 Adiabatic Flame Temperature ### Solving Energy Balances for Implicit Terms

So far, we have solved energy balances for Q or Ws. In reacting systems, it is common to measure the heat and infer from this the extent of reaction or the outlet condition. There is no new principle involved in doing this, but sometimes it makes solution a little more complicated. Consider for example the case of finding the temperature of the hottest part of a propane flame. If we try to find the maximum possible temperature of that flame due to combustion of propane in air, we would want to assume adiabatic conditions. This seems rather strange, because we have all stood around a fire and felt the heat given off. However, if no heat were to be lost, then we would have the highest possible temperature, so it is a useful calculation to set an upper limit on the temperature of a flame. In practice, it represents the hottest part of a flame better than what we might at first conjecture. This is because the cooler parts of the flame will insulate the center, hottest part. Most of the heat will be lost from the outer parts of the flame where the temperature difference is greatest.

Consider the combustion reaction C3H8(g) + 5O2(g) = 3CO2(g) + 4H2O(v) where air is used as the oxidant. To find the maximum possible flame temperature, assume Q = 0. We will feed the fuel, propane, and air into the combustion unit at a temperature To. Now, regardless of the actual pathway or mechanism for the combustion process, we can evaluate the process along a path of our own choosing. Remember that all we need to do is find the DH between the initial and final state for any path since the value will be the same for all path between those two end points. We will choose the path of combusting the fuel at To and then heating all of the products up to the final temperature, T. This is diagrammed in the figure below. In this figure, the actual path between the inlet and outlet conditions might be represented by the hypotenuse of this triangle. We, instead, will use the two-step pathway to get from the inlet to outlet conditions. The horizontal arrow represents the reaction at constant temperature, and the vertical arrow represents the change in temperature from To to T with no reaction.

The energy balance for this adiabatic process is

0 = Hout - Hin = xDHr(To) + DHsensible

In these kinds of problems, we know the extent of reaction and the heat of reaction. We can then use this equation to solve for the upper temperature limit in the integral over the heat capacities that occurs in the sensible enthalpy change. This is best done in Mathcad because temperature-dependent heat capacities usually result in this being a fourth or fifth order equation. The concrete example given below will help illustrate this.

### Example of Adiabatic Flame Temperature Calculation

Problem Statement:
Calculate the adiabatic flame temperature of liquid butane burned with 30% excess air. Both the air and liquid butane enter at 25oC. Solution:
First we will perform the mass balance calculations so that we know the number of moles of each component in the inlet and outlet streams.

Mass Balances
in:      noO2 = (1 mol)(6.5)(1.3) = 8.45 mol         note: 1.3 is the 30% excess O2
noN2 = (79/21)noO2 = 31.79 mol

out:   nO2 = 8.45 mol - (1 mol)(6.5) = 1.95 mol
nCO2 = (1 mol)(4) = 4 mol
nH2O = (1 mol)(5) = 5 mol
nN2 = noN2 = 31.79 mol

Energy Balances
Next we will set up the energy balance.

Q = Hout - Hin

We will calculate the change in enthalpy from the inlet to the outlet stream by using a path of our choice. As was explained at the top of this page, the easiest path is to react the  materials at the inlet temperature and then heat everything in the outlet stream up to the final temperature.  Thus, we write,
Since adiabatic, where Q has been set to zero since we are finding the maximum or adiabatic flame temperature.

The heat of reaction at 25oC is easily found from the heat of combustion of propane. Note that it is not exactly the heat of combustion because the water formed by the reaction is vapor whereas it must be in the liquid phase for the heat of combustion to apply directly.  By combining the combustion reaction with the "reaction" of liquid water going to vapor so that the liquid terms cancel, we find that the heat of reaction for this process is:

DHr = DHc + 5DHvap,H2O(25oC) = -2855.6 kJ/mol + 5(44.01 kJ/mol) = -2635.5 kJ/mol

The sensible enthalpy term is evaluated from the heat capacities of everything in the outlet stream. Don't forget to include the nitrogen in the air and the excess oxygen. All of these will be in the product stream and will absorb some of the heat produced by the combustion process, thereby affecting the final computed adiabatic flame temperature. Combining the heat of reaction term with the heat capacity terms shown in the energy balance gives, The heat capacities will be a function of temperature. The final temperature, the upper limit for the integrals, will therefore be the only remaining unknown. We can therefore solve for the adiabatic flame temperature using an appropriate solver. The solution obtained in Mathcad is shown below.

Problem Statement:
Using Mathcad, find the adiabatic flame temperature of liquid butane when burned with 30% excess air.

Solution:  Some notes about these example
We can make some intereseting qualitative observations about the results of the above adiabatic flame temperature calculations. For example, suppose we were to substitute pure O2 as the oxidant in place of air, what would happen to the adiabatic flame temperature? From the energy balance, we can see that N2 acts as an inert when air is fed and absorbs some of the heat of combustion as sensible heat. In fact, a large fraction of the combustion gas effluent is nitrogen, and so a large fraction of the sensible heat goes into heating the nitrogen up from 25oC to the outlet temperature. Thus, a flame using pure O2 as the oxidant would be much hotter than one for which air is the oxidizing stream. If we perform the above calculations for a variety of molar ratios of air to fuel, we obtain a curve like that at the right. Can you explain the general shape of this curve? Where do you think the maximum occurs? Why does the temperature drop off as the number of moles of air is increased?

One final comment concerns terminology that is commonly used when speaking about fuels. Fuels are often described in terms of their heating value. In the previous example, H2O in the product stream was vapor. Notice that we had to add the heat of vaporization to the heat of combustion to get the heat of reaction. Some of the released heat was used to vaporize the water. The negative of this heat of reaction, when water is a vapor in the flue gas, is called the lower heating value (LHV) of the fuel. The higher heating value (HHV) of a fuel is the negative of the heat of combustion, i.e., H2O is a liquid in the flue gas. The difference between HHV and LHV is the heat of vaporization of the water, multiplied by the appropriate stoichiometric coefficient. Thus,

• HHV = -DHc                                       ------ implies liquid H2O in flue gas
• LHV = -(DHc + nwaterDHvap,water)     ------ implies vapor H2O in flue gas

The LHV is the value that you would realize in an actual furnace or combustor where water is a vapor in the hot flue gas.