Fuels, Combustion, and Flue Gas Analysis Chapter 3 PDF

Chapter 3 ' Fuels, Combustion, and Flue Gas Analysis OBJECTIVE 10 Calculate quantities of oxygen, air, and excess air from a given flue gas analysis. Explain the analysis offlue gas for the measurement of oxygen (Oy), carbon monoxide (CO), and carbon dioxide (COy) in relation to combustion efficien...

Chapter 3 ' Fuels, Combustion, and Flue Gas Analysis OBJECTIVE 10 Calculate quantities of oxygen, air, and excess air from a given flue gas analysis. Explain the analysis offlue gas for the measurement of oxygen (Oy), carbon monoxide (CO), and carbon dioxide (COy) in relation to combustion efficiency. Combustion gas analysis is the process of analyzing the total flue gas that leaves a boiler or other industrial furnace to determine and measure the various individual components of the gas. The process provides a quantitative analysis of the products of combustion in the flue gas. The two primary purposes of this process are: 1. The analysis offlue gas: This analysis looks specifically at the products of combustion in the flue gas and the combustion air that is supplied. It concentrates on the efficiency and completeness of the combustion process. 2. The monitoring of stack emissions: This monitoring looks specifically for the components in the flue gas that are emitted to atmosphere and are undesirable above certain limits. The limits are government-regulated environmental standards. While some of the components of an analysis maybe of interest for both flue gas analysis and stack emissions monitoring (such as COz and N0^), stack emissions monitoring includes additional components such as particulates, SO-^, and mercury. UNITS FOR FLUE GAS ANALYSIS AND EMISSIONS MONITORING Flue gas constituents are most often listed by percent volume. Smaller concentrations, such as airborne pollutants, are measured in parts per million (ppm). For an ideal gas, parts per million by volume or by mole are practically identical. This applies for most flue gases at atmospheric pressure. The conversion factor to change between units is 1% by volume =10 000 ppm. To convert percent by volume to parts per million (ppm), multiply the percentage by 10 000. To convert parts per million to percent by volume, divide by 10 000. Example 21 A flue gas sample contains 0.01% CO and 0.05% N0^:. Convert these amounts to parts per million. Solution 21 Multiply percent by volume by 10 000 to obtain parts per mUlion (ppm). 0.01% CO = 0.01 x 10 000 ppm = 100 ppm CO (Ans.) 0.05% NO;, = 0.05 x 10 000 ppm = 500 ppm NO^; (Ans.) Example 22 A flue gas sample contains 200 ppm N0^;. What is this reading as percent by volume? Solution 22 Divide ppm by 10 000 to obtain percent by volume. 200 ppm N0^ = ^200^ = 0.02% N0^ by volume (Ans.) 158 3rd Class Edition 3 • Part A2 Fuels, Combustion, and Flue Gas Analysis • Chapter 3 ^ FLUE GAS ANALYSIS For combustion efficiency purposes, the most common components measured in a flue gas analysis are carbon dioxide (€02), carbon monoxide (CO), and oxygen (02). The importance of each is as follows: 1. Carbon dioxide (€02) is the product of complete combustion. Therefore, the higher the C02 content of the flue gas, the more complete the combustion. 2. Carbon monoxide (CO) is a product of incomplete combustion. Therefore, any measurement of CO in the flue gas indicates that the fuel is not burning completely. The presence of CO indicates that the furnace is starved of air or the flame is not proper. If CO is a product of combustion, the operator should investigate the operation of the burners because there may be a mechanical problem, a fuel-air mixing problem, or insufficient air to the furnace. 3. Oxygen (02) in the flue gas is an indication of excess air. Too much excess air in the flue gas is inefficient as it carries excess heat out of the furnace. Too little excess air may cause incomplete combustion. In general, the most efficient furnace operation occurs when the flue gas analysis indicates maximum CO^, minimum 02, and zero CO. Gas Measurements for Combustion Analysis Figure 12 shows flue gas curves for natural gas. The graph illustrates how the flue gas readings vary with the amount of combustion air supplied to the burner. At the stoichiometric line, 100% theoretical air is supplied, which results in perfect combustion. The oxygen percentage is zero, which indicates that no excess air is supplied. The carbon monoxide percentage is zero since enough air is supplied to burn all the fuel. The carbon dioxide percentage is at the maximum since no excess air is supplied. Figure 12 - Flue Gas Analysis for Natural Gas Gas percent by volume ^02 10% co' 02 a? c 0 t. •l-l a? 0 5% .c 0 5 0) Excess fuel Excess air 50% 100% 150% % Theoretical air 3rd Class Edition 3 - Part A2 159 ?& Chapter 3 • Fuels, Combustion, and Flue Gas Analysis The most reliable method for combustion analysis is to measure 0^ directly. The oxygen readings follow a simple pattern and closely follow excess air trends. At the stoichiometric line or in the excess fuel region, the 02 measurement is zero percent. However, when combustion has excess air, the 03 reading increases. Another advantage to measuring the 0^ directly is that the accuracy of oxygen percentage analyzers is better than those that measure the percentage of carbon dioxide (C02). Although 002 can be measured alone, there are some advantages to using the 02 concentration to determine the C02. As shown on the CO^ curve in Figure 12, the same CO^ measurement is possible on both sides of the peak. When CO^ is measured alone, it is not possible to tell if the mixture of exhaust gases indicates an excess in fuel or an excess in air. With 02 measurements, this ambiguity does not occur. Determining Excess Air To determine the amount of excess air, a sample of the flue gas is analyzed for the percentages of C02 or 02 content. Even though the amount of CO^ produced per unit mass of fuel burned is constant, the percentage ofCO^ in the flue gas varies depending on the amount of air supplied to the burner. For typical fuel compositions, excess air charts are available to relate flue gas CO^ or 02 to excess air. Table 13 is an example of a chart that shows the percentages ofC02 and 02 that can be expected in the flue gas analysis at different percentages of excess air. These percentages are listed for the efficient burning of coal, oil, and natural gas. For example, a furnace burning fuel oil with 40% excess air has flue gas that is 10.6% C02 and 5.2% 02. If the flue gas analysis of a gas-fired boiler shows 3.9% 0^, this chart suggests that the excess air is 20% and the C02 in the flue gas is about 9.9%. A noticeable trend from Table 13, regardless of the fuel, is that as the excess air increases, the flue gas 02 increases and the flue gas CO^ decreases. Table 13 - Typical COz and Oz Percentages in Flue Gas Coal (bituminous) Fuel oil (light) Natural gas Percentage of excess air Combustion products 0 10 20 30 40 C02 18.6 16.9 15.5 14.3 13.2 12.2 02 0.0 2.0 3.5 5.0 6.1 7.1 C02 15.4 13.8 12.6 11.5 10.6 02 0.0 2.0 3.7 5.2 5.2 6.3 C02 12.2 10.9 9.9 9.1 8.4 7.9 02 0.0 2.1 3.9 5.4 6.4 80 100 11.3 10.2 9.2 8.0 9.4 10.6 10.0 ; 9.3 8.2 7.4 7.3 8.2 9.6 7.3 6.4 5.7 7.5 ; 8.4 9.8 10.9 50 60 Example 23 A coal-fired plant has excess air calculated at 60%. What are the CO^ and 0^ percentages by volume? Solution 23 From Table 13, CO^ = 11.3% and 0^ = 8%. (Ans.) 160 3rd Class Edition 3 • Part A2 Fuels, Combustion, and Flue Gas Analysis • Chapter 3 CALCULATE QUANTITIES OF OXYGEN, AIR, AND EXCESS AIR FROM A GIVEN FLUE GAS ANALYSIS The PanGlobal Academic Supplement gives two formulas which are required for flue gas calculations. The theoretical air is calculated from composition of the fuel, while the actual air supplied to the burner can be determined from the flue gas analysis. These formulas are: Theoretical (stoichiometric) air required to burn fuel: Air(kg/kgoffuel) = Jc+S^-^) +S^ Where C, N3, 02, and S are the percent composition by mass entered as decimal fractions. Air supplied from analysis offlue gases: N-. Air(kg/kgoffuel) = —————xC 33(002 + CO) Where: C is the percentage of carbon in fuel by mass N3 is the percentage of nitrogen in flue gas by volume C02 is the percentage of carbon dioxide in flue gas by volume CO is the percentage of carbon monoxide in flue gas by volume Note: The value obtained from this formula is the air-fuel ratio (AF) by mass, which is covered earlier in this chapter. Example 24 A coal-fired boiler uses coal with an analysis by mass of C = 78%, N2 = 6%, 0^ = 3%, and non-combustibles = 13%. The percentage analysis by volume of a flue gas sample is 002 = 10.4%, CO = 0.2%, 02 = 7.8%, and N2 = 81.6%. Calculate the percent of excess air that is used. Solution 24 Using formulas from the PanGlobal Academic Supplement, calculate theoretical air and actual air supplied. Also determine the percentage of excess air. Calculate the theoretical (stoichiometric) air required to burn fuel: Note: Values entered into the formula are percentages converted to decimals. Air(kg/kgoffuel) = Jc+S^H;-^) +SJ 100 23 - [txo-s(o.o.-<f).o]f 8 x 0.78 +8 (0.05625)1 ^)0 = (2.08+0.45) 2.53 x 100 23 100 23 = 11.0 kg air/kg fuel (Ans.) 3rd Class Edition 3 • Part A2 161 Chapter 3 • Fuels, Combustion, and Flue Gas Analysis Calculate the air supplied from the analysis offlue gases: Note: Values for this formula are entered as percentages by volume for N3, C02, and CO. The value for C is percentage by mass. N-, Air(kg/kgoffuel) = ——— x C 33(002 + CO) 8L6 .x 78 33(10.4+0.2) 81.6 ' ^8X78 = 18.2 kg air/kg fuel (Ans.) This data is now used to find the percentage of excess air. Excess air equals air supplied minus theoretical air = 18.2 - 11.0 = 7.2 kg air/ kg fuel. Percent excess air equals „ , = ( —^ )x 100% = 40% (Ans.) air supplied V 18.^ Example 25 The dry flue gas analysis by volume for a hydrocarbon fuel is the following: CO^ = 8.85 %, CO = 1.2%, 02 = 6.8%, and N2 = 83.15%. The fuel has the following composition by mass: C = 84%, H2 = 14%, and 0^ = 2%. Calculate the theoretical air, the mass of air supplied per kilogram of fuel, the mass of oxygen supplied per kilogram of fuel, and the percent of excess air. Solution 25 Using formulas from the PanGlobal Academic Supplement, calculate the theoretical air and actual air supplied. Calculate the theoretical (stoichiometric) air required to burn fuel: Air(kg/kgoffuel) = [jc+S^-^)+S^ ^x0.84+8fo.l4-^2^o1^° [3"""" ' "V'" 8 / ' "J 23 = |8 x 0.84+8 (0.1375) I = (2.24+1.1) 100 100 23 100 23 = 3-34X^ = 14.52 kg air/kg fuel (Ans.) Calculate the air supplied from the analysis offlue gases: N^ Air(kg/kgoffuel) = ——^—xC 33(C02 + CO) 83.15 33(8.85 + 1.2) x 84 = -^X84 331.65 = 21.06 kg air/kg fuel (Ans.) 162 3rd Class Edition 3' Part A2 Fuels, Combustion, and Flue Gas Analysis • Chapter 3 This data can be used to find the other required quantities: 23 Mass of oxygen supplied =21.06 kg air x —— = 4.84 kg 02/kg fuel (Ans.) Excess air equals supplied air minus theoretical air = 21.06 - 14.52 = 6.54 kg air/ kg fuel. excess air / 6.54 excess air = , "— — ^ ^ ( —lr^ I x 100% = 45% (AnsJ theoretical air M4.52/ COMBUSTION EFFICIENCY In general, efficiency means the ratio of output to input expressed as a percentage. Combustion efficiency measures how effectively the energy of the fuel has been converted into useful heat. The useful heat acquired by the boiler or furnace is calculated indirectly using the boiler or furnace stack heat loss. Combustion efficiency is usually calculated as follows: useful heat in boiler or furnace fuel heating value fuel heating value - heat loss in stack gases fuel heating value This equation assumes that the heat produced from combustion has either transferred to the working fluid (steam and water) or gone up the stack. This equation is a simplification that does not include smaller losses due to radiation, blowdown, or leaks. Stack heat losses are calculated using gas concentration (usually 02 but sometimes €02), stack temperature measurements, and the fuel specifications for the chemical composition and heat content. Example 26 If the perfect combustion of a fuel releases 55 MJ/kg of fuel, and 8 MJ/kg of fuel is lost up the stack, what is the combustion efficiency? Solution 26 Combustion efficiency = fuel heating value - heat loss in stack gases fuel heating value 55 x 100% = 85.5% (Ans.) 3rd Class Edition 3 • Part A2 163 ?& Chapter 3 • Fuels, Combustion, and Flue Gas Analysis Figure 13 shows there is an optimum range of excess air where combustion ef&ciency is at its maximum. For natural gas, the maximum combustion efficiency occurs in the range of 10%-15% excess air. The upper graph of the figure shows that the maximum combustion efficiency coincides with the peak of the curve. The lower graph indicates that combustion heat losses are caused by either unburned fuel or too much excess air. The left leg of the curve indicates heat loss due to the lack of excess air, which results in unburned fuel. The right side of the curve shows that a heat loss also occurs as the excess air increases. At high volumes of excess air, velocities of combustion gas in the furnace cause reduced heat transfer through boiler heating surfaces. Combustion control attempts to keep excess air within the optimum range where heat losses are at a minimum and combustion efficiency is highest. Figure 13 - Combustion Efficiency for Natural Gas > 0 c <u 0 sj= LU Max combustion efficiency at excess air 10% -15% w V) b •<-• (0 0) I Stack heat loss Heat loss due to unburned fuel due to excess air 60 80 100 120 140 160 % Theoretical air ® Side Track Besides combustion efficiency, there are other types of efficiency used to describe furnace or boiler performance. An example is boiler efficiency (for a steam or hot water boiler), which requires a more rigorous set of performance tests than combustion efficiency. Boiler efficiency accounts for all boiler losses and is therefore more accurate. Combustion efficiency is simpler and less costly to determine. Boiler efficiency is calculated as follows: Boiler efficiency = 164 heat transferred to working fluid heating value of fuel 3rd Class Edition 3 - Part A2 Fuels, Combustion, and Hue Gas Analysis • Chapter 3 f£ The net stack temperature and flue gas oxygen concentration are the standard measurements used to calculate combustion efficiency. The net stack temperahire is the difference between flue gas temperature and the temperature of the air supplied for combustion. Typically, combustion ef&ciency is read directly from a combustion analyzer. Efficiency can also be determined from a chart or a table, such as Table 14. To read this table, use the excess air or oxygen reading to find the correct row. Move right along the row to the correct temperature column. For example, 5% excess 0^ and net stack temperature 200 "C gives a combustion efficiency of 79.8%. Table 14 - Combustion Efficiency Table for Natural Gas Combustion Efficiency % % Excess Net Stack Temperature °C Air Oxygen 100 150 200 250 300 9.5 2.0 85.1 83.1 81.0 78.8 76.7 15.0 3.0 84.9 82.8 80.6 78.3 76.1 28.1 5.0 84.4 82.1 79.8 77.2 74.8 44.9 7.0 83.7 81.2 78.5 75.8 73.0 81.6 10.0 82.4 79.3 76.0 72.6 69.3 Example 27 A boiler operates at 2.0% excess oxygen and a stack temperature of 250°C. What is the plant combustion efficiency? Solution 27 Using the first row of Table 14 (where oxygen is 2.0%), move across to the column for 250°C net stack temperature to obtain a combustion efficiency of 78.8%. (Ans.) 3rd Class Edition 3 ' Part A2 165