05 - 1 thermodynamics of combustion

8/3/2019 05 - 1 Thermodynamics of Combustion

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2M TF Thermodynamics Sec. 5 2001-02

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5. THERMODYNAMICS OF COMBUSTION

The relevant part of engel and Boles is Chapter 14, Chemical Reactions

5.1 INTRODUCTION

In 1M Thermodynamics and in the first parts of 2M Thermodynamics we have considered non-

reacting thermodynamic systems only. Here, the analysis will be extended from non-reacting

systems to reacting systems, i.e. the species composition will change during the thermodynamic

process. In general, the known principles of conservation of mass and the first and second laws of

thermodynamics can still be applied. However, conservation of mass needs to be extended to the

principles of conservation of elements and the conservation of energy equation needs to include

the effects of chemical energy associated with changes in chemical composition. Inclusion of

these additional parameters will help us to determine the reactant and product species

composition, their state, the heat released, the work done, the efficiency of the combustion

process and its irreversibilities. In particular, the accurate product species composition that

determines pollutant emissions from a combustion process has become increasingly importantover the last two decades.

5.2 PROBLEM DESCRIPTION

Fig 1: Combustion chamber and steam power cycle

Only non-reacting systems have been subject of previous lectures. The heat flow Q& was assumed

to be known and the thermodynamic analysis has focused on non-reacting system analysis. The

next four lectures will discuss the reacting system and we will develop the tools of how to

calculate the state of the products and the heat flow rate from the combustion chamber to the

power generation cycle.

?=Q&

Combustion

chamber

turbine

condenser

Products

composition=?

T=?, p=?

Reactants

T, p, n&

heat exchanger

Reacting system

Non-reacting system

pump

pumpW&

netturbW ,&


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5.3 CHEMICAL REACTIONS

Chemical bonds between the atoms of a molecule store chemical energy. During a combustion

process the composition of the species changes (e.g. H2+0.5O2H2O), i.e. chemical bonds breakup and new bonds are formed. During break-up chemical energy is released and some of the

energy is bound in the newly formed chemical bonds. Generally, energy release and

consumption do not balance and the reaction is said to be exothermic when energy is set free.

The reaction is endothermic when energy is needed to drive the chemical process. How to account

for chemical energies will be topic of section 5.5.

For a reaction to take place, fuel and oxidiser need to be mixed on the molecular level in the right

proportions. Mixtures of fuel and oxidiser can be called reactive mixtures. Extremely fuel rich or

fuel lean mixtures will not burn. In addition, most reactive mixtures do not react automatically.

The mixture must be brought locally above a certain ignition temperature (e.g. lighter, matches,

spark plug) to start chemical reaction.

The activation energy Eactivation must be provided to the system to trigger the onset of the chemical

reaction. If E is negative, the reaction will proceed automatically until conversion of fuel andoxidiser into products is complete. When E is positive, the reaction is endothermic and it willstop if insufficient energy is provided to keep the reaction alive.

5.3.1 Fuels

For boilers, one main objective of the thermodynamic analysis of reactive systems is to calculate

the heat flux from the combustion chamber to the heat exchanger. In gas turbine combustors, the

aim is an enthalpy increase in the stream and in a reciprocating engine the work output due to

increase pressure and temperature is the quantity of interest. Heat flux, change in enthalpy and

work output can be associated with the energy release from the combustion process. The energy

release strongly depends on destruction and formation of chemical bonds and hence, the chemical

mechanisms and the fuel composition must be known. Reaction mechanisms usually have a large

number of intermediate steps but in 2M we are only concerned with the overall result and willrepresent chemical reaction of fuel and oxidizer by one single (global) reaction step. Most

commercial fuels are complex mixtures of a variety of chemical species. However, they can

usually be represented adequately by higher order hydrocarbons. Hydrocarbons can be described

by the chemical formula CxHy. Other fuels with oxygen content will be considered in Section

5.11. Typical hydrocarbon fuels are listed in Table 1.

Table 1: Simplified fuel compositions

Approximated as Chemical formulaGasoline (petrol) Octane C8H18

Diesel oil Dodecane C12

H26

Ethanol C2H5OH

Natural gas Methane CH4

fuel

products

Eactivation

E

E

Fig. 2: Activation energy and

energy release of a

chemical reaction


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5.3.2 Oxidiser

The most common oxidiser (sometimes called oxidant) is air since it is free and readily available.

Air consists of mainly oxygen and nitrogen with some traces of carbon monoxide, Argon etc. The

trace species and nitrogen do not react and can be grouped together as atmospheric nitrogen (AN).

The molar composition of air can then be represented by 21% O2 and 79% AN. In general, all

trace elements can be neglected and the simplified representation of air is 21% O2 and 79% N2 ona molar basis (nN2:nO2 = 79:21). The representation of air on a mass basis can then be written as

(see Section 2 on Mixtures of Gases)

Table 2: Conversion table (quantities per 100 kmol)

ni (no. of kmol) Mi (molar mass) mi = Mi ni (mass) Yi= mi / m (mass fraction)

O2 21 32 672 0.232

N2 79 28 2212 0.768

total 100 2884 1.000

The composition of air on a mass basis is mO2:mN2 = 23.2:76.8.

Pure oxygen (O2) is only used in special applications (e.g. rocket engines). Pure oxygen provides

higher combustion temperatures since no energy is used for the heating of inert N2.

Note: The assumption of inert N2 is not quite correct. At high temperatures some N2 will react to

form nitrogen oxides (NO and NO2) and nitrous oxide (N2O). All these compounds are harmful

pollutants and need to be considered in a detailed combustion analysis. However, reactions

involving nitrogen are slow in comparison to the main reactions for the conversion of

hydrocarbons into products and concentrations of NOx are generally in the range of ppm (parts per

million). Concentrations of N2O are even smaller. These small quantities do hardly influence theglobal thermodynamic quantities and N2 can assumed to be inert for the present analysis.

Conversion of N2 and oxygen into nitrogen oxides will be a topic of the combustion module in

3M Energy Engineering.

5.3.3 Products

Products of complete hydrocarbon combustion with air are CO2 and H2O. Complete combustion

means that enough oxygen is provided to fully convert all fuel-carbon atoms into CO2 and all

fuel-hydrogen atoms into H2O. If not enough oxygen is available during combustion some CO

will form. Oxygen will leave the combustion chamber as a product if more oxygen is provided

than needed for complete combustion.

Note: Depending on the combustion process, fuel composition temperature, some species maydissociate into their elements. In particular, if exit temperatures are above 1500K, dissociation

products like atomic H and O will form and the assumption of complete combustion with the only

products being CO2 and H2O is not accurate. Detailed analysis of the combustion kinetics will be

part of 3M Energy Engineering and 4M Combustion.

The product analysis leads to the following definitions:


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5.3.4 Some Definitions

Stoichiometric Combustion:

Stoichiometric combustion occurs when the minimum amount of air needed for complete

combustion of fuel is available. The only products will be CO2, H2O and N2 .The latter does not

take part in the reaction but cannot be ignored in the thermodynamics.

Excess fuel (fuel rich mixture):

Insufficient O2 supply will result in incomplete conversion of hydrocarbons to CO2 and some CO

will form.

Excess air (fuel lean mixture):

All fuel is converted to CO2 and H2O. Excess O2 will leave the system. In reality, some excess air

is needed due to imperfect mixing of fuel and air.

Obviously, the fuel composition and the proportion of fuel to oxidiser need to be known for a

complete analysis of the combustion process. The fuel composition is uniquely given by the

chemical formula. However, there are three common ways of how to express the proportion of

fuel to oxidiser:

1. Air/fuel ratio (AFR) = Fair mm && , (sometimes FAR, = airF mm && )

2. Theoretical air: e.g. 150% theoretical air means 1.5 times that of stoichiometric combustion

3. Excess air: e.g. 50% excess air means the same as 150% theoretical air

Now, the combustion equations can be written as

5.3.5 Combustion Equations:

Combustion of hydrocarbon fuel with pure oxygen:

CxHy + nox O2 nCO2 CO2 + nH2O H2O + nO2 O2 + nCO CO + ...

Combustion of hydrocarbon fuel with air:

CxHy +nox (O2+79/21N2) nCO2 CO2 + nH2O H2O + nN2 N2 + nO2 O2 + nCO CO + Eq. (1)

Equations are typically written per kmol of fuel and nox depends on the proportion of fuel to air.

5.4 CONSERVATION EQUATIONS

5.4.1 Conservation of mass:

In a steady state combustion process as shown in Fig. 1 no mass is accumulated inside the

combustion chamber, hence =reactantsproducts

mm && . However, no conclusions can be drawn about the

products composition. Therefore, the mass conservation analysis needs to be extended to the

analysis of conservation of elements.


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5.4.2 Conservation of elements:

The number of atoms of each element j must be conserved, i. e.

=reactants

ji,i

products

ji,i lnln

with ni being the number of moles of species i and li,j being the number of atoms of element j in amolecule of species i.

We obtain one equation for each element, e.g. 4 equations for hydrocarbon combustion with air.

The four elements are C, H, O, N.

The chemical reaction for stoichiometric hydrocarbon combustion with air can be written as

CxHy + nox (O2+79/21N2) nCO2 CO2 + nH2O H2O + nN2 N2

Element conservation:

C: x = 1nCO2 x elements of C in fuel, 1 element of C in CO2

H: y = 2nH2O y elements of H in fuel, 2 elements of H in H2O

O: 2nox = 2nCO2+1nH2O

N: 79/21nox = nN2

The system has four unknowns (nox, nCO2, nH2O, nN2) and four equations and will therefore have a

unique solution.

Note: The number of moles is not conserved (e.g. 2H2+O2 2H2O)

5.4.3 Examples

Examples for lean and fuel rich mixtures will be calculated in the lecture.

5.5 ENERGY

Previous thermodynamic analysis has accounted for sensible and latent energy only. Sensible

energy is associated with a change of state of a species (e.g. h = cp T). Latent energy isassociated with a phase change of the species. However, molecules also possess chemical and

nuclear energies. Nuclear energy will not be topic of this lecture and will not be discussed.

Chemical energy is associated with chemical bonds between molecules. If a chemical reaction

occurs, these bonds are broken and energy is released. New bonds form and this formation

process is in general associated with consumption of energy, i.e. energy is bound in the newly

formed molecular bonds. This change in chemical energy needs

1) to be quantified and

2) to be included in the overall energy balance.

5.5.1 Chemical Energy

Inclusion in the overall energy balance is relatively easy to achieve. Changes in specific internal

energy of a particular species can be written as

pekehe t ++= with

sct hhh += ht, hc and hs denote the total enthalpy, chemical enthalpy and sensible enthalpy, respectively. We

will see in Section 5.8 that latent heat can be included into the enthalpy of formation (for a

definition of enthalpy of formation, see below). Note that by convention total enthalpy now meansthe total of sensible and chemical energy but it excludes kinetic and potential energies. This is

unlike the use of the term total energy in the compressible flow theory.

Eq. (2)


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To evaluate the change in chemical energy, the experiment shown in Fig. 3 is conducted:

Fig 3: Steady flow combustion process

Carbon (e.g. graphite that consists of 99% carbon atoms) and oxygen are fed into the combustion

chamber at 25C and 1 atm. They react and form CO2 as product. The product is cooled to 25C

and the heat flow can be measured: kJ/s-393,520Q =&

Neglecting kinetic and potential energy changes, First Law analysis of the combustion chamberyields

[ ] kJ/kmol][andkJ/kgwith

====

=

tttt

inout

hhhmhnH

HHWQ

&&&

&&&&

Note that enthalpy flows can be written as the product of mass flow multiplied by mass specific

enthalpy ht (units kJ/kg) or as molar flow multiplied by mole specific enthalpy th (unit kJ/kmol).

ht and th are related by tt Mhh = .No work is done, hence equation (3) results in

2

2

2

2

2

2

22

22

,,,

,,,

Ot

CO

OCt

CO

CCOt

CO

OtOCtCCOtCO

hn

nh

n

nh

n

Q

hnhnhnQ

&

&

&

&

&

&

&&&&

=

=

CO2nQ && =-393,520 kJ/kmolCO2 is a measure of the enthalpy difference between CO2 and its

elements C and O2 at the same temperature T and p. This experiment leads to the following

definitions:

Enthalpy of Formation: The enthalpy of formation of a substance, hf0, is defined as the heat

evolved when one kmol of the substance is formed from its elements at their standard states.

Standard State: The standard state is defined as 25C (298.15 K) and 1 atm. This is the referencestate of a substance. By convention, each element in its standard state is assigned an enthalpy of

zero (or has an enthalpy of formation of zero). The standard states of some elements are, e.g.

H2(g), O2(g), N2(g), Hg(l), C(s,graphite).

In particular the second part of the definition of standard state is of significance. Remember that

in previous first law analyses of non-reacting systems only enthalpy differences have been used.

However, in the presence of combustion, species are not conserved and representation of energy

differences in the form of ( ))(Th)(ThnH 12 = && is no longer of any use. Therefore, absolute

values of enthalpy are needed to solve equation (4). Applying above convention the absolute totalenthalpy of any species i can be written as

Q&

kmol/s1atm1C,25,CO2

2= COn&

Combustion cooling

kmol/s1atm1C,25,O22=

On&

C, 25C, 1 atm kmol/s1=Cn&

Eq. (4)

Eq. (3)


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)25()()( 00,, CThThhTh iiifit =+=

where the subscript s for sensible enthalpy has been dropped. Enthalpy at standard state reduces to

enthalpy of formation. Deviations from the standard state can be associated with a change of state

of species i and are accounted for by the second and third RHS terms. Hence, 0if,h accounts for

chemical energies and )(Th-(T)h 0ii accounts for changes in sensible energies. One of the majorconclusions is that

SFEE can be equally used for non-reacting and reacting processes when total enthalpies asdefined in equation (5) are used. The use of total enthalpies automatically accounts for the

presence of chemical reactions.

If absolute enthalpy values need to be used, the enthalpy of formation must be known. If perfect

gas assumptions do not hold, sensible energy changes must also be listed (i.e. hscpT). Sensibleenthalpies of some major species are given in Table 7 in the Data and Formula book. Table 3 of

these notes gives enthalpies of formation of the most common species.

Table 3: Enthalpies of formation at 25C and 1 atm (from engel and Boles)

Substance Formula 0fh [kJ/kmol]

Carbon C(s) 0

Hydrogen H2(g) 0

Nitrogen N2(g) 0

Oxygen O2(g) 0

Carbon monoxide CO(g) -110,530

Carbon dioxide CO2(g) -393,520

Water vapour H2O(g) -241,820Water (liquid) H2O(l) -285,830

Hydrogen peroxide H2O2(g) -136,310

Ammonia NH3(g) -46,190

Methane CH4(g) -74,850

Acetylene C2H2(g) +226,730

Ethylene C2H4(g) +52,280

Ethane C2H6(g) -84,680

Propylene C3H6(g) +20,410

Propane C3H8(g) -103,850

n-Butane C4H10(g) -126,150n-Octane vapour C8H18(g) -208,450

n-Octane liquid C8H18(l) -249,950

n-Dodecane C12H26(g) -291,010

Benzene C6H6(g) +82,930

Methanol vapour CH3OH(g) -200,670

Methanol liquid CH3OH(l) -238,660

Ethanol vapour C2H5OH(g) -235,310

Ethanol liquid C2H5OH(l) -277,690

Oxygen O(g) +249,190

Hydrogen H(g) +218,000

Nitrogen N(g) +472,650

Hydroxyl OH(g) +39,460

Eq. (5)


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Table 3 above gives enthalpies of formation at standard state. However, inlet and outlet conditions

of a combustion chamber may not be at standard state but at an arbitrary temperature T.

T 1 2

hs

T00

fh

Reactants Products

Fig 4: Two different temperature-reaction paths

Initial and end states are state 1 and 2, respectively. The process path is of no importance since

enthalpies (temperatures) and molar (mass) concentrations are properties. Hence, path A is

equivalent to path (B+C) and the difference in enthalpies between state 1 and 2 can be expressed

as function of enthalpy of formation at standard state and sensible enthalpies, viz,

[ ]44444 344444 21

444 3444 21 C

compoundf

B

elements

i ThThThThThnTh )()()())()(()( 000

0 ++=

The first term accounts for changes in sensible enthalpies of all reactants when cooled from T to

T0. The second term accounts for change in chemical energy at T0 and change in sensible enthalpy

when the product is heated to temperature T.

Note: It can also be seen from First Law analysis (equation (4)) that enthalpies (and internal

energies) of reactants and products are not identical although they are at identical thermodynamic

states (equal T and p). Obviously, two properties are not sufficient to fully describe the state of themixture, but the exact composition must also be known.

5.5.2 Examples:

Examples for absolute enthalpy of CO2 at 800 K and enthalpy change of the reaction C+O2CO2 with reactants and products at 800 K are given in the lecture.

Answer: kJ/kmol195,380)800(hkJ/kmol,718,356)800(2 == KKhCO

A

B C

Eq. (6)


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5.6 STEADY FLOW COMBUSTION PROCESS

Fig. 5: Steady flow process

In most cases, the fuel composition, air/fuel ratio, inlet and outlet temperatures and pressures are

given. To obtain the heat flow rate the following steps should be carried out:

1. find the appropriate chemical equation,

2. write element balances for all elements,

3. apply first law and4. apply continuity of mass whenever necessary

Hint: First law analysis may be simpler on a molar basis.

5.6.1 Example Problem:

Methane and air enter the combustion chamber in separate streams at 25C and 1 atm. The flow

rate of CH4 is 1 kmol/s. The heat flow from the walls of the combustor is measured to be 20 MJ/s.

Calculate the amount of excess air needed to keep the product temperature down to 1100 K.

This example is calculated in the lecture.Answer: 131%

5.6.2 Adiabatic Flame Temperature

The combustion process is said to be adiabatic when the combustion chamber is isolated such that

no heat is lost to the surroundings. All the chemical energy released during the combustion

process is used to raise the temperature of the gases. In the absence of any work done by the

system, the adiabatic flame temperature is the maximum temperature that can be achieved and it

is a characteristic quantity for a certain fuel/oxidiser mixture. The adiabatic temperature is of high

significance in the design of combustion chambers since it gives a first estimate of the

temperatures that can be expected. Metallurgical constraints may limit the maximum allowabletemperature and adiabatic flame temperatures can be taken as a measure of the degree to which

additional excess air needs to be supplied to reduce the maximum temperature to acceptable

levels.

The first law (eq. 3) for a steady adiabatic flow process reduces to

=tsreacproducts

hnhntan

0 &&

Example:Find the adiabatic flame temperature of stoichiometric combustion of methane with air. Inlet fuel

flow rate, temperature and pressure are 1 kmol/s, 25C and 1 atm, respectively.This is example is given in the lecture.Answer: T=2126 K

Q&

productsn&p,T,Products,Combustion (cooling)

oxn&p,T,oxidizer,

Fuel, T, p, fueln&

Eq. (7)


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5.7 COMBUSTION PROCESS IN A CLOSED SYSTEM

Fig. 6: Combustion process in a cylinder

Combustion in closed systems can be treated similarly to combustion in flow processes. First Law

analysis for non-reacting and reacting processes do not differ if the correct internal energies are

taken. The total internal energy can be defined as

)( 1,25, atmCpTo

f uuuu +=

Overbar denotes molar quantities and again, molar based and mass based quantities are related by

)()( TMuTu = . The internal energy of formation at standard conditions, 0fu , can be obtained fromthe experiment shown in Fig. 6. If no work is done, if the vessel is cooled to T0=25C and the

final pressure is 1atm, then Q is a measure of 0fu .

The first law for closed reacting systems can then be written as

=

=

reactantsproducts

reactantsproducts

mumuWQ

ununWQ

In general, only enthalpy of formation is given in tables. However, internal energy of formation

can be recovered from enthalpy of formation using

Mpvhu

pvhu

=

=

and for ideal gases these relationships yield

TRhu

RThu

=

=

In particular

0

00

0

00

RThu

TRhu

ff

ff

=

=

p1, T1p2, T2

Q-W = ?

Q

W

Eq. (8)

Eq. (9)

Eq. (10)


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5.8 ENTHALPY OF REACTION & CALORIFIC VALUES

Enthalpy of formation refers to a particular species i and is needed in order to refer properties to

the same datum state when chemical reactions occur. In other words, enthalpy of formation

enables us to assign absolute enthalpy values to each substance at each temperature.

However, if species i is a fuel,

0

fh does notindicate how good a fuel is, i.e. how much energy isreleased as a result of combustion. For that purpose we define the

5.8.1 Standard Enthalpy of Reaction:

The standard enthalpy of reaction, 0h , is the difference in enthalpy of the products and reactantsat the standard state of 25C and 1 atm when complete combustion occurs in a flow process. The

standard enthalpy of reaction per kmol fuel is then

fuel

tsreacproducts

nhnhnh &&&

=

tan

0

Since 0h is evaluated at standard state all enthalpies equal 0h)(Th0

f0t += for both products and

reactants and 0h can be simplified as

fuel

tsreac

f

products

f nhnhnh &&&

=

tan

000

The enthalpy of reaction can easily be calculated for a particular fuel ifthe chemical composition

is known. For complex fuels, however, it is easier to find 0h experimentally.

Note: Values for standard enthalpy of reaction are often given per kg rather than per kmol.

5.8.2 Fuels Containing H

If complete combustion occurs all hydrogen content in the fuel will be converted to H2O. In most

cases temperatures are high enough that all H2O leaves the combustor as water vapour. If the

water condenses latent heat is set free and 0h will increase. It then follows from equation (11)that enthalpy of formation of liquid water needs to be smaller than enthalpy of formation of water

vapour. The difference equals the heat that is released during condensation. In general we can

deduce the relationship between enthalpies of formation of the same species in liquid and gaseous

phase as

Cfg,25

0

gf,

0

lf, hhh Note that this is not an exact equality. The liquid product water is subcooled (pressure generally

about 1 atm) and its enthalpy is belowCf,25

h o which assumes a saturation pressure of 0.032 bar.

Therefore, equation (12) underpredicts the difference between lf,h and gf,h

Some enthalpies of formation for species in liquid and gas phase are listed in Table 3 of these

notes.

Example: H2O

CfgOHlfgf hMhh + 25,20,

0,

-241820 kJ/kmol -285830 kJ/kmol +18 kg/kmol2441.5 kJ/kg = -241883 kJ/kmol

The error introduced is 0.02%.

Eq. (11)

Eq. (12)


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Note: Often, water should not be allowed to condense since liquid water may cause corrosion or

the impact of water droplets may cause damage to the casing if velocities are high. Therefore, all

products should be kept in the gas phase, i.e. above dew point. The partial pressure can be

obtained from the product analysis. For example, methane/air combustion with 200% excess air

yields nH2O=2 and nproducts=29.571. Therefore

0676.022 ==total

OH

products

OH

pp

nn

and Tsat(0.0676 bar)=36C. The products should be kept above 36C.

Traditionally, engineers dislike negative numbers ( 0h is negative for exothermic reactions).Therefore, two new quantities are commonly in use.

5.8.3 Net Calorific Value

The Net Calorific Value (NCV) (or lower heating value, LHV, in the U.S.) is the negative of

0h when the product H2O is vapour (used mainly in the UK and Europe)

5.8.4 Gross Calorific Value

The Gross Calorific Value (GCV) (or higher heating value, HHV, in the U.S.) is the negative of

0h when the product H2O is liquid (used mainly in the U.S.)

5.9 SFEE USING STANDARD ENTHALPY OF REACTION

This is an alternative method to using equation (3) provided that the standard enthalpy of reaction

of the fuel is known and the combustion is complete. The reaction can be split into three sub-processes, see Fig. 7.

A B C

Reactants Reactants Products Products

p1, T1 p0, T0 p0, T0 p2, T2

Fig. 7: Combustion process in three stages

First the reactants are cooled to standard temperature (A). The combustion takes place in B understandard conditions and in C, the products are heated to the exit temperature. Processes A and C

involve pure substances only, i.e. enthalpy changes are differences in sensible enthalpy. Enthalpy

changes in B correspond to 0h provided the combustion is complete. Then the SFEE can bewritten as

++=products

fuelfuel

tsreac

hhnhnhhnWQ )()( 020tan

10&&&&&

Note: Compare this process with Fig. 4.

Eq. (13)


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5.10 COMBUSTION EFFICIENCY

The combustion efficiency compares the heat released in a combustion process with the heat that

is potentially available from complete combustion with products and reactants all at 25C, 1 atm.The efficiency is defined as

( )0hm

Q

fuel

comb

&

&

h0 may be based on liquid or vapour H2O, i.e. h0 is either NCV or GCV. In 2M you shouldalways assume that h0 is based on NCV, however, it should usually be stated if NCV or GCVwas used to compute the efficiency.

The thermal efficiency of a power plant has been defined as the cycle efficiency

in

net

Q

W

&

&

The use of the heat transfer to the working fluid, Q& , did not take into account of any loss of

energy during the combustion process which is necessary to provide Q& . To compare netW& with

the energy which is potentially available from complete combustion of a fuel, we define the

thermal efficiency of the plant as

fuelfuel

netth

NCVm

W

&

&

The NCV is usually taken here since water vapour is not normally allowed to condense. As an

alternative to thermal efficiency it is common to use the specific fuel consumption, sfc, which is

net

fuel

W

mnconsumptiofuelspecific

&

&=

5.11 REAL FUELS

Application of the first law of thermodynamics to combustion of real fuels is best demonstrated

by an example.

5.11.1 Problem Description

Determine the heat transfer from a combustion chamber for a combustion process of 100kg/s coalwith 30% excess air. Heat losses are 2% of the net calorific value and products leave the

combustion chamber at 500 C.

Fig. 7: Steady state coal combustion with excess air.

Eq. (14)

Eq. (15)

Eq. (16)

Combustion

Coal, 25C, 1atm

Air, 25C, 1atm

CO2, H2O, N2, O2, 500C, 1atm

lossesQ&

outQ&


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As outlined in Section 5.6 the following steps are necessary for a complete analysis of the present

problem.

1. find the appropriate chemical equation

2. write the element balance for all elements

3. apply first law

First, we need to know the fuel composition. Table 5 lists the composition of different fuels by

mass of carbon, hydrogen, oxygen and water.

Table 5: Typical fuel composition of different fuels by mass, H2O is in the form of moisture

Note: Fuel composition and NCV are given on a mass basis and none of the ni,fnor nox is known.

This is a typical example where mass based quantities should be used throughout.

Conversion into molar-based units is not necessary and the following approach will

facilitate the analysis. However, conversion can be quickly done by division/multiplication

with the molar mass of species i.

Due to the excess air in the inlet stream it can be assumed that combustion is complete and all C is

converted into CO2 and all H is converted into H2O.

1. The chemical equation can be written as

nC,fC + nH.fH + nO,fO + nH2O,fH2O +nox (O2+79/21 N2)

nCO2,out CO2 + nH2O,out H2O + nO2,out O2 + nN2,out N2

2. Element balance on a mass basis

C:

Conversion of C in fuel to CO2: C + O2 CO2

Mass balance for 1 kmol of CO2: 12 + 32 = 44

Mass balance for 1 kg of C atoms: 1 + 32/12 = 44/12 (A)

H:

Conversion of H in fuel to H2O: 2H + 0.5 O2 H2O

Mass balance for 1 kmol of H2O: 2 + 16 = 18

Mass balance for 1 kg of H atoms: 1 + 16/2 = 18/2 (B)

Content in mass % Wood Peat Coal Oil Natural gas

C 42.5 47.2 79.8 86.8 74.9

H 5.1 4.8 4.8 13.2 25.1

O 37.4 28.0 10.4 0 0H2O 15 20 5.0 0 0

Total 100.0 100.0 100.0 100.0 100.0

NCV [kJ/kg] 13,500 16,300 30,900 42,720 50,020

GCV [kJ/kg] 15,000 17,850 32,050 45,600 55,500


15/21


15

During every single chemical reaction, total mass must be conserved. Hence, the mass balance for

carbon [see (A)] results in fCCCOCO mmmm ,2,212

32&&&& == . From the mass balance for hydrogen

[see (B)] we get fHHOHHO mmmm ,2,22

16&&&& == . The second subscripts C and H denote that this

oxygen mass flow is needed for conversion of C to CO2 and H to H2O, respectively.

O:

Mass flow rate of O2 for stoichiometric combustion: HOCOinO mmm ,2,2,2 &&& +=

Mass flow rate of O2 in air for stoich. combustion: fOHOCOoxO mmmm ,,2,2,2 &&&& +=

Mass flow rate of O2 in air with 30% excess air: )(3.1 ,,2,2,2 fOHOCOoxO mmmm &&&& +=

N:

Mass flux of N2 for stoichiometric combustion: oxOoutNinN mmm ,2,2,2 2.23/8.76 &&& ==

Mass flux of N2 for combustion with excess air: oxOoutNinN mmm ,2,2,2 2.23/8.76 &&& == using

corresponding value for oxOm ,2& from O-balance.

The fuel water is assumed not to take part in the reaction. The liquid water will evaporate and

form water vapour.

From above element mass balances all mass flows in and out of the combustion chamber can be

calculated. The fuel mass flow rates H,fO,fH,fC,f mandm,m,m &&&& are known quantities and

oxO,N2outO,outH2O,outCO2, mandm,m,m,m &&&&& can be found from

kg/s2.10362.23

8.76

kg/s0.3132

16

12

323.1

kg/s2.722

16

12

323.0

kg/s2.482

18

kg/s6.29212

44

,2

,,,,

,,,,

,2,,2

,,2

==

=

+=

=

+=

=+=

==

oxON

fOfHfCoxO

fOfHfCoutO

fOHfHoutOH

fCoutCO

mm

mmmm

mmmm

mmm

mm

&&

&&&&

&&&&

&&&

&&

3. SFEE

We can either apply equation (3)

++=treac

atmCinf

productsatmCoutf

hThhmhThhmWQ

tan1,25

0

1,25

0 )()( oo &&&&

or equation (13)

++=products

outfuelfuel

tsreac

in hThmhmThhmWQ ))(())(( 00tan

0&&&&&

The latter may be more suitable for the current problem since h0 is known. Equation (13) canthen be written as


16/21


16

) ) ++=products

atmCatmCcoalcoal

tsreac

atmCatmClossesouthhmhmhhmWQQ

1,251,500,0

tan

1,251,25 oooo&&&&&&

where Q& is split into outQ& and lossesQ

& and W& equals zero. The standard enthalpy of reaction is

kJ/kg900,300 == NCVh

and the heat losses are

kJ/s800,61%2 == NCVmQ fuellosses &&

The change of sensible enthalpy of the reactants (first RHS term) is zero and the change in

sensible enthalpy of the products is

kJ746,719kJ/s6.5082.10360.4652.726.9432.487.4806.292, =+++=prodsH&

Finally, the heat transfer from the combustion chamber can be calculated as

MJ/s2281

kJ/s746719)900,30(100800,61

=++=Q&

5.12 SECOND LAW ANALYSIS OF COMBUSTION PROCESSES

5.12.1 Second Law for Steady Flow Processes

The second law of thermodynamics for reversible non-reacting steady flow processes can be

expressed as

==surr

surriiee

revT

QSsmsm

T

Q &&&&&

and,

where e denotes the exit state, i denotes the inlet state and surrrefers to the surroundings. If the

process is irreversible then the equality needs to be replaced by the inequality

=+

iieegen

iiee

smsmT

QS

smsmTQ

&&&

&

&&&

The positive entropy generation term, genS& , has been introduced. This term quantifies the

irreversibilities of the process1. Equations (17-19) are essentially identical to equation (7-3) in the

1M Thermodynamics lecture notes.

Similar to equation (19), the second law can be written for a steady combustion process as

1Example 1: Consider a process in an adiabatic and reversible turbine. No heat is exchanged over the boundaries and

the RHS of equation (18) is zero (adiabatic and reversible = isentropic). Consider now a process in an adiabatic

turbine with irreversibilities due to friction. No heat is exchange but s e>si. This change in entropy equals the entropygeneration.

Eq. (17)

Eq. (18)

Eq. (19)


17/21


17

=+tsreac

rr

products

ppgen smsmT

QS

tan

&&&

&

Remember that in 1M Thermodynamics only entropy changes have been calculated, e.g. for ideal

gases ( )1212 lnln ppRTTcmS p = && . However, in the presence of combustion entropy changes

cannot be expressed in the form of smS = && since species are not conserved. Absolute entropyneeds to be introduced to solve equation (20). This is similar to the definition of absolute enthalpy

that was needed to solve equation (3) (see Section 5.5 Definition of Standard State).

5.12.2 Absolute Entropy

For the definition of absolute entropy a common base is needed. You may remember that entropy

can be associated with order. Lower entropy is associated with a higher degree of order 2.

Therefore, the common base is at T = 0K. At absolute zero, the oscillations of the molecules

vanish and the molecules reach a state of maximum order.

Definition (engel and Boles): The entropy of a pure crystalline substance at absolute zero

temperature is zero.

The entropy with respect to the common base at T = 0K is called absolute entropy. Absolute

entropy values ),( 00

pTs are given in Table 6 for the reference pressure p= 1atm. The absolute

entropy for ideal gases at arbitrary T and p can be obtained from

0

0

0 ln),(),(p

pRpTspTs u=

Table 6: Absolute entropy for selected substances at p=1atm in kJ/kmol K

Temperature C(s) H2(g) N2(g) O2(g) CO(g) CO2(g) H2O(g) CH4(g)

0K 0 0 0 0 0 0220K 107.15 182.64 196.17 188.68 202.97 178.57

298K 5.74 130.68 191.61 205.04 197.65 213.80 188.83 186.16

300K 130.75 191.68 205.21 197.72 213.92 188.93

400K 139.11 200.07 213.76 206.13 225.23 198.67

500K 145.63 206.63 220.59 212.72 234.81 206.41

600K 150.97 212.07 226.35 218.20 243.20 212.92

700K 155.49 216.76 231.36 222.95 250.66 218.61

800K 159.44 220.91 235.81 227.16 257.41 223.69

900K 162.94 224.65 239.82 230.96 263.56 228.32

1000K 166.11 228.06 243.47 234.42 269.22 232.60

1100K 169.00 231.20 246.82 237.61 274.44 236.58

1200K 171.68 234.12 249.91 240.66 279.31 240.33

1500K 178.72 241.77 257.97 248.31 292.11 250.45

2000K 188.30 251.97 268.66 258.60 309.21 264.57

2500K 196.13 260.07 277.21 266.76 322.81 276.28

2Consider adiabatic mixing of two substances with identical physical properties in a closed container of volume, V.

At time t=t1, both substances are separated by a membrane and occupy exactly half the volume. The membrane isremoved and at time t=t2, both substances are perfectly mixed. Initial pressures and final total pressure are 1 bar.

Entropy changes can be obtained from ( )1,212 lnln2 ppRTTcSSS ipBAsys =+= . From the ideal gaslaw we obtain T2=T1 since piVi and mi are constant. Hence, 05.0ln2 >= RS sys which is consistent with theconcept that mixing of two substances increases disorder.

Eq. (20)

Eq. (21)


18/21


18

5.12.3 IrreversibilitiesThe first law for a fully reversible flow process is

dhwq revrev =

where d denotes the exact differential of a property and is the differential in a path function

( =2

1

12 hhdh but 2

1

12 www since work is not a property and it does not have a specific

value at a specific state).

Similarly, the first law for an irreversible process is

dhwq irrevirrev =

From the second law we get

T

q

T

qds

rev

=

and

Tdsqrev =

Dividing eq. (19) by m& gives for single inlet and single exit

TdsqsT irrevgen =+

If inlet and exit states are fixed (Tds = const) then qirrev < qrev and therefore it follows from thefirst law (equations 22-23) that

revirrev ww <

We need to introduce now a measure for the irreversibilities. The difference between the work

done by the reversible and the irreversible process, w = (wrev-wirrev), quantifies the work thatis lost due to irreversibilities. Subtracting eq. (23) from (22) we get

w=q

and q can be obtained from the difference of eqs. (25) and (26).

gensTqw ==

Obviously, the entropy generation term quantifies the losses due to irreversibilities. Sgen canbe obtained from equation (19).

What does it mean for a combustion process? Effects of irreversibilities due to combustion can be

best illustrated with an example:

Eq. (22)

Eq. (23)

Eq. (24)

Eq. (27)

Eq. (26)

Eq. (25)

Eq. (28)

Eq. (29)


19/21


19

5.12.4 Irreversibilities due to combustion of methane with oxygen

Fig 8: Entropy changes in steady state combustion process

The global reaction mechanism is CH4 + 2O2 CO2 + 2H2O and the molar flow rate iskmol/s1nCH4 =&

First, we need to know how much water condenses.

967.1033.0)1(032.0032.0 )(2)(2)(2)(225@)(2 ==+=== lOHgOHgOHgOH

tot

Csat

tot

gOH nnnnp

pn

n o

From a first law analysis we get the heat flow across the system boundaries.

[ ] [ ][ ]

[ ] [ ]( ) ( ) ( ) kJ/s878,888kJ/s02850,741)830,285(967.1820,241033.0520,3931

)()()()(

)()(

)()()()(

01

0

,01

0

,

0)(2)(

0

)(,)(

0)(2)(0

)(,)(020

,

)()()()(

22224444

2222

22222222

2244222222

-Q

ThThhnThThhn

ThThhn

ThThhnThThhn

hnhnhnhnhnWQ

OOOfOCHCHCHfCH

lOHlOHlOHflOH

gOHgOHgOHfgOHCOCOCOfCO

OOCHCHlOHgOHgOHgOHCOCO

=++=

++

++

+++=

++=

&

&&

&

&&

&&&&&&&

Using equation (19) and KkJ/kmol92.690 )(2 =lOHs

( ) ( ) ( )

( ) ( )

( ) ( )( )

( )

skJ/K8.2743

skJ/K15.298

878,88804.205216.1861

skJ/K92.69967.1032.0ln314.883.188033.0968.0ln314.880.2131

1,2981,298

1,298)033.1/033.0(,298)033.1/1(,298

2244

222222 )()()()(

=

++

++=

+

++=

+=

=

surrOOCHCH

lOHlOHgOHgOHCOCO

surrsys

sysgen

TQbKsnbKsn

bKsnbKsnbKsn

SS

T

QSS

&&&

&&&

&&

&&&

Here we have made use of the fact that the heat flow across the system boundaries is equivalent to

the heat absorbed by the surroundings. Therefore surrSTQ&& = . Since no work is done during

this process and the process is irreversible the work losses due to irreversilibilities amount to

kJ/s060,818kJ/s8.274315.298 === genSTW &&

This also means that the process could have provided 818,060 kJ/s work output if it was fullyreversible.

SsysCombustion

CH4, 25C, 1atm

O2, 25C, 1atmCO2, H2O, 25C, 1atm

Q&Surroundings

Ssurr


20/21


20

We will now consider an adiabatic process with identical inlet conditions as pictured in Fig. 9.

Fig 9: Entropy changes in adiabatic steady state combustion process

The exit temperature can be obtained from first law analysis:

[ ] [ ][ ] [ ]

( ) ( ) ( )

C3852K4150

)(2)(kJ/s310802

kJ/s02850,741)(820,2412)(520,39310

)()()()(

)()()()(

2

22

22

01

0

,01

0

,

0)(2)(

0

)(,)(02

0

,

)()(

22

22

22224444

22222222

22442222

o

&&

&&

&&&&&&

==

+=

+++=

++

+++=

+=

T

ThTh,

ThTh

ThThhnThThhn

ThThhnThThhn

hnhnhnhnWQ

OHCO

OHCO

OOOfOCHCHCHfCH

gOHgOHgOHfgOHCOCOCOfCO

OOCHCHgOHgOHCOCO

This temperature T2 has been obtained by trial and error. The entropy generation for the adiabatic

flow process can be obtained from the second law (eq. 19) as

( ) ( )

( ) ( )[ ] [ ]( )

skJ/K7.385

skJ/K1.20522.186167.0ln314.89.305233.0ln314.84.3541

01,K9821,K982

67.0,K415033.0,K4150

2244

2222

=

+=+

+=

=

bsnbsn

bsnbsn

T

QSS

OOCHCH

OHOHCOCO

sysgen

&&

&&

&&&

Then, the work losses due to irreversibilities are

kJ/s989,114kJ/s7.38515.298 === genSTW &&

These are the losses that can be associated with the combustion process itself.

Combustion irreversibilities are visualised in Fig. 10. If a fully reversible process existed that

converted the reactants to products at 25C then 818,060 kJ/s work could be gained. However,irreversibilities of the combustion process amount to 114,989 kJ/s and the amount of work that isavailable from the hot product gases is 818,060-114,989 kJ/s = 703,071 kJ/s. Now it becomes

clear why the second law analysis is needed to quantify losses of a combustion process. Using the

first law of thermodynamics we have no means of quantifying the loss of potential reversible

work due to combustion.

CH4, 25C, 1atm

O2, 25C, 1atm

CO2, H2O, ?? C, 1atm

Combustion

Ssurr=0


21/21


Fig 10: Theoretical available work from reactants and combustion products

Note: Combustion of methane with pure oxygen has been chosen as example problem due to its

simplicity. However, the resulting high exit temperatures result in dissociation of the combustion

products and the assumption of CO2 and H2O as the only product gases introduces significant

errors. In addition, losses in the order of 14% do not seem very large. The tutorial problem of

methane/air combustion is much more realistic and shows a dramatic increase in losses due to

irreversibilities of combustion.

Combustion

Reactants, CH4,O225C, 1atmTheoretical reversible

work888,878 kJ/s

Products, CO2,,H2O3852C, 1atm

Theoretical reversiblework

703,071 kJ/s

05 - 1 thermodynamics of combustion

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