Problem statement: Repeat Example 2.1 for a VC heater behaving as a Helmholtz resonator with a furnace height of 40 ft and a stack height of 20 ft. Presume that the stack diameter is 20% of the furnace diameter.
Solution: The speed of sound will be identical to our first exam- ple: 2165 ft/sec. Then Equation 2.11 gives
2.3.6 Mechanism for Thermoacoustic Coupling
Earlier, we stated that the heat must add in phase with the pressure wave in order for a burner to couple thermoacoustically with a furnace. Upon reflection, this seems counterintuitive. It seems that an increase in pressure near the burner would attenuate the flow of air, and thus the heat release
ν
π
γ
π
=
⎛
⎝ ⎜
⎞
⎠ ⎟
⎛
⎝ ⎜
⎞
⎠ ⎟ =
1 2 2 A V L RT W c A V L S F S S F Sν
π
γ
π
=
⎛
⎝ ⎜
⎞
⎠ ⎟
=
⎛
⎝ ⎜
⎞
⎠ ⎟
1 2 2 L L D D RT W c L L D D F s S F F s S Fν
π
=
⎡
⎣
⎢
⎤
⎦
⎥
⎡⎣ ⎤⎦ ⎡⎣ ⎤⎦
( )=
2164 2 40 20 0 2 2 4 ft s ft ft . .⎡⎣ ⎤⎦
HzzIntroduction to Combustion 133 would be out of phase with an acoustic pressure wave. (We can usually neglect any variation in fuel pressure because the fuel flow across an orifice is usually in critical flow — also termed sonic or choked flow. It is impossible for a pressure wave to propagate upstream of a critical flow nozzle.) To understand the mechanism for thermoacoustic coupling, we must examine the transfer of heat.
The velocity wave must be out of phase with the pressure wave because when a high velocity encounters a wall, the velocity must drop to zero (velocity node) and the pressure must rise to maximum (pressure antinode). Therefore, the waves are 90° out of phase. Now, imagine we are at the position of the flame and pressure and velocity are fluctuating there. Let us focus on the velocity for a moment.
The average velocity is the mean flow rate through the heater divided by the cross-sectional area. An acoustic wave superimposed on the mean flow will alternately enhance or attenuate the flow as the velocity alternates. Now in the case of enhanced flow, the flue gas is colder than the mean temperature because it comprises a greater portion of cool influent air. In the case of
attenuated flow, the flue gas temperature is warmer than average because less cold air flows into the furnace during that period. Clearly, we transfer more heat when the thermal gradient is larger, that is, when the flowing fluid is cooler than average. In other words, more heat transfer results during the enhanced flow cycle than during the attenuated flow cycle.
Now since pressure and velocity are not in phase, we can find a position in the heater for the flame where the increase in heat transfer will coincide with an increase in the pressure wave. For a furnace resonating in quarter- wave mode, that position is L/4. That is, if the position of our heat source (flame) is at L/4, then we will enhance thermoacoustic coupling. Not only can we add heat at L/4, but also, if we remove heat at 3L/4, we will subtract heat during falling pressure, which amounts to the same thing, as Rijke showed.4 So in fact, having a flame at L/4 and a convection section at 3L/4
reinforces thermoacoustic coupling. The minor miracle seems not to be why some heaters experience instability, but why not all heaters experience insta- bility. The answer seems to lie in the following facts:
1. Flames are not concentrated at L/4.
2. Surfaces and volumes within the heater can cause destructive inter- ference of the fundamental resonant mode.
3. Resistances in the heater, such as convection tubes, etc., reduce the efficiency of thermoacoustic coupling.
2.3.7 Comments Regarding Thermoacoustic Resonance
Burner–furnace interactions are a cause for concern if the test furnace is acoustically similar to the field unit — even if the burner itself is not the
134 Modeling of Combustion Systems: A Practical Approach problem. However, the test unit almost never bears acoustic similarity to the field unit. Therefore, thermoacoustic resonance in the test furnace does not usually signal a concern in the field, but burner instability does. How can we know the difference?
To be sure, one must construct an acoustically similar system, either phys- ically or mathematically. This is not trivial and requires expert assistance. However, if the following remedies are effective at eliminating or attenuating the instability, then thermoacoustic coupling is the likely culprit.
1. Reduce the stack damper opening and compensate with a steam sparger or induced draft (ID) fan, if available. By closing the damper, one may possibly convert the furnace from quarter-wave to half- wave mode and decouple the burner and the furnace.
2. Switch test furnaces. If the same burner is stable in a furnace with different dimensions, the culprit is likely thermoacoustic resonance. 3. Measure the frequency of the instability. Calculate the expected fre- quencies of the resonant phenomenon for quarter-wave and Helm- holtz behavior. If the frequencies match, this is a strong indication that thermoacoustic behavior is an issue.
4. Is there an induction period? Does the problem build? Resonance is the constructive addition of multiple pressure waves. Theoretically, the pressure waves add forever, becoming infinitely strong. How- ever, nonlinear behavior and attenuating mechanisms put a ceiling on the maximum amplitude. Notwithstanding, during the induction phase, pressure waves grow over time. Usually some seconds are required to reach the maximum. Although one cannot hear such fre- quencies by ear, the “huffing” of air in the 2- to 20-Hz range can be heard or felt. Theoretically, the velocity lags the pressure wave by 90°, but the frequency is the same and thus an appropriate indicator.
2.3.7.1 Resonance in the Field
Sometimes, heaters resonate. Even if due to thermoacoustic resonance, it is easier to redesign the burner rather than the furnace, although the addition of quarter-wave tubes on the furnace roof can help. These reflect the pressure wave 180° out of phase with the resonant frequency. The destructive inter- ference eliminates the resonance. The author has used this method success- fully for a process heater,* and the literature reports successful application to boilers.5 Redesign of the burners could include changing the flame dimen-
sions. Sometimes, shortening of the flame below the L/4 criterion can be effective; however, shorter flames may elevate NOx.
* Together with Mahmoud Fliefil, acoustic engineer, John Zink LLC, Tulsa, OK.
Introduction to Combustion 135
2.4
Mass Balance for Combustion in Air
With this brief discussion of combustion equipment concluded, we turn our attention to the combustion reaction itself. Reaction 2.12 shows the combus- tion reaction for methane:
CH4 + 2O2
→
CO2 + 2H2O (2.12)The single arrow (
→
) indicates a reaction proceeding unilaterally from reactants to products at right. Reaction 2.12 encodes many important aspects of the combustion reaction. First, the stoichiometry (law of proportions) is apparent: one volume of methane (CH4) reacts with two volumes of oxygen(O2) to form one volume of carbon dioxide (CO2) and two volumes of water (H2O). Second, the combustion reaction alters molecular entities, but not atomic ones. That is, reactants and products each have the same number of C, H, and O atoms. However, the identity of the molecular entities has changed from CH4 and O2 to CO2 and H2O. Therefore, combustion reactions
conserve mass: 1 kg of reactants will yield 1 kg of products without fail. We may augment Reaction 2.12 to account specifically for the nitrogen in the air. Nitrogen does not take part in combustion chemistry to any appreciable extent. Combustion in air does form some nitrogen oxides in part per million quantities, but they are a trivial part of the mass balance, although they form an important class of regulated compounds, which we discuss later. Reaction 2.13 augments the combustion reaction to account for nitrogen. The nitrogen dilutes both reactants and products. The ratio of nitrogen to oxygen in air is approximately 79/21 by volume.
CH4 + 2O2 + 2(79/21) N2
→
CO2 + 2H2O + 2(79/21) N2 (2.13)Reaction 2.13 gives the stoichiometric amount of air required for combus- tion of one volume of CH4. However, the typical industrial practice is to use
some quantity of extra air to ensure complete combustion. In principle, if one were to mix the air and fuel thoroughly during the combustion process, then no excess air would be required. However, adding a little excess air is the most cost-effective way to ensure complete combustion. Reaction 2.14 accounts for this excess air (
ε
):CH4 + 2(1 +
ε
)O2 + 2(79/21)(1 +ε
) N2→
CO2 + 2H2O + 2(79/21)(1 +
ε
)N2 + 2ε
O2 (2.14)For the first time, oxygen appears in the products. The reader may verify that atomic entities are still equal for both products and reactants.
136 Modeling of Combustion Systems: A Practical Approach One final step will transform Reaction 2.14 to a general equation for hydro- carbon combustion: we rewrite CH4 as CHψ
,
whereψ
[ ] is a generalized subscript used to account for any desired H/C ratio (ψ
):CHψ + (1 +
ψ
/4)(1 +ε
) O2 + (79/21)(1 +ψ
/4)(1 +ε
) N2→
CO2 + (