3–54

The compressibility factor can also be

determined from a knowledge of PR

and vR.

H 2 O

T = 600°F

v = 0.51431 ft^3 /lbm

P =?

FIGURE 3–55

Schematic for Example 3–12.

EXAMPLE 3–12 Using Generalized Charts to Determine Pressure

Determine 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 com-

pressibility chart.

Solution The pressure of water vapor is to be determined in three different

ways.

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 determined

from Table A–1E to be

(a) The pressure at the specified state is determined from Table A–6E to be

This is the experimentally determined value, and thus it is the most

accurate.

(b) The pressure of steam under the ideal-gas assumption is determined

from the ideal-gas relation to be

Therefore, 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 Zfrom the compressibility chart (Fig.

A–15), we first need to calculate the pseudo-reduced specific volume and

the reduced temperature:

vR

vactual

RTcr>Pcr

1 0.51431 ft^3 >lbm 21 3200 psia 2

1 0.5956 psia#ft^3 >lbm#R 21 1164.8 R 2

2.372

TR

T

Tcr

1060 R

1164.8 R

0.91

∂¬PR
0.33

P

RT

v

1 0.5956 psia#ft^3 >lbm#R 21 1060 R 2

0.51431 ft^3 >lbm

1228 psia

v
0.51431 ft^3 >lbm

T
600°F

f¬P
1000 psia

Tcr
1164.8 R

Pcr
3200 psia

R
0.5956 psia#ft^3 >lbm#R