Fuels, Combustion, and Flue Gas Analysis PDF

Fuels, Combustion, and Flue Gas Analysis • Chapter 3 OBJECTIVE 3 Explain the purpose and benefits of excess air, and calculate the theoretical and excess air required for the complete combustion of a given fuel. THEORETICAL AND EXCESS AIR The oxygen required for complete combustion comes from the...

Fuels, Combustion, and Flue Gas Analysis • Chapter 3 OBJECTIVE 3 Explain the purpose and benefits of excess air, and calculate the theoretical and excess air required for the complete combustion of a given fuel. THEORETICAL AND EXCESS AIR The oxygen required for complete combustion comes from the air supplied to the furnace. The amount of air required to supply just enough oxygen for complete combustion is called the theoretical air. In practice, more than this theoretical amount of air must be supplied to ensure that every particle of fuel comes into contact with oxygen. Excess air is the amount of air supplied to a combustion process in addition to the theoretical air required for perfect combustion. The amounts of excess and theoretical air are usually expressed as mass (kg of air/kg of fuel) but can also be expressed as volumes or kilomoles. Excess air is calculated using the following equation: excess air excess air = __^rr^i^n— x 100% theoretical air For example, if the theoretical air required for the complete combustion of 1 kg of coal is 12 kg, but the actual air supplied to the furnace is 18 kg per 1 kg of coal, then the excess air is 6 kg of air per 1 kg of coal. As a percentage, this is -;- x 100 = 50%. The total air supplied for combustion is defined as: Total air supplied = excess air + theoretical air In the example above, the total air supplied is 6 kg + 12 kg = 18 kg. The required excess air depends on the fuel, the method of firing, the burner and furnace design, and the load on the boiler. The following are typical excess air percentages required (by mass) when burning gas, oil, and coal. Natural gas 5% to 10% Oil 5% to 15% Coal (pulverized) 15% to 30% Coal (stoker fire) 25% to 50% It is the responsibility of the manufacturer to design the boiler, furnace, and firing equipment for efficient combustion. It is the responsibility of the operator to operate the equipment to obtain complete combustion with a minimum of excess air. 3rd Class Edition 3 • Part A2 119 Chapter 3 • Fuels, Combustion, and Flue Gas Analysis Effects of Incorrect Excess Air Too much excess air may have the following effects: a) Lower furnace temperature, which results in less heat transfer. b) Reduced combustion efficiency due to less retention time in the combustion zone. c) The extra oxygen and nitrogen leaving the stack carry away sensible heat, which results in reduced efficiency. d) A pulsating flame develops, the flame is pulled too far away from the burner, or both. e) Temperatures at the back of the furnace or further along the flue gas path rise due to increased velocities of the flue gas. f) The draft fans require extra power to handle the extra air volumes. Not enough excess air may have the following effects: a) Incomplete combustion due to inadequate mixing of fuel and air. b) Deposits ofunburned solid carbon (as soot) and the production of carbon monoxide. c) Lower furnace temperature because less heat is liberated from the fuel. d) Smoky flame and grey or black flue gas leaving the stack, which is an environmental concern. e) Reduced thermal efficiency due to less heat energy released during combustion. CALCULATING THEORETICAL AIR An analysis of a fuel gives the percentage, by mass, of each element in the fuel, including the combustible elements. Basic combustion equations are used to determine the mass (in kilograms) of oxygen required for each of the three combustible elements (carbon, hydrogen, and sulfur). These individual masses can then be totaled to obtain the required oxygen. Once the required oxygen is known, the required air can be determined. The mass of air required to supply a specific quantity of oxygen maybe calculated for a combustion 23 process. Since dry air is 23% oxygen by mass, each kilogram of air contains —— kg of oxygen. 23 1kg air = -^ kg oxygen 100,_ _ 1kg oxygen = ^ kg air Therefore, —— or 4.35 kg of air contains 1 kg of oxygen. For every 1 kg of oxygen required for I', there must be at least 4.35 kg of air supplied. To derive a formula for the required theoretical air, the following procedure is followed: 1. Write out the equation for perfect combustion. 2. Construct a table for each constituent with the headings n (number of kilomoles), M (molar mass), m (mass), and an additional column to obtain 1 kg of fuel constituent. 3. Convert mass of oxygen to mass of combustion required. 4. Add required air for carbon, hydrogen, and sulfur to form a single equation. 120 3rd Class Edition 3 - Part A2 Fuels, Combustion, and Flue Gas Analysis • Chapter 3 ^ The molar mass values shown in Table 5 are required for combustion calculations. Table 5 - Molar Mass Values for Various Substances Molar mass M (kg/kmol) H2 2 H20 18 c 12 C02 44 02 32 S02 64 s 32 co 28 Air Required for the Combustion of Carbon The combustion equation for carbon is C + 0^-> CO^. The equation coefficients can be specified as number of kilomoles. The analysis that follows determines the mass of air required for the complete combustion of 1 kg of hydrogen (Hy. The column headings are n (number of kilomoles) from the equation coefficients, M (molar mass), m (mass), and division to obtain 1 kg of carbon. This analysis is shown in Table 6. Table 6 - Air Required for the Combustion of Carbon 1 4 Substance 2 n (kmol) 3 M (kg/kmol) c 1 12 02 1 32 \ 32 C02 1 44 ; 5 m = M x n (kg) | divide by 12 12 44 1kgC 32 _ 8 TJ'ikg02 ^ = 3.67 kg C02 The analysis shows that 1 kg C requires — kg 0^ for complete combustion. Oxygen is supplied by combustion air, so the next step is to find the mass of air containing — kg 0^. g 8, ^ 1(^0 kg air O? = ^ kg O? x 3"&"2 ~ 3JV&"2" 23kg02 8.. 100,_,, = ix^kgair = 11.59 kg air (Ans.) Therefore, 1 kg C requires 11.59 kg air. (Ans.) 3rd Class Edition 3 • Part A2 121 ^ Chapter 3 • Fuels, Combustion, and Flue Gas Analysis Air Required for the Combustion of Hydrogen The combustion equation for hydrogen is 2 H^ + 02 -> 2 I-^O. The same analysis as before can be used to determine the amount of air for complete combustion of 1 kg of hydrogen. This process is shown in Table 7. Table 7 - Air Required for Combustion of Hydrogen 4 Substance 2 n (kmol) 3 M (kg/kmol) m = M x n (kg) 5 divide by 4 HS 2 2 4 1kgH2 02 1 32 32 ^ = 8 kg Os H20 2 18 36 ^ = 9 kg H20 1 The analysis shows that 1 kg H^ requires 8 kg 0^ for complete combustion. Oxygen is supplied by combustion air, so the next step is to find the mass of air containing 8 kg 02. 8 kg 02 = 8 kg 02 x 100 kg air 23 kg 02 100, . = Sx^kgair = 34.78 kg air (Ans.) Therefore, 1 kg H^ requires 34.78 kg air. (Ans.) If the fuel itself contains oxygen, that oxygen is assumed to be from the water in the fuel. Therefore, some of the hydrogen in the fuel is bound up with the oxygen as N20, so is not available for combustion. From the above, we know that each kilogram of oxygen requires l/s kg of hydrogen to form water. Therefore, the hydrogen available for combustion is the total hydrogen (N2) minus °2^.. ., , , ., ,, . , .... 02 the hydrogen in the water, which is —. That is, the hydrogen available for combustion is N2 - —^-. 8 Therefore, the amount of air to burn the available N2 = 8 ( H2 - -^- j x —— kg Air Required for the Combustion of Sulfur The combustion equation for sulfur is S + 0^-> SO^. The same analysis as before can be used to determine the amount of air required for complete combustion of 1 kg of sulfur. This analysis is shown in Table 8. Table 8 - Air Required for Combustion of Hydrogen 1 2 Substance n (kmol) S 5 divide by 32 1 32 32 1 kg S 1 32 32 1 kg02 S02 1 64 64 2 kg S02 02 122 3 4 M (kg/kmol) m =M x n (kg) 3rd Class Edition 3 • Part A2 Fuels, Combustion, and Flue Gas Analysis • Chapter 3 The analysis shows that 1 kg S requires 1 kg 0^ for complete combustion. Oxygen is supplied by combustion air, so the next step is to find mass of air containing 1 kg 02. 100 kg air 1kg 02 = 1 kg 02 x '2 ~ 1^^2" 23kg02 100 ,„ _ = ^ kg air = 4.35 kg air (Ans.) Therefore, 1 kg S requires 4.35 kg air. (Ans.) Theoretical (Stoichiometric) Air for Combustibles The next step is to add the air required for the combustion of carbon, hydrogen, and sulfur, as calculated above. 8^.. 100 , ^(^ Oi\ ^ 100 , ^^ 100 air = ^ C x -^ + 8 ( H-? - —1 ) x -^-lr + S x 3 ~ 23 ~ \"^ 8/23 ' ~ 23 Factoring out —— simplifies the equation to: 100.. |8 ^ . „/'„ Ol\ . J, , Theoretical air = -^- x |^-C+8(H2--^j+S|kg air/kg fuel In this equation, the chemical symbols have the following meanings: C = kg carbon per kg of fuel = percent carbon H2 = kg hydrogen per kg of fuel = percent hydrogen 02 = kg oxygen per kg of fuel = percent oxygen S = kgsulfurperkgoffuel = percent sulfur Note that the percentage values are entered into the equation as decimals. For example, if the carbon content of the fuel is 75%, then enter C as 0.75 in the equation. The previous calculations are all based on the equations for complete combustion of carbon, hydrogen, and sulfur. Remember that incomplete combustion of carbon produces carbon monoxide. This type of combustion should be avoided, as carbon monoxide is a combustible and toxic gas. 3rd Class Edition 3 • Part A2 123 Chapter 3 • Fuels, Combustion, and Flue Gas Analysis Example 8 If the analysis of any fuel is provided, the above equation can be applied to determine the theoretical air required. For example, given the following analysis of coal, calculate the theoretical mass of air required to burn each kilogram of the coal. Note that ash and nitrogen are not combustible. nitrogen 2% ash 7.0% oxygen 7.9% sulfur 0.5% carbon 76.7% hydrogen 4.9% Solution 8 TheoreticalAir.tc+T-fJ+SJT = |8C+sfH,^+sl^ 8x0.767+8fo.049-ao79^0.005lwo 23 100 [2.045 + 0.313 + 0.005] 23 100 = (2.363)^ 23 = 10.27 kg air/kg fuel (Ans.) Therefore, each kilogram of fuel burned requires a theoretical air supply of 10.27 kg. (Ans.) CALCULATING EXCESS AIR If the percentage of excess air required to operate a furnace is given, then the exact mass of excess air and the total air required (theoretical plus excess) can be calculated. The amount of excess and theoretical air are usually expressed as mass (kg) but can also be expressed in kilomoles of air. / excess air Percent excess air == ( . —"'" "" — ) x 100% \ theoretical air, mass of air supplied - mass of theoretical air mass ot theoretical air x 100% Example 9 If the safe operation of a furnace requires that 20% excess air be admitted, and the theoretical air required is 10.27 kg per kg of fuel, what is the excess air in kg/kg fuel? What is the total amount of air supplied? Solution 9 Calculate excess air: Excess air = 0.20 x 10.27 kg/kg fuel = 2.054 kg/kg fuel (Ans.) The total air required to operate the furnace would then be calculated as follows: Total air = excess air + theoretical air = 2.054 + 10.27 kg/kg fuel 12.324 kg/kg fuel (Ans.) 124 3rd Class Edition 3 • Part A2 Fuels, Combustion, and Flue Gas Analysis • Chapter 3 Note: Excess air affects the amount of oxygen and nitrogen in the flue gases. For every kilogram of excess air there is an extra 0.23 kg of oxygen (since air is 23% oxygen by mass) and an extra 0.77 kg of nitrogen (since air is 77% nitrogen by mass). In the above example, 2.054 kg of excess air would result in an extra 2.054 x 0.23 = 0.472 kg of oxygen and 2.054 x 0.77 = 1.582 kg of nitrogen. AIR-FUEL RATIO The air-fuel ratio (AF) is defined as the ratio of the mass of combustion air to the mass of fuel. This ratio can be written as the following equation: mass of combustion air mass of fuel The percent of excess air can also be calculated in terms of the air-fuel ratio for the air supplied versus the theoretical air: supplied — Ar theoretical Percent excess air = —'— — x 100% Example 10 Calculate the theoretical air-fuel ratio for the perfect combustion of methane (CH4). Solution 10 Write out the combustion equation €N4 + 2 0^ -^ CO^ + 21-^O in tabular form. CH4 + 202 -> C02 + 2H20 1 kmol + 2 kmol -^ 1 kmol + 2 kmol 16kg + 64kg -» 44kg + 36kg Taking the mass percentage of oxygen in air to be 23.2%, the following can be calculated: 64 kg of 02 is supplied by 64 x —— kg of air = 275.86 kg of air To burn 16 kg of methane requires 275.86 kg of air. Calculate the air-fuel ratio: AF = mass air mass fuel 275.86 16 = 17.2 (Ans.) The air-fuel ratio for perfect combustion (stoichiometric ratio) for methane is 17.2 (Ans.) 3rd Class Edition 3 • Part A2 125 r®- Chapter 3 • Fuels, Combustion, and Flue Gas Analysis Example 11 If the air-fael ratio for methane combustion in a particular boiler is 20.64, calculate the excess air supplied. Solution 11 Calculate excess air as follows: supplied AF = 20.64-17.2' 17.2 3.44 J7.2, x 100% x 100% x 100% = 20%(Ans.) Control ofAir-Fuel Ratio Comprehension of the air-fuel ratio is essential for understanding boiler combustion control systems. A boiler combustion control system is designed to maintain a proper air-fuel mixture under varying load conditions and within safe limits. The best combustion efficiency occurs at the optimum air-to-fuel ratio, and controlling this ratio provides the highest efficiency. The ideal air-to-fuel relationship varies at different operating loads. Tuning is the process of establishing the desired air-to-fuel relationship under various operating conditions. Proper tuning is accomplished by evaluating readings such as temperature, oxygen concentration, carbon monoxide, and nitrogen oxide (NOx) emissions in the boiler furnace stack. 126 3rd Class Edition 3 • Part A2