Fuels, Combustion, and Flue Gas Analysis (3rd Class Edition 3, Part A2) PDF

Fuels, Combustion, and Flue Gas Analysis • Chapter 3 ^ INTRODUCTION TO FUELS, COMBUSTION, AND FLUE GAS ANALYSIS This chapter expands on the principles of fuels, combustion, and flue gas analysis covered at the Fourth Class level. The chapter begins with an in-depth description of basic chemistry...

Fuels, Combustion, and Flue Gas Analysis • Chapter 3 ^ INTRODUCTION TO FUELS, COMBUSTION, AND FLUE GAS ANALYSIS This chapter expands on the principles of fuels, combustion, and flue gas analysis covered at the Fourth Class level. The chapter begins with an in-depth description of basic chemistry principles that are relevant to combustion and gas analysis. Then the chapter moves into the concept of molar mass, which is the basis for many types of combustion calculations. It is suggested that the learner work through combustion calculations using simple tables to keep the data organized. This tabular format helps with checking over molar and mass balances after the calculation is complete. These tables are a useful tool when tackling complex combustion problems. The concepts of theoretical and excess air are also reviewed. The Power Engineer should begin to understand how combustion theory is relevant to practical applications. This theory is the basis for boiler combustion control, which is studied later in the Third Class course. Traditional and emerging fuel types are considered next. These fuel types are important to the modern industrial economy as it diversifies to meet technological and environmental challenges. Finally, the chapter examines flue gas analysis and how it applies to combustion efficiency and emissions control. The principles of operation and the specific technologies used for flue gas analysis are examined. Some additional calculations are also covered, using flue gas analysis as a basis to determine air supplied for combustion. 3rd Class Edition 3' Part A2 105 ^ Chapter 3 • Fuels, Combustion, and Flue Gas Analysis OBJECTIVE 1 Calculate the mass of combustion products using molarmass. CHEMISTRY PRINCIPLES RELATED TO COMBUSTION This objective begins with a review of some first principles related to combustion chemistry, such as the periodic table, the concepts of atomic numbers, atomic mass, formula mass, and molar mass, and the laws which pertain to gases and combustion. Mass Units In the International System of Units (SI units), mass is measured in grams or kilograms. For an unknown mass, the symbol m is used. The atomic mass unit (AMU) is used to express atomic mass and molecular mass. One AMU is equal to one-twelfth of the mass of a carbon-12 atom. An AMU is roughly equal to the sum of the number of protons and neutrons in the atomic nucleus (electrons have a relatively small mass and are not included). This unit is mentioned for reference only and is not used in combustion calculations. Number of Atoms or Molecules A mole is a number used to count particles such as atoms, molecules, and ions. One mole is equal to 6.022 x 1023 particles. This number is called Avogadro's number. For combustion calculations, n represents the number of moles. In the same way a dozen is always 12 of something, a mole is always 6.022 x 1023 particles, no matter what type of particle is in question. The mole is a huge number. For example, one mole ofjellybeans, which is 6.022 x 1023 beans, would be the size of planet Earth. However, one mole of sulfur atoms, which is 6.022 x 1023 atoms, would fit in a small dish. Due to its size, the mole unit is useful to keep track of large quantities of atoms and molecules. Atomic Mass This section refers to the periodic table in the PanGlobal Academic Supplement. Each element has an entry on the periodic table with two numbers listed on it: the atomic number and the atomic mass. See Figure 1 for an example. The atomic number is the number of protons that an atom of an element contains. Atomic mass is the mass of one atom. For example, the atomic number for carbon is 6 and its atomic mass is 12.0107. For combustion calculations, this number can round to 12. Figure 1 - Atomic Mass Atomic mass Atomic weight ->12.0107 6<- Atomic number Chemical symbol Carbon 106 3rd Class Edition 3 • Part A2 Fuels, Combustion, and Flue Gas Analysis • Chapter 3 Terminology Atomic mass is the term used in Sl units; for imperial units, the term is atomic weight. 7® 0 Formula Mass Since most compounds in nature exist as molecules rather than individual atoms, the mass of these compounds can be calculated. A formula mass is the mass of a single molecule of a substance in atomic mass units (AMU). To calculate the formula mass of a molecular substance, add together the individual atomic masses. For example, the formula mass of water (I-^O) is calculated as follows: Mass of hydrogen = number of H atoms x atomic mass =2x1 =2 Mass of oxygen = number of 0 atoms x atomic mass = 1 x 16 = 16 Mass of water molecule = mass of H plus 0 = 2+16 = 18 The atomic and formula masses for some common substances are shown in Table 1. Table 1 - Atomic and Formula Masses Element Atomic Mass Molecule Formula Mass HsO 18 Carbon (C) 12 C02 44 Oxygen (0) 16 co 28 Sulfur (S) 32 S02 64 Hydrogen (H) \ 1 Molar Mass Since an AMU is an extremely small unit and impractical to use, the molar mass unit is used for combustion calculations. Molar mass is the mass (in grams) of one mole of a substance. Molar mass uses the same number as atomic mass or formula mass but is measured in grams per mole (g/mol) or kilograms per kilomole (kg/kmol). Molar mass is related to the atomic mass as follows: one mole of an element weighs M. grams, where M is the molar mass. For example, in Figure 2, hydrogen has an atomic mass of 1.01 on the periodic table. Hydrogen therefore has a molar mass (M) of 1.01 g/mol, meaning that 1 mole of hydrogen has a mass of 1.01 grams. Note: For combustion calculations, the atomic mass from the periodic table are rounded to the nearest whole number. Figure 2 - Molar Mass 1.01 1 32.07 16 10.81 5 H B s Hydrogen Boron Sulfur 1 mole (6.02 x1023) of 1 mole (6.02 x 1023) of boron weighs 10.81 grams. hydrogen weighs 1.01 grams. 1 mole (6.02 x1023) of sulfur 3rd Class Edition 3 • Part A2 weighs 32.07 grams. 107 r5> Chapter 3 • Fuels, Combustion, and Flue Gas Analysis Self-Test 1 Refer to Figure 2 and answer the following questions: a) What is the b) What is the atomic mass of boron? c) What is the d) atomic number of sulfur? What is the atomic mass ofsulfur? g) What is the h) molar mass of boron? What is the mass of 1 mol of boron? e) What is the f) atomic number of boron? molar mass of sulfur? What is the mass of 1 mol of sulfur? Formula mass is also related to molar mass as follows: one mole of a molecular substance weighs M grams, where M is the molar mass. For example, to find the molar mass of carbon dioxide {CO^), the formula mass is first calculated, as shown in Table 2. Table 2 - Determining COz Formula Mass Element Number of Atoms Atomic Mass Mass Carbon (C) 1 12 12 Oxygen (0) 2 16 32 44 Total (formula mass) Therefore, carbon dioxide (€02) has a formula mass of 44. The calculated value for the formula mass is also equal to the molar mass. For the molar mass, the units are grams per mole (g/mol) or kilograms per kilomole (kg/kmol). For 002, the molar mass M = 44 g/mol or kg/kmol. Molar Mass Formula m The quantities of mass, number of moles, and molar mass are related with the formula M = -—. Where n = number of moles. Units (mol) m = mass of the substance in grams. Units (g) M = molar mass (i.e., the mass in grams of one mole of the substance. Units (g/mol) 108 3rd Class Edition 3 - Part A2 Fuels, Combustion, and Flue Gas Analysis • Chapter 3 Example 1 Determine the number of moles contained in 36 g of carbon. Solution 1 Given m = 36 g, find n = number of moles. For M, we use the atomic mass from the periodic table. For carbon, M = 12 g/mol. w. Apply the formula, M = -— m Rearranging, n = — 36 g 12 g/mol n = 3 mol (Ans.) Therefore, 36 g of carbon contains 3 mol. (Ans.) For combustion calculations, the kilomole (kmol) is most convenient and more commonly used than the mole. For carbon, the kilomole is defined as the number of particles in 12 kg. The kilomole is related to the mole as follows: 1 kmol = 1000 mol. Combustion calculations use the following units: kg (mass), kmol, and kg/kmol (molar mass). Molar mass is the number of kilograms in one kilomole of a substance. For example, since the molar mass of oxygen is 32, then 1 kmol of oxygen has a mass of 32 kg. YVl The molar mass (M) is also calculated using the formula M = -—. Where m = mass in kg M = molar mass in kg/kmol n = number of kmol Example 2 Determine the number ofkilomoles in 176 kg of carbon dioxide. Solution 2 Given m = 176 kg, find n = number ofkilomoles. For carbon dioxide (002), M = 44 kg/kmol, as calculated earlier. The formula M = — be rearranged to get n = —. ^ can _-.__-__^__.-_-^_ ^---^- m " = 'M 176kg 44 kg/kmol = 4 kmol (Ans.) Therefore, 176 kg of carbon dioxide contains 4 kmol. (Ans.) 3rd Class Edition 3 • Part A2 109 ^r Chapter 3 • Fue/s, Combustion, and Flue Gas Analysis GAS LAWS An ideal gas is a simplified physical description of real gases found in nature. An ideal gas obeys the law PV = mRT. The constant R is the characteristic gas constant for the particular gas (-R = Cp - Cy\ P is pressure (kPa), V is volume (m3), T is temperature (K), and m is mass (kg). An ideal gas also has constant specific heat capacities that make use of relationships such as R = Cp- Cy (where the c variables stand in for specific heat capacities). Ideal gases obey the laws in the following sections. AS The General Gas Law By combining any two of Boyles, Gay-Lussacs, or Charles' laws, the general gas law is obtained ^V-i ?2^2 as — — = —=:—. TI T, Ideal Gas Equation As shown in the PanGlobal Academic Supplement, there are several versions of the general gas law. The first is = . The second equation listed is PV = mRT. The third equation is Tl TI PV= nRyT where RQ is the universal gas constant 8.314 kJ/kmol-K; P, V, and T are the same; and n is the number ofkilomoles of gas. The third equation is derived as follows: (eqn. 1) For a particular gas, PV = mRT By definition, the gas constant R = — mRoT Substituting into equation 1, PV = But, n M m = ^ M Substituting n form— gives PV = nRnT (eqn. 2) M Note: The equation PV = mRT is stated with volume as V and m as the mass of gas in kilograms. It is also possible to write this equation as Pv = RT where v is the specific volume (volume divided by mass). Example 3 Given 10 kg ofCO^ at temperature 273 K and absolute pressure 101.3 kPa, determine the volume of gas ifRo = 8.314 kJ/kmol-K. Solution 3 Use ideal gas equation PV = mRoT. For 002, molar mass M = 44 kg/kmol. TYl Find the number ofkmol, n = — M 10kg 44 kg/kmol 0.227 kmol Rearrange gas equation, V = nRoT 0.227 kmol x 8.314 kJ/kmol-K x 273 K 101.3 kPa = 5.09m3(Ans.) 110 3rd Class Edition 3 • Part A2 Fuels, Combustion, and Flue Gas Analysis • Chapter 3 ^ Dalton's Law Daltons law states that the pressure of a mixture of gases is equal to the sum of the partial pressures of its constituents. In equation form, PT = -PI + -P2 + J^3 where PT is the total pressure and the partial pressures are ?i, ?2> and P^. Avogadro's Law Avogadros law states that equal volumes of all gases, at the same temperature and pressure, have the same number of molecules. By extension, it can be proven that, for ideal gases, the mole percentage is the same as the volume percentage. Example 4 Air is a mixture of gas that contains 23.2% 02 by mass. If the formula mass of air is 28.96 kg/kmol, what is the percent of 0^ by volume in air? Solution 4 This solution illustrates a very important concept: how to change mass percentage to volume percentage. Base the analysis on 100 kg of air. If air is 23.2% 02 by mass, then 100 kg of air contains 23.2 kg 02. Calculate the number ofkilomoles ofOz and air. Air Oxygen(03) m = 23.2 kg m= 100kg M = 32 kg/kmol M = 28.96 kg/kmol n= m 23.2 kg M 32 kg/kmol = 0.725 kmol 03 The kilomole fraction of oxygen in air is n= ,77 100kg M 28.96 kg/kmol 0.725 kmol 0^ 3.453 kmol air = 3.453 kmol air = 0.21. From Avogadros law, the kilomole fraction is equal to the percent composition by volume. Therefore, the percent composition by volume of oxygen in air is 21.0%. (Ans.) COMBUSTION EQUATIONS A complete combustion reaction or equation occurs when a fuel reacts quickly with oxygen (02) and produces carbon dioxide (C02) and water {H^0~). The general equation for a complete combustion reaction is the following: Fuel+02 -> C02+H20 Stoichiometry The term stoichiometry is derived from the Greek stoiechion (element) and metron (to measure) and has the literal meaning to measure the elements. In chemistry, stoichiometry means to determine the amount of reactants and products in a reaction. The principles of stoichiometry are based on the law of conservation of mass. This law states that mass is neither created nor destroyed in chemical reactions. The mass of any one element at the beginning of a reaction equals the mass of that element at the end of the reaction. In a typical combustion reaction, the fuel combines with oxygen in the air to produce carbon dioxide and water. The following shows an equation for a combustion reaction in which propane (CsHg) is the fuel: C3Hg+502 ^ 3C02+4H20 3rd Class Edition 3 - Part A2 111 Chapter 3 • Fuels, Combustion, and Flue Gas Analysis The numbers 5,3, and 4 before 02, CO-^, andH^O are called coefficients. There is also a coefficient 1 before C3Hg that is left out. In a chemical equation, the coefficients express the number of molecules in each compound—or, less frequently, atoms, ions, or other particles. For combustion calculations, the coefficients correspond to the number of kilomoles of each compound. The coefficients can also represent the volumes of the reactants and products provided they are all at the same temperature and pressure. Balancing Equations An important principle of stoichiometry is that equations must balance (i.e., the number of atoms of each element must balance between the reactants and the products). When performing combustion calculations, always verify that the equations are balanced. Take the equation for the combustion ofpropane again: CsHg+SOz -» 3C02+4H20 The schematic in Figure 3 shows a colour-coded illustration of the equation with carbon in red, oxygen in white, and hydrogen in blue. The number of atoms of each element is the same on both sides of the equation, which means the equation is balanced. Figure 3 - Balancing Equations 00 A ^ 00 . ^ ^ 00 ^ ^ 00 00 (X) ^ CsHg + 5 02 ^ 3 COs + 4 H20 Example 5 Verify that the equation €^N22011 + 1202 -» 12C02 + HHzO is balanced. Solution 5 To verify a balanced equation, count the number of atoms for each element on each side of the equation. Ci2H220n + 12 02 -> 12C02+11H20 Number of carbon atoms = 12 Number of hydrogen atoms = 22 Number of oxygen atoms = 35 The number of atoms for each element is the same for the reactants and the products. This verifies that the equation is balanced. (Ans.) 112 3rd Class Edition 3 - Part A2 Fuels, Combustion, and Flue Gas Analysis • Chapter 3 Self-Test 2 Determine the coefficients required to balance the following equations: a) _CzHe + _0-i -> ___ COz +_HzO b) _NN3 + _02 -» _NOz +_HzO c) _C4Hio + _02 ^ _C02 +_HzO There are two other laws that describe how reactants and products combine in simple ratios: the law of combining weights, and the law of combining volumes. It is important to note the significance of the coefficients in the combustion equation. These coefficients represent numbers. The numbers occur in simple whole number ratios and can indicate the number of molecules, number of moles, or other aspects in a compound. Law of Combining Weights All substances combine in accordance with simple, definite weight relationships. These relationships are exactly proportional to the molecular weights of the constituents. For example, in the equation for the combustion of carbon, the coefficients are 1:1:1. These coefficients are assigned the kilomole unit, and then the molar mass formula, m = n x M, is applied. To obtain mass (m)> the number ofkilomoles (n) is multiplied by the molar mass (M). C+02 ^ COz 1 kmol C + 1 kmol 02 -> 1 kmol 002 1 kmol x 12 kg/kmol C + 1 kmol x 32 kg/kmol 0^ -> 1 kmol x 44 kg/kmol COz 12kgC+32kg02 ^ 44 kg C02 The mass amounts follow simple arithmetic and conform with the law of conservation of mass for a balanced equation: 12 + 32 =44. Law of Combining Volumes The coefficients in an equation can also represent volumes of gases, provided the volumes of gases are at the same temperature and pressure. For example, when nitrogen and hydrogen gases combine to form ammonia, they do so in the ratio of 1:3:2. N2(g)+3H2(g) -> 2NHs(g) The coefficients represent volumes if the gases are all at the same temperature and pressure. One volume of N2 reacts with three volumes of N2 to produce two volumes of NN3. Note that molar and volume numbers do not follow simple arithmetic. There is a volumetric contraction during the reaction. 3rd Class Edition 3 • Part A2 113 r®- Chapter 3 • Fuels, Combustion, and Flue Gas Analysis MOLAR MASS CALCULATIONS The following equation shows the combustion of methane gas (€N4): CH4+202 ^ C02+2H20 This equation contains all the information needed to carry out a basic combustion calculation. The results of the calculation can be compiled in a simple balance sheet based on the principles reviewed above. Recall that M = molar mass, m = mass, and n = number of moles. The law of combining weights shows that the substances in this equation combine in simple whole number kilomoles. The ratios of CH4, 0^, CO-^, and H^O are 1:2:1:2. For the complete combustion of 1 kmol of €N4, 2 kmol of 03 are required. For 1 kmol of €N4 burned, the reaction produces 1 kmol ofCOz and 2 kmol ofI-^O. A balance sheet can be used to keep track of all the reactants and products. Columns two and three are M (kg/kmol) and n (kmol). The mass of each substance is found using the equation m= Mx n. The multiplication of M x n is written in column four. Lastly, the calculated mass is entered in column five. The overall results are shown in Table 3. Table 3 - Calculations for the Combustion of Methane 2 M (kg/kmol) 3 n (kmol) 4 5 Substance Mx n m (kg) CH4 16 1 16x 1 16 02 32 2 32x2 64 C02 44 1 44 x 1 44 HgO 18 2 18x2 36 1 Note: Table 3 can be considered a template for calculating molar mass when the values and substances are removed. The results of problems are verified by checking the mass of products and reactants, and by ensuring that the equation is balanced. 1. The mass of the reactants (€N4 and 02) is 16 +64= 80 kg. The mass of the products (C02 and H20) is 44 + 36 = 80 kg. Therefore, the calculations comply with the law of conservation of mass. 2. On the left side of the equation are one carbon, four hydrogen, and four oxygen atoms. On the right side of the equation are one carbon, four hydrogen, and four oxygen atoms. Therefore, the equation is balanced. Example 6 The combustion ofpropane ^Hg) follows the equation C3Hg + 502 -> 3C02 + 4H20. If 176 kg ofpropane is burned, calculate the mass of oxygen (Oz) required, and the mass of carbon dioxide (C02) and water vapour (H20) produced. Solution 6 It is suggested to do the calculation in tabular form. The molar mass in column two is simply the formula mass with units in kg/kmol. The equation coefficients are placed in column three. Write out the multiplication M x n in column four. Put the result m for each multiplication in column five. 114 3rd Class Edition 3 - Part A2 Fuels, Combustion, and Hue Gas Analysis • Chapter 3 f£ The problem listed 176 kg of C^H.s' but column five currently has 44 kg of CsHg. Determine the 176kg required multiplication factor: " = 4. Therefore, multiply all column five values by four to 44kg obtain column six. Substance 2 M (kg/kmol) 3 n (kmol) Mxn 5 m (kg) 6 m x 4 (kg) CsHs 44 1 44 x 1 44 176 02 32 5 32x5 160 640 COs 44 3 44x3 132 528 HzO 18 4 18x4 72 288 1 4 The calculated values are as follows: 02 required: 640 kg (Ans.) C02 produced: 528 kg (Ans.) H:20 produced: 288 kg (Ans.) The balanced equation reads as follows: 176 kg CsHg + 640kg 02 ^ 528 kg CO^ + 288 kgHzO Example 7 The combustion ofethane (CzHg) follows the equation 2 C^ + 7 Oz -> 4 002 + 6 HzO. If 30 kg of ethane is burned, calculate the mass of oxygen required, and the mass of carbon dioxide and water vapour produced. Solution 7 It is suggested to do the calculation in tabular form. The molar mass in column two is simply the formula mass with units in kg/kmol. The equation coefficients are placed in column three. Write out the multiplication M x n in column four. Put the result m for each multiplication in column five. The problem gave 30 kg ofC2H<3. In column five, there are 60 kg ofC2Ho. Determine the required 30kg multiplication factor: ——^— = 0.5. Therefore, multiply all column five values by 0.5 to obtain column six. 60 2 M (kg/kmol) 3 n(kmol) 4 5 6 Substance Mx n m (kg) m x 0.5 (kg) C2He 30 2 30x2 60 30 02 32 7 32x7 224 112 C02 77 4 44x4 176 88 HsO 18 6 18x6 108 54 1 The calculated values are as foUows: 30kgC2H6+112kg02 -> 88 kg €03 + 54 kg H^O (Ans.) 3rd Class Edition 3 • Part A2 115