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

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13.3 Determining the Adiabatic Flame Temperature 803

which corresponds to Eq. 13.21b. Since the reactants enter at 25°C, the

¢h

terms on the right side vanish, and

the energy rate equation becomes

1¢h2

(2)

(a) For combustion of liquid octane with the theoretical amount of air, the chemical equation is

1l21 12.5O

1 47N

8CO

1 9H

O1g21 47N

Introducing the coefficients of this equation, Eq. (2) takes the form

81¢h2

1 91¢h2

O1g2

1 471¢h2

5 31h8

1l2

1 12.51h8

1 471h8

2 381h8

1 91h8

O1g2

1 471h8

The right side of the above equation can be evaluated with enthalpy of formation data from Table A-25,

giving

81¢h2

1 91¢h2

O1g2

1 471¢h2

5 5,074,630 kJ

kmol 1fuel2

Each ¢

term on the left side of this equation depends on the temperature of the products,

. This temperature

can be determined by an iterative procedure.

The following table gives a summary of the iterative procedure for three trial values of T

. Since the summa-

tion of the enthalpies of the products equals 5,074,630 kJ/kmol, the actual value of T

is in the interval from

2350 to 2400 K. Interpolation between these temperatures gives T

5 2395 K.

2500 K 2400 K 2350 K

81¢h2

975,408 926,304 901,816

91¢h2

O1g2

890,676 842,436 818,478

471¢h2

3,492,664 3,320,597 3,234,869

1¢h2

5,358,748 5,089,337 4,955,163

Alternative Solution:

The following IT code can be used as an alternative to iteration with table data, where hN2_R and hN2_P denote

the enthalpy of N

in the reactants and products, respectively, and so on. In the Units menu, select temperature

in K and amount of substance in moles.

TR 5 25 1 273.15 // K

// Evaluate reactant and product enthalpies, hR and hP, respectively

hR 5 hC8H18 1 12.5 * hO2_R 1 47 * hN2_R

hP 5 8 * hCO2_P 1 9 * hH2O_P 1 47 * hN2_P

hC8H18 5 2249910 // kj/kmol (value from Table A-25)

hO2_R 5 h_T("O2",TR)

hN2_R 5 h_T("N2",TR)

hCO2_P 5 h_T("CO2",TP)

hH2O_P 5 h_T("H2O",TP)

hN2_P 5 h_T("N2",TP)

// Energy balance

hP 5 hR

Using the Solve button, the result is TP 5 2394 K, which agrees closely with the result obtained above.

(b) For complete combustion of liquid octane with 400% theoretical air, the chemical equation is

1l21 50O

1 188N

8CO

1 9H

O1g21 37.5O

1 188N

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804 Chapter 13 Reacting Mixtures and Combustion

The energy rate balance, Eq. (2), reduces for this case to

81¢h2

1 91¢h2

O1g2

1 37.51¢h2

1 1881¢h2

5 5,074,630 kJ

kmol 1fuel2

➊ Observe that the right side has the same value as in part (a). Proceeding

iteratively as above, the temperature of the products is T

5 962 K. The use

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

➊ The temperature determined in part (b) is considerably lower than the value

found in part (a). This shows that once enough oxygen has been provided

for complete combustion, bringing in more air dilutes the combustion prod-

ucts, lowering their temperature.

Ability to…

❑

apply the control volume

energy balance to calculate

the adiabatic flame

temperature.

❑

evaluate enthalpy values

appropriately.

✓

Skills Developed

If octane gas entered instead of liquid octane, would the adiabatic

flame temperature increase, decrease, or stay the same? Ans. Increase.

13.3.3

Closing Comments

For a specified fuel and specified temperature and pressure of the reactants, the

maximum adiabatic flame temperature is for complete combustion with the theo-

retical amount of air. The measured value of the temperature of the combustion

products may be several hundred degrees below the calculated maximum adiabatic

flame temperature, however, for several reasons:

c Once adequate oxygen has been provided to permit complete combustion, bringing

in more air dilutes the combustion products, lowering their temperature.

c Incomplete combustion also tends to reduce the temperature of the products, and

combustion is seldom complete (see Sec. 14.4).

c Heat losses can be reduced but not altogether eliminated.

c As a result of the high temperatures achieved, some of the combustion products

may dissociate. Endothermic dissociation reactions lower the product tempera-

ture. The effect of dissociation on the adiabatic flame temperature is considered

in Sec. 14.4.

13.4 Fuel Cells

A fuel cell is an electrochemical device in which fuel and an oxidizer (normally oxy-

gen from air) undergo a chemical reaction, providing electrical current to an external

circuit and producing products. The fuel and oxidizer react catalytically in stages on

separate electrodes: the anode and the cathode. An electrolyte separating the two

electrodes allows passage of ions formed by reaction. Depending on the type of fuel

cell, the ions may be positively or negatively charged. Individual fuel cells are con-

nected in parallel or series to form fuel cell stacks to provide the desired level of

power output.

With today’s technology, the preferred fuel for oxidation at the fuel cell anode

is hydrogen because of its exceptional ability to produce electrons when suitable

catalysts are used, while producing no harmful emissions from the fuel cell itself.

Depending on the type of fuel cell, methanol (CH

OH) and carbon monoxide (CO)

can be oxidized at the anode in some applications, but often with performance

penalties.

fuel cell

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Since hydrogen is not naturally occurring, it must be produced. Production meth-

ods include electrolysis of water (see Sec. 2.7) and chemically reforming hydrogen-

bearing fuels, predominantly hydrocarbons. See the following box.

Hydrogen Production by Reforming of Hydrocarbons

Reforming reactions to produce hydrogen include endothermic steam reforming of

hydrocarbons. For example, steam reforming of methane takes the form

1 H

O S CO 1 3H

This often is accompanied by the exothermic water–gas shift reaction

1 H

O S CO

1 H

to produce additional hydrogen and eliminate from the hydrogen stream carbon mon-

oxide, which poisons platinum catalysts used to promote reaction rates in some fuel

cells. Other reforming techniques include dry reforming (carbon dioxide reforming) of

hydrocarbons, partial oxidation of hydrocarbons, pyrolysis (thermal cracking) of hydro-

carbons, and steam reforming of alcohols.

Hydrocarbon reforming can occur either separately or within the fuel cell (depending

on type). When hydrogen is produced by reforming fuel separately from the fuel cell

itself, this is known as external reforming. If not fed directly from the reformer to a fuel

cell, hydrogen can be stored as a compressed gas, a cryogenic liquid, or atoms absorbed

within metallic structures, and then provided to fuel cells from storage, when required.

Internal reforming refers to applications where hydrogen production by reforming fuel

is integrated within the fuel cell. Owing to limitations of current technology, internal

reforming is feasible only in fuel cells operating at temperatures above about 6008C.

Rates of reaction in fuel cells are limited by the time it takes for diffusion of

chemical species through the electrodes and the electrolyte and by the speed of the

chemical reactions themselves. The reaction in a fuel cell is not a combustion process.

These features result in fuel cell internal irreversibilities that are inherently less sig-

nificant than those encountered in power systems employing combustion. Thus, fuel

cells have the potential of providing more power from a given supply of fuel and

oxidizer than conventional internal combustion engines and gas turbines.

Fuel cells do not operate as thermodynamic power cycles, and thus the notion of

a limiting thermal efficiency imposed by the second law is not applicable. However,

as for all power systems, the power provided by fuel cell systems is eroded by inef-

ficiencies in auxiliary equipment. For fuel cells this includes heat exchangers, com-

pressors, and humidifiers. Irreversibilities and losses inherent in hydrogen production

also can be greater than those seen in production of more conventional fuels.

In comparison to reciprocating internal combustion engines and gas turbines that

incorporate combustion, fuel cells typically produce relatively few damaging emissions

as they develop power. Still, such emissions accompany production of fuels consumed

by fuel cells as well as the manufacture of fuel cells and their supporting components.

See Horizons, Sec. 5.3.3 for additional discussion.

Despite potential thermodynamic advantages, widespread fuel cell use has not

occurred thus far owing primarily to cost. Table 13.2 summarizes the most promising

fuel cell technologies currently under investigation. Included are potential applica-

tions and other characteristics.

Cooperative efforts by government and industry have fostered advances in both

proton exchange membrane fuel cells and solid oxide fuel cells, which appear to

provide the greatest range of potential applications in transportation, portable power,

and stationary power. The proton exchange membrane fuel cell and the solid oxide

fuel cell are discussed next.

13.4 Fuel Cells 805

TAKE NOTE...

As discussed in the

Horizons in Sec. 5.3.3,

hydrogen production by

electrolysis of water and

by reforming of hydrocar-

bons are each burdened by

the second law. Significant

exergy destruction is

observed with each method

of production.

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806 Chapter 13

Reacting Mixtures and Combustion

13.4.1

Proton Exchange Membrane Fuel Cell

The fuel cells shown in Fig. 13.3 are proton exchange membrane fuel cells (PEMFCs).

At the anode, hydrogen ions (H

) and electrons (e

) are produced. At the cathode,

oxygen, hydrogen ions, and electrons react to produce water.

c The fuel cell shown schematically in Fig. 13.3a operates with hydrogen (H

) as the

fuel and oxygen (O

) as the oxidizer. The reactions at these electrodes and the

overall cell reaction are labeled on the figure. The only products of this fuel cell

are water, power generated, and waste heat.

Characteristics of Major Fuel Cell Types

Proton Exchange

Membrane Fuel Phosphoric Acid Molten Carbonate Solid Oxide Fuel

Cell (PEMFC) Fuel Cell (PAFC) Fuel Cell (MCFC) Cell (SOFC)

Transportation Automotive Large vehicle None Vehicle auxiliary

application power power power

Heavy vehicle

propulsion

Other applications Portable power On-site On-site On-site

Small-scale cogeneration cogeneration cogeneration

stationary power Electric power Electric power Electric power

generation generation generation

Electrolyte Ion exchange Liquid phosphoric Liquid molten Solid oxide

membrane acid carbonate ceramic

Charge carrier H

Operating 60–808C 150–2208C 600–7008C 600–10008C

temperature

Fuel oxidized H

or methanol H

or CO

at anode

Fuel reforming External External Internal or external Internal or external

Fuels typically None None CO Light hydrocarbons

used for internal Light hydrocarbons (e.g., methane,

reforming (e.g., methane, propane)

propane) Synthetic diesel

Methanol and gasoline

Sources: Fuel Cell Handbook, Seventh Edition, 2004, EG&G Technical Services, Inc., DOE Contract No. DE-AM26-99FT40575. Larminie, J., and Dicks, A.,

2000, Fuel Cell Systems Explained, John Wiley & Sons, Ltd., Chichester, West Sussex, England.

TABLE 13.2

Fig. 13.3 Proton exchange membrane fuel cell. (a) Hydrogen fueled. (b) Humidified-methanol

fueled.

Fuel

Product

Anode

Electrolyte

Cathode

Oxidizer

–

External electric circuit

Overall cell reaction: H

+ O

→ H

–

→ 2H

+ 2e

–

+ 2H

+ 2e

–

→

–

(a)

Overall cell reaction: CH

OH + O

→ 2H

O + CO

–

Fuel

OH + H

Product

Anode

Electrolyte

Cathode

Oxidizer

–

OCO

–

External electric circuit

OH + H

O → 6H

+ 6e

–

+ CO

+ 6H

+ 6e

–

→

–

(b)

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c The fuel cell shown schematically in Fig. 13.3b operates with humidified methanol

(CH

OH 1 H

O) as the fuel and oxygen (O

) as the oxidizer. This type of PEMFC

is a direct-methanol fuel cell. The reactions at these electrodes and the overall cell

reaction are labeled on the figure. The only products of this fuel cell are water,

carbon dioxide, power generated, and waste heat.

For PEMFCs, charge-carrying hydrogen ions are conducted through the electro-

lytic membrane. For acceptable ion conductivity, high membrane water content is

required. This requirement restricts the fuel cell to operating below the boiling point

of water, so PEMFCs typically operate at temperatures in the range 60–808C. Cooling

is generally needed to maintain the fuel cell at the operating temperature.

Owing to the relatively low-temperature operation of proton exchange membrane

fuel cells, hydrogen derived from hydrocarbon feedstock must be produced using

external reforming, while costly platinum catalysts are required at both the anode

and cathode to increase ionization reaction rates. Due to extremely slow reaction rate

at the anode, the direct-methanol fuel cell requires almost ten times as much platinum

catalyst as the hydrogen-fueled PEMFC to improve anode reaction rate. Catalytic

activity is more important in lower-temperature fuel cells because rates of reaction

at the anode and cathode tend to decrease with decreasing temperature.

Automakers are introducing hydrogen-powered proton exchange membrane fuel

cell vehicles. Fuel cell vehicles are undergoing large-scale market testing (United

States) and are being offered by limited lease (Japan). Both hydrogen-fueled PEMFCs

and direct-methanol fuel cells have potential to replace batteries in portable devices

such as cellular phones, laptop computers, and video players (see Horizons, above).

Hurdles to wider deployment of PEMFCs include extending stack life, simplifying

system integration, and reducing costs.

13.4 Fuel Cells 807

Power needs of cellular phones, laptops, and other

portable electronic devices are increasing so rapidly

that the battery industry is struggling to keep up. Some

observers say that today’s batteries won’t be able to provide

enough power, are too heavy, and don’t last long enough to meet

the needs of quickly evolving electronics. Pocket-size fuel cells

might prove to be a viable alternative.

To meet consumer needs, companies are rushing to develop

small fuel cells that provide as much, or more, power on a single

charge than conventional batteries, and can be charged instantly

just by adding more fuel.

Battery companies are fighting back with a new generation of

batteries. One of these, the lithium-ion battery, is already used in

watches, flash cameras, and rechargeable power packs. Lithium-

ion batteries provide several times the output of similar-size

alkaline batteries and can be recharged numerous times.

To compete, fuel cells must prove themselves as reliable and

versatile as batteries, and stakes are high in the consumer elec-

tronics market. Still, many think fuel cells for portable electron-

ics will be the first fuel cells most of us will actually use because

of strong consumer demand for cost-competitive, longer-lasting,

instantly rechargeable power.

Goodbye Batteries, Hello Fuel Cells?

Fuel

Product

Anode

Electrolyte

Cathode

Oxidizer

–

External electric circuit

Overall cell reaction: H

+ O

→ H

–

+ O

→ H

O + 2e

–

+ 2e

–

→

–

(b)(a)

Fig. 13.4 Solid oxide fuel cell. (a) Module. (b) Schematic.

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808 Chapter 13

Reacting Mixtures and Combustion

13.4.2

Solid Oxide Fuel Cell

For scale, Fig. 13.4a shows a solid oxide fuel cell (SOFC) module. The fuel cell sche-

matic shown in Fig. 13.4b operates with hydrogen (H

) as the fuel and oxygen (O

)

as the oxidizer. At the anode, water (H

O) and electrons (e

) are produced. At the

cathode, oxygen reacts with electrons (e

) to produce oxygen ions (O

) that migrate

through the electrolyte to the anode. The reactions at these electrodes and the overall

cell reaction are labeled on the figure. The only products of this fuel cell are water,

power generated, and waste heat.

For SOFCs, an alternative fuel to hydrogen is carbon monoxide (CO) that produces

carbon dioxide (CO

) and electrons (e

) during oxidation at the anode. The cathode

reaction is the same as that in Fig. 13.4(b). Due to their high-temperature operation,

solid oxide fuel cells can incorporate internal reforming of various hydrocarbon fuels

to produce hydrogen and/or carbon monoxide at the anode.

Since waste heat is produced at relatively high temperature, solid oxide fuel cells

can be used for cogeneration of power and process heat or steam. SOFCs also can

be used for distributed (decentralized) power generation and for fuel cell–microturbine

hybrids. These technologies are in the proof-of-concept testing and demonstration

phases of development. Such applications are very attractive because they achieve

objectives without using highly irreversible combustion.

For instance, a solid oxide fuel cell replaces the combustor in the gas turbine shown

in the fuel cell–microturbine schematic in Fig. 13.5. The fuel cell produces electric

power while its high-temperature exhaust expands through the microturbine, produc-

ing shaft power W

. By producing power electrically and mechanically without com-

bustion, fuel cell–microturbine hybrids have the potential of significantly improving

effectiveness of fuel utilization over that achievable with comparable conventional

gas turbine technology and with fewer harmful emissions.

Solid oxide

fuel cell

Electric power produced

Fuel

+–

TurbineCompressor

ExhaustAir

Exhaust

Air

Fig. 13.5 Solid oxide fuel cell–microturbine hybrid.

13.5 Absolute Entropy and the Third Law

of Thermodynamics

Thus far our analyses of reacting systems have been conducted using the conserva-

tion of mass and conservation of energy principles. In the present section some of

the implications of the second law of thermodynamics for reacting systems are

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considered. The discussion continues in the second part of this chapter dealing with

the exergy concept, and in the next chapter where the subject of chemical equilib-

rium is taken up.

13.5.1

Evaluating Entropy for Reacting Systems

The property entropy plays an important part in quantitative evaluations using the

second law of thermodynamics. When reacting systems are under consideration, the

same problem arises for entropy as for enthalpy and internal energy: A common

datum must be used to assign entropy values for each substance involved in the reac-

tion. This is accomplished using the third law of thermodynamics and the absolute

entropy concept.

The third law deals with the entropy of substances at the absolute zero of tempera-

ture. Based on empirical evidence, this law states that the entropy of a pure crystalline

substance is zero at the absolute zero of temperature, 0 K or 08R. Substances not hav-

ing a pure crystalline structure at absolute zero have a nonzero value of entropy at

absolute zero. The experimental evidence on which the third law is based is obtained

primarily from studies of chemical reactions at low temperatures and specific heat

measurements at temperatures approaching absolute zero.

ABSOLUTE ENTROPY. For present considerations, the importance of the third

law is that it provides a datum relative to which the entropy of each substance par-

ticipating in a reaction can be evaluated so that no ambiguities or conflicts arise. The

entropy relative to this datum is called the absolute entropy. The change in entropy

of a substance between absolute zero and any given state can be determined from

precise measurements of energy transfers and specific heat data or from procedures

based on statistical thermodynamics and observed molecular data.

Tables A-25 and A-25E give the value of the absolute entropy for selected substances

at the standard reference state, T

ref

5 298.15 K, p

ref

5 1 atm, in units of kJ/kmol ? K

and Btu/lbmol ? 8R, respectively. Two values of absolute entropy for water are provided.

One is for liquid water and the other is for water vapor. As for the case of the enthalpy

of formation of water considered in Sec. 13.2.1, the vapor value listed is for a hypo-

thetical ideal gas state in which water is a vapor at 258C (778F) and p

ref

5 1 atm.

Tables A-23 and A-23E give tabulations of absolute entropy versus temperature at a

pressure of 1 atm for selected gases. In these tables, the absolute entropy at 1 atm and

temperature T is designated s8

, and ideal gas behavior is assumed for the gases.

USING ABSOLUTE ENTROPY. When the absolute entropy is known at the

standard state, the specific entropy at any other state can be found by adding the

specific entropy change between the two states to the absolute entropy at the stan-

dard state. Similarly, when the absolute entropy is known at the pressure p

ref

and

temperature T, the absolute entropy at the same temperature and any pressure p can

be found from

T, p

5 s

T, p

ref

T, p

2 s

T, p

ref

For the ideal gases listed in Tables A-23, the first term on the right side of this

equation is s8

, and the second term on the right can be evaluated using Eq. 6.18.

Collecting results, we get

s1T, p25 s81T22 R ln

ref

(ideal gas)

(13.22)

To reiterate, s8

is the absolute entropy at temperature T and pressure p

ref

5 1 atm.

The entropy of the ith component of an ideal gas mixture is evaluated at the mix-

ture temperature T and the partial pressure p

: s

T, p

. The partial pressure is given

third law of

thermodynamics

absolute entropy

13.5 Absolute Entropy and the Third Law of Thermodynamics 809

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810 Chapter 13

Reacting Mixtures and Combustion

by p

5 y

p, where y

is the mole fraction of component i and p is the mixture pres-

sure. Thus, Eq. 13.22 takes the form

1T, p

25 s8

1T22 R ln

ref

1T, p

25 s8

1T22 R ln

ref



component i of an

ideal gas mixture

(13.23)

where s8

is the absolute entropy of component i at temperature T and p

ref

5 1 atm.

Equation 13.23 corresponds to Eq. (b) of Table 13.1.

Finally, note that Interactive Thermodynamics (IT ) returns absolute entropy directly

and does not use the special function s8.

13.5.2

Entropy Balances for Reacting Systems

Many of the considerations that enter when energy balances are written for reacting

systems also apply to entropy balances. The writing of entropy balances for reacting

systems will be illustrated by referring to special cases of broad interest.

CONTROL VOLUMES AT STEADY STATE. Let us begin by reconsidering the

steady-state reactor of Fig. 13.2, for which the combustion reaction is given by Eq.

13.11. The combustion air and the products of combustion are each assumed to form

ideal gas mixtures, and thus Eq. 12.26 from Table 13.1 for mixture entropy is applicable

to them. The entropy rate balance for the two-inlet, single-exit reactor can be expressed

on a per mole of fuel basis as

0 5

1 s

1 caa 1

1 aa 1

b3.76s

2 cas

1 aa 1

b 3.76s

(13.24)

where n

is the molar flow rate of the fuel and the coefficients appearing in the

underlined terms are the same as those for the corresponding substances in the reac-

tion equation.

All entropy terms of Eq. 13.24 are absolute entropies. The first underlined term

on the right side of Eq. 13.24 is the entropy of the combustion air per mole of fuel.

The second underlined term is the entropy of the exiting combustion products per

mole of fuel. In accord with Table 13.1, the entropies of the air and combustion prod-

ucts are evaluated by adding the contribution of each component present in the

respective gas mixtures. For instance, the specific entropy of a substance in the com-

bustion products is evaluated from Eq. 13.23 using the temperature of the combustion

products and the partial pressure of the substance in the combustion product mixture.

Such considerations are illustrated in Example 13.9.

Evaluating Entropy Production for a Reactor Fueled by Liquid Octane

c c c c EXAMPLE 13.9 c

Liquid octane at 258C, 1 atm enters a well-insulated reactor and reacts with air entering at the same temperature

and pressure. The products of combustion exit at 1 atm pressure. For steady-state operation and negligible effects

of kinetic and potential energy, determine the rate of entropy production, in kJ/K per kmol of fuel, for complete

combustion with (a) the theoretical amount of air, (b) 400% theoretical air.

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SOLUTION

Known:

Liquid octane and air, each at 258C and 1 atm, burn completely within a well-insulated reactor operating

at steady state. The products of combustion exit at 1 atm pressure.

Find: Determine the rate of entropy production, in kJ/K per kmol of fuel, for combustion with (a) the theoreti-

cal amount of air, (b) 400% theoretical air.

Schematic and Given Data:

Analysis: The temperature of the exiting products of combustion T

was evaluated in Example 13.8 for each of

the two cases. For combustion with the theoretical amount of air, T

5 2395 K. For complete combustion with

400% theoretical air, T

5 962 K.

(a) For combustion of liquid octane with the theoretical amount of air, the chemical equation is

1 12.5O

1 47N

8CO

1 9H

1 47N

With assumptions 1 and 3, the entropy rate balance on a per mole of fuel basis, Eq. 13.24, takes the form

0 5

1 s

1 112.5s

1 47s

22 18s

1 9s

O1g2

1 47s

or on rearrangement

5 18s

1 9s

O1g2

1 47s

22 s

2 112.5s

1 47s

(1)

Each coefficient of this equation is the same as for the corresponding term of the balanced chemical equation.

The fuel enters the reactor separately at T

ref

, p

ref

. The absolute entropy of liquid octane required by the entropy

balance is obtained from Table A-25 as 360.79 kJ/kmol ? K.

The oxygen and nitrogen in the combustion air enter the reactor as components of an ideal gas mixture at

ref

, p

ref

. With Eq. 13.23 and absolute entropy data from Table A-23

5 s 8

ref

22 R ln

ref

5 2

14 ln

.21 5 21

1 k

kmol ? K

5 s 8

ref

22 R ln

ref

5 191.5 2 8.314 ln 0.79 5 193.46 kJ

kmol ? K

Engineering Model:

The control volume shown on the accompa-

nying figure by a dashed line operates at

steady state and without heat transfer with

its surroundings.

2. Combustion is complete. Each mole of oxy-

gen in the combustion air is accompanied

by 3.76 moles of nitrogen, which is inert.

3. The combustion air can be modeled as an

ideal gas mixture, as can the products of

combustion.

4. The reactants enter at 258C, 1 atm. The

products exit at a pressure of 1 atm.

Air

25°C, 1 atm

Combustion products,

1 atm

(l)

25°C, 1 atm

= 2395 K (part a)

= 962 K (part b)

Insulation

Fig. E13.9

13.5 Absolute Entropy and the Third Law of Thermodynamics 811

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812 Chapter 13

Reacting Mixtures and Combustion

The product gas exits as an ideal gas mixture at 1 atm, 2395 K with the following composition: y

5 8/64 5

0.125, y

O1g2

5 9/64 5 0.1406, y

5 47/64 5 0.7344. With Eq. 13.23 and absolute entropy data at 2395 K from

Tables A-23

5 s8

2 R ln y

.17

14 ln

.12

7.4

kmol ? K

5 273.986 2 8.314 ln 0.1406 5 290.30 kJ

kmol ? K

5 258.503 2 8.314 ln 0.7344 5 261.07 kJ

kmol ? K

Inserting values into Eq. (1), the expression for the rate of entropy production, we get

5 81337.4621 91290.3021 471261.072

2 360.79 2 12.5

218.01

2 47

193.46

5 5404 kJ

kmol

octane

? K

Alternative Solution:

As an alternative, the following IT code can be used to determine the entropy production per mole of fuel entering,

where sigma denotes s

, and sN2_R and sN2_P denote the entropy of

in the reactants and products, respec-

tively, and so on. In the Units menu, select temperature in K, pressure in bar, and amount of substance in moles.

TR = 25 1 273.15 // K

p = 1.01325 // bar

TP = 2394 // K (Value from the IT alternative solution of Example 13.8)

// Determine the partial pressures

pO2_R = 0.21 * p

pN2_R = 0.79 * p

pCO2_P = (8/64) * p

pH2O_P = (9/64) * p

pN2_P = (47/64) * p

// Evaluate the absolute entropies

sC8H18 = 360.79 // kJ/kmol ? K (from Table A-25)

sO2_R = s_TP(“O2”, TR, pO2_R)

sN2_R = s_TP(“N2”, TR, pN2_R)

sCO2_P = s_TP(“CO2”, TP, pCO2_P)

sH2O_P = s_TP(“H2O”, TP, pH2O_P)

sN2_P = s_TP(“N2”, TP, pN2_P)

// Evaluate the reactant and product entropies, sR and sP, respectively

sR = sC8H18 1 12.5 * sO2_R 1 47 * sN2_R

sR = 8 * sCO2_P 1 9 * sH2O_P 1 47 * sN2_P

// Entropy balance, Eq. (1)

sigma = sP – sR

Using the Solve button, the result is sigma 5 5404 kJ/kmol (octane) ? K, which agrees with the result obtained above.

(b) The complete combustion of liquid octane with 400% theoretical air is described by the following chemical

equation:

1 50O

1 188N

8CO

1 9H

1 37.5O

1 188N

The entropy rate balance on a per mole of fuel basis takes the form

5 18s

1 9s

O1g2

1 37.5s

1 188s

22 s

2 150s

1 188s

The specific entropies of the reactants have the same values as in part (a). The product gas exits as an ideal gas

mixture at 1 atm, 962 K with the following composition: y

5 8

242.5 5 0.033, y

O1g2

5 9

242.5 5 0.0371,

5 37.5

242.5 5 0.1546, y

5 0.7753. With the same approach as in part (a)

➊

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