EGR 334 Thermodynamics
Chapter 12: Sections 1-4
Lecture 38:
Ideal Gas Mixtures
Quiz Today?
Today’s main concepts:
• Be able to describe ideal gas mixture composition in terms of
mass fractions and mole fractions.
• explain use of the Dalton model to relate pressure, volume,
and temperature and to calculate changes in U, H, and S for
ideal gas mixtures.
• Be able to apply mass, energy, and entropy balances to
systems involving ideal gas mixtures, including mixing
processes.
Reading Assignment:
Read Chapter 12, Sections 5-8
Homework Assignment:
Problems from Chap 12: 4, 10, 22, 28
Sec 12.1 : Describing Mixture Composition
3
Thus far we have been working with pure fluids
• Nitrogen
• Water
• Oxygen
• Air
• Ammonia
• R22
For gas mixtures, in addition to the normal to parameters (T, p), we
also need to know the mixture composition.
Number of moles
ni 
mi
M
i
mass of i

Molecular
Weight
Mass fraction mf  m i  mass of i
i
mT
total mass
where
Mole fraction y  n i  moles of i
i
nT
total moles
where
Avg. Molecular Weight
M 
mT
nT

of i
1
 mf
i
i
to tal m ass
to tal m o les
1
nM
i

i
nT

yi
i
i


i
yi M
i
Sec 12.2 : Relating P, V, and T for Ideal Gas Mixtures
Ideal gas
pV  nRT
4
This refers to the overall mixture, on average.
How do we describe the relationship between
p, V, and T for each component?
Dalton Model: Assumes that each component behaves as an ideal
gas at the specified T and V of the mixture. This assumes that the
gases do not interact with each other.
Partial Pressure
and
pi

p
pi 
ni R T
V
ni R T V

nT R T V
ni
n
 yi
thus
pi  yi p
Amagat Model: Assumes that each component behaves as an ideal
gas at the specified T and p of the mixture.
Partial Volume
Vi 
ni R T
p
and
V i  y iV
5
Example (12.6): Natural gas at 23°C, 1 bar enters a furnace with the
following molar analysis:
40% propane (C3H8), 40% ethane (C2H6), and 20% methane (CH4).
Determine
(a) The analysis in terms of mass fractions
(b) The partial pressure of each component, in bar
(c) The mass flow rate, in kg/s, for a volumetric flow rate of 20 m3/s.
i
C3H8
C2H6
CH4
Total
T (°C)
23
p (bar)
1
yi
Mi
mi
mfi
mi (kg/s)
0.40
0.40
0.20
1.0
yi is the mole fraction of
each component
6
Example (12.6):
40% propane (C3H8),
40% ethane (C2H6), and 20% methane (CH4).
(a) The analysis in terms of mass fractions
(b) The partial pressure of each component, in bar
(c) The mass flow rate, in kg/s, for a volumetric flow rate of 20 m3/s.
Molecular weights of each component are found in Table A-1:
i
C33H88
C22H66
CH44
Total
T (°C)
23
p (bar)
1
yii
0.40
0.40
0.20
Mii
44.09
30.07
16.04
mii
17.636
12.028
3.208
1.0
32.872
mfii
mii(kg/s)
Units are effectively kg/kmol, thus mt = M
The mass of each component is
found
m  y M
i
i
i
On a 1 kmol basis then
m P   0 . 4  44 . 09   17 . 636
m E   0 . 4 30 . 09   12 . 028
m M   0 . 2 16 . 04   3 . 208
m to ta l 
m
i
i
 3 2 .8 7 2
7
Example (12.6):
(b) The partial pressure of each component, in bar
(c) The mass flow rate, in kg/s, for a volumetric flow rate of 20 m3/s.
To find Partial Pressure:
pi  yi p
Total
p P   0 . 4 1bar

0 . 4 bar
T (°C)
23
1

0 . 4 bar
p (bar)
p E   0 . 4 1bar
i
C3H8
C2H6
CH4
yi
0.40
0.40
0.20
Mi
44.09
30.07
16.04
mi
17.636
12.028
3.208
32.872
mfi
0.5365
0.3659
0.0976
1.0
mi(kg/s)
1.0
p M   0 . 2 1bar

0 . 2 bar
8
Example (12.6):
(c) The mass flow rate, in kg/s, for a volumetric flow rate of 20 m3/s.
First convert volumetric
flow rate to mass flow rate:
m 
 AV 

 AV  P
v
RT
3
m 
i
C3H8
C2H6
CH4
T (°C)
R
2
M T
2
( 2 0 m / s )(1 0 kN / m )
 8 .3 1 4 kJ / km o l  K 

  296 K
 3 2 .8 7 2 kg / km o l 
Total
23
p (bar)
0.40
0.40
0.20
1.0
yi
0.40
0.40
0.20
1.0
Mi
44.09
30.07
16.04
mi
17.636
12.028
3.208
32.872
mfi
0.5365
0.3659
0.0976
1.0
mi(kg/s)

 AV  p
 26.71
kg
s
kJ

kN  m
9
Example (12.6):
(c) The mass flow rate, in kg/s, for a volumetric flow rate of 20 m3/s.
Can also find the mass flow rate of each component
mi 
 m fi 
m
m P   0 .5 3 6 5   2 6 .7 1 kg / s 
i
C3H8
C2H6
CH4
T (°C)
Total
23
P (bar)
0.40
0.40
0.20
1.0
yi
0.40
0.40
0.20
1.0
Mi
44.09
30.07
16.04
mi
17.636
12.028
3.208
32.872
mfi
0.5365
0.3659
0.0976
1.0
mi(kg/s)
14.33
9.77
2.61
26.71
 14.33
kg
s
m E   0 .3 6 5 9   2 6 .7 1 kg / s 
 9.77
kg
s
m M   0 .0 9 7 6   2 6 .7 1 kg / s 
 2.61
kg
s
Sec 12.3 : Evaluating U, H, S and specific heats
10
The properties of U, H, S, and c of a mixture are additive.
Mass of system: m  m  m  m  ...... 
T
1
2
3
m
i
i
Internal energy
U T  U 1  U 2  U 3  ...... 
U
i
i
mu
 ......   n u
 ......   y u
mu T  m 1u 1  m 2 u 2  m 3 u 3  ...... 
n u T  n1u 1  n 2 u 2  n 3 u 3
u T  y 1u 1  y 2 u 2  y 3 u 3
i
i
i
i
i
i
i
i
i
Where a similar set of equations can be written for H and S
For specific heat c P  y1 c P ,1  y 2 c P , 2  y 3 c P , 3  ...... 
cV  y1 cV ,1  y 2 cV , 2  y 3 cV , 3  ...... 

y i c P ,i
i

i
y i cV , i
Sec 12.4 : Analyzing Systems Involving Mixtures
11
For mixtures with constant
composition (no chemical reaction):
U 2 U1 
 n u @ T   u @ T 
i
i
2
i
1
i
u 
 y u @ T   u @ T 
i
i
2
i
1
i
For constant specific heats,
 u  c v T 2  T1 
with
 u i  c v , i T 2  T1 
Equations for enthalpy (H) are similar to those for internal energy (U), but
uses cp.
Sec 12.4 : Analyzing Systems Involving Mixtures
12
For entropy, is also dependent upon pressure changes.
S 2  S1 
 n  s @ T
i
i
s 

i
2
, p 2   s i  @ T1 , p1  

y i s i  @ T 2 , p i , 2   s i  @ T1 , p i ,1 
i
 pi ,2
 s  s  i  @ T 2   s  i  @ T1   R ln 
 p
 i ,1
For constant specific heats,
 T2 
 p2 
 s  c P ln 
  R ln 

T
p
 1
 1 
T 
 p 
w ith  s i  c P , i ln  2   R ln  2 
 T1 
 p1 
or
T 
V 
 s  cV ln  2   R ln  2 
 T1 
 V1 
w ith
T 
V 
 s i  cV , i ln  2   R ln  2 
 T1 
 V1 


 yi p2 
  s  i  @ T 2   s  i  @ T1   R ln 

y
p
 i 1 

13
Example (12.17): A mixture of 2 kg of H2 and 4 kg of N2 is compressed in a
piston-cylinder assembly in a polytropic process for which n = 1.2. The
temperature increases from 22 to 150°C. Using constant values for the
specific heats, determine
(a) The heat transfer, in kJ.
(b) The entropy change, in kJ/K.
Principles to be applied:
Closed Ideal Gas system:
1st Law of Thermodynamics:
pV  nRT
p iV  n i R T
and
 U  U 2  U 1  Q1 2  W 1 2
where for polytropic process
W12 

2
2
2nd
Law of thermodynamics:
 S  S 2  S1 
pdV 
1
Q
T
1

m R  T 2  T1 
1 n
14
Example (12.17): A mixture of 2 kg of H2 and 4 kg of N2 is compressed in a
piston-cylinder assembly in a polytropic process for which n = 1.2. The
temperature increases from 22 to 150°C. Using constant values for the
specific heats, determine
(a) The heat transfer, in kJ.
(b) The entropy change, in kJ/K.
Find the moles of each component and the Mixtures Molecular Weight:
nH 2 
mH2
M
2 kg

H2
nN2 
2 .0 1 8 kg / km o l
 0.991km ol
yH2 
nH 2
n to ta l

m N2
M
N2
4 kg

2 8 .0 1kg / km o l
 0.143km ol
0 .9 9 1km o l
yN2 
(0 .9 9 1  0 .1 4 3) km o l
 0 .8 7 4
nN2
n to ta l

0 .1 4 3 km o l
(0 .9 9 1  0 .1 4 3) km o l
 0.126
Molecular
Weight:
M  y H M H  y N M N   0.874   2.018    0.126   28.01   5.29
2
2
2
2
15
Example (12.17): A mixture of 2 kg of H2 and 4 kg of N2 is compressed in a
piston-cylinder assembly in a polytropic process for which n = 1.2. The
temperature increases from 22 to 150°C. Using constant values for the
specific heats, determine
(a) The heat transfer, in kJ.
(b) The entropy change, in kJ/K.
To Evaluate work for a polytropic process.
W 12 
mR T 2  T1 
1 n

m R M
T
2
 T1 
1 n
 8 .3 1 4 kJ / km o l  K 
 2  4  kg 
 1 5 0  2 2  K
5 .2 9 kg / km o l



 1  1 .2 
  6035.1kJ
16
Example (12.17):
(a) The heat transfer, in kJ.
(b) The entropy change, in kJ/K.
Evaluate change in internal energy
Assuming constant heat capacity from Table A-20 at average temperature (359 K)
 U  m H 2 c v , H 2  T 2  T1   m N 2 c v , N 2  T 2  T1 
  2 kg   1 0 .3 1 1 kJ / kg  K
 1 5 0  2 2  K
  4 kg   0 .7 4 5 kJ / kg  K
 1 5 0  2 2  K
 3021.1kJ
Then using the energy balance to evaluate Q
Q1 2   U 1 2  W 1 2
 3021.1  6035.1   3014 kJ
17
Example (12.17):
(b) The entropy change, in kJ/K.

 T2 
 V2  
 s   cV ln    R ln 

T
V
 1
 1 


 T2 
 T2
R
  cV ln   
ln 
 T1   n  1   T1





1  n 1


 T2 
 T2 
 cV ln    R ln  



 T1 
 T1 



R   T2 
  cV 
 ln  
 n  1    T1 

Using M = 5.29 kg/kmol,
The mixture’s cv (avg. heat capacity) needs to be found.
cV 

m f i cV , i  m f H 2 cV , H 2  m f N 2 cV , N 2
i
 2 kg 
 4 kg 

10.311
kJ
/
kg

K

 

  0.745 kJ / kg  K
 6 kg 
 6 kg 
so
cV 
  3.933 kJ
cV
M
cV  cV M  (3 .9 3 3 kJ / kg  K )(5 .2 9 kg / km o l )
 20.8 kJ / km ol  K
/ kg  K
18
Example (12.17):
(b) The entropy change, in kJ/K.
Therefore:


 T2 
R
 s   cV 
 ln 

 n  1    T1 


 8.314 kJ / km ol  K    423 K 
   20.8 kJ / km ol  K  
 ln 

(1.2  1)

  395 K 
  1 .4 2 2 kJ / km o l  K
s 
s
M

 1 .4 2 2 kJ / km o l  K
  1 .4 2 2 kJ / km o l  K
  0 .2 6 9 kJ / kg  K
5 .2 9 kg / km o l
 S  m  s  (6 kg )(  0 .2 6 9 kJ / kg  K )   1 .6 2 3 kJ / K
Sec 12.4.2 : Mixing of Ideal gases
19
The previous example considered a mixture that had already been formed.
How is a process different when a mixture is formed from two individual
gases which might originally be a different temperatures and pressures?
Whenever two highly ordered substances are mixed, entropy is expected to
increase. This is because
-- Gases are initially at different temperature.
-- Gases are initially at different pressures
-- Gases are distinguishable from one another.
irrev.
process
Sec 12.4.2 : Mixing of Ideal gases
20
Example 3:
Consider a canister that is initially divided into two sections.
One side contains 2 lbmol of Nitrogen (N2) at 500°R and 2 atm
One side contains 3 lbmol of Oxygen (O2) at 300°R and 1 atm.
Determine final temperature, pressure, and entropy production when mixed.
Start with Energy Balance: Assume no heat lost from system
No change of total volume  no work
U  Q  W  0
where
U 1  n N 2 u N 2 (T N 2 )  n O 2 u O 2 (TO 2 )
U 2  n N 2 u N 2 (T f )  n O 2 u O 2 (T f )
so
or

 
0  n N 2 u N 2 (T f )  n O 2 u O 2 (T f )  n N 2 u N 2 (T N 2 )  n O 2 u O 2 (T O 2 )
0  n N 2  u N 2 (T f )  u N 2 (T N 2 )   n O 2  u O 2 (T f )  u O 2 (TO 2 ) 
Assuming constant specific heat, c v _ N
2
and c v _ O 2
0  n N 2  c v _ N 2 (T f  T N 2 )   n O 2  c v _ O 2 (T f  TO 2 ) 

Sec 12.4.2 : Mixing of Ideal gases
21
Example 3:
Consider a canister that is initially divided into two equal sized sections. One
side contains 2 lbmol of Nitrogen (N2) at 500°R and 2 atm
One side contains 3 lbmol of Oxygen (O2) at 300°R and 1 atm.
Determine final temperature, pressure, and entropy production when mixed.
Solving for final temperature Tf
Tf 
n N 2 c v _ N 2 T N 2  n O 2 c v _ O 2 TO 2
n N 2 cv _ N 2  nO2 cv _ O2
A good choice to use for the temperature to find the cv is the average
temperature of the initial gases (Tave = 400 R). From Table A-20E.
M
c v _ O 2 (@ 400 R )  0.168 B tu / lb m  R
M O 2  32.0 lb m / lb m ol
o
and
so
 28.01lb m / lb m ol
c v _ N 2 (@ 400 R )  0.180 B tu / lb m  R
o
cv _ N 2  cv _ N 2 M
N2
N2
 (0.180 B tu / lb m  R )(28.01lb m / lb m ol )  5.042 B tu / lb m ol  R
c v _ O 2  c v _ O 2 M O 2  (0.168 B tu / lb m  R )(32.0 lb m / lb m ol )  5.376 B tu / lb m ol  R
Sec 12.4.2 : Mixing of Ideal gases
22
Example 3:
Consider a canister that is initially divided into two sections.
One side contains 2 lbmol of Nitrogen (N2) at 500°R and 2 atm
One side contains 3 lbmol of Oxygen (O2) at 300°R and 1 atm.
Determine final temperature, pressure, and entropy production when mixed.
therefore:
Tf 

n N 2 c v _ N 2 T N 2  n O 2 c v _ O 2 TO 2
n N 2 cv _ N 2  nO2 cv _ O2
(2 lb m ol )(5.042 B tu / lb m ol  R )(500 R )  (3 lb m ol )(5.376 B tu / lb m ol  R )(300 R )
(2 lb m ol )(5.042 B tu / lb m ol  R )  (3 lb m ol )(5.376 B tu / lb m ol  R )
 377 R
o
Sec 12.4.2 : Mixing of Ideal gases
23
Example 3:
Consider a canister that is initially divided into two sections.
One side contains 2 lbmol of Nitrogen (N2) at 500°R and 2 atm
One side contains 3 lbmol of Oxygen (O2) at 300°R and 1 atm.
Determine final temperature, pressure, and entropy production when mixed.
To find the final pressure first find the volume of the total original gases.
V  V N 2  VO2
where
p O 2 V O 2  n O 2 R TO 2
p N 2V N 2  n N 2 R TN 2
VN2 

n N 2 R TN 2
VO2 
pN2
(2 lb m ol )(1.986 B tu / lb m ol  R )(500 R ) 1 ft
29.4 lb f / in
 365 ft
2
3
778 lbf  ft
2
144 in
2
B tu

n O 2 R TO 2
p O2
(3 lb m ol )(1.986 B tu / lb m ol  R )(300 R ) 1 ft
14.7 lb f / in
 657 ft
total volume is then
V  V N 2  V O 2  365  657  1022 ft
3
3
2
778 lbf  ft
2
144 in
2
B tu
Sec 12.4.2 : Mixing of Ideal gases
24
Example 3:
Consider a canister that is initially divided into two equal sized sections. One
side contains 2 lbmol of Nitrogen (N2) at 500°R and 2 atm
One side contains 3 lbmol of Oxygen (O2) at 300°R and 1 atm.
Determine final temperature, pressure, and entropy production when mixed.
The mixed gases have a combined Molecular weight given by
M  yN2 M

2 lb m ol
5 lb m ol
N2
 y O2 M O2
(28.01 lb m / lb m ol ) 
3 lb m ol
5 lb m ol
(32.00 lb m / lb m ol )  30.40 lb m / lb m ol
using Ideal Gas equation:
pf 
nt R T f

(5 lb m ol )(1.986 B tu / lb m ol  R )(377 R ) 778 lb f  ft
V
 19.79 psi
1022 ft

1.346 atm
3
B tu
1 ft
2
144 in
2
Sec 12.4.2 : Mixing of Ideal gases
25
Example 3:
Consider a canister that is initially divided into two sections.
One side contains 2 lbmol of Nitrogen (N2) at 500°R and 2 atm
One side contains 3 lbmol of Oxygen (O2) at 300°R and 1 atm.
Determine final temperature, pressure, and entropy production when mixed.
To find the change in entropy:
f
 S  S f  S1 

1
where
0
Q

T
S 1  n N 2 s N 2 (T N 2 , p N 2 )  n O 2 s O 2 (TO 2 , p N 2 )
S f  n N 2 s N 2 (T f , y N 2 p f )  n O 2 s O 2 (T f , y O 2 p f )
therefore
  n N  s N (T f , y N p f )  s N (T N , p N )   n O  s O (T f , y O p f )  s O (TO , p N ) 
2
2
2
2
2
2
2
2
2
2
2
2
then using the form based on ideal gas behavior with constant specific heat
  nN
2

 Tf
 c p _ N 2 ln 
T

 N2

 yN p f
2
  R ln 

 p
N2



   nO2

 

 Tf
 c p _ O 2 ln 
T

 O2

 yO p f
2
  R ln 

 p
O2





 
Sec 12.4.2 : Mixing of Ideal gases
26
Example 3:
Consider a canister that is initially divided into two sections.
One side contains 2 lbmol of Nitrogen (N2) at 500°R and 2 atm
One side contains 3 lbmol of Oxygen (O2) at 300°R and 1 atm.
Determine final temperature, pressure, and entropy production when mixed.
To find the change in entropy:
  nN
2

 Tf
 c p _ N 2 ln 
T

 N2

 yN p f
2
  R ln 

 p
N2



   nO2

 

 Tf
 c p _ O 2 ln 
T

 O2

 yO p f
2
  R ln 

 p
O2



 377 
 (0 .4 )(1 .3 4 6 a tm )
 2 lb m o l  (5 .0 4 2 B tu / lb m o l  R ) ln 

(1
.9
8
6
B
tu
/
lb

R
)
ln
m ol


2 a tm
 500 





 




 377 K 
 (0 .6 )(1 .3 4 6 a tm )
 3 lb m o l  ( 4 .3 7 6 B tu / lb m o l  R ) ln 

(1
.9
8
6
B
tu
/
lb

R
)
ln
m ol


3
0
0
K
1a tm




 2.365  4.272

6.637 B tu / R



27
end of slides Lecture 38