Modern Engineering Thermodynamics

434 CHAPTER 12: Mixtures of Gases and Vapors
EXAMPLE 12.15 (Continued )
First, we find the mixture composition and molecular mass. The mass of the entire mixture is m m = m ammonia + m chlorine +
m nitrous oxide = 1.00 + 1.00 + 1.00 = 3.00 kg. The mass fractions are
w ammonia = w chlorine = w nitrous oxide = 1:00=3:00 = 0:333
The molecular masses of the components are found in Table C.12b as
Equation (12.12) gives the molecular mass of the mixture as
m m =
∑ 3
w i
i=1
M ammonia = 17:030 kg=kgmole
M chlorine = 70:906 kg=kgmole
M nitrous oxide = 44:013 kg=kgmole
1
1
=
w ammonia
+ w chlorine
+ w 1
kg
=
nitrous oxide 0:333
M ammonia M chlorine M nitrous oxide 17:030 + 0:333
70:906 + 0:333 = 31:4
kgmole
44:013
M i
The Amagat compressibility factors for the components, Z Ai , are found from the reduced pressure and reduced temperature
for each component. From Table C.12b, we find that
ðp c Þ ammonia = 11:280 MPa and ðT c Þ ammonia = 405:5K
ðp c Þ chlorine
= 7:710 MPa and ðT c Þ chlorine
= 417:0K
ðp c Þ nitrous oxide
= 7:270 MPa and ðT c Þ nitrous oxide
= 309:7K
and the component gas constants can be computed as
R ammonia = R=M ammonia = 8:3143 kJ=ðkgmole.KÞ=17:030 kg=kgmole = 0:488 kJ=kg .K
R chlorine = R=M chlorine = 8:3143 kJ=ðkgmole.KÞ=70:906 kg=kgmole = 0:117 kJ=kg .K
R nitrous oxide = R=M nitrous oxide = 8:3143 kJ=ðkgmole.KÞ=44:013 kg=kgmole = 0:189 kJ=kg .K
Then, the reduced temperatures and pressures are
ðT R Þ ammonia
=
ðp R Þ ammonia
=
ðT R Þ chlorine
=
ðp R Þ chlorine
=
ðT R Þ nitrous oxide
=
ðp R Þ nitrous oxide
=
T m
ðT c Þ ammonia
= 500: K
405:5K = 1:23
p m 20:0 MPa
=
ðp c Þ ammonia
11:280 MPa = 1:77
T m
ðT c Þ chlorine
= 500: K
417:0K = 1:20
p m 20:0 MPa
=
ðp c Þ chlorine
7:710 MPa = 2:59
T m
ðT c Þ nitrous oxide
= 500: K
309:7K = 1:61
p m
=
20:0 MPa
ðp c Þ nitrous oxide
7:270 MPa = 2:75
and using these values on Figure 7.6 gives the following Amagat compressibility factors:
ðZ A Þ ammonia = 0:64
ðZ A Þ chlorine
= 0:55
ðZ A Þ nitrous oxide
= 0:86