Fuels are generally comprised of a number of different elements in a vast array of molecular forms. The main elements in flammable materials are Carbon C, Hydrogen, H, Oxygen, O, Nitrogen, N, and Sulphur with other trace elements to boot.

The components of a mixture (whether liquid, solid or gas) can take be expressed in many ways: volumetric, gravimetric, dry, wet, ash free, as supplied etc. etc. It is important right from the start to know exactly what these mean so that, when combustion calculations are carried out, we have the right numbers.


Gravimetric and Volumetric Ratios

The two fundamental methods of defining the quantities of mixture are by weight (gravimetric) or by volume (molar). For clarity, the following symbols are often (but not exclusively) used.

c i = the gravimetric fraction (mass fraction) of element / molecule i

y i =the volumetric fraction (molar fraction) of element / molecule i

Of course, the sum of mass or mole fractions for all species in a mixture will be unity:

To convert from mole fraction to mass fraction we must know the molecular weight, M (kg/kmol), of the individual chemical species. We then calculate the mass of all species by multiplying the volume fraction by the molecular weight. Then we divide the mass of the individual element/molecule by the total mass.

- Molar to mass


- Mass to molar

So, if we are given a coal of composition C 95%, H 4%, Cl 0.5%, S 0.5% by weight. Then the molar composition is calculated thus:


c i (mass)

c i / Mi

y i (volumetric)





















The result doesn't quite add up to 100% due to rounding errors. This should be sorted out manually (i.e. fiddle the figures).

IMPORTANT: It is virtually always the volumetric compositions that are used in chemical equations since these are related to the number of molecules of the particular species involved. However, it is virtually always the gravimetric composition that is given, since this is the easiest to measure. Before attempting to do any further calculation make sure that you are clear which fraction you require.



A mixture ratio is specified wet or dry. For example, combustion products are often quoted dry since many analysers do not operate in the presence of water vapour which is therefore condensed out of the sample. To calculate a dry mixture ratio, the sum of the mixture ratio of all other molecules except water is calculated. Each mixture ratio is then re-evaluated by division with this number.

Example: A measurement of combustion gases reveals the following ratios: CO2 - 12% vol wet, H2O - 10% vol wet, O2 - 10% vol wet, N2 balance.

Calculate the volume fractions on a dry basis.

The summation of all dry mixture ratios is 12 +10 + 68 = 90%

Thus the dry mixture ratios are:

Molecule, i

y i dry

y i wet



12/90 = 13.33%



10/90 = 11.11%



68/90 = 75.55%




It should be noted that once the mixture is quoted on a dry basis, from that point all information about water in the stream is lost and further, wet mass / volumetric conversions are impossible unless you know how much water was present before it was removed.


Density of a Mixture

Density of any mixture (solid, liquid or gas) is defined as the weight per unit volume. For a gas mixture, we can approximate the density using the ideal gas law:

Where Ro is the Universal gas constant (8314.4J/kmolK), V is volume (m3), n is no of kmoles of gas.This may also be expressed in terms of density.

If we have a gas mixture, such as combustion products, we must calculate the average molecular weight in order to perform the calculation.

Note the use of molar / volumetric fractions here.


Mixture Ratios

One of the most vital operating parameters of a combustion system is the mixture ratio. That is the ratio, either locally or overall, in which the fuel and air are present in the system. This will effect the furnace performance, life, efficiency and pollution characteristics.

In the case where air and fuel are initially separated (a diffusion flame or non-premixed combustion) there is an infinite range of ratios ranging from pure fuel to pure air. In the case where fuel and air are initially mixed, there is only one mixture ratio throughout. How do we define this mixture ratio?

Answer. In many different ways.

1) Air / Fuel Ratio (AFR)

This is the most common method of mixture definition and is simply the ratio by mass of the air and fuel at the point of interest. It is expressed as a number e.g. 16.1 or as a ratio e.g. 16.1:1.

To give an indication of the kind of values expected. The stoichiometric AFR's for some fuels is given below.

Methane » 17:1

Kerosine / Gasoline » 15:1

Hydrogen » 35:1

If the actual value is higher than this there is more air than required for combustion and the system is said to be "lean". If the value is less there is not enough air and the system is said to be "rich".

2) Fuel / Air Ratio (FAR)

This is the inverse of the AFR but is not so common because in the majority of cases, it is significantly less than unity.

Stoichiometric FAR's.

Methane » 0.0588:1

Kero / Gaso » 0.0666:1

Hydrogen » 0.0286:1

3) Equivalence Ratio, f

This is also a very common one. It is defined as the ratio between the actual FAR and the stoichiometric FAR at a given location. Hence, if we have a stoichiometric mixture ratio, then the equivalence ratio = 1. If the mixture is lean, the value is less than one. If the mixture is rich, the value is greater than one. This value is somewhat more difficult to calculate than others but is ideal for situations where the fuel might change (dual fuel burners for example). It's value is independent of the fuel unlike AFR or FAR where the stoichiometric value is a function also of the fuel itself.

Equivalence Ratio, f = FAR / FARstoich = AFRstoich / AFR


4) Excess Air, XSA

The majority of combustion systems operate with a slight excess of air. This prevents the formation of products of incomplete combustion (also termed unburned hydrocarbons, UHC, or volatile organic components, VOC) which can be very toxic and are legislated against. The "excess air" is defined as the percentage of air in the system by mass which is surplus to requirements for complete combustion.

This property has no meaning if the system is rich

5) Mixture Fraction, x

This is a relative newcomer to the field of combustion and has been derived by combustion modellers who need to have property which is bounded. The problem with all the parameters so far is that they have a value of infinity at either the pure air or pure fuel side (AFR, FAR, f ) or they do not map the complete mixture spectrum (XSA). This is no good to computer modellers since computers get very upset when they try to calculate infinity.

Mixture fraction is defined as the ratio by mass of mixture which originated from the fuel stream as opposed to the oxidant stream.

In the air stream, AFR ® ¥ and so x = 0

In the fuel stream AFR = 0 and so x = 1

Stoichiometric values of mixture fraction are as follows.

Methane = 0.0555

Kerosine = 0.0625

Hydrogen = 0.0278

These values are close but not the same as FAR.



Pollutant emissions are heavily legislated against and their concentrations are reported in a variety of different and sometimes confusing ways:

Normalised Emissions

The difficulty with gaseous emissions is the effect of dilution. An unscrupulous operator could invest in a large fan to dilute the combustion products to effectively lower the pollutant level instead of expensive abatement equipment. Emissions are therefore normalised to a standard oxygen level (3, 6, 11, or 15%) to prevent this from happening.

To correct an emission to a standard requires knowledge of the excess oxygen already present. Imagine 1m3 of the products. It contains [O2]m m3 of oxygen and [P] m3 of pollutant. We must dilute it with an extra volume V of air to normalise it to an oxygen level of [O2]n.

The new oxygen level may be calculated as follows:

Hence the new volume of gas = 1+V :

and so the corrected pollutant concentration [P]n may be calculated as:


Emissions are often measured and reported in parts per million (ppm) by volume but some standards require a measurement in mg/m3 at standard conditions. To convert we must realise that 1m3 of gas at STP contains 1/22.4 kmoles of gas (pV=nRoT). Hence if the volumetric concentration is given by e (m3/m3) and the mass emission is given by E (kg/m3). Then:

NB: mass emission is dependent on temperature and pressure since mass remains the same whilst volume changes. The following correction must be used to convert from one set of conditions to another:



Combustion is simply an exothermic chemical reaction between air and fuel to form CO2 and H2O. The reaction may be written in the form of a chemical equation like any other reaction. The chemical equation describes the relationship between the molecules of fuel and oxidant and hence volumetric ratios must be used and not gravimetric.

Simple Case - the reaction between propane and air.

fuel air products

The value of A determines the ratio between air and fuel (AFR). a, b, c, d are unknowns which may be calculated by conservation of mass. The number of atoms of each element on the right and left hand side of the arrows must balance.


3 = a


8 = 2b so b = 4


2A = 2a + b + 2c


2A = 6+4+2c so c = A - 5


2 x 3.76 x A = 2d so d = 3.76A

The resulting equation therefore becomes:

A mixture is said to be "stoichiometric" if there is exactly enough (but no more) oxidant to fully oxidise the fuel. This point is unique to each fuel and in practice elusive since at flame temperatures we also have the phenomenon of dissociation (see later). Theoretically at the point of stoichiometric there are no UHC's in the products (not shown in the above equation) and there is no oxygen. Hence:

A - 5 = 0 Þ A = 5

The stoichiometric equation will thus be as follows:


In practice, stoichiometric is not the best operating condition for a combustion chamber due to dissociation effects. If we operate rich of stoichiometric, we have poisonous UHC's in the exhaust and waste a lot of good fuel. If we run lean, the exhaust is clear of toxic substances but we are effectively heating up a lot of excess air. In practice, furnaces are run with a small amount of excess air to give harmless but still hot products. Typically values may vary from 1 to 6% depending on the burner, fuel and application. It is very important (financially as well as from an engineering standpoint) to know how much excess air is present.

How do we calculate the excess air levels from the exhaust composition of a burner?

The Relationship between Excess Oxygen and Mixture Ratio

The exhaust oxygen concentration is one of the easiest and most common exhaust gases to measure (see later). Due to the analysers used it is measured dry (i.e. the water is removed from the sample).

Let's say we measure the % oxygen in the exhaust by volume = [O2] %vol dry

Now: [O2] = No of moles of oxygen / Total No. of molecules.

- Water not included because it's dry.

Rearranging we get:

- This gives us the number of "molecules" of air

In order to convert it into a meaningful value we must do further algebra. For example AFR:

AFR = Air fuel ratio by weight, hence:

So, from the measurement of oxygen in the exhaust we can determine the operating AFR of the furnace. This is a much more accurate measure of the stoichiometry of a flame because if flow metering is used, errors on both air and fuel meters effectively double when calculating the ratio. The measurement of oxygen also takes into account the fact that the air and fuel may not be perfectly mixed.

A similar calculation may be done with the concentration of CO2.



A combustion system may be treated thermodynamically like any other. It obeys the same laws and is under the same restrictions. The main difference, however, comes about from the enthalpy of formation of the species, D Hf . When we consider systems with no chemical change, this value is not an issue since the material under consideration remains the same. In this case we only need to consider the "sensible enthalpy" of the system. I.E. the energy required to take it from one temperature state to another. With chemical reaction, the internal energy of the molecular bonds of the molecule must also be considered.

Consider the simplest case of a combustion process in which heat and work are zero.

The fuel and air at energy state, 1, react to from product at energy state, 2. In the process there is a change of chemical structure and bonds are broken and rejoined.

The first law of thermodynamics states that energy cannot be created or destroyed.

Q + W = 0

In this system this is upheld since Q = W = 0

In order to determine the physical state of the reactants we must consider what is actually happening. There is a procedure of doing this to prevent endless thermodynamic tables being calculated. We consider the reaction proceding in stages:

a) The reactants are taken to a datum temperature, To. Usually 298K

b) The reactants are broken down to their constituent elements in their natural forms.

c) The products are formed from their constituent elements in their natural forms.

d) Take products are brought to their final temperature, T.

This is easier to understand in graphical form:


The enthalpy exchanges are shown. The enthalpy of reactants and products of course must be equal to obey the first law. However, there is a difference between the heats of formation of reactants and products. It is lower for the products than reactants (i.e. the reaction is exothermic). The balance is therefore given as sensible enthalpy of the products which, thus, end up at state 2 at a higher temperature. The opposite is true for an endothermic reaction.

NB: It is VERY important to know the physical state of the products and reactants as this will influennce the thermodynamics. For example the heat of formation of products will be less if the water is in the liquid phase, likewise so will the heat of formation of the reactants.

We can write the heat balance based on enthalpies as follows:

In reality, the reactants and products are composed of a mixture of species. We must therefore calculate the above as a summation over all constituents.

However, the heats of formation of every possible molecule available are not always to hand. If you consider the energy diagram above, a more easily measureable quantity would be the difference between the overall heats of formation (assuming complete combustion). This would then represent the sensible enthalpy rise of the products. This quantity is the most commonly used in combustion and is referred to as the Calorific Value (CV).

There are two values commonly used:

Gross Calorific Value (GCV) - Water in products in liquid phase.

Lower Calorific Value (LCV) - Water in products in the vapour phase.

The advantage of this parameter is that the heats of formation of all reactants and products are combined in a single value. For complete combustion, this simplifies the. Vis:

If the products also contain combustibles, then an additional term must be included in the right hand side of the equation to include product calorific value.

Calorific value is easily measured as the heat loss required to reduce the products of complete combustion of the fuel of interest to 298K. Since the heats of formation of Oxygen and Nitrogen are 0, the amout or composition of oxidant is unimportant. The experiments are thus performed with excess oxygen. Again, it is important to know what phase the fuel is in.


In order to proceed we must be able to calculate the sensible enthalpy of the combustion mixtures as a function of temperature. The values are tabulated for each gaseous species in most thermodynamic tables books (Steam tables). However, this is not the best form to use. It is easier to use curve fitted values of the tabulated data.

The following equations will be of great use in combustion calculations.

H (CO2) = 0.00458 (D T)2 + 45.59 D T

H (H2O) = 0.00502 (D T)2 + 34.03 D T

H (H2) = 0.00172 (D T)2 + 28.24 D T

H (CO) = 0.00188 (D T)2 + 30.03 D T

H (O2) = 0.00196(D T)2 + 31.38 D T

H (N2) = 0.00190 (D T)2 + 29.66 D T


D T is the temperature with relation to 298.15K D T = T -298.15 (K)

H is the enthalpy of the species (in vapour form) (kJ/kmole)

IMPORTANT: The relations are only valid between temperatures of 298.15 to 2600 K


Calculate the adiabatic flame temperature of propane and air at normal temperature and pressure when mixed to stoichiometric proportions given the following data. CV of propane = 46.4 MJ /kg

We have the stoichiometric equation from before

Remeber the energy balance

The first thing to do is to calculate the left hand side of the equation. This is easy since we know the CV and the reactants are at 298K. Hence the sensible enthalpy = 0. We must ensure that units are maintained. The CV is given in MJ/kg but we must have it terms of MJ/kmole. Hence multiply by the molecular mass of propane = 44 kg/kmol

LHS = 44 ´ 46.4 = 2041.6 MJ


We now have to balance this with the right hand side of the equation. The products are at a higher temperature than 298 K so we must use the enthalpy relations. However, we are assuming complete combustion so the CV of the products = 0. Make a table as below:

Enthapies in kJ/kmol






0.00458 (D T)2 + 45.59 D T



0.00502 (D T)2 + 34.03 D T



0.00190 (D T)2 + 29.66 D T


Multiply through to get a general quadratic equation:

RHS = 0.0695 (D T)2 + 830.5D T kJ

We can now balance the equation: LHS = RHS. Be careful with the units.

0.0695 (D T)2 + 830.5D T - 2.0416´ 106 = 0

Calculate the roots of the quadratic equation:

D T = 2091 K

So the temperature of the products = 2091 +298 = 2389 K

You can see the importance of the amount of air now. The 18.8 moles of Nitrogen play an very important role in keeping the temperature down. If pure oxygen were used instead of air, the temperatures would be huge.

The method above works very well in all circumstances.



The flame temperature calculations above give the maximum temperature rise thermodynamically possible for that system. In practice this is never achived. Firstly, there are the usual heat losses which cause problems but secondly and more importantly, there is the problem of dissociation. That is, the reaction does not go to completion but finds an equilibrium point.

To understand this, we must consider the chemical aspects of combustion. When fuel and air mix, they do not instantaneously convert but do so at a reaction rate which in the simplest terms may be described by the Arhennius rate equation:

Where E is the minimum joint energy of the colliding molecules for a reaction to ensue. In the constant, A, there are factors relating to the chance of two molecules colliding and others on the particular orientation which may reduce the chance of an effective collision.

In full, the reaction rate is governed by the frequency at which the two species collide (which in turn depends on their partial pressures) and the proportion of collisions which are fruitful which is descibed by the exponential. Hence:

NB: partial pressure = mole fraction x overall pressure.

In a chemical reaction such as the oxidation of CO

there is a reaction rate associated with the reverse (oxidation) reaction, but there is also a reaction rate for the forward (reduction) reaction. The chemistry dictates that the reverse reaction will be far stronger than the forward but the forward still occurs. In practice every chemical system reaches a stable point when the forward and reverse reactions balance - chemical equilibrium. NB: the reactions are still occuring on a microscopic scale and will continue to do so, but macroscopically the concentrations remain constant.

Each reaction has a unique equilibrium constant which describes as a function of temperature the relative importance of the forward and reverse reactions in terms of the partial pressures of each component.

For this reaction, the equilibrium constant is defined as follows:

The oxygen is taken to the power 1/2 because only 1/2 a molecule of oxygen is required for the reaction. This is generally true for all reactions and numbers of molecules.

For each reaction, K is tabulated against temperature. It is independent of pressure. The other equilibria important with combustion gases are given below.

The final equilibrium reaction is known as the "water gas" reaction and is a combination of the CO2 and H2O equilibria above. It only occurs in fuel rich conditions where there is a shortage of oxygen. The oxygen liberated in the CO2 dissociation is instantly used in the creation of H2O. Goodger gives an excellent method of calculating general flame equilibria. Below is only a very simple calculation for a CO flame to demonstate the method.


The products of stoichiometric combustion of carbon monoxide and oxygen are held at a temperature of 2400K. Calculate the equilibrium composition of the products at pressures of 1atm and 100atm.

We have the chemical equation:

Total no of moles in exhaust = 1+a/2

Hence pCO2 = P x (1 - a) / (1+a/2)

pCO = P x a/(1+a/2)

pO2 = P x (a/2) / (1+a/2)


Based on the equilibrium between CO and CO2 (shown above) we have:

Using algebra, this simplifies to:

Hence, we can make an iterative equation as follows

First look up the value of K at 2400K. From the tables:

Hence, K = 47.73

First consider the 1atm case, P = 1atm.

So we begin with an assumption and iterate to find the right answer.

Guess a

New a







Covergence at a value of a of 0.0912

Hence the carbon dioxide in the exhaust has dissociated by almost 10% vol into CO.

Repeat for 100atm

Same procedure as before

Guess a

New a







Convergence at a = 0.0204. Hence at the higher pressure there is less dissociation.

The cause of this is the effect of pressure forcing the reaction over to the side of fewest molecules (ie carbon dioxide). If you look at the dissociation reaction, the number of molecules on the RHS is 1, on the LHS, 1.5. As pressure increases the reaction is forced over to the side which has the least molecules and hence volume.

In either case, though, a substantial part of the fuel has not been converted to products. This results in a higher heat of formation of the products and hence a lower sensible enthalpy (i.e. lower temperature).

It is important to note that the effects of dissociation cannot be ignored at temperatures in excess of 1600K.

In this example, the temperature of the mixture was specified previously. In reality we would not know it and would have to perform an iteration as before. This is extremely time consuming. Also, the above example is very simple in terms of the fuel and oxidant. The calculation can involve many different equilibria at a time and is extremely complex. Any practical equilibrium flame temperature calculation is done by computer and there a number of packages available.



It has been shown above how to calculate theoretical temperature rises and percentage dissociation for combustion products. The effect of dissociation is to effectively decrease the amount of fuel burned or the degree to which it is burned. We must have some method of defining the efficiency of combustion to take this into account. This is achieved via the derivation of combustion efficiciency. This parameter describes the efficiency of conversion of the chemical energy of the fuel into sensible enthalpy in the products and should under no circumstance be confused with thermal efficiency in work transfer applications.

Unfortunately, unlike, thermal efficiency, this is an ill defined parameter. Different authors / institutions use different definitions. The two most common are listed below:

1) Based on Heat Release:

This method relates the theoretical heat release to the actual heat release:

The maximum heat release is simply related to the calorific value (Net usually since the combustion gases are hot). Actual heat release is difficult to measure but we can measure exhaust gas composition which gives information on products of incomplete combustion. The main components are CO and UHC's. The quantity of these gives indication of the actual heat release.

actual heat release = theoretical heat release - losses from prods of incomplete comb.

Consider carbon monoxide:

If we have the presence of CO in the exhaust, the extra sensible enthalpy released if it were to react to completion would be:

Now the maximum heat release =

So the efficiency of combustion may be calculated as 1 minus the ratio:

Using exactly the same technique we can calculate the value for any other product of incomplete combustion. It should also be noted that the value of AFR above may be calculated based of [O2] or [CO2]. This makes a very convenient method of determining the efficiency just based on gas analysis. NB: the unburned products will affect the calculation of AFR in this way but this will be minor for values normally experienced (i.e. less than 1%vol)


2) Based on Temperature Rise

An alternative method is to base the calculation on the difference between theoretical and actual temperature rise. This method is only applicable to devices where there is a well defined stoichiometry and where there is no heat transfer before the temperature measurement stage.

This method requires knowledge of the theoretical temperature rise which can be quite complicated to calculate if dissociation is important. Hence tables of heat release vs. FAR are often used.


This section of the course is the very basics of combustion thermochemistry which has been understood for 100 years. Most good engineering thermodynamics textbooks contain the information. Some particularly good ones are listed below.


Goodger E.M. (1977) "Combustion calculations, Theory, worked examples and problems", Macmillan Press, London

Rogers and Mayhew (1980) "Engineering Thermodynamics work and heat transfer", 3rd Ed. , Longman.

Glassman I (1977), "Combustion", Academic Press

Chigier N (1981) "Energy, Combustion and Environment", McGraw Hill


Rogers and Mayhew "Thermodynamics and Transport Properties of Fluids", Blackwell

JANNAF Thermochemical data , Dow Chemical Co. , Michegan. National Bureau of Standards Circ. C461 (1947).