Thermodynamics

Thermodynamic analysis in combustion and fire systems describes the equilibrium conditions for the system.  This means that there is no specification of how long it takes to go from the initial state (reactants) to the final state (products).  In other words, there is no rate information.  Thermodynamics can still be very useful in fire analysis.  As an example, imagine if a fixed volume of natural gas is released into a closed space which contained air.  Using thermodynamic principles we can calculate a final equilibrium pressure in the chamber and the maximum temperature of the flame.

Flame Temperatures

Use the Flame Calculators below to get some approximate numbers for the peak pressure and temperatures. Remember, thermodynamics assumes that the system reaches equilibrium, which can in reality take 1 millisecond or 4 days.  The first calculator is for contained flames.  The mixture of gas and fuel is held at a constant volume, and the pressure in the system is allowed to increase.  The second calculator is for open flames, where the pressure is held constant at
atmospheric and the temperature is allowed to increase. 

Contained Flame Calculator (Constant volume)

Choose Fuel   Volume % fuel  Volume % air   Peak Temperature (—C)   Peak Pressure (atm)

Open Flame Calculator (Constant pressure)

Choose Fuel   Volume % fuel      Volume % air   Peak Temperature (—C)   

The following tutorial gives you a better idea of what is required to get the numbers in the calculators.  The peak temperature found above is the temperature of a flame burning in a perfectly insulated container, which is known as the adiabatic flame temperature.  To find the adiabatic flame temperature, you basically apply the first law of thermodynamics to a specified fuel and air mixture and find out what the final equilibrium temperature is.  Mathematically, that is:

Sum over all reactants of [(mass of each reactant) * (enthalpy of each reactant)] =
Sum over all products of [(mass of each product) * (enthalpy of each product)]
where: enthalpy = enthalpy of formation + sensible enthalpy

sensible enthalpy =

Because the enthalpy of formation of the fuel gas is much higher than the enthalpy of formation of the products (i.e. there is much more chemical energy stored in the fuel than in the products), the temperature of the products is much higher than that of the reactants. 

The enthalpies of formation of the various chemical species are available in tables. 
T is temperature.  Tref  is the reference temperature used for the enthalpy of formation, generally 298 K.  Tactual is the temperature of the reactant or product.  For the reactants, use the inlet temperature.  You then have to solve the equation for the actual temperature of the products.  This value is the adiabatic flame temperature.  c is the specific heat of the gas.  For a constant pressure flame, use cp, the constant pressure specific heat.  For a constant volume flame, use cv.  (Recall that the difference between cp and cv is R, the ideal gas constant.)  The simplest way to do this calculation is to assume constant specific heats, so that the sensible enthalpy becomes c(Tactual+ Tref).  Then the equation is easily solved for Tactual, products:

(1)

There are higher levels of sophistication at which this calculation can be performed.  The next step from equation 1 is allowing the specific heat to be variable with temperature, and not simplifying the sensible enthalpy integral.  A simple root finding algorithm or mathematical analysis program is needed to solve these problems.  The most complete and accurate method for finding the adiabatic flame temperature is to find the equilibrium products of combustion at the same time as you find the combustion temperature.  As flame temperatures increase, various gases can form, including carbon monoxide, oxides of nitrogen and various ionized gases.  Computing the adiabatic flame temperature with full dissociation requires a specialized code, such as the NASA code CET89. 

For more detail on adiabatic flame calculations, you can look at almost any introductory combustion text, such as:

Glassman, I. (1996) Combustion, third ed., Academic Press, San Diego. 

Kuo, K. K. (1986) Principles of Combustion, John Wiley and Sons, New York.

Table 1 compares the final temperature for stoichiometric constant pressure combustion of various fuels using different methods of calculation. 

Reviewing the values in this table, several points jump out.  First, taking the step from constant to variable cp corrects a lot of the error in the simplest level of calculation for many fuels.  The temperature calculated from the variable cp calculation is low enough that the formation of ions and products of non-ideal combustion is restricted.  The exception to this is acetylene, which tends to burn hotter due to the high-energy triple bond in its structure.  For that fuel, the variable cp temperature is high enough that formation of other chemical species is important.  The second point is that most common hydrocarbon fuels have approximately the same flame temperature, independent of the weight of the fuel molecule.  Finally, keep in mind that these results are for premixed flames.  For various reasons, the temperature of the products of non-premixed flames, where the fuel and air never touch until they are near the flame, is lower and less easy to calculate.  Non-premixed flames are the most common practical configuration.  Non-premixed flame temperatures are often around 1500 K, but can vary quite a bit.  Note also that this is an adiabatic calculations, which means that heat losses from the flame are not accounted for.  Heat losses will reduce the flame temperature.  The adiabatic flame temperature is then

As mentioned above, this calculation can be done with constant volume containers (such as an enclosed room) as well as constant pressure.  The constant volume specific heat, cv, is lower than cp, so the final temperatures are higher.  The final pressure is allowed to vary, since the volume is constant, and is calculated from the ideal gas law as shown here:

The n values are the initial and final numbers of moles, which changes due to the chemical reaction which occurs.  This is generally not a huge effect, but it can be significant especially with larger fuel molecules.  Also keep in mind that the temperatures are absolute temperatures, i.e. Kelvin or Rankine rather than Celsius or Fahrenheit degrees.  As a rule of thumb, the final pressure is generally around 10 times the initial pressure. 

Thermodynamics is also used to examine the vaporization of solid and liquid fuels.  Take, for example, a container that is partially full of a liquid fuel.  This might be a tank of gasoline or diesel fuel.  We want to know if there will be a flammable mixture in the air above the fuel. 

First, it is important to understand the concept of vapor pressure.  Any liquid which is in contact with a gas will produce some vapor, as molecules of the liquid escape the gas and exist in a gaseous state.  Different liquids will produce different amounts of vapor at the same temperature.  As described by Boyle's law, the total pressure of the gases in the mixture above the fuel will be the sum of the partial pressures of the various constituent gases (oxygen, nitrogen, and some trace elements in the air, and the fuel vapor).  The partial pressure of the fuel is determined by the Clausius-Clapeyron equation, one form of which is:

You can solve this equation for the partial pressure of the vapor, pº, which is in mm Hg, (1 mm Hg = 133.3 Pa), as a function of temperature, in Kelvin.  E and F are constants that depend on the fuel.  Some examples:

The ratio of the partial pressure to the total pressure in the chamber is then equal to the mole fraction of fuel in the gas.  You can compare this fraction to experimentally determined upper and lower flammability limits.  These limits are given as a volume fraction of fuel gas and air mixed at the same pressure.  Percent volume at constant pressure and percent pressure at constant volume (the ratio of partial pressure to total pressure in a closed chamber) are both equal to the mole fraction.  The flammability limits for the compounds listed above are:

So, to find the range of temperatures over which the gas mixture above a pool of fuel is flammable at a given pressure, you set the partial pressure divided by the total pressure equal to the lower flammability limit.  If Patm is the total pressure of the gas mixture, then

Note that the upper and lower flammability limits in the chart are at 298 K, and there are changes in these limits with temperature.  However, these changes are small for temperatures near 298 K and have been ignored here for simplicity.

If Patm is 1 atm, (760 mm Hg), then the temperatures where the three above fuels produces flammable mixtures are: 

This means that the fuel-air mixture above a pool of methanol in a closed chamber at room temperature is flammable.  However, the mixture above a pool of n-decane would not be flammable, due to a lack of fuel in the air, and the mixture above a pool of n-pentane would not be flammable because there is too much fuel in the air. 

This last condition is fairly surprising, but true.  There can in fact be too much fuel mixed in air to have a flammable mixture.  However, don't test this at home.  To understand why not, consider that you have an empty tank which you half-fill with pentane.  One of two things are likely to happen.  First, you might fill it and then try to light it immediately, before enough liquid has evaporated to reach an equilibrium state.  There would be less fuel vapor in the mixture, possibly leading to a flammable condition. 

A second possibility is this - you fill the container with fuel and let it sit for a long time to allow everything to come to equilibrium, but then you open the cap to put the flame in.  In the tank you have too much fuel vapor, outside the tank you have essentially none, but somewhere near the cap, the outside air is diluting the fuel-air mixture from the tank, leading to a flammable mixture.  Either way, the tank might blow up, doing serious damage.  To see a description of the Bureau of Mines apparatus used to find the flammability limits, go to the experimental equipment section (link this). 

For a general discussion of the Clausius-Clapeyron equation, look at a thermodynamics text such as:

Van Wylen, G. J., and Sonntag, R. E. (1986) Fundamentals of Classical Thermodynamics, third ed. John Wiley and Sons, New York.

For a more specific look at vapor pressure and flammability limits in fire science, as well as charts of flammability limits and the values of E and F from the Clausius-Clapeyron equation, see:

Drysdale, D. (1998) An Introduction to Fire Dynamics, second ed. John Wiley and Sons, New York.

Return to top