cen84959_ch15

cen84959_ch15 - cen84959_ch15.qxd 9/15/06 6:06 AM Page 773...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: cen84959_ch15.qxd 9/15/06 6:06 AM Page 773 Chapter 15 CHEMICAL REACTIONS n the preceding chapters we limited our consideration to nonreacting systems—systems whose chemical composition remains unchanged during a process. This was the case even with mixing processes during which a homogeneous mixture is formed from two or more fluids without the occurrence of any chemical reactions. In this chapter, we specifically deal with systems whose chemical composition changes during a process, that is, systems that involve chemical reactions. When dealing with nonreacting systems, we need to consider only the sensible internal energy (associated with temperature and pressure changes) and the latent internal energy (associated with phase changes). When dealing with reacting systems, however, we also need to consider the chemical internal energy, which is the energy associated with the destruction and formation of chemical bonds between the atoms. The energy balance relations developed for nonreacting systems are equally applicable to reacting systems, but the energy terms in the latter case should include the chemical energy of the system. In this chapter we focus on a particular type of chemical reaction, known as combustion, because of its importance in engineering. But the reader should keep in mind, however, that the principles developed are equally applicable to other chemical reactions. We start this chapter with a general discussion of fuels and combustion. Then we apply the mass and energy balances to reacting systems. In this regard we discuss the adiabatic flame temperature, which is the highest temperature a reacting mixture can attain. Finally, we examine the second-law aspects of chemical reactions. I Objectives The objectives of Chapter 15 are to: • Give an overview of fuels and combustion. • Apply the conservation of mass to reacting systems to determine balanced reaction equations. • Define the parameters used in combustion analysis, such as air–fuel ratio, percent theoretical air, and dew-point temperature. • Apply energy balances to reacting systems for both steadyflow control volumes and fixed mass systems. • Calculate the enthalpy of reaction, enthalpy of combustion, and the heating values of fuels. • Determine the adiabatic flame temperature for reacting mixtures. • Evaluate the entropy change of reacting systems. • Analyze reacting systems from the second-law perspective. | 773 cen84959_ch15.qxd 9/15/06 6:06 AM Page 774 774 | Thermodynamics 15–1 Gasoline Kerosene CRUDE OIL Diesel fuel Fuel oil FUELS AND COMBUSTION FIGURE 15–1 Most liquid hydrocarbon fuels are obtained from crude oil by distillation. Any material that can be burned to release thermal energy is called a fuel. Most familiar fuels consist primarily of hydrogen and carbon. They are called hydrocarbon fuels and are denoted by the general formula CnHm. Hydrocarbon fuels exist in all phases, some examples being coal, gasoline, and natural gas. The main constituent of coal is carbon. Coal also contains varying amounts of oxygen, hydrogen, nitrogen, sulfur, moisture, and ash. It is difficult to give an exact mass analysis for coal since its composition varies considerably from one geographical area to the next and even within the same geographical location. Most liquid hydrocarbon fuels are a mixture of numerous hydrocarbons and are obtained from crude oil by distillation (Fig. 15–1). The most volatile hydrocarbons vaporize first, forming what we know as gasoline. The less volatile fuels obtained during distillation are kerosene, diesel fuel, and fuel oil. The composition of a particular fuel depends on the source of the crude oil as well as on the refinery. Although liquid hydrocarbon fuels are mixtures of many different hydrocarbons, they are usually considered to be a single hydrocarbon for convenience. For example, gasoline is treated as octane, C8H18, and the diesel fuel as dodecane, C12H26. Another common liquid hydrocarbon fuel is methyl alcohol, CH3OH, which is also called methanol and is used in some gasoline blends. The gaseous hydrocarbon fuel natural gas, which is a mixture of methane and smaller amounts of other gases, is often treated as methane, CH4, for simplicity. Natural gas is produced from gas wells or oil wells rich in natural gas. It is composed mainly of methane, but it also contains small amounts of ethane, propane, hydrogen, helium, carbon dioxide, nitrogen, hydrogen sulfate, and water vapor. On vehicles, it is stored either in the gas phase at pressures of 150 to 250 atm as CNG (compressed natural gas), or in the liquid phase at 162°C as LNG (liquefied natural gas). Over a million vehicles in the world, mostly buses, run on natural gas. Liquefied petroleum gas (LPG) is a byproduct of natural gas processing or the crude oil refining. It consists mainly of propane and thus LPG is usually referred to as propane. However, it also contains varying amounts of butane, propylene, and butylenes. Propane is commonly used in fleet vehicles, taxis, school buses, and private cars. Ethanol is obtained from corn, grains, and organic waste. Methonal is produced mostly from natural gas, but it can also be obtained from coal and biomass. Both alcohols are commonly used as additives in oxygenated gasoline and reformulated fuels to reduce air pollution. Vehicles are a major source of air pollutants such as nitric oxides, carbon monoxide, and hydrocarbons, as well as the greenhouse gas carbon dioxide, and thus there is a growing shift in the transportation industry from the traditional petroleum-based fuels such as gaoline and diesel fuel to the cleaner burning alternative fuels friendlier to the environment such as natural gas, alcohols (ethanol and methanol), liquefied petroleum gas (LPG), and hydrogen. The use of electric and hybrid cars is also on the rise. A comparison of some alternative fuels for transportation to gasoline is given in Table 15–1. Note that the energy contents of alternative fuels per unit volume are lower than that of gasoline or diesel fuel, and thus the driving range of a cen84959_ch15.qxd 9/15/06 6:06 AM Page 775 Chapter 15 TABLE 15–1 A comparison of some alternative fuels to the traditional petroleum-based fuels used in transportation Fuel Gasoline Light diesel Heavy diesel LPG (Liquefied petroleum gas, primarily propane) Ethanol (or ethyl alcohol) Methanol (or methyl alcohol) CNG (Compressed natural gas, primarily methane, at 200 atm) LNG (Liquefied natural gas, primarily methane) Energy content kJ/L 31,850 33,170 35,800 23,410 29,420 18,210 8,080 20,490 Gasoline equivalence,* L/L-gasoline 1 0.96 0.89 1.36 1.08 1.75 3.94 1.55 | 775 *Amount of fuel whose energy content is equal to the energy content of 1-L gasoline. vehicle on a full tank is lower when running on an alternative fuel. Also, when comparing cost, a realistic measure is the cost per unit energy rather than cost per unit volume. For example, methanol at a unit cost of $1.20/L may appear cheaper than gasoline at $1.80/L, but this is not the case since the cost of 10,000 kJ of energy is $0.57 for gasoline and $0.66 for methanol. A chemical reaction during which a fuel is oxidized and a large quantity of energy is released is called combustion (Fig. 15–2). The oxidizer most often used in combustion processes is air, for obvious reasons—it is free and readily available. Pure oxygen O2 is used as an oxidizer only in some specialized applications, such as cutting and welding, where air cannot be used. Therefore, a few words about the composition of air are in order. On a mole or a volume basis, dry air is composed of 20.9 percent oxygen, 78.1 percent nitrogen, 0.9 percent argon, and small amounts of carbon dioxide, helium, neon, and hydrogen. In the analysis of combustion processes, the argon in the air is treated as nitrogen, and the gases that exist in trace amounts are disregarded. Then dry air can be approximated as 21 percent oxygen and 79 percent nitrogen by mole numbers. Therefore, each mole of oxygen entering a combustion chamber is accompanied by 0.79/0.21 3.76 mol of nitrogen (Fig. 15–3). That is, 1 kmol O2 3.76 kmol N2 4.76 kmol air (15–1) FIGURE 15–2 Combustion is a chemical reaction during which a fuel is oxidized and a large quantity of energy is released. © Reprinted with special permission of King Features Syndicate. AIR ( 21% O2 79% N2 ) 1 kmol O2 3.76 kmol N 2 During combustion, nitrogen behaves as an inert gas and does not react with other elements, other than forming a very small amount of nitric oxides. However, even then the presence of nitrogen greatly affects the outcome of a combustion process since nitrogen usually enters a combustion chamber in large quantities at low temperatures and exits at considerably higher temperatures, absorbing a large proportion of the chemical energy released during combustion. Throughout this chapter, nitrogen is assumed to remain perfectly FIGURE 15–3 Each kmol of O2 in air is accompanied by 3.76 kmol of N2. cen84959_ch15.qxd 9/15/06 6:06 AM Page 776 776 | Thermodynamics inert. Keep in mind, however, that at very high temperatures, such as those encountered in internal combustion engines, a small fraction of nitrogen reacts with oxygen, forming hazardous gases such as nitric oxide. Air that enters a combustion chamber normally contains some water vapor (or moisture), which also deserves consideration. For most combustion processes, the moisture in the air and the H2O that forms during combustion can also be treated as an inert gas, like nitrogen. At very high temperatures, however, some water vapor dissociates into H2 and O2 as well as into H, O, and OH. When the combustion gases are cooled below the dew-point temperature of the water vapor, some moisture condenses. It is important to be able to predict the dew-point temperature since the water droplets often combine with the sulfur dioxide that may be present in the combustion gases, forming sulfuric acid, which is highly corrosive. During a combustion process, the components that exist before the reaction are called reactants and the components that exist after the reaction are called products (Fig. 15–4). Consider, for example, the combustion of 1 kmol of carbon with 1 kmol of pure oxygen, forming carbon dioxide, C O2 S CO2 (15–2) Reactants Reaction chamber Products FIGURE 15–4 In a steady-flow combustion process, the components that enter the reaction chamber are called reactants and the components that exit are called products. 2 kg hydrogen 16 kg oxygen H2 1 O → H2O 22 2 kg hydrogen 16 kg oxygen FIGURE 15–5 The mass (and number of atoms) of each element is conserved during a chemical reaction. Here C and O2 are the reactants since they exist before combustion, and CO2 is the product since it exists after combustion. Note that a reactant does not have to react chemically in the combustion chamber. For example, if carbon is burned with air instead of pure oxygen, both sides of the combustion equation will include N2. That is, the N2 will appear both as a reactant and as a product. We should also mention that bringing a fuel into intimate contact with oxygen is not sufficient to start a combustion process. (Thank goodness it is not. Otherwise, the whole world would be on fire now.) The fuel must be brought above its ignition temperature to start the combustion. The minimum ignition temperatures of various substances in atmospheric air are approximately 260°C for gasoline, 400°C for carbon, 580°C for hydrogen, 610°C for carbon monoxide, and 630°C for methane. Moreover, the proportions of the fuel and air must be in the proper range for combustion to begin. For example, natural gas does not burn in air in concentrations less than 5 percent or greater than about 15 percent. As you may recall from your chemistry courses, chemical equations are balanced on the basis of the conservation of mass principle (or the mass balance), which can be stated as follows: The total mass of each element is conserved during a chemical reaction (Fig. 15–5). That is, the total mass of each element on the right-hand side of the reaction equation (the products) must be equal to the total mass of that element on the left-hand side (the reactants) even though the elements exist in different chemical compounds in the reactants and products. Also, the total number of atoms of each element is conserved during a chemical reaction since the total number of atoms is equal to the total mass of the element divided by its atomic mass. For example, both sides of Eq. 15–2 contain 12 kg of carbon and 32 kg of oxygen, even though the carbon and the oxygen exist as elements in the reactants and as a compound in the product. Also, the total mass of reactants is equal to the total mass of products, each being 44 kg. (It is common practice to round the molar masses to the nearest integer if great accuracy is cen84959_ch15.qxd 9/15/06 6:06 AM Page 777 Chapter 15 not required.) However, notice that the total mole number of the reactants (2 kmol) is not equal to the total mole number of the products (1 kmol). That is, the total number of moles is not conserved during a chemical reaction. A frequently used quantity in the analysis of combustion processes to quantify the amounts of fuel and air is the air–fuel ratio AF. It is usually expressed on a mass basis and is defined as the ratio of the mass of air to the mass of fuel for a combustion process (Fig. 15–6). That is, AF m air m fuel (15–3) Fuel 1 kg Air 17 kg Combustion chamber AF = 17 | Products 18 kg 777 FIGURE 15–6 The air–fuel ratio (AF) represents the amount of air used per unit mass of fuel during a combustion process. The mass m of a substance is related to the number of moles N through the relation m NM, where M is the molar mass. The air–fuel ratio can also be expressed on a mole basis as the ratio of the mole numbers of air to the mole numbers of fuel. But we will use the former definition. The reciprocal of air–fuel ratio is called the fuel–air ratio. EXAMPLE 15–1 Balancing the Combustion Equation C8H18 1 kmol Combustion chamber One kmol of octane (C8H18) is burned with air that contains 20 kmol of O2, as shown in Fig. 15–7. Assuming the products contain only CO2, H2O, O2, and N2, determine the mole number of each gas in the products and the air–fuel ratio for this combustion process. AIR x y z w CO2 H2O O2 N2 Solution The amount of fuel and the amount of oxygen in the air are given. The amount of the products and the AF are to be determined. Assumptions The combustion products contain CO2, H2O, O2, and N2 only. Properties The molar mass of air is Mair 28.97 kg/kmol 29.0 kg/kmol (Table A–1). Analysis The chemical equation for this combustion process can be written as FIGURE 15–7 Schematic for Example 15–1. C8H18 20 1 O2 3.76N2 2 S xCO2 yH2O zO2 wN2 where the terms in the parentheses represent the composition of dry air that contains 1 kmol of O2 and x, y, z, and w represent the unknown mole numbers of the gases in the products. These unknowns are determined by applying the mass balance to each of the elements—that is, by requiring that the total mass or mole number of each element in the reactants be equal to that in the products: C: H: O: N2: Substituting yields 8 18 20 2 1 20 2 1 3.76 2 2x y x S x 2y S y 2z S z w S w 8 9 7.5 75.2 C8H18 20 1 O2 3.76N2 2 S 8CO2 9H2O 7.5O2 75.2N2 Note that the coefficient 20 in the balanced equation above represents the number of moles of oxygen, not the number of moles of air. The latter is obtained by adding 20 3.76 75.2 moles of nitrogen to the 20 moles of cen84959_ch15.qxd 9/15/06 6:06 AM Page 778 778 | Thermodynamics oxygen, giving a total of 95.2 moles of air. The air–fuel ratio (AF) is determined from Eq. 15–3 by taking the ratio of the mass of the air and the mass of the fuel, AF m air m fuel 1 8 kmol 2 1 12 kg> kmol 2 24.2 kg air/kg fuel 1 20 1 NM 2 C 1 NM 2 air 4.76 kmol 2 1 29 kg> kmol 2 1 NM 2 H2 1 9 kmol 2 1 2 kg> kmol 2 That is, 24.2 kg of air is used to burn each kilogram of fuel during this combustion process. 15–2 Fuel CnHm AIR Combustion chamber n CO2 m HO 22 Excess O2 N2 THEORETICAL AND ACTUAL COMBUSTION PROCESSES FIGURE 15–8 A combustion process is complete if all the combustible components of the fuel are burned to completion. CH4 + 2(O2 + 3.76N2) → CO2 + 2H2O + 7.52N2 • no unburned fuel • no free oxygen in products FIGURE 15–9 The complete combustion process with no free oxygen in the products is called theoretical combustion. It is often instructive to study the combustion of a fuel by assuming that the combustion is complete. A combustion process is complete if all the carbon in the fuel burns to CO2, all the hydrogen burns to H2O, and all the sulfur (if any) burns to SO2. That is, all the combustible components of a fuel are burned to completion during a complete combustion process (Fig. 15–8). Conversely, the combustion process is incomplete if the combustion products contain any unburned fuel or components such as C, H2, CO, or OH. Insufficient oxygen is an obvious reason for incomplete combustion, but it is not the only one. Incomplete combustion occurs even when more oxygen is present in the combustion chamber than is needed for complete combustion. This may be attributed to insufficient mixing in the combustion chamber during the limited time that the fuel and the oxygen are in contact. Another cause of incomplete combustion is dissociation, which becomes important at high temperatures. Oxygen has a much greater tendency to combine with hydrogen than it does with carbon. Therefore, the hydrogen in the fuel normally burns to completion, forming H2O, even when there is less oxygen than needed for complete combustion. Some of the carbon, however, ends up as CO or just as plain C particles (soot) in the products. The minimum amount of air needed for the complete combustion of a fuel is called the stoichiometric or theoretical air. Thus, when a fuel is completely burned with theoretical air, no uncombined oxygen is present in the product gases. The theoretical air is also referred to as the chemically correct amount of air, or 100 percent theoretical air. A combustion process with less than the theoretical air is bound to be incomplete. The ideal combustion process during which a fuel is burned completely with theoretical air is called the stoichiometric or theoretical combustion of that fuel (Fig. 15–9). For example, the theoretical combustion of methane is CH4 2 1 O2 3.76N2 2 S CO2 2H2O 7.52N2 Notice that the products of the theoretical combustion contain no unburned methane and no C, H2, CO, OH, or free O2. cen84959_ch15.qxd 9/15/06 6:06 AM Page 779 Chapter 15 In actual combustion processes, it is common practice to use more air than the stoichiometric amount to increase the chances of complete combustion or to control the temperature of the combustion chamber. The amount of air in excess of the stoichiometric amount is called excess air. The amount of excess air is usually expressed in terms of the stoichiometric air as percent excess air or percent theoretical air. For example, 50 percent excess air is equivalent to 150 percent theoretical air, and 200 percent excess air is equivalent to 300 percent theoretical air. Of course, the stoichiometric air can be expressed as 0 percent excess air or 100 percent theoretical air. Amounts of air less than the stoichiometric amount are called deficiency of air and are often expressed as percent deficiency of air. For example, 90 percent theoretical air is equivalent to 10 percent deficiency of air. The amount of air used in combustion processes is also expressed in terms of the equivalence ratio, which is the ratio of the actual fuel–air ratio to the stoichiometric fuel–air ratio. Predicting the composition of the products is relatively easy when the combustion process is assumed to be complete and the exact amounts of the fuel and air used are known. All one needs to do in this case is simply apply the mass balance to each element that appears in the combustion equation, without needing to take any measurements. Things are not so simple, however, when one is dealing with actual combustion processes. For one thing, actual combustion processes are hardly ever complete, even in the presence of excess air. Therefore, it is impossible to predict the composition of the products on the basis of the mass balance alone. Then the only alternative we have is to measure the amount of each component in the products directly. A commonly used device to analyze the composition of combustion gases is the Orsat gas analyzer. In this device, a sample of the combustion gases is collected and cooled to room temperature and pressure, at which point its volume is measured. The sample is then brought into contact with a chemical that absorbs the CO2. The remaining gases are returned to the room temperature and pressure, and the new volume they occupy is measured. The ratio of the reduction in volume to the original volume is the volume fraction of the CO2, which is equivalent to the mole fraction if ideal-gas behavior is assumed (Fig. 15–10). The volume fractions of the other gases are determined by repeating this procedure. In Orsat analysis the gas sample is collected over water and is maintained saturated at all times. Therefore, the vapor pressure of water remains constant during the entire test. For this reason the presence of water vapor in the test chamber is ignored and data are reported on a dry basis. However, the amount of H2O formed during combustion is easily determined by balancing the combustion equation. EXAMPLE 15–2 Dew-Point Temperature of Combustion Products C2H6 Combustion chamber AIR (20% excess) 100 kPa | 779 BEFORE AFTER 100 kPa 25°C Gas sample including CO2 1 liter 100 kPa 25°C Gas sample without CO2 0.9 liter y CO = 2 VCO V 2 = 0.1 = 0.1 1 FIGURE 15–10 Determining the mole fraction of the CO2 in combustion gases by using the Orsat gas analyzer. Ethane (C2H6) is burned with 20 percent excess air during a combustion process, as shown in Fig. 15–11. Assuming complete combustion and a total pressure of 100 kPa, determine (a) the air–fuel ratio and (b) the dew-point temperature of the products. CO2 H2O O2 N2 Solution The fuel is burned completely with excess air. The AF and the dew point of the products are to be determined. FIGURE 15–11 Schematic for Example 15–2. cen84959_ch15.qxd 9/15/06 6:06 AM Page 780 780 | Thermodynamics Assumptions 1 Combustion is complete. 2 Combustion gases are ideal gases. Analysis The combustion products contain CO2, H2O, N2, and some excess O2 only. Then the combustion equation can be written as C2H6 1.2ath 1 O2 3.76N2 2 S 2CO2 3H2O 0.2athO2 1 1.2 3.76 2 athN2 where a th is the stoichiometric coefficient for air. We have automatically accounted for the 20 percent excess air by using the factor 1.2ath instead of a th for air. The stoichiometric amount of oxygen (a thO2) is used to oxidize the fuel, and the remaining excess amount (0.2athO2) appears in the products as unused oxygen. Notice that the coefficient of N2 is the same on both sides of the equation, and that we wrote the C and H balances directly since they are so obvious. The coefficient ath is determined from the O2 balance to be O2: 1.2ath Substituting, 2 1.5 0.2ath S ath 3.5 C2H6 4.2 1 O2 3.76N2 2 S 2CO2 1 4.2 3H2O 0.7O2 15.79N2 (a) The air–fuel ratio is determined from Eq. 15–3 by taking the ratio of the mass of the air to the mass of the fuel, AF m air m fuel 1 2 kmol 2 1 12 kg> kmol 2 4.76 kmol 2 1 29 kg> kmol 2 1 3 kmol 2 1 2 kg> kmol 2 19.3 kg air/kg fuel That is, 19.3 kg of air is supplied for each kilogram of fuel during this combustion process. (b) The dew-point temperature of the products is the temperature at which the water vapor in the products starts to condense as the products are cooled at constant pressure. Recall from Chap. 14 that the dew-point temperature of a gas–vapor mixture is the saturation temperature of the water vapor corresponding to its partial pressure. Therefore, we need to determine the partial pressure of the water vapor Pv in the products first. Assuming ideal-gas behavior for the combustion gases, we have Pv Thus, a Nv b 1 Pprod 2 N prod Tdp a 3 kmol b 1 100 kPa 2 21.49 kmol 13.96 kPa Tsat @ 13.96 kPa 52.3°C (Table A–5) EXAMPLE 15–3 FUEL CH4, O2, H2, N2, CO2 AIR 20°C, φ = 80% Combustion of a Gaseous Fuel with Moist Air Combustion chamber 1 atm CO2 H2O N2 A certain natural gas has the following volumetric analysis: 72 percent CH4, 9 percent H2, 14 percent N2, 2 percent O2, and 3 percent CO2. This gas is now burned with the stoichiometric amount of air that enters the combustion chamber at 20°C, 1 atm, and 80 percent relative humidity, as shown in Fig. 15–12. Assuming complete combustion and a total pressure of 1 atm, determine the dew-point temperature of the products. FIGURE 15–12 Schematic for Example 15–3. Solution A gaseous fuel is burned with the stoichiometric amount of moist air. The dew point temperature of the products is to be determined. cen84959_ch15.qxd 9/15/06 6:06 AM Page 781 Chapter 15 Assumptions 1 The fuel is burned completely and thus all the carbon in the fuel burns to CO2 and all the hydrogen to H2O. 2 The fuel is burned with the stoichiometric amount of air and thus there is no free O2 in the product gases. 3 Combustion gases are ideal gases. Properties The saturation pressure of water at 20°C is 2.3392 kPa (Table A–4). Analysis We note that the moisture in the air does not react with anything; it simply shows up as additional H2O in the products. Therefore, for simplicity, we balance the combustion equation by using dry air and then add the moisture later to both sides of the equation. Considering 1 kmol of fuel, | 781 The unknown coefficients in the above equation are determined from mass balances on various elements, Next we determine the amount of moisture that accompanies 4.76a th (4.76)(1.465) 6.97 kmol of dry air. The partial pressure of the moisture in the air is Assuming ideal-gas behavior, the number of moles of the moisture in the air is which yields The balanced combustion equation is obtained by substituting the coefficients determined earlier and adding 0.131 kmol of H2O to both sides of the equation: 1 0.72CH 4 The dew-point temperature of the products is the temperature at which the water vapor in the products starts to condense as the products are cooled. Again, assuming ideal-gas behavior, the partial pressure of the water vapor in the combustion gases is 1555255553 155555555555552555555555555553 155525553 1555555555555552555555555555553 1 0.72CH 4 0.09H 2 0.14N2 0.02O2 0.03CO2 2 a th 1 O2 xCO2 yH 2O C: H: O2 : N2: 0.72 0.02 4 0.03 0.14 0.72 0.09 a th 0.03 2 x x S x 2y S y y S a th 2 z S z 0.75 1.53 1.465 5.648 3.76a th Pv,air fair Psat @ 20°C 1 0.80 2 1 2.3392 kPa 2 1.871 kPa N v,air a Pv,air Ptotal b N total N v,air a 1.871 kPa b 1 6.97 101.325 kPa 0.131 kmol N v,air 2 fuel dry air fuel dry air 3.76N2 2 S zN2 0.09H 2 0.14N2 0.02O2 0.03CO2 2 1.465 1 O2 3.76N2 2 5.648N2 moisture includes moisture 0.131H 2O S 0.75CO2 1.661H 2O Pv,prod a N v,prod N prod b Pprod a 1.661 kmol b 1 101.325 kPa 2 8.059 kmol 14243 14243 20.88 kPa cen84959_ch15.qxd 9/15/06 6:06 AM Page 782 782 | Thermodynamics Thus, Tdp Tsat @ 20.88 kPa 60.9°C Discussion If the combustion process were achieved with dry air instead of moist air, the products would contain less moisture, and the dew-point temperature in this case would be 59.5°C. EXAMPLE 15–4 C8H18 Combustion chamber 10.02% CO2 5.62% O2 0.88% CO 83.48% N2 Reverse Combustion Analysis Octane (C8H18) is burned with dry air. The volumetric analysis of the products on a dry basis is (Fig. 15–13) AIR FIGURE 15–13 Schematic for Example 15–4. CO2: O2: CO: N2: 10.02 percent 5.62 percent 0.88 percent 83.48 percent Determine (a) the air–fuel ratio, (b) the percentage of theoretical air used, and (c) the amount of H2O that condenses as the products are cooled to 25°C at 100 kPa. Solution Combustion products whose composition is given are cooled to 25°C. The AF, the percent theoretical air used, and the fraction of water vapor that condenses are to be determined. Assumptions Combustion gases are ideal gases. Properties The saturation pressure of water at 25°C is 3.1698 kPa (Table A–4). Analysis Note that we know the relative composition of the products, but we do not know how much fuel or air is used during the combustion process. However, they can be determined from mass balances. The H2O in the combustion gases will start condensing when the temperature drops to the dewpoint temperature. For ideal gases, the volume fractions are equivalent to the mole fractions. Considering 100 kmol of dry products for convenience, the combustion equation can be written as xC8H 18 a 1O2 3.76N2 2 S 10.02CO2 3.76a 8x 0.88CO 5.62O2 83.48N2 22.20 1.36 12.24 22.20 bH 2O The unknown coefficients x, a, and b are determined from mass balances, N2: C: H: O2: a 10.02 83.48 S a 0.88 S x 2b S b b S 22.20 2 10.02 18x 0.44 5.62 The O2 balance is not necessary, but it can be used to check the values obtained from the other mass balances, as we did previously. Substituting, we get 1.36C8H18 22.2 1 O2 3.76N2 2 S 10.02CO2 0.88CO 5.62O2 83.48N2 12.24H2O cen84959_ch15.qxd 9/15/06 6:06 AM Page 783 Chapter 15 The combustion equation for 1 kmol of fuel is obtained by dividing the above equation by 1.36, | 783 C8H18 16.32 1 O2 3.76N2 2 S 7.37CO2 0.65CO 4.13O2 61.38N2 9H2O (a) The air–fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel (Eq. 15–3), AF m air m fuel 1 8 kmol 2 1 12 kg> kmol 2 1 16.32 4.76 kmol 2 1 29 kg> kmol 2 1 9 kmol 2 1 2 kg> kmol 2 19.76 kg air/kg fuel (b) To find the percentage of theoretical air used, we need to know the theoretical amount of air, which is determined from the theoretical combustion equation of the fuel, C8H18 O2: Then, ath 1 O2 ath 8 3.76N2 2 S 8CO2 4.5 S ath 12.5 9H2O 3.76athN2 Percentage of theoretical air m air,act m air,th N air,act N air,th 1 16.32 2 1 4.76 2 kmol 131% 1 12.50 2 1 4.76 2 kmol That is, 31 percent excess air was used during this combustion process. Notice that some carbon formed carbon monoxide even though there was considerably more oxygen than needed for complete combustion. 0.65 4.13 61.38 9 (c) For each kmol of fuel burned, 7.37 82.53 kmol of products are formed, including 9 kmol of H2O. Assuming that the dew-point temperature of the products is above 25°C, some of the water vapor will condense as the products are cooled to 25°C. If Nw kmol of H2O condenses, there will be (9 Nw) kmol of water vapor left in the products. The mole number of the products in the gas phase will also decrease to 82.53 Nw as a result. By treating the product gases (including the remaining water vapor) as ideal gases, Nw is determined by equating the mole fraction of the water vapor to its pressure fraction, Nv N prod,gas 9 Nw 82.53 N w Nw Pv Pprod 3.1698 kPa 100 kPa 6.59 kmol Therefore, the majority of the water vapor in the products (73 percent of it) condenses as the product gases are cooled to 25°C. cen84959_ch15.qxd 9/15/06 6:06 AM Page 784 784 | Thermodynamics 15–3 Nuclear energy Chemical energy Latent energy Sensible energy ATOM MOLECULE ATOM ENTHALPY OF FORMATION AND ENTHALPY OF COMBUSTION MOLECULE FIGURE 15–14 The microscopic form of energy of a substance consists of sensible, latent, chemical, and nuclear energies. We mentioned in Chap. 2 that the molecules of a system possess energy in various forms such as sensible and latent energy (associated with a change of state), chemical energy (associated with the molecular structure), and nuclear energy (associated with the atomic structure), as illustrated in Fig. 15–14. In this text we do not intend to deal with nuclear energy. We also ignored chemical energy until now since the systems considered in previous chapters involved no changes in their chemical structure, and thus no changes in chemical energy. Consequently, all we needed to deal with were the sensible and latent energies. During a chemical reaction, some chemical bonds that bind the atoms into molecules are broken, and new ones are formed. The chemical energy associated with these bonds, in general, is different for the reactants and the products. Therefore, a process that involves chemical reactions involves changes in chemical energies, which must be accounted for in an energy balance (Fig. 15–15). Assuming the atoms of each reactant remain intact (no nuclear reactions) and disregarding any changes in kinetic and potential energies, the energy change of a system during a chemical reaction is due to a change in state and a change in chemical composition. That is, ¢ Esys ¢ Estate ¢ Echem (15–4) Sensible energy ATOM ATOM ATOM Broken chemical bond FIGURE 15–15 When the existing chemical bonds are destroyed and new ones are formed during a combustion process, usually a large amount of sensible energy is absorbed or released. 393,520 kJ 1 kmol C 25°C, 1 atm 1 kmol O2 25°C, 1 atm Combustion chamber CO2 25°C, 1 atm FIGURE 15–16 The formation of CO2 during a steadyflow combustion process at 25 C and 1 atm. Therefore, when the products formed during a chemical reaction exit the reaction chamber at the inlet state of the reactants, we have Estate 0 and the energy change of the system in this case is due to the changes in its chemical composition only. In thermodynamics we are concerned with the changes in the energy of a system during a process, and not the energy values at the particular states. Therefore, we can choose any state as the reference state and assign a value of zero to the internal energy or enthalpy of a substance at that state. When a process involves no changes in chemical composition, the reference state chosen has no effect on the results. When the process involves chemical reactions, however, the composition of the system at the end of a process is no longer the same as that at the beginning of the process. In this case it becomes necessary to have a common reference state for all substances. The chosen reference state is 25°C (77°F) and 1 atm, which is known as the standard reference state. Property values at the standard reference state are indicated by a superscript (°) (such as h° and u°). When analyzing reacting systems, we must use property values relative to the standard reference state. However, it is not necessary to prepare a new set of property tables for this purpose. We can use the existing tables by subtracting the property values at the standard reference state from the values at the specified state. The ideal-gas enthalpy of N2 at 500 K relative to the standard refer– – ence state, for example, is h500 K h° 14,581 8669 5912 kJ/kmol. Consider the formation of CO2 from its elements, carbon and oxygen, during a steady-flow combustion process (Fig. 15–16). Both the carbon and the oxygen enter the combustion chamber at 25°C and 1 atm. The CO2 formed during this process also leaves the combustion chamber at 25°C and 1 atm. The combustion of carbon is an exothermic reaction (a reaction dur- cen84959_ch15.qxd 9/15/06 6:06 AM Page 785 Chapter 15 ing which chemical energy is released in the form of heat). Therefore, some heat is transferred from the combustion chamber to the surroundings during this process, which is 393,520 kJ/kmol CO2 formed. (When one is dealing with chemical reactions, it is more convenient to work with quantities per unit mole than per unit time, even for steady-flow processes.) The process described above involves no work interactions. Therefore, from the steady-flow energy balance relation, the heat transfer during this process must be equal to the difference between the enthalpy of the products and the enthalpy of the reactants. That is, Q Hprod Hreact 393,520 kJ> kmol (15–5) | 785 Since both the reactants and the products are at the same state, the enthalpy change during this process is solely due to the changes in the chemical composition of the system. This enthalpy change is different for different reactions, and it is very desirable to have a property to represent the changes in chemical energy during a reaction. This property is the enthalpy of reaction hR, which is defined as the difference between the enthalpy of the products at a specified state and the enthalpy of the reactants at the same state for a complete reaction. For combustion processes, the enthalpy of reaction is usually referred to as the enthalpy of combustion hC, which represents the amount of heat released during a steady-flow combustion process when 1 kmol (or 1 kg) of fuel is burned completely at a specified temperature and pressure (Fig. 15–17). It is expressed as hR hC Hprod Hreact (15–6) hC = Q = –393,520 kJ/kmol C 1 kmol C 25°C, 1 atm 1 kmol O2 25°C, 1 atm Combustion process 1 kmol CO2 25°C, 1 atm which is 393,520 kJ/kmol for carbon at the standard reference state. The enthalpy of combustion of a particular fuel is different at different temperatures and pressures. The enthalpy of combustion is obviously a very useful property for analyzing the combustion processes of fuels. However, there are so many different fuels and fuel mixtures that it is not practical to list hC values for all possible cases. Besides, the enthalpy of combustion is not of much use when the combustion is incomplete. Therefore a more practical approach would be to have a more fundamental property to represent the chemical energy of an element or a compound at some reference state. This property – is the enthalpy of formation hf , which can be viewed as the enthalpy of a substance at a specified state due to its chemical composition. To establish a starting point, we assign the enthalpy of formation of all a stable elements (such as O2, N2, H2, and C) – value of zero at the standard 0 for all stable elements. reference state of 25°C and 1 atm. That is, hf (This is no different from assigning the internal energy of saturated liquid water a value of zero at 0.01°C.) Perhaps we should clarify what we mean by stable. The stable form of an element is simply the chemically stable form of that element at 25°C and 1 atm. Nitrogen, for example, exists in diatomic form (N2) at 25°C and 1 atm. Therefore, the stable form of nitrogen at the standard reference state is diatomic nitrogen N2, not monatomic nitrogen N. If an element exists in more than one stable form at 25°C and 1 atm, one of the forms should be specified as the stable form. For carbon, for example, the stable form is assumed to be graphite, not diamond. FIGURE 15–17 The enthalpy of combustion represents the amount of energy released as a fuel is burned during a steady-flow process at a specified state. cen84959_ch15.qxd 9/15/06 6:06 AM Page 786 786 | Thermodynamics h f = Q = –393,520 kJ/kmol CO2 1 kmol C 25°C, 1 atm 1 kmol O2 25°C, 1 atm Combustion chamber 1 kmol CO2 25°C, 1 atm Now reconsider the formation of CO2 (a compound) from its elements C and O2 at 25°C and 1 atm during a steady-flow process. The enthalpy change during this process was determined to be 393,520 kJ/kmol. How0 since both reactants are elements at the standard reference ever, Hreact state, and the products consist of 1 kmol of CO2 at the same state. Therefore, the enthalpy of formation of CO2 at the standard reference state is 393,520 kJ/kmol (Fig. 15–18). That is, h°,CO2 f 393,520 kJ> kmol FIGURE 15–18 The enthalpy of formation of a compound represents the amount of energy absorbed or released as the component is formed from its stable elements during a steady-flow process at a specified state. LHV = Qout Fuel 1 kg Air (mhfg) H 2O Combustion chamber Products (vapor H2O) The negative sign is due to the fact that the enthalpy of 1 kmol of CO2 at 25°C and 1 atm is 393,520 kJ less than the enthalpy of 1 kmol of C and 1 kmol of O2 at the same state. In other words, 393,520 kJ of chemical energy is released (leaving the system as heat) when C and O2 combine to form 1 kmol of CO2. Therefore, a negative enthalpy of formation for a compound indicates that heat is released during the formation of that compound from its stable elements. A positive value indicates heat is absorbed. – You will notice that two hf° values are given for H2O in Table A–26, one for liquid water and the other for water vapor. This is because both phases of H2O are encountered at 25°C, and the effect of pressure on the enthalpy of formation is small. (Note that under equilibrium conditions, water exists only as a liquid at 25°C and 1 atm.) The difference between the two enthalpies of formation is equal to the hfg of water at 25°C, which is 2441.7 kJ/kg or 44,000 kJ/kmol. Another term commonly used in conjunction with the combustion of fuels is the heating value of the fuel, which is defined as the amount of heat released when a fuel is burned completely in a steady-flow process and the products are returned to the state of the reactants. In other words, the heating value of a fuel is equal to the absolute value of the enthalpy of combustion of the fuel. That is, Heating value 0 hC 0 1 kJ> kg fuel 2 Products (liquid H2O) HHV = LHV + (mhfg) H 2O The heating value depends on the phase of the H2O in the products. The heating value is called the higher heating value (HHV) when the H2O in the products is in the liquid form, and it is called the lower heating value (LHV) when the H2O in the products is in the vapor form (Fig. 15–19). The two heating values are related by HHV LHV 1 mhfg 2 H2O 1 kJ> kg fuel 2 (15–7) FIGURE 15–19 The higher heating value of a fuel is equal to the sum of the lower heating value of the fuel and the latent heat of vaporization of the H2O in the products. where m is the mass of H2O in the products per unit mass of fuel and hfg is the enthalpy of vaporization of water at the specified temperature. Higher and lower heating values of common fuels are given in Table A–27. The heating value or enthalpy of combustion of a fuel can be determined from a knowledge of the enthalpy of formation for the compounds involved. This is illustrated with the following example. EXAMPLE 15–5 Evaluation of the Enthalpy of Combustion Determine the enthalpy of combustion of liquid octane (C8H18) at 25°C and 1 atm, using enthalpy-of-formation data from Table A–26. Assume the water in the products is in the liquid form. cen84959_ch15.qxd 9/15/06 6:06 AM Page 787 Chapter 15 Solution The enthalpy of combustion of a fuel is to be determined using enthalpy of formation data. Properties The enthalpy of formation at 25°C and 1 atm is 393,520 kJ/kmol for CO2, 285,830 kJ/kmol for H2O( ), and 249,950 kJ/kmol for C8H18( ) (Table A–26). Analysis The combustion of C8H18 is illustrated in Fig. 15–20. The stoichiometric equation for this reaction is C8H18 ( ) 25°C, 1 atm AIR 25°C, 1 atm | 787 hC = Hprod – Hreact 25°C 1 atm CO2 H2O( ) N2 C8H18 ath 1 O2 3.76N2 2 S 8CO2 9H2O 1 2 3.76athN2 Both the reactants and the products are at the standard reference state of 25°C and 1 atm. Also, N2 and O2 are stable elements, and thus their enthalpy of formation is zero. Then the enthalpy of combustion of C8H18 becomes (Eq. 15–6) FIGURE 15–20 Schematic for Example 15–5. hC Hprod Hreact f a N r h°,r f a N ph°,p 1 Nh°2 CO2 f 1 Nh°2 H2O f 1 Nh°2 C8H18 f Substituting, hC 1 1 kmol 2 1 249,950 kJ> kmol 2 5,471,000 kJ> kmol C8H18 1 8 kmol 2 1 393,520 kJ> kmol 2 1 9 kmol 2 1 285,830 kJ> kmol 2 47,891 kJ> kg C8H18 which is practially identical to the listed value of 47,890 kJ/kg in Table A–27. Since the water in the products is assumed to be in the liquid phase, this hC value corresponds to the HHV of liquid C8H18. Discussion It can be shown that the result for gaseous octane is 5,512,200 kJ/kmol or 48,255 kJ/kg. When the exact composition of the fuel is known, the enthalpy of combustion of that fuel can be determined using enthalpy of formation data as shown above. However, for fuels that exhibit considerable variations in composition depending on the source, such as coal, natural gas, and fuel oil, it is more practical to determine their enthalpy of combustion experimentally by burning them directly in a bomb calorimeter at constant volume or in a steady-flow device. 15–4 FIRST-LAW ANALYSIS OF REACTING SYSTEMS The energy balance (or the first-law) relations developed in Chaps. 4 and 5 are applicable to both reacting and nonreacting systems. However, chemically reacting systems involve changes in their chemical energy, and thus it is more convenient to rewrite the energy balance relations so that the changes in chemical energies are explicitly expressed. We do this first for steady-flow systems and then for closed systems. Steady-Flow Systems Before writing the energy balance relation, we need to express the enthalpy of a component in a form suitable for use for reacting systems. That is, we need to express the enthalpy such that it is relative to the standard reference cen84959_ch15.qxd 9/15/06 6:06 AM Page 788 788 | Thermodynamics state and the chemical energy term appears explicitly. When expressed prop– erly, the enthalpy term should reduce to the enthalpy of formation h° at the f standard reference state. With this in mind, we express the enthalpy of a component on a unit mole basis as (Fig. 15–21) Enthalpy h° f 1h h° 2 1 kJ> kmol 2 Enthalpy at 25°C, 1 atm H = N (hf°+ h – h°) Sensible enthalpy enthalpy relative to to 25°C, 1 atm where the term in the parentheses represents the sensible enthalpy relative to – the standard reference state, which is the difference between h the sensible – enthalpy at the specified state) and h° (the sensible enthalpy at the standard reference state of 25°C and 1 atm). This definition enables us to use enthalpy values from tables regardless of the reference state used in their construction. When the changes in kinetic and . potential energies are negligible, the . steady-flow energy balance relation Ein Eout can be expressed for a chemically reacting steady-flow system more explicitly as # # # Qin Win a nr 1 h° h h° 2 r f 1555555552555555553 Rate of net energy transfer in by heat, work, and mass FIGURE 15–21 The enthalpy of a chemical component at a specified state is the sum of the enthalpy of the component at 25 C, 1 atm (hf°), and the sensible enthalpy of the component relative to 25 C, 1 atm. # # # Qout Wout a np 1 h° h h° 2 p f 15555555525555555553 Rate of net energy transfer out by heat, work, and mass (15–8) . . where np and nr represent the molal flow rates of the product p and the reactant r, respectively. In combustion analysis, it is more convenient to work with quantities expressed per mole of fuel. Such a relation is obtained by dividing each term of the equation above by the molal flow rate of the fuel, yielding Q in Win a N r 1 h° h h° 2 r f 1555555552555555553 Energy transfer in per mole of fuel by heat, work, and mass Q out Wout a N p 1 h° h h° 2 p f 15555555525555555553 Energy transfer out per mole of fuel by heat, work, and mass (15–9) where Nr and Np represent the number of moles of the reactant r and the product p, respectively, per mole of fuel. Note that Nr 1 for the fuel, and the other Nr and Np values can be picked directly from the balanced combustion equation. Taking heat transfer to the system and work done by the system to be positive quantities, the energy balance relation just discussed can be expressed more compactly as Q W f a Np 1 h° h h° 2 p f a Nr 1 h° h h° 2 r (15–10) or as Q W Hprod Hreact 1 kJ> kmol fuel 2 h h h° 2 p 1 kJ> kmol fuel 2 (15–11) where Hprod Hreact f a N r 1 h° f a N p 1 h° ° If the enthalpy of combustion hC for a particular reaction is available, the steady-flow energy equation per mole of fuel can be expressed as Q W hC ° a Np 1 h h° 2 p a N r 1 h h° 2 r 1 kJ> kmol 2 (15–12) – h° 2 r 1 kJ> kmol fuel 2 The energy balance relations above are sometimes written without the work term since most steady-flow combustion processes do not involve any work interactions. cen84959_ch15.qxd 9/15/06 6:06 AM Page 789 Chapter 15 A combustion chamber normally involves heat output but no heat input. Then the energy balance for a typical steady-flow combustion process becomes Qout f a Nr 1 h° h h° 2 r 15555255553 f a N p 1 h° h h ° 2 p 15555255553 | 789 (15–13) Energy in by mass per mole of fuel Energy out by mass per mole of fuel It expresses that the heat output during a combustion process is simply the difference between the energy of the reactants entering and the energy of the products leaving the combustion chamber. Closed Systems The general closed-system energy balance relation Ein Eout Esystem can be expressed for a stationary chemically reacting closed system as 1 Qin Qout 2 1 Win Wout 2 Uprod Ureact 1 kJ> kmol fuel 2 (15–14) where Uprod represents the internal energy of the products and Ureact represents the internal energy of the reactants. To avoid using another property— – the internal energy of formation –u°—we utilize the definition of enthalpy f – – – – or u ° u – – – –° u h° h h° Pv) and express the above (u h Pv f f equation as (Fig. 15–22) Q W f a N p 1 h° U = H – PV – – – = N(h° + h – h°) – PV f – = N(h° + h – h° – Pv ) f – – – h h° Pv 2 p f a N r 1 h° h h° Pv 2 r (15–15) where we have taken heat transfer to the system and work done by the sys– tem to be positive quantities. The Pv terms are negligible for solids and liquids, and can be replaced by RuT for gases that behave as an ideal gas. Also, – if desired, the h Pv terms in Eq. 15–15 can be replaced by u. The work term in Eq. 15–15 represents all forms of work, including the H for nonreactboundary work. It was shown in Chap. 4 that U Wb ing closed systems undergoing a quasi-equilibrium P constant expansion or compression process. This is also the case for chemically reacting systems. There are several important considerations in the analysis of reacting systems. For example, we need to know whether the fuel is a solid, a liquid, or a gas since the enthalpy of formation h° of a fuel depends on the phase of f the fuel. We also need to know the state of the fuel when it enters the combustion chamber in order to determine its enthalpy. For entropy calculations it is especially important to know if the fuel and air enter the combustion chamber premixed or separately. When the combustion products are cooled to low temperatures, we need to consider the possibility of condensation of some of the water vapor in the product gases. EXAMPLE 15–6 First-Law Analysis of Steady-Flow Combustion FIGURE 15–22 An expression for the internal energy of a chemical component in terms of the enthalpy. Q=? C3H8 ( ) 25°C, 0.05 kg/min AIR 7°C H2O CO2 1500 K Combustion CO chamber O2 N2 · Liquid propane (C3H8) enters a combustion chamber at 25°C at a rate of 0.05 kg/min where it is mixed and burned with 50 percent excess air that enters the combustion chamber at 7°C, as shown in Fig. 15–23. An analysis of the combustion gases reveals that all the hydrogen in the fuel burns to H2O but only 90 percent of the carbon burns to CO2, with the remaining 10 percent forming CO. If the exit temperature of the combustion gases is FIGURE 15–23 Schematic for Example 15–6. cen84959_ch15.qxd 9/15/06 6:06 AM Page 790 790 | Thermodynamics 1500 K, determine (a) the mass flow rate of air and (b) the rate of heat transfer from the combustion chamber. Solution Liquid propane is burned steadily with excess air. The mass flow rate of air and the rate of heat transfer are to be determined. Assumptions 1 Steady operating conditions exist. 2 Air and the combustion gases are ideal gases. 3 Kinetic and potential energies are negligible. Analysis We note that all the hydrogen in the fuel burns to H2O but 10 percent of the carbon burns incompletely and forms CO. Also, the fuel is burned with excess air and thus there is some free O2 in the product gases. The theoretical amount of air is determined from the stoichiometric reaction to be C3H8 1 2 ath 1 O2 3.76N2 2 S 3CO2 3 2 5 4H2O 3.76athN2 O2 balance: ath Then the balanced equation for the actual combustion process with 50 percent excess air and some CO in the products becomes C3H 8 1 2 7.5 1O2 3.76N2 2 S 2.7CO2 1 7.5 0.3CO 4H 2O 2.65O2 28.2N2 (a) The air–fuel ratio for this combustion process is AF 25.53 kg air> kg fuel Thus, m air m fuel 1 3 kmol 2 1 12 kg> kmol 2 4.76 kmol 2 1 29 kg> kmol 2 1 4 kmol 2 1 2 kg> kmol 2 # m air 1 23.53 kg air> kg fuel 2 1 0.05 kg fuel> min 2 1.18 kg air/min # 1 AF 2 1 m fuel 2 (b) The heat transfer for this steady-flow combustion process is determined from the steady-flow energy balance Eout Ein applied on the combustion chamber per unit mole of the fuel, Qout or f a Np 1 h° f a Nr 1 h° h h h° 2 p h° 2 r f a Nr 1 h° f a Np 1 h° h h h° 2 r h° 2 p Qout Assuming the air and the combustion products to be ideal gases, we have h h(T ), and we form the following minitable using data from the property tables: – h f° kJ/kmol 118,910 0 0 241,820 393,520 110,530 – h 280 K kJ/kmol — 8150 8141 — — — – h 298 K kJ/kmol — 8682 8669 9904 9364 8669 – h 1500 K kJ/kmol — 49,292 47,073 57,999 71,078 47,517 Substance C3H8( ) O2 N2 H2O(g) CO2 CO cen84959_ch15.qxd 9/15/06 6:06 AM Page 791 Chapter 15 – – The h f° of liquid propane is obtained by subtracting the h fg of propane at – 25°C from the h f° of gas propane. Substituting gives | 791 Q out 1 1 kmol C3H 8 2 3 1 118,910 1 7.5 kmol O2 2 3 1 0 1 28.2 kmol N2 2 3 1 0 h 298 8150 8141 8682 2 kJ> kmol O2 4 71,078 47,517 57,999 h 298 2 kJ> kmol C3H 8 4 1 2.7 kmol CO2 2 3 1 393,520 1 0.3 kmol CO 2 3 1 110,530 1 4 kmol H 2O 2 3 1 241,820 1 2.65 kmol O2 2 3 1 0 1 28.2 kmol N2 2 3 1 0 8669 2 kJ> kmol N2 4 49,292 47,073 8682 2 kJ> kmol O2 4 8669 2 kJ> kmol N2 4 9904 2 kJ> kmol H 2O 4 8669 2 kJ> kmol CO 4 9364 2 kJ> kmol CO2 4 363,880 kJ> kmol of C3H 8 Thus 363,880 kJ of heat is transferred from the combustion chamber for each kmol (44 kg) of propane. This corresponds to 363,880/44 8270 kJ of heat loss per kilogram of propane. Then the rate of heat transfer for a mass flow rate of 0.05 kg/min for the propane becomes # Q out # mqout 1 0.05 kg> min 2 1 8270 kJ> kg 2 413.5 kJ> min 6.89 kW EXAMPLE 15–7 First-Law Analysis of Combustion in a Bomb BEFORE REACTION 1 lbmol CH4 3 lbmol O2 77°F 1 atm AFTER REACTION CO2 H2O O2 1800 R P2 The constant-volume tank shown in Fig. 15–24 contains 1 lbmol of methane (CH4) gas and 3 lbmol of O2 at 77°F and 1 atm. The contents of the tank are ignited, and the methane gas burns completely. If the final temperature is 1800 R, determine (a) the final pressure in the tank and (b) the heat transfer during this process. Solution Methane is burned in a rigid tank. The final pressure in the tank and the heat transfer are to be determined. Assumptions 1 The fuel is burned completely and thus all the carbon in the fuel burns to CO2 and all the hydrogen to H2O. 2 The fuel, the air, and the combustion gases are ideal gases. 3 Kinetic and potential energies are negligible. 4 There are no work interactions involved. Analysis The balanced combustion equation is FIGURE 15–24 Schematic for Example 15–7. CH4 1 g 2 3O2 S CO2 2H2O O2 (a) At 1800 R, water exists in the gas phase. Using the ideal-gas relation for both the reactants and the products, the final pressure in the tank is determined to be PreactV PprodV Substituting, we get NreactRuTreact f Pprod NprodRuTprod 1 1 atm 2 a Preact a Nprod Nreact ba Tprod Treact b Pprod 4 lbmol 1800 R ba b 4 lbmol 537 R 3.35 atm cen84959_ch15.qxd 9/15/06 6:06 AM Page 792 792 | Thermodynamics (b) Noting that the process involves no work interactions, the heat transfer during this constant-volume combustion process can be determined from the energy balance Ein Eout Esystem applied to the tank, Qout f a Np 1 h° h h° Pv 2 p f a Nr 1 h° h h° Pv 2 r Since both the reactants and the products are assumed to be ideal gases, all – the internal energy and enthalpies depend on temperature only, and the P v terms in this equation can be replaced by RuT. It yields Qout f a Nr 1 h° Ru T 2 r f a Np 1 h° h 1800 R h 537 R RuT 2 p since the reactants are at the standard reference temperature of 537 R. – From h f° and ideal-gas tables in the Appendix, – h f° Btu/lbmol 32,210 0 169,300 104,040 – h 537 R Btu/lbmol — 3725.1 4027.5 4258.0 – h 1800 R Btu/lbmol — 13,485.8 18,391.5 15,433.0 Substance CH4 O2 CO2 H2O(g) Substituting, we have Q out 1 1 lbmol CH 4 2 3 1 32,210 1 3 lbmol O2 2 3 1 0 1.986 1.986 1 1 lbmol CO2 2 3 1 169,300 Btu> lbmol CO2 4 1 2 lbmol H 2O 2 3 1 104,040 Btu> lbmol H 2O 4 11 lbmol O2 2 3 1 0 13,485.8 537 2 Btu> lbmol O2 4 18,391.5 4027.5 537 2 Btu> lbmol CH 4 4 1.986 1800 2 1800 2 15,433.0 4258.0 1.986 3725.1 1.986 1800 2 Btu> lbmol O2 4 308,730 Btu/lbmol CH4 Discussion On a mass basis, the heat transfer from the tank would be 308,730/16 19,300 Btu/lbm of methane. Insulation Fuel Combustion chamber Products Tmax 15–5 ADIABATIC FLAME TEMPERATURE Air FIGURE 15–25 The temperature of a combustion chamber becomes maximum when combustion is complete and no heat is lost to the surroundings (Q 0). In the absence of any work interactions and any changes in kinetic or potential energies, the chemical energy released during a combustion process either is lost as heat to the surroundings or is used internally to raise the temperature of the combustion products. The smaller the heat loss, the larger the temperature rise. In the limiting case of no heat loss to the surroundings (Q 0), the temperature of the products reaches a maximum, which is called the adiabatic flame or adiabatic combustion temperature of the reaction (Fig. 15–25). cen84959_ch15.qxd 9/15/06 6:06 AM Page 793 Chapter 15 The adiabatic flame temperature of a steady-flow combustion process is determined from Eq. 15–11 by setting Q 0 and W 0. It yields Hprod Hreact f a Nr 1 h° | 793 (15–16) or f a Np 1 h° h h° 2 p h h° 2 r (15–17) Once the reactants and their states are specified, the enthalpy of the reactants Hreact can be easily determined. The calculation of the enthalpy of the products Hprod is not so straightforward, however, because the temperature of the products is not known prior to the calculations. Therefore, the determination of the adiabatic flame temperature requires the use of an iterative technique unless equations for the sensible enthalpy changes of the combustion products are available. A temperature is assumed for the product gases, and the Hprod is determined for this temperature. If it is not equal to Hreact, calculations are repeated with another temperature. The adiabatic flame temperature is then determined from these two results by interpolation. When the oxidant is air, the product gases mostly consist of N2, and a good first guess for the adiabatic flame temperature is obtained by treating the entire product gases as N2. In combustion chambers, the highest temperature to which a material can be exposed is limited by metallurgical considerations. Therefore, the adiabatic flame temperature is an important consideration in the design of combustion chambers, gas turbines, and nozzles. The maximum temperatures that occur in these devices are considerably lower than the adiabatic flame temperature, however, since the combustion is usually incomplete, some heat loss takes place, and some combustion gases dissociate at high temperatures (Fig. 15–26). The maximum temperature in a combustion chamber can be controlled by adjusting the amount of excess air, which serves as a coolant. Note that the adiabatic flame temperature of a fuel is not unique. Its value depends on (1) the state of the reactants, (2) the degree of completion of the reaction, and (3) the amount of air used. For a specified fuel at a specified state burned with air at a specified state, the adiabatic flame temperature attains its maximum value when complete combustion occurs with the theoretical amount of air. EXAMPLE 15–8 Adiabatic Flame Temperature in Steady Combustion Heat loss Fuel • Incomplete combustion Air • Dissociation Products Tprod < Tmax FIGURE 15–26 The maximum temperature encountered in a combustion chamber is lower than the theoretical adiabatic flame temperature. Liquid octane (C8H18) enters the combustion chamber of a gas turbine steadily at 1 atm and 25°C, and it is burned with air that enters the combustion chamber at the same state, as shown in Fig. 15–27. Determine the adiabatic flame temperature for (a) complete combustion with 100 percent theoretical air, (b) complete combustion with 400 percent theoretical air, and (c) incomplete combustion (some CO in the products) with 90 percent theoretical air. C8H18 25°C, 1 atm Air 25°C, 1 atm Combustion chamber TP 1 atm CO2 H2O N2 O2 Solution Liquid octane is burned steadily. The adiabatic flame temperature is to be determined for different cases. FIGURE 15–27 Schematic for Example 15–8. cen84959_ch15.qxd 9/15/06 6:06 AM Page 794 794 | Thermodynamics Assumptions 1 This is a steady-flow combustion process. 2 The combustion chamber is adiabatic. 3 There are no work interactions. 4 Air and the combustion gases are ideal gases. 5 Changes in kinetic and potential energies are negligible. Analysis (a) The balanced equation for the combustion process with the theoretical amount of air is C8H18 1 2 12.5 1 O2 h 3.76N2 2 S 8CO2 f a N r h°,r 9H2O 1 Nh°2 C8H18 f 47N2 The adiabatic flame temperature relation Hprod f a N p 1 h° – since all the reactants are at the standard reference state and h f° – and N2. The h f° and h values of various components at 298 K are – h f° kJ/kmol 249,950 0 0 241,820 393,520 – h 298 K kJ/kmol — 8682 8669 9904 9364 h° 2 p Hreact in this case reduces to 0 for O2 Substance C8H18( ) O2 N2 H2O(g) CO2 Substituting, we have 1 8 kmol CO2 2 3 1 393,520 1 47 kmol N2 2 3 1 0 1 9 kmol H 2O 2 3 1 241,820 h N2 h CO2 9364 2 kJ> kmol CO2 4 1 1 kmol C8H 18 2 1 249,950 kJ> kmol C8H 18 2 which yields 8669 2 kJ> kmol N2 4 h H2O 9904 2 kJ> kmol H 2O 4 8h CO2 9h H2O 47h N2 5,646,081 kJ It appears that we have one equation with three unknowns. Actually we have h(T ) only one unknown—the temperature of the products Tprod—since h for ideal gases. Therefore, we have to use an equation solver such as EES or a trial-and-error approach to determine the temperature of the products. A first guess is obtained by dividing the right-hand side of the equation by the total number of moles, which yields 5,646,081/(8 + 9 + 47) 88,220 kJ/kmol. This enthalpy value corresponds to about 2650 K for N2, 2100 K for H2O, and 1800 K for CO2. Noting that the majority of the moles are N2, we see that Tprod should be close to 2650 K, but somewhat under it. Therefore, a good first guess is 2400 K. At this temperature, 8h CO2 9h H2O 47h N2 8 125,152 9 103,508 47 79,320 5,660,828 kJ This value is higher than 5,646,081 kJ. Therefore, the actual temperature is slightly under 2400 K. Next we choose 2350 K. It yields 8 122,091 9 100,846 47 77,496 5,526,654 cen84959_ch15.qxd 9/15/06 6:06 AM Page 795 Chapter 15 which is lower than 5,646,081 kJ. Therefore, the actual temperature of the products is between 2350 and 2400 K. By interpolation, it is found to be Tprod 2395 K. (b) The balanced equation for the complete combustion process with 400 percent theoretical air is | 795 C8H18 1 2 50 1 O2 3.76N2 2 S 8CO2 9H2O 37.5O2 188N2 By following the procedure used in (a), the adiabatic flame temperature in this case is determined to be Tprod 962 K. Notice that the temperature of the products decreases significantly as a result of using excess air. (c) The balanced equation for the incomplete combustion process with 90 percent theoretical air is C8H18 1 2 11.25 1 O2 3.76N2 2 S 5.5CO2 2.5CO 9H2O 42.3N2 Following the procedure used in (a), we find the adiabatic flame temperature in this case to be Tprod 2236 K. Discussion Notice that the adiabatic flame temperature decreases as a result of incomplete combustion or using excess air. Also, the maximum adiabatic flame temperature is achieved when complete combustion occurs with the theoretical amount of air. 15–6 ENTROPY CHANGE OF REACTING SYSTEMS So far we have analyzed combustion processes from the conservation of mass and the conservation of energy points of view. The thermodynamic analysis of a process is not complete, however, without the examination of the second-law aspects. Of particular interest are the exergy and exergy destruction, both of which are related to entropy. The entropy balance relations developed in Chap. 7 are equally applicable to both reacting and nonreacting systems provided that the entropies of individual constituents are evaluated properly using a common basis. The entropy balance for any system (including reacting systems) undergoing any process can be expressed as Sin Sout 15253 Sgen 123 ¢ Ssystem 1 kJ> K 2 123 (15–18) Surroundings Net entropy transfer Entropy by heat and mass generation Change in entropy Using quantities per unit mole of fuel and taking the positive direction of heat transfer to be to the system, the entropy balance relation can be expressed more explicitly for a closed or steady-flow reacting system as (Fig. 15–28) Qk aT k S gen S prod S react 1 kJ> K 2 (15–19) Reactants Sreact Reaction chamber ΔSsys Products Sprod where Tk is temperature at the boundary where Qk crosses it. For an adiabatic process (Q 0), the entropy transfer term drops out and Eq. 15–19 reduces to S gen,adiabatic S prod S react 0 (15–20) FIGURE 15–28 The entropy change associated with a chemical relation. cen84959_ch15.qxd 9/15/06 6:06 AM Page 796 796 | Thermodynamics The total entropy generated during a process can be determined by applying the entropy balance to an extended system that includes the system itself and its immediate surroundings where external irreversibilities might be occurring. When evaluating the entropy transfer between an extended system and the surroundings, the boundary temperature of the extended system is simply taken to be the environment temperature, as explained in Chap. 7. The determination of the entropy change associated with a chemical reaction seems to be straightforward, except for one thing: The entropy relations for the reactants and the products involve the entropies of the components, not entropy changes, which was the case for nonreacting systems. Thus we are faced with the problem of finding a common base for the entropy of all substances, as we did with enthalpy. The search for such a common base led to the establishment of the third law of thermodynamics in the early part of this century. The third law was expressed in Chap. 7 as follows: The entropy of a pure crystalline substance at absolute zero temperature is zero. Therefore, the third law of thermodynamics provides an absolute base for the entropy values for all substances. Entropy values relative to this base are called the absolute entropy. The –° values listed in Tables A–18 through s A–25 for various gases such as N2, O2, CO, CO2, H2, H2O, OH, and O are the ideal-gas absolute entropy values at the specified temperature and at a pressure of 1 atm. The absolute – entropy values for various fuels are listed in Table A–26 together with the h° values at the standard reference state of f 25°C and 1 atm. Equation 15–20 is a general relation for the entropy change of a reacting system. It requires the determination of the entropy of each individual component of the reactants and the products, which in general is not very easy to do. The entropy calculations can be simplified somewhat if the gaseous components of the reactants and the products are approximated as ideal gases. However, entropy calculations are never as easy as enthalpy or internal energy calculations, since entropy is a function of both temperature and pressure even for ideal gases. When evaluating the entropy of a component of an ideal-gas mixture, we should use the temperature and the partial pressure of the component. Note that the temperature of a component is the same as the temperature of the mixture, and the partial pressure of a component is equal to the mixture pressure multiplied by the mole fraction of the component. 1 atm for any temAbsolute entropy values at pressures other than P0 perature T can be obtained from the ideal-gas entropy change relation written for an imaginary isothermal process between states (T,P0) and (T,P), as illustrated in Fig. 15–29: s 1 T,P 2 s ° 1 T,P0 2 R u ln P P0 (15–21) T P P0 =1 atm T s (T,P) s °(T,P0 ) (Tabulated) Δ s = – Ru ln P P0 s FIGURE 15–29 At a specified temperature, the absolute entropy of an ideal gas at pressures other than P0 1 atm can be determined by subtracting Ru ln (P/P0) from the tabulated value at 1 atm. For the component i of an ideal-gas mixture, this relation can be written as s i 1 T,Pi 2 s ° 1 T,P0 2 i R u ln y i Pm 1 kJ> kmol # K 2 P0 (15–22) where P0 1 atm, Pi is the partial pressure, yi is the mole fraction of the component, and Pm is the total pressure of the mixture. cen84959_ch15.qxd 9/15/06 6:06 AM Page 797 Chapter 15 If a gas mixture is at a relatively high pressure or low temperature, the deviation from the ideal-gas behavior should be accounted for by incorporating more accurate equations of state or the generalized entropy charts. | 797 15–7 SECOND-LAW ANALYSIS OF REACTING SYSTEMS Exergy ts Once the total entropy change or the entropy generation is evaluated, the exergy destroyed Xdestroyed associated with a chemical reaction can be determined from Xdestroyed T0 Sgen 1 kJ 2 (15–23) R c ea tan Reversible work Pro duc ts where T0 is the thermodynamic temperature of the surroundings. When analyzing reacting systems, we are more concerned with the changes in the exergy of reacting systems than with the values of exergy at various states (Fig. 15–30). Recall from Chap. 8 that the reversible work Wrev represents the maximum work that can be done during a process. In the absence of any changes in kinetic and potential energies, the reversible work relation for a steady-flow combustion process that involves heat transfer with only–the surroundings at T0 can be obtained by replacing the enthalpy – – terms by hf° h h°, yielding Wrev f a Nr 1 h° T, P State h h° T0 s 2 r f a Np 1 h° h h° T0 s 2 p (15–24) An interesting situation arises when both the reactants and the products are – – – – at the temperature of the surroundings T0. In that case, h T0 s (h T0 s )T0 – , which is, by definition, the Gibbs function of a unit mole of a subg0 stance at temperature T0. The Wrev relation in this case can be written as Wrev a N r g 0,r g° 2 r a N p g 0,p f a Np 1 g° FIGURE 15–30 The difference between the exergy of the reactants and of the products during a chemical reaction is the reversible work associated with that reaction. (15–25) or Wrev f a Nr 1 g° g T0 g T0 g° 2 p T0 = 25°C (15–26) Stable elements Compound 0 for stable elements like N2 and O2 at the standard reference state of 25°C and 1 atm, just like the – – g° represents the value of the sensible enthalpy of formation) and gT0 Gibbs function of a substance at temperature T0 relative to the standard reference state. For the very special case of Treact Tprod T0 25°C (i.e., the reactants, the products, and the surroundings are at 25°C) and the partial pressure Pi 1 atm for each component of the reactants and the products, Eq. 15–26 reduces to Wrev f a N r g °,r f a n p g °,p 1 kJ 2 – – where gf° is the Gibbs function of formation (gf° C + O2 → CO2 25°C, 1 atm 25°C, 1 atm 25°C, 1 atm –, Wrev = – gf° CO2 = 394,360 k J/ kmol 394,360 CO (15–27) We can conclude from the above equation that the of the Gibbs function of formation at 25°C and 1 atm) of a compound represents the reversible work associated with the formation of that compound from its stable elements at 25°C and 1 atm in an environment at – 25°C and 1 atm (Fig. 15–31). The gf° values of several substances are listed in Table A–26. – gf° value (the negative FIGURE 15–31 The negative of the Gibbs function of formation of a compound at 25 C, 1 atm represents the reversible work associated with the formation of that compound from its stable elements at 25 C, 1 atm in an environment that is at 25 C, 1 atm. cen84959_ch15.qxd 9/15/06 6:06 AM Page 798 798 | Thermodynamics T0 = 77°F P0 = 1 atm EXAMPLE 15–9 Reversible Work Associated with a Combustion Process C 77°F, 1 atm O2 77°F, 1 atm Combustion chamber CO2 77°F, 1 atm One lbmol of carbon at 77°F and 1 atm is burned steadily with 1 lbmol of oxygen at the same state as shown in Fig. 15–32. The CO2 formed during the process is then brought to 77°F and 1 atm, the conditions of the surroundings. Assuming the combustion is complete, determine the reversible work for this process. FIGURE 15–32 Schematic for Example 15–9. Solution Carbon is burned steadily with pure oxygen. The reversible work associated with this process is to be determined. Assumptions 1 Combustion is complete. 2 Steady-flow conditions exist during combustion. 3 Oxygen and the combustion gases are ideal gases. 4 Changes in kinetic and potential energies are negligible. Properties The Gibbs function of formation at 77°F and 1 atm is 0 for C and O2, and 169,680 Btu/lbmol for CO2. The enthalpy of formation is 0 for C and O2, and 169,300 Btu/lbmol for CO2. The absolute entropy is 1.36 Btu/lbmol · R for C, 49.00 Btu/lbmol · R for O2, and 51.07 Btu/lbmol · R for CO2 (Table A–26E). Analysis The combustion equation is C O2 S CO2 The C, O2, and CO2 are at 77°F and 1 atm, which is the standard reference state and also the state of the surroundings. Therefore, the reversible work in this case is simply the difference between the Gibbs function of formation of the reactants and that of the products (Eq. 15–27): Wrev f f a N r g °,r a N p g °,p 0 0 N C g °,C¡ N O2 g °,O¡ N CO2 g °,CO2 f f2 f 1 1 lbmol 2 1 169,680 Btu> lbmol 2 N CO2 g °,CO2 f 169,680 Btu – since the g f° of stable elements at 77°F and 1 atm is zero. Therefore, 169,680 Btu of work could be done as 1 lbmol of C is burned with 1 lbmol of O2 at 77°F and 1 atm in an environment at the same state. The reversible work in this case represents the exergy of the reactants since the product (the CO2) is at the state of the surroundings. Discussion We could also determine the reversible work without involving the Gibbs function by using Eq. 15–24: Wrev f a N r 1 h° f a N r 1 h° h T0 s ° 2 C N C 1 h° f T0 s 2 r h° N O2 1 h° f f a N p 1 h° f a N p 1 h° T0 s 2 p T0 s 2 r h h° T0 s 2 p T0 s ° 2 CO2 T0 s ° 2 O2 N CO2 1 h° f Substituting the enthalpy of formation and absolute entropy values, we obtain Wrev 1 1 lbmol C 2 3 0 1 1 lbmol O2 2 3 0 1 537 R 2 1 1.36 Btu> lbmol # R 2 4 11 lbmol CO2 2 3 169,300 Btu> lbmol 1 537 R 2 1 49.00 Btu> lbmol # R 2 4 1 537 R 2 1 51.07 Btu> lbmol # R 24 169,680 Btu which is identical to the result obtained before. cen84959_ch15.qxd 9/15/06 6:06 AM Page 799 Chapter 15 EXAMPLE 15–10 Second-Law Analysis of Adiabatic Combustion CH4 25°C, 1 atm AIR 25°C, 1 atm Adiabatic combustion chamber T0 = 25°C | 799 Methane (CH4) gas enters a steady-flow adiabatic combustion chamber at 25°C and 1 atm. It is burned with 50 percent excess air that also enters at 25°C and 1 atm, as shown in Fig. 15–33. Assuming complete combustion, determine (a) the temperature of the products, (b) the entropy generation, and (c) the reversible work and exergy destruction. Assume that T0 298 K and the products leave the combustion chamber at 1 atm pressure. CO2 H2O O2 N2 Solution Methane is burned with excess air in a steady-flow combustion chamber. The product temperature, entropy generated, reversible work, and exergy destroyed are to be determined. Assumptions 1 Steady-flow conditions exist during combustion. 2 Air and the combustion gases are ideal gases. 3 Changes in kinetic and potential energies are negligible. 4 The combustion chamber is adiabatic and thus there is no heat transfer. 5 Combustion is complete. Analysis (a) The balanced equation for the complete combustion process with 50 percent excess air is FIGURE 15–33 Schematic for Example 15–10. CH4 1 g 2 3 1 O2 3.76N2 2 S CO2 h° 2 p 2H2O O2 11.28N2 Under steady-flow conditions, the adiabatic flame temperature is determined from Hprod Hreact, which reduces to – O for O2 since all the reactants are at the standard reference state and h f° – and N2. Assuming ideal-gas behavior for air and for the products, the h f° and h values of various components at 298 K can be listed as – h f° kJ/kmol 74,850 0 0 241,820 393,520 – h 298 K kJ/kmol — 8682 8669 9904 9364 f a N p 1 h° h f a N r h°,r 1 Nh°2 CH4 f Substance CH4(g) O2 N2 H2O(g) CO2 Substituting, we have 1 1 kmol CO2 2 3 1 393,520 1 11.28 kmol N2 2 3 1 0 1 1 kmol O2 2 3 1 0 1 2 kmol H 2O 2 3 1 241,820 h O2 h CO2 9364 2 kJ> kmol CO2 4 h H2O h N2 1 1 kmol CH 4 2 1 74,850 kJ> kmol CH 4 2 which yields 8682 2 kJ> kmol O2 4 8669 2 kJ> kmol N2 4 9904 2 kJ> kmol H 2O 4 h CO2 2h H2O h O2 Tprod 11.28h N2 1789 K 937,950 kJ By trial and error, the temperature of the products is found to be cen84959_ch15.qxd 9/15/06 6:06 AM Page 800 800 | Thermodynamics (b) Noting that combustion is adiabatic, the entropy generation during this process is determined from Eq. 15–20: Sgen Sprod Sreact a Np s p a Nr s r – The CH4 is at 25°C and 1 atm, and thus its absolute entropy is s CH4 186.16 kJ/kmol K (Table A–26). The entropy values listed in the ideal-gas tables are for 1 atm pressure. Both the air and the product gases are at a total pressure of 1 atm, but the entropies are to be calculated at the partial pressure of the components, which is equal to Pi yiPtotal, where yi is the mole fraction of component i. From Eq. 15–22: Si N i s i 1 T, Pi 2 yi 1.00 0.21 0.79 0.0654 0.1309 0.0654 0.7382 N i 3 s ° 1 T, P0 2 i – s i°(T, 1 atm) 186.16 205.04 191.61 302.517 258.957 264.471 247.977 R u ln y iPm 4 Ru ln yiPm — 12.98 1.96 Sreact 22.674 16.905 22.674 2.524 Sprod – Ni s i 186.16 654.06 2183.47 3023.69 325.19 551.72 287.15 2825.65 3989.71 The entropy calculations can be represented in tabular form as follows: Ni CH4 O2 N2 CO2 H2O O2 N2 1 3 11.28 1 2 1 11.28 Thus, S gen S prod S react 966.0 kJ/kmol # K 1 3989.71 3023.69 2 kJ> kmol # K CH 4 (c) The exergy destruction or irreversibility associated with this process is determined from Eq. 15–23, X destroyed T0 S gen 1 298 K 2 1 966.0 kJ> kmol # K 2 288 MJ/kmol CH4 That is, 288 MJ of work potential is wasted during this combustion process for each kmol of methane burned. This example shows that even complete combustion processes are highly irreversible. This process involves no actual work. Therefore, the reversible work and exergy destroyed are identical: Wrev T0 = 25°C CH4 25°C, 1 atm AIR 25°C, 1 atm Combustion chamber CO2 H2O 1 atm O2 N2 25°C, 288 MJ/kmol CH4 That is, 288 MJ of work could be done during this process but is not. Instead, the entire work potential is wasted. EXAMPLE 15–11 FIGURE 15–34 Schematic for Example 15–11. Second-Law Analysis of Isothermal Combustion Methane (CH4) gas enters a steady-flow combustion chamber at 25°C and 1 atm and is burned with 50 percent excess air, which also enters at 25°C and 1 atm, as shown in Fig. 15–34. After combustion, the products are allowed to cool to 25°C. Assuming complete combustion, determine cen84959_ch15.qxd 9/15/06 6:06 AM Page 801 Chapter 15 (a) the heat transfer per kmol of CH4, (b) the entropy generation, and (c) the reversible work and exergy destruction. Assume that T0 298 K and the products leave the combustion chamber at 1 atm pressure. | 801 Solution This is the same combustion process we discussed in Example 15–10, except that the combustion products are brought to the state of the surroundings by transferring heat from them. Thus the combustion equation remains the same: CH4 1 g 2 3 1 O2 3.76N2 2 S CO2 2H2O O2 11.28N2 At 25°C, part of the water will condense. The amount of water vapor that remains in the products is determined from (see Example 15–3) Nv N gas and Pv Ptotal 3.1698 kPa 101.325 kPa 0.03128 Nv a Pv Ptotal b N gas 1 0.03128 2 1 13.28 Nv 2 S Nv 0.43 kmol Therefore, 1.57 kmol of the H2O formed is in the liquid form, which is removed at 25°C and 1 atm. When one is evaluating the partial pressures of the components in the product gases, the only water molecules that need to be considered are those that are in the vapor phase. As before, all the gaseous reactants and products are treated as ideal gases. (a) Heat transfer during this steady-flow combustion process is determined from the steady-flow energy balance Eout Ein on the combustion chamber, Q out f a N p h °,p f a N r h °,r since all the reactants and products are at the standard reference of 25°C and the enthalpy of ideal gases depends on temperature only. Solving for – Qout and substituting the h f° values, we have Q out 1 1 kmol CO2 2 1 393,520 kJ> kmol CO2 2 1 1 kmol CH 4 2 1 74,850 kJ> kmol CH 4 2 3 0.43 kmol H 2O 1 g 2 4 3 241,820 kJ> kmol H 2O 1 g 2 4 3 1.57 kmol H 2O 1 2 4 3 285.830 kJ> kmol H 2O 1 2 4 871,400 kJ/kmol CH4 (b) The entropy of the reactants was evaluated in Example 15–10 and was 3023.69 kJ/kmol · K CH4. By following a similar determined to be Sreact approach, the entropy of the products is determined to be Ni H2O( ) H2O CO2 O2 N2 1.57 0.43 1 1 11.28 yi 1.0000 0.0314 0.0729 0.0729 0.8228 – s i°(T, 1 atm) 69.92 188.83 213.80 205.04 191.61 Ru ln yiPm — 28.77 21.77 21.77 1.62 Sprod – Ni s i 109.77 93.57 235.57 226.81 2179.63 2845.35 cen84959_ch15.qxd 9/15/06 6:06 AM Page 802 802 | Thermodynamics Then the total entropy generation during this process is determined from an entropy balance applied on an extended system that includes the immediate surroundings of the combustion chamber S gen S prod S react Q out Tsurr 3023.69 2 kJ> kmol 1 2845.35 871,400 kJ> kmol 298 K 2746 kJ/kmol # K CH4 (c) The exergy destruction and reversible work associated with this process are determined from X destroyed T0 S gen 1 298 K 2 1 2746 kJ> kmol # K 2 818 MJ/kmol CH4 and Wrev X destroyed 818 MJ/kmol CH4 since this process involves no actual work. Therefore, 818 MJ of work could be done during this process but is not. Instead, the entire work potential is wasted. The reversible work in this case represents the exergy of the reactants before the reaction starts since the products are in equilibrium with the surroundings, that is, they are at the dead state. Discussion Note that, for simplicity, we calculated the entropy of the product gases before they actually entered the atmosphere and mixed with the atmospheric gases. A more complete analysis would consider the composition of the atmosphere and the mixing of the product gases with the gases in the atmosphere, forming a homogeneous mixture. There is additional entropy generation during this mixing process, and thus additional wasted work potential. TOPIC OF SPECIAL INTEREST* Fuel Cells Fuels like methane are commonly burned to provide thermal energy at high temperatures for use in heat engines. However, a comparison of the reversible works obtained in the last two examples reveals that the exergy of the reactants (818 MJ/kmol CH4) decreases by 288 MJ/kmol as a result of the irreversible adiabatic combustion process alone. That is, the exergy of the hot combustion gases at the end of the adiabatic combustion process is 818 288 530 MJ/kmol CH4. In other words, the work potential of the hot combustion gases is about 65 percent of the work potential of the reactants. It seems that when methane is burned, 35 percent of the work potential is lost before we even start using the thermal energy (Fig. 15–35). Thus, the second law of thermodynamics suggests that there should be a better way of converting the chemical energy to work. The better way is, of course, the less irreversible way, the best being the reversible case. In chemi*This section can be skipped without a loss in continuity. cen84959_ch15.qxd 9/15/06 6:06 AM Page 803 Chapter 15 cal reactions, the irreversibility is due to uncontrolled electron exchange between the reacting components. The electron exchange can be controlled by replacing the combustion chamber by electrolytic cells, like car batteries. (This is analogous to replacing unrestrained expansion of a gas in mechanical systems by restrained expansion.) In the electrolytic cells, the electrons are exchanged through conductor wires connected to a load, and the chemical energy is directly converted to electric energy. The energy conversion devices that work on this principle are called fuel cells. Fuel cells are not heat engines, and thus their efficiencies are not limited by the Carnot efficiency. They convert chemical energy to electric energy essentially in an isothermal manner. A fuel cell functions like a battery, except that it produces its own electricity by combining a fuel with oxygen in a cell electrochemically without combustion, and discards the waste heat. A fuel cell consists of two electrodes separated by an electrolyte such as a solid oxide, phosphoric acid, or molten carbonate. The electric power generated by a single fuel cell is usually too small to be of any practical use. Therefore, fuel cells are usually stacked in practical applications. This modularity gives the fuel cells considerable flexibility in applications: The same design can be used to generate a small amount of power for a remote switching station or a large amount of power to supply electricity to an entire town. Therefore, fuel cells are termed the “microchip of the energy industry.” The operation of a hydrogen–oxygen fuel cell is illustrated in Fig. 15–36. Hydrogen is ionized at the surface of the anode, and hydrogen ions flow through the electrolyte to the cathode. There is a potential difference between the anode and the cathode, and free electrons flow from the anode to the cathode through an external circuit (such as a motor or a generator). Hydrogen ions combine with oxygen and the free electrons at the surface of the cathode, forming water. Therefore, the fuel cell operates like an electrolysis system working in reverse. In steady operation, hydrogen and oxygen continuously enter the fuel cell as reactants, and water leaves as the product. Therefore, the exhaust of the fuel cell is drinkable quality water. The fuel cell was invented by William Groves in 1839, but it did not receive serious attention until the 1960s, when they were used to produce electricity and water for the Gemini and Apollo spacecraft during their missions to the moon. Today they are used for the same purpose in the space shuttle missions. Despite the irreversible effects such as internal resistance to electron flow, fuel cells have a great potential for much higher conversion efficiencies. Currently fuel cells are available commercially, but they are competitive only in some niche markets because of their higher cost. Fuel cells produce high-quality electric power efficiently and quietly while generating low emissions using a variety of fuels such as hydrogen, natural gas, propane, and biogas. Recently many fuel cells have been installed to generate electricity. For example, a remote police station in Central Park in New York City is powered by a 200-kW phosphoric acid fuel cell that has an efficiency of 40 percent with negligible emissions (it emits 1 ppm NOx and 5 ppm CO). 25°C REACTANTS (CH4, air) Exergy = 818 MJ (100%) | 803 Adiabatic combustion chamber 1789 K PRODUCTS Exergy = 530 MJ (65%) FIGURE 15–35 The availability of methane decreases by 35 percent as a result of irreversible combustion process. 2e– Load H2 2e– H2 2H+ 2e– O2 O2 Porous anode H2O Porous cathode FIGURE 15–36 The operation of a hydrogen–oxygen fuel cell. cen84959_ch15.qxd 9/15/06 6:06 AM Page 804 804 | Thermodynamics Hybrid power systems (HPS) that combine high-temperature fuel cells and gas turbines have the potential for very high efficiency in converting natural gas (or even coal) to electricity. Also, some car manufacturers are planning to introduce cars powered by fuel-cell engines, thus more than doubling the efficiency from less than 30 percent for the gasoline engines to up to 60 percent for fuel cells. In 1999, DaimlerChrysler unveiled its hydrogen fuel-cell powered car called NECAR IV that has a refueling range of 280 miles and can carry 4 passengers at 90 mph. Some research programs to develop such hybrid systems with an efficiency of at least 70 percent by 2010 are under way. SUMMARY Any material that can be burned to release energy is called a fuel, and a chemical reaction during which a fuel is oxidized and a large quantity of energy is released is called combustion. The oxidizer most often used in combustion processes is air. The dry air can be approximated as 21 percent oxygen and 79 percent nitrogen by mole numbers. Therefore, 1 kmol O2 3.76 kmol N2 4.76 kmol air standard reference state for all substances, which is chosen to be 25°C (77°F) and 1 atm. The difference between the enthalpy of the products at a specified state and the enthalpy of the reactants at the same state for a complete reaction is called the enthalpy of reaction hR. For combustion processes, the enthalpy of reaction is usually referred to as the enthalpy of combustion hC, which represents the amount of heat released during a steadyflow combustion process when 1 kmol (or 1 kg) of fuel is burned completely at a specified temperature and pressure. The enthalpy of a substance at a specified state due to its chemical composition is called the enthalpy of formation hf . The enthalpy of formation of all stable elements is assigned a value of zero at the standard reference state of 25°C and 1 atm. The heating value of a fuel is defined as the amount of heat released when a fuel is burned completely in a steadyflow process and the products are returned to the state of the reactants. The heating value of a fuel is equal to the absolute value of the enthalpy of combustion of the fuel, Heating value During a combustion process, the components that exist before the reaction are called reactants and the components that exist after the reaction are called products. Chemical equations are balanced on the basis of the conservation of mass principle, which states that the total mass of each element is conserved during a chemical reaction. The ratio of the mass of air to the mass of fuel during a combustion process is called the air–fuel ratio AF: AF m air m fuel (Ni Mi)fuel. where m air (NM )air and m fuel A combustion process is complete if all the carbon in the fuel burns to CO2, all the hydrogen burns to H2O, and all the sulfur (if any) burns to SO2. The minimum amount of air needed for the complete combustion of a fuel is called the stoichiometric or theoretical air. The theoretical air is also referred to as the chemically correct amount of air or 100 percent theoretical air. The ideal combustion process during which a fuel is burned completely with theoretical air is called the stoichiometric or theoretical combustion of that fuel. The air in excess of the stoichiometric amount is called the excess air. The amount of excess air is usually expressed in terms of the stoichiometric air as percent excess air or percent theoretical air. During a chemical reaction, some chemical bonds are broken and others are formed. Therefore, a process that involves chemical reactions involves changes in chemical energies. Because of the changed composition, it is necessary to have a 0 hC 0 1 kJ> kg fuel 2 Taking heat transfer to the system and work done by the system to be positive quantities, the conservation of energy relation for chemically reacting steady-flow systems can be expressed per unit mole of fuel as Q W f a Np 1 h° h h° 2 p f a Nr 1 h° h h° 2 r where the superscript ° represents properties at the standard reference state of 25°C and 1 atm. For a closed system, it becomes Q W f a Np 1 h° h h° P v 2 p f a Nr 1 h° h h° P v 2 r The P v terms are negligible for solids and liquids and can be replaced by RuT for gases that behave as ideal gases. cen84959_ch15.qxd 9/15/06 6:06 AM Page 805 Chapter 15 In the absence of any heat loss to the surroundings (Q 0), the temperature of the products will reach a maximum, which is called the adiabatic flame temperature of the reaction. The adiabatic flame temperature of a steady-flow combustion process is determined from Hprod Hreact or f a Np 1 h° | 805 s 1 T, P 2 s ° 1 T, P0 2 Ru ln P P0 For component i of an ideal-gas mixture, this relation can be written as s i 1 T, Pi 2 s ° 1 T, P0 2 i R u ln y i Pm P0 h h° 2 p f a Nr 1 h° h h° 2 r Taking the positive direction of heat transfer to be to the system, the entropy balance relation can be expressed for a closed system or steady-flow combustion chamber as Qk aT k S gen,adiabatic Sgen Sprod Sreact where Pi is the partial pressure, yi is the mole fraction of the component, and Pm is the total pressure of the mixture in atmospheres. The exergy destruction and the reversible work associated with a chemical reaction are determined from X destroyed Wrev T0 s 2 r For an adiabatic process it reduces to S prod S react 0 and Wrev Wact T0 S gen T0 s 2 p The third law of thermodynamics states that the entropy of a pure crystalline substance at absolute zero temperature is zero. The third law provides a common base for the entropy of all substances, and the entropy values relative to this base are called the absolute entropy. The ideal-gas tables list the absolute entropy values over a wide range of temperatures but at a fixed pressure of P0 1 atm. Absolute entropy values at other pressures P for any temperature T are determined from f a N r 1 h° h h° f a N p 1 h° h h° When both the reactants and the products are at the temperature of the surroundings T0, the reversible work can be expressed in terms of the Gibbs functions as Wrev f a Nr 1 g° g T0 g° 2 r f a Np 1 g° g T0 g° 2 p REFERENCES AND SUGGESTED READINGS 1. S. W. Angrist. Direct Energy Conversion. 4th ed. Boston: Allyn and Bacon, 1982. 2. I. Glassman. Combustion. New York: Academic Press, 1977. 3. R. Strehlow. Fundamentals of Combustion. Scranton, PA: International Textbook Co., 1968. PROBLEMS* Fuels and Combustion 15–1C What are the approximate chemical compositions of gasoline, diesel fuel, and natural gas? *Problems designated by a “C” are concept questions, and students are encouraged to answer them all. Problems designated by an “E” are in English units, and the SI users can ignore them. Problems with the icon are solved using EES, and complete solutions together with parametric studies are included on the enclosed DVD. Problems with the icon are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text. 15–2C How does the presence of N2 in air affect the outcome of a combustion process? 15–3C How does the presence of moisture in air affect the outcome of a combustion process? 15–4C What does the dew-point temperature of the product gases represent? How is it determined? 15–5C Is the number of atoms of each element conserved during a chemical reaction? How about the total number of moles? 15–6C What is the air–fuel ratio? How is it related to the fuel–air ratio? cen84959_ch15.qxd 9/15/06 6:06 AM Page 806 806 | Thermodynamics 15–18 Acetylene (C2H2) is burned with 10 percent excess oxygen in a cutting torch. Determine the mass fraction of each of the products. Calculate the mass of oxygen used per unit mass of acetylene burned. 15–19 Determine the fuel-air ratio when coal from Colorado which has an ultimate analysis (by mass) as 79.61 percent C, 4.66 percent H2, 4.76 percent O2, 1.83 percent N2, 0.52 percent S, and 8.62 percent ash (non-combustibles) is burned with 50 percent excess air. Answer: 0.0576 kg fuel/kg air Coal Combustion chamber CO2, H2O SO2, O2, N2 15–7C Is the air–fuel ratio expressed on a mole basis identical to the air–fuel ratio expressed on a mass basis? 15–8 Trace amounts of sulfur (S) in coal are burned in the presence of diatomic oxygen (O2) to form sulfur dioxide (SO2). Determine the minimum mass of oxygen required in the reactants and the mass of sulfur dioxide in the products when 1 kg of sulfur is burned. 15–9E Methane (CH4) is burned in the presence of diatomic oxygen. The combustion products consist of water vapor and carbon dioxide gas. Determine the mass of water vapor generated when 1 lbm of methane is burned. Answer: 2.25 lbm H2O/lbm fuel Theoretical and Actual Combustion Processes 15–10C What are the causes of incomplete combustion? 15–11C Which is more likely to be found in the products of an incomplete combustion of a hydrocarbon fuel, CO or OH? Why? 15–12C What does 100 percent theoretical air represent? 15–13C Are complete combustion and theoretical combustion identical? If not, how do they differ? 15–14C Consider a fuel that is burned with (a) 130 percent theoretical air and (b) 70 percent excess air. In which case is the fuel burned with more air? 15–15 Propane fuel (C3H8) is burned in the presence of air. Assuming that the combustion is theoretical—that is, only nitrogen (N2), water vapor (H2O), and carbon dioxide (CO2) are present in the products—determine (a) the mass fraction of carbon dioxide and (b) the mole and mass fractions of the water vapor in the products. C3H8 Combustion chamber CO2 H2O N2 Air 50% excess FIGURE P15–19 15–20 Propane (C3H8) is burned with 75 percent excess air during a combustion process. Assuming complete combustion, determine the air–fuel ratio. Answer: 27.5 kg air/kg fuel 15–21 Acetylene (C2H2) is burned with stoichiometric amount of air during a combustion process. Assuming complete combustion, determine the air–fuel ratio on a mass and on a mole basis. 15–22E Ethylene (C2H4) is burned with 200 percent theoretical air during a combustion process. Assuming complete combustion and a total pressure of 14.5 psia, determine (a) the air–fuel ratio and (b) the dew-point temperature of the products. Answers: (a) 29.6 lbm air/lbm fuel, (b) 101°F 15–23 Propylene (C3H6) is burned with 50 percent excess air during a combustion process. Assuming complete combustion and a total pressure of 105 kPa, determine (a) the air–fuel ratio and (b) the temperature at which the water vapor in the products will start condensing. 15–24 Butane (C4H10) is burned in 200 percent theoretical air. For complete combustion, how many kmol of water must be sprayed into the combustion chamber per kmol of fuel if the products of combustion are to have a dew-point temperature of 60°C when the product pressure is 100 kPa? 15–25 A fuel mixture of 20 percent by mass methane (CH4) and 80 percent by mass ethanol (C2H6O), is burned completely with theoretical air. If the total flow rate of the fuel is 31 kg/s, determine the required flow rate of air. Answer: 330 kg/s 15–26 Octane (C8H18) is burned with 250 percent theoretical air, which enters the combustion chamber at 25°C. Assuming complete combustion and a total pressure of 1 atm, determine (a) the air–fuel ratio and (b) the dew-point temperature of the products. Theoretical air FIGURE P15–15 15–16 5.9 The fuel mixer in a natural gas burner mixes methane (CH4) with air to form a combustible mixture at the outlet. Determine the mass flow rates at the two inlets needed to produce 0.01 kg/s of an ideal combustion mixture at the outlet. 15–17 n-Octane (C8H18) is burned with stoichiometric amount of oxygen. Calculate the mass fractions of each of the products and the mass of water in the products per unit mass of fuel burned. cen84959_ch15.qxd 9/15/06 6:06 AM Page 807 Chapter 15 C8H18 Combustion chamber AIR 25°C P = 1 atm Products | 807 15–35 A gaseous fuel with a volumetric analysis of 60 percent CH4, 30 percent H2, and 10 percent N2 is burned to completion with 130 percent theoretical air. Determine (a) the air–fuel ratio and (b) the fraction of water vapor that would condense if the product gases were cooled to 20°C at 1 atm. Answers: (a) 18.6 kg air/kg fuel, (b) 88 percent FIGURE P15–26 15–27 Gasoline (assumed C8H18) is burned steadily with air in a jet engine. If the air–fuel ratio is 18 kg air/kg fuel, determine the percentage of theoretical air used during this process. 15–28 n-Butane fuel (C4H10) is burned with stoichiometric amount of air. Determine the mass fraction of each product. Also, calculate the mass of carbon dioxide in the products and mass of air required per unit of fuel mass burned. 15–29 A coal from Pennsylvania which has an ultimate analysis (by mass) as 84.36 percent C, 1.89 percent H2, 4.40 percent O2, 0.63 percent N2, 0.89 percent S, and 7.83 percent ash (non-combustibles) is burned with stoichiometric amount of air. Calculate the mole fractions of the products, the apparent molecular weight of the product gas, and the air-fuel ratio required neglecting the ash constituent. 15–30 Methyl alcohol (CH3OH) is burned with stoichiometric amount of air. Calculate the mole fractions of each of the products, and the apparent molar mass of the product gas. Also, calculate the mass of water in the products per unit mass of fuel burned. Answers: 0.116 (CO2), 0.231 (H2O), 0.653 (N2), 27.5 kg/kmol, 1.13 kg H2O/kg fuel Reconsider Prob. 15–35. Using EES (or other) software, study the effects of varying the percentages of CH4, H2, and N2 making up the fuel and the product gas temperature in the range 5 to 150°C. 15–37 Carbon (C) is burned with dry air. The volumetric analysis of the products is 10.06 percent CO2, 0.42 percent CO, 10.69 percent O2, and 78.83 percent N2. Determine (a) the air–fuel ratio and (b) the percentage of theoretical air used. 15–38 Methane (CH4) is burned with dry air. The volumetric analysis of the products on a dry basis is 5.20 percent CO2, 0.33 percent CO, 11.24 percent O2, and 83.23 percent N2. Determine (a) the air–fuel ratio and (b) the percentage of theoretical air used. Answers: (a) 34.5 kg air/kg fuel, (b) 200 percent 15–39 Methyl alcohol (CH3OH) is burned with 50 percent excess air. The combustion is incomplete with 10 percent of the carbon in the fuel forming carbon monoxide. Calculate the mole fraction of carbon monoxide and the apparent molecular weight of the products. CH3OH Combustion chamber CO2, CO H2O, O2, N2 15–36 15–31 A coal from Utah which has an ultimate analysis (by mass) as 61.40 percent C, 5.79 percent H2, 25.31 percent O2, 1.09 percent N2, 1.41 percent S, and 5.00 percent ash (noncombustibles) is burned with stoichiometric amount of air, but the combustion is incomplete with 5 percent of the carbon in the fuel forming carbon monoxide. Calculate the mass fraction and the apparent molecular weight of the products, and the mass of air required per unit mass of fuel burned. 15–32 Propane fuel (C3H8) is burned with 30 percent excess air. Determine the mole fractions of each of the products. Also, calculate the mass of water in the products per unit mass of the fuel and the air-fuel ratio. 15–33 A certain natural gas has the following volumetric analysis: 65 percent CH4, 8 percent H2, 18 percent N2, 3 percent O2, and 6 percent CO2. This gas is now burned completely with the stoichiometric amount of dry air. What is the air–fuel ratio for this combustion process? 15–34 Repeat Prob. 15–33 by replacing the dry air by moist air that enters the combustion chamber at 25°C, 1 atm, and 85 percent relative humidity. Air 50% excess FIGURE P15–39 15–40 n-Octane (C8H18) is burned with 100 percent excess air with 15 percent of the carbon in the fuel forming carbon monoxide. Calculate the mole fractions of the products and the dew-point temperature of the water vapor in the products when the products are at 1 atm pressure. Answers: 0.0548 (CO2), 0.0097 (CO), 0.0725 (H2O), 0.1056 (O2), 0.7575 (N2), 39.9 C Enthalpy of Formation and Enthalpy of Combustion 15–41C What is enthalpy of combustion? How does it differ from the enthalpy of reaction? 15–42C What is enthalpy of formation? How does it differ from the enthalpy of combustion? 15–43C What are the higher and the lower heating values of a fuel? How do they differ? How is the heating value of a fuel related to the enthalpy of combustion of that fuel? cen84959_ch15.qxd 9/15/06 6:06 AM Page 808 808 | Thermodynamics cal air, and (c) the chemically correct amount of pure oxygen. For which case will the amount of heat transfer be the lowest? Explain. 15–58 Propane fuel (C3H8) is burned with an air-fuel ratio of 18 in an atmospheric pressure heating furnace. Determine the heat transfer per kilogram of fuel burned when the temperature of the products is such that liquid water just begins to form in the products. 15–59 n-Octane gas (C8H18) is burned with 100 percent excess air in a constant pressure burner. The air and fuel enter this burner steadily at standard conditions and the products of combustion leave at 257 C. Calculate the heat transfer, in kJ/kg fuel, during this combustion. Qout C8H18 25 C 100% excess air 25 C Combustion chamber P 1 atm 15–44C When are the enthalpy of formation and the enthalpy of combustion identical? 15–45C Does the enthalpy of formation of a substance change with temperature? 15–46C The h° of N2 is listed as zero. Does this mean that f N2 contains no chemical energy at the standard reference state? 15–47C Which contains more chemical energy, 1 kmol of H2 or 1 kmol of H2O? 15–48 Determine the enthalpy of combustion of methane (CH4) at 25°C and 1 atm, using the enthalpy of formation data from Table A–26. Assume that the water in the products is in the liquid form. Compare your result to the value listed in Table A–27. Answer: –890,330 kJ/kmol 15–49 Reconsider Prob. 15–48. Using EES (or other) software, study the effect of temperature on the enthalpy of combustion. Plot the enthalpy of combustion as a function of temperature over the range 25 to 600°C. 15–50 15–51 Repeat Prob. 15–48 for gaseous ethane (C2H6). Repeat Prob. 15–48 for liquid octane (C8H18). Products 257 C 15–52 Ethane (C2H6) is burned at atmospheric pressure with stoichiometric amount of air as the oxidizer. Determine the heat rejected, in kJ/kmol fuel, when the products and reactants are at 25 C, and the water appears in the products as water vapor. 15–53 What is the minimum pressure of the products of Prob. 15–52 which will assure that the water in the products will be in vapor form? 15–54 Calculate the higher and lower heating values of a coal from Utah which has an ultimate analysis (by mass) as 61.40 percent C, 5.79 percent H2, 25.31 percent O2, 1.09 percent N2, 1.41 percent S, and 5.00 percent ash (non-combustibles). The enthalpy of formation of SO2 is 297,100 kJ/kmol. Answers: 30,000 kJ/kg, 28,700 kJ/kg FIGURE P15–59 15–60 A coal from Texas which has an ultimate analysis (by mass) as 39.25 percent C, 6.93 percent H2, 41.11 percent O2, 0.72 percent N2, 0.79 percent S, and 11.20 percent ash (non-combustibles) is burned steadily with 40 percent excess air in a power plant boiler. The coal and air enter this boiler at standard conditions and the products of combustion in the smokestack are at 127 C. Calculate the heat transfer, in kJ/kg fuel, in this boiler. Include the effect of the sulfur in the energy analysis by noting that sulfur dioxide has an enthalpy of formation of –297,100 kJ/kmol and an average specific heat at constant pressure of cp 41.7 kJ/kmol # K. 15–61 Methane (CH4) is burned completely with the stoichiometric amount of air during a steady-flow combustion process. If both the reactants and the products are maintained at 25°C and 1 atm and the water in the products exists in the liquid form, determine the heat transfer from the combustion chamber during this process. What would your answer be if combustion were achieved with 100 percent excess air? Answer: 890,330 kJ/kmol First-Law Analysis of Reacting Systems 15–55C Derive an energy balance relation for a reacting closed system undergoing a quasi-equilibrium constant pressure expansion or compression process. 15–56C Consider a complete combustion process during which both the reactants and the products are maintained at the same state. Combustion is achieved with (a) 100 percent theoretical air, (b) 200 percent theoretical air, and (c) the chemically correct amount of pure oxygen. For which case will the amount of heat transfer be the highest? Explain. 15–57C Consider a complete combustion process during which the reactants enter the combustion chamber at 20°C and the products leave at 700°C. Combustion is achieved with (a) 100 percent theoretical air, (b) 200 percent theoreti- 15–62 Liquid propane (C3H8) enters a combustion chamber at 25°C at a rate of 1.2 kg/min where it is mixed and burned with 150 percent excess air that enters the combustion chamber at 12°C. If the combustion is complete and the exit temperature of the combustion gases is 1200 K, determine (a) the mass flow rate of air and (b) the rate of heat transfer from the combustion chamber. Answers: (a) 47.1 kg/min, (b) 5194 kJ/min cen84959_ch15.qxd 9/15/06 6:06 AM Page 809 Chapter 15 · Q C3H8 25°C AIR 12°C Combustion chamber Products 1200 K out | 809 Qout C2H6 25°C AIR 500 K H2O CO2 Combustion CO chamber 800 K O 2 N2 · FIGURE P15–62 15–63E Liquid octane (C8H18) at 77°F is burned completely during a steady-flow combustion process with 180 percent theoretical air that enters the combustion chamber at 77°F. If the products leave at 2500 R, determine (a) the air–fuel ratio and (b) the heat transfer from the combustion chamber during this process. 15–64 Benzene gas (C6H6) at 25°C is burned during a steady-flow combustion process with 95 percent theoretical air that enters the combustion chamber at 25°C. All the hydrogen in the fuel burns to H2O, but part of the carbon burns to CO. If the products leave at 1000 K, determine (a) the mole fraction of the CO in the products and (b) the heat transfer from the combustion chamber during this process. Answers: (a) 2.1 percent, (b) 2,112,800 kJ/kmol C6H6 15–65E Diesel fuel (C12H26) at 77°F is burned in a steadyflow combustion chamber with 20 percent excess air that also enters at 77°F. The products leave the combustion chamber at 800 R. Assuming combustion is complete, determine the required mass flow rate of the diesel fuel to supply heat at a rate of 1800 Btu/s. Answer: 0.1 lbm/s Octane gas (C8H18) at 25°C is burned steadily with 30 percent excess air at 25°C, 1 atm, and 60 percent relative humidity. Assuming combustion is complete and the products leave the combustion chamber at 600 K, determine the heat transfer for this process per unit mass of octane. 15–67 Reconsider Prob. 15–66. Using EES (or other) software, investigate the effect of the amount of excess air on the heat transfer for the combustion process. Let the excess air vary from 0 to 200 percent. Plot the heat transfer against excess air, and discuss the results. 15–68 Ethane gas (C2H6) at 25°C is burned in a steady-flow combustion chamber at a rate of 5 kg/h with the stoichiometric amount of air, which is preheated to 500 K before entering the combustion chamber. An analysis of the combustion gases reveals that all the hydrogen in the fuel burns to H2O but only 95 percent of the carbon burns to CO2, the remaining 5 percent forming CO. If the products leave the combustion chamber at 800 K, determine the rate of heat transfer from the combustion chamber. Answer: 200,170 kJ/h 15–66 15–69 FIGURE P15–68 A constant-volume tank contains a mixture of 120 g of methane (CH4) gas and 600 g of O2 at 25°C and 200 kPa. The contents of the tank are now ignited, and the methane gas burns completely. If the final temperature is 1200 K, determine (a) the final pressure in the tank and (b) the heat transfer during this process. 15–70 Reconsider Prob. 15–69. Using EES (or other) software, investigate the effect of the final temperature on the final pressure and the heat transfer for the combustion process. Let the final temperature vary from 500 to 1500 K. Plot the final pressure and heat transfer against the final temperature, and discuss the results. 15–71 Propane fuel (C3H8) is burned in a space heater with 50 percent excess air. The fuel and air enter this heater steadily at 1 atm and 17 C, while the combustion products leave at 1 atm and 97 C. Calculate the heat transferred in this heater, in kJ/kmol fuel. Answer: 1,953,000 kJ/kmol fuel 15–72E One lbmol of methane (CH4) undergoes complete combustion with stoichiometric amount of air in a rigid container. Initially, the air and methane are at 14.4 psia and 77 F. The products of combustion are at 840 F. How much heat is rejected from the container, in Btu/lbmol fuel? Qout CH4 Theoretical air 14.4 psia, 77 F 840 F FIGURE P1572E 15–73 The effective mass-specific heat product (mcv) of a constant volume bomb calorimeter is 100 kJ/K. Wheat straw that is being considered as an alternative fuel is tested in the calorimeter. Ten grams of this straw are placed in the calorimeter. After charging the calorimeter with oxygen and burning the straw, it is found that the temperature of the calorimeter has increased by 1.8 C. Determine the heating value of this straw. How does this compare to the higher heating value of propane fuel? cen84959_ch15.qxd 9/15/06 6:06 AM Page 810 810 | Thermodynamics H2 7°C AIR 7°C Combustion chamber Products Tprod 15–74 A constant-volume tank contains a mixture of 1 kmol of benzene (C6H6) gas and 30 percent excess air at 25°C and 1 atm. The contents of the tank are now ignited, and all the hydrogen in the fuel burns to H2O but only 92 percent of the carbon burns to CO2, the remaining 8 percent forming CO. If the final temperature in the tank is 1000 K, determine the heat transfer from the combustion chamber during this process. Qout C6H6 30% excess air 25°C 1 atm FIGURE P15–79 15–80 Reconsider Prob. 15–79. Using EES (or other) software, modify this problem to include the fuels butane, ethane, methane, and propane as well as H2; to include the effects of inlet air and fuel temperatures; and the percent theoretical air supplied. Select a range of input parameters and discuss the results for your choices. 15–81E Hydrogen (H2) at 40°F is burned with 20 percent excess air that is also at 40°F during an adiabatic steady-flow combustion process. Assuming complete combustion, find the exit temperature of the product gases. 15–82 What is the adiabatic flame temperature of methane (CH4) when it is burned with 30 percent excess air? 15–83 Liquid octane (C8H18) is burned in the constant pressure, adiabatic combustor of an aircraft engine with 40 percent excess air. The air enters this combustor at 600 kPa and 307 C, and the fuel is injected into the combustor at 25 C. Estimate the temperature at which the products of combustion leave the combustor. Answer: 2113 K 15–84 A large railroad has experimented with burning powdered coal in a gas turbine combustor. Fifty percent excess air was introduced to the combustor at 1380 kPa and 127 C while the powdered coal was injected at 25 C. The combustion was adiabatic and at constant pressure. Based on a coal from Colorado which has an ultimate analysis (by mass) as 79.61 percent C, 4.66 percent H2, 4.76 percent O2, 1.83 percent N2, 0.52 percent S, and 8.62 percent ash (noncombustibles), what is the estimated temperature of the combustion products? Neglect the effect of the sulfur in the energy balance. FIGURE P15–74 15–75E A constant-volume tank contains a mixture of 1 lbmol of benzene (C6H6) gas and 30 percent excess air at 77°F and 1 atm. The contents of the tank are now ignited, and all the hydrogen in the fuel burns to H2O but only 92 percent of the carbon burns to CO2, the remaining 8 percent forming CO. If the final temperature in the tank is 1800 R, determine the heat transfer from the combustion chamber during this process. Answer: 946,870 Btu 15–76 To supply heated air to a house, a high-efficiency gas furnace burns gaseous propane (C3H8) with a combustion efficiency of 96 percent. Both the fuel and 140 percent theoretical air are supplied to the combustion chamber at 25°C and 100 kPa, and the combustion is complete. Because this is a high-efficiency furnace, the product gases are cooled to 25°C and 100 kPa before leaving the furnace. To maintain the house at the desired temperature, a heat transfer rate of 31,650 kJ/h is required from the furnace. Determine the volume of water condensed from the product gases per day. Answer: 8.7 L/day Adiabatic Flame Temperature 15–77C A fuel is completely burned first with the stoichiometric amount of air and then with the stoichiometric amount of pure oxygen. For which case will the adiabatic flame temperature be higher? 15–78C A fuel at 25°C is burned in a well-insulated steady-flow combustion chamber with air that is also at 25°C. Under what conditions will the adiabatic flame temperature of the combustion process be a maximum? Hydrogen (H2) at 7°C is burned with 20 percent excess air that is also at 7°C during an adiabatic steady-flow combustion process. Assuming complete combustion, determine the exit temperature of the product gases. Answer: 2251 K 15–79 Coal 25 C 50% excess air 127 C Combustion chamber 1380 kPa CO2, H2O SO2, O2, N2 Tprod FIGURE P15–84 15–85 The combustion products of Prob. 15–84 are expanded in an isentropic turbine to 140 kPa. Calculate the work produced by this turbine, in kJ/kg fuel. cen84959_ch15.qxd 9/15/06 6:06 AM Page 811 Chapter 15 15–86 Ethyl alcohol (C2H5OH (g)) is burned with 200 percent excess air in an adiabatic, constant volume container. Initially, air and ethyl alcohol are at 100 kPa and 25 C. Assuming complete combustion, determine the final temperature and pressure of the products of combustion. Answers: 1435 K, 493 kPa | 811 that T0 298 K and the products leave the combustion chamber at 1 atm pressure. 15–94E Benzene gas (C6H6) at 1 atm and 77°F is burned during a steady-flow combustion process with 95 percent theoretical air that enters the combustion chamber at 77°F and 1 atm. All the hydrogen in the fuel burns to H2O, but part of the carbon burns to CO. Heat is lost to the surroundings at 77°F, and the products leave the combustion chamber at 1 atm and 1500 R. Determine (a) the heat transfer from the combustion chamber and (b) the exergy destruction. Liquid propane (C3H8) enters a steady-flow combustion chamber at 25°C and 1 atm at a rate of 0.4 kg/min where it is mixed and burned with 150 percent excess air that enters the combustion chamber at 12°C. If the combustion products leave at 1200 K and 1 atm, determine (a) the mass flow rate of air, (b) the rate of heat transfer from the combustion chamber, and (c) the rate of 25°C. entropy generation during this process. Assume T0 Answers: (a) 15.7 kg/min, (b) 1732 kJ/min, (c) 34.2 kJ/min · K 15–87 An adiabatic constant-volume tank contains a mixture of 1 kmol of hydrogen (H2) gas and the stoichiometric amount of air at 25°C and 1 atm. The contents of the tank are now ignited. Assuming complete combustion, determine the final temperature in the tank. 15–95 Entropy Change and Second-Law Analysis of Reacting Systems 15–88C Express the increase of entropy principle for chemically reacting systems. 15–89C How are the absolute entropy values of ideal gases at pressures different from 1 atm determined? 15–90C What does the Gibbs function of formation g° of a f compound represent? 15–91 One kmol of H2 at 25°C and 1 atm is burned steadily with 0.5 kmol of O2 at the same state. The H2O formed during the process is then brought to 25°C and 1 atm, the conditions of the surroundings. Assuming combustion is complete, determine the reversible work and exergy destruction for this process. 15–92 Ethylene (C2H4) gas enters an adiabatic combustion chamber at 25°C and 1 atm and is burned with 20 percent excess air that enters at 25°C and 1 atm. The combustion is complete, and the products leave the combustion chamber at 1 atm pressure. Assuming T0 25°C, determine (a) the temperature of the products, (b) the entropy generation, and (c) the exergy destruction. Answers: (a) 2269.6 K, (b) 1311.3 kJ/kmol · K, (c) 390,760 kJ/kmol 15–96 Reconsider Prob. 15–95. Using EES (or other) software, study the effect of varying the surroundings temperature from 0 to 38°C on the rate of exergy destruction, and plot it as a function of surroundings temperature. 15–97 n-Octane (C8H18) is burned in an automobile engine with 200 percent excess air. Air enters this engine at 1 atm and 25 C. Liquid fuel at 25 C is mixed with this air before combustion. The exhaust products leave the exhaust system at 1 atm and 77 C. What is the maximum amount of work, in kJ/kg fuel, that can be produced by this engine? Take T0 25 C. 15–98 The automobile engine of Prob. 15-97 is to be converted to natural gas (methane, CH4) fuel. Assuming that all factors remain the same, what is the maximum work that can be produced by the modified engine, in kJ/kg fuel? Answer: 51,050 kJ/kg fuel 15–93 Liquid octane (C8H18) enters a steady-flow combustion chamber at 25°C and 1 atm at a rate of 0.25 kg/min. It is burned with 50 percent excess air that also enters at 25°C and 1 atm. After combustion, the products are allowed to cool to 25°C. Assuming complete combustion and that all the H2O in the products is in liquid form, determine (a) the heat transfer rate from the combustion chamber, (b) the entropy generation rate, and (c) the exergy destruction rate. Assume T0 = 298 K Qout C8H18( ) 25°C AIR 25°C Combustion chamber 1 atm Products 25°C Qout CH4 25 C 200% excess air 25 C Combustion chamber 1 atm Products 77 C · FIGURE P15–98 15–99 n-Octane (C8H18) is burned in the constant pressure combustor of an aircraft engine with 30 percent excess air. Air enters this combustor at 600 kPa and 327 C, liquid fuel is injected at 25 C, and the products of combustion leave at 600 kPa and 1227 C. Determine the entropy generation and FIGURE P15–93 cen84959_ch15.qxd 9/15/06 6:06 AM Page 812 812 | Thermodynamics chimney is 1 atm. Presuming complete combustion, determine the temperature of the combustion products at which liquid water will begin to form in the chimney. 15–107 Calculate the higher and lower heating values of gaseous methane fuel (CH4). Compare your results with the values in Table A–27. 15–108 Develop an expression for the higher heating value of a gaseous alkane CnH2n+2 in terms of n. 15–109 A coal from Colorado which has an ultimate analysis (by mass) as 79.61 percent C, 4.66 percent H2, 4.76 percent O2, 1.83 percent N2, 0.52 percent S, and 8.62 percent ash (non-combustibles) is burned in an industrial boiler with 10 percent excess air. The temperature and pressure in the smokestack are 50 C and 1 atm, respectively. Calculate the fraction of the water in the combustion products that is liquid and the fraction that is vapor. 15–110 n-Butane (C4H10) is burned with stoichiometric amount of air in a cook stove. The products of combustion are at 1 atm pressure and 40 C. What fraction of the water in these products is liquid? C4H10 80% CH4 15% N2 5% O2 AIR 3.36% CO2 0.09% CO 14.91% O2 81.64% N2 exergy destruction per unit mass of fuel during this combustion process. Take T0 25 C. Review Problems 15–100 A 1-g sample of a certain fuel is burned in a bomb calorimeter that contains 2 kg of water in the presence of 100 g of air in the reaction chamber. If the water temperature rises by 2.5°C when equilibrium is established, determine the heating value of the fuel, in kJ/kg. 15–101E Hydrogen (H2) is burned with 100 percent excess air that enters the combustion chamber at 90°F, 14.5 psia, and 60 percent relative humidity. Assuming complete combustion, determine (a) the air–fuel ratio and (b) the volume flow rate of air required to burn the hydrogen at a rate of 25 lbm/h. 15–102 A gaseous fuel with 80 percent CH4, 15 percent N2, and 5 percent O2 (on a mole basis) is burned with dry air that enters the combustion chamber at 25°C and 100 kPa. The volumetric analysis of the products on a dry basis is 3.36 percent CO2, 0.09 percent CO, 14.91 percent O2, and 81.64 percent N2. Determine (a) the air–fuel ratio, (b) the percent theoretical air used, and (c) the volume flow rate of air used to burn fuel at a rate of 1.4 kg/min. Combustion chamber 1 atm Combustion chamber Theoretical air CO2, H2O, N2 40 C FIGURE P15–110 15–111 A steady-flow combustion chamber is supplied with CO gas at 37°C and 110 kPa at a rate of 0.4 m3/min and air at 25°C and 110 kPa at a rate of 1.5 kg/min. The combustion products leave the combustion chamber at 900 K. Assuming combustion is complete, determine the rate of heat transfer from the combustion chamber. 15–112 Methane gas (CH4) at 25°C is burned steadily with dry air that enters the combustion chamber at 17°C. The volumetric analysis of the products on a dry basis is 5.20 percent CO2, 0.33 percent CO, 11.24 percent O2, and 83.23 percent N2. Determine (a) the percentage of theoretical air used and (b) the heat transfer from the combustion chamber per kmol of CH4 if the combustion products leave at 700 K. 15–113 Propane gas (C3H8) enters a steady-flow combustion chamber at 1 atm and 25°C and is burned with air that enters the combustion chamber at the same state. Determine the adiabatic flame temperature for (a) complete combustion with 100 percent theoretical air, (b) complete combustion with 300 percent theoretical air, and (c) incomplete combustion (some CO in the products) with 95 percent theoretical air. FIGURE P15–102 15–103 Propane (C3H8) fuel is burned with stoichiometric amount of oxygen. Determine the mass fractions of carbon dioxide and water in the products. Also, calculate the mass of water in the products per unit of fuel mass burned. 15–104E A coal from Utah which has an ultimate analysis (by mass) as 61.40 percent C, 5.79 percent H2, 25.31 percent O2, 1.09 percent N2, 1.41 percent S, and 5.00 percent ash (non-combustibles) is burned with 20 percent excess air in an atmospheric pressure boiler. Calculate the mass of water in the products per unit mass of coal burned, and the dew-point temperature of the water vapor in the products. Answers: 0.549 lbm H2O/lbm coal, 108 F 15–105 n-Octane (C8H18) is burned with 60 percent excess air in an automobile engine. Assuming complete combustion and that the pressure in the exhaust system is 1 atm, determine the minimum temperature of the combustion products before liquid water will begin to form in the exhaust system. 15–106 Methane fuel (CH4) is burned with 50 percent excess air in a space-heating furnace. The pressure in the cen84959_ch15.qxd 9/15/06 6:06 AM Page 813 Chapter 15 15–114 Determine the highest possible temperature that can be obtained when liquid gasoline (assumed C8H18) at 25°C is burned steadily with air at 25°C and 1 atm. What would your answer be if pure oxygen at 25°C were used to burn the fuel instead of air? 15–115 The combustion of a fuel usually results in an increase in pressure when the volume is held constant, or an increase in volume when the pressure is held constant, because of the increase in the number of moles and the temperature. The increase in pressure or volume will be maximum when the combustion is complete and when it occurs adiabatically with the theoretical amount of air. Consider the combustion of methyl alcohol vapor (CH3OH(g)) with the stoichiometric amount of air in an 0.8-L combustion chamber. Initially, the mixture is at 25°C and 98 kPa. Determine (a) the maximum pressure that can occur in the combustion chamber if the combustion takes place at constant volume and (b) the maximum volume of the combustion chamber if the combustion occurs at constant pressure. 15–116 Reconsider Prob. 15–115. Using EES (or other) software, investigate the effect of the initial volume of the combustion chamber over the range 0.1 to 2.0 liters on the results. Plot the maximum pressure of the chamber for constant volume combustion or the maximum volume of the chamber for constant pressure combustion as functions of the initial volume. 15–117 Repeat Prob. 15–115 using methane (CH4(g)) as the fuel instead of methyl alcohol. 15–118 A mixture of 40 percent by volume methane (CH4), and 60 percent by volume propane (C3H8), is burned completely with theoretical air and leaves the combustion chamber at 100°C. The products have a pressure of 100 kPa and are cooled at constant pressure to 39°C. Sketch the T-s diagram for the water vapor that does not condense, if any. How much of the water formed during the combustion process will be condensed, in kmol H2O/kmol fuel? Answer: 1.96 15–119 A steam boiler heats liquid water at 200 C to superheated steam at 4 MPa and 400 C. Methane fuel (CH4) is burned at atmospheric pressure with 50 percent excess air. The fuel and air enter the boiler at 25 C and the products of combustion leave at 227 C. Calculate (a) the amount of steam generated per unit of fuel mass burned, (b) the change in the exergy of the combustion streams, in kJ/kg fuel, (c) the change in the exergy of the steam stream, in kJ/kg steam, and 25 C. (d) the lost work potential, in kJ/kg fuel. Take T0 Answers: (a) 18.72 kg steam/kg fuel, (b) 49,490 kJ/kg fuel, (c) 1039 kJ/kg steam, (d) 30,040 kJ/kg fuel | 813 15–121 n-Octane (C8H18(g)) is burned with stoichiometric amount of air. Determine the maximum work that can be produced, in kJ/kg fuel, when the air, fuel, and products are all at 25 C and 1 atm. 15–122 How much will the maximum work of Prob. 15– 121 change when 100 percent excess air is used for the combustion? Answer: 45,870 kJ/kg fuel 15–123E Methane (CH4) is burned with 100 percent excess air, with 10 percent of the carbon forming carbon monoxide. Determine the maximum work that can be produced, in Btu/lbm fuel, when the air, fuel, and products are all at 77 F and 1 atm. 15–124 A gaseous fuel mixture of 30 percent propane (C3H8), and 70 percent butane (C4H10), on a volume basis is burned in air such that the air–fuel ratio is 20 kg air/kg fuel when the combustion process is complete. Determine (a) the moles of nitrogen in the air supplied to the combustion process, in kmol/kmol fuel; (b) the moles of water formed in the combustion process, in kmol/kmol fuel; and (c) the moles of oxygen in the product gases. Answers: (a) 29.41, (b) 4.7, (c) 1.77 15–125 The furnace of a particular power plant can be considered to consist of two chambers: an adiabatic combustion chamber where the fuel is burned completely and adiabatically, and a heat exchanger where heat is transferred to a Carnot heat engine isothermally. The combustion gases in the heat exchanger are well-mixed so that the heat exchanger is at a uniform temperature at all times that is equal to the temperature of the exiting product gases, Tp. The work output of the Carnot heat engine can be expressed as W QhC Qa1 T0 b Tp where Q is the magnitude of the heat transfer to the heat engine and T0 is the temperature of the environment. The work output of the Carnot engine will be zero either when Taf (which means the product gases will enter and exit Tp the heat exchanger at the adiabatic flame temperature Taf, and thus Q 0) or when Tp T0 (which means the temperature of the product gases in the heat exchanger will be T0, and 0), and will reach a maximum somewhere in thus hC between. Treating the combustion products as ideal gases with constant specific heats and assuming no change in their composition in the heat exchanger, show that the work output of the Carnot heat engine will be maximum when Tp 2TafT0 T0 2 b B Taf 15–120 Repeat Prob. 15–119 using a coal from Utah which has an ultimate analysis (by mass) as 61.40 percent C, 5.79 percent H2, 25.31 percent O2, 1.09 percent N2, 1.41 percent S, and 5.00 percent ash (non-combustibles). Neglect the effect of the sulfur in the energy and entropy balances. Also, show that the maximum work output of the Carnot engine in this case becomes Wmax CTaf a 1 cen84959_ch15.qxd 9/15/06 6:06 AM Page 814 814 | Thermodynamics That is, the work output of the reversible engine would be the same if the furnace above is considered to be an isothermal furnace at a constant temperature Te. Fuel Adiabatic combustion chamber T0 Air where C is a constant whose value depends on the composition of the product gases and their specific heats. Fuel T0 Air Adiabatic combustion chamber Heat exchanger Tp = const. Tp Heat exchanger T0 Taf T0 Q W W Surroundings T0 FIGURE P15–126 Surroundings T0 15–127 FIGURE P15–125 15–126 The furnace of a particular power plant can be considered to consist of two chambers: an adiabatic combustion chamber where the fuel is burned completely and adiabatically and a counterflow heat exchanger where heat is transferred to a reversible heat engine. The mass flow rate of the working fluid of the heat engine is such that the working fluid is heated from T0 (the temperature of the environment) to Taf (the adiabatic flame temperature) while the combustion products are cooled from Taf to T0. Treating the combustion products as ideal gases with constant specific heats and assuming no change in their composition in the heat exchanger, show that the work output of this reversible heat engine is W CT0 a Taf T0 1 ln Taf b T0 Using EES (or other) software, determine the effect of the amount of air on the adiabatic flame temperature of liquid octane (C8H18). Assume both the air and the octane are initially at 25°C. Determine the adiabatic flame temperature for 75, 90, 100, 120, 150, 200, 300, 500, and 800 percent theoretical air. Assume the hydrogen in the fuel always burns H2O and the carbon CO2, except when there is a deficiency of air. In the latter case, assume that part of the carbon forms CO. Plot the adiabatic flame temperature against the percent theoretical air, and discuss the results. 15–128 Using EES (or other) software, write a general program to determine the adiabatic flame temperature during the complete combustion of a hydrocarbon fuel (CnHm) at 25°C in a steady-flow combustion chamber when the percent of excess air and its temperature are specified. As a sample case, determine the adiabatic flame temperature of liquid propane (C3H8) as it is burned steadily with 50 percent excess air at 25°C. Using EES (or other) software, determine the minimum percent of excess air that needs to be used for the fuels CH4(g), C2H2(g), CH3OH(g), C3H8(g), C8H18( ) if the adiabatic flame temperature is not to exceed 1500 K. Assume both the fuel and the air enter the steadyflow combustion chamber at 25°C. 15–130 Using EES (or other) software, repeat Prob. 15–129 for adiabatic flame temperatures of (a) 1200 K, (b) 1750 K, and (c) 2000 K. 15–129 where C is a constant whose value depends on the composition of the product gases and their specific heats. Also, show that the effective flame temperature Te of this furnace is Te Taf T0 ln 1 Taf> T0 2 cen84959_ch15.qxd 9/15/06 6:06 AM Page 815 Chapter 15 15–131 Using EES (or other) software, determine the adiabatic flame temperature of CH4(g) when both the fuel and the air enter the combustion chamber at 25°C for the cases of 0, 20, 40, 60, 80, 100, 200, 500, and 1000 percent excess air. 15–132 (a) 890 MJ/kg (d ) 56 MJ/kg (b) 802 MJ/kg (e) 50 MJ/kg | 815 (c) 75 MJ/kg 15–139 The higher heating value of a hydrocarbon fuel CnHm with m 8 is given to be 1560 MJ/kmol of fuel. Then its lower heating value is (a) 1384 MJ/kmol (c) 1402 MJ/kmol (e) 1551 MJ/kmol (b) 1208 MJ/kmol (d ) 1514 MJ/kmol Using EES (or other) software, determine the fuel among CH4(g), C2H2(g), C2H6(g), C3H8(g), C8H18( ) that gives the highest temperature when burned completely in an adiabatic constant-volume chamber with the theoretical amount of air. Assume the reactants are at the standard reference state. Fundamentals of Engineering (FE) Exam Problems 15–133 A fuel is burned with 90 percent theoretical air. This is equivalent to (a) 10% excess air (c) 10% deficiency of air (e) stoichiometric amount of air (b) 90% excess air (d ) 90% deficiency of air 15–140 Acetylene gas (C2H2) is burned completely during a steady-flow combustion process. The fuel and the air enter the combustion chamber at 25°C, and the products leave at 1500 K. If the enthalpy of the products relative to the standard reference state is 404 MJ/kmol of fuel, the heat transfer from the combustion chamber is (a) 177 MJ/kmol (d ) 631 MJ/kmol (b) 227 MJ/kmol (e) 751 MJ/kmol (c) 404 MJ/kmol 15–134 Propane (C3H8) is burned with 150 percent theoretical air. The air–fuel mass ratio for this combustion process is (a) 5.3 (d ) 23.4 (b) 10.5 (e) 39.3 (c) 15.7 15–141 Benzene gas (C6H6) is burned with 90 percent theoretical air during a steady-flow combustion process. The mole fraction of the CO in the products is (a) 1.6% (d ) 10% (b) 4.4% (e) 16.7% (c) 2.5% 15–135 One kmol of methane (CH4) is burned with an unknown amount of air during a combustion process. If the combustion is complete and there are 2 kmol of free O2 in the products, the air–fuel mass ratio is (a) 34.3 (d ) 14.9 (b) 17.2 (e) 12.1 (c) 19.0 15–142 A fuel is burned during a steady-flow combustion process. Heat is lost to the surroundings at 300 K at a rate of 1120 kW. The entropy of the reactants entering per unit time is 17 kW/K and that of the products is 15 kW/K. The total rate of exergy destruction during this combustion process is (a) 520 kW (d ) 340 kW (b) 600 kW (e) 739 kW (c) 1120 kW 15–136 A fuel is burned steadily in a combustion chamber. The combustion temperature will be the highest except when (a) the fuel is preheated. (b) the fuel is burned with a deficiency of air. (c) the air is dry. (d ) the combustion chamber is well insulated. (e) the combustion is complete. 15–137 An equimolar mixture of carbon dioxide and water vapor at 1 atm and 60°C enter a dehumidifying section where the entire water vapor is condensed and removed from the mixture, and the carbon dioxide leaves at 1 atm and 60°C. The entropy change of carbon dioxide in the dehumidifying section is (a) 2.8 kJ/kg · K (c) 0 (e) 2.8 kJ/kg · K (b) 0.13 kJ/kg · K (d ) 0.13 kJ/kg · K Design and Essay Problems 15–143 The safe disposal of hazardous waste material is a major environmental concern for industrialized societies and creates challenging problems for engineers. The disposal methods commonly used include landfilling, burying in the ground, recycling, and incineration or burning. Incineration is frequently used as a practical means for the disposal of combustible waste such as organic materials. The EPA regulations require that the waste material be burned almost completely above a specified temperature without polluting the environment. Maintaining the temperature above a certain level, typically about 1100°C, necessitates the use of a fuel when the combustion of the waste material alone is not sufficient to obtain the minimum specified temperature. A certain industrial process generates a liquid solution of ethanol and water as the waste product at a rate of 10 kg/s. The mass fraction of ethanol in the solution is 0.2. This solution is to be burned using methane (CH4) in a steady-flow combustion chamber. Propose a combustion process that will accomplish this task with a minimal amount of methane. State your assumptions. 15–138 Methane (CH4) is burned completely with 80 percent excess air during a steady-flow combustion process. If both the reactants and the products are maintained at 25°C and 1 atm and the water in the products exists in the liquid form, the heat transfer from the combustion chamber per unit mass of methane is cen84959_ch15.qxd 9/15/06 6:06 AM Page 816 816 | Thermodynamics anything else that will have to be changed to use the new coal? 15–146 A steam boiler is designed to convert saturated liquid water at 3500 kPa to steam at 3450 kPa and 400 C. This boiler burns natural gas (i.e., methane) at 1 atm and 25 C. Combustion air is also supplied at 1 atm and 25 C. The flow of combustion air is adjusted to maintain the temperature of the combustion products at 300 C. On a common basis of 1 kilogram of steam produced, calculate the change in the exergy of the combustion and water streams for excess air amounts ranging from 0 to 200 percent. Also calculate the exergy lost in this boiler as the excess air amount is varied. 15–147 Reconsider Prob. 15–146. Now an economizer (a heat exchanger that preheats the combustion air by cooling the combustion products) is added to the boiler, and thus the temperature of the combustion products is reduced to 200 C. All other factors remain the same. Air at 1 atm and 25 C passes through the economizer where it is heated before entering the boiler. Compare the loss of exergy of the economizer-boiler combination to that of the boiler only in the previous problem as the amount of excess air is varied. 15–144 Obtain the following information about a power plant that is closest to your town: the net power output; the type and amount of fuel; the power consumed by the pumps, fans, and other auxiliary equipment; stack gas losses; and the rate of heat rejection at the condenser. Using these data, determine the rate of heat loss from the pipes and other components, and calculate the thermal efficiency of the plant. 15–145 An electrical utility uses a Pennsylvania coal which has an ultimate analysis (by mass) as 84.36 percent C, 1.89 percent H2, 4.40 percent O2, 0.63 percent N2, 0.89 percent S, and 7.83 percent ash (non-combustibles) as fuel for its boilers. The utility is changing from the Pennsylvania coal to an Illinois coal which has an ultimate analysis (by mass) as 67.40 percent C, 5.31 percent H2, 15.11 percent O2, 1.44 percent N2, 2.36 percent S, and 8.38 percent ash (non-combustibles) as fuel for its boilers. With the Pennsylvania coal, the boilers used 15 percent excess air. Develop a schedule for the new coal showing the heat released, the smokestack dewpoint temperature, adiabatic flame temperature, and carbon dioxide production for various amount of excess air. Use this schedule to determine how to operate with the new coal as closely as possible to the conditions of the old coal. Is there ...
View Full Document

This note was uploaded on 03/09/2009 for the course ME 430 taught by Professor Y during the Spring '09 term at CUNY City.

Ask a homework question - tutors are online