3.1 SECTION 3 COMBUSTION Combustion Calculations Using the Million BTU (1.055 MJ) Method 3.1 Savings Produced by Preheating Combustion Air 3.4 Combustion of Coal Fuel in a Furnace 3.5 Percent Excess Air While Burning Coke 3.8 Combustion of Fuel Oil in a Furnace 3.9 Combustion of Natural Gas in a Furnace 3.11 Combustion of Wood Fuel in a Furnace 3.17 Molal Method of Combustion Analysis 3.19 Final Combustion Products Temperature Estimate 3.22 COMBUSTION CALCULATIONS USING THE MILLION BTU (1.055MJ) METHOD The energy absorbed by a steam boiler fired by natural gas is 100-million Btu /hr (29.3 MW). Boiler efficiency on a higher heating value (HHV) basis is 83 percent. If 15 percent excess air is used, determine the total air and flue-gas quantities produced. The approximate HHV of the natural gas is 23,000 Btu /lb (53,590 kJ/ kg). Ambient air temperature is 80 Њ F (26.7 Њ C) and relative humidity is 65 percent. How can quick estimates be made of air and flue-gas quantities in boiler operations when the fuel analysis is not known? Calculation Procedure: 1. Determine the energy input to the boiler The million Btu (1.055MJ) method combustion calculations is a quick way of estimating air and flue-gas quantities generated in boiler and heater operations when the ultimate fuel analysis is not available and all the engineer is interested in is good estimates. Air and flue-gas quantities determined may be used to calculate the size of fans, ducts, stacks, etc. It can be shown through comprehensive calculations that each fuel such as coal, oil, natural gas, bagasse, blast-furnace gas, etc. requires a certain amount of dry stochiometric air per million Btu (1.055MJ) fired on an HHV basis and that this quantity does not vary much with the fuel analysis. The listing below gives the dry air required per million Btu (1.055MJ) of fuel fired on an HHV basis for various fuels. Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. Source: HANDBOOK OF MECHANICAL ENGINEERING CALCULATIONS 3.2 POWER GENERATION Combustion Constants for Fuels Fuel Constant, lb dry air per million Btu (kg/MW) Blast furnace gas 575 (890.95) Bagasse 650 (1007.2) Carbon monoxide gas 670 (1038.2) Refinery and oil gas 720 (1115.6) Natural gas 730 (1131.1) Furnace oil and lignite 745–750 (1154.4–1162.1) Bituminous coals 760 (1177.6) Anthracite coal 780 (1208.6) Coke 800 (1239.5) To determine the energy input to the boiler, use the relation Q f ϭ (Q s )/E h , where energy input by the fuel, Btu/ hr (W); Q s ϭ energy absorbed by the steam in the boiler, Btu /Hr (W); Q s ϭ energy absorbed by the steam, Btu/hr (W); E h ϭ effi- ciency of the boiler on an HHV basis. Substituting for this boiler, Q f ϭ 100/ 0.83 ϭ 120.48 million Btu/ hr on an HHV basis (35.16 MW). 2. Estimate the quantity of dry air required by this boiler The total air required T a ϭ (Q f )(Fuel constant from list above). For natural gas, T a ϭ (120.48)(730) ϭ 87,950 lb/hr (39,929 kg/ hr). With 15 percent excess air, total air required ϭ (1.15)(87,950) ϭ 101,142.5 lb/ hr (45,918.7 kg /hr). 3. Compute the quantity of wet air required Air has some moisture because of its relative humidity. Estimate the amount of moisture in dry air in M lb /lb (kg /kg) from, M ϭ 0.622 ( p w )/(14.7 Ϫ p w ), where 0.622 is the ratio of the molecular weights of water vapor and dry air; p w ϭ partial pressure of water vapor in the air, psia (kPa) ϭ saturated vapor pressure (SVP) ϫ relative humidity expressed as a decimal; 14.7 ϭ atmospheric pressure of air at sea level (101.3 kPa). From the steam tables, at 80 F (26.7 C), SVP ϭ 0.5069 psia (3.49 kPa). Substituting, M ϭ 0.622 (0.5069 ϫ 0.65)/(14.7 Ϫ [0.5069 ϫ 0.65]) ϭ 0.01425 lb of moisture/lb of dry air (0.01425 kg/kg). The total flow rate of the wet air then ϭ 1.0142 (101,142.5) ϭ 102,578.7 lb/ hr (46,570.7 kg/ hr). To convert to a volumetric-flow basis, recall that the density of air at 80 Њ F (26.7 Њ C) and 14.7 psia (101.3 kPa) ϭ 39/(480 ϩ 80) ϭ 0.0722 lb/ cu ft (1.155 kg /cu m). In this relation, 39 ϭ a constant and the temperature of the air is converted to degrees Rankine. Hence, the volumetric flow ϭ 102,578.7/(60 min/hr)(0.0722) ϭ 23,679.3 actual cfm (670.1 cm m/min). 4. Estimate the rate of fuel firing and flue-gas produced The rate of fuel firing ϭ Q f /HHV ϭ (120.48 ϫ 10 6 )/23,000 ϭ 5238 lb/hr (2378 kg/hr). Hence, the total flue gas produced ϭ 5238 ϩ 102,578 ϭ 107,816 lb/hr (48,948 kg/ hr). If the temperature of the flue gas is 400 Њ F (204.4 Њ C) (a typical value for a natural- gas fired boiler), then the density, as in Step 3 is: 39 /(400 ϩ 460) ϭ 0.04535 lb / cu ft (0.7256 kg/ cu m). Hence, the volumetric flow ϭ (107,816)/(60 min /hr ϫ 0.04535) ϭ 39,623.7 actual cfm (1121.3 cu m/min). Related Calculations. Detailed combustion calculations based on actual fuel gas analysis can be performed to verify the constants given in the list above. For example, let us say that the natural-gas analysis was: Methane ϭ 83.4 percent; Ethane ϭ 15.8 percent; Nitrogen ϭ 0.8 percent by volume. First convert the analysis to a percent weight basis: Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. COMBUSTION COMBUSTION 3.3 Fuel Percent volume MW Col. 2 ϫ Col. 3 Percent weight Methane 83.4 16 1334.4 72.89 Ethane 15.8 30 474 25.89 Nitrogen 0.8 28 22.4 1.22 Note that the percent weight in the above list is calculated after obtaining the sum under Column 2 ϫ Column 3. Thus, the percent methane ϭ (1334.4)/(1334.4 ϩ 474 ϩ 22.4) ϭ 72.89 percent. From a standard reference, such as Ganapathy, Steam Plant Calculations Man- ual, Marcel Dekker, Inc., find the combustion constants, K, for various fuels and use them thus: For the air required for combustion, A c ϭ (K for methane)(percent by weight methane from above list) ϩ (K for ethane)(percent by weight ethane); or A c ϭ (17.265)(0.7289) ϩ (16.119)(0.2589) ϭ 16.76 lb/ lb (16.76 kg /kg). Next, compute the higher heating value of the fuel (HHV) using the air constants from the same reference mentioned above. Or HHV ϭ (heat of combustion for methane)(percent by weight methane) ϩ (heat of combustion of ethane)(percent by weight ethane) ϭ (23,879)(0.7289) ϩ (22,320)(0.2589) ϭ 23,184 Btu/lb (54,018.7 kJ/kg). Then, the amount of fuel equivalent to 1,000,000 Btu (1,055,000 kJ) ϭ (1,000,000)/23,184 ϭ 43.1 lb (19.56 kg), which requires, as computed above, (43.1)(16.76) ϭ 722.3 lb dry air (327.9 kg), which agrees closely with the value given in Step 1, above. Similarly, if the fuel were 100 percent methane, using the steps given above, and suitable constants from the same reference work, the air required for combus- tion is 17.265 lb/ lb (7.838 kg /kg) of fuel. HHV ϭ 23,879 Btu /lb (55,638 kJ/kg). Hence, the fuel in 1,000,000 Btu (1,055,000 kJ) ϭ (1,000,000)/(23,879) ϭ 41.88 lb (19.01 kg). Then, the dry air per million Btu (1.055 kg) fired ϭ (17.265) (41.88) ϭ 723 lb (328.3 kg). Likewise, for propane, using the same procedure, 1 lb (0.454 kg) requires 15.703 lb (7.129 kg) air and 1 million Btu (1,055,000 kJ) has (1,000,000)/21,661 ϭ 46.17 lb (20.95 kg) fuel. Then, 1 million Btu (1,055,000 kJ) requires (15.703)(46.17) ϭ 725 lb (329.2 kg) air. This general approach can be used for various fuel oils and solid fuels—coal, coke, etc. Good estimates of excess air used in combustion processes may be obtained if the oxygen and nitrogen in dry flue gases are measured. Knowledge of excess air amounts helps in performing detailed combustion and boiler efficiency calculations. Percent excess air, EA ϭ 100(O 2 –CO2) /[0.264 ϫ N 2 –(O 2 –CO /2)], where O 2 ϭ oxygen in the dry flue gas, percent volume; CO ϭ percent volume carbon mon- oxide; N 2 ϭ percent volume nitrogen. You can also estimate excess air from oxygen readings. Use the relation, EA ϭ (constant from list below)((O 2 )/(21–O 2 ). Constants for Excess Air Calculations Fuel Constant Carbon 100 Hydrogen 80 Carbon monoxide 121 Sulfur 100 Methane 90 Oil 94.5 Coal 97 Blast furnace gas 223 Coke oven gas 89.3 Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. COMBUSTION 3.4 POWER GENERATION If the percent volume of oxygen measured is 3 on a dry basis in a natural-gas (methane) fired boiler, the excess air, EA ϭ (90)[3/(21–3)] ϭ 15 percent. This procedure is the work of V. Ganapathy, Heat Transfer Specialist, ABCO Industries. SAVINGS PRODUCED BY PREHEATING COMBUSTION AIR A 20,000 sq ft (1858 sq m) building has a calculated total seasonal heating load of 2,534,440 MBH (thousand Btu) (2674 MJ). The stack temperature is 600 Њ F (316 Њ C) and the boiler efficiency is calculated to be 75 percent. Fuel oil burned has a higher heating value of 140,000 Btu/ gal (39,018 MJ /L). A preheater can be purchased and installed to reduce the breeching discharge combustion air temper- ature by 250 Њ F (139 Њ C) to 350 Њ F (177 Њ C) and provide the burner with preheated air. How much fuel oil will be saved? What will be the monetary saving if fuel oil is priced at 80 cents per gallon? Calculation Procedure: 1. Compute the total combustion air required by this boiler A general rule used by design engineers is that 1 cu ft (0.0283 cu m) of combustion air is required for each 100 Btu (105.5 J) released during combustion. To compute the combustion air required, use the relation CA ϭ H/100 ϫ Boiler efficiency, expressed as a decimal, where CA ϭ annual volume of combustion air, cu ft (cu m); H ϭ total seasonal heating load, Btu/ yr (kJ /yr). Substituting for this boiler, CA ϭ (2,534,400)(1000)/100 ϫ 0.75 ϭ 33,792,533 cu ft /yr (956.329 cu m). 2. Calculate the annual energy savings The energy savings, ES ϭ (stack temperature reduction, deg F)(cu ft air per yr)(0.018), where the constant 0.018 is the specific heat of air. Substituting, ES ϭ (250)(33,792,533)(0.018) ϭ 152,066,399 Btu/ yr (160,430 kJ/yr). With a boiler efficiency of 75 percent, each gallon of oil releases 0.75 ϫ 140,000 Btu/gal ϭ 105,000 Btu (110.8 jk). Hence, the fuel saved, FS ϭ ES/usuable heat in fuel, Btu /gal. Or, FS ϭ 152,066,399/105,000 ϭ 1448.3 gal/ yr (5.48 cu m/yr). With fuel oil at $1.10 per gallon, the monetary savings will be $1.10 (1448.3) ϭ $1593.13. If the preheater cost $6000, the simple payoff time would be $6000/ 1593.13 ϭ 3.77 years. Related Calculations. Use this procedure to determine the potential savings for burning any type of fuel—coal, oil, natural gas, landfill gas, catalytic cracker offgas, hydrogen purge gas, bagesse, sugar cane, etc. Other rules of thumb used by designers to estimate the amount of combustion air required for various fuels are: 10 cu ft of air (0.283 cu m) per 1 cu ft (0.0283 cu m) of natural gas; 1300 cu ft of air (36.8 cu m) per gal (0.003785 cu m) of No. 2 fuel oil; 1450 cu ft of air (41 cu m) per gal of No. 5 fuel oil; 1500 cu ft of air (42.5 cu m) per gal of No. 6 fuel oil. These values agree with that used in the above computation—i.e. 100 cu ft per 100 Btu of 140,000 Btu per gal oil ϭ 140,000/100 ϭ 1400 cu ft per gal (39.6 cu m/0.003785 cu m). This procedure is the work of Jerome F. Mueller, P.E. of Mueller Engineering Corp. Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. COMBUSTION COMBUSTION 3.5 COMBUSTION OF COAL FUEL IN A FURNACE A coal has the following ultimate analysis (or percent by weight): C ϭ 0.8339; H 2 ϭ 0.0456; O 2 ϭ 0.0505; N 2 ϭ 0.0103; S ϭ 0.0064; ash ϭ 0.0533; total ϭ 1.000 lb (0.45 kg). This coal is burned in a steam-boiler furnace. Determine the weight of air required for theoretically perfect combustion, the weight of gas formed per pound (kilogram) of coal burned, and the volume of flue gas, at the boiler exit temperature of 600 Њ F (316 Њ C) per pound (kilogram) of coal burned; air required with 20 percent excess air, and the volume of gas formed with this excess; the CO 2 percentage in the flue gas on a dry and wet basis. Calculation Procedure: 1. Compute the weight of oxygen required per pound of coal To find the weight of oxygen required for theoretically perfect combustion of coal, set up the following tabulation, based on the ultimate analysis of the coal: Note that of the total oxygen needed for combustion, 0.0505 lb (0.023 kg), is furnished by the fuel itself and is assumed to reduce the total external oxygen required by the amount of oxygen present in the fuel. The molecular-weight ratio is obtained from the equation for the chemical reaction of the element with oxygen in combustion. Thus, for carbon C ϩ O 2 → CO 2 ,or12 ϩ 32 ϭ 44, where 12 and 32 are the molecular weights of C and O 2 , respectively. 2. Compute the weight of air required for perfect combustion Air at sea level is a mechanical mixture of various gases, principally 23.2 percent oxygen and 76.8 percent nitrogen by weight. The nitrogen associated with the 2.5444 lb (1.154 kg) of oxygen required per pound (kilogram) of coal burned in this furnace is the product of the ratio of the nitrogen and oxygen weights in the air and 2.5444, or (2.5444)(0.768/ 0.232) ϭ 8.4228 lb (3.820 kg). Then the weight of air required for perfect combustion of 1 lb (0.45 kg) of coal ϭ sum of nitrogen and oxygen required ϭ 8.4228 ϩ 2.5444 ϭ 10.9672 lb (4.975 kg) of air per pound (kilogram) of coal burned. 3. Compute the weight of the products of combustion Find the products of combustion by addition: Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. COMBUSTION 3.6 POWER GENERATION 4. Convert the flue-gas weight to volume Use Avogadro’s law, which states that under the same conditions of pressure and temperature, 1 mol (the molecular weight of a gas expressed in lb) of any gas will occupy the same volume. At 14.7 lb /in 2 (abs) (101.3 kPa) and 32 Њ F(0 Њ C), 1 mol of any gas occupies 359 ft 3 (10.2 m 3 ). The volume per pound of any gas at these conditions can be found by dividing 359 by the molecular weight of the gas and correcting for the gas temperature by multiplying the volume by the ratio of the absolute flue-gas temperature and the atmospheric temperature. To change the weight analysis (step 3) of the products of combustion to volumetric analysis, set up the calculation thus: In this calculation, the temperature correction factor 2.15 ϭ absolute flue-gas tem- perature, Њ R/absolute atmospheric temperature, Њ R ϭ (600 ϩ 460)/(32 ϩ 460). The total weight of N 2 in the flue gas is the sum of the N 2 in the combustion air and the fuel, or 8.4228 ϩ 0.0103 ϭ 8.4331 lb (3.8252 kg). The value is used in com- puting the flue-gas volume. 5. Compute the CO 2 content of the flue gas The volume of CO 2 in the products of combustion at 600 Њ F (316 Њ C) is 53.6 ft 3 (1.158 m 3 ), as computed in step 4; and the total volume of the combustion products is 303.85 ft 3 (8.604 m 3 ). Therefore, the percent CO 2 on a wet basis (i.e., including the moisture in the combustion products) ϭ ft 3 CO 2 /total ft 3 ϭ 53.6/303.85 ϭ 0.1764, or 17.64 percent. The percent CO 2 on a dry, or Orsat, basis is found in the same manner, except that the weight of H 2 O in the products of combustion, 17.6 lb (7.83 kg) from step 4, is subtracted from the total gas weight. Or, percent CO 2 , dry, or Orsat basis ϭ (53.6)/(303.85 Ϫ 17.6) ϭ 0.1872, or 18.72 percent. Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. COMBUSTION COMBUSTION 3.7 6. Compute the air required with the stated excess flow With 20 percent excess air, the air flow required ϭ (0.20 ϩ 1.00)(air flow with no excess) ϭ 1.20 (10.9672) ϭ 13.1606 lb (5.970 kg) of air per pound (kilogram) of coal burned. The air flow with no excess is obtained from step 2. 7. Compute the weight of the products of combustion The excess air passes through the furnace without taking part in the combustion and increases the weight of the products of combustion per pound (kilogram) of coal burned. Therefore, the weight of the products of combustion is the sum of the weight of the combustion products without the excess air and the product of (per- cent excess air)(air for perfect combustion, lb); or, given the weights from steps 3 and 2, respectively, ϭ 11.9139 ϩ (0.20)(10.9672) ϭ 14.1073 lb (6.399 kg) of gas per pound (kilogram) of coal burned with 20 percent excess air. 8. Compute the volume of the combustion products and the percent CO 2 The volume of the excess air in the products of combustion is obtained by con- verting from the weight analysis to the volumetric analysis and correcting for tem- perature as in step 4, using the air weight from step 2 for perfect combustion and the excess-air percentage, or (10.9672)(0.20)(359/ 28.95)(2.15) ϭ 58.5 ft 3 (1.656 m 3 ). In this calculation the value 28.95 is the molecular weight of air. The total volume of the products of combustion is the sum of the column for perfect com- bustion, step 4, and the excess-air volume, above, or 303.85 ϩ 58.5 ϭ 362.35 ft 3 (10.261 m 3 ). By using the procedure in step 5, the percent CO 2 , wet basis ϭ 53.6/362.35 ϭ 14.8 percent. The percent CO 2 , dry basis ϭ 53.8/(362.35 Ϫ 17.6) ϭ 15.6 percent. Related Calculations. Use the method given here when making combustion calculations for any type of coal—bituminous, semibituminous, lignite, anthracite, cannel, or cooking—from any coal field in the world used in any type of furnace—boiler, heater, process, or waste-heat. When the air used for combustion contains moisture, as is usually true, this moisture is added to the combustion- formed moisture appearing in the products of combustion. Thus, for 80 Њ F (26.7 Њ C) air of 60 percent relative humidity, the moisture content is 0.013 lb/ lb (0.006 kg / kg) of dry air. This amount appears in the products of combustion for each pound of air used and is a commonly assumed standard in combustion calculations. Fossil-fuel-fired power plants release sulfur emissions to the atmosphere. In turn, this produces sulfates, which are the key ingredient in acid rain. The federal Clean Air Act regulates sulfur dioxide emissions from power plants. Electric utilities which burn high-sulfur coal are thought to produce some 35 percent of atmospheric emissions of sulfur dioxide in the United States. Sulfur dioxide emissions by power plants have declined some 30 percent since passage of the Clean Air Act in 1970, and a notable decline in acid rain has been noted at a number of test sites. In 1990 the Acid Rain Control Program was created by amendments to the Clean Air Act. This program further reduces the allowable sulfur dioxide emissions from power plants, steel mills, and other industrial facil- ities. The same act requires reduction in nitrogen oxide emissions from power plants and industrial facilities, so designers must keep this requirement in mind when designing new and replacement facilities of all types which use fossil fuels. Coal usage in steam plants is increasing throughout the world. An excellent example of this is the New England Electric System (NEES). This utility has been converting boiler units from oil to coal firing. Their conversions have saved cus- Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. COMBUSTION 3.8 POWER GENERATION FIGURE 1 Energy Independence transports coal to central stations. (Power.) tomers more than $60-million annually by displacing about 14-million bbl (2.2 million cu m) of oil per year. To reduce costs, the company built the first coal-fired collier, Fig. 1, in more than 50 years in the United States, and assumed responsibility for coal transpor- tation to its stations, cutting operating costs by more than $2-million per year. The collier makes economic sense because the utility stations in the system are not accessible by rail. This ship, the Energy Independence, has been an economic suc- cess for the utility. Measuring 665 ft (203 m) long by 95 ft (29 m) wide by 56 ft (17 m) deep with a 34-ft (10-m) draft, the vessel discharges a typical 40,000-ton load in 12 hours. Data in these two paragraphs and Fig. 1 are from Power magazine. PERCENT EXCESS AIR WHILE BURNING COAL A certain coal has the following composition by weight percentages: carbon 75.09, nitrogen 1.56, ash 3.38, hydrogen, 5.72, oxygen 13.82, sulfur 0.43. When burned in an actual furnace, measurements showed that there was 8.93 percent combustible in the ash pit refuse and the following Orsat analysis in percentages was obtained: carbon dioxide 14.2, oxygen 4.7, carbon monoxide 0.3. If it can be assumed that there was no combustible in the flue gas other that the carbon monoxide reported, calculate the percentage of excess air used. Calculation Procedure: 1. Compute the amount of theoretical air required per lb m (kg) of coal Theoretical air required per pound (kilogram) of coal, w ta ϭ 11.5C Ј ϩ 34.5[H Ј 2 Ϫ O Ј 2 /8)] ϩ 4.32S Ј , where C Ј ,H Ј 2 ,O Ј 2 , and S Ј represent the percentages by weight, expressed as decimal fractions, of carbon, hydrogen, oxygen, and sulfur, respec- Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. COMBUSTION COMBUSTION 3.9 tively. Thus, w ta ϭ 11.5(0.7509) ϩ 34.5[0.0572 Ϫ (0.1382/8)] ϩ 4.32(0.0043) ϭ 10.03 lb (4.55 kg) of air per lb (kg) of coal. The ash and nitrogen are inert and do not burn. 2. Compute the correction factor for combustible in the ash The correction factor for combustible in the ash, C 1 ϭ (w f C f Ϫ w r C r )/(w f ϫ 100), where the amount of fuel, w f ϭ 1 lb (0.45 kg) of coal; percent by weight, expressed as a decimal fraction, of carbon in the coal, C f ϭ 75.09; percent by weight of the ash and refuse in the coal, w r ϭ 0.0338; percent by weight of combustible in the ash, C r ϭ 8.93. Hence, C 1 ϭ [(1 ϫ 75.09) Ϫ (0.0338 ϫ 8.93)]/(1 ϫ 100) ϭ 0.748. 3. Compute the amount of dry flue gas produced per lb (kg) of coal The lb (kg) of dry flue gas per lb (kg) of coal, w dg ϭ C 1 (4CO 2 ϩ O 2 ϩ 704)/ [3(CO 2 ϩ CO)], where the Orsat analysis percentages are for carbon dioxide, CO 2 ϭ 14.2; oxygen, O 2 ϭ 4.7; carbon monoxide, CO ϭ 0.3. Hence, w dg ϭ 0.748 ϫ [(4 ϫ 14.2) ϩ 4.7 ϩ 704)]/[3(14.2 ϩ 0.3)] ϭ 13.16 lb/ lb (5.97 kg /kg). 4. Compute the amount of dry air supplied per lb (kg) of coal The lb (kg) of dry air supplied per lb (kg) of coal, w da ϭ w dg Ϫ C 1 ϩ 8[H Ј 2 Ϫ (O Ј 2 /8)] Ϫ (N Ј 2 /N), where the percentage by weight of nitrogen in the fuel, N Ј 2 ϭ 1.56, and ‘‘atmospheric nitrogen’’ in the supply air, N 2 ϭ 0.768; other values are as given or calculated. Then, w da ϭ 13.16 Ϫ 0.748 ϩ 8[0.0572 Ϫ (0.1382/8)] Ϫ (0.0156/0.768) ϭ 12.65 lb/ lb (5.74 kg /kg). 5. Compute the percent of excess air used Percent excess air ϭ (w da Ϫ w ta )/w ta ϭ (12.65 Ϫ 10.03)/10.03 ϭ 0.261, or 26.1 percent. Related Calculations. The percentage by weight of nitrogen in ‘‘atmospheric air’’ in step 4 appears in Principles of Engineering Thermodynamics, 2nd edition, by Kiefer et al., John Wiley & Sons, Inc. COMBUSTION OF FUEL OIL IN A FURNACE A fuel oil has the following ultimate analysis: C ϭ 0.8543; H 2 ϭ 0.1131; O 2 ϭ 0.0270; N 2 ϭ 0.0022; S ϭ 0.0034; total ϭ 1.0000. This fuel oil is burned in a steam-boiler furnace. Determine the weight of air required for theoretically perfect combustion, the weight of gas formed per pound (kilogram) of oil burned, and the volume of flue gas, at the boiler exit temperature of 600 Њ F (316 Њ C), per pound (kilogram) of oil burned; the air required with 20 percent excess air, and the volume of gas formed with this excess; the CO 2 percentage in the flue gas on a dry and wet basis. Calculation Procedure: 1. Compute the weight of oxygen required per pound (kilogram) of oil The same general steps as given in the previous calculation procedure will be followed. Consult that procedure for a complete explanation of each step. Using the molecular weight of each element, we find Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. COMBUSTION 3.10 POWER GENERATION 2. Compute the weight of air required for perfect combustion The weight of nitrogen associated with the required oxygen ϭ (3.1593)(0.768/ 0.232) ϭ 10.458 lb (4.706 kg). The weight of air required ϭ 10.4583 ϩ 3.1593 ϭ 13.6176 lb/ lb (6.128 kg /kg) of oil burned. 3. Compute the weight of the products of combustion As before, 4. Convert the flue-gas weight to volume As before, In this calculation, the temperature correction factor 2.15 ϭ absolute flue-gas temperature, Њ R/absolute atmospheric temperature, Њ R ϭ (600 ϩ 460)/(32 ϩ 460). Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies. All rights reserved. Any use is subject to the Terms of Use as given at the website. COMBUSTION [...]... lb of moisture per lb (0.006 kg / kg) of air Calculation Procedure: 1 Convert the ultimate analysis to moles A mole of any substance is an amount of the substance having a weight equal to the molecular weight of the substance Thus, 1 mol of carbon is 12 lb (5.4 kg) of carbon, because the molecular weight of carbon is 12 To convert an ultimate analysis of a fuel to moles, assume that 100 lb (45 kg) of. .. with the stated excess flow With 20 percent excess air, (1.20)(16.132) ϭ 19.3584 lb of air per lb (8.71 kg / kg) of natural gas, or 19.3584 / 22.1 ϭ 0.875 lb of air per ft3 (13.9 kg / m3) of natural gas See step 4 for an explanation of the value 22.1 7 Compute the weight of the products of combustion Weight of the products of combustion ϭ product weight for perfect combustion, lb ϩ (percent excess air)... (7.353 kg) of air per pound (kilogram) of oil burned 7 Compute the weight of the products of combustion The weight of the products of combustion ϭ product weight for perfect combustion, lb ϩ (percent excess air)(air for perfect combustion, lb) ϭ 14.6173 ϩ (0.20)(13.6176) ϭ 17.3408 lb (7.803 kilogram) of flue gas per pound (kilogram) of oil burned with 20 percent excess air 8 Compute the volume of the combustion... The composition of the gas is given on a volumetric basis, which is the usual way of expressing a fuel-gas analysis To use the volumetric-analysis data in combustion calculations, they must be converted to a weight basis This is done by dividing the weight of each component by the total weight of the gas A volume of 1 ft3 (1 m3) of the gas is used for this computation Find the weight of each component... ϭ 6.443 lb / lb (2.899 kg / kg) of wood burned, if the air is dry But the air contains 0.013 lb of moisture per lb (0.006 kg / kg) of air Hence, the total weight of the air ϭ 6.443 ϩ (0.013)(6.443) ϭ 6.527 lb (2.937 kg) 3 Compute the weight of the products of combustion Use the following relation: Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright... (2.2 kg) is 1 / 0.04517 ϭ 22.1 ft3 (0.626 m3) Therefore, the weight of N2 per ft3 of fuel burned ϭ 12.39 / 22.1 ϭ 0.560 lb (0.252 kg) This, plus the weight of N2 in the fuel, step 1, is 0.560 ϩ 0.0025 ϭ 0.5625 lb (0.253 kg) of N2 in the products of combustion Next, find the total weight of the products of combustion by taking the sum of the CO2, H2O, and N2 weights, or 0.11688 ϩ 0.09332 ϩ 0.5625 ϭ 0.7727... Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies All rights reserved Any use is subject to the Terms of Use as given at the website COMBUSTION 3.20 POWER GENERATION 2 Compute the mols of oxygen for complete combustion From Table 2, the burning of carbon to carbon dioxide requires 1 mol of carbon and 1 mol of oxygen, yielding 1 mol of. .. total moles of CO2 is obtained from step 2 The moles of H2 in 100 lb (45 kg) of the fuel, 2.280, is assumed to form H2O In addition, the air from step 4, 47.24 mol, contains 0.013 lb of moisture per lb (0.006 kg / kg) of air This moisture is converted to moles by dividing the molecular weight of air, 28.95, by the molecular weight of water, 18, and multiplying the result by the moisture content of the air,... content of the air, or (28.95 / 18)(0.013) ϭ 0.0209, say 0.021 mol of water per mol of air The product of this and the moles of air gives the total moles of Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies All rights reserved Any use is subject to the Terms of Use as given at the website COMBUSTION 3.22 POWER GENERATION... relative weight of the reactants equal that of the products at 177.4 Downloaded from Digital Engineering Library @ McGraw-Hill (www.digitalengineeringlibrary.com) Copyright © 2006 The McGraw-Hill Companies All rights reserved Any use is subject to the Terms of Use as given at the website COMBUSTION COMBUSTION 3.23 4 Compute the relative weights of the products of combustion on the basis of a per unit . weight of the products of combustion per pound (kilogram) of coal burned. Therefore, the weight of the products of combustion is the sum of the weight of the. reserved. Any use is subject to the Terms of Use as given at the website. Source: HANDBOOK OF MECHANICAL ENGINEERING CALCULATIONS 3.2 POWER GENERATION Combustion