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

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636 Chapter 13 Reacting Mixtures and Combustion

Consider first the reactants. With the enthalpy of formation for methane from Table A-25, the given value for methane,

and enthalpy values for nitrogen and oxygen from Table A-23

Next consider the products. With enthalpy of formation values for CO

, CO, and H

O(g) from Table A-25 and enthalpy

values from Table A-23

Inserting values into the expression for the rate of heat transfer

 h

 h

228,079  168442221,235 kJ/kmol 1CH

228,079 kJ/kmol 1CH

 2 3241,820  172,513  990424

 0.289160,371  86822 8.515157,651  86692

 0.049 3110,530  158,191  866924

 0.9513393,520  188,806  936424

6844

kJ/kmol 1CH

 374,850  381400  29824 2.265314,770  86824 8.515314,581  86694

 1h°

 c

¢T 2

 2.2651¢h2

 8.5151¢h2

CONDENSATION OF COMBUSTION PRODUCTS. When the combustion products of a hy-

drocarbon fuel are cooled sufficiently, condensation of a portion of the water formed will

occur. The energy balance must then be written to account for the presence of water in the

products as both a liquid and a vapor. For cooling of combustion products at constant pressure,

the dew point temperature method of Example 13.2 is used to determine the temperature at

the onset of condensation.

CLOSED SYSTEMS

Let us consider next a closed system involving a combustion process. In the absence of ki-

netic and potential energy effects, the appropriate form of the energy balance is

where U

denotes the internal energy of the reactants and U

denotes the internal energy of

the products. If the reactants and products form ideal gas mixtures, the energy balance can

be expressed as

(13.16)

where the coefficients n on the left side are the coefficients of the reaction equation giving

the moles of each reactant or product.

Since each component of the reactants and products behaves as an ideal gas, the respec-

tive specific internal energies of Eq. 13.16 can be evaluated as so the equation

becomes

(13.17a)

Q  W 

n1h  RT

2

n1h  RT

u  h  RT,

nu 

nu  Q  W

 U

 Q  W

13.2 Conservation of Energy—Reacting Systems 637

where T

and T

denote the temperature of the products and reactants, respectively. With ex-

pressions of the form of Eq. 13.13 for each of the reactants and products, Eq. 13.17a can be

written alternatively as

(13.17b)

The enthalpy of formation terms are obtained from Table A-25. The terms are evaluated

as discussed above.

The foregoing concepts are illustrated in Example 13.6, where a gaseous mixture burns

in a closed, rigid container.

¢h



n1h°

 ¢h2

n1h°

 ¢h2 RT

n  RT

Q  W 

n1h°

 ¢h  RT

2

n1h°

 ¢h  RT

EXAMPLE 13.6 Analyzing Combustion at Constant Volume

A mixture of 1 kmol of gaseous methane and 2 kmol of oxygen initially at 25C and 1 atm burns completely in a closed, rigid

container. Heat transfer occurs until the products are cooled to 900 K. If the reactants and products each form ideal gas mix-

tures, determine (a) the amount of heat transfer, in kJ, and (b) the final pressure, in atm.

SOLUTION

Known: A mixture of gaseous methane and oxygen, initially at 25C and 1 atm, burns completely within a closed rigid con-

tainer. The products are cooled to 900 K.

Find: Determine the amount of heat transfer, in kJ, and the final pressure of the combustion products, in atm.

Schematic and Given Data:

1kmol CH

(g)

2 kmol O

= 25°C

= 1 atm

State 1

Products of

combustion

State 2

 Figure E13.6

Assumptions:

1. The contents of the closed, rigid container are taken as the system.

2. Kinetic and potential energy effects are absent, and W  0.

3. Combustion is complete.

4. The initial mixture and the products of combustion each form ideal gas mixtures.

5. The initial and final states are equilibrium states.

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

 2O

S CO

 2H

638 Chapter 13 Reacting Mixtures and Combustion

(a) With assumptions 2 and 3, the closed system energy balance takes the form

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

Since each reactant and product behaves as an ideal gas, the respective specific internal energies can be evaluated as

. The energy balance then becomes

where T

and T

denote, respectively, the initial and final temperatures. Collecting like terms

The specific enthalpies are evaluated in terms of the respective enthalpies of formation to give

Since the methane and oxygen are initially at 25C, for each of these reactants. Also, for oxygen.

With enthalpy of formation values for CO

O(g), and CH

(g) from Table A-25 and enthalpy values for H

O and CO

from Table A-23

(b) By assumption 3, the initial mixture and the products of combustion each form ideal gas mixtures. Thus, for the

reactants

where n

is the total number of moles of reactants and p

is the initial pressure. Similarly, for the products

where n

is the total number of moles of products and p

is the final pressure.

Since n

 n

 3 and volume is constant, these equations combine to give



 a

900 K

298 K

b 11 atm2 3.02 atm

V  n

745,436 kJ

 174,8502 318.31421298  9002

Q  3393,520  137,405  936424 23241,820  131,828  990424

h°

 0¢h  0

 1h

 ¢h

1 g2

 21h°

 ¢h

4 3R1T

 T

Q  31h°

 ¢h2

 21h°

 ¢h2

O 1 g2

Q  1h

 2h

O 1 g2

 h

1 g2

 2h

2 3R1T

 T

Q  311h

 RT

2 21h

O 1g2

 RT

24 311h

1g2

 RT

2 21h

 RT

u  h  RT

Q  U

 U

 11u

 2u

O 1 g2

2 11u

1 g2

 2u

 U

 Q  W

 13.2.3 Enthalpy of Combustion and Heating Values

Although the enthalpy of formation concept underlies the formulations of the energy bal-

ances for reactive systems presented thus far, the enthalpy of formation of fuels is not always

tabulated.  for example. . . fuel oil and coal are normally composed of several indi-

vidual chemical substances, the relative amounts of which may vary considerably, depend-

ing on the source. Owing to the wide variation in composition that these fuels can exhibit,

we do not find their enthalpies of formation listed in Tables A-25 or similar compilations of

thermophysical data.  In many cases of practical interest, however, the enthalpy of com-

bustion, which is accessible experimentally, can be used to conduct an energy analysis when

enthalpy of formation data are lacking.

13.2 Conservation of Energy—Reacting Systems 639

The enthalpy of combustion is defined as the difference between the enthalpy of the

products and the enthalpy of the reactants when complete combustion occurs at a given tem-

perature and pressure. That is

(13.18)

where the n’s correspond to the respective coefficients of the reaction equation giving the

moles of reactants and products per mole of fuel. When the enthalpy of combustion is ex-

pressed on a unit mass of fuel basis, it is designated h

. Tabulated values are usually given

at the standard temperature T

ref

and pressure p

ref

introduced in Sec. 13.2.1. The symbol

or is used for data at this temperature and pressure.

The heating value of a fuel is a positive number equal to the magnitude of the enthalpy

of combustion. Two heating values are recognized by name: the higher heating value (HHV)

and the lower heating value (LHV). The higher heating value is obtained when all the wa-

ter formed by combustion is a liquid; the lower heating value is obtained when all the water

formed by combustion is a vapor. The higher heating value exceeds the lower heating value

by the energy that would be required to vaporize the liquid formed. Values for the HHV and

LHV also depend on whether the fuel is a liquid or a gas. Heating value data for several

hydrocarbons are provided in Tables A-25.

The calculation of the enthalpy of combustion, and the associated heating value, using

table data is illustrated in the next example.

h°





enthalpy of combustion

EXAMPLE 13.7 Calculating Enthalpy of Combustion

Calculate the enthalpy of combustion of gaseous methane, in kJ per kg of fuel, (a) at 25C, 1 atm with liquid water in the

products, (b) at 25C, 1 atm with water vapor in the products. (c) Repeat part (b) at 1000 K, 1 atm.

SOLUTION

Known: The fuel is gaseous methane.

Find: Determine the enthalpy of combustion, in kJ per kg of fuel, (a) at 25C, 1 atm with liquid water in the products,

(b) at 25C, 1 atm with water vapor in the products, (c) at 1000 K, 1 atm with water vapor in the products.

Assumptions:

1. Each mole of oxygen in the combustion air is accompanied by 3.76 moles of nitrogen, which is inert.

2. Combustion is complete, and both reactants and products are at the same temperature and pressure.

3. The ideal gas model applies for methane, the combustion air, and the gaseous products of combustion.

Analysis: The combustion equation is

The enthalpy of combustion is, from Eq. 13.18

Introducing the coefficients of the combustion equation and evaluating the specific enthalpies in terms of the respective

enthalpies of formation

 1h

 ¢h2

 21h°

 ¢h2

 1h°

 ¢h2

1g2

 21h°

 ¢h2

 h

 2h

 h

1 g2

 2h



1h°

 ¢h2



1h°

 ¢h2

 2O

 7.52N

S CO

 2H

O  7.52N

higher and lower

heating values

640 Chapter 13 Reacting Mixtures and Combustion

For nitrogen, the enthalpy terms of the reactants and products cancel. Also, the enthalpy of formation of oxygen is zero by

definition. On rearrangement, the enthalpy of combustion expression becomes

The values for and ( ) depend on whether the water in the products is a liquid or a vapor.

(a) Since the reactants and products are at 25C, 1 atm in this case, the terms drop out of the above expression for .

Thus, for liquid water in the products, the enthalpy of combustion is

With enthalpy of formation values from Table A-25

Dividing by the molecular weight of methane places this result on a unit mass of fuel basis

which agrees with the higher heating value of methane given in Table A-25.

(b) As in part (a), the terms drop out of the above expression for which for water vapor in the products reduces to

, where

With enthalpy of formation values from Table A-25

On a unit of mass of fuel basis, the enthalpy of combustion for this case is

which agrees with the lower heating value of methane given in Table A-25.

value determined in part (b): (fuel), and the terms for O

O(g), and CO

can be evaluated

using specific enthalpies at 298 and 1000 K from Table A-23. The results are

For methane, the expression of Table A-21 can be used to obtain

Substituting values into the expression for the enthalpy of combustion

800,522 kJ/kmol 1fuel2

802,310  333,405  2125,9782 38,189  2122,70724

 38,189 kJ/kmol 1fuel2

 R

a3.826 T 

3.979



24.558



22.733



6.963

1000

298

1¢h2

1 g2





1000

298

1¢h2

 42,769  9364  33,405 kJ/kmol

1¢h

O 1 g2

 35,882  9904  25,978 kJ/kmol

1¢h

 31,389  8682  22,707 kJ/kmol

¢h

h°

802,310 kJ/kmol

h°



802,310

16.04

50,019 kJ/kg

fuel

393,520  21241,8202 174,8502802,310 kJ/kmol 1fuel2

 1h°

 21h°

O1 g2

 1h°

1 g2

h°

,¢h

h°



890,330 kJ/kmol 1fuel2

16.04 kg 1fuel2



kmol 1fuel2

55,507 kJ/kg 1fuel2

393,520  21285,8302 174,8502890,330 kJ/kmol 1fuel2

 1h°

 21h°

O1l2

 1h°

1g2

¢h

¢hh°

 h°

 31¢h2

 21¢h2

 1¢h2

1g2

 21¢h2

 1h°

 21h°

 1h°

1g2

 31¢h2

 21¢h2

 1¢h2

1g2

 21¢h2

❶

13.3 Determining the Adiabatic Flame Temperature 641



13.3 Determining the Adiabatic

Flame Temperature

Let us reconsider the reactor at steady state pictured in Fig. 13.2. In the absence of work

and appreciable kinetic and potential energy effects, the energy liberated on combustion is trans-

ferred from the reactor in two ways only: by energy accompanying the exiting combustion

products and by heat transfer to the surroundings. The smaller the heat transfer, the greater the

When enthalpy of formation data are available for all the reactants and products, the enthalpy

of combustion can be calculated directly from Eq. 13.18, as illustrated in Example 13.7. Other-

wise, it must be obtained experimentally using devices known as calorimeters. Both constant-

volume (bomb calorimeters) and flow-through devices are employed for this purpose. Consider

as an illustration a reactor operating at steady state in which the fuel is burned completely with

air. For the products to be returned to the same temperature as the reactants, a heat transfer from

the reactor would be required. From an energy rate balance, the required heat transfer is

(13.19)

where the symbols have the same significance as in previous discussions. The heat transfer

per mole of fuel, would be determined from measured data. Comparing Eq. 13.19

with the defining equation, Eq. 13.18, we have In accord with the usual sign

convention for heat transfer, the enthalpy of combustion would be negative.

As noted previously, the enthalpy of combustion can be used for energy analyses of re-

acting systems.  for example. . . consider a control volume at steady state for which

the energy rate balance takes the form

All symbols have the same significance as in previous discussions. This equation can be re-

arranged to read

For a complete reaction, the underlined term is just the enthalpy of combustion , at T

ref

and p

ref

. Thus, the equation becomes

(13.20)

The right side of Eq. 13.20 can be evaluated with an experimentally determined value for

and values for the reactants and products determined as discussed previously. ¢hh°



 h°



1¢h2



1¢h2

h°





1h°



1h°



1¢h2



1¢h2





1h°

 ¢h2



1h°

 ¢h2

 Q







On a unit mass basis

Using Interactive Thermodynamics: IT, we get 38,180 kJ/kmol (fuel).

Comparing the values of parts (b) and (c), the enthalpy of combustion of methane is seen to vary little with temperature.

The same is true for many hydrocarbon fuels. This fact is sometimes used to simplify combustion calculations.



800,552

16.04

49,910 kJ/kg 1fuel2

❶

❷

642 Chapter 13 Reacting Mixtures and Combustion

energy carried out with the combustion products and thus the greater the temperature of the prod-

ucts. The temperature that would be achieved by the products in the limit of adiabatic operation

of the reactor is called the adiabatic flame temperature or adiabatic combustion temperature.

The adiabatic flame temperature can be determined by use of the conservation of mass

and conservation of energy principles. To illustrate the procedure, let us suppose that the

combustion air and the combustion products each form ideal gas mixtures. Then, with the

other assumptions stated above, the energy rate balance on a per mole of fuel basis, Eq. 13.12b,

reduces to the form —that is

(13.21a)

where i denotes the incoming fuel and air streams and e the exiting combustion products.

With this expression, the adiabatic flame temperature can be determined using table data or

computer software, as follows.

USING TABLE DATA. When using Eq. 13.9 with table data to evaluate enthalpy terms,

Eq. 13.21a takes the form

(13.21b)

The n’s are obtained on a per mole of fuel basis from the balanced chemical reaction equation.

The enthalpies of formation of the reactants and products are obtained from Table A-25 or

A-25E. Enthalpy of combustion data might be employed in situations where the enthalpy of

formation for the fuel is not available. Knowing the states of the reactants as they enter the re-

actor, the terms for the reactants can be evaluated as discussed previously. Thus, all terms

on the right side of Eq. 13.21b can be evaluated. The terms on the left side account for

the changes in enthalpy of the products from T

ref

to the unknown adiabatic flame temperature.

Since the unknown temperature appears in each term of the sum on the left side of the equa-

tion, determination of the adiabatic flame temperature requires iteration: A temperature for the

products is assumed and used to evaluate the left side of Eq. 13.21b. The value obtained is

compared with the previously determined value for the right side of the equation. The proce-

dure continues until satisfactory agreement is attained. Example 13.8 gives an illustration.

USING COMPUTER SOFTWARE. Thus far we have emphasized the use of Eq. 13.9 together

with table data when evaluating the specific enthalpies required by energy balances for re-

acting systems. Such enthalpy values also can be retrieved using Interactive Thermodynam-

ics: IT. With IT, the quantities on the right side of Eq. 13.9 are evaluated by software, and

data are returned directly.  for example. . . consider CO

at 500 K modeled as an

ideal gas. The specific enthalpy is obtained from IT as follows:

T = 500 // K

h = h_T(“CO

”, T)

Choosing K for the temperature unit and moles for the amount under the Units menu, IT re-

turns h 3.852  10

kJ/kmol.

This value agrees with the value calculated from Eq. 13.9 using enthalpy data for CO

from Table A-23, as follows



3.852  10

kJ/kmol

393,520  317,678  9364 4

h  h°

 3h1500 K2 h1298 K24

(¢h)

¢h

1¢h2



1¢h2



h°



h°

1h°

 ¢h2



1h°

 ¢h2



 h

adiabatic flame

temperature

13.3 Determining the Adiabatic Flame Temperature 643

As suggested by this discussion, IT is also useful for analyzing reacting systems. In par-

ticular, the equation solver and property retrieval features of IT allow the adiabatic flame

temperature to be determined without the iteration required when using table data. This is

illustrated in Example 13.8.

EXAMPLE 13.8 Determining the Adiabatic Flame Temperature

Liquid octane at 25C, 1 atm enters a well-insulated reactor and reacts with air entering at the same temperature and pressure.

For steady-state operation and negligible effects of kinetic and potential energy, determine the temperature of the combustion

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

SOLUTION

Known: Liquid octane and air, each at 25C and 1 atm, burn completely within a well-insulated reactor operating at steady

state.

Find: Determine the temperature of the combustion products for (a) the theoretical amount of air and (b) 400% theoretical

air.

Schematic and Given Data:

Air

25°C, 1 atm

Combustion products,

(l)

25°C, 1 atm

Insulation

 Figure E13.8

Assumptions:

1. The control volume indicated on the accompanying figure by a dashed line operates at steady state.

2. For the control volume, , and kinetic and potential effects are negligible.

3. The combustion air and the products of combustion each form ideal gas mixtures.

4. Combustion is complete.

5. Each mole of oxygen in the combustion air is accompanied by 3.76 moles of nitrogen, which is inert.

Analysis: At steady state, the control volume energy rate balance Eq. 13.12b reduces with assumptions 2 and 3 to give

Eq. 13.21a

(1)

When Eq. 13.9 and table data are used to evaluate the enthalpy terms, Eq. (1) is written as

On rearrangement, this becomes

1¢h2



1¢h2



h°



h°

1h°

 ¢h2



1h°

 ¢h2



 0, W

 0

644 Chapter 13 Reacting Mixtures and Combustion

which corresponds to Eq. 13.21b. Since the reactants enter at 25C, the ( )

terms on the right side vanish, and the energy

rate equation becomes

(2)

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

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

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

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

. This temperature can be de-

termined by an iterative procedure.

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

. Since the summation of the en-

thalpies 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

 2395 K.

2500 K 2400 K 2350 K

975,408 926,304 901,816

890,676 842,436 818,478

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

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 = 25 + 273.15 // K

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

hR = hC8H18 + 12.5 * hO2_R + 47 * hN2_R

hP = 8 * hCO2_P + 9 * hH2O_P + 47 * hN2_P

hC8H18 = –249910 // kJ/kmol (Value from Table A-25)

hO2_R = h_T(“O2’’,TR)

hN2_R = h_T(“N2’’,TR)

hCO2_P = h_T(“CO2’’,TP)

hH2O_P = h_T(“H2O’’,TP)

hN2_P = h_T(“N2’’,TP)

// Energy balance

hP = hR

Using the Solve button, the result is TP = 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

1l2 50O

 188N

S 8CO

 9H

O  37.5O

 188N

1¢h2

471¢h2

91¢h2

O1g2

81¢h2

¢h

81¢h2

 91¢h2

O1g2

 471¢h2

 5,074,630 kJ/kmol 1fuel2

 381h°

 91h°

O1g2

 471h°

 31h°

1l2

 12.51h°

 471h°

81¢h2

 91¢h2

O1g2

 471¢h2

1l2 12.5O

 47N

S 8CO

 9H

O1g2 47N

1¢h2



h°



h°

¢h

13.4 Fuel Cells 645

Equation (2), the energy rate balance, reduces for this case to

Observe that the right side has the same value as in part (a). Proceeding iteratively as above, the temperature of the products

is T

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

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 products, lowering their

temperature.

81¢h

 91¢h2

O1g2

 37.51¢h2

 1881¢h2

 5,074,630 kJ/kmol 1fuel2

❶

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 the-

oretical 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:

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

more air dilutes the combustion products, lowering their temperature.

 Incomplete combustion also tends to reduce the temperature of the products, and com-

bustion is seldom complete (see Sec. 14.4).

 Heat losses can be reduced but not altogether eliminated.

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

dissociate. Endothermic dissociation reactions lower the product temperature. 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 oxygen from

air) undergo a chemical reaction, providing electrical current to an external circuit and pro-

ducing 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. The reaction is not a combustion process.

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. These features, together with other aspects of fuel cell operation, result in in-

ternal irreversibilities that are inherently less significant than encountered in power producing

devices relying on combustion.

By avoiding highly irreversible combustion, fuel cells have the potential of providing more

power from a given supply of fuel and oxidizer, while forming fewer undesirable products,

than internal combustion engines and gas turbines. In contrast to power plants studied in pre-

vious chapters, fuel cells can produce electric power without moving parts or utilizing

intermediate heat exchangers. 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.

Despite these thermodynamic advantages, widespread use of fuel cells has not occurred thus

far owing primarily to cost.

For scale, Fig. 13.3a shows a solid oxide fuel cell module. Figure 13.3b gives the

schematic of a proton exchange membrane fuel cell, which is discussed next as a repre-

sentative case.

fuel cell