```13-24
13-46 The volumetric analysis of a mixture of gases is given. The volumetric and mass flow rates are to be determined
using three methods.
Properties The molar masses of O2, N2, CO2, and CH4 are 32.0, 28.0, 44.0, and 16.0 kg/kmol, respectively (Table A-1).
Analysis (a) We consider 100 kmol of this mixture. Noting that volume fractions are equal to the mole fractions, mass of
each component are
m O2  N O2 M O2  (30 kmol)(32 kg/kmol)  960 kg
m N2  N N2 M N2  (40 kmol)(28 kg/kmol)  1120 kg
30% O2
40% N2
10% CO2
20% CH4
(by volume)
m CO2  N CO2 M CO2  (10 kmol)(44 kg/kmol)  440 kg
m CH4  N CH4 M CH4  (20 kmol)(16 kg/kmol)  320 kg
The total mass is
m m  m O2  m N2  m CO2  m CH4
 960  1120  440  320  2840 kg
The apparent molecular weight of the mixture is
Mm 
Mixture
8 MPa, 15C
mm
2840 kg

 28.40 kg/kmol
N m 100 kmol
The apparent gas constant of the mixture is
R
Ru
8.314 kJ/kmol  K

 0.2927 kJ/kg  K
28.40 kg/kmol
Mm
The specific volume of the mixture is
v
RT (0.2927 kPa  m 3 /kg  K)(288 K)

 0.01054 m 3 /kg
P
8000 kPa
The volume flow rate is
V  AV 
D 2
V
4
 (0.02 m) 2
4
(3 m/s)  0.0009425 m 3 /s
and the mass flow rate is
m 
V 0.0009425 m 3 /s

 0.08942 kg/s
v
0.01054 m 3 /kg
(b) To use the Amagat’s law for this real gas mixture, we first need the mole fractions and the Z of each component at the
mixture temperature and pressure. The compressibility factors are obtained using Fig. A-15 to be



 Z O2  0.95
8 MPa

 1.575 

5.08 MPa

T R ,O2 
Tm
288 K

 1.860
Tcr,O2 154.8 K
T R , N2 
PR ,O2 
Pm
Pcr,O2
PR , N2
288 K
 0.947
304.2 K
8 MPa

 1.083
7.39 MPa
T R ,CO2 
PR ,CO2


 Z CO2  0.199


288 K
 2.282
126.2 K
8 MPa

 2.360
3.39 MPa
288 K
 1.507
191.1 K
8 MPa

 1.724
4.64 MPa
T R ,CH4 
PR ,CH4


 Z N2  0.99




 Z CH4  0.85


and
Zm 
y Z
i
i
 y O2 Z O2  y O2 Z O2  y CO2 Z CO2  y CH4 Z CH4
 (0.30)(0.95)  (0.40)(0.99)  (0.10)(0.199)  (0.20)(0.85)  0.8709
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
13-25
Then,
v
Z m RT (0.8709)(0.2927 kPa  m 3 /kg  K)(288 K)

 0.009178 m 3 /kg
P
8000 kPa
V  0.0009425 m 3 /s
m 
V 0.0009425 m 3 /s

 0.10269 kg/s
v 0.009178 m 3 /kg
(c) To use Kay's rule, we need to determine the pseudo-critical temperature and pseudo-critical pressure of the mixture
using the critical point properties of mixture gases.
Tcr , m 
yT
i cr ,i
 y O2 Tcr ,O2  y N2 Tcr , N2  y CO2 Tcr ,CO2  y CH4 Tcr ,CH4
 (0.30)(154.8 K)  (0.40)(126.2 K)  (0.10)(304.2 K)  (0.20)(191.1 K)  165.6 K
Pcr , m 
y P
i cr ,i
 y O2 Pcr ,O2   y N2 Pcr , N2  y CO2 Pcr ,CO2  y CH4 Pcr ,CH4
 (0.30)(5.08 MPa)  (0.40)(3.39 MPa)  (0.10)(7.39 MPa)  (0.20)(4.64 MPa)  4.547 MPa
and
TR 
PR 
Tm
'
Tcr,
m
Pm
'
Pcr,
m



 Z m  0.92
8 MPa

 1.759 

4.547 MPa


288 K
 1.739
165.6 K
(Fig. A-15)
Then,
v
Z m RT (0.92)(0.2927 kPa  m 3 /kg  K)(288 K)

 0.009694 m 3 /kg
P
8000 kPa
V  0.0009425 m 3 /s
m 
V 0.0009425 m 3 /s

 0.009723 kg/s
v
0.09694 m 3 /kg
PROPRIETARY MATERIAL. © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
preparation. If you are a student using this Manual, you are using it without permission.
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