section 1 energy conversion engineering

SECTION 1 

ENERGY CONVERSION 

ENGINEERING 

Combustion Calculation Parameters 1.1 

Coal Fuel Combustion in a Furnace 1.1 

Percent Excess Air while Burning Coal 1.4 

Fuel Oil Combustion in a Furnace 1.5 

Natural Gas Fuel Combustion in a 

Furnace 1.7 

Wood Fuel Combustion in a Furnace 1.11 

Combustion CalCulation Parameters 

Combustion Analysis by the Molal 

Method 1.13 

Estimating the Temperature of the Final 

Products of Combustion 1.15 

Determination of the Savings Produced by 

Preheating Combustion Air 1.16 

Combustion Calculations Using the Million 

Btu (1.055 MJ) Method 1.17 

The focus in this section of the handbook is on external combustion of fossil fuels—such as in a boiler 

furnace. A large portion of fossil-fuel energy use takes place in the form of external combustion to 

generate steam for power generation and space heating. Internal combustion of fossil fuels—such as 

in gas turbines, and diesel and gas engines—is discussed within sections 3 and 4 of this handbook. 

In today’s world the greatest emphasis is on complete combustion to reduce greenhouse gases 

and the ensuing pollution when such gases are emitted into the atmosphere. Thus, in the midwestern 

United States, combustion of coal in steam power plants produces some 210 lb (95.3 kg) of CO 2 per 

million Btu (1.055 MJ) of coal used in a steam boiler. Contrast this with 120 lb (54.4 kg) of CO 2 

produced per million Btu (1.055 MJ) of natural gas used in a steam boiler. 

On the basis of power generated, 1.9 lb (0.86 kg) of CO 2 is produced per kWh generated by a 

coal-fired boiler and turbine. With natural-gas firing of a steam boiler, 1.25 lb (0.57 kg) of CO 2 is 

produced per kWh generated. These statistics indicate that more efficient combustion of fossil fuels 

reduces greenhouse gases and operating costs. Hence design and operating engineers are making 

strenuous efforts to improve combustion efficiency to comply with ever stricter regulations while 

reducing plant operating costs. 

As indicated later in this section of the handbook, some states and countries are considering banning 

coal-fired power plants unless they meet much more stringent emission requirements. At this writing, 

the Environmental Protection Agency seeks to establish ozone levels of 0.060 to 0.070 ppm. 

Coal Fuel Combustion in a FurnaCe 

A coal fuel has the following ultimate analysis (or percent by weight): C = 0.8339; H 2 = 0.0456; 

O 2 = 0.0505; N 2 = 0.0103; S = 0.0064; ash = 0.0533; total = 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°F (316°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. 

1.1

1.2 SECTION 1 

Calculation Procedure: 

1. Compute the weight of oxygen required per pound (kilogram) 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: 

Element × 

C; 0.8339 

H 2 ; 0.0456 

O 2 ; 0.0505; decreases external O 2 required 

N 2 ; 0.0103 is inert in combustion and is 

ignored 

S; 0.0064 

Ash 0.0533 is inert in combustion and is 

ignored 

× 

× 

× 

Molecular-weight 

ratio = 

32 /12 

16 /2 

32 /32 

= 

= 

= 

= 

lb (kg) O 2 

required 

2.2237 (1.009) 

0.3648 (0.165) 

−0.0505 (0.023) 

0.0064 (0.003) 

Total 1.0000 

lb (kg) external O 2 per lb (kg) fuel = 2.5444 (1.154) 

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 + O 2 → CO 2 , or 12 + 32 = 44, where 12 

and 32 are the molecular weights of C and O 2 , respectively. 

2. Compute the weight of air required for perfect combustion 

Air at sea level is a mechanical mixture of various gases, principally 23.2 percent oxygen and 

76.8 percent nitrogen by weight. The nitrogen associated with the 2.5444 lb (1.154 kg) of oxygen 

required per pound (kilogram) of coal burned in this furnace is the product of the ratio of the nitrogen 

and oxygen weights in the air and 2.5444, or (2.5444)(0.768/0.232) = 8.4228 lb (3.820 kg). 

Then the weight of air required for perfect combustion of 1 lb (0.45 kg) of coal = sum of nitrogen 

and oxygen required = 8.4228 + 2.5444 = 10.9672 lb (4.975 kg) of air per pound (kilogram) of 

coal burned. 

3. Compute the weight of the products of combustion 

Find the products of combustion by addition: 

Products of combustion 

Fuel constituents + Oxygen → lb kg 

C; 0.8339 + 2.2237 → CO 2 = 3.0576 1.387 

H; 0.0465 + 0.3648 → H 2 O = 0.4104 0.186 

O 2 ; 0.0505; this is not a product combustion 

N 2 ; 0.0103; inert but passes through furnace = 0.0103 0.005 

S; 0.0064 + 0.0064 → SO 2 = 0.0128 0.006 

Outside nitrogen from step 2 = N 2 = 8.4228 3.820 

lb (kg) of flue gas lb (kg) of coal burned = 11.9139 5.404 

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.

ENERGY CONVERSION ENGINEERING 1.3 

At 14.7 lb/in 2 (abs) (101.3 kPa) and 32°F (0°C), 1 mol of any gas occupies 359 ft 3 (10.2 m 3 ). 

The volume per pound (kilogram) 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: 

Weight 

Molecular 

Volume at 

Products lb kg weight Temperature correction 600°F, ft3 316°C, m3 CO 2 3.0576 1.3869 44 (359/44)(3.0576)(2.15) = 53.6 1.518 

H 2 O 0.4104 0.1862 18 (359/18)(0.4104)(2.15) = 17.6 0.498 

Total N 2 8.4331 3.8252 28 (359/28)(8.4331)(2.15) = 232.5 6.584 

SO 4 0.0128 0.0058 64 (359/64)(0.0128)(2.15) = 0.15 0.004 

ft 3 (m 3 ) of flue gas per lb (kg) of coal burned = 303.85 8.604 

In this calculation, the temperature correction factor 2.15 = absolute flue-gas temperature, °R/absolute 

atmospheric temperature, °R = (600 + 460)/(32 + 460). The total weight of N 2 in the flue gas is 

the sum of the N 2 in the combustion air and the fuel, or 8.4228 + 0.0103 = 8.4331 lb (3.8252 kg). 

The value is used in computing the flue-gas volume. 

5. Compute the CO 2 content of the flue gas 

The volume of CO 2 in the products of combustion at 600°F (316°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) = ft 3 CO 2 / 

total ft 3 = 53.6 / 303.85 = 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 2 O in the products of combustion, 17.6 lb (7.83 kg) from step 4, is subtracted from the total gas 

weight. Or, percent CO 2 , dry, or Orsat basis = (53.6)/(303.85 − 17.6) = 0.1872, or 18.72 percent. 

6. Compute the air required with the stated excess flow 

With 20 percent excess air, the air flow required = (0.20 + 1.00)(air flow with no excess) = 1.20 

(10.9672) = 13.1606 lb (5.970 kg) of air per pound (kilogram) of coal burned. The air flow with no 

excess is obtained from step 2. 


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 (percent excess air)(air for perfect combustion, lb); or, given the weights from 

steps 3 and 2, respectively, = 11.9139 + (0.20)(10.9672) = 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 converting from the weight 

analysis to the volumetric analysis and correcting for temperature 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) = 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 combustion, step 4, 

and the excess-air volume, above, or 303.85 + 58.5 = 362.35 ft 3 (10.261 m 3 ). 

By using the procedure in step 5, the percent CO 2 , wet basis = 53.6/362.35 = 14.8 percent. The 

percent CO 2 , dry basis = 53.8/(362.35 − 17.6) = 15.6 percent. 

Related Calculations. Use the method given here when making combustion calculations for any 

type of coal—bituminous, semibituminous, lignite, anthracite, cannel, or cooking—from any coal

1.4 SECTION 1 

field in the world used in any type of furnace—boiler, heater, process, or waste-heat. When the air 

used for combustion contains moisture, as is usually true, this moisture is added to the combustionformed 

moisture appearing in the products of combustion. Thus, for 80°F (26.7°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 facilities. 

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. 

At the time of this writing, there are a number of states and countries considering phasing out 

coal-burning power plants at the end of their useful commercial lives if they do not install means to 

capture carbon dioxide and other greenhouse gases. These entities are also urging a shift to naturalgas 

fuel. Further, combined-cycle natural-gas-fueled power plants are being urged as replacements 

of coal-fired power plants. 

PerCent eXCess air WHile burninG Coal 


A certain coal has the following composition by weight percentages: carbon 75.09, nitrogen 1.56, 

ash 3.38, hydrogen 5.72, oxygen 13.82, sulfur 0.43. When burned in an actual furnace, measurements 

showed that there was 8.93 percent combustible in the ash pit refuse and the following Orsat 

analysis in percentages was obtained: carbon dioxide 14.2, oxygen 4.7, carbon monoxide 0.3. If 

it can be assumed that there was no combustible in the flue gas other that the carbon monoxide 

reported, calculate the percentage of excess air used. 

1. Compute the amount of theoretical air required per lb m (kg) of coal 

Theoretical air required per pound (kilogram) of coal, w ta = 11.5C′ + 34.5[H′ 2 − O′ 2 /8)] + 4.32S′, 

where C′, H′ 2 , O′ 2 , and S′ represent the percentages by weight, expressed as decimal fractions, 

of carbon, hydrogen, oxygen, and sulfur, respectively. Thus, w ta = 11.5(0.7509) + 34.5[0.0572 − 

(0.1382/8)] + 4.32(0.0043) = 10.03 lb (4.55 kg) of air per lb (kg) of coal. The ash and nitrogen are 

inert and do not burn. 

2. Compute the correction factor for combustible in the ash 

The correction factor for combustible in the ash, C 1 = (w t C f − w r C r )/(w f × 100), where the amount 

of fuel, w f = 1 lb (0.45 kg) of coal; percent by weight, expressed as a decimal fraction, of carbon 

in the coal, C f = 75.09; percent by weight of the ash and refuse in the coal, w r = 0.0338; percent 

by weight of combustible in the ash, C r = 8.93. Hence, C 1 = [(1 × 75.09) − (0.0338 × 8.93)]/ 

(l × 100) = 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 = C 1 (4CO 2 + O 2 + 704)/[3(CO 2 + CO)], where 

the Orsat analysis percentages are for carbon dioxide, CO 2 = 14.2; oxygen, O 2 = 4.7; carbon


monoxide, CO = 0.3. Hence, w dg = 0.748 × [(4 × 14.2) + 4.7 + 704)]/[3(14.2 + 0.3)] = 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 = w dg – C 1 + 8[H′ 2 − (O′ 2 / 8)] − (N′ 2 / N), 

where the percentage by weight of nitrogen in the fuel, N′ 2 = 1.56, and “atmospheric nitrogen” 

in the supply air, N 2 = 0.768; other values are as given or calculated. Then, w da = 13.16 − 0.748 + 

8[0.0572 − (0.1382 / 8)] − (0.0156 / 0.768) = 12.65 lb / lb (5.74 kg/kg). 

5. Compute the percent of excess air used 

Percent excess air = (w da − w ta ) / w ta = (12.65 − 10.03) / 10.03 = 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. 

Fuel oil Combustion in a FurnaCe 


A fuel oil has the following ultimate analysis: C = 0.8543; H 2 = 0.1131; O 2 = 0.0270; N 2 = 0.0022; 

S = 0.0034; total = 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°F (316°C), per pound 

(kilogram) of oil burned; the 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. 

1. Compute the weight of oxygen required per pound (kilogram) of oil 

The same general steps as given in the previous calculation procedure will be followed. Consult that 

procedure for a complete explanation of each step. 

Using the molecular weight of each element, we find 

Element × Molecular-weight ratio = lb (kg) O 2 required 

C; 0.8543 × 32 / 12 = 2.2781 (1.025) 

H2 ; 0.1131 

O2 ; 0.0270; decreases external O2 required 

N2 ; 0.0022 is inert in combustion 

and is ignored 

× 16 /2 = 

= 

0.9048 

− 0.0270 

(0.407) 

(− 0.012) 

S; 0.0034 × 32 / 32 = 0.0034 (0.002) 

Total 1.0000 



The weight of nitrogen associated with the required oxygen = (3.1593)(0.768 / 0.232) = 10.458 lb 

(4.706 kg). The weight of air required = 10.4583 + 3.1593 = 13.6176 lb / lb (6.128 kg/kg) of oil 

burned.

1.6 SECTION 1 


As before, 

Fuel constituents + Oxygen = Products of combustion 

C; 0.8543 + 2.2781 = 3.1324 = CO 2 

H 2 ; 0.1131 + 0.9148 = 1.0179 = H 2 O 

O 2 ; 0.270; not a product of combustion 

N 2 ; 0.0022; inert but passes through furnace = 0.0022 = N 2 

S; 0.0034 + 0.0034 = 0.0068 = SO 2 

Outside N 2 from step 2 = 10.458 = N 2 

lb (kg) of flue gas per lb (kg) of oil burned = 14.6173 (6.578) 


As before, 

Weight 

Molecular 

Volume at 

Products lb kg weight Temperature correction 600°F, ft3 316°C, m3 CO 2 3.1324 1.4238 44 (359 / 44)(3.1324)(2.15) = 55.0 1.557 

H 2 O 1.0179 0.4626 18 (359 / 18)(1.0179)(2.15) = 43.5 1.231 

N 2 (total) 10.460 4.7545 28 (359 / 28)(10.460)(2.15) = 288.5 8.167 

SO 2 0.0068 0.0031 64 (359 / 64)(0.0068)(2.15) = 0.82 0.023 

ft 3 (m 3 ) of flue gas per lb (kg) of oil burned = 387.82 10.978 

In this calculation, the temperature correction factor 2.15 = absolute flue-gas temperature, °R/ 

absolute atmospheric temperature, °R = (600 + 460)/(32 + 460). The total weight of N 2 in the flue 

gas is the sum of the N 2 in the combustion air and the fuel, or 10.4580 + 0.0022 = 10.4602 lb 

(4.707 kg). 


CO 2 , wet basis = 55.0/387.82 = 0.142, or 14.2 percent. CO 2 , dry basis = 55.0/(387.2 − 43.5) = 0.160, 

or 16.0 percent. 

6. Compute the air required with stated excess flow 

The pounds (kilograms) of air per pound (kilogram) of oil with 20 percent excess air = (1.20) 

(13.6176) = 16.3411 lb (7.353 kg) of air per pound (kilogram) of oil burned. 


The weight of the products of combustion = product weight for perfect combustion, lb + (percent 

excess air)(air for perfect combustion, lb) = 14.6173 + (0.20)(13.6176) = 17.3408 lb (7.803 kilogram) 

of flue gas per pound (kilogram) of oil burned with 20 percent excess air. 


The volume of excess air in the products of combustion is found by converting from the weight to 

the volumetric analysis and correcting for temperature as in step 4, using the air weight from step 2 

for perfect combustion and the excess-air percentage, or (13.6176)(0.20)(359/28.95)(2.15) = 72.7 ft 3 

(2.058 m 3 ). Add this to the volume of the products of combustion found in step 4, or 387.82 + 72.70 = 

460.52 ft 3 (13.037 m 3 ). 

By using the procedure in step 5, the percent CO 2 , wet basis = 55.0/460.52 = 0.1192, or 11.92 percent. 

The percent CO 2 , dry basis = 55.0/(460.52 − 43.5) = 0.1318, or 13.18 percent.



type of fuel oil—paraffin-base, asphalt-base, Bunker C, no. 2, 3, 4, or 5—from any source, domestic 

or foreign, in any type of furnace—boiler, heater, process, or waste-heat. When the air used for 

combustion contains moisture, as is usually true, this moisture is added to the combustion-formed 

moisture appearing in the products of combustion. Thus, for 80°F (26.7°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 (kilogram) of air used and is a commonly assumed standard 

in combustion calculations. 

natural Gas Fuel Combustion in a FurnaCe 


A natural gas has the following volumetric analysis at 60°F (15.5°C): CO 2 = 0.004; CH 4 = 0.921; 

C 2 H 6 = 0.041; N 2 = 0.034; total = 1.000. This natural gas 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 natural gas burned, and the volume of the flue gas, at the boiler exit temperature 

of 650°F (343°C), per pound (kilogram) of natural gas burned; air required with 20 percent 

excess air, and the volume of gas formed with this excess: CO 2 percentage in the flue gas on a dry 

and wet basis. 

1. Compute the weight of oxygen required per pound of gas 

The same general steps as given in the previous calculation procedures will be followed, except that 

they will be altered to make allowances for the differences between natural gas and coal. 

The composition of the gas is given on a volumetric basis, which is the usual way of expressing a 

fuel-gas analysis. To use the volumetric-analysis data in combustion calculations, they must be converted 

to a weight basis. This is done by dividing the weight of each component by the total weight 

of the gas. A volume of 1 ft 3 (1 m 3 ) of the gas is used for this computation. Find the weight of each 

component and the total weight of 1 ft 3 (1 m 3 ) as follows, using the properties of the combustion 

elements and compounds given in Table 1: 

Density 

Component weight = 

column 1 × column 2 

Component Percent by volume lb/ft 3 kg/m 3 lb/ft 3 kg/m 3 

CO 2 0.004 0.1161 1.859 0.0004644 0.007 

CH 4 0.921 0.0423 0.677 0.0389583 0.624 

C 2 H 6 0.041 0.0792 1.268 0.0032472 0.052 

N 2 0.034 0.0739 0.094 0.0025026 0.040 

Total 1.000 0.0451725 0.723 

Percent CO 2 = 0.0004644/0.0451725 = 0.01026, or 1.03 percent 

Percent CH 4 by weight = 0.0389583/0.0451725 = 0.8625, or 86.25 percent 

Percent C 2 H 6 by weight = 0.0032472/0.0451725 = 0.0718, or 7.18 percent 

Percent N 2 by weight = 0.0025026/0.0451725 = 0.0554, or 5.54 percent 

The sum of the weight percentages = 1.03 + 86.25 + 7.18 + 5.54 = 100.00. This sum checks the 

accuracy of the weight calculation, because the sum of the weights of the component parts should 

equal 100 percent.

1.8 

TABLE 1 Properties of Combustion Elements* 

Heat value, Btu (kJ) 

Nature 

At 14.7 lb / in2 (abs) (101.3 kPa), 

60°F (15.6°C) 

Per ft3 (m3 ) at 

14.7 lb/in2 (abs) 

(101.3 kPa), 60°F 

(15.6°C) Per mole 

Per lb (kg) 

Gas or 

solid Combustible 

Volume, ft3 /lb 

(m3 /kg) 

Weight, lb/ft3 (kg/m3 ) 

— 

0.0053 (0.0849) 

— 

0.0739 (1.183) 

0.0423 (0.677) 

0.0686 (1.098) 

0.0739 (1.183) 

0.0792 (1.268) 

0.0844 (1.351) 

0.0739 (1.183) 

0.0765 (1.225) 

0.1161 (1.859) 

0.0475 (0.760) 

Molecular 

weight 

Element or compound Formula 

174,500 

123,100 

129,600 

122,400 

384,000 

562,000 

622,400 

668,300 

— 

325 (12,109) 

— 

323 (12,035) 

1,012 (37,706) 

1,483 (55,255) 

1,641 (61,141) 

1,762 (65,650) 

14,540 (33,820) 

61,000 (141,886) 

4,050 (9,420) 

4,380 (10,187) 

24,000 (55,824) 

21,500 (50,009) 

22,200 (51,637) 

22,300 (51,870) 

Yes 

Yes 

Yes 

Yes 

Yes 

Yes 

Yes 

Yes 

S 

G 

S 

G 

G 

G 

G 

G 

G 

G 

G 

G 

G 

— 

188 (11.74) 

— 

13.54 (0.85) 

23.69 (1.48) 

14.58 (0.91) 

13.54 (0.85) 

12.63 (0.79) 

11.84 (0.74) 

13.52 (0.84) 

13.07 (0.82) 

8.61 (0.54) 

21.06 (1.31) 

12 

2.02 † 

32 

28 

16 

26 

28 

30 

32 

28 

29 

44 

18 

C 

H2 S 

CO 

CH4 C2H2 C2H4 C2H6 O2 N2 — 

CO2 H2O Carbon 

Hydrogen 

Sulfur 

Carbon monoxide 

Methane 

Acetylene 

Ethylene 

Ethane 

Oxygen 

Nitrogen 

Air ‡ 

Carbon dioxide 

Water 

*P. W. Swain and L. N. Rowley, “Library of Practical Power Engineering” (collection of articles published in Power). 

† For most practical purposes, the value of 2 is sufficient. 

‡ The molecular weight of 29 is merely the weighted average of the molecular weight of the constituents.

TABLE 2 Chemical Reactions 


Next, find the oxygen required for combustion. Since both the CO 2 and N 2 are inert, they do not 

take part in the combustion; they pass through the furnace un changed. Using the molecular weights 

of the remaining components in the gas and the weight percentages, we have 

Compound × Molecular-weight ratio = lb (kg) O 2 required 

CH 4 ; 0.0718 

C 2 H 6 ; 0.0718 

lb (kg) external O 2 required 

per lb (kg) fuel 

× 

× 

64/16 

112/30 

= 

= 

= 

3.4500 (1.553) 

0.2920 (0.131) 

3.7420 (1.684) 

In this calculation, the molecular-weight ratio is obtained from the equation for the combustion 

chemical reaction, or CH 4 + 2O 2 = CO 2 + 2H 2 O, that is, 16 + 64 = 44 + 36, and C 2 H 6 + 7/2O 2 = 

2CO 2 + 3H 2 O 2 , that is 30 + 112 = 88 + 54. See Table 2 from these and other useful chemical 

reactions in combustion. 

Combustible substance Reaction Mols lb (kg)* 

Carbon to carbon monoxide C + ½O 2 = CO 1 + ½ = 1 12 + 16 = 28 

Carbon to carbon dioxide C + O 2 = CO 2 1 + 1 = 1 12 + 16 = 28 

Carbon monoxide to carbon dioxide CO + ½O 2 = CO 2 1 + ½ = 1 28 + 16 = 44 

Hydrogen H 2 + ½O 2 = H 2 O 1 + ½ = 1 2 + 16 = 18 

Sulfur to sulfur dioxide S + O 2 = SO 2 1 + 1 = 1 32 + 32 = 64 

Sulfur to sulfur trioxide S + 3/2O 2 = SO 3 1 + 2/2 = 1 32 + 48 = 80 

Methane CH 4 + 2O 2 = CO 2 + 2H 2 O 1 + 2 = 1 + 2 16 + 64 = 44 + 36 

Ethane C 2 H 6 + 7/2O 2 = 2CO 2 + 3H 2 O 1 + 7/2 = 2 + 3 30 + 112 = 88 + 54 

Propane C 3 H 8 + 5O 2 = 3CO 2 + 4H 2 O 1 + 5 = 3 + 4 44 + 160 = 132 + 72 

Butane C 4 H 10 + 13/2O 2 = 4CO 2 + 5H 2 O 1 + 12/2 = 4 + 5 58 + 208 = 176 + 90 

Acetylene C 2 H 2 + 5/2O 2 = 2CO 2 + H 2 O 1 + 5/2 = 2 + 2 26 + 80 = 88 + 18 

Ethylene C 2 H 4 + 3O 2 = 2CO 2 + 2H 2 O 1 + 3 = 2 + 2 28 + 96 = 88 + 36 

*Substitute the molecular weights in the reaction equation to secure lb (kg). The lb (kg) on each side of the equation must balance. 


The weight of nitrogen associated with the required oxygen = (3.742)(0.768/0.232) = 12.39 lb (5.576 kg). 

The weight of air required = 12.39 + 3.742 = 16.132 lb/lb (7.259 kg/kg) of gas burned. 

3. Compute the weight or the products of combustion 

Products of combustion 

Fuel constituents + Oxygen = lb kg 

CO2 ; 0.0103: inert but passes 

through the furnace 

= 0.010300 0.005 

CH4 ; 0.8625 

C2H6 ; 0.003247 

N2 ; 0.0554; inert but passes 

through the furnace 

+ 

+ 

3.45 

0.2920 

= 

= 

= 

4.312500 

0.032447 

0.055400 

1.941 

0.015 

0.025 

Outside N2 from step 2 

lb (kg) of flue gas per lb (kg) of 

natural gas burned 

= 

= 

12.390000 

16.800347 

5.576 

7.562

1.10 SECTION 1 


The products of complete combustion of any fuel that does not contain sulfur are CO 2 , H 2 O, and 

N 2 . Using the combustion equation in step 1, compute the products of combustion thus: CH 4 + 2O 2 = 

CO 2 + H 2 O; 16 + 64 = 44 + 36; or the CH 4 burns to CO 2 in the ratio of 1 part CH 4 to 44/16 parts CO 2 . 

Since, from step 1, there is 0.03896 lb CH 4 per ft 3 (0.624 kg/m 3 ) of natural gas, this forms (0.03896) 

(44/16) = 0.1069 lb (0.048 kg) of CO 2 . Likewise, for C 2 H 6 , (0.003247)(88/30) = 0.00952 lb (0.004 

kg). The total CO 2 in the combustion products = 0.00464 + 0.1069 + 0.00952 = 0.11688 lb (0.053 

kg), where the first quantity is the CO 2 in the fuel. 

Using a similar procedure for the H 2 O formed in the products of combustion by CH 4 , we find 

(0.03896)(36/16) = 0.0875 lb (0.039 kg). For C 2 H 6 , (0.003247)(54/30) = 0.005816 lb (0.003 kg). The 

total H 2 O in the combustion products = 0.0875 + 0.005816 = 0.093316 lb (0.042 kg). 

Step 2 shows that 12.39 lb (5.58 kg) of N 2 is required per lb (kg) of fuel. Since 1 ft 3 (0.028 m 3 ) of 

the fuel weights 0.04517 lb (0.02 kg), the volume of gas which weighs 1 lb (2.2 kg) is 1/0.04517 = 

22.1 ft 3 (0.626 m 3 ). Therefore, the weight of N 2 per ft 3 of fuel burned = 12.39/22.1 = 0.560 lb (0.252 kg). 

This, plus the weight of N 2 in the fuel, step 1, is 0.560 + 0.0025 = 0.5625 lb (0.253 kg) of N 2 in the 

products of combustion. 

Next, find the total weight of the products of combustion by taking the sum of the CO 2 , H 2 O, and 

N 2 weights, or 0.11688 + 0.09332 + 0.5625 = 0.7727 lb (0.35 kg). Now convert each weight to ft 3 at 

650°F (343°C), the temperature of the combustion products, or: 

Weight 

Molecular 

Volume at 


H 2 O 0.09332 0.04233 18 (379/18)(0.09332)(2.255) = 4.425 0.1252 

N 2 (total) 0.5625 0.25515 28 (379/28)(0.5625)(2.255) = 17.190 0.4866 

ft 3 (m 3 ) of flue gas per ft 3 (m 3 ) of natural-gas fuel = 23.880 0.6759 

In this calculation, the value of 379 is used in the molecular-weight ratio because at 60°F (15.6°C) 

and 14.7 lb/in 2 (abs) (101.3 kPa), the volume of 1 lb (0.45 kg) of any gas = 379/gas molecular weight. 

The fuel gas used is initially at 60°F (15.6°C) and 14.7 lb/in 2 (abs) (101.3 kPa). The ratio 2.255 = 

(650 + 460)/(32 + 460). 


CO 2 , wet basis = 2.265/23.88 = 0.947, or 9.47 percent. CO 2 , dry basis = 2.265/(23.88 − 4.425) = 0.1164, 

or 11.64 percent. 


With 20 percent excess air, (1.20)(16.132) = 19.3584 lb of air per lb (8.71 kg/kg) of natural gas, or 

19.3584/22.1 = 0.875 lb of air per ft 3 (13.9 kg/m 3 ) of natural gas. See step 4 for an explanation of 

the value 22.1. 


Weight of the products of combustion = product weight for perfect combustion, lb + (percent excess 

air) (air for perfect combustion, lb) = 16.80 + (0.20)(16.132) = 20.03 lb (9.01 kg). 


The volume of excess air in the products of combustion is found by converting from the weight to 

the volumetric analysis and correcting for temperature as in step 4, using the air weight from step 2 for 

perfect combustion and the excess-air percentage, or (16.132/22.l)(0.20)(379/28.95)(2.255) = 4.31 ft 3 


28.19 ft 3 (0.798 m 3 ). 

By the procedure in step 5, the percent CO 2 , wet basis = 2.265/28.19 = 0.0804, or 8.04 percent. 

The percent CO 2 , dry basis = 2.265/(28.19 − 4.425) = 0.0953, or 9.53 percent.



type of gas used as a fuel—natural gas, blast-furnace gas, coke-oven gas, producer gas, water gas, 

sewer gas—from any source, domestic or for eign, in any type of furnace—boiler, heater, process, or 

waste-heat. When the air used for combustion contains moisture, as is usually true, this moisture is 

added to the combustion-formed moisture appearing in the products of combustion. Thus, for 80°F 

(26.7°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. 

WooD Fuel Combustion in a FurnaCe 


The weight analysis of a yellow-pine wood fuel is: C = 0.490; H 2 = 0.074; O 2 = 0.406; N 2 = 0.030. 

Determine the weight of oxygen and air required with perfect combustion and with 20 percent excess 

air. Find the weight and volume of the products of combustion under the same conditions, and the 

wet and dry CO 2 . The flue-gas temperature is 600°F (316°C). The air supplied for combustion has a 

mois ture content of 0.013 lb/lb (0.006 kg/kg) of dry air. 

1. Compute the weight of oxygen required per pound (kilogram) of wood 

The same general steps as given in earlier calculation procedures will be followed; consult them for 

a complete explanation of each step. Using the molecular weight of each element, we have 

Element × Molecular-weight ratio = lb (kg) O 2 required 

C; 0.490 × 32 /12 = 1.307 (0.588) 

H 2 ; 0.074 × 16 / 2 = 0.592 (0.266) 

O 2 ; 0.406; decreases external O 2 required = −0.406 (−0.183) 

N 2 ; 0.030 inert in combustion 

Total 1.000 


2. Compute the weight of air required for complete combustion 

The weight of nitrogen associated with the required oxygen = (1.493)(0.768 / 0.232) = 4.95 lb (2.228 kg). 

The weight of air required = 4.95 + 1.493 = 6.443 lb / lb (2.899 kg / kg) of wood burned, if the air is 

dry. But the air contains 0.013 lb of moisture per lb (0.006 kg / kg) of air. Hence, the total weight of 

the air = 6.443 + (0.013)(6.443) = 6.527 lb (2.937 kg). 


Use the following relation: 

Fuel constituents + Oxygen = Products of combustion, lb (kg) 

C; 0.490 + 1.307 = 1.797 (0.809) = CO 2 

H 2 ; 0.074 + 0.592 = 0.666 (0.300) = H 2 O 

O 2 ; not a product of combustion 

N 2 ; inert but passes through the furnace = 0.030 (0.014) = N 2 

Outside N 2 from step 2 = 4.950 (2.228) = N 2 

Outside moisture from step 2 = 0.237 (0.107) 

lb (kg) of flue gas per lb (kg) of wood burned = 7.680 (3.458)



Use, as before, the following tabulation: 

Weight 

Molecular 

Volume at 


H 2 O (fuel) 0.666 0.300 18 (359/18)(0.666)(2.15) = 28.6 0.810 

N 2 (total) 4.980 2.241 28 (359/28)(4.980)(2.15) = 137.2 3.884 

H 2 O (outside air) 0.837 0.377 18 (359/18)(0.837)(2.15) = 35.9 10.16 

Cu ft (m 3 ) of flue gas per lb (kg) of oil 233.2 6.602 

In this calculation the temperature correction factor 2.15 = (absolute flue-gas tem perature, °R)/ 

(absolute atmospheric temperature, °R) = (600 + 460)/(32 + 460). The total weight of N 2 is the sum 

of the N 2 in the combustion air and the fuel. 


The CO 2 , wet basis = 31.5/233.2 = 0.135, or 13.5 percent. The CO 2 , dry basis = 31.5/(233.2 − 28.6 − 

35.9) = 0.187, or 18.7 percent. 


With 20 percent excess air, (1.20)(6.527) = 7.832 lb (3.524 kg) of air per lb (kg) of wood burned. 


The weight of the products of combustion = product weight for perfect combustion, lb + (percent 

excess air)(air for perfect combustion, lb) = 8.280 + (0.20)(6.527) = 9.585 lb (4.313 kg) of flue gas 

per lb (kg) of wood burned with 20 percent excess air. 


The volume of the excess air in the products of combustion is found by converting from the weight 

to the volumetric analysis and correcting for temperature as in step 4, using the air weight from step 2 

for perfect combustion and the excess-air per centage, or (6.527)(0.20)(359/28.95)(2.15) = 34.8 ft 3 


268.0 ft 3 (7.587 m 3 ). 

By using the procedure in step 5, the percent CO 2 , wet basis = 31.5/268 = 0.1174, or 11.74 percent. 

The percent CO 2 , dry basis = 31.5/(268 − 28.6 − 35.9 − 0.20 × 0.837) = 0.155, or 15.5 percent. In 

the dry-basis calculation, the factor (0.20)(0.837) is the outside moisture in the excess air. 


type of wood or woodlike fuel—spruce, cypress, maple, oak, sawdust, wood shavings, tanbark, 

bagesse, peat, charcoal, redwood, hemlock, fir, ash, birch, cottonwood, elm, hickory, walnut, 

chopped trimmings, hogged fuel, straw, corn, cottonseed hulls, city refuse—in any type of furnace— 

boiler, heating, process, or waste-heat. Most of these fuels contain a small amount of ash—usually 

less than 1 percent. This was ignored in this calculation procedure because it does not take part in 

the combustion. 

Industry is making greater use of discarded process waste to generate electricity and steam by 

burning the waste in a steam boiler. An excellent example is that of Agrilectric Power Partners Ltd., 

Lake Charles, LA. This plant burns rice hulls from its own process and buys other producers’ surplus 

rice hulls for continuous opera tion. Their plant is reported as the first small-power-production facility 

to operate on rice hulls. 

By burning the waste rice hulls, Agrilectric is confronting, and solving, an environmental nuisance 

often associated with rice processing. When rice hulls are disposed of by being spread on land 

adjacent to the mill, they often smolder, cre ating continuous, uncontrolled burning. Installation of 

its rice-hull burning, electric-generating plant has helped Agrilectric avoid the costs associated with 

landfilling and disposal, as well as potential environmental problems.


The boiler supplies steam for a turbine-generator with an output ranging from 11.2 to 11.8 MW. 

Excess power that cannot be used in the plant is sold to the local utility at a negotiated price. Thus, 

the combustion of an industrial waste is produc ing useful power while eliminating the environmental 

impact of the waste. The advent of PURPA (Public Utility Regulatory Policies Act) requiring local 

utilities to purchase power from such plants has been a major factor in the design, devel opment, and 

construction of many plants by food processors to utilize waste ma terials for combustion and power 

production. 

Combustion analYsis bY tHe molal metHoD 


A coal fuel has this ultimate analysis: C = 0.8339; H 2 = 0.0456; O 2 = 0.0505; N 2 = 0.0103; S = 0.0064; 

ash = 0.0533; total = 1.000. This coal is completely burned in a boiler furnace. Using the molal method, 

determine the weight of air required per lb (kg) of coal with complete combustion. How much air 

is needed with 25 percent excess air? What is the weight of the combustion products with 25 percent 

excess air? The combustion air contains 0.013 lb of moisture per lb (0.006 kg/kg) of air. 

1. Convert the ultimate analysis to moles 

A mole of any substance is an amount of the substance having a weight equal to the molecular weight 

of the substance. Thus, 1 mol of carbon is 12 lb (5.4 kg) of carbon, because the molecular weight 

of carbon is 12. To convert an ultimate anal ysis of a fuel to moles, assume that 100 lb (45 kg) of the 

fuel is being considered. Set up a tabulation thus: 

Weight 

Ultimate analysis, % lb kg Molecular weight Moles per 100-lb (45-kg) fuel 

C = 0.8339 83.39 37.526 12 6.940 

H 2 = 0.0456 4.56 2.052 2 2.280 

O 2 = 0.0505 5.05 2.678 32 0.158 

N 2 = 0.0103 1.03 0.464 28 0.037 

S = 0.0064 0.64 0.288 32 

Ash = 0.0533 5.33 2.399 Inert 

Total 100.00 45.407 9.435 

2. Compute the mols of oxygen for complete combustion 

From Table 2, the burning of carbon to carbon dioxide requires 1 mol of carbon and 1 mol of oxygen, 

yielding 1 mol of CO 2 . Using the molal equations in Table 2 for the other elements in the fuel, set 

up a tabulation thus, entering the product of columns 2 and 3 in column 4: 

(1) 

Element 

(2) 

Moles per 100-lb 

(45-kg) fuel 

(3) 

Moles O 2 per 100-lb 

(45-kg) fuel 

(4) 

Total moles O2 C 6.940 1.00 6.940 

H 2 2.280 0.5 1.140 

O 2 0.158 Reduces O 2 required −0.158 

N 2 0.037 Inert in combustion 

S 0.020 1.00 0.020 

Total moles of O 2 required — — 7.942


3. Compute the mols of air for complete combustion 

Set up a similar tabulation for air, thus: 

(1) 

Element 

(2) 

Moles per 100-lb 

(45-kg) fuel 

(3) 

Moles air per 100-lb 

(45-kg) fuel 

(4) 

Total moles air 

C 6.940 4.76 33.050 

H 2 2.280 2.38 5.430 

O 2 0.158 Reduces O 2 required −0.752 

N 2 0.037 Inert in combustion 

S 0.020 4.76 0.095 

Total moles of air required — 37.823 

In this tabulation, the factors in column 3 are constants used for computing the total moles of air 

required for complete combustion of each of the fuel elements listed. These factors are given in the 

Babcock & Wilcox Company—Steam: Its Generation and Use and similar treatises on fuels and 

their combustion. A tabu lation of these factors is given in Table 3. 

TABLE 3 Molal Conversion Factors 

Mol/mol of combustible for complete combustion; 

no excess air 

For combustion Combustion products 

Element or compound O 2 N 2 Air CO 2 H 2 O N 2 

Carbon,* C 1.0 3.76 4.76 1.0 — 3.76 

Hydrogen, H 2 0.5 0.188 2.38 — 1.0 1.88 

Oxygen, O 2 

Nitrogen, N 2 

Carbon monoxide, CO 0.5 1.88 2.38 1.0 — 1.88 

Carbon dioxide, CO 2 

Sulfur,* S 1.0 3.76 4.76 1.0 — 3.76 

Methane, CH 4 2.0 7.53 0.53 1.0 2.0 7.53 

Ethane, C 2 H 6 3.5 13.18 16.68 2.0 3.0 13.18 

*In molal calculations, carbon and sulfur are considered as gases. 

An alternative, and simpler, way of computing the moles of air required is to convert the required 

O 2 to the corresponding N 2 and find the sum of the O 2 and N 2 . Or, 376O 2 = N 2 ; N 2 + O 2 = moles of 

air required. The factor 3.76 converts the required O 2 to the corresponding N 2 . These two relations 

were used to convert the 0.158 mol of O 2 in the above tabulation to moles of air. 

Using the same relations and the moles of O 2 required from step 2, we get (3.76)(7.942) = 29.861 

mol of N 2 . Then 29.861 + 7.942 = 37.803 mol of air, which agrees closely with the 37.823 mol computed 

in the tabulation. The difference of 0.02 mol is traceable to roundings. 

4. Compute the air required with the stated excess air 

With 25 percent excess, the air required for combustion = (125/100)(37.823) = 47.24 mol.


5. Compute the mols of combustion products 

Using data from Table 3, and recalling that the products of combustion of a sulfur-containing fuel are 

CO 2 , H 2 O, and SO 2 , and that N 2 and excess O 2 pass through the furnace, set up a tabulation thus: 

(1) 

Moles per 100-lb (45-kg) fuel 

(2) 

Mol/mol of 

combustible 

(3) 

Moles of combustion 

product per 100-lb 

(45-kg) fuel 

CO 2 ; 6.940 1 6.940 

H 2 O; 2.280 + (47.24)(0.021 + 0.158) — 3.430 

SO 2 ; 0.020 1 0.202 

N 2 ; (47.24)(0.79) — 37.320 

Excess O 2 ; (1.25)(7.942) − 7.942 — 1.986 

Total moles, wet combustion products = 49.878 

Total moles, dry combustion products = 49.878 − 3.232 

= 46.646 

In this calculation, the total moles of CO 2 is obtained from step 2. The moles of H 2 in 100 lb (45 kg) of 

the fuel, 2.280, is assumed to form H 2 O. In addition, the air from step 4, 47.24 mol, contains 0.013 lb 

of moisture per lb (0.006 kg/kg) of air. This moisture is converted to moles by dividing the molecular 

weight of air, 28.95, by the molecular weight of water, 18, and multiplying the result by the moisture 

content of the air, or (28.95/18)(0.013) = 0.0209, say 0.021 mol of water per mol of air. The product 

of this and the moles of air gives the total moles of moisture (water) in the combustion products per 

100 lb (45 kg) of fuel fired. To this is added the moles of O 2 , 0.158, per 100 lb (45 kg) of fuel, because 

this oxygen is assumed to unite with hydrogen in the air to form water. The nitrogen in the products of 

combustion is that portion of the moles of air required, 47.24 mol from step 4, times the proportion of 

N 2 in the air, or 0.79. The excess O 2 passes through the furnace and adds to the combustion products 

and is computed as shown in the tabulation. Subtracting the total moisture, 3430 mol, from the total (or 

wet) com bustion products gives the moles of dry combustion products. 

Related Calculations. Use this method for molal combustion calculations for all types of fuels— 

solid, liquid, and gaseous—burned in any type of furnace—boiler, heater, process, or waste-heat. 

Select the correct factors from Table 3. 

estimatinG tHe temPerature oF tHe Final 

ProDuCts oF Combustion 


Pure carbon is burned to carbon dioxide at constant pressure in an insulated cham ber. An excess air 

quantity of 20 percent is used and the carbon and the air are both initially at 77°F (25°C). Assume 

that the reaction goes to completion and that there is no dissociation. Calculate the final product’s 

temperature using the following constants: Heating value of carbon, 14,087 Btu/lb (32.74 × 10 3 kJ/kg); 

constant-pressure specific heat of oxygen, nitrogen, and carbon dioxide are 0.240 Btu /lb m (0.558 kJ/kg), 

0.285 Btu/lb m (0.662 kJ/kg), and 0.300 Btu/lb (0.697 kJ/kg), respectively. 

1. Establish the chemical equation for complete combustion with 100 percent air 

With 100 percent air: C + O 2 + 3.78N 2 → CO 2 + 3.78N 2 , where approximate molecular weights are: 

for carbon, MC = 12; oxygen, MO 2 = 32; nitrogen, MN 2 = 28; carbon dioxide, MCO 2 = 44. See the 

Related Calculations of this procedure for a general description of the 3.78 coefficient for N 2 . 

2. Establish the chemical equation for complete combustion with 20 percent excess air 

With 20 percent excess air: C + 1.2O 2 + (1.2 × 3.78)N 2 → CO 2 + 0.2O 2 + (1.2 × 3.78)N 2 .


3. Compute the relative weights of the reactants and products of the combustion process 

Relative weight = moles × molecular weight. Coefficients of the chemical equation in step 2 represent 

the number of moles of each component. Hence, for the reactants, the relative weights are: for C = 

1 × MC = 1 × 12 = 12; O 2 = 1.2 × MO 2 = 1.2 × 32 = 38.4; N 2 = (1.2 × 3.78)MN 2 = (1.2 × 3.78 × 28) = 127. 

For the products, relative weights are: for CO 2 = 1 × MCO 2 = 1 × 44 = 44; O 2 , = 0.2 × MO 2 = 

0.2 × 32 = 6.4; N 2 = 127, unchanged. It should be noted that the total relative weight of the reactants 

equal that of the products at 177.4. 

4. Compute the relative weights of the products of combustion on the basis of a per unit relative 

weight of carbon 

Since the relative weight of carbon, C = 12 in step 3; hence, on the basis of a per unit relative 

weight of carbon, the corresponding relative weights of the products are: for carbon dioxide, wCO 2 = 

MCO 2 /12 = 44/12 = 3.667; oxygen, wO 2 = MO 2 /12 = 6.4/12 = 0.533; nitrogen, wN 2 = MN 2 /12 = 

127/12 = 10.58. 

5. Compute the final product’s temperature 

Since the combustion chamber is insulated, the combustion process is considered adiabatic. Hence, 

on the basis of a per unit mass of carbon, the heating value (HV) of the carbon = the corresponding 

heat content of the products. Thus, relative to a temperature base of 77°F (25°C), 1 × HVC = 

[(wCO 2 × c pCO 2 ) + (wO 2 × c p O 2 ) + (wH 2 + c p N 2 )](t 2 − 77), where the heating value of carbon, HVC = 

14,087 Btu/lb m (32.74 × 10 3 kJ/kg); the constant-pressure specific heat of carbon dioxide, oxygen, 

and nitrogen are c p CO 2 = 0.300 Btu/lb (0.697 kJ/kg), cpO 2 = 0.240 Btu/lb (0.558 kJ/kg), and c p N 2 = 

0.285 Btu/lb (0.662 kJ/kg), respectively; final product temperature is t 2 ; other values as before. Then, 

1 × 14,087 = [(3.667 × 0.30) + (0.533 × 0.24) + (10.58 × 0.285)(t 2 − 77)]. Solving, t 2 = 3320 + 77 = 

3397°F (1869°C). 

Related Calculations. In the above procedure it is assumed that the carbon is burned in dry air. Also, 

the nitrogen coefficient of 3.78 used in the chemical equation in step 1 is based on a theoretical composition 

of dry air as 79.1 percent nitrogen and 20.9 percent oxygen by volume, so that 79.1/20.9 = 3.78. 

For a more detailed description of this coefficient see Section 3 of this handbook. 

Determination oF tHe saVinGs ProDuCeD bY 

PreHeatinG Combustion air 


A 20,000 ft 2 (1858 m 2 ) building has a calculated total seasonal heating load of 2,534,440 MBH (thousand 

Btu) (2674 MJ). The stack temperature is 600°F (316°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 temperature 

by 250°F (139°C) to 350°F (177°C) and provide the burner with preheated air. How much fuel oil will 

be saved? What will be the monetary saving if fuel oil is priced at $1.10 per gallon? 

1. Compute the total combustion air required by this boiler 

A general rule used by design engineers is that 1 ft 3 (0.0283 m 3 ) of combustion air is required for 

each 100 Btu (105.5 J) released during combustion. To compute the combustion air required, use 

the relation CA = H/100 × Boiler efficiency, expressed as a decimal, where CA = annual volume of 

combustion air, ft 3 (m 3 ); H = total seasonal heating load, Btu/yr (kJ/yr). Substituting for this boiler, 

CA = (2,534,400)(1000)/100 × 0.75 = 33,792,533 ft 3 /yr (956.329 m 3 /yr). 

2. Calculate the annual energy savings 

The energy savings, ES = (stack temperature reduction, deg F)(ft 3 air per year)(0.018), where the 

constant 0.018 is the specific heat of air. Substituting, ES = (250)(33,792,533)(0.018) = 152,066,399 

Btu/yr (160,430 kJ/yr).


With a boiler efficiency of 75 percent, each gallon of oil releases 0.75 × 140,000 Btu/gal = 

105,000 Btu (110.8 jk). Hence, the fuel saved, FS = ES/usable heat in fuel, Btu/gal. Or, FS = 

152,066,399/105,000 = 1448.3 gal/yr (5.48 m 3 /yr). 

With fuel oil at $1.10 per gallon, the monetary savings will be $1.10 (1448.3) = $1593.13. If the 

preheater cost $6000, the simple payoff time would be $6000/1593.13 = 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 ft 3 of air (0.283 m 3 ) per 1 ft 3 (0.0283 m 3 ) of natural gas; 1300 ft 3 

of air (36.8 m 3 ) per gal (0.003785 m 3 ) of No. 2 fuel oil; 1450 ft 3 of air (41 m 3 ) per gal of No. 5 fuel 

oil; 1500 ft 3 of air (42.5 m 3 ) per gal of No. 6 fuel oil. These values agree with that used in the 

above computation—i.e. 100 ft 3 per 100 Btu of 140,000 Btu per gal oil = 140,000/100 = 1400 ft 3 

per gal (39.6 m 3 /0.003785 m 3 ). 

This procedure is the work of Jerome F. Mueller, P.E. of Mueller Engineering Corp. 

Combustion CalCulations usinG tHe million btu 

(1.055 mJ) metHoD 


The energy absorbed by a steam boiler fired by natural gas is 100-million Btu/h (29.3 MW). Boiler 

efficiency on a higher heating value (HHV) basis is 83 percent. If 15 percent excess air is used, 

determine the total air and flue-gas quantities produced. The approximate HHV of the natural gas 

is 23,000 Btu/lb (53,590 kJ/kg). Ambient air temperature is 80°F (26.7°C) and relative humidity 

is 65 percent. How can quick estimates be made of air and flue-gas quantities in boiler operations 

when the fuel analysis is not known? 

1. Determine the energy input to the boiler 

The million Btu (1.055 MJ) method of combustion calculations is a quick way of estimating air and 

flue-gas quantities generated in boiler and heater operations when the ultimate fuel analysis is not 

available and all the engineer is interested in is good estimates. Air and flue-gas quantities determined 

may be used to calculate the size of fans, ducts, stacks, etc. 

It can be shown through comprehensive calculations that each fuel such as coal, oil, natural gas, 

bagasse, blast-furnace gas, etc. requires a certain amount of dry stoichiometric air per million Btu 

(1.055 MJ) fired on an HHV basis and that this quantity does not vary much with the fuel analysis. 

The listing below gives the dry air required per million Btu (1.055 MJ) of fuel fired on an HHV 

basis for various fuels. 

Combustion Constants for Fuels 

Fuel 

Blast furnace gas 

Bagasse 

Carbon monoxide gas 

Refinery and oil gas 

Natural gas 

Furnace oil and lignite 

Bituminous coals 

Anthracite coal 

Coke 

Constant, lb dry air per million Btu 

(kg/MW) 

575 (890.95) 

650 (1007.2) 

670 (1038.2) 

720 (1115.6) 

730 (1131.1) 

745 − 750 (1154.4 − 1162.1) 

760 (1177.6) 

780 (1208.6) 

800 (1239.5)


To determine the energy input to the boiler, use the relation Q f = (Q s )/E h , where Q f = energy input 

by the fuel, Btu/h (W); Q s = energy absorbed by the steam in the boiler, Btu/h (W); E h = efficiency 

of the boiler on an HHV basis. Substituting for this boiler, Q f = 100/0.83 = 120.48 million Btu/h on 

an HHV basis (35.16 MW). 

2. Estimate the quantity of dry air required by this boiler 

The total air required T a = (Q f )(Fuel constant from list above). For natural gas, T a = (120.48)(730) = 

87,950 lb/h (39,929 kg/h). With 15 percent excess air, total air required = (1.15)(87,950) = 101,142.5 lb/h 

(45,918.7 kg/h). 

3. Compute the quantity of wet air required 

Air has some moisture because of its relative humidity. Estimate the amount of moisture in dry air in 

M lb/lb (kg/kg) from M = 0.622 (p w )/(14.7 − p w ), where 0.622 is the ratio of the molecular weights 

of water vapor and dry air; p w = partial pressure of water vapor in the air, psia (kPa) = saturated 

vapor pressure (SVP) × relative humidity expressed as a decimal; 14.7 = atmospheric pressure of 

air at sea level (101.3 kPa). From the steam tables, at 80°F (26.7°C), SVP = 0.5069 psia (3.49 kPa). 

Substituting, M = 0.622 (0.5069 × 0.65)/(14.7 − [0.5069 × 0.65]) = 0.01425 lb of moisture/lb of dry 

air (0.01425 kg/kg). 

The total flow rate of the wet air then = 1.0142 (101,142.5) = 102,578.7 lb/h (46,570.7 kg/h). 

To convert to a volumetric-flow basis, recall that the density of air at 80°F (26.7°C) and 14.7 psia 

(101.3 kPa) = 39/(480 + 80) = 0.0722 lb/ft 3 (1.155 kg/m 3 ). In this relation, 39 = a constant and the 

temperature of the air is converted to degrees Rankine. Hence, the volumetric flow = 102,578.7/ 

(60 min/h)(0.0722) = 23,679.3 actual cfm (670.1 m 3 /min). 

4. Estimate the rate of fuel firing and flue-gas produced 

The rate of fuel firing = Qf / HHV = (120.48 × 106 )/23,000 = 5238 lb/h (2378 kg/h). Hence, the total 

flue gas produced = 5238 + 102,578 = 107,816 lb/h (48,948 kg/h). 

If the temperature of the flue gas is 400°F (204.4°C) (a typical value for a natural-gas fired 

boiler), then the density, as in step 3, is: 39/(400 + 460) = 0.04535 lb/ft 3 (0.7256 kg/m 3 ). Hence, the 

volumetric flow = (107,816)/(60 min/h × 0.04535) = 39,623.7 actual cfm (1121.3 m 3 /min). 

Related Calculations. Detailed combustion calculations based on actual fuel gas analysis can be 

performed to verify the constants given in the list above. For example, let us say that the natural-gas 

analysis was: methane = 83.4 percent; ethane = 15.8 percent; nitrogen = 0.8 percent by volume. First 

convert the analysis to a percent weight basis. 

Fuel Percent volume MW Col. 2 × Col. 3 Percent weight 

Methane 83.4 16 1334.4 72.89 

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 × 

Column 3. Thus, the percent methane = (1334.4)/(1334.4 + 474 + 22.4) = 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 = (K for methane)(percent by weight methane from above list) + 

(K for ethane)(percent by weight ethane); or A c = (17.265)(0.7289) + (16.119)(0.2589) = 16.76 lb/lb 

(16.76 kg/kg). 

Next, compute the higher heating value of the fuel (HHV) using the air constants from the 

same reference mentioned above. Or HHV = (heat of combustion for methane)(percent by 

weight methane) + (heat of combustion of ethane)(percent by weight ethane) = (23,879)(0.7289) + 

(22,320)(0.2589) = 23,184 Btu/lb (54,018.7 kJ/kg). Then, the amount of fuel equivalent to 

1,000,000 Btu (1,055,000 kJ) = (1,000,000)/23,184 = 43.1 lb (19.56 kg), which requires, as 

computed above, (43.1)(16.76) = 722.3-lb dry air (327.9 kg), which agrees closely with the value 

given in step 1, above.


Similarly, if the fuel were 100 percent methane, using the steps given above, and suitable constants 

from the same reference work, the air required for combus tion is 17.265 lb/lb (7.838 kg/kg) 

of fuel. HHV = 23,879 Btu/lb (55,638 kJ/kg). Hence, the fuel in 1,000,000 Btu (1,055,000 kJ) = 

(1,000,000)/(23,879) = 41.88 lb (19.01 kg). Then, the dry air per million Btu (1.055 kg) fired = 

(17.265) (41.88) = 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 = 46.17-lb (20.95-kg) fuel. Then, 1 million 

Btu (1,055,000 kJ) requires (15.703)(46.17) = 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 = 100(O 2 − CO 2 )/ 

[0.264 × N 2 – (O 2 – CO/2)], where O 2 = oxygen in the dry flue gas, percent volume; CO = percent 

volume carbon mon oxide; N 2 = percent volume nitrogen. 

You can also estimate excess air from oxygen readings. Use the relation, EA = (constant from 

list below)(O 2 )/(21 − O 2 ). 

Constants for Excess-Air Calculations 

Fuel Constant 

Carbon 100 

Hydrogen 80 

Carbon monoxide 121 

Sulfur 100 

Methane 90 

Oil 94.5 

Coal 97 

Blast furnace gas 223 

Coke oven gas 89.3 

If the percent volume of oxygen measured is 3 on a dry basis in a natural-gas (methane) fired boiler, 

the excess air, EA = (90)[3/(21 – 3)] = 15 percent. 

This procedure is the work of V. Ganapathy, Heat Transfer Specialist, ABCO Industries.

section 1 energy conversion engineering

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