Adiabatic flame temperature

In the study of combustion, there are two types of adiabatic flame temperature depending on how the process is completed, constant volume and constant pressure, describing the temperature the combustion products theoretically reach if no energy is lost to the outside environment.

The constant volume adiabatic flame temperature is the temperature that results from a complete combustion process that occurs without any work, heat transfer or changes in kinetic or potential energy. The constant pressure adiabatic flame temperature is the temperature that results from a complete combustion process that occurs without any heat transfer or changes in kinetic or potential energy. Its temperature is lower than the constant volume process because some of the energy is utilized to change the volume of the system (i.e., generate work).

Common flames

File:Propan Lewis.svg

Propane

File:Octane vector.svg

Octane

In daily life, the vast majority of flames one encounters are those of organic compounds including wood, wax, fat, common plastics, propane, and gasoline. The constant-pressure adiabatic flame temperature of such substances in air is in a relatively narrow range around 1950 °C. This is because, in terms of stoichiometry, the combustion of an organic compound with n carbons involves breaking roughly 2n C–H bonds, n C–C bonds, and 1.5n O₂ bonds to form roughly n CO₂ molecules and n H₂O molecules.

Because most combustion processes that happen naturally occur in the open air, there is nothing that confines the gas to a particular volume like the cylinder in an engine. As a result, these substances will burn at a constant pressure allowing the gas to expand during the process.

Common flame temperatures

Assuming initial atmospheric conditions (1 bar and 20 °C), the following table list the adiabatic flame temperature for various gases under constant pressure conditions. The temperatures mentioned here are for a stoichiometric fuel-oxidizer mixture (i.e. equivalence ratio φ = 1).

Note these are theoretical, not actual, flame temperatures produced by a flame that loses no heat. The closest will be the hottest part of a flame, where the combustion reaction is most efficient. This also assumes complete combustion (e.g. perfectly balanced, non-smokey, usually bluish flame)

Adiabatic flame temperature (constant pressure) of common gases/materials

Fuel	Oxidizer	$T_{ad}$ (°C)	$T_{ad}$ (°F)
Acetylene (C2H2)	Air	2500	4532
Acetylene (C2H2)	Oxygen	3480	6296
Butane (C4H10)	Air	1970	3578
Cyanogen (C2N2)	Oxygen	4525	8177
Dicyanoacetylene (C4N2)	Oxygen	4990	9010
Ethane (C2H6)	Air	1955	3551
Ethanol (C 2H 5OH)	Air	2082	3779
Gasoline	Air	2138	3880
Hydrogen (H2)	Air	2254	4089
Hydrogen (H2)	Oxygen	3200	5792
Methane (CH4)	Air	1950	3542
Natural gas	Air	1960	3562
Propane (C3H8)	Air	1980	3596
Propane (C3H8)	Oxygen	2526	4579
MAPP gas Methylacetylene (C3H4)	Air	2010	3650
MAPP gas Methylacetylene (C3H4)	Oxygen	2927	5301
Wood	Air	1980	3596
Kerosene	Air	2093	3801
Light fuel oil	Air	2104	3820
Medium fuel oil	Air	2101	3815
Heavy fuel oil	Air	2102	3817
Bituminous Coal	Air	2172	3943
Anthracite	Air	2180	3957
Anthracite	Oxygen	≈2900	≈5255

Thermodynamics

File:Adiabatic flame temperature (diagram).jpg

First law of thermodynamics for a closed reacting system

From the first law of thermodynamics for a closed reacting system we have,

${}_RQ_P - {}_RW_P = U_P - U_R$

where, ${}_RQ_P$ and ${}_RW_P$ are the heat and work transferred during the process respectively, and $U_R$ and $U_P$ are the internal energy of the reactants and products respectively. In the constant volume adiabatic flame temperature case, the volume of the system is held constant hence there is no work occurring,

${}_RW_P = \int\limits_R^P {pdV} = 0$

and there is no heat transfer because the process is defined to be adiabatic: ${}_RQ_P = 0$. As a result, the internal energy of the products is equal to the internal energy of the reactants: $U_P = U_R$. Because this is a closed system, the mass of the products and reactants is constant and the first law can be written on a mass basis,

$U_P = U_R \Rightarrow m_P u_P = m_R u_R \Rightarrow u_P = u_R$.

File:Urup.jpg

Internal energy versus temperature diagram illustrating closed system calculation

File:Hrhp.jpg

Enthalpy versus temperature diagram illustrating closed system calculation

In the constant pressure adiabatic flame temperature case, the pressure of the system is held constant which results in the following equation for the work,

${}_RW_P = \int\limits_R^P {pdV} = p\left( {V_P - V_R } \right)$

Again there is no heat transfer occurring because the process is defined to be adiabatic: ${}_RQ_P = 0$. From the first law, we find that,

$- p\left( {V_P - V_R } \right) = U_P - U_R \Rightarrow U_P + pV_P = U_R + pV_R$

Recalling the definition of enthalpy we recover: $H_P = H_R$. Because this is a closed system, the mass of the products and reactants is constant and the first law can be written on a mass basis,

$H_P = H_R \Rightarrow m_P h_P = m_R h_R \Rightarrow h_P = h_R$.

We see that the adiabatic flame temperature of the constant pressure process is lower than that of the constant volume process. This is because some of the energy released during combustion goes into changing the volume of the control system. One analogy that is commonly made between the two processes is through combustion in an internal combustion engine. For the constant volume adiabatic process, combustion is thought to occur instantaneously when the piston reaches the top of its apex (Otto cycle or constant volume cycle). For the constant pressure adiabatic process, while combustion is occurring the piston is moving in order to keep the pressure constant (Diesel cycle or constant pressure cycle).

File:Flametemp.jpg

Adiabatic flame temperatures and pressures as a function of stoichiometry (relative ratio of fuel and air)

File:Cvft.jpg

Constant volume flame temperature of a number of fuels

If we make the assumption that combustion goes to completion (i.e. $CO_2$ and $H_2O$), we can calculate the adiabatic flame temperature by hand either at stoichiometric conditions or lean of stoichiometry (excess air). This is because there are enough variables and molar equations to balance the left and right hand sides,

${\rm{C}}_\alpha {\rm{H}}_\beta {\rm{O}}_\gamma {\rm{N}}_\delta + \left( {a{\rm{O}}_{\rm{2}} + b{\rm{N}}_{\rm{2}} } \right) \to \nu _1 {\rm{CO}}_{\rm{2}} + \nu _2 {\rm{H}}_{\rm{2}} {\rm{O}} + \nu _3 {\rm{N}}_{\rm{2}} + \nu _4 {\rm{O}}_{\rm{2}}$

Rich of stoichiometry there are not enough variables because combustion cannot go to completion with at least $CO$ and $H_2$ needed for the molar balance (these are the most common incomplete products of combustion),

${\rm{C}}_\alpha {\rm{H}}_\beta {\rm{O}}_\gamma {\rm{N}}_\delta + \left( {a{\rm{O}}_{\rm{2}} + b{\rm{N}}_{\rm{2}} } \right) \to \nu _1 {\rm{CO}}_{\rm{2}} + \nu _2 {\rm{H}}_{\rm{2}} {\rm{O}} + \nu _3 {\rm{N}}_{\rm{2}} + \nu _5 {\rm{CO}} + \nu _6 {\rm{H}}_{\rm{2}}$

However, if we include the Water gas shift reaction,

${\rm{CO}}_{\rm{2}} + H_2 \Leftrightarrow {\rm{CO}} + {\rm{H}}_{\rm{2}} {\rm{O}}$

and use the equilibrium constant for this reaction, we will have enough variables to complete the calculation.

Different fuels with different levels of energy and molar constituents will have different adiabatic flame temperatures.

File:Constant pressure flame temperature of a number of fuels.jpg

Constant pressure flame temperature of a number of fuels

File:Nitro.jpg

Nitromethane versus isooctane flame temperature and pressure

We can see by the following figure why nitromethane (CH₃NO₂) is often used as a power boost for cars. Since each mole of nitromethane contains two moles of oxygen, it can burn much hotter because it provides its own oxidant along with fuel. This in turn allows it to build up more pressure during a constant volume process. The higher the pressure, the more force upon the piston creating more work and more power in the engine. It is interesting to note that it stays relatively hot rich of stoichiometry because it contains its own oxidant. However, continual running of an engine on nitromethane will eventually melt the piston and/or cylinder because of this higher temperature.

Effects of dissociation on adiabatic flame temperature

In real world applications, complete combustion does not typically occur. Chemistry dictates that dissociation and kinetics will change the relative constituents of the products. There are a number of programs available that can calculate the adiabatic flame temperature taking into account dissociation through equilibrium constants (Stanjan, NASA CEA, AFTP). The following figure illustrates that the effects of dissociation tend to lower the adiabatic flame temperature. This result can be explained through Le Chatelier's principle.

See also

• Flame speed

External links

General information

• Computation of adiabatic flame temperature
• Adiabatic flame temperature

Tables

• adiabatic flame temperature of hydrogen, methane, propane and octane with oxygen or air as oxidizers
• Temperature of a blue flame and common materials

Calculators

• Online adiabatic flame temperature calculator using Cantera
• Adiabatic flame temperature program
• Gaseq, program for performing chemical equilibrium calculations.
• Flame Temperature Calculator - Constant pressure bipropellant adiabatic combustion
• Adiabatic Flame Temperature calculator