1. Ideal Gas Mixture and Psychrometric Applications
Chapter 12
Ideal Gas Mixture and Psychrometric Applications

2. Learning Outcomes
Describe ideal gas mixture composition in terms of mass fractions and mole fractions.
Use the Dalton model to relate pressure, volume, and temperature and to calculate changes in U, H, and S for ideal gas mixtures.
Apply mass, energy, and entropy balances to systems involving ideal gas mixtures, including mixing processes.

3. Learning Outcomes, cont.
Demonstrate understanding of psychrometric terminology, including humidity ratio, relative humidity, mixture enthalpy, and dew point temperature.
Use the psychrometric chart to represent common air-conditioning processes and to retrieve data.
Apply mass, energy, and entropy balances to analyze air-conditioning processes and cooling towers.

4. Describing Mixture Composition
Consider a system consisting of a number of gases within a container of volume V. The temperature and pressure of the gas mixture are T and p, respectively.
The composition of the mixture can be described by giving the mass mi or the number of moles ni for each component present.
Gas 1: n1, m1
Gas 2: n2, m2
Gas j: nj, mj
Sum: n m …
The mass mi, number of moles ni, and molecular weight Mi of component i are related by
ni is in kmol when mi is in kg and Mi is in kg/kmol.
ni is in lbmol when mi is in lb and Mi is in lb/lbmol.
(Eq. 12.1)

5. Describing Mixture Composition
The mass fraction is the relative amount of each component in the mixture. The mass fraction mfi of component i is
(Eq. 12.3)
where m is the total mass of mixture.
The sum of the mass fractions of all components in a mixture equals unity.

6. Describing Mixture Composition
Alternatively, the mole fraction can be used to describe the relative amount of each component in the mixture. The mole fraction yi of component i is
(Eq. 12.6)
where n is the total moles of mixture.
The sum of the mole fractions of all components in a mixture equals unity.
The apparent (or average) molecular weight M of a mixture is determined as a mole-fraction average of the component molecular weights:
(Eq. 12.9)

7. Describing Mixture Composition
Example: The molar analysis of a gas mixture is 50% N2, 35% CO2, and 15% O2. Determine
(a) the apparent molecular weight of the mixture and
(b) the analysis in terms of mass fractions.
Solution:
(a) The apparent molecular weight of the mixture is found using Eq and molecular weights (rounded) from Table A-1
= 34.2 kg/kmol

8. Describing Mixture Composition
(b) Although the actual amount of mixture is not known, the calculations can be based on any convenient amount. We use 1 kmol of mixture.
Then, the amount ni of each component, in kmol, is equal to its mole fraction, as shown in column (ii).
Column (iii) lists the respective molecular weights.
Column (iv) gives the mass mi of each component, in kg per kmole of mixture, obtained using mi = niMi (Eq. 12.1).
The mass fractions, listed as percentages in column (v), are obtained by dividing the values in column (iv) by the column total and multiplying by 100.
(i)
Component
(ii) ni ×
(iii) Mi =
(iv) mi
(v)
mfi % N2
0.50 28 14
40.94
CO2
0.35 44
15.4
45.03 O2
0.15 32
4.8
14.04
1.00
34.2
100

9. Relating p, V, and T for Ideal Gas Mixtures
Many systems of practical interest involve mixtures where the overall mixture and each of its components can be modeled as ideal gases. For such mixtures the Dalton mixture model is commonly used.
The overall mixture is considered an ideal gas
(Eq )
The Dalton model also assumes each component behaves as an ideal gas as if it were alone at temperature T and volume V.

10. Relating p, V, and T for Ideal Gas Mixtures
Accordingly, with the Dalton model the individual components do not exert the mixture pressure p but rather a partial pressure denoted by pi:
(Eq )
By combining Eqs and the partial pressure pi can be determined alternatively from
(Eq )
where the sum of the partial pressures equals the mixture pressure
(Eq )

11. Evaluating U, H, and S for Ideal Gas Mixtures
For an ideal gas mixture, the values of U, H, and S are evaluated by adding the contribution of each component at the condition at which the component exists in the mixture.
Evaluation of the specific internal energy or specific enthalpy of a mixture component i requires only a single intensive property: the mixture temperature, T.
Evaluation of the specific entropy of a mixture component i requires two intensive properties. We will use the mixture temperature, T, and the partial pressure, pi.

12. Evaluating U, H, and S for Ideal Gas Mixtures (Molar Basis)
Accordingly, when working on a molar basis expressions for U, H, and S of a mixture consisting of several components are:

13. Evaluating U, H, and S for Ideal Gas Mixtures (Molar Basis)
The mixture specific heats and are mole-fraction averages of the respective component specific heats.
(Eq )
(Eq )
See Sec for applications using these expressions for U, H, S, and the specific heats.

14. Evaluating U, H, and S for Ideal Gas Mixtures (Mass Basis)
When working on a mass basis the expressions for U, H, S, and specific heats of a mixture consisting of two components – a binary mixture – are:
Table 12.2

15. Engineering Applications of Ideal Gas Mixtures
We encounter ideal gas mixtures in many important areas of application. Two of these are:
Systems involving chemical reactions and, in particular, combustion. For these applications we typically work on a molar basis. Combustion systems are considered in Chapter 13.
Systems for air-conditioning and other applications requiring close control of water vapor in gas mixtures. For these applications we typically work on a mass basis. Systems of this type are considered in the second part of Chapter 12.

16. Psychrometric Applications
The remainder of this presentation centers on systems involving moist air. A condensed water phase may also be present in such systems.
The term moist air refers to a mixture of dry air and water vapor in which the dry air is treated as a pure component.
The Dalton model applies to moist air.
By identifying gas 1 with dry air and gas 2 with water vapor, Table 12.2 gives moist air property relations on a mass basis.
The study of systems involving moist air is known as psychrometrics.

17. Moist Air
Consider a closed system consisting of moist air occupying a volume V at mixture pressure p and mixture temperature T.
In moist air the amount of water vapor present is much less than the amount of dry air:
mv << ma nv << na.
The Dalton model applies to the mixture of dry air and water vapor:

18. Moist Air
1. The overall mixture and each component, dry air and water vapor, obey the ideal gas equation of state.
2. Dry air and water vapor within the mixture are considered as if they each exist alone in volume V at the mixture temperature T while each exerts part of the mixture pressure.
3. The partial pressures pa and pv of dry air and water vapor are, respectively
pa = ya p
pv = yv p
(Eq b)
where ya and yv are the mole fractions of the dry air and water vapor, respectively. These moist air expressions conform to Eqs. (c) of Table 12.2.

19. Moist Air
4. The mixture pressure is the sum of the partial pressures of the dry air and the water vapor:
p = pa + pv
Mixture pressure, p , T
5. A typical state of water vapor in moist air is fixed using partial pressure pv and the mixture temperature T.
The water vapor is superheated at this state.
Typical state of
the water vapor
in moist air

20. Moist Air
6. When pv corresponds to pg at temperature T, the mixture is said to be saturated.
7. The ratio of pv and pg is called the relative humidity, f:
(Eq )
Relative humidity is usually expressed as a percent and ranges as
dry air only
(pv = 0)
0 ≤ f ≤ 100%
saturated air
(pv = pg)

21. Humidity Ratio
The humidity ratio w of a moist air sample is the ratio of the mass of the water vapor to the mass of the dry air.
(Eq )
Since mv << ma, the value of w is typically << 1.
Using the ideal gas equation of state and the relationship pa = p – pv
18.02/28.97 = 0.622
(Eq )

22. Mixture Enthalpy (Eq. 12.45) (Eq. 12.46) (Eq. 12.47)
Values for U, H, and S for moist air can be found by adding contributions of each component.
For example, the enthalpy H is
which conforms to Eq. (d) in Table 12.2.
Dividing by ma and introducing w, the mixture enthalpy per unit mass of dry air is
For moist air, the enthalpy hv is very closely given by the saturated vapor value corresponding to the given temperature.
(Eq )
(Eq )
(Eq )

23. Heating Moist Air in a Duct
Example: Moist air enters a duct at 10oC, 80% relative humidity, is heated as it flows through the duct, and exits at 30oC. No moisture is added or removed and the mixture pressure remains constant at 1 bar. For steady-state operation and ignoring kinetic and potential energy changes, determine
(a) the humidity ratio, w2, and
(b) the rate of heat transfer, in kJ per kg of dry air.

24. Heating Moist Air in a Duct
Solution:
(a) At steady state, mass rate balances for the dry air and water vapor read:
Since the mass flow rates of the dry air and water vapor do not change from inlet to exit, they are denoted for simplicity as ma and mv. Moreover, since no moisture is added or removed, the humidity ratio does not change from inlet to exit: w1 = w2. The common humidity ratio is denoted by w. ∙

25. Heating Moist Air in a Duct
The humidity ratio is evaluated using data at the inlet:
The partial pressure of the water vapor at the inlet, pv1, can be evaluated from the given inlet relative humidity f1 and the saturated pressure pg1 at 10oC from Table A-2:
pv1 = f1pg1 = 0.8( bar) = bar
The humidity ratio can be found from:

26. Heating Moist Air in a Duct
(b) The steady-state form of the energy rate balance reduces to:
Solving for Qcv ∙ ∙ ∙
Noting that mv = wma, we get

27. Heating Moist Air in a Duct
For the dry air, ha1 and ha2 are obtained from ideal gas table Table A-22 at 10oC and 30oC, respectively.
For the water vapor, hv1 and hv2 are obtained from steam table Table A-2 at 10oC and 30oC, respectively, using hv ≈ hg
The contribution of the water vapor to the heat transfer magnitude is relatively minor.

28. Dew Point Temperature
When moist air is cooled, partial condensation of the water vapor initially present can occur. This is observed in condensation of vapor on window panes, pipes carrying cold water, and formation of dew on grass.
An important special case is cooling of moist air at constant mixture pressure, p.
The figure shows a sample of moist air, initially at State 1,
where the water vapor is superheated. The accompanying T-v
diagram locates
states of water.
Let’s study this system as it is cooled in stages from its initial temperature.

29. Dew Point Temperature
In the first part of the cooling process, the mixture pressure and water vapor mole fraction remain constant.
Since pv = yv p, the partial pressure of the water vapor remains constant.
Accordingly, the water vapor cools at constant pv from state 1 to state d, called the dew point.
The temperature at state d is called the dew point temperature.
As the system cools below the dew point temperature, some of the water vapor initially present condenses. The rest remains a vapor.

30. Dew Point Temperature mw = mv1 – mv2 At the final temperature,
the system consists of the dry air
initially present plus saturated water vapor
and saturated liquid.
Since some of the water vapor initially present has condensed, the partial pressure of the water vapor at the final state, pg2, is less than the partial pressure initially, pv1.
The amount of water that condenses, mw, equals the difference in the initial and final amounts of water vapor:
mw = mv1 – mv2

31. Dew Point Temperature
Using mv = wma and the fact that the amount of dry air remains constant, the amount of water condensed per unit mass of dry air is
where
and p denotes the mixture pressure, which remains constant while cooling occurs.

32. Dry-bulb Temperature and Wet-bulb Temperature
In engineering applications involving moist air, two readily-measured temperatures are commonly used: the dry-bulb and wet-bulb temperatures.
The dry-bulb temperature, Tdb, is simply the temperature measured by an ordinary thermometer placed in contact with the moist air.
The wet-bulb temperature, Twb, is the temperature measured by a thermometer whose bulb is enclosed by a wick moistened with water.

33. Dry-bulb Temperature and Wet-bulb Temperature
The figure shows wet-bulb and dry-bulb thermometers mounted on an instrument called a psychrometer. Flow of moist air over the two thermometers is induced by a battery-operated fan.
Owing to evaporation from the wet wick to the moist air, the wet-bulb temperature reading is less than the dry-bulb temperature: Twb < Tdb.
Each temperature is easily read from its respective thermometer.
Moist
Air in

34. Psychrometric Chart
Graphical representations of moist-air data are provided by psychrometric charts.
Psychrometric charts in SI and English units are given in Figs. A-9 and A-9E, respectively. These charts are constructed for a moist air mixture pressure of 1 atm.
Several important features of the psychrometric chart are discussed in Sec. 12.7, including

35. Psychrometric Chart
Dry-bulb temperature, Tdb.
Moist air
state
Tdb

36. Psychrometric Chart
Humidity ratio, w.
Moist air
state w

37. Psychrometric Chart Dew point temperature, Tdp.
Since the dew point is the state where moist air becomes saturated when cooled at constant pressure, the dew point for a given state is determined from the chart by following a line of constant w (constant pv) to the saturation line where f = 100%.
Moist air
state
Tdp

38. Psychrometric Chart
Relative humidity, f.
Moist air
state

39. The value of (ha + whv) is calculated using
Psychrometric Chart
Mixture enthalpy per unit mass of dry air, (ha + whv).
ha = cpaT
Fig. 12.9: T in oC, cpa = kJ/kg-K
Fig. 12.9E: T in oF, cpa = 0.24 Btu/lb-R
The value of (ha + whv) is calculated using
(ha + whv)
Moist air
state

40. Psychrometric Chart Wet-bulb temperature, Twb.
Lines of constant wet-bulb temperature are approximately lines of constant mixture enthalpy.
Twb
Moist air
state

41. Psychrometric Chart Volume per unit mass of dry air, V/ma.
Lines giving V/ma can be interpreted as the volume of dry air or of water vapor (each per unit mass of dry air) because in keeping with the Dalton model each component is considered to fill the entire volume.
Moist air
state
V/ma

42. Psychrometric Chart
Example: Using Fig. A-9, determine relative humidity, humidity ratio, and mixture enthalpy, in kJ/kg (dry air) corresponding to dry-bulb and wet-bulb temperatures of 30oC and 25oC, respectively.

43. Psychrometric Chart Solution: (ha + whv) = 76 kJ/kg dry air f = 67%
25oC
w = kg water/kg dry air

44. Analyzing Air-Conditioning Systems
The next series of slides demonstrates the application of mass and energy rate balances together with property data to typical air-conditioning systems using the psychrometric principles introduced thus far.
Featured applications include
Dehumidification
Humidification
Mixing of two moist air streams
An application of psychrometric principles to a cooling tower is also considered.

45. Dehumidification
The aim of a dehumidifier is to remove some of the water vapor in the moist air passing through the unit.
This is achieved by allowing the moist air to flow across a cooling coil carrying a refrigerant at a temperature low enough that some water vapor condenses.

46. Dehumidification
The figure shows a control volume enclosing a dehumidifier operating at steady state.
Moist air enters at state 1.
As the moist air flows over the cooling coil, some water vapor condenses.
Saturated moist air exits at state 2 (T2 < T1).
Condensate exits as saturated liquid at state 3. Here, we take T3 = T2. 1 2
f2 = 100%,
T2 < T1,
w2 < w1 ∙
ma, T1, w1 3 ∙ mw
T3 = T2

47. Dehumidification For the control volume, let us evaluate
The amount of condensate exiting per unit mass of dry air: mw/ma and
The rate of heat transfer between the moist air and cooling coil, per unit mass of dry air: Qcv/ma. ∙ ∙ 1 2
f2 = 100%,
T2 < T1,
w2 < w1 ∙
ma, T1, w1 ∙ ∙ 3 ∙ mw
T3 = T2

48. Dehumidification (dry air) (water)
Mass rate balances. At steady state, mass rate balances for the dry air and water are, respectively
(dry air)
(water) 1 2
Solving for the mass flow rate of the condensate
f2 = 100%,
T2 < T1,
w2 < w1 ∙
ma, T1, w1
Then, with mv1 = w1ma and mv2 = w2ma, where ma denotes the common mass flow rate of the dry air, we get the following expression for the amount of water condensed per unit mass of dry air
(1) ∙ 3 ∙ mw
T3 = T2

49. Dehumidification ∙
Energy rate balance. With Wcv = 0 and no significant kinetic and potential energy changes, the energy rate balance for the control volume reduces at steady state to
(2)
With mv1 = w1ma, mv2 = w2ma, and Eq. (1), Eq. (2) becomes ∙
(3)
Since heat transfer occurs from the moist air to the cooling coil, Qcv/ma will be negative in value. ∙

50. Dehumidification
(3)
For the condensate, hw = hf (T2), where hf is obtained from Table A-2.
Options for evaluating the underlined terms of Eq. (3) include
w1 and w2 are known. Since T1 and T2 are also known, ha1 and ha2 can be obtained from ideal gas table Table A-22, while hv1 and hv2 can be obtained from steam table Table A-2 using hv = hg.
(ha + whv)1
(ha + whv)2
Alternatively, using the respective temperature and humidity ratio values to fix the states, (ha + whv) at states 1 and 2 can be read from a psychrometric chart. T2 T1

51. Humidification
The aim of a humidifier is to increase the amount of water vapor in the moist air passing through the unit.
This is achieved by injecting steam or liquid water.

52. Humidification
The figure shows a control volume enclosing a humidifier operating at steady state.
Moist air enters at state 1.
Steam or liquid water is injected.
Moist air exits at state 2 with greater humidity ratio, w2 > w1.
ma1 ∙ 3 m3
h3,
Wcv = 0, Qcv = 0

53. Humidification
For adiabatic operation, the accompanying psychrometric charts show states 1 and 2 for each case.
With relatively high-temperature steam injection, the temperature of the moist air increases.
With liquid injection the temperature of the moist air may decrease because the liquid is vaporized by the moist air into which it is injected.
ma1 ∙ 3 m3
h3,
Wcv = 0, Qcv = 0

54. Humidification For the control volume, let us evaluate
The humidity ratio, w2, and
The temperature, T2.

55. Humidification (dry air) (water)
Mass rate balances. At steady state, mass rate balances for the dry air and water are, respectively
(dry air)
(water)
Then, since mv1 = w1ma and mv2 = w2ma, where ma denotes the common mass flow rate of the dry air, we get
(1) ∙
Since w1, ma, and m3 are specified, the humidity ratio w2 can be calculated from Eq. (1) ∙

56. Humidification
Energy rate balance. With no significant kinetic and potential energy changes, the energy rate balance for the control volume reduces to
Since Wcv and Qcv are each zero in this case ∙
(2)
With mv1 = w1ma and mv2 = w2ma, Eq. (2) becomes ∙
(3)
Solving Eq. (3)
(4)

57. Humidification (4) Options for determining T2 from Eq. (4) includew
Specific enthalpy of moist air,
in kJ/kg (dry air)
Options for determining T2 from Eq. (4) include
Use the psychrometric chart:
The first term on the right side of Eq. (4) can be read from the chart using T1 and w1 to fix the state.
Since the second term on the right is known, the value of (ha2 + w2hv2) can be calculated.
This value together with w2 fixes the exit state, which allows T2 to be determined by inspection.
(ha2 + w2hv2)
(ha1 + w1hv1) 2 w2 w1 1 T1 T2

58. Humidification (4) Options for determining T2 from Eq. (4) include
An iterative solution using data from Table A-22: ha(T) for the dry air and Table A-2: hv = hg(T) for the water vapor:
The value of the right side of Eq. (4) is known because the data are either known or can be obtained from the indicated tables using T1.
On the left side of Eq. (4), w2 is known from the mass rate balance.
Accordingly, the only unknown is T2, which can be found iteratively:
For each assumed value of T2, Table A-22 gives ha2 and Table A-2 gives hv2.
This allows the left side to be calculated.
Iteration with T2 continues until the calculated value on the left agrees with the known value on the right.

59. Adiabatic Mixing of Two Moist Air Streams
In air-conditioning systems, a frequent component is one that mixes moist air streams as shown in the figure:
For the case of adiabatic mixing, let us consider how the following quantities at the exit of the control volume, ma3, w3, and T3, can be evaluated knowing the respective quantities at the inlets. ∙

60. Adiabatic Mixing of Two Moist Air Streams
Mass rate balances. At steady state, mass rate balances for the dry air and water vapor are, respectively
(dry air)
(water vapor)
With mv = wma, these equations combine to give ∙
Alternatively
(1)
These equations can be solved for w3 using known values of w1, w2, ma1, and ma2. ∙

61. Adiabatic Mixing of Two Moist Air Streams
Energy rate balance. Ignoring the effects of kinetic and potential energy, the energy rate balance for the control volume reduces at steady state to
Since Wcv and Qcv are each zero in this case ∙
(Eq c)
The enthalpies of the water vapor are evaluated using hv = hg. With ma3 = ma1 + ma2, Eq c can be solved to give an expression with the same form as Eq. (1) ∙ ∙ ∙
(2)
Using known data, this equation can be solved for (ha + whg)3, from which T3 can be evaluated.

62. Adiabatic Mixing of Two Moist Air Streams
From study of Eqs. (1) and (2) we conclude that on a psychrometric chart state 3 lies on a straight line connecting states 1 and 2, as shown in the figure
(1)
(2)

63. Adiabatic Mixing of Two Moist Air Streams
Example: For adiabatic mixing of two moist air streams with the data provided in the table below, use the psychrometric chart to determine
(a) w3, in kg (vapor)/kg (dry air), and
(b) T3 in oC.
State 1 2 T
(oC) 24 5 w
(kg (dry air)/kg (vapor))
0.0094
0.002 ma
(kg (dry air)/min)
497
180
(ha + whg)*
(kJ/kg (dry air)) 48 10
*The values of (ha + whg) are read from Fig. A-9 using the respective temperature and humidity ratio values. ∙

64. Adiabatic Mixing of Two Moist Air Streams
Solution:
(a) Inserting known values in Eq. (1),
we get w3 = kg (vapor)/kg (dry air).
(b) Then from Fig. A-9
T3 = 19oC

65. Adiabatic Mixing of Two Moist Air Streams
Solution:
(a) Inserting known values in Eq. (1),
we get w3 = kg (vapor)/kg (dry air).
(b) Then from Fig. A-9
Alternatively, Eq. (2) can be used to determine
(ha + whg)3 = 38 kJ/kg (dry air).
Then, from Fig. A-9
T3 = 19oC

66. Cooling Towers
Moist air principles also play a role in the analysis of cooling towers such as shown in the figure.
Major events occurring within the control volume enclosing the tower include the following:
The warm water to be cooled enters at 1 and is sprayed from the top of the tower.
Atmospheric air enters at 3 and flows counter to the falling water.

67. Cooling Towers
As the liquid water and moist air interact within the tower, a fraction of the liquid evaporates, resulting in
Liquid water that exits the tower at 2 with a lower temperature than the water entering at 1, which is the objective.
Moist air that exits the tower at 4 with a greater humidity ratio than the air entering at 3.
Since some of the incoming water has evaporated, an equivalent amount of makeup water is added at 5 so that the return mass flow rate equals the mass flow rate entering at 1.

68. Cooling Towers (dry air) (water)
Mass rate balances. To evaluate the mass flow rate of the makeup water, apply mass rate balances to the control volume at steady state to get
(dry air)
(water)
With mv3 = w3ma and mv4 = w4ma, where ma is the common mass flow rate of the dry air, this becomes ∙
Energy rate balance. Application of the energy rate balance to such a cooling tower is demonstrated in Example