Part 28

BY VOLUME

_______________________________________________________________ | | | | | | | | 1 | 2 | 11 | 12 | 13 | 14 | |_______________|______|____|__|______________|______| | | | | | | | | Carbon | C | 1C | 2 | 2CO{2} | 14.95 | | Carbon | C | 1C | 1 | 2CO | 14.95 | | Carbon Monoxide | CO | 2CO | 1 | 2CO{2} | 12.80 | | Hydrogen | H | 2H | 1 | 2H{2}O | 179.32 | | Methane | CH{4} | 1C4H | 4 | 1CO{2} 2H{2}O| 22.41 | | Sulphur | S | 1S | 2 | 1SO{2} | 5.60 | |________________|______|____|__|______________|_______|

____________________________________________________________ | | | | | | | | 1 | 2 | 15 | 16 | 17 | 18 | |_______________|______|_____|______|______|______| | | | | | | | | Carbon | C | 29.89 | 29.89 | 112.98 | 142.87 | | Carbon | C | 14.95 | 29.89 | 56.49 | 86.38 | | Carbon Monoxide | CO | 6.40 | 12.80 | 24.20 | 37.00 | | Hydrogen | H | 89.66 | 179.32 | 339.09 | 518.41 | | Methane | CH{4} | 44.83 | 67.34 | 169.55 | 236.89 | | Sulphur | S | 11.21 | 11.21 | 42.39 | 53.60 | |________________|______|_____|______|______|_______|

It will be seen from this table that a pound of carbon will unite with 2-2/3 pounds of oxygen to form carbon dioxide, and will evolve 14,600 B. t. u. As an intermediate step, a pound of carbon may unite with 1-1/3 pounds of oxygen to form carbon monoxide and evolve 4450 B. t. u., but in its further conversion to CO{2} it would unite with an additional 1-1/3 times its weight of oxygen and evolve the remaining 10,150 B. t. u. When a pound of CO burns to CO{2}, however, only 4350 B. t. u. are evolved since the pound of CO contains but 3/7 pound carbon.

Air Required for Combustion--It has already been shown that each combustible element in fuel will unite with a definite amount of oxygen. With the ultimate analysis of the fuel known, in connection with Table 31, the theoretical amount of air required for combustion may be readily calculated.

Let the ultimate analysis be as follows:

Per Cent Carbon 74.79 Hydrogen 4.98 Oxygen 6.42 Nitrogen 1.20 Sulphur 3.24 Water 1.55 Ash 7.82 ------ 100.00

When complete combustion takes place, as already pointed out, the carbon in the fuel unites with a definite amount of oxygen to form CO{2}. The hydrogen, either in a free or combined state, will unite with oxygen to form water vapor, H{2}O. Not all of the hydrogen shown in a fuel analysis, however, is available for the production of heat, as a portion of it is already united with the oxygen shown by the analysis in the form of water, H{2}O. Since the atomic weights of H and O are respectively 1 and 16, the weight of the combined hydrogen will be 1/8 of the weight of the oxygen, and the hydrogen available for combustion will be H - 1/8 O. In complete combustion of the sulphur, sulphur dioxide SO{2} is formed, which in solution in water forms sulphuric acid.

Expressed numerically, the theoretical amount of air for the above analysis is as follows:

0.7479 C × 2-2/3 = 1.9944 O needed ( 0.0642 ) ( 0.0498 - -------) H × 8 = 0.3262 O needed ( 8 ) 0.0324 S × 1 = 0.0324 O needed ------ Total 2.3530 O needed

One pound of oxygen is contained in 4.32 pounds of air.

The total air needed per pound of coal, therefore, will be 2.353 × 4.32 = 10.165.

The weight of combustible per pound of fuel is .7479 + .0418 + .0324 + .012 = .83 pounds, and the air theoretically required per pound of combustible is 10.165 ÷ .83 = 12.2 pounds.

The above is equivalent to computing the theoretical amount of air required per pound of fuel by the formula:

( O) Weight per pound = 11.52 C + 34.56 (H - -) + 4.32 S (10) ( 8)

where C, H, O and S are proportional parts by weight of carbon, hydrogen, oxygen and sulphur by ultimate analysis.

In practice it is impossible to obtain perfect combustion with the theoretical amount of air, and an excess may be required, amounting to sometimes double the theoretical supply, depending upon the nature of the fuel to be burned and the method of burning it. The reason for this is that it is impossible to bring each particle of oxygen in the air into intimate contact with the particles in the fuel that are to be oxidized, due not only to the dilution of the oxygen in the air by nitrogen, but because of such factors as the irregular thickness of the fire, the varying resistance to the passage of the air through the fire in separate parts on account of ash, clinker, etc. Where the difficulties of drawing air uniformly through a fuel bed are eliminated, as in the case of burning oil fuel or gas, the air supply may be materially less than would be required for coal. Experiment has shown that coal will usually require 50 per cent more than the theoretical net calculated amount of air, or about 18 pounds per pound of fuel either under natural or forced draft, though this amount may vary widely with the type of furnace, the nature of the coal, and the method of firing. If less than this amount of air is supplied, the carbon burns to monoxide instead of dioxide and its full heat value is not developed.

An excess of air is also a source of waste, as the products of combustion will be diluted and carry off an excessive amount of heat in the chimney gases, or the air will so lower the temperature of the furnace gases as to delay the combustion to an extent that will cause carbon monoxide to pass off unburned from the furnace. A sufficient amount of carbon monoxide in the gases may cause the action known as secondary combustion, by igniting or mingling with air after leaving the furnace or in the flues or stack. Such secondary combustion which takes place either within the setting after leaving the furnace or in the flues or stack always leads to a loss of efficiency and, in some instances, leads to overheating of the flues and stack.

Table 32 gives the theoretical amount of air required for various fuels calculated from formula (10) assuming the analyses of the fuels given in the table.

The process of combustion of different fuels and the effect of variation in the air supply for their combustion is treated in detail in the chapters dealing with the various fuels.

TABLE 32

CALCULATED THEORETICAL AMOUNT OF AIR REQUIRED PER POUND OF VARIOUS FUELS

___________________________________________________________ | |Weight of Constituents in One |Air Required| | Fuel |Pound Dry Fuel |per Pound | | |____________________________|of Fuel | | | Carbon | Hydrogen| Oxygen |Pounds | | | Per Cent| Per Cent| Per Cent | | |______________|_______|_______|________|__________| |Coke | 94.0 | . | . | 10.8 | |Anthracite Coal | 91.5 | 3.5 | 2.6 | 11.7 | |Bituminous Coal | 87.0 | 5.0 | 4.0 | 11.6 | |Lignite | 70.0 | 5.0 | 20.0 | 8.9 | |Wood | 50.0 | 6.0 | 43.5 | 6.0 | |Oil | 85.0 | 13.0 | 1.0 | 14.3 | |______________|_______|_______|________|___________|

ANALYSIS OF FLUE GASES

The object of a flue gas analysis is the determination of the completeness of the combustion of the carbon in the fuel, and the amount and distribution of the heat losses due to incomplete combustion. The quantities actually determined by an analysis are the relative proportions by volume, of carbon dioxide (CO_{2}), oxygen (O), and carbon monoxide (CO), the determinations being made in this order.

The variations of the percentages of these gases in an analysis is best illustrated in the consideration of the complete combustion of pure carbon, a pound of which requires 2.67 pounds of oxygen, or 32 cubic feet at 60 degrees Fahrenheit. The gaseous product of such combustion will occupy, when cooled, the same volume as the oxygen, namely, 32 cubic feet. The air supplied for the combustion is made up of 20.91 per cent oxygen and 79.09 per cent nitrogen by volume. The carbon united with the oxygen in the form of carbon dioxide will have the same volume as the oxygen in the air originally supplied. The volume of the nitrogen when cooled will be the same as in the air supplied, as it undergoes no change. Hence for complete combustion of one pound of carbon, where no excess of air is supplied, an analysis of the products of combustion will show the following percentages by volume:

Actual Volume for One Pound Carbon Per Cent Cubic Feet by Volume Carbon Dioxide 32 = 20.91 Oxygen 0 = 0.00 Nitrogen 121 = 79.09 --- ------ Air required for one pound Carbon 153 = 100.00

For 50 per cent excess air the volume will be as follows: