3.1
SECTION 3
COMBUSTION
Combustion Calculations Using the
Million BTU (1.055 MJ) Method
3.1
Savings Produced by Preheating
Combustion Air
3.4
Combustion of Coal Fuel in a
Furnace
3.5
Percent Excess Air While Burning
Coke
3.8
Combustion of Fuel Oil in a
Furnace
3.9
Combustion of Natural Gas in a
Furnace
3.11
Combustion of Wood Fuel in a
Furnace
3.17
Molal Method of Combustion
Analysis
3.19
Final Combustion Products Temperature
Estimate
3.22
COMBUSTION CALCULATIONS USING THE

POWER GENERATION
Combustion Constants for Fuels
Fuel
Constant, lb dry air per
million Btu (kg/MW)
Blast furnace gas 575 (890.95)
Bagasse 650 (1007.2)
Carbon monoxide gas 670 (1038.2)
Refinery and oil gas 720 (1115.6)
Natural gas 730 (1131.1)
Furnace oil and lignite 745–750 (1154.4–1162.1)
Bituminous coals 760 (1177.6)
Anthracite coal 780 (1208.6)
Coke 800 (1239.5)
To determine the energy input to the boiler, use the relation Q
f
ϭ
(Q
s
)/E
h
, where
energy input by the fuel, Btu/ hr (W); Q
s
ϭ
energy absorbed by the steam in the
boiler, Btu /Hr (W); Q
s
ϭ
energy absorbed by the steam, Btu/hr (W); E

moisture in dry air in M lb /lb (kg /kg) from, M
ϭ
0.622 ( p
w
)/(14.7
Ϫ
p
w
), where
0.622 is the ratio of the molecular weights of water vapor and dry air; p
w
ϭ
partial
pressure of water vapor in the air, psia (kPa)
ϭ
saturated vapor pressure (SVP)
ϫ
relative humidity expressed as a decimal; 14.7
ϭ
atmospheric pressure of air at sea
level (101.3 kPa). From the steam tables, at 80 F (26.7 C), SVP
ϭ
0.5069 psia
(3.49 kPa). Substituting, M
ϭ
0.622 (0.5069
ϫ
0.65)/(14.7
Ϫ
[0.5069

4. Estimate the rate of fuel firing and flue-gas produced
The rate of fuel firing
ϭ
Q
f
/HHV
ϭ
(120.48
ϫ
10
6
)/23,000
ϭ
5238 lb/hr (2378
kg/hr). Hence, the total flue gas produced
ϭ
5238
ϩ
102,578
ϭ
107,816 lb/hr
(48,948 kg/ hr).
If the temperature of the flue gas is 400
Њ
F (204.4
Њ
C) (a typical value for a natural-
gas fired boiler), then the density, as in Step 3 is: 39 /(400
ϩ
460)

Ethane 15.8 30 474 25.89
Nitrogen 0.8 28 22.4 1.22
Note that the percent weight in the above list is calculated after obtaining the sum
under Column 2
ϫ
Column 3. Thus, the percent methane
ϭ
(1334.4)/(1334.4
ϩ
474
ϩ
22.4)
ϭ
72.89 percent.
From a standard reference, such as Ganapathy, Steam Plant Calculations Man-
ual, Marcel Dekker, Inc., find the combustion constants, K, for various fuels and
use them thus: For the air required for combustion, A
c
ϭ
(K for methane)(percent
by weight methane from above list)
ϩ
(K for ethane)(percent by weight ethane);
or A
c
ϭ
(17.265)(0.7289)
ϩ
(16.119)(0.2589)
ϭ

ϭ
(1,000,000)/(23,879)
ϭ
41.88
lb (19.01 kg). Then, the dry air per million Btu (1.055 kg) fired
ϭ
(17.265)
(41.88)
ϭ
723 lb (328.3 kg).
Likewise, for propane, using the same procedure, 1 lb (0.454 kg) requires 15.703
lb (7.129 kg) air and 1 million Btu (1,055,000 kJ) has (1,000,000)/21,661
ϭ
46.17
lb (20.95 kg) fuel. Then, 1 million Btu (1,055,000 kJ) requires (15.703)(46.17)
ϭ
725 lb (329.2 kg) air. This general approach can be used for various fuel oils and
solid fuels—coal, coke, etc.
Good estimates of excess air used in combustion processes may be obtained if
the oxygen and nitrogen in dry flue gases are measured. Knowledge of excess air
amounts helps in performing detailed combustion and boiler efficiency calculations.
Percent excess air, EA
ϭ
100(O
2
–CO2) /[0.264
ϫ
N
2
–(O

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COMBUSTION
3.4
POWER GENERATION
If the percent volume of oxygen measured is 3 on a dry basis in a natural-gas
(methane) fired boiler, the excess air, EA
ϭ
(90)[3/(21–3)]
ϭ
15 percent.
This procedure is the work of V. Ganapathy, Heat Transfer Specialist, ABCO
Industries.
SAVINGS PRODUCED BY PREHEATING
COMBUSTION AIR
A 20,000 sq ft (1858 sq m) building has a calculated total seasonal heating load
of 2,534,440 MBH (thousand Btu) (2674 MJ). The stack temperature is 600
Њ
F
(316
Њ
C) and the boiler efficiency is calculated to be 75 percent. Fuel oil burned has
a higher heating value of 140,000 Btu/ gal (39,018 MJ /L). A preheater can be
purchased and installed to reduce the breeching discharge combustion air temper-
ature by 250
Њ
F (139
Њ
C) to 350
Њ

(stack temperature reduction, deg F)(cu ft air per
yr)(0.018), where the constant 0.018 is the specific heat of air. Substituting, ES
ϭ
(250)(33,792,533)(0.018)
ϭ
152,066,399 Btu/ yr (160,430 kJ/yr).
With a boiler efficiency of 75 percent, each gallon of oil releases 0.75
ϫ
140,000
Btu/gal
ϭ
105,000 Btu (110.8 jk). Hence, the fuel saved, FS
ϭ
ES/usuable heat
in fuel, Btu /gal. Or, FS
ϭ
152,066,399/105,000
ϭ
1448.3 gal/ yr (5.48 cu m/yr).
With fuel oil at $1.10 per gallon, the monetary savings will be $1.10 (1448.3)
ϭ
$1593.13. If the preheater cost $6000, the simple payoff time would be $6000/
1593.13
ϭ
3.77 years.
Related Calculations. Use this procedure to determine the potential savings
for burning any type of fuel—coal, oil, natural gas, landfill gas, catalytic cracker
offgas, hydrogen purge gas, bagesse, sugar cane, etc. Other rules of thumb used by
designers to estimate the amount of combustion air required for various fuels are:
10 cu ft of air (0.283 cu m) per 1 cu ft (0.0283 cu m) of natural gas; 1300 cu ft

0.0103; S
ϭ
0.0064; ash
ϭ
0.0533; total
ϭ
1.000
lb (0.45 kg). This coal is burned in a steam-boiler furnace. Determine the weight
of air required for theoretically perfect combustion, the weight of gas formed per
pound (kilogram) of coal burned, and the volume of flue gas, at the boiler exit
temperature of 600
Њ
F (316
Њ
C) per pound (kilogram) of coal burned; air required
with 20 percent excess air, and the volume of gas formed with this excess; the CO
2
percentage in the flue gas on a dry and wet basis.
Calculation Procedure:
1. Compute the weight of oxygen required per pound of coal
To find the weight of oxygen required for theoretically perfect combustion of coal,
set up the following tabulation, based on the ultimate analysis of the coal:
Note that of the total oxygen needed for combustion, 0.0505 lb (0.023 kg), is
furnished by the fuel itself and is assumed to reduce the total external oxygen
required by the amount of oxygen present in the fuel. The molecular-weight ratio
is obtained from the equation for the chemical reaction of the element with oxygen
in combustion. Thus, for carbon C
ϩ
O
2

3. Compute the weight of the products of combustion
Find the products of combustion by addition:
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COMBUSTION
3.6
POWER GENERATION
4. Convert the flue-gas weight to volume
Use Avogadro’s law, which states that under the same conditions of pressure and
temperature, 1 mol (the molecular weight of a gas expressed in lb) of any gas will
occupy the same volume.
At 14.7 lb /in
2
(abs) (101.3 kPa) and 32
Њ
F(0
Њ
C), 1 mol of any gas occupies
359 ft
3
(10.2 m
3
). The volume per pound of any gas at these conditions can be
found by dividing 359 by the molecular weight of the gas and correcting for the
gas temperature by multiplying the volume by the ratio of the absolute flue-gas
temperature and the atmospheric temperature. To change the weight analysis (step
3) of the products of combustion to volumetric analysis, set up the calculation thus:
In this calculation, the temperature correction factor 2.15
ϭ

F (316
Њ
C) is 53.6 ft
3
(1.158 m
3
), as computed in step 4; and the total volume of the combustion products
is 303.85 ft
3
(8.604 m
3
). Therefore, the percent CO
2
on a wet basis (i.e., including
the moisture in the combustion products)
ϭ
ft
3
CO
2
/total ft
3
ϭ
53.6/303.85
ϭ
0.1764, or 17.64 percent.
The percent CO
2
on a dry, or Orsat, basis is found in the same manner, except
that the weight of H

The excess air passes through the furnace without taking part in the combustion
and increases the weight of the products of combustion per pound (kilogram) of
coal burned. Therefore, the weight of the products of combustion is the sum of the
weight of the combustion products without the excess air and the product of (per-
cent excess air)(air for perfect combustion, lb); or, given the weights from steps 3
and 2, respectively,
ϭ
11.9139
ϩ
(0.20)(10.9672)
ϭ
14.1073 lb (6.399 kg) of gas
per pound (kilogram) of coal burned with 20 percent excess air.
8. Compute the volume of the combustion products and the percent CO
2
The volume of the excess air in the products of combustion is obtained by con-
verting from the weight analysis to the volumetric analysis and correcting for tem-
perature as in step 4, using the air weight from step 2 for perfect combustion and
the excess-air percentage, or (10.9672)(0.20)(359/ 28.95)(2.15)
ϭ
58.5 ft
3
(1.656
m
3
). In this calculation the value 28.95 is the molecular weight of air. The total
volume of the products of combustion is the sum of the column for perfect com-
bustion, step 4, and the excess-air volume, above, or 303.85
ϩ
58.5

C)
air of 60 percent relative humidity, the moisture content is 0.013 lb/ lb (0.006 kg /
kg) of dry air. This amount appears in the products of combustion for each pound
of air used and is a commonly assumed standard in combustion calculations.
Fossil-fuel-fired power plants release sulfur emissions to the atmosphere. In turn,
this produces sulfates, which are the key ingredient in acid rain. The federal Clean
Air Act regulates sulfur dioxide emissions from power plants. Electric utilities
which burn high-sulfur coal are thought to produce some 35 percent of atmospheric
emissions of sulfur dioxide in the United States.
Sulfur dioxide emissions by power plants have declined some 30 percent since
passage of the Clean Air Act in 1970, and a notable decline in acid rain has been
noted at a number of test sites. In 1990 the Acid Rain Control Program was created
by amendments to the Clean Air Act. This program further reduces the allowable
sulfur dioxide emissions from power plants, steel mills, and other industrial facil-
ities.
The same act requires reduction in nitrogen oxide emissions from power plants
and industrial facilities, so designers must keep this requirement in mind when
designing new and replacement facilities of all types which use fossil fuels.
Coal usage in steam plants is increasing throughout the world. An excellent
example of this is the New England Electric System (NEES). This utility has been
converting boiler units from oil to coal firing. Their conversions have saved cus-
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COMBUSTION
3.8
POWER GENERATION
FIGURE 1 Energy Independence transports coal to central stations.
(Power.)
tomers more than $60-million annually by displacing about 14-million bbl (2.2

Ϫ
O
Ј
2
/8)]
ϩ
4.32S
Ј
, where C
Ј
,H
Ј
2
,O
Ј
2
, and S
Ј
represent the percentages by weight,
expressed as decimal fractions, of carbon, hydrogen, oxygen, and sulfur, respec-
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COMBUSTION
COMBUSTION
3.9
tively. Thus, w
ta
ϭ
11.5(0.7509)

as a decimal fraction, of carbon in the coal, C
f
ϭ
75.09; percent by weight of the
ash and refuse in the coal, w
r
ϭ
0.0338; percent by weight of combustible in the
ash, C
r
ϭ
8.93. Hence, C
1
ϭ
[(1
ϫ
75.09)
Ϫ
(0.0338
ϫ
8.93)]/(1
ϫ
100)
ϭ
0.748.
3. Compute the amount of dry flue gas produced per lb (kg) of coal
The lb (kg) of dry flue gas per lb (kg) of coal, w
dg
ϭ
C

704)]/[3(14.2
ϩ
0.3)]
ϭ
13.16 lb/ lb (5.97 kg /kg).
4. Compute the amount of dry air supplied per lb (kg) of coal
The lb (kg) of dry air supplied per lb (kg) of coal, w
da
ϭ
w
dg
Ϫ
C
1
ϩ
8[H
Ј
2
Ϫ
(O
Ј
2
/8)]
Ϫ
(N
Ј
2
/N), where the percentage by weight of nitrogen in the fuel, N
Ј
2

(12.65
Ϫ
10.03)/10.03
ϭ
0.261, or 26.1
percent.
Related Calculations. The percentage by weight of nitrogen in ‘‘atmospheric
air’’ in step 4 appears in Principles of Engineering Thermodynamics, 2nd edition,
by Kiefer et al., John Wiley & Sons, Inc.
COMBUSTION OF FUEL OIL IN A FURNACE
A fuel oil has the following ultimate analysis: C
ϭ
0.8543; H
2
ϭ
0.1131; O
2
ϭ
0.0270; N
2
ϭ
0.0022; S
ϭ
0.0034; total
ϭ
1.0000. This fuel oil is burned in a
steam-boiler furnace. Determine the weight of air required for theoretically perfect
combustion, the weight of gas formed per pound (kilogram) of oil burned, and the
volume of flue gas, at the boiler exit temperature of 600
Њ