Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
MED - 06109 NTA - 06 Power Production - 1 COMBUSTION The objective of this chapter is to study systems involving chemical reactions. Since the combustion of hydrocarbon fuels occurs in most power-producing devices combustion is emphasized. COMBUSTION FUNDAMENTALS When a chemical reaction occurs, the bonds within molecules of the reactants are broken, and atoms and electrons rearrange to form products. In combustion reactions, rapid oxidation of combustible elements of the fuel results in energy release as combustion products are formed. The three major combustible chemical elements in most common fuels are carbon, hydrogen, and sulfur. Sulfur is usually a relatively unimportant contributor to the energy released, but it can be a significant cause of pollution and corrosion problems. Combustion is complete when all the carbon present in the fuel is burned to carbon dioxide, all the hydrogen is burned to water, all the sulfur is burned to sulfur dioxide, and all other combustible elements are fully oxidized. When these conditions are not fulfilled, combustion is incomplete. Combustion reactions expressed by chemical equations of the form When dealing with chemical reactions, it is necessary to remember that mass is conserved, so the mass of the products equals the mass of the reactants. The total mass of each chemical element must be the same on both sides of the equation, even though the elements exist in different chemical compounds in the reactants and products. However, the number of moles of products may differ from the number of moles of reactants. For example... consider the complete combustion of hydrogen with oxygen In this case, the reactants are hydrogen and oxygen. Hydrogen is the fuel and oxygen is the oxidizer. Water is the only product of the reaction. The numerical coefficients in the equation, which precede the chemical symbols to give equal amounts of each chemical element on both sides of the equation, are called stoichiometric coefficients. In words, states Note that the total numbers of moles on the left and right sides of are not equal. However, because mass is conserved, the total mass of reactants must equal the total mass of products. Since 1 kmol of H2 equals 2 kg, kmol of O2 equals 16 kg, and 1 kmol of H2O equals 18 kg, this can be interpreted as stating Ref: Michael J. Moran & Howard N. Shapiro. Fundamentals of Engineering Thermodynamics Page 1 MED - 06109 NTA - 06 Power Production - 1 In the remainder of this lecture, consideration is given to the makeup of the fuel, oxidizer, and combustion products typically involved in engineering combustion applications. FUELS A fuel is simply a combustible substance. In this lecture emphasis is on hydrocarbon fuels, which contain hydrogen and carbon. Sulfur and other chemical substances also may be present. Hydrocarbon fuels can exist as liquids, gases, and solids. Liquid hydrocarbon fuels are commonly derived from crude oil through distillation and cracking processes. Examples are gasoline, diesel fuel, kerosene, and other types of fuel oils. Most liquid fuels are mixtures of hydrocarbons for which compositions are usually given in terms of mass fractions. For simplicity in combustion calculations, gasoline is often modeled as octane, C8H18, and diesel fuel as dodecane, C12H26. Gaseous hydrocarbon fuels are obtained from natural gas wells or are produced in certain chemical processes. Natural gas normally consists of several different hydrocarbons, with the major constituent being methane, CH4. The compositions of gaseous fuels are usually given in terms of mole fractions. Both gaseous and liquid hydrocarbon fuels can be synthesized from coal, oil shale, and tar sands. Coal is a familiar solid fuel. Its composition varies considerably with the location from which it is mined. For combustion calculations, the composition of coal is usually expressed as an ultimate analysis. The ultimate analysis gives the composition on a mass basis in terms of the relative amounts of chemical elements (carbon, sulfur, hydrogen, nitrogen, oxygen) and ash. MODELING COMBUSTION AIR Oxygen is required in every combustion reaction. Pure oxygen is used only in special applications such as cutting and welding. In most combustion applications, air provides the needed oxygen.. For the combustion calculations of this lecture, the following model is used for simplicity: All components of air other than oxygen are lumped together with nitrogen. Accordingly, air is considered to be 21% oxygen and 79% nitrogen on a molar basis. With this idealization the molar ratio of the nitrogen to the oxygen is 0.79/0.21 = 3.76. When air supplies the oxygen in a combustion reaction, therefore, every mole of oxygen is accompanied by 3.76 moles of nitrogen. We also assume that the nitrogen present in the combustion air does not undergo chemical reaction. That is, nitrogen is regarded as inert. The nitrogen in the products is at the same temperature as the other products, however, so the nitrogen undergoes a change of state if the products are at a temperature other than the air temperature before combustion. If high enough temperatures are attained, nitrogen can form compounds such as nitric oxide and nitrogen dioxide. Even trace amounts of oxides of nitrogen appearing in the exhaust of internal combustion engines can be a source of air pollution. Ref: Michael J. Moran & Howard N. Shapiro. Fundamentals of Engineering Thermodynamics Page 2 MED - 06109 NTA - 06 Power Production - 1 AIR–FUEL RATIO Two parameters that are frequently used to quantify the amounts of fuel and air in a particular combustion process are the air–fuel ratio and its reciprocal, the fuel–air ratio. The air–fuel ratio is simply the ratio of the amount of air in a reaction to the amount of fuel. The ratio can be written on a molar basis (moles of air divided by moles of fuel) or on a mass basis (mass of air divided by mass of fuel). Conversion between these values is accomplished using the molecular weights of the air, Mair, and fuel, Mfuel, , where is the air–fuel ratio on a molar basis and AF is the ratio on a mass basis. For the combustion calculations of this lecture the molecular weight of air is taken as 28.97. THEORETICAL AIR The minimum amount of air that supplies sufficient oxygen for the complete combustion of all the carbon, hydrogen, and sulfur present in the fuel is called the theoretical amount of air. For complete combustion with the theoretical amount of air, the products would consist of carbon dioxide, water, sulfur dioxide, the nitrogen accompanying the oxygen in the air, and any nitrogen contained in the fuel. No free oxygen would appear in the products. For example, let us determine the theoretical amount of air for the complete combustion of methane. For this reaction, the products contain only carbon dioxide, water, and nitrogen. The reaction is where a, b, c, and d represent the numbers of moles of oxygen, carbon dioxide, water, and nitrogen. In writing the left side of the equation 3.76 moles of nitrogen are considered to accompany each mole of oxygen. Applying the conservation of mass principle to the carbon, hydrogen, oxygen, and nitrogen, respectively, results in four equations among the four unknowns Solving these equations, the balanced chemical equation is Ref: Michael J. Moran & Howard N. Shapiro. Fundamentals of Engineering Thermodynamics Page 3 MED - 06109 NTA - 06 Power Production - 1 The coefficient 2 before the term (O2 + 3.76N2) in the above equation is the number of moles of oxygen in the combustion air, per mole of fuel, and not the amount of air. The amount of combustion air is 2 moles of oxygen plus 2 × 3.76 moles of nitrogen, giving a total of 9.52 moles of air per mole of fuel. Thus, for the reaction above the air–fuel ratio on a molar basis is 9.52. To calculate the air–fuel ratio on a mass basis, we write Normally the amount of air supplied is either greater or less than the theoretical amount. The amount of air actually supplied is commonly expressed in terms of the percent of theoretical air. For example, 150% of theoretical air means that the air actually supplied is 1.5 times the theoretical amount of air. The amount of air supplied can be expressed alternatively as a percent excess or a percent deficiency of air. Thus, 150% of theoretical air is equivalent to 50% excess air, and 80% of theoretical air is the same as a 20% deficiency of air. For example, consider the complete combustion of methane with 150% theoretical air (50% excess air). The balanced chemical reaction equation is In this equation, the amount of air per mole of fuel is 1.5 times the theoretical amount. Accordingly, the air–fuel ratio is 1.5 times the air–fuel ratio. Since complete combustion is assumed, the products contain only carbon dioxide, water, nitrogen, and oxygen. The excess air supplied appears in the products as uncombined oxygen and a greater amount of nitrogen based on the theoretical amount of air. The equivalence ratio is the ratio of the actual fuel–air ratio to the fuel–air ratio for complete combustion with the theoretical amount of air. The reactants are said to form a lean mixture when the equivalence ratio is less than unity. When the ratio is greater than unity, the reactants are said to form a rich mixture. EXAMPLE: Determining the Air–Fuel Ratio Determine the air–fuel ratio on both a molar and mass basis for the complete combustion of octane, C8H18, with (a) the theoretical amount of air, (b) 150% theoretical air (50% excess air). SOLUTION Known: Octane, C8H18, is burned completely with (a) the theoretical amount of air, (b) 150% theoretical air. Find: Determine the air–fuel ratio on a molar and a mass basis. Ref: Michael J. Moran & Howard N. Shapiro. Fundamentals of Engineering Thermodynamics Page 4 MED - 06109 NTA - 06 Power Production - 1 Assumptions: 1. Each mole of oxygen in the combustion air is accompanied by 3.76 moles of nitrogen. 2. The nitrogen is inert. 3. Combustion is complete. Analysis: (a) For complete combustion of C8H18 with the theoretical amount of air, the products contain carbon dioxide, water, and nitrogen only. That is Applying the conservation of mass principle to the carbon, hydrogen, oxygen, and nitrogen, respectively, gives Solving these equations, a=12.5, b=8, c=9, d=47. The balanced chemical equation is The air–fuel ratio on a molar basis is The air–fuel ratio expressed on a mass basis is (b) For 150% theoretical air, the chemical equation for complete combustion takes the form Ref: Michael J. Moran & Howard N. Shapiro. Fundamentals of Engineering Thermodynamics Page 5 MED - 06109 NTA - 06 Power Production - 1 DETERMINING PRODUCTS OF COMBUSTION In each of the illustrations given above, complete combustion is assumed. For a hydrocarbon fuel, this means that the only allowed products are CO2, H2O, and N2, with O2 also present when excess air is supplied. If the fuel is specified and combustion is complete, the respective amounts of the products can be determined by applying the conservation of mass principle to the chemical equation. The procedure for obtaining the balanced reaction equation of an actual reaction where combustion is incomplete is not always so straightforward. Combustion is the result of a series of very complicated and rapid chemical reactions, and the products formed depend on many factors. When fuel is burned in the cylinder of an internal combustion engine, the products of the reaction vary with the temperature and pressure in the cylinder. In combustion equipment of all kinds, the degree of mixing of the fuel and air is a controlling factor in the reactions that occur once the fuel and air mixture is ignited. Although the amount of air supplied in an actual combustion process may exceed the theoretical amount, it is not uncommon for some carbon monoxide and unburned oxygen to appear in the products. This can be due to incomplete mixing, insufficient time for complete combustion, and other factors. When the amount of air supplied is less than the theoretical amount of air, the products may include both CO2 and CO, and there also may be unburned fuel in the products. Unlike the complete combustion cases considered above, the products of combustion of an actual combustion process and their relative amounts can be determined only by measurement. Among several devices for measuring the composition of products of combustion are the Orsat analyzer, gas chromatograph, infrared analyzer, and flame ionization detector. Data from these devices can be used to determine the mole fractions of the gaseous products of combustion. The analyses are often reported on a “dry” basis. In a dry product analysis, the mole fractions are given for all gaseous products except the water vapor. Ref: Michael J. Moran & Howard N. Shapiro. Fundamentals of Engineering Thermodynamics Page 6 MED - 06109 NTA - 06 Power Production - 1 Since water is formed when hydrocarbon fuels are burned, the mole fraction of water vapor in the gaseous products of combustion can be significant. If the gaseous products of combustion are cooled at constant pressure, the dew point temperature is reached when water vapor begins to condense. Since water deposited on duct work, mufflers, and other metal parts can cause corrosion, knowledge of the dew point temperature is important. EXAMPLE: Using a Dry Product Analysis Methane, CH4, is burned with dry air. The molar analysis of the products on a dry basis is CO2, 9.7%; CO, 0.5%; O2, 2.95%; and N2, 86.85%. Determine (a) the air–fuel ratio on both a molar and a mass basis, (b) the percent theoretical air, (c) the dew point temperature of the products, in oC, if the mixture were cooled at 1 atm. SOLUTION Known: Methane is burned with dry air. The molar analysis of the products on a dry basis is provided. Find: Determine (a) the air–fuel ratio on both molar and mass basis, (b) the percent theoretical air, and (c) the dew point temperature of the products, in oC, if cooled at 1 atm. Assumptions: 1. Each mole of oxygen in the combustion air is accompanied by 3.76 moles of nitrogen, which is inert. 2. The products form an ideal gas mixture. Analysis: (a) The solution is conveniently conducted on the basis of 100 kmol of dry products. The chemical equation then reads In addition to the assumed 100 kmol of dry products, water must be included as a product. Applying conservation of mass to carbon, hydrogen, and oxygen, respectively Solving this set of equations gives a=10.2, b=23.1, c=20.4. The balanced chemical equation is On a molar basis, the air–fuel ratio is Ref: Michael J. Moran & Howard N. Shapiro. Fundamentals of Engineering Thermodynamics Page 7 MED - 06109 NTA - 06 Power Production - 1 On a mass basis (b) To determine the dew point temperature requires the partial pressure of the water vapor pv. The partial pressure pv is found from pv=yvp, where yv is the mole fraction of the water vapor in the combustion products and p is 1 atm. Referring to the balanced chemical equation of part (a), the mole fraction of the water vapor is Ref: Michael J. Moran & Howard N. Shapiro. Fundamentals of Engineering Thermodynamics Page 8 MED - 06109 NTA - 06 Power Production - 1 EXAMPLE: Burning Natural Gas with Excess Air Ref: Michael J. Moran & Howard N. Shapiro. Fundamentals of Engineering Thermodynamics Page 9 MED - 06109 NTA - 06 Power Production - 1 Ref: Michael J. Moran & Howard N. Shapiro. Fundamentals of Engineering Thermodynamics Page 10