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CHAPTER 44
COMBUSTION
Richard J. Reed
North American Manufacturing Company
Cleveland, Ohio
44.3 BURNERS 14 39
44.3.1 Burners for Gaseous Fuels 1439
44.3.2 Burners for Liquid Fuels 1441
44.1 FUNDAMENTALS OF
COMBUSTION 14 31
44.1.1 Air-Fuel Ratios 14 31
44.1.2 Fuels
14 33
44.4 SAFETY CONSIDERATIONS 1442
44.2 PURPOSES OF COMBUSTION 1435
44.5 OXY-FUEL FIRING
14 47
44.1 FUNDAMENTALS OF COMBUSTION
44.1.1 Air-Fuel Ratios
Combustion is rapid oxidation, usually for the purpose of changing chemical energy into thermal
energy—heat. This energy usually comes from oxidation of carbon, hydrogen, sulfur, or compounds
containing C, H, and/or S. The oxidant is usually O2—molecular oxygen from the air.
The stoichiometry of basic chemical equation balancing permits determination of the air required
to burn a fuel. For example,
1CH4 + 202 — 1CO2 + 2H2O
where the units are moles or volumes; therefore, 1 ft3 of methane (CH4) produces 1 ft3 of CO2; or
1000 m3 CH4 requires 2000 m3 O2 and produces 2000 m3 H2O. Knowing that the atomic weight of
C is 12, H is 1, N is 14, O is 16, and S is 32, it is possible to use the balanced chemical equation
to predict weight flow rates: 16 Ib/hr CH4 requires 64 Ib/hr O2 to burn to 44 Ib/hr CO2 and 36 lb/
hr H2O.
If the oxygen for combustion comes from air, it is necessary to know that air is 20.99% O2 by
volume and 23.20% O2 by weight, most of the remainder being nitrogen.
It is convenient to remember the following ratios:
air/02 - 100/20.99 = 4.76 by volume
N2/O2 = 3.76 by volume
air/02 - 100/23.20 - 4.31 by weight
N2/O2 - 3.31 by weight
Rewriting the previous formula for combustion of methane,
1CH4 + 2O2 4- 2(3.76)N2 — 1CO2 + 2H2O + 2(3.76)N2
or
1CH4 + 2(4.76)air —> 1CO2 + 2H2O + 2(3.76)N2
Table 44.1 lists the amounts of air required for stoichiometric (quantitatively and chemically
Mechanical Engineers' Handbook, 2nd ed., Edited by Myer Kutz.
ISBN 0-471-13007-9 © 1998 John Wiley & Sons, Inc.
815044117.004.png 815044117.005.png 815044117.006.png
Table 44.1 Proper Combining Proportions for Perfect Combustion3
vol air
vol fuel
11.9
35.7
31.0
wtO2
wt fuel
3.08
3.08
3.59
2.67
0.571
3.73
8.00
1.41
4.00
3.00
3.51
3.64
3.43
1.00
wt air
wt fuel
13.3
13.3
15.5
11.5
2.46
16.1
34.5
6.08
17.2
12.9
15.1
15.7
14.8
4.31
ft3O2
Ib fuel
36.5
36.5
42.5
31.6
6.76
44.2
94.7
16.7
47.4
35.5
41.6
43.1
40.6
11.8
ft3 air
Ib fuel
174
174
203
150
32.2
210
451
79.5
226
169
198
205
193
56.4
vol O2
vol fuel
2.50
7.50
6.50
m2O2
kg fuel
2.28
2.28
2.65
1.97
0.422
2.76
5.92
1.04
2.96
2.22
2.60
2.69
2.54
0.74
m3 air
kg fuel
10.8
10.8
12.6
9.39
2.01
13.1
28.2
4.97
14.1
10.6
12.4
12.8
12.1
3.52
Fuel
Acetylene, C2H2
Benzene, C6H6
Butane, C4H10
Carbon, C
Carbon monoxide, CO
Ethane, C2H6
Hydrogen, H2
Hydrogen sulfide, H2S
Methane, CH4
Naphthalene, C10H8
Octane, C8H18
Propane, C3H8
Propylene, C3H6
Sulfur, S
a Reproduced with permission from Combustion Handbook.1 (See Ref. 1)
2.38
16.7
2.38
7.15
9.53
0.50
3.50
0.50
1.50
2.00
5.00
4.50
23.8
21.4
815044117.007.png
correct) combustion of a number of pure fuels, calculated by the above method. (Table 46. Ic lists
similar information for typical fuels that are mixtures of compounds, calculated by the above method,
but weighted for the percentages of the various compounds in the fuels.)
The stoichiometrically correct (perfect, ideal) air/fuel ratio from the above formula is therefore
2 + 2(3.76) = 9.52 volumes of air per volume of the fuel gas. More than that is called a "lean"
ratio, and includes excess air and produces an oxidizing atmosphere. For example, if the actual air/
fuel ratio were 10:1, the %excess air would be
1°^ X 100 - 5.04%
Communications problems sometimes occur because some people think in terms of air/fuel ratios,
others in fuel/air ratios; some in weight ratios, others in volume ratios; and some in mixed metric
units (such as normal cubic meters of air per metric tonne of coal), others in mixed American units
(such as ft3 air/gal of oil). To avoid such confusions, the following method from Ref. 1 is
recommended.
It is more convenient to specify air/fuel ratio in unitless terms such as %air (%aeration), %excess
air, %deficiency of air, or equivalence ratio. Those experienced in this field prefer to converse in
terms of %excess air. The scientific community favors equivalence ratio. The %air is easiest to use
and explain to newcomers to the field: "100% air" is the correct (stoichiometric) amount; 200% air
is twice as much as necessary, or 100% excess air. Equivalence ratio, widely used in combustion
research, is the actual amount of fuel expressed as a fraction percent of the stoichiometrically correct
amount of fuel. The Greek letter phi, 4>, is usually used: 4> = 0.9 is lean; <|> = 1.1 is rich; and 4> —
1.0 is "on-ratio."
Formulas relating %air, 4>, %excess air (%XS), and %deficiency of air (%def) are
%air - 100/cf> = %XS + 100 = 100 - %def
100 1
* ~ %XS + 100 ~ 1 - (%def/100)
%XS - %air - 100 - —— X 100
$
%def = 100 - %air - ^— x 100
4>
Table 44.2 lists a number of equivalent terms for convenience in converting values from one
"language" to another.
Excess air is undesirable, because, like N2, it passes through the combustion process without
chemical reaction; yet it absorbs heat, which it carries out the flue. The percent available heat (best
possible fuel efficiency) is highest with zero excess air. (See Fig. 44.1.)
Excess fuel is even more undesirable because it means there is a deficiency of air and some of
the fuel cannot be burned. This results in formation of soot and smoke. The accumulation of unburned
fuel or partially burned fuel can represent an explosion hazard.
Enriching the oxygen content of the combustion "air" above the normal 20.9% reduces the
nitrogen and thereby reduces the loss due to heat carried up the stack. This also raises the flame
temperature, improving heat transfer, especially that by radiation.
Vitiated air (containing less than the normal 20.9% oxygen) results in less fuel efficiency, and
may result in flame instability. Vitiated air is sometimes encountered in incineration of fume streams
or in staged combustion, or with flue gas recirculation.
44.1.2 Fuels
Fuels used in practical industrial combustion processes have such a major effect on the combustion
that they must be studied simultaneously with combustion. Fuels are covered in detail in later chap-
ters, so the treatment here is brief, relating only to the aspects having direct bearing on the combustion
process.
Gaseous fuels are generally easier to burn, handle, and control than are liquid or solid fuels.
Molecular mixing of a gaseous fuel with oxygen need not wait for vaporization nor mass transport
within a solid. Burning rates are limited only by mixing rates and the kinetics of the combustion
reactions; therefore, combustion can be compact and intense. Reaction times as short as 0.001 sec
and combustion volumes from 104 to 107 Btu/hr • ft3 are possible at atmospheric pressure.2 Gases
of low calorific value may require such large volumes of air that their combustion rates will be
limited by the mixing time.
Combustion stability means that a flame lights easily and then burns steadily and reliably after
the pilot (or direct spark) is programmed off. Combustion stability depends on burner geometry, plus
815044117.001.png
Table 44.2 Equivalent Ways to Express Fuel-to-Air or Air-to-Fuel Ratios1
%air
40
60
80
90
95
100
105
110
120
130
140
160
180
200
250
300
400
500
600
1100
2100
%def
60
40
20
10
5
0
%xs
*
2.50
1.67
1.25
1.11
1.05
1.00
0.95
0.91
0.83
0.78
0.71
0.62
0.56
0.50
0.40
0.33
0.25
0.20
0.167
0.091
0.048
Fuel rich
(air lean)
Stoiehiometric
Fuel lean
(air rich)
0
5
10
20
30
40
60
80
100
150
200
300
400
500
1000
2000
Fig. 44.1 Percent available heat (best possible efficiency) peaks at Stoiehiometric air/fuel ratio.1
815044117.002.png
air and fuel flow controls that maintain the point(s) of flame initiation (a) above the fuel's minimum
ignition temperature, (b) within the fuel's flammability limits, and (c) with feed speed equal to flame
speed—throughout the burner's full range of firing rates and conditions. (Fuel properties are discussed
and tabulated in Chapters 46 and 47.)
Liquid fuels are usually not as easily burned, handled, or controlled as are gaseous fuels. Mixing
with oxygen can occur only after the liquid fuel is evaporated; therefore, burning rates are limited
by vaporization rates. In practice, combustion intensities are usually less with liquid fuels than with
high calorific gaseous fuels such as natural gas.
Because vaporization is such an integral part of most liquid fuel burning processes, much of the
emphasis in evaluating liquid fuel properties is on factors that relate to vaporization, including vis-
cosity, which hinders good atomization, the primary method for enhancing vaporization. Much con-
cern is also devoted to properties that affect storage and handling because, unlike gaseous fuels that
usually come through a public utility's mains, liquid fuels must be stored and distributed by the user.
The stability properties (ignition temperature, flammability limits, and flame velocity) are not
readily available for liquid fuels, but flame stability is often less critical with liquid fuels.
Solid fuels are frequently more difficult to burn, handle, and control than liquid or gaseous fuels.
After initial volatilization, the combustion reaction rate depends on diffusion of oxygen into the
remaining char particle, and the diffusion of carbon monoxide back to its surface, where it burns as
a gas. Reaction rates are usually low and required combustion volumes high, even with pulverized
solid fuels burned in suspension. Some fluidized bed and cyclone combustors have been reported to
reach the intensities of gas and oil flames.2
Most commonly measured solid fuel properties apply to handling in stokers or pulverizers. See
Chapter 48.
Wastes, by-product fuels, and gasified solids are being used more as fuel costs rise. Operations
that produce such materials should attempt to consume them as energy sources. Handling problems,
the lack of a steady supply, and pollution problems often complicate such fuel usage.
For the precise temperature control and uniformity required in many industrial heating processes,
the burning of solids, especially the variable quality solids found in wastes, presents a critical problem.
Such fuels are better left to very large combustion chambers, particularly boilers. When solids and
wastes must be used as heat sources in small and accurate heating processes, a better approach is to
convert them to low-Btu (producer) gas, which can be cleaned and then controlled more precisely.
44.2 PURPOSES OF COMBUSTION
The purposes of combustion, for the most part, center around elevating the temperature of something.
This includes the first step in all successive combustion processes—the pilot flame—and, similarly,
the initiation of incineration. Elevating the temperature of something can also make it capable of
transmitting light or thermal energy (radiation and convection heat transfer), or it can cause chemical
dissociation of molecules in the products of combustion to generate a special atmosphere gas for
protection of materials in industrial heat processing.
All of the above functions of combustion are minor in comparison to the heating of air, water
and steam, metals, nonmetallic minerals, and organics for industrial processing, and for space comfort
conditioning. For all of these, it is necessary to have a workable method for evaluating the heat
available from a combustion process.
Available heat is the heat accessible for the load (useful output) and to balance all losses other
than stack losses. (See Fig. 44.2.) The available heat per unit of fuel is
AH = HHV - total stack gas loss = LHV - dry stack gas loss
% available heat - 100(AH/HHV)
where AH = available heat, HHV = higher heating value, and LHV = lower heating value, as
defined in Chapter 47. Figure 44.3 shows values of % available heat for a typical natural gas; Fig.
44.4 for a typical residual oil; and Fig. 53.2 in Chapter 53, for a typical distillate oil.
Example 44.1
A process furnace is to raise the heat content of 10,000 Ib/hr of a load from 0 to 470 Btu/lb in a
continuous furnace (no wall storage) with a flue gas exit temperature of 1400°F. The sum of wall
loss and opening loss is 70,000 Btu/hr. There is no conveyor loss. Estimate the fuel consumption
using 1000 Btu/ft3 natural gas with 10% excess air.
Solution: From Fig. 44.3, % available heat = 58.5%. In other words, the flue losses are
100% - 58.5% = 41.5%. The sum of other losses and useful output = 70,000 + (10,000)(470) =
4,770,000 Btu/hr. This constitutes the "available heat" required. The required gross input is therefore
4,770,000/0.585 - 8,154,000 Btu/hr, of 8154 ft3/hr of natural gas (and about 81.540 ft3/hr of air).
The use of the above precalculated % available heats has proved to be a practical way to avoid
long iterative methods for evaluating stack losses and what is therefore left for useful heat output
815044117.003.png
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