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

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CLOSED SYSTEMS. Next consider an entropy balance for a process of a closed

system during which a chemical reaction occurs

2 S

1 s

(13.25)

and S

denote, respectively, the entropy of the reactants and the entropy of the

products.

When the reactants and products form ideal gas mixtures, the entropy balance can

be expressed on a per mole of fuel basis as

ns 2

ns 5

(13.26)

where the coefficients n on the left are the coefficients of the reaction equation giv-

ing the moles of each reactant or product per mole of fuel. The entropy terms of Eq.

13.26 are evaluated from Eq. 13.23 using the temperature and partial pressures of the

reactants or products, as appropriate. In any such application, the fuel is mixed with

the oxidizer, so this must be taken into account when determining the partial pressures

of the reactants.

Example 13.10 provides an illustration of the evaluation of entropy change for

combustion at constant volume.

13.5 Absolute Entropy and the Third Law of Thermodynamics 813

5 267.12 2 8.314 ln 0.033 5 295.481 kJ

kmol ? K

5 231.01 2 8.314 ln 0.0371 5 258.397 kJ

kmol ? K

5 242.12 2 8.314 ln 0.1546 5 257.642 kJ

kmol ? K

5 226.795 2 8.314 ln 0.7753 5 228.911 kJ

kmol ?

Inserting values into the expression for the rate of entropy production

5 81295.48121 91258.39721 37.51257.64221 1881228.9112

2 360.79 2 50

218.01

2 188

193.46

5 9754 kJ

kmol

octane

? K

The use of IT to solve part (b) is left as an exercise.

➊ For several gases modeled as ideal gases, IT directly returns the absolute

entropies required by entropy balances for reacting systems. The entropy

data obtained from IT agree with values calculated from Eq. 13.23 using

table data.

➋ Although the rates of entropy production calculated in this example are

positive, as required by the second law, this does not mean that the proposed

reactions necessarily would occur, for the results are based on the assump-

tion of complete combustion. The possibility of achieving complete combus-

tion with specified reactants at a given temperature and pressure can be

investigated with the methods of Chap. 14, dealing with chemical equilib-

rium. For further discussion, see Sec. 14.4.1.

Ability to…

❑

apply the control volume

entropy balance to a

reacting system.

❑

evaluate entropy values

appropriately based on

absolute entropies.

✓

Skills Developed

How do combustion product temperature and rate of entropy

production vary, respectively, as percent excess air increases? Assume

complete combustion. Ans. Decrease, increase.

➋

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814 Chapter 13 Reacting Mixtures and Combustion

Determining Entropy Change for Combustion of Gaseous

Methane with Oxygen at Constant Volume

c c c c EXAMPLE 13.10 c

Determine the change in entropy of the system of Example 13.6 in kJ/K.

SOLUTION

Known:

A mixture of gaseous methane and oxygen, initially at 258C and 1 atm, burns completely within a closed

rigid container. The products are cooled to 900 K, 3.02 atm.

Find: Determine the change in entropy for the process in kJ/K.

Schematic and Given Data: See Fig. E13.6.

Engineering Model:

The contents of the container are taken as the system.

2. The initial mixture can be modeled as an ideal gas mixture, as can the products of combustion.

3. Combustion is complete.

Analysis: The chemical equation for the complete combustion of methane with oxygen is

1 2O

1 2H

The change in entropy for the process of the closed system is ¢S 5 S

2 S

, where S

and S

denote, respec-

tively, the initial and final entropies of the system. Since the initial mixture forms an ideal gas mixture (assump-

tion 2), the entropy of the reactants can be expressed as the sum of the contributions of the components, each

evaluated at the mixture temperature and the partial pressure of the component. That is

5 1s

, y

21 2s

, y

where y

5 1

3 and y

5 2

3 denote, respectively, the mole fractions of the methane and oxygen in the initial

mixture. Similarly, since the products of combustion form an ideal gas mixture (assumption 2)

5 1s

, y

21 2s

, y

where y

5 1

3 and y

5 2

3 denote, respectively, the mole fractions of the carbon dioxide and water vapor

in the products of combustion. In these equations, p

and p

denote the pressure at the initial and final states,

respectively.

The specific entropies required to determine S

can be calculated from Eq. 13.23. Since T

5 T

ref

and p

5 p

ref

absolute entropy data from Table A-25 can be used as follows

, y

25 s 8

ref

22 R ln

ref

5 186.16 2 8.314 ln

5 195.294 kJ

kmol ? K

Similarly

, y

25 s8

ref

22 R ln

ref

5 205.03 2 8.314 ln

5 208.401 kJ

kmol ? K

At the final state, the products are at T

5 900 K and p

5 3.02 atm. With Eq. 13.23 and absolute entropy

data from Tables A-23

, y

25 s8

22 R ln

ref

5 263.559 2 8.314 ln

3.02

5 263.504 kJ

kmol ? K

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13.5.3

Evaluating Gibbs Function for Reacting Systems

The thermodynamic property known as the Gibbs function plays a role in the second

part of this chapter dealing with exergy analysis. The specific Gibbs function

, intro-

duced in Sec. 11.3, is

5 h 2 T s (13.27)

The procedure followed in setting a datum for the Gibbs function closely parallels

that used in defining the enthalpy of formation: To each stable element at the stan-

dard state is assigned a zero value of the Gibbs function. The Gibbs function of formation

of a compound,

, equals the change in the Gibbs function for the reaction in which

the compound is formed from its elements, the compound and the elements all being

at T

ref

5 258C (778F) and p

ref

5 1 atm. Tables A-25 and A-25E give the Gibbs function

of formation,

, for selected substances.

The Gibbs function at a state other than the standard state is found by adding to

the Gibbs function of formation the change in the specific Gibbs function ¢g between

the standard state and the state of interest

1T, p25

1 3

1T, p22

ref

, p

ref

245

1 ¢

(13.28a)

With Eq. 13.27, ¢g can be written as

5 3h1T, p22 h1T

ref

, p

ref

242 3T s 1T, p22 T

ref

s 1T

ref

, p

ref

24 (13.28b)

The Gibbs function of component i in an ideal gas mixture is evaluated at the partial

pressure of component i and the mixture temperature.

The procedure for determining the Gibbs function of formation is illustrated in

the next example.

, y

25 s8

22 R ln

ref

5 228.321 2 8.314 ln

3.02

5 222.503 kJ

kmol ? K

Finally, the entropy change for the process is

¢S 5 S

2 S

5 3263.504 1 21222.503242 3195.294 1 21208.40124

5 96.414 kJ

Ability to…

❑

apply the closed system

entropy balance to a

reacting system.

❑

evaluate entropy values

appropriately based on

absolute entropies.

✓Skills Developed

Applying the entropy balance, Eq. 13.25, is s greater than,

less than, or equal to DS? Ans. Greater than.

Gibbs function of

formation

TAKE NOTE...

Gibbs function is introduced

here because it contributes

to subsequent developments

of this chapter.

Gibbs function is a prop-

erty because it is defined

in terms of properties.

Like enthalpy, introduced

as a combination of proper-

ties in Sec. 3.6.1, Gibbs

function has no physical

significance—in general.

Determining the Gibbs Function of Formation for Methane

c c c c EXAMPLE 13.11 c

Determine the Gibbs function of formation of methane at the standard state, 258C and 1 atm, in kJ/kmol, and

compare with the value given in Table A-25.

13.5 Absolute Entropy and the Third Law of Thermodynamics 815

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816 Chapter 13 Reacting Mixtures and Combustion

Chemical Exergy

The objective of this part of the chapter is to extend the exergy concept introduced in

Chap. 7 to include chemical exergy. Several important exergy aspects are listed in Sec.

7.3.1. We suggest you review this material before continuing the current discussion.

A key aspect carried from Chap. 7 is that exergy is a measure of the departure of

the state of a system from that of a thermodynamic model of the Earth and its atmo-

sphere called the exergy reference environment, or simply the environment. In the

current discussion, the departure of the system state from the environment centers

on the respective temperature, pressure, and composition, for composition now plays

a key role. If one or more of system temperature, pressure, and composition differs

from that of the environment, the system has exergy.

Exergy is the maximum theoretical work obtainable from an overall system of sys-

tem plus environment as the system passes from a specified state to equilibrium with

the environment. Alternatively, exergy is the minimum theoretical work input required

to form the system from the environment and bring it to the specified state.

For conceptual and computational ease, we think of the system passing to equilib-

rium with the environment in two steps. With this approach, exergy is the sum of two

contributions: thermomechanical, developed in Chap. 7, and chemical, developed in

this chapter.

SOLUTION

Known:

The compound is methane.

Find: Determine the Gibbs function of formation at the standard state, in kJ/kmol, and compare with the Table

A-25 value.

Assumptions: In the formation of methane from carbon and hydrogen (H

), the carbon and hydrogen are each

initially at 258C and 1 atm. The methane formed is also at 258C and 1 atm.

Analysis: Methane is formed from carbon and hydrogen according to C 1 2H

S CH

. The change in the Gibbs

function for this reaction is

5 1h 2 T s2

2 1h 2 T s2

2 21h 2 T s2

5 1h

2 h

2 2h

22 T1s

2 s

2 2s

2 (1)

where g

and g

denote, respectively, the Gibbs functions of the products and reactants, each per kmol of methane.

In the present case, all substances are at the same temperature and pressure, 258C and 1 atm, which correspond

to the standard reference state values. At the standard reference state, the enthalpies and Gibbs functions for

carbon and hydrogen are zero by definition. Thus, in Eq. (1),

5 h

5 0. Also,

5 1

. Eq. (1) then

reads

5 1h8

2 T

ref

1s8

2 s8

2 2

2 (2)

where all properties are at T

ref

, p

ref

. With enthalpy of formation and absolute entropy data from Table A-25,

Eq. (2) gives

1g 8

5274,850 2 298.153186.16 2 5.74 2 21130.57245250,783 kJ

kmol

The slight difference between the calculated value for the Gibbs function of

formation of methane and the value from Table A-25 can be attributed to

round-off.

Ability to…

❑

apply the definition of Gibbs

function of formation to

calculate

✓

Skills Developed

Using the method applied in the example, calculate g8

for

monatomic oxygen at the standard state, in kJ/kmol. Begin by writing

O. Ans. 231,750 kJ/kmol, which agrees with Table A-25.

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Set of Substances Represented by C

C H

CO CO

O(liq.)

a 1 0 a 1 1 0

b 0 2 b 0 0 2

c 0 0 0 1 2 1

TABLE 13.3

Exergy Reference Environment Used in Sec. 13.6

Gas phase at T

5 298.15 K (258C), p

5 1 atm

Component y

(%)

75.67

20.35

O(g) 3.12

0.03

Other 0.83

TABLE 13.4

13.6 Conceptualizing Chemical Exergy

In this section, we consider a thought experiment to bring out impor-

tant aspects of chemical exergy. This involves

c a set of substances represented by C

(see Table 13.3),

c an environment modeling Earth’s atmosphere (see Table 13.4),

and

c an overall system including a control volume (see Fig. 13.6).

Referring to Table 13.4, the exergy reference environment considered

in the present discussion is an ideal gas mixture modeling the Earth’s

atmosphere. T

and p

denote the temperature and pressure of the envi-

ronment, respectively. The composition of the environment is given in

terms of mole fractions denoted by y

, where superscript e is used to

signal the mole fraction of an environmental component. The values of

these mole fractions, and the values of T

and p

, are specified and

remain unchanged throughout the development to follow. The gas mix-

ture modeling the atmosphere adheres to the Dalton model (Sec. 12.2).

Considering Fig. 13.6, a substance represented by C

enters

the control volume at T

, p

. Depending on the particular substance,

compounds present in the environment enter (O

) and exit (CO

and

O(g)) at T

and their respective partial pressures in the environ-

ment. All substances enter and exit with negligible effects of motion

and gravity. Heat transfer between the control volume and environ-

ment occurs only at temperature T

. The control volume operates at

steady state, and the ideal gas model applies to all gases. Finally, for

the overall system whose boundary is denoted by the dotted line,

total volume is constant and there is no heat transfer across the

boundary.

Next, we apply conservation of mass, an energy balance, and an

entropy balance to the control volume of Fig. 13.6 with the objective of determin-

ing the maximum theoretical work per mole of substance C

entering—namely,

the maximum theoretical value of W

. This value is the molar chemical exergy of

the substance. The chemical exergy is given by

5 ch

1 aa 1

2 ah

2 T

1 aa 1

2 as

(13.29)

where the superscript ch is used to distinguish this contribution to the exergy mag-

nitude from the thermomechanical exergy introduced in Chap. 7. The subscript F

Fig. 13.6

Illustration used to conceptualize

chemical exergy.

Boundary of

overall syste

Boundary of the

control volume

Heat transfer

with environment

Environment at T

, p

at T

, y

+ [a + b/4 − c/2]O

→ aCO

+ b/2 H

O(g)

at T

, y

O at T

, y

13.6 Conceptualizing Chemical Exergy 817

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818 Chapter 13

Reacting Mixtures and Combustion

denotes the substance represented by C

. The other molar enthalpies and entro-

pies appearing in Eq. 13.29 refer to the substances entering and exiting the control

volume, each evaluated at the state at which it enters or exits. See the following box

for the derivation of Eq. 13.29.

TAKE NOTE...

Observe that the approach

used here to evaluate

chemical exergy parallels

those used in Secs. 7.3

and 7.5 to evaluate exergy

of a system and flow

exergy. In each case, energy

and entropy balances are

applied to evaluate maxi-

mum theoretical work in

the limit as entropy pro-

duction tends to zero.

Evaluating Chemical Exergy

Although chemical reaction does not occur in every case we will be considering, con-

servation of mass is accounted for generally by the following expression

1 3a 1 b

4 2 c

24O

S aCO

1 b

O1g2 (13.30)

which assumes that when reaction occurs, the reaction is complete.

For steady-state operation, the energy rate balance for the control volume of Fig. 13.6

reduces to give

1 h

1 aa 1

2 ah

(13.31)

where the subscript F denotes a substance represented by C

(Table 13.3). Since the

control volume is at steady state, its volume does not change with time, so no portion

of W

is required to displace the environment. Thus, in keeping with all specified

idealizations, Eq. 13.31 also gives the work developed by the overall system of control

volume plus environment whose boundary is denoted on Fig. 13.6 by a dotted line. The

potential for such work is the difference in composition between substance C

and

the environment.

Heat transfer is assumed to occur with environment only at temperature T

. An entropy

balance for the control volume takes the form

0 5

1 s

1 aa 1

2 as

(13.32)

Eliminating the heat transfer rate between Eqs. 13.31 and 13.32 gives

5 ch

1 aa 1

2 a

2 T

1 aa 1

2 as

d2 T

(13.33)

In Eq. 13.33, the specific enthalpy h

and specific entropy s

are evaluated at T

and

. Since the ideal gas model applies to the environment (Table 13.4), the specific enthal-

pies of the first underlined term of Eq. 13.33 are determined knowing only the temperature

. Further, the specific entropy of each substance of the second underlined term is deter-

mined by temperature T

and the partial pressure in the environment of that substance.

Accordingly, since the environment is specified, all enthalpy and entropy terms of Eq.

13.33 are known and independent of the nature of the processes occurring within the

control volume.

The term T

depends explicitly on the nature of such processes, however. In accor-

dance with the second law, T

is positive whenever irreversibilities are present, van-

ishes in the limiting case of no irreversibilities, and is never negative. The maximum

theoretical value for the work developed is obtained when no irreversibilities are pres-

ent. Setting T

to zero in Eq. 13.33, yields the expression for chemical exergy given

by Eq. 13.29.

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13.6.1

Working Equations for Chemical Exergy

For computational convenience, the chemical exergy given by Eq. 13.29 is written as

Eqs. 13.35 and 13.36. The first of these is obtained by recasting the specific entropies

of O

, CO

, and H

O using the following equation obtained by application of Eq. (a)

of Table 13.1:

, y

25 s

, p

22 R ln y

(13.34)

The first term on the right is the absolute entropy at T

and p

, and y

is the mole

fraction of component i in the environment.

Applying Eq. 13.34, Eq. 13.29 becomes

5 ch

1 aa 1

b h

2 ah

O1g2

d 1T

, p

2 T

1 aa 1

b s

2 as

O1g2

d 1T

, p

(13.35)

1 RT

ln c

a1b

42c

where the notation (T

, p

) signals that the specific enthalpy and entropy terms of

Eq. 13.35 are each evaluated at T

and p

, although T

suffices for the enthalpy of

substances modeled as ideal gases.

Recognizing the Gibbs function in Eq. 13.35—

5 h

2 T

, for instance—

Eq. 13.35 can be expressed alternatively in terms of the Gibbs functions of the several

substances as

5 c

1 aa 1

2 a

O1g2

d 1T

, p

1 RT

ln c

a1b

42c

(13.36)

The logarithmic term common to Eqs. 13.35 and 13.36 typically contributes only a

few percent to the chemical exergy magnitude. Other observations follow:

c The specific Gibbs functions of Eq. 13.36 are evaluated at the temperature T

and

pressure p

of the environment. These terms can be determined with Eq. 13.28a as

, p

1 3

, p

ref

, p

ref

24 (13.37)

where g8

is the Gibbs function of formation and T

ref

5 258C (778F), p

ref

5 1 atm.

c For the special case where T

and p

are the same as T

ref

and p

ref

, respectively, the

second term on the right of Eq. 13.37 vanishes and the specific Gibbs function is

just the Gibbs function of formation. That is, the Gibbs function values of Eq. 13.36

can be simply read from Tables A-25 or similar compilations.

c Finally, note that the underlined term of Eq. 13.36 can be written more compactly

as 2¢G: the negative of the change in Gibbs function for the reaction, Eq. 13.30,

regarding each substance as separate at temperature T

and pressure p

13.6.2

Evaluating Chemical Exergy for Several Cases

Cases of practical interest corresponding to selected values of a, b, and c in the repre-

sentation C

can be obtained from Eq. 13.36. For example, a 5 8, b 5 18, c 5 0

13.6 Conceptualizing Chemical Exergy 819

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820 Chapter 13

Reacting Mixtures and Combustion

corresponds to octane, C

. An application of Eq. 13.36 to evaluate the chemical

exergy of octane is provided in Example 13.12. Further special cases follow:

c Consider the case of pure carbon monoxide at T

, p

. For CO we have a 5 1, b 5 0,

c 5 1. Accordingly, Eq. 13.30 reads CO 1

S CO

, and the chemical exergy

obtained from Eq. 13.36 is

5 3

41T

, p

21 RT

ln c

(13.38)

If carbon monoxide is not pure but a component of an ideal gas mixture at T

, p

each component i of the mixture enters the control volume of Fig. 13.6 at tem-

perature T

and the respective partial pressure y

. The contribution of carbon

monoxide to the chemical exergy of the mixture, per mole of CO, is then given by

Eq. 13.38, but with the mole fraction of carbon monoxide in the mixture,

appearing in the numerator of the logarithmic term that then reads ln3y

This becomes important when evaluating the exergy of combustion products

involving carbon monoxide.

c Consider the case of pure water at T

and p

. Water is a liquid when at T

, p

, but

is a vapor within the environment of Table 13.4. Thus water enters the control

volume of Fig. 13.6 as a liquid and exits as a vapor at T

, y

, with no chemical

reaction required. In this case, a 5 0, b 5 2, and c 5 1. Equation 13.36 gives the

chemical exergy as

5 3

O1l2

O1g2

41T

, p

21 RT

ln a

(13.39)

c Consider the case of pure carbon dioxide at T

, p

. Like water, carbon dioxide is

present within the environment and thus requires no chemical reaction to evaluate

its chemical exergy. With a 5 1, b 5 0, c 5 2, Eq. 13.36 gives the chemical exergy

simply in terms of a logarithmic expression of the form

5 RT

ln a

(13.40)

Provided the appropriate mole fraction y

is used, Eq. 13.40 also applies to

other substances that are gases in the environment, in particular to O

and N

Moreover, Eqs. 13.39 and 13.40 reveal that a chemical reaction does not always

play a part when conceptualizing chemical exergy. In the cases of liquid water,

, O

, N

, and other gases present in the environment, we think of the work

that could be done as the particular substance passes by diffusion from the dead

state, where its pressure is p

, to the environment, where its pressure is the partial

pressure, y

c Finally, for an ideal gas mixture at T

, p

consisting only of substances present as

gases in the environment, the chemical exergy is obtained by summing the contri-

butions of each of the components. The result, per mole of mixture, is

5 RT

i51

ln a

(13.41a)

where y

and y

denote, respectively, the mole fraction of component i in the

mixture at T

, p

and in the environment.

TAKE NOTE...

For liquid water, we think

only of the work that could

be developed as water

expands through a turbine,

or comparable device, from

pressure p

to the partial

pressure of the water

vapor in the environment:

O (l)

, p

O(g)

, y

, p

, y

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Expressing the logarithmic term as 1ln11

21 ln y

2 and introducing a relation

like Eq. 13.40 for each gas i, Eq. 13.41a can be written alternatively as

i51

1 RT

i51

ln y

(13.41b)

The development of Eqs. 13.41a and 13.41b is left as an exercise.

13.6.3

Closing Comments

The approach introduced in this section for conceptualizing the chemical exergy of

the set of substances represented by C

can also be applied, in principle, for

other substances. In any such application, the chemical exergy is the maximum theo-

retical work that could be developed by a control volume like that considered in Fig. 13.6

where the substance of interest enters the control volume at T

, p

and reacts com-

pletely with environmental components to produce environmental components. All

participating environmental components enter and exit the control volume at their

conditions within the environment. By describing the environment appropriately, this

approach can be applied to many substances of practical interest.

13.7 Standard Chemical Exergy

While the approach used in Sec. 13.6 for conceptualizing chemical exergy can be

applied to many substances of practical interest, complications are soon encountered.

For one thing, the environment generally must be extended; the simple environment

of Table 13.4 no longer suffices. In applications involving coal, for example, sulfur

dioxide or some other sulfur-bearing compound must appear among the environmen-

tal components. Furthermore, once the environment is determined, a series of calcu-

lations are required to obtain exergy values for the substances of interest. These

complexities can be sidestepped by using a table of standard chemical exergies.

Standard chemical exergy values are based on a standard exergy reference environ-

ment exhibiting standard values of the environmental temperature T

and pressure

such as 298.15 K (536.678R) and 1 atm, respectively. The exergy reference environ-

ment also consists of a set of reference substances with standard concentrations

reflecting as closely as possible the chemical makeup of the natural environment. To

exclude the possibility of developing work from interactions among parts of the envi-

ronment, these reference substances must be in equilibrium mutually.

The reference substances generally fall into three groups: gaseous components of

the atmosphere, solid substances from the Earth’s crust, and ionic and nonionic sub-

stances from the oceans. A common feature of standard exergy reference environ-

ments is a gas phase, intended to represent air, that includes N

, O

, CO

, H

O(g),

and other gases. The ith gas present in this gas phase is assumed to be at temperature

and the partial pressure p

5 y

Two standard exergy reference environments are considered in this book, called

Model I and Model II. For each of these models, Table A-26 gives values of the standard

chemical exergy for several substances, in units of kJ/kmol, together with a brief descrip-

tion of the underlying rationale. The methods employed to determine the tabulated

standard chemical exergy values are detailed in the references accompanying the tables.

Only one of the two models should be used in a particular analysis.

The use of a table of standard chemical exergies often simplifies the application

of exergy principles. However, the term “standard” is somewhat misleading, for there

is no one specification of the environment that suffices for all applications. Still,

chemical exergies calculated relative to alternative specifications of the environment

are generally in good agreement. For a broad range of engineering applications, the

For further discussion see M. J. Moran, Availability Analysis: A Guide to Efficient Energy Use, ASME Press, New

York, 1989, pp. 169–170.

standard chemical

exergy

TAKE NOTE...

Standard exergy Model II is

commonly used in practice.

Model I is provided to show

that other standard refer-

ence environments can at

least be imagined.

13.7 Standard Chemical Exergy 821

TAKE NOTE...

Equation 13.41b is also

applicable for mixtures

containing gases other

than those present in the

reference environment, for

example gaseous fuels.

Moreover, this equation can

be applied to mixtures that

do not adhere to the ideal

gas model. In all such appli-

cations, the terms

may

be selected from a table of

standard chemical exergies,

introduced in Sec. 13.7 to

follow.

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822 Chapter 13 Reacting Mixtures and Combustion

convenience of using standard values generally outweighs the slight lack of accuracy

that might result. In particular, the effect of slight variations in the values of T

and

about their standard values normally can be neglected.

13.7.1

Standard Chemical Exergy of a Hydrocarbon: C

In principle, the standard chemical exergy of a substance not present in the environment

can be evaluated by considering a reaction of the substance with other substances for

which the chemical exergies are known.

To illustrate this for the case of a pure hydrocarbon fuel

at T

, p

, refer to

the control volume at steady state shown in Fig. 13.7 where the fuel reacts completely

with oxygen to form carbon dioxide and liquid water. All substances are assumed to

enter and exit at T

, p

and heat transfer occurs only at temperature T

Assuming no irreversibilities, an exergy rate balance for the control volume reads

0 5

c1 2

b2 a

int

rev

1 e

1 aa 1

2 ae

2 a

O1l2

2 E

where the subscript F denotes C

. Solving for the chemical exergy e

, we get

5 a

int

rev

1 ae

1 a

b e

O1l2

2 aa 1

b e

(13.42)

Applying energy and entropy balances to the control volume, as in the development

in the box on p. 818, we get

int

rev

5 ch

1 aa 1

b h

2 ah

O1l2

d1T

, p

2 T

1 aa 1

b s

2 as

O1l2

d1T

, p

(13.43)

The underlined term in Eq. 13.43 is recognized from Sec. 13.2.3 as the molar higher

heating value HH

, p

). Substituting Eq. 13.43 into Eq. 13.42, we obtain

5 HHV1T

, p

22 T

1 aa 1

2 as

O1l2

d1T

, p

1 ae

1 a

b e

O1l2

2 aa 1

b e

(13.44a)

at T

, p

at T

, p

at T

, p

O (l)

at T

, p

(

a +

)

→ aCO

+ H

O(l)

Fig. 13.7 Reactor used to introduce the standard chemical exergy of C

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