Modern Engineering Thermodynamics

396 CHAPTER 11: More Thermodynamic Relations
EXAMPLE 11.16 (Continued )
where, for a constant specific heat ideal gas,
and, since T 2 = T 1 here, this becomes
s 2
* − s 1
* = c p ln ðT 2 /T 1 Þ − Rlnðp 2 /p 1 Þ − s 1 *
s* 2
− s* 1
= 0 − 55:1ft
.lbf/ðlbm.RÞ
ln
778:16 ft .lbf/Btu
15:0 × 103
150:
= −0:326 Btu
lbm.R
Using p R1 = 0.202 and T R1 = 1.06, Figure 11.11 gives the entropy correction for state 1 as
kJ
ðs* − sÞ 1
= 1:50
kgmole.K = 1:50 kJ 1 Btu/ðlbmole.RÞ
Btu
= 0:360
kgmole.K 4:1865 kJ/ðkgmole.KÞ lbmole.R
and using p R2 = 20.2 and T R2 = 1.06, Figure 11.11 gives the entropy correction for state 2 as
kJ
ðs* − sÞ 2
= 22:2
kgmole.K = 22:2 kJ 1 Btu/ðlbmole.RÞ
Btu
= 5:30
kgmole.K 4:1865 kJ/ðkgmole.KÞ lbmole.R
Then, Eq. (11.42) gives
s 2 − s 1 = ðs 1
* 2 − s*Þ 1
− ðs* − sÞ 2
− ðs* − sÞ 1
M
= −0:326 Btu h
− 5:30 − 0:360
Btu i
1
lbm.R lbm.R 28:05 lbm/lbmole
= −0:500 Btu
lbm.R
The following exercises illustrate some of the elements of Example 11.16.
Exercises
45. Determine the values of Z 1 and Z 2 in Example 11.16 if the isothermal compression occurs at 1060°F instead of 80.0°F
and all the remaining variables are unchanged. Answer: Z 1 = 1.0 and Z 2 = 2.0.
46. Determine the changes in specific enthalpy and specific entropy in Example 11.16 if the final pressure is 7500. psia instead of
15.0 × 10 3 psia and all the remaining variables are unchanged. Answer: h 2 – h 1 = –145 Btu/lbm and s 2 – s 1 = –0.509 Btu/(lbm·R).
47. Rework Example 11.16 when the ethylene is replaced by air but all the other variables are unchanged. Answer: h 2 – h 1 =
–4.85 Btu/lbm, u 2 – u 1 = –51.1 Btu/lbm, and s 2 – s 1 = –0.357 Btu/(lbm· R).
11.11 IS STEAM EVER AN IDEAL GAS?
One of the most unforgivable mistakes that a thermodynamics student can make is to use the ideal gas
equations to calculate the values of the properties u, h, ands of superheated steam. Yet we do just that in the
next chapter when we discuss the thermodynamics of water vapor and air mixtures (i.e., humidity). Where did
this great academic fear of steam as an ideal gas come from, and is it really justified?
When the term steam vapor was introduced into engineering jargon in the 19th century, it originally meant only
visible, or “wet” steam. That is, steam whose state was far enough under the vapor dome to contain tiny visible
foglike liquid water droplets (sometimes called water dust). As soon as the steam became a saturated or superheated
vapor, it became invisible to the naked eye and was called steam gas, and for a long time, its properties
were actually calculated from the ideal gas equations. By the end of the 19th century, it had become clear to
thermodynamicists that the ideal gas equations did not accurately describe the behavior of high-pressure steam,
and during the first half of the 20th century, considerable effort was devoted to developing new empirical
equations for steam properties over the full range of pressures and temperatures of industrial interest.
However, these empirical equations were generally too complex and time consuming for ordinary engineering work,
so they were used instead to generate elaborate saturation and superheated steam tables that were accurate to within
1% or less over their full range. Those tables could be used easily and quickly by working engineers with at most a
simple linear interpolation required between table entries. These tables were widely distributed and continuously
improved through a series of annual International Conferences on the Properties of Steam, which began in 1929.
Today, the full steam tables have small pressure and temperature increment listings and fill an entire book.
To provide engineering students with a working knowledge of these new tables, a condensed version was
appended to all thermodynamics textbooks. Authors and professors attempted to encourage the use of these
tables and discourage the use of ideal gas equations for steam by extending the definition of a vapor to include