1. EGR 334 Thermodynamics Chapter 12: Sections 1-4
Lecture 38:
Ideal Gas Mixtures
Quiz Today?

2. 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

3. Thus far we have been working with pure fluids Water Air R22 Nitrogen
Sec 12.1 : Describing Mixture Composition
Thus far we have been working with pure fluids
Water
Air
R22
Nitrogen
Oxygen
Ammonia
For gas mixtures, in addition to the normal to parameters (T, p), we also need to know the mixture composition.
Number of moles
Mass fraction
where
Mole fraction
where
Avg. Molecular Weight

4. This refers to the overall mixture, on average.
Sec 12.2 : Relating P, V, and T for Ideal Gas Mixtures
Ideal gas
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
thus
Amagat Model: Assumes that each component behaves as an ideal gas at the specified T and p of the mixture.
Partial Volume
and

5. 40% propane (C3H8), 40% ethane (C2H6), and 20% methane (CH4).
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
The analysis in terms of mass fractions
The partial pressure of each component, in bar
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
0.40
0.20
1.0 Mi mi
mfi
mi (kg/s) i
C3H8
C2H6
CH4
Total
T (°C) 23
p (bar) 1 yi Mi mi
mfi
mi (kg/s)
yi is the mole fraction of each component

6. 40% propane (C3H8), 40% ethane (C2H6), and 20% methane (CH4).
Example (12.6):
40% propane (C3H8), 40% ethane (C2H6), and 20% methane (CH4).
The analysis in terms of mass fractions
The partial pressure of each component, in bar
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:
The mass of each component is found i
C3H8
C2H6
CH4
Total
T (°C) 23
p (bar) 1 yi
0.40
0.20
1.0 Mi
44.09
30.07
16.04 mi
17.636
12.028
3.208
32.872
mfi
mi(kg/s) i
C3H8
C2H6
CH4
Total
T (°C) 23
p (bar) 1 yi
0.40
0.20
1.0 Mi
44.09
30.07
16.04 mi
mfi
mi(kg/s) i
C3H8
C2H6
CH4
Total
T (°C) 23
p (bar) 1 yi
0.40
0.20
1.0 Mi
44.09
30.07
16.04 mi
17.636
12.028
3.208
mfi
mi(kg/s)
On a 1 kmol basis then
Units are effectively kg/kmol, thus mt = M

7. (b) The partial pressure of each component, in bar
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: i
C3H8
C2H6
CH4
Total
T (°C) 23
p (bar) 1 yi
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
mi(kg/s)

8. First convert volumetric flow rate to mass flow rate:
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: i
C3H8
C2H6
CH4
Total
T (°C) 23
p (bar)
0.40
0.20
1.0 yi Mi
44.09
30.07
16.04 mi
17.636
12.028
3.208
32.872
mfi
0.5365
0.3659
0.0976
mi(kg/s)

9. Can also find the mass flow rate of each component
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 i
C3H8
C2H6
CH4
Total
T (°C) 23
P (bar)
0.40
0.20
1.0 yi Mi
44.09
30.07
16.04 mi
17.636
12.028
3.208
32.872
mfi
0.5365
0.3659
0.0976
mi(kg/s)
26.71 i
C3H8
C2H6
CH4
Total
T (°C) 23
P (bar)
0.40
0.20
1.0 yi Mi
44.09
30.07
16.04 mi
17.636
12.028
3.208
32.872
mfi
0.5365
0.3659
0.0976
mi(kg/s)
14.33
9.77
2.61
26.71

10. The properties of U, H, S, and c of a mixture are additive.
Sec 12.3 : Evaluating U, H, S and specific heats
The properties of U, H, S, and c of a mixture are additive.
Mass of system:
Internal energy
Where a similar set of equations can be written for H and S
For specific heat

11. For mixtures with constant composition (no chemical reaction):
Sec 12.4 : Analyzing Systems Involving Mixtures
For mixtures with constant composition (no chemical reaction):
For constant specific heats,
with
Equations for enthalpy (H) are similar to those for internal energy (U), but uses cp.

12. For entropy, is also dependent upon pressure changes.
Sec 12.4 : Analyzing Systems Involving Mixtures
For entropy, is also dependent upon pressure changes.
For constant specific heats, or

13. The entropy change, in kJ/K.
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
The heat transfer, in kJ.
The entropy change, in kJ/K.
Principles to be applied:
Closed Ideal Gas system:
and
1st Law of Thermodynamics:
where for polytropic process
2nd Law of thermodynamics:

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
The heat transfer, in kJ.
The entropy change, in kJ/K.
Find the moles of each component and the Mixtures Molecular Weight:
Molecular
Weight:

15. The entropy change, in kJ/K.
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
The heat transfer, in kJ.
The entropy change, in kJ/K.
To Evaluate work for a polytropic process.

16. The entropy change, in kJ/K.
Example (12.17):
The heat transfer, in kJ.
The entropy change, in kJ/K.
Evaluate change in internal energy
Assuming constant heat capacity from Table A-20 at average temperature (359 K)
Then using the energy balance to evaluate Q

17. Example (12.17):
(b) The entropy change, in kJ/K.
Using M = 5.29 kg/kmol,
The mixture’s cv (avg. heat capacity) needs to be found. so

18. Example (12.17):
(b) The entropy change, in kJ/K.
Therefore:

19. -- Gases are initially at different temperature.
Sec : Mixing of Ideal gases
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

20. Sec 12.4.2 : Mixing of Ideal gases
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
where so or
Assuming constant specific heat,

21. Sec 12.4.2 : Mixing of Ideal gases
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
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.
and so

22. Sec 12.4.2 : Mixing of Ideal gases
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:

23. Sec 12.4.2 : Mixing of Ideal gases
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.
where
total volume is then

24. Sec 12.4.2 : Mixing of Ideal gases
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
using Ideal Gas equation:

25. Sec 12.4.2 : Mixing of Ideal gases
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:
where
therefore
then using the form based on ideal gas behavior with constant specific heat

26. Sec 12.4.2 : Mixing of Ideal gases
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:

27. end of slides Lecture 38