Combustion and Incineration Processes
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2 Stoichiometry Stoichiometry is the discipline of tracking matter (particularly the elements, partitioned in accord with the laws of chemical combining weights and proportions) and energy (in all its forms) in either batch or continuous processes. Since these quantities are conserved in the course of any process, the engineer can apply the principle of conservation to follow the course of combustion and flow processes and to compute air requirements, flow volumes, and velocities, temperatures, and other useful quantities. As a refinement, the engineer should acknowledge the fact that some reactions and heat transfer processes sometimes stop short of ‘‘completion’’ because of equilibrium limitations. Also, for some situations, the chemical reaction rate may limit the degree of completeness, especially when system residence time is short or temperatures are low. I. UNITS AND FUNDAMENTAL RELATIONSHIPS A. Units In analyzing combustion problems, it is advantageous to use the molecular (atomic) weight expressed in kilograms (the kilogram mol or kilogram atom) as the unit quantity. This advantage derives from the facts that one molecular (atomic) weight of any compound (element) contains the same number of molecules (atoms) and that each mol of gas, at standard pressure and temperature, occupies the same volume if the gases are ‘‘ideal.’’ The concept of an ideal gas arises in the course of thermodynamic analysis and describes a gas for which intermolecular attractions are negligibly small, in which the actual volume of the molecules is small in comparison with the space they inhabit and where intermolecular collisions are perfectly elastic. B. Gas Laws 1. The Perfect Gas Law The behavior of ‘‘ideal gases’’ is described by Eq. (1a), the perfect gas law: PV ¼ nRT ð1aÞ SOFTbank E-Book Center Tehran, Phone: 66403879,66493070 For Educational Use. In this relationship P is the absolute pressure of the gas, V its volume, n the number of mols of gas, R the universal gas constant, and T the absolute temperature. Note that 273.15 must be added to the Celsius temperature and 459.69 to the Fahrenheit temperature to get the absolute temperature in degrees Kelvin (K) or degrees Rankine (R), respectively. The perfect gas law was developed from a simplified model of the kinetic behavior of molecules. The relationship becomes inaccurate at very low temperatures, at high pressures, and in other circumstances when intermolecular forces become significant. In the analysis of combustion systems at elevated temperatures and at atmospheric pressure, the assumption of ideal gas behavior is sound. The same value of the universal gas constant R is used for all gases. Care must be given to assure compatibility of the units of R with those used for P; V; n,andT. CommonlyusedvaluesofRaregiveninTable1. In the mechanical engineering literature, one often finds a gas law in use where the numerical value of the gas constant (say, R0) is specific to the gas under consideration. The gas constant in such relationships is usually found to be the universal gas constant divided by the molecular weight of the compound. In this formulation of the gas law, the weight w rather than the number of mols of gas is used: PV ¼ wR0T ð1bÞ EXAMPLE 1. Ten thousand kg=day of a spent absorbent containing 92% carbon, 6% ash, and 2% moisture is to be burned completely to generate carbon dioxide for process use. The exit temperature of the incinerator is 1000C. How many kilogram mols and how many kilograms of CO2 will be formed per minute? How many cubic meters per minute at a pressure of 1.04 atm? One must first determine the number of kilogram atoms per minute of carbon (atomic weight ¼ 12.01) flowing in the waste feed: ð0:92  10;000Þ=ð12:01  24  60Þ¼0:532 kg atoms=min Table 1 Values of the Universal Gas Constant R for Ideal Gases Pressure Volume Mols Temperature Gas constant Energy ðPÞðVÞðnÞðTÞðRÞ m3 atm — atm m3 kg mol K0:08205 kg mol K kPa m3 —kPam3 kg mol K8:3137 kg mol K kcal kcal — — kg mol K1:9872 kg mol K joules joules (abs) — — g mol K8:3144 g mol K ft lb ft-lb psia ft3 lb mol R 1545:0 lb mol R Btu Btu — — lb mol R1:9872 lb mol R ft3 atm —atmft3 lb mol R0:7302 lb mol R SOFTbank E-Book Center Tehran, Phone: 66403879,66493070 For Educational Use. Noting that with complete combustion each atom of carbon yields one molecule of carbon = dioxide, the generation rate of CO2 is 0.532 kg mol min. The weight flow of CO2 = (molecular weight ¼ 44.01) will be (0.532)(44.01) ¼ 23.4 kg min of CO2. The tempera- ture (K) is 1000 þ 273:15 ¼ 1273:15, and from Eq. 1. nRT V ¼ P 0:532  0:08206  1273:15 ¼ 1:04 : 3= ¼ 53 4m min CO2 In combustion calculations, one commonly knows the number of mols and the temperature and needs to compute the volume. For these calculations, it is convenient to obtain the answer by adjusting a unit volume at a specified standard condition to the conditions of interest. For such calculations one can use the gas laws expressed in terms of the volume of 1 kg or lb mol of an ideal gas at the standard conditions of 0Cor273:15 K (32For 492R) and 1 atm. The molecular volume is 22:4m3 (359:3ft3). If we denote the molecular volume as V0, and the pressure and temperature at standard conditions as P0 and T0, respectively, the gas law then yields: P0 T V ¼ n  V0   ð2Þ P T0 where P0 and T0 may be expressed in any consistent absolute units. For example, the gas volume from the preceding example could be calculated as 1:00 1273:15 V ¼ 0:532  22:4   1:04 273:15 : 3= ¼ 53 41 m min CO2 Dalton’s law (1801) of partial pressures is another useful identity derived from the ideal gas law. Dalton’s law states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the constituent gases, where the partial pressure is defined as the pressure each gas would exert if it alone occupied the volume of the mixture at the same temperature. For a perfect gas, Dalton’s law equates the volume percent with the mol percent and the partial pressure: Volume percent ¼ 100 ðmol fractionÞ partial pressure ¼ 100 ð3Þ total pressure EXAMPLE 2. Flue gases from combustion of 14.34 kg of graphite (essentially, pure carbon) with oxygen are at a temperature of 1000C and a pressure of 1.04 atm. What is 3 the volume in m of the CO2 formed? What is its density? 14:34 1:0 ð1000 þ 273:15Þ V ¼  22:4   ¼ 119:87 m3 12:01 1:04 273:15 SOFTbank E-Book Center Tehran, Phone: 66403879,66493070 For Educational Use. For each mol of CO2 formed, the mass is 12.01 þ 32.00, or 44.01 kg (see Table 1 in Appendix C for atomic weights). The gas density under these conditions is 44:12 r ¼ ¼ 0:438 kg=m3 CO2 1:0 ð1000 þ 273:15Þ 22:4   1:04 273:15 EXAMPLE 3. The carbon monoxide (CO) concentration in a flue gas stream with 10% moisture at 500C and 1.05 atm is measured at 88.86 parts per million by volume. Does this meet the regulatory limit of 100 mg per normal cubic meter (Nm3), established by the regulation as being a dry basis at 0C and 1.0 atm? Adjustment to the temperature or pressure conditions of the regulatory limit does not change the mol fraction of CO in the gas. The CO concentration corresponding to the regulatory limit under dry conditions is calculated as (88.86)(1.0 À 0.1), or 79.97 parts per million, dry volume (ppmdv). The volume of one kilogram mol of any gas at 0C and 1.0 atm is 22.4 m3. For CO, with a molecular weight of 28.01, the mass of one mol is 28:01  106 mg. Therefore, the mass concentration of CO in the flue gases is given by ð79:97  10À6Þð28:01  106Þ ½CO¼ ¼ 100:0mg=Nm3 22:4 2. Standard Conditions Throughout the published literature, in regulatory language, in industrial data sheets, etc., one frequently finds references to the term standard conditions. While the words imply standardization, it should be cautioned that the meaning is not at all consistent. For example, flow calculations by U.S. fan manufacturers and the U.S. natural gas industry are referenced to 60F ð15:6C), and 1 atm (29.92 in Hg or 14:7lb=in:2 absolute). The manufactured gas industry uses 60F, saturated with water vapor at 30 in. of Hg absolute, for marketing but 60F dry at 1 atm for combustion calculations. Other important appearances of the term ‘‘standard conditions’’ are found in the calculations and reports associated with permits for atmospheric discharges. The standard conditions in which to report stack gas flows in the United States are often based on 20C (68F) and 1 atm. The reference states used to specify and report the concentration of pollutants in air pollution regulations and permits often generate another set of standard conditions. In the United States, this may involve the calculation of volume in ‘‘standard cubic feet (scf)’’ at a reference temperature of 32F and 1 atm. In Europe, the metric equivalent (0C and 1.0 atm) is used to characterize the ‘‘normal cubic meter,’’ or Nm3.