Dr. Artemio J. Benítez Fundora Fundora
Combustion Calculations for Solid and Liquid Fuels Solid and Liquid Fuels Elementary Elementary Composi tion also known as Ultimate Analysis of Composition, or Composition of Fuel as Received is expressed in gravimetric fractions of the chemical elements present on them. This elementary composition is normally associated mainly with Carbon, Hydrogen, Oxygen, Nitrogen, Sulphur, A shes and Water contents in percent ( CHONSAW). From the Ultimate Analysis of a Fuel, is possible to estimate the theoretical quantity of Oxygen to make the complete combustion of it, liberating all the potential energy of the reaction chemical reaction into heat. To accomplish this purpose the only thing that is needed to know, are the elementary reactions of combustion of the elements that composes the fuel. Not all of them reacts with oxygen, only Carbon, Hydrogen and Sulphur; the rest of the elements except the Oxygen present in the fuel, are considered as ballast because they only absorbs part of the heat produced for the combustion of the other elements during the burning of the fuel. So, Nitrogen, Ashes and Water are ballast or parasite elements that do not produce any heat of reaction. Some examples of Ultimate Composition for Liquid and Solid Fuels are referred in the following table: Table Table I: Elementary Elementary Composition o f som e liquid and so lid fuels ULTIMATE ANALYSIS (Mass Fraction) kg/kg of Fuel of Fuel
FUEL LIQUID FUELS
Carb Carbon on Hydr Hydrog ogen en
Sulf Sulfur ur
Nitrog Nitrogen en Oxygen Oxygen
Ash
Moisture
Gasoline
0.862
0.128
0.01
0
0
0
0
Diesel
0.838
0.121
0.035
0
0
0.006
0
Light Fuel Oil (Bunker A)
0.834
0.117
0.04
0
0
0.009
0
Heavy Fuel Oil (Bunker B)
0.829
0.114
0.045
0
0
0.012
0
Residual Fuel Oil (Bunker C)
0.883
0.095
0.012
0
0
0.01
0
Coal No. 1
0.847
0.026
0.011
0.016
0.01
0.04
0.05
Coal No. 2
0.785
0.047 0.047
0.019
0.042
0.017 0.017
0.04
0.05
Coal No. 3
0.63
0.039
0.018
0.09
0.013
0.16
0.05
Coal No. 4
0.59
0.037
0.017
0.084
0.012 0.012
0.18
0.08
Wood
0.335
0.06
0
0.3
0
0.3
0.005
Rice Hull
0.37 0.3781 81
0.04 0.0468 68
0
0.33 0.335 5
0.00 0.0028 28
0
0.23 0.2368 68
SOLID FUELS
The total sum of the elements in the composition must be 1 if it is expressed in mass fractions or 100 if it is expressed in percentage. 1
Dr. Artemio J. Benítez Fundora Fundora
Perfect Combustion Reactions or Stoichiometric Reactions:
→ → → These are molar stoichiometric reactions of combustion for Carbon, Hydrogen and Sulphur; the combustible elements in the fuel composition, and the first one states that 1 mole of Carbon requires 1 mole of Oxygen to produce 1 mole of Carbon Dioxide; and so on. The molar relationship of the need of Oxygen to complete each ea ch reaction is as follows:
1 mole of molecular Oxygen is required to burn 1 mole of Carbon ( 1 mole O2/mole of C) ½ mole of molecular Oxygen is required to burn 1 mole of molecular Hydrogen ( ½ mole O2/mole of H2) 1 mole of molecular Oxygen is required to burn 1 mole of Sulphur ( 1 mole O2/mole of S)
The molar relationship of the reaction products obtained at the end of each reaction is as follows:
1 mole of Carbon Dioxide is produced when 1 mole of Carbon is burnt ( 1 mole CO2/mole of C) 1 mole of Water is produced when 1 mole of molecular Hydrogen is burnt ( 1 mole H2O/mole O/mole of H2) 1 mole of Sulphur Dioxide is produced when 1 mole of Sulphur is burnt ( 1 mole SO2/mole of S)
Due to the fact that the elementary compositions of solid and liquid fuels are normally expressed in gravimetric form, we need to convert these molar proportions to mass relationships, using the molecular weights of each component in the above equations. By the Periodic Table we can know that:
1 mole of atomic Carbon weights 12,0107 grams 1 mole of molecular Oxygen weights 31,9988 grams 1 mole of molecular Hydrogen weights 2,01588 grams 1 mole of atomic Sulphur weights 32,065 grams
Using the Mass Conservation Law, we can re-write the above stoichiometric equation in mass form as:
2
Dr. Artemio J. Benítez Fundora
12,0107 31,9988 → 44,0095 2,01588 ∙ 31,9988 → 18,01528 32,065 31,9988 → 64,0638 The mass relationship of the need of Oxygen to complete each reaction is as follows:
2,6641911 gram of molecular Oxygen is required to burn 1 gram of Carbon (2,6641911 grams O2/gram of C) 7.93668274 grams of molecular Oxygen is required to burn 1 gram of molecular Hydrogen (7,93668274 grams O2/gram of H2) 0.99793544 grams of molecular Oxygen is required to burn 1 gram of Sulphur (0,99793544 grams O2/gram of S)
The molar relationship of the reaction products obtained at the end of each reaction is as follows:
3,6641911 gram of Carbon Dioxide is produced when 1 gram of Carbon is burnt (3,6641911 grams CO2/gram of C) 8,93668274 grams of Water is produced when 1 mole of molecular Hydrogen is burnt (8,93668274 grams H2O/gram of H2) 1,99793544 grams of Sulphur Dioxide is produced when 1 mole of Sulphur is burnt (1,99793544 grams SO2/grams of S)
Departing from these relationships we can calculate the Oxygen demand of a liquid or solid fuel knowing only its elementary composition, for example: Fuel Light Fuel Oil (Bunker A)
C
H
O
N
S
A
W
0.834
0.117
0
0
0.04
0.009
0
∙ 2,6641911 0,117 0,834 0,04 ∙7,93668274 0 ∙0,99793544 3.1904447 Symbolically, this expression could be written as:
2,6641911∙ 7,93668274∙ 0,99793544∙ , , 3
Dr. Artemio J. Benítez Fundora
Where C, H, S and O are the mass fractions of these elements in the Fuel composition, as was shown in Table I, and the Oxygen content on Fuel should be subtracted in order not to alter the theoretical quantity of supplied oxygen to make a perfect combustion. Using the former equation, it is possible to calculate the theoretical oxygen required for the stoichiometric reaction of every fuel, as shown in the table below: Table II: Theoretical Oxigen required fo r the co mbustio n of the fuels ind icated in Table I Theoretical Oxygen Required
LIQUID FUELS
Gasoline
3.322
Diesel
3.228
Light Fuel Oil (Bunker A)
3.190
Heavy Fuel Oil (Bunker B)
3.158
Residual Fuel Oil (Bunker C)
3.118
SOLID FUELS Coal No. 1
2.474
Coal No. 2
2.483
Coal No. 3
2.006
Coal No. 4
1.882
Wood
1.369
Rice Hull
1.379
But the actual combustion of a fuel is not made with pure oxygen, instead, dry air is used to supply the quantities of oxygen that are required for the theoretical combustion. The Standard Molar Composition of Dry Air in the atmosphere is:
Standard Atmosphere
Mixing ratio Molecular (mol/mol) Weight
Nitrogen (N2)
0,78
28,0134
Oxygen (O2)
0,21
31,9988
Argon (Ar)
0,0093
39,9480
Carbon dioxide (CO2)
365e
-6
44,0095
Neon (Ne)
-6
18e
20,1797 -6
Ozone (O3)
0,01-10e
Helium (He)
5,2e
Methane (CH4)
1.7e
Krypton (Kr)
1,1e
47,9982
-6
4,002602
-6
16,04246
-6
83,7980
4
Dr. Artemio J. Benítez Fundora
-9
2,01588
-9
44,0128
Hydrogen (H2)
500x10
Nitrous oxide (N2O)
320x10
Dry Air
-
Expressing the upper table in gravimetric form:
Standard Atmosphere
Mixing ratio (kg/kg)
Molecular Weight
Nitrogen (N2)
0,7545364
28,0134
Oxygen (O2)
0,2320453
31,9988
Argon (Ar)
0,0128291
39,9480
Carbon dioxide (CO2)
5,55E-04
44,0095
Neon (Ne)
1,25E-05
20,1797
Ozone (O3)
1,66E-05
47,9982
Helium (He)
7,19E-07
4,0026
Methane (CH4)
9,42E-07
16,0425
Krypton (Kr) Hydrogen (H2) Nitrous oxide (N2O) Dry Air
3,18E-06 3,48E-08 4,86E-07 1
83,7980 2,0159 44,0128 28,958778
11,4813405∙ 34,203161 ∙ 4,300606∙ 4.309503 , , ∙ Considering that 4,3006606 are approximately 4,309503 (0,21 % of error), then:
,∙,∙,∙ , , If the Fuel composition is supplied in %, then:
% % ,∙ ,∙ , % % ∙ 5
Dr. Artemio J. Benítez Fundora
Or also can be written as:
, ∙ % , ∙ % , ∙ %% , , Instead of considering the Theoretical Oxygen required for combustion, using the theoretical dry air required expressions, we can recalculate the new values for the Table II as: Table III: Theoretical Oxigen and Theoretical Dry Air required for the combu stion of the fu els indicated in Table I
FUEL
Theoretical Oxygen
Theoretical Air
LIQUID FUELS
kg O2/kg Fuel
kg dry air/kg Fuel
Gasoline
3.322
14.318
Diesel
3.228
13.910
Light Fuel Oil (Bunker A)
3.190
13.749
Heavy Fuel Oil (Bunker B)
3.158
13.611
Residual Fuel Oil (Bunker C)
3.118
13.439
Coal No. 1
2.474
10.661
Coal No. 2
2.483
10.702
Coal No. 3
2.006
8.645
Coal No. 4
1.882
8.113
Wood
1.369
5.898
Rice Hull
1.379
5.942
SOLID FUELS
Normal Conditions of Environmental Air But normally the air that is supplied to a combustion process, in boilers and steam generators, is taking directly from the surrounding environment, and IS NOT DRY , because it contents a small amount of water vapor called humidity. The air humidity (d ) can be obtained by knowing the local conditions of the weather variables in the location, by the dry and wet bulb temperatures or by the dry bulb temperature and the relative humidity in percent of the place; and using a psychrometric 6
Dr. Artemio J. Benítez Fundora
chart or by the psychrometric thermodynamic relationships of ASHRAE, it value can be estimated accurately. Example: Normal Weather Cond itio ns for Coatzacoalcos, VER
Tdb = 35 °C and φ = 84 % For this condition by the psychrometric chart:
The absolute humidity of air is 30 grams of water vapor per kilogram of dry air . Also can be determined using the psychrometric thermodynamic relationships, as follows: Barometric Pressure (PT) = 101325 Pa Saturation Water Vapor Pressure (P SW) calculated by the dry bulb temperature (Tdb) as:
,∙ 610,78 ∙ ,
And using the following relationship:
∙ ∙ 100
7
Dr. Artemio J. Benítez Fundora
Above we calculate the molecular weight of both components, so:
∙ 18,01528 100 28,958778 ∙
Solving:
,∙ , 610,78 ∙ 5576,5866
84 ∙5576,5866 18,01528 100 28,958778 ∙ 1013255576,5866 0,03043521 30,43521 A little more accurate result than using the psychrometric chart.
Theoretical Humid Air requir ed for the combust ion of 1 kg of Fuel: They can be re-written as:
∙ ,∙,∙,∙ % ,∙ % , ∙,∙ % % ∙ ∙ ,∙%,∙%, ∙ %% , , But
must be used in these equations, expressed in
8
Dr. Artemio J. Benítez Fundora
Ac tual Humi d Air requi red fo r the com bustio n of 1 kg of Fuel: The actual quantity of air supplied for combustion always is greater than the theoretical humid air calculated by the fuel composition, due to different facts that difficult the correct mixture air-fuel with an homogeneous distribution of the temperature in every place in which the combustion process is done; mainly due to the fuel preparation before be burnt, the combustion chamber design, the air distribution inside of it, etc. All these facts obligate for granting a complete liberation of heat from the fuel reacting with air, that an excess of oxidant must be supplied over the theoretical one.
This value is known as air excess coefficient and will be symbolized by the Greek letter Normally the values of
α varies
α .
according to the type of fuel to burn, and also taking into
account the conditions of the combustion chamber design; solid fuels are more difficult to burn effectively and will need higher values of
α
(from 1,5 to 2,0) and because that is so
easy to pulverize liquid fuels by spraying, the liquid fuels will require lower values of
α
(from 1,15 to 1,45), and finally, gaseous fuels easily mixtures with the combustion humid air and the need of excess of air is minimum ( α from 1,03 to 1,08). The best recommend practice is to adjust this air excess coefficient in the boiler, steam generator or combustion system directly, by minimizing progressively the formation of carbon monoxide (CO) in the combustion gases exhausted that leave from these equipment’s, until it almost disappears in the stream. Then the actual humid air required for the combustion of 1 kg of fuel, is only equal to:
∙ Or as:
∙ ∙ ,∙,∙,∙ , ,
9
Dr. Artemio J. Benítez Fundora
Theoretical Combustion Products generated in the combustion of 1 kg of Fuel: According to the Mass Conservation Law, if a quantity of Oxygen is required to burn 1 unit of mass of a fuel, the combustion products generated are the sum of the quantity of Oxygen required plus the part of the fuel elements that are converted to gaseous products; the ash content of the fuel is excluded due to the fact that ashes generated during the combustion are normally in solid form. By a simple mass balance we obtain:
1 If the combustion is made using the theoretical dry air, instead of pure Oxygen:
1 If the combustion is made using the theoretical humid air, instead of dry air:
1 And finally, if the combustion is made using the actual humid air, instead of theoretical humid air:
1 These relationships are accurate, but do not offer any information about the chemical composition of the combustion gases, very often needed to make corrections in excess air coefficient for an optimal combustion process and to improve energy savings in boilers, steam generators or combustion systems like fired heaters and incinerators. Using the stoichiometric proportions obtained before for the quantity of combustion products, we can write for 1 kg of fuel burned:
3,6641911 ∙ For the combustion of Hydrogen present in 1 kg of Fuel:
∙ 8,93668274 For the combustion of Sulphur present in 1 kg of Fuel:
1,99793544 ∙ 10
Dr. Artemio J. Benítez Fundora
For the Nitrogen present in 1 kg of Fuel and for the Nitrogen supplied with the theoretical combustion air:
0,7545364 ∙ For the Water present in 1 kg of Fuel:
Then, the quantity of combustion products in the theoretical reaction of 1 kg of Fuel with air:
It is to note, that there is no Oxygen in the combustion products, because all the Oxygen supplied with the theoretical air have been consumed to generate CO 2, SO2 and part of the water present in gaseous form. Because Hydrogen always reacts faster than Carbon and Sulphur, is quite common in Combustion Technology to make an special group with the products formed by these two last components of the fuel; named it as Tri-Atomic Gases ( RO2):
So, we can write again, the equation for the sum of all combustion products as:
But all the flue gas analyzers, that are the type of equipment that measure the composition of the combustion products that leaves the combustion system in a boiler, fired heater, etc.; use to work with the combustion products in dry form, and gives the proportion of each component in dry basis and in volumetric or molar form. So it is very interesting to know the dry composition expected in a theoretical combustion of 1 kg of Fuel, as fired, that can be calculated in mass basis by:
And the mass fractions are calculated simply by proportion:
11
Dr. Artemio J. Benítez Fundora
And the sum of these mass fractions must equal:
1 The
molecular
weight
of
dry
combustion
gases
can
be
calculated
in
1 44,0095 64,0638 28,0134 by:
And the volumetric or molar fractions of these gases can be calculated as:
, ∙ 44,0095 . 64,0638 ∙ , . 28,0134 ∙ , .
And again the sum of these molar or volumetric fractions must equal:
1 1 Due to the fact that a theoretical combustion is what is analyzed, for this condition, the quantity of combustion gases is the minimum possible obtained, and the volumetric concentration of RO2 is the maximum possible obtained for a fuel:
, . 12
Dr. Artemio J. Benítez Fundora
This value only depends of the Fuel Composition or Ultimate Analysis, so there is a unique value for each Fuel, and is obtained for the theoretical combustion, also known as perfect combustion. Example:
For the Light Fuel Oil (Bunker A) of the example before: Fuel Light Fuel Oil (Bunker A)
C
H
O
N
S
A
W
0.834
0.117
0
0
0.04
0.009
0
The theoretical quantity of dry air is 13,749 kg of dry air/ kg of Fuel , from Table III, and according to the fuel composition, when 1 kg is burned, only 0,991 kg is transformed to gaseous form, because the ash content (0,009 kg Ash/kg of Fuel), is separated during combustion in solid form and precipitates to the bottom of the combustion chamber. Then, the quantity of combustion products formed by kg of Fuel is:
1 13,7490,991 14,740 Using the individual contributions of each component in the Fuel and in the Theoretical Air supplied for the combustion process:
3,6641911 ∙0,834 3,0559353774 1,99793544 ∙0,04 0,0799174176 00,7545364∙ 13,749 10,3741209636 ∙0,117 1,04559188058 8,93668274 0 Then:
3,05593537740,07991741761,045591880580 10,3741209636 14,55556563918 13
Dr. Artemio J. Benítez Fundora
Relative Error in %:
| ∙ 100 ≅ 1,27 % |14,7414,55556563918 14,55556563918 This admissible error is due to the approximation of the coefficients that multiply the composition terms in the individual equations for the calculation of every component of the combustion gases. The Dry Combustion Gases formed are:
3,0559353774 0,0799174176 10,3741209636 13,5099737586 And the mass fractions in dry gases are:
3,0559353774 0,2262 13,5099737586 0,0799174176 13,5099737586 0,00592 10,3741209636 13,5099737586 0,76788
And the sum of the mass fractions in dry gases is:
0,22620,005920,76788 1 It can be observed that the great part of the theoretical combustion gases is due to the presence of Nitrogen, thus the molecular weight of the combustion products will be closer to the molecular weight of Nitrogen (28,0134):
1 30,6341 0,2262 0,00592 0,76788 44,0095 64,0638 28,0134 Using this value, the volumetric fractions can be calculated as:
30,6341 ∙0,2262 ∙ 44,0095 44,0095 0,15745 . 14
Dr. Artemio J. Benítez Fundora
30,6341 ∙0,00592 ∙ 64,0638 64,0638 0.00283, . 30,6341 28,0134 ∙ 28,0134 ∙0,76788 0.83972 .
And the sum of the volumetric fractions must be:
0,15745 0,00283 0,83972 1 And the maximum molar or volumetric content of Tri-Atomic Gases in the combustion products of Light Fuel Oil (Bunker A) is:
0,16028 Or:
% ∙ 100 16,028 % Actu al Comb usti on Product s generated in th e combusti on of 1 kg of Fuel: The actual combustion process always is made with environmental air from the surroundings of the plant and this atmospheric air is always humid in tropical zones or in other latitudes during the spring and summer seasons diminishing in water content and in temperature as winter season approaches. Using the theoretical humid air in a combustion process generates an incomplete burning of the fuel, due to the fact that the 3 “T’s” of the combustion ideal conditions are never reached. For an ideal combustion it is necessary to make it at the highest possible Temperature, because the combustion reactions accelerates exponentially with the temperature, at the same time, there must be an intimate mixture between the fuel components (C, H and S) with the Oxygen in the combustion zone at the level of a molecular mixture that requires of a level of Turbulence that is almost impossible to archive; and finally, the fuel and the oxidant (O) must be kept under this conditions mentioned before enough Time to complete all the chemical reactions that will be needed to obtain the complete combustion of the Carbon, Hydrogen an Sulphur presents in the Fuel.
15
Dr. Artemio J. Benítez Fundora
Due to these reasons, it will be necessary to supply an extra quantity of oxygen to the actual combustion process to grant the total burning of the fuel, and it is traduced in an increase of the air needed above the theoretical air value. The relationship between the actual air and the theoretical one was stated before as:
Then the quantity of actual air needed to burn effectively 1 kg of fuel is calculated by:
∙ ∙ When the quantity of air supplied is increased, also the quantity of combustion gases will increases too, and a quantity of oxygen appears as a new combustion product, and also the fraction of nitrogen increases above the value obtained by a theoretical combustion; water in flue gases also increases due to the presence of humidity (moisture) in the air that is introduced from the environment to the combustion process. Thus the equations for calculate the combustion gases produced must be modified to accomplish these new conditions. These three equations remain the same:
, ∙ ∙ , , ∙ They are not affected either by the excess of air of by the moisture content (humidity) of the air; but the rest must be modified as:
, ∙ ∙ To take into account the air excess, and:
∙ ∙ To take into account the water introduced as humidity with the actual air, and a new one:
16
Dr. Artemio J. Benítez Fundora
, ∙ ∙ To take into account the excess of oxygen that is not used in the combustion process which was introduced with the actual air. And the new equation to calculate the quantity of combustion gases formed by burning 1 kg of fuel will be:
∙ ∙ Or in individual products contribution:
The actual dry gases formed in this condition are:
And the new mass fractions on dry conditions:
And the sum of these mass fractions must equal:
17
Dr. Artemio J. Benítez Fundora
The molecular weight of the actual dry combustion gases can be calculated in
, , , , by:
And the volumetric or molar fractions of these gases can be calculated as:
∙ , , . , ∙ , . , ∙ , . ∙ , , . And again the sum of these molar or volumetric fractions must equal:
The actual volumetric concentration of RO2 is:
, . . For the example seen before, and considering a local absolute humidity in the atmospheric air of 30 grams of water vapor per kilogram of dry air, and a coefficient of air excess of 1,18: New data:
, , 18
Dr. Artemio J. Benítez Fundora
For these conditions:
3,6641911 ∙0,834 3,0559353774 1,99793544 ∙0,04 0,0799174176 ∙0,117 1,04559188058 8,93668274 And:
0 0,7545364 ∙ 1,18∙ 13,749 12,241462737048 0 0, 0 3 ∙ 1, 1 8 ∙ 13,749 0,4867146 Finally:
0,2320453∙ 1,181 ∙ 13,749 0,574270349346 The quantity of actual combustion gases formed by burning 1 kg of Light Fuel Oil (Bunker A) is:
3,05593537740,07991741761,045591880580,4867146 12,2414627370480,574270349346 17,483892361974 And the quantity of actual dry combustion gases is:
3,05593537740,079917417612,241462737048 0,574270349346 15,951585881394 The mass fraction for each component in the actual dry combustion gases is therefore:
3,0559353774 0,19158 15,951585881394 0,0799174176 0,00501 15,951585881394 19
Dr. Artemio J. Benítez Fundora
12,241462737048 0,76741 15,951585881394 0,574270349346 15,951585881394 0,03600 Checking the sum:
0,191580,005010,767410,03600 1 The Molecular Weight of the actual dry flue gases is:
0,19158 0,00501 1 0,76741 0,03600 44,0095 64,0638 28,0134 31,9988 30,63374 Then, the volumetric concentrations that will be expected for each component of the actual dry flue gases are:
30,63374 ∙ 0, 1 9158 0, 1 3335 . 44,0095 30,63374 ∙ 0, 0 0501 0, 0 0240 . 64,0638 30,63374 ∙ 0, 7 6741 0, 8 3919 . 28,0134 30,63374 ∙ 0, 0 3600 0,03446 . 31,9988 Checking the sum:
0,13335 0,00240 0,83919 0,03446 1,00094 ≅ 1 And finally, the RO2 content in the actual dry flue gas is:
0,133350,00240 0,13575, . 20
Dr. Artemio J. Benítez Fundora
An d in tr od uc in g a new rel ati on sh ip :
, , ≅ , , The error in this estimation is:
| | 1,18071,18 1,18 ∙ 100 ≅ 0,0006 %
And therefore, it is possible to calculate the value of
using the following equation:
Or conveniently simplified as:
, , , But when the combustion process is not too complete, in flue gases appears a small quantity of CO as a volumetric fraction , due to the incomplete burning of C that must
be added to obtain the true value of the air excess coefficient:
21
