Moran M.J., Shapiro H.N. Fundamentals of Engineering Thermodynamics

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704

Library stacks require careful temperature and humidity control using air-conditioning

processes considered in Sec. 12.8.

ENGINEERING CONTEXT Many systems of interest involve gas mixtures of two or more compo-

nents. To apply the principles of thermodynamics introduced thus far to these systems requires that we eval-

uate properties of the mixtures. Means are available for determining the properties of mixtures from the

mixture composition and the properties of the individual pure components from which the mixtures are

formed. Methods for this purpose are discussed both in Chap. 11 and in the present chapter.

The objective of the present chapter is to study mixtures where the overall mixture and each of its compo-

nents can be modeled as ideal gases. General ideal gas mixture considerations are provided in the first part

of the chapter. Understanding the behavior of ideal gas mixtures of dry air and water vapor is prerequisite

to considering air-conditioning processes in the second part of the chapter, which is identified by the head-

ing, Psychrometric Applications. In those processes, we sometimes must consider the presence of liquid

water as well. We will also need to know how to handle ideal gas mixtures when we study the subjects of

combustion and chemical equilibrium in Chaps. 13 and 14, respectively.

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Ideal Gas Mixture

and Psychrometric

Applications

When you complete your study of this chapter, you will be able to...

describe ideal gas mixture composition in terms of mass fractions or 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.

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.

LEARNING OUTCOMES

705

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706 Chapter 12

Ideal Gas Mixture and Psychrometric Applications

Ideal Gas Mixtures: General

Considerations

12.1 Describing Mixture Composition

To specify the state of a mixture requires the composition and the values of two

independent intensive properties such as temperature and pressure. The object of the

present section is to consider ways for describing mixture composition. In subsequent

sections, we show how mixture properties other than composition can be evaluated.

Consider a closed system consisting of a gaseous mixture of two or more compo-

nents. The composition of the mixture can be described by giving the mass or the

number of moles of each component present. With Eq. 1.8, the mass, the number of

moles, and the molecular weight of a component i are related by

(12.1)

where m

is the mass, n

is the number of moles, and M

is the molecular weight of

component i, respectively. When m

is expressed in terms of the kilogram, n

is in

kmol. When m

is in terms of the pound mass, n

is in lbmol. However, any unit of

mass can be used in this relationship.

The total mass of the mixture, m, is the sum of the masses of its components

m 5 m

1 m

. . .

1 m

i51

(12.2)

The relative amounts of the components present in the mixture can be specified

in terms of mass fractions. The mass fraction mf

of component i is defined as

(12.3)

A listing of the mass fractions of the components of a mixture is sometimes referred

to as a

gravimetric analysis.

Dividing each term of Eq. 12.2 by the total mass of mixture m and using Eq. 12.3

1 5

i51

(12.4)

That is, the sum of the mass fractions of all the components in a mixture is equal to

unity.

The total number of moles in a mixture, n, is the sum of the number of moles of

each of its components

n 5 n

1 n

. . .

1 n

i51

(12.5)

The relative amounts of the components present in the mixture also can be described

in terms of mole fractions. The mole fraction y

of component i is de fined as

(12.6)

mass fractions

gravimetric analysis

mole fractions

TAKE NOTE...

In Secs. 12.1–12.3, we

introduce mixture concepts

required for study of psy-

chrometrics in the second

part of this chapter and

combustion in Chap. 13.

In Sec. 12.4, we extend

the discussion of mixtures

and provide several solved

examples illustrating

important types of mixture

applications. For economy

of effort, some readers may

elect to defer Sec. 12.4

and proceed directly to

content having more

immediate interest for

them: psychrometrics

beginning in Sec. 12.5 or

combustion beginning in

Sec. 13.1.

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A listing of the mole fractions of the components of a mixture may be called a molar

analysis

. An analysis of a mixture in terms of mole fractions is also called a

volumetric analysis.

Dividing each term of Eq. 12.5 by the total number of moles of mixture n and

using Eq. 12.6

1 5

i51

(12.7)

That is, the sum of the mole fractions of all the components in a mixture is equal to unity.

The apparent (or average) molecular weight of the mixture, M, is defined as the ratio

of the total mass of the mixture, m, to the total number of moles of mixture, n

M 5

(12.8)

Equation 12.8 can be expressed in a convenient alternative form. With Eq. 12.2, it

becomes

M 5

1 m

. . .

1 m

Introducing m

5 n

from Eq. 12.1

M 5

1 n

. . .

1 n

Finally, with Eq. 12.6, the apparent molecular weight of the mixture can be calculated

as a mole-fraction average of the component molecular weights

M 5

i51

(12.9)

consider the case of air. A sample of atmospheric air contains sev-

eral gaseous components including water vapor and contaminants such as dust, pollen,

and pollutants. The term dry air refers only to the gaseous components when all water

vapor and contaminants have been removed. The molar analysis of a typical sample of

dry air is given in Table 12.1. Selecting molecular weights for nitrogen, oxygen, argon,

and carbon dioxide from Table A-1, and neglecting the trace substances neon, helium,

etc., the apparent molecular weight of dry air obtained from Eq. 12.9 is

M < 0.7808128.0221 0.2095132.0021 0.0093139.9421 0.0003144.012

5 28.97 kg

kmol 5 28.97 lb

lbmol

This value, which is the entry for air in Tables A-1, would not be altered significantly

if the trace substances were also included in the calculation. b b b b b

molar analysis

volumetric analysis

apparent molecular weight

dry air

Approximate Composition of Dry Air

Mole Fraction

Component (%)

Nitrogen 78.08

Oxygen 20.95

Argon 0.93

Carbon dioxide 0.03

Neon, helium, methane, and others 0.01

TABLE 12.1

12.1 Describing Mixture Composition 707

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708 Chapter 12 Ideal Gas Mixture and Psychrometric Applications

Next, we consider two examples illustrating, respectively, the conversion from an analy-

sis in terms of mole fractions to an analysis in terms of mass fractions, and conversely.

Converting Mole Fractions to Mass Fractions

c c c c EXAMPLE 12.1 c

Determine the mass, in kg, of CO

in 0.5 kmol of mixture.

Ans. 1.76 kg.

The molar analysis of the gaseous products of combustion of a certain hydrocarbon fuel is CO

, 0.08; H

O, 0.11;

, 0.07; N

, 0.74. (a) Determine the apparent molecular weight of the mixture. (b) Determine the composition

in terms of mass fractions (gravimetric analysis).

SOLUTION

Known:

The molar analysis of the gaseous products of combustion of a hydrocarbon fuel is given.

Find: Determine (a) the apparent molecular weight of the mixture, (b) the composition in terms of mass fractions.

Analysis:

(a) Using Eq. 12.9 and molecular weights (rounded) from Table A-1

M 5 0.0814421 0.1111821 0.0713221 0.741282

5 28.46 kg

kmol 5 28.46 lb

lbmol

(b) Equations 12.1, 12.3, and 12.6 are the key relations required to determine the composition in terms of mass

fractions.

➊ Although the actual amount of mixture is not known, the calculations can be based on any convenient

amount. Let us base the solution on 1 kmol of mixture. Then, with Eq. 12.6 the amount n

of each component

present in kmol is numerically equal to the mole fraction, as listed in column (ii) of the accompanying table.

Column (iii) of the table gives the respective molecular weights of the components.

Column (iv) of the table gives the mass m

of each component, in kg per kmol of mixture, obtained with

Eq. 12.1 in the form m

5 M

. The values of column (iv) are obtained by multiplying each value of column

(ii) by the corresponding value of column (iii). The sum of the values in column (iv) is the mass of the mixture:

kg of mixture per kmol of mixture. Note that this sum is just the apparent mixture molecular weight determined

in part (a). Finally, using Eq. 12.3, column (v) gives the mass fractions as a percentage. The values of column

(v) are obtained by dividing the values of column (iv) by the column (iv) total and multiplying by 100.

➊ If the solution to part (b) were conducted on the basis of some other assumed amount of mixture—for exam-

ple, 100 kmol or 100 lbmol—the same result for the mass fractions would be obtained, as can be verified.

Ability to…

❑

calculate the apparent

molecular weight with known

mole fractions.

❑

determine the gravimetric

analysis given the molar analysis.

✓

Skills Developed

(i) (ii)* (iii) (iv)** (v)

Component n

3 M

5 m

0.08 3 44 5 3.52 12.37

O 0.11 3 18 5 1.98 6.96

0.07 3 32 5 2.24 7.87

0.74 3 28 5 20.72 72.80

1.00 28.46 100.00

*Entries in this column have units of kmol per kmol of mixture. For example, the first entry

is 0.08 kmol of CO

per kmol of mixture.

**Entries in this column have units of kg per kmol of mixture. For example, the first entry

is 3.52 kg of CO

per kmol of mixture. The column sum, 28.46, has units of kg of mixture

per kmol of mixture.

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Converting Mass Fractions to Mole Fractions

c c c c EXAMPLE 12.2 c

A gas mixture has the following composition in terms of mass fractions: H

, 0.10; N

, 0.60; CO

, 0.30. Determine

(a) the composition in terms of mole fractions and (b) the apparent molecular weight of the mixture.

SOLUTION

Known:

The gravimetric analysis of a gas mixture is known.

Find: Determine the analysis of the mixture in terms of mole fractions (molar analysis) and the apparent molec-

ular weight of the mixture.

Analysis:

(a) Equations 12.1, 12.3, and 12.6 are the key relations required to determine the composition in terms of mole

fractions.

➊

Although the actual amount of mixture is not known, the calculation can be based on any convenient

amount. Let us base the solution on 100 kg. Then, with Eq. 12.3, the amount m

of each component present,

in kg, is equal to the mass fraction multiplied by 100 kg. The values are listed in column (ii) of the accom-

panying table. Column (iii) of the table gives the respective molecular weights of the components.

Column (iv) of the table gives the amount n

of each component, in kmol per 100 kg of mixture, obtained

using Eq. 12.1. The values of column (iv) are obtained by dividing each value of column (ii) by the corre-

sponding value of column (iii). The sum of the values of column (iv) is the total amount of mixture, in kmol

per 100 kg of mixture. Finally, using Eq. 12.6, column (v) gives the mole fractions as a percentage. The values

of column (v) are obtained by dividing the values of column (iv) by the column (iv) total and multiplying

by 100.

(b) The apparent molecular weight of the mixture can be found by using Eq. 12.9 and the calculated mole frac-

tions. The value can be determined alternatively by using the column (iv) total giving the total amount of mixture

in kmol per 100 kg of mixture. Thus, with Eq. 12.8

M 5

100 k

7.82 kmol

5 12.79

kmol

5 12.79

lbmol

➊ If the solution to part (a) were conducted on the basis of some other assumed

amount of mixture, the same result for the mass fractions would be obtained,

as can be verified.

➋ Although H

has the smallest mass fraction, its mole fraction is the largest.

Ability to…

❑

determine the molar analysis

given the gravimetric analysis.

✓

Skills Developed

How many kmol of H

would be present in 200 kg of mixture?

Ans. 10 kmol.

(i) (ii)* (iii) (iv)** (v)

Component m

4 M

5 n

10 4 2 5 5.00 63.9

60 4 28 5 2.14 27.4

30 4 44 5 0.68 8.7

100 7.82 100.0

*Entries in this column have units of kg per 100 kg of mixture. For example, the first entry

is 10 kg of H

per 100 kg of mixture.

**Entries in this column have units of kmol per 100 kg of mixture. For example, the first

entry is 5.00 kmol of H

per 100 kg of mixture. The column sum, 7.82, has units of kmol

of mixture per 100 kg of mixture.

➋

12.1 Describing Mixture Composition 709

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710 Chapter 12

Ideal Gas Mixture and Psychrometric Applications

12.2 Relating p, V, and T for Ideal

Gas Mixtures

The definitions given in Sec. 12.1 apply generally to mixtures. In the present sec-

tion we are concerned only with ideal gas mixtures and introduce a model com-

monly used in conjunction with this idealization: the Dalton model.

Consider a system consisting of a number of gases contained within a closed ves-

sel of volume V as shown in Fig. 12.1. The temperature of the gas mixture is T and

the pressure is p. The overall mixture is considered an ideal gas, so p, V, T, and the

total number of moles of mixture n are related by the ideal gas equation of state

p 5 n

(12.10)

With reference to this system let us consider the Dalton model.

The Dalton model is consistent with the concept of an ideal gas as being made

up of molecules that exert negligible forces on one another and whose volume is

negligible relative to the volume occupied by the gas (Sec. 3.12.3). In the absence

of significant intermolecular forces, the behavior of each component is unaffected

by the presence of the other components. Moreover, if the volume occupied by

the molecules is a very small fraction of the total volume, the molecules of each gas

present may be regarded as free to roam throughout the full volume. In keeping with

this simple picture, the Dalton model assumes that each mixture component behaves

as an ideal gas as if it were alone at the temperature T and volume V of the mixture.

It follows from the Dalton model that the individual components would not exert

the mixture pressure p but rather a partial pressure. As shown below, the sum of the

partial pressures equals the mixture pressure. The partial pressure of component i, p

is the pressure that n

moles of component i would exert if the component were alone

in the volume V at the mixture temperature T. The partial pressure can be evaluated

using the ideal gas equation of state

(12.11)

Dividing Eq. 12.11 by Eq. 12.10

5 y

Thus, the partial pressure of component i can be evaluated in terms of its mole frac-

tion y

and the mixture pressure p

5 y

p (12.12)

To show that the sum of partial pressures equals the mixture pressure, sum both

sides of Eq. 12.12 to obtain

i51

p 5 p

i51

Since the sum of the mole fractions is unity (Eq. 12.7), this becomes

p 5

i51

(12.13)

The Dalton model is a special case of the additive pressure rule for relating the

pressure, specific volume, and temperature of gas mixtures introduced in Sec. 11.8.

The concept of an ideal gas mixture is a special case of the ideal solution concept introduced in Sec. 11.9.5.

Dalton model

Gas 1 : n

Gas 2 : n

n moles

mixture

Pressure = p

Temperature = T

Gas j : n

Volume = VBoundary

Fig. 12.1 Mixture of several

gases.

partial pressure

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Among numerous other mixture rules found in the engineering literature is the

Amagat model considered in the box that follows.

Introducing the Amagat Model

The underlying assumption of the Amagat model is that each mixture component

behaves as an ideal gas as if it existed separately at the pressure p and temperature T

of the mixture.

The volume that n

moles of component i would occupy if the component existed at p

and T is called the partial volume, V

, of component i. As shown below, the sum of the

partial volumes equals the total volume. The partial volume can be evaluated using the

ideal gas equation of state

(12.14)

Dividing Eq. 12.14 by the total volume V

5 y

Thus, the partial volume of component i also can be evaluated in terms of its mole

fraction y

and the total volume

5 y

V (12.15)

This relationship between volume fraction and mole fraction underlies the use of the term

volumetric analysis as signifying an analysis of a mixture in terms of mole fractions.

To show that the sum of partial volumes equals the total volume, sum both sides of

Eq. 12.15 to obtain

i51

V 5 V

i51

Since the sum of the mole fractions equals unity, this becomes

V 5

i51

(12.16)

Finally, note that the Amagat model is a special case of the additive volume model

introduced in Sec. 11.8.

12.3 Evaluating U, H, S, and Specific Heats

To apply the conservation of energy principle to a system involving an ideal gas

mixture requires evaluation of the internal energy, enthalpy, or specific heats of the

mixture at various states. Similarly, to conduct an analysis using the second law normally

requires the entropy of the mixture. The objective of the present section is to develop

means to evaluate these properties for ideal gas mixtures.

12.3.1 Evaluating U and H

Consider a closed system consisting of an ideal gas mixture. Extensive properties of

the mixture, such as U, H, or S, can be found by adding the contribution of each

component at the condition at which the component exists in the mixture. Let us apply

this model to internal energy and enthalpy.

12.3 Evaluating U, H, S, and Specific Heats 711

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712 Chapter 12

Ideal Gas Mixture and Psychrometric Applications

Since the internal energy and enthalpy of ideal gases are functions of temperature

only, the values of these properties for each component present in the mixture are

determined by the mixture temperature alone. Accordingly

U 5 U

1 U

. . .

1 U

i51

(12.17)

H 5 H

1 H

. . .

1 H

i51

(12.18)

where U

and H

are the internal energy and enthalpy, respectively, of component i

evaluated at the mixture temperature.

Equations 12.17 and 12.18 can be rewritten on a molar basis as

nu 5 n

1 n

. . .

1 n

i51

(12.19)

and

nh 5 n

1 n

. . .

1 n

i51

(12.20)

where u and h are the specific internal energy and enthalpy of the mixture per mole

of mixture, and u

and h

are the specific internal energy and enthalpy of component

i per mole of i. Dividing by the total number of moles of mixture, n, gives expressions

for the specific internal energy and enthalpy of the mixture per mole of mixture,

respectively

u 5

i51

(12.21)

h 5

i51

(12.22)

Each of the molar internal energy and enthalpy terms appearing in Eqs. 12.19 through

12.22 is evaluated at the mixture temperature only.

12.3.2 Evaluating c

and c

Differentiation of Eqs. 12.21 and 12.22 with respect to temperature results, respec-

tively, in the following expressions for the specific heats c

and c

of the mixture on

a molar basis

i51

y,i

(12.23)

i51

p,i

(12.24)

That is, the mixture specific heats c

and c

are mole-fraction averages of the respec-

tive component specific heats. The specific heat ratio for the mixture is k 5 c

12.3.3 Evaluating S

The entropy of a mixture can be found, as for U and H, by adding the contribution

of each component at the condition at which the component exists in the mixture.

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