Department of Mechanical Engineering
ME 322 – Mechanical Engineering
Thermodynamics
Lecture 31
Ideal Gas Mixtures
Mixtures in Engineering Applications
• Natural gas
– Methane, ethane, propane, butane, nitrogen,
hydrogen, carbon dioxide, and others
• Refrigerants
– Zeotropes - True mixture behavior
• Example: R407c - R32/125/134a (23/25/52 by mass)
– Azeotropes - Mixtures that behave as a pure fluid
• Example: R507A - R125/143a (50/50 by mass)
• Air and water vapor
– Psychrometric analysis
• Air conditioning applications
2
Pure Fluid vs. Mixture Behavior
R22
103
R407C
103
200°F
150°F
150°F
100°F
P [psia]
P [psia]
100°F
50°F
102
200°F
50°F
102
0°F
0°F
-50°F
101
8x100
-50°F
101
60
80
100
120
140
160
180
200
220
h [Btu/lbm]
R22: a pure fluid; a
halogenated methane molecule
(chlorodiflouromethane)
3
-20
0
20
40
60
80
100
120
140
160
h [Btu/lbm]
R407C: a mixture of R32, R125,
and R134a
Thermodynamic Properties of Mixtures
• Real mixture behavior
– Real mixture model
• Very complex to describe analytically
– Topic for an advanced course
• EES can calculate real-properties of common mixtures!
• Low-pressure, moderate density
– Ideal solution model
• Gases are treated as real fluids with idealized mixing
– Topic for an advanced course
• Low-pressure, low density
– Ideal gas mixing model
• Gases are treated as ideal gases with idealized mixing
– ME 322!!
4
Ideal Gas Mixture Models
Even though the ideal gas mixing model is simplified, it turns
out to be fairly accurate for two important processes that
mechanical engineers deal with ...
• Air Conditioning
– Water vapor + air mixtures
• Conditions are suitable for ideal gas property
estimation – even for water vapor!
• Combustion Analysis
– Products of combustion are often at high
temperatures and low pressure
The rest of ME 322 deals with these two processes
5
An Example – Gas Turbines
Air into the
combustion
chamber
Products of combustion leaving
the combustion chamber
Combustion products can
contain CO2, H2O, O2, N2,
CO, NO2, and others!
In order to get a better
estimate of the performance
of the gas turbine, we need
to be able to determine the
properties of the mixture
passing through the turbine
6
Properties of Ideal Gas Mixtures
Consider any property, B (extensive) or b (intensive). For a
mixture,
Molar Basis
Mass Basis*
N
N
k 1
k 1
N
k 1
k 1
Bm   Bk   mk bk
Bm   Bk   nk bk
N  m 
N
Bm
k
bm 
 
 bk   wk bk

mm k 1  m m 
k 1
N
Bm N  n k 
bm 
    bk   yk bk
nm k 1  nm 
k 1
wk 
mk
mm
mass fraction
*Other common words: weight
basis or gravimetric basis
7
N
yk 
nk
nm
mole fraction**
**Note: The text uses ci for
mole fractions
Mass/Mole Fraction Conversion
In some instances, a conversion between mass fraction and
mole fraction is needed. The mass of a substance is related
to the number of moles through the molecular mass, Mi ,
mi
Mi 
ni
Considering the mass fraction,
 ni 
  Mi
nm 
ni M i
yi M i
mi




wi 
 nk 
yk M k
mm  nk M k

M
k  n  k k
k
 m
8
Mass/Mole Fraction Conversion
A similar analysis for the mole fraction reveals,
 mi  1
mi


ni
m
Mi
wi / M i
m  Mi

yi 



mk
nm
k M   m k  1 k  wk / M k 
k
k  mm  M k
Summary of findings ...
yi M i
wi 
 yk M k
k
9
wi / M i
yi 
  wk / M k 
k

mi 
M

 i

n
i 

Example
Given: A mixture of ideal gases has the following molar
composition; Argon (yAr = 0.20), helium (yHe = 0.54), and the
balance is carbon monoxide.
Find: (a) mole fraction of carbon monoxide
(b) the molecular mass of the mixture
(c) the gravimetric (mass) composition of the mixture
Note: The molecular mass of the mixture can be found by,
Mm 
10
mm
nm
1

nm
1
k mk  n
m
n M
k
k
k
  yk M k
k
Example
The mole fractions of the argon and helium are given.
Therefore, the mole fraction of carbon monoxide can be found,
nm   nk
k
y

nk
k n  1 
m
y
k
1
k
 yAr  yHe  yCO  1  yCO  1  yAr  yHe  1  0.20  0.54  0.26
k
k
Now, the molecular mass of the mixture can be found,
M m   yk M k  yAr M Ar  yHe M He  yCO M CO
k
lbm 
lbm 
lbm 
lbm



M m   0.20   39.94

0.54
4.003

0.26
28.01

17.43









lbmol 
lbmol 
lbmol 
lbmol



Table C.13a
11
Table C.13a
Table C.13a
Example
The mass fraction composition of the mixture can be found by,
yi M i
yi M i

wi 
 yk M k M m
k
Therefore,
wAr 
lbm 

0.20
39.94
 

lbmol 

Table C.13a
lbm
17.43
lbmol
 0.458
wHe 
 0.26   28.01
wCO 
12

lbm 

0.54
4.003
 

lbmol 

lbm 

lbmol 
Table C.13a
lbm
17.43
lbmol
Table C.13a
lbm
17.43
lbmol
 0.418
 0.124
Example
Comparison of mole fractions and mass fractions for this
mixture ...
Component
Ar
He
CO
y
w
0.20
0.54
0.26
0.458
0.124
0.418
  1.00
It is always a good idea to check if the
calculated fractions sum up to one!
13
  1.00
Ideal Gas Mixture Properties
We have previously seen that,
bm   wk bk
k
or bm   yk bk
k
Consider the internal energy and enthalpy of an ideal gas
mixture. The components of the mixture exist at the same
temperature as the mixture. Therefore, according to the
expressions above,
um Tm    wk uk Tk  or um   yk uk Tk 
Tm  Tk 
hm Tm    wk hk Tk  or hm   yk hk Tk 
Tm  Tk 
k
k
14
k
k
Another Example
Tm1  20F Pm1  15 psia
Given: A mixture of ideal gases is
Tm 2  300F Pm 2  60 psia
contained in a closed, rigid container
that has a volume of 2 ft3. The mixture
3
V

2
ft
m
is an equimolar binary mixture of
Q12
methane and ethane. The mixture is
initially at 15 psia, 20°F. Heat is now
transferred to the mixture pressure and
temperature become 60 psia, 300°F.
Find: The amount of heat transferred in
this process.
15