FLUE GAS -Flue gas composition

Flue gas is the combustion product gas from a fireplace, oven, furnace, boiler or steam generator that exits to the atmosphere via a ''flue'' which may be a pipe, channel or chimney. The flue is most commonly referred to as a "flue gas stack" by engineers or a "smokestack" by lay people.

Flue gases are produced when natural gas, fuel oil, coal, wood or any other fuel is combusted in an industrial furnace, a steam generator in a fossil fuel power plant or other combustion sources.

Flue gas composition

Flue gas is usually composed of carbon dioxide (CO2) and water vapor as well as nitrogen and excess oxygen remaining from the intake combustion air. It may also contain a small percentage of air pollutants such as particulate matter, carbon monoxide, nitrogen oxides, sulfur oxides and mercury. Typically, more than two-thirds of the flue gas is nitrogen.

The table below provides the amount of flue gas (on a dry basis as well as a wet basis) generated by burning a typical fuel gas, fuel oil or coal. The flue gas amounts were obtained by stoichiometric calculations using the indicated typical excess combustion air percentages:
Flue_Gas_Generation.png
Given the amount of gas, oil or coal fuel burned in a combustion device, then the flue gas generation data (i.e., m³/GJ of fuel) in the above table provides a basis for estimating the amount of flue gas generated.

Flue gas treatment

At power plants, flue gas is often treated with a series of chemical processes and scrubbers, which remove pollutants. Electrostatic precipitators or fabric filters remove particulate matter and flue gas desulfurization removes the sulfur dioxide (SO2) produced by burning fossil fuels, particularly coal.

Nitrogen oxides emissions are reduced either by modifications to the combustion process to prevent their formation, or by catalytic reaction with ammonia or urea. In either case, the aim is to produce nitrogen gas, rather than nitrogen oxides.

In the United States, there is also a rapid deployment of technologies to remove mercury from flue gas - typically by adsorption on sorbents or by capture in inert solids as part of the flue gas desulfurization product.

Technologies for the removal and capture of carbon dioxide from flue gases are now under active research and development as a means of reducing the emissions of so-called ''greenhouse gas''. 


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 (CO2), 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,[28] 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
Cubic Feet
Per Cent
by Volume
Carbon Dioxide  32=  20.91
Oxygen    0=    0.00
Nitrogen121=  79.09
–––––––––––––––––
Air required for one pound Carbon153=100.00
For 50 per cent excess air the volume will be as follows:
153 × 1½ = 229.5 cubic feet of air per pound of carbon.
Actual Volume
for One Pound Carbon
Cubic Feet
Per Cent
by Volume
Carbon Dioxide  32    =  13.91}=20.91 per cent
Oxygen  16    =    7.00
Nitrogen181.5=  79.09
–––––––––––––––––––––
Air required for one pound Carbon229.5=100.00
For 100 per cent excess air the volume will be as follows:
153 × 2 = 306 cubic feet of air per pound of carbon.
Actual Volume
for One Pound Carbon
Cubic Feet
Per Cent
by Volume
Carbon Dioxide  32=  10.45}=20.91 per cent
Oxygen  32=  10.45
Nitrogen242=  79.09
–––––––––––––––––
Air required for one pound Carbon306=100.00
In each case the volume of oxygen which combines with the carbon is equal to (cubic feet of air × 20.91 per cent)—32 cubic feet.

It will be seen that no matter what the excess of air supplied, the actual amount of carbon dioxide per pound of carbon remains the same, while the percentage by volume decreases as the excess of air increases. The actual volume of oxygen and the percentage by volume increases with the excess of air, and the percentage of oxygen is, therefore, an indication of the amount of excess air. In each case the sum of the percentages of CO2 and O is the same, 20.9. Although the volume of nitrogen increases with the excess of air, its percentage by volume remains the same as it undergoes no change while combustion takes place; its percentage for any amount of air excess, therefore, will be the same after combustion as before, if cooled to the same temperature. It must be borne in mind that the above conditions hold only for the perfect combustion of a pound of pure carbon.
Carbon monoxide (CO) produced by the imperfect combustion of carbon, will occupy twice the volume of the oxygen entering into its composition and will increase the volume of the flue gases over that of the air supplied for combustion in the proportion of
1 to
100 + ½ the per cent CO
–––––––––––––––––––––––––––––––––––––––
100
When pure carbon is the fuel, the sum of the percentages by volume of carbon dioxide, oxygen and one-half of the carbon monoxide, must be in the same ratio to the nitrogen in the flue gases as is the oxygen to the nitrogen in the air supplied, that is, 20.91 to 79.09. When burning coal, however, the percentage of nitrogen is obtained by subtracting the sum of the percentages by volume of the other gases from 100. Thus if an analysis shows 12.5 per cent CO2, 6.5 per cent O, and 0.6 per cent CO, the percentage of nitrogen which ordinarily is the only other constituent of the gas which need be considered, is found as follows:
100 - (12.5 + 6.5 + 0.6) = 80.4 per cent.
The action of the hydrogen in the volatile constituents of the fuel is to increase the apparent percentage of the nitrogen in the flue gases. This is due to the fact that the water vapor formed by the combustion of the hydrogen will condense at a temperature at which the analysis is made, while the nitrogen which accompanied the oxygen with which the hydrogen originally combined maintains its gaseous form and passes into the sampling apparatus with the other gases. For this reason coals containing high percentages of volatile matter will produce a larger quantity of water vapor, and thus increase the apparent percentage of nitrogen.
Air Required and Supplied—When the ultimate analysis of a fuel is known, the air required for complete combustion with no excess can be found as shown in the chapter on combustion, or from the following approximate formula:

Pounds of air required per pound of fuel = 34.56(
C
––––
3
 + (H - 
O
––––
8
) + 
S
––––
8
)[29]            (11)
where C, H and O equal the percentage by weight of carbon, hydrogen and oxygen in the fuel divided by 100.

When the flue gas analysis is known, the total, amount of air supplied is:

Pounds of air supplied per pound of fuel = 3.036(
N
–––––––––––––––––
CO2 + CO
) × C[30]            (12)
where N, CO2 and CO are the percentages by volume of nitrogen, carbon dioxide and carbon monoxide in the flue gases, and C the percentage by weight of carbon which is burned from the fuel and passes up the stack as flue gas. This percentage of C which is burned must be distinguished from the percentage of C as found by an ultimate analysis of the fuel. To find the percentage of C which is burned, deduct from the total percentage of carbon as found in the ultimate analysis, the percentage of unconsumed carbon found in the ash. This latter quantity is the difference between the percentage of ash found by an analysis and that as determined by a boiler test. It is usually assumed that the entire combustible element in the ash is carbon, which assumption is practically correct. Thus if the ash in a boiler test were 16 per cent and by an analysis contained 25 per cent of carbon, the percentage of unconsumed carbon would be 16 × .25 = 4 per cent of the total coal burned. If the coal contained by ultimate analysis 80 per cent of carbon the percentage burned, and of which the products of combustion pass up the chimney would be 80 - 4 = 76 per cent, which is the correct figure to use in calculating the total amount of air supplied by formula (12).
The weight of flue gases resulting from the combustion of a pound of dry coal will be the sum of the weights of the air per pound of coal and the combustible per pound of coal, the latter being equal to one minus the percentage of ash as found in the boiler test. The weight of flue gases per pound of dry fuel may, however, be computed directly from the analyses, as shown later, and the direct computation is that ordinarily used.
The ratio of the air actually supplied per pound of fuel to that theoretically required to burn it is:

3.036(
N
–––––––––––––––––
CO2 + CO
) × C
            (13)
–––––––––––––––––––––––––––––––––––––––––––––
34.56(
C
–––
3
 + H - 
O
–––
8
)
in which the letters have the same significance as in formulae (11) and (12).
The ratio of the air supplied per pound of combustible to the amount theoretically required is:

N
––––––––––––––––––––––––––––––––––
N - 3.782(O - ½CO)
            (14)
which is derived as follows:
The N in the flue gas is the content of nitrogen in the whole amount of air supplied. The oxygen in the flue gas is that contained in the air supplied and which was not utilized in combustion. This oxygen was accompanied by 3.782 times its volume of nitrogen. The total amount of excess oxygen in the flue gases is (O - ½CO); hence N - 3.782(O - ½CO) represents the nitrogen content in the air actually required for combustion and N ÷ (N - 3.782[O - ½CO]) is the [Pg 158] ratio of the air supplied to that required. This ratio minus one will be the proportion of excess air.

The heat lost in the flue gases is L = 0.24 W (T - t)            (15)
WhereL=B. t. u. lost per pound of fuel,
W=weight of flue gases in pounds per pound of dry coal,
T=temperature of flue gases,
t=temperature of atmosphere,
0.24=specific heat of the flue gases.
The weight of flue gases, W, per pound of carbon can be computed directly from the flue gas analysis from the formula:

11 CO2 + 8 O + 7 (CO + N)
–––––––––––––––––––––––––––––––––––––––––––––
3 (CO2 + CO)
            (16)
where CO2, O, CO, and N are the percentages by volume as determined by the flue gas analysis of carbon dioxide, oxygen, carbon monoxide and nitrogen.
The weight of flue gas per pound of dry coal will be the weight determined by this formula multiplied by the percentage of carbon in the coal from an ultimate analysis.
Footnote31Graph of Heat Loss

Fig. 20. Loss Due to Heat Carried Away by Chimney Gases for Varying Percentages of Carbon Dioxide.
Based on Boiler Room Temperature = 80 Degrees Fahrenheit.
Nitrogen in Flue Gas = 80.5 Per Cent. Carbon Monoxide in Flue Gas = 0. Per Cent

Fig. 20 represents graphically the loss due to heat carried away by dry chimney gases for varying percentages of CO2, and different temperatures of exit gases.
[Pg 159]
The heat lost, due to the fact that the carbon in the fuel is not completely burned and carbon monoxide is present in the flue gases, in B. t. u. per pound of fuel burned is:

L' = 10,150 × (
CO
–––––––––––––––––
CO + CO2
)            (17)
where, as before, CO and CO2 are the percentages by volume in the flue gases and C is the proportion by weight of carbon which is burned and passes up the stack.
Fig. 21 represents graphically the loss due to such carbon in the fuel as is not completely burned but escapes up the stack in the form of carbon monoxide.
Footnote32Graph of Heat Loss

Fig. 21. Loss Due to Unconsumed Carbon Contained in the CO in the Flue Gases

Apparatus for Flue Gas Analysis—The Orsat apparatus, illustrated in Fig. 22, is generally used for analyzing flue gases. The burette A is graduated in cubic centimeters up to 100, and is surrounded by a water jacket to prevent any change in temperature from affecting the density of the gas being analyzed.
For accurate work it is advisable to use four pipettes, B, C, D, E, the first containing a solution of caustic potash for the absorption of carbon dioxide, the second an alkaline solution of pyrogallol for the absorption of oxygen, and the remaining two an acid solution of cuprous chloride for absorbing the carbon monoxide. Each pipette contains a number of glass tubes, to which some of the solution clings, thus facilitating [Pg 160] the absorption of the gas. In the pipettes D and E, copper wire is placed in these tubes to re-energize the solution as it becomes weakened. The rear half of each pipette is fitted with a rubber bag, one of which is shown at K, to protect the solution from the action of the air. The solution in each pipette should be drawn up to the mark on the capillary tube.
Orsat Apparatus

Fig. 22. Orsat Apparatus

The gas is drawn into the burette through the U-tube H, which is filled with spun glass, or similar material, to clean the gas. To discharge any air or gas in the apparatus, the cock G is opened to the air and the bottle F is raised until the water in the burette reaches the 100 cubic centimeters mark. The cock G is then turned so as to close the air opening and allow gas to be drawn through H, the bottle F being lowered for this purpose. The gas is drawn into the burette to a point below the zero mark, the cock G then being opened to the air and the excess gas expelled until the level of the water in F and in A are at the zero mark. This operation is necessary in order to obtain the zero reading at atmospheric pressure.
The apparatus should be carefully tested for leakage as well as all connections leading thereto. Simple tests can be made; for example: If after the cock G is closed, the bottle F is placed on top of the frame for a short time and again brought to the zero mark, the level of the water in A is above the zero mark, a leak is indicated.
Before taking a final sample for analysis, the burette A should be filled with gas and emptied once or twice, to make sure that all the apparatus is filled with the new gas. The cock G is then closed and the cock I in the pipette B is opened and the gas driven over into B by raising the bottle F. The gas is drawn back into A by lowering F and when the solution in B has reached the mark in the capillary tube, the cock I is closed and a reading is taken on the burette, the level of the water in the bottle F being brought to the same level as the water in A. The operation is repeated until a constant reading is obtained, the number of cubic centimeters being the percentage of CO2 in the flue gases.
The gas is then driven over into the pipette C and a similar operation is carried out. The difference between the resulting reading and the first reading gives the percentage of oxygen in the flue gases.
The next operation is to drive the gas into the pipette D, the gas being given a final wash in E, and then passed into the pipette C to neutralize any hydrochloric acid fumes which may have been given off by the cuprous chloride solution, which, especially if it be old, may give off such fumes, thus increasing the volume of the gases and making the reading on the burette less than the true amount.
The process must be carried out in the order named, as the pyrogallol solution will also absorb carbon dioxide, while the cuprous chloride solution will also absorb oxygen.
[Pg 161]
As the pressure of the gases in the flue is less than the atmospheric pressure, they will not of themselves flow through the pipe connecting the flue to the apparatus. The gas may be drawn into the pipe in the way already described for filling the apparatus, but this is a tedious method. For rapid work a rubber bulb aspirator connected to the air outlet of the cock G will enable a new supply of gas to be drawn into the pipe, the apparatus then being filled as already described. Another form of aspirator draws the gas from the flue in a constant stream, thus insuring a fresh supply for each sample.
The analysis made by the Orsat apparatus is volumetric; if the analysis by weight is required, it can be found from the volumetric analysis as follows:
Multiply the percentages by volume by either the densities or the molecular weight of each gas, and divide the products by the sum of all the products; the quotients will be the percentages by weight. For most work sufficient accuracy is secured by using the even values of the molecular weights.
The even values of the molecular weights of the gases appearing in an analysis by an Orsat are:
Carbon Dioxide    44
Carbon Monoxide    28
Oxygen    32
Nitrogen    28
Table 33 indicates the method of converting a volumetric flue gas analysis into an analysis by weight.
TABLE 33

CONVERSION OF A FLUE GAS ANALYSIS BY VOLUME TO ONE BY WEIGHT
GasAnalysis by Volume
Per Cent
Molecular WeightVolume times
Molecular Weight
Analysis by Weight
Per Cent
Carbon DioxideCO2  12.212+(2×16)  536.8
  536.8
–––––––––––
3022.8
 =   17.7    
Carbon MonoxideCO      .412+16    11.2
    11.2
–––––––––––
3022.8
 =       .4    
OxygenO    6.92×16  220.8
  220.8
–––––––––––
3022.8
 =     7.3    
NitrogenN  80.52×142254.0
2254.0
–––––––––––
3022.8
 =   74.6    
Total100.03022.8
100.0    
Application of Formulae and Rules—Pocahontas coal is burned in the furnace, a partial ultimate analysis being:
Per Cent
Carbon82.1  
Hydrogen4.25
Oxygen2.6  
Sulphur1.6  
Ash6.0  
B. t. u., per pound dry14500  
[Pg 162]
The flue gas analysis shows:
Per Cent
CO210.7  
O9.0  
CO0.0  
N (by difference)80.3  
Determine: The flue gas analysis by weight (see Table 33), the amount of air required for perfect combustion, the actual weight of air per pound of fuel, the weight of flue gas per pound of coal, the heat lost in the chimney gases if the temperature of these is 500 degrees Fahrenheit, and the ratio of the air supplied to that theoretically required.
Solution: The theoretical weight of air required for perfect combustion, per pound of fuel, from formula (11) will be,
W = 34.56
(
.821
–––––––
3
 + (.0425 - 
.026
–––––––
8
) + 
.016
–––––––
8
)
 = 10.88 pounds.
If the amount of carbon which is burned and passes away as flue gas is 80 per cent, which would allow for 2.1 per cent of unburned carbon in terms of the total weight of dry fuel burned, the weight of dry gas per pound of carbon burned will be from formula (16):
W = 
11 × 10.7 + 8 × 9.0 + 7(0 + 80.3)
–––––––––––––––––––––––––––––––––––––––––––––––––––––
3 (10.7 + 0)
 = 23.42 pounds
and the weight of flue gas per pound of coal burned will be .80 × 23.42 = 18.74 pounds.
The heat lost in the flue gases per pound of coal burned will be from formula (15) and the value 18.74 just determined.
Loss = .24 × 18.74 × (500 - 60) = 1979 B. t. u.
The percentage of heat lost in the flue gases will be 1979 ÷ 14500 = 13.6 per cent.
The ratio of air supplied per pound of coal to that theoretically required will be 18.74 ÷ 10.88 = 1.72 per cent.
The ratio of air supplied per pound of combustible to that required will be from formula (14):
.803
–––––––––––––––––––––––––––––––––––––––
.803 - 3.782(.09 - ½ × 0)
 = 1.73
The ratio based on combustible will be greater than the ratio based on fuel if there is unconsumed carbon in the ash.
Unreliability of CO2 Readings Taken Alone—It is generally assumed that high CO2 readings are indicative of good combustion and hence of high efficiency. This is true only in the sense that such high readings do indicate the small amount of excess air that usually accompanies good combustion, and for this reason high CO2 readings alone are not considered entirely reliable. Wherever an automatic CO2 recorder is used, it should be checked from time to time and the analysis carried further with a view to ascertaining whether there is CO present. As the percentage of CO2 in these gases increases, there is a tendency toward the presence of CO, which, of course, cannot be shown by a CO2 recorder, and which is often difficult to detect with an Orsat apparatus. The greatest care should be taken in preparing the cuprous chloride solution in making analyses and it must be known to be fresh and capable of absorbing CO. [Pg 163] In one instance that came to our attention, in using an Orsat apparatus where the cuprous chloride solution was believed to be fresh, no CO was indicated in the flue gases but on passing the same sample into a Hempel apparatus, a considerable percentage was found. It is not safe, therefore, to assume without question from a high CO2 reading that the combustion is correspondingly good, and the question of excess air alone should be distinguished from that of good combustion. The effect of a small quantity of CO, say one per cent, present in the flue gases will have a negligible influence on the quantity of excess air, but the presence of such an amount would mean a loss due to the incomplete combustion of the carbon in the fuel of possibly 4.5 per cent of the total heat in the fuel burned. When this is considered, the importance of a complete flue gas analysis is apparent.
Table 34 gives the densities of various gases together with other data that will be of service in gas analysis work.
TABLE 34

DENSITY OF GASES AT 32 DEGREES FAHRENHEIT AND ATMOSPHERIC PRESSURE
ADAPTED FROM SMITHSONIAN TABLES
GasChemical
Symbol
Specific Gravity
Air=1
Weight of
One Cubic Foot
Pounds
Volume of
One Pound
Cubic Feet
Relative Density, Hydrogen = 1
ExactApproximate
OxygenO1.053  .08922    11.20815.8716
NitrogenN0.9673.07829    12.77313.9214
HydrogenH0.0696.005621177.90    1.00  1
Carbon DioxideCO21.5291.12269      8.15121.8322
Carbon MonoxideCO0.9672.07807    12.80913.8914
MethaneCH40.5576.04470    22.371  7.95  8
EthaneC2H61.075  .08379    11.93514.9115
AcetyleneC2H20.920  .07254    13.78512.9113
Sulphur DioxideSO22.2639.17862      5.59831.9632
Air1.0000.08071    12.390

[Pg 164] [Pl 164]



FOOTNOTES

[28] See Table 31, page 151.
[29] This formula is equivalent to (10) given in chapter on combustion. 34.56 = theoretical air required for combustion of one pound of H (see Table 31).
[30] For degree of accuracy of this formula, see Transactions, A. S. M. E., Volume XXI, 1900, page 94.
[31] For loss per pound of coal multiply by per cent of carbon in coal by ultimate analysis.
[32] For loss per pound of coal multiply by per cent of carbon in coal by ultimate analysis.

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