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In combustion processes the oxidizer is usually air but could be pure oxygen an. oxygen mixture or a substance involving some other oxidizing element such as. fluorine Here we will limit our attention to combustion of a fuel with air or pure. Chemical fuels exist in gaseous liquid or solid form Natural gas gasoline and. coal perhaps the most widely used examples of these three forms are each a complex. mixture of reacting and inert compounds We will consider each more closely later in. the chapter First let s review some important fundamentals of mixtures of gases such. as those involved in combustion reactions,Mass and Mole Fractions. The amount of a substance present in a sample may be indicated by its mass or by the. number of moles of the substance A mole is defined as the mass of a substance equal to. its molecular mass or molecular weight A few molecular weights commonly used in. combustion analysis are tabulated below For most combustion calculations it is. sufficiently accurate to use integer molecular weights The error incurred may easily be. evaluated for a given reaction and should usually not be of concern Thus a gram mole. of water is 18 grams a kg mole of nitrogen is 28 kg and a pound mole of sulfur is 32. Molecule Molecular Weight, The composition of a mixture may be given as a list of the fractions of each of the. substances present Thus we define the mass fraction of a component i mfi as the. ratio of the mass of the component mi to the mass of the mixture m. It is evident that the sum of the mass fractions of all the components must be 1 Thus. Analogous to the mass fraction we define the mole fraction of component i xi as. the ratio of the number of moles of i ni to the total number of moles in the mixture n. The total number of moles n is the sum of the number of moles of all the components. of the mixture, It follows that the sum of all the mole fractions of the mixture must also equal 1. The mass of component i in a mixture is the product of the number of moles of i and its. molecular weight Mi The mass of the mixture is therefore the sum m n1M1 n2M2. over all components of the mixture Substituting xin for ni the total mass becomes. m x1M1 x2M2 n, But the average molecular weight of the mixture is the ratio of the total mass to the. total number of moles Thus the average molecular weight is. M m n x1M1 x2M2,EXAMPLE 3 1, Express the mass fraction of component 1 of a mixture in terms of a the number of.
moles of the three components of the mixture n1 n2 and n3 and b the mole fractions. of the three components c If the mole fractions of carbon dioxide and nitrogen in a. three component gas containing water vapor are 0 07 and 0 38 respectively what are. the mass fractions of the three components, a Because the mass of i can be written as mi niMi the mass fraction of component. i can be written as,mfi niMi n1M1 n2M2 dl, For the first of the three components i 1 this becomes. mf1 n1M1 n1M1 n2M2 n3M3,Similarly for i 2 and i 3,mf2 n2M2 n1M1 n2M2 n3M3. mf3 n3M3 n1M1 n2M2 n3M3, b Substituting n1 x1 n n2 x2 n etc in the earlier equations and simplifying we. obtain for the mass fractions,mf1 x1M1 x1M1 x2M2 x3M3.
mf2 x2M2 x1M1 x2M2 x3M3,mf3 x3M3 x1M1 x2M2 x3M3, c Identifying the subscripts 1 2 and 3 with carbon dioxide nitrogen and water. vapor respectively we have x1 0 07 x2 0 38 and x3 1 0 07 0 038 0 55. mf1 0 07 44 0 07 44 0 38 28 0 55 18,0 07 44 23 62 0 1304. mf2 0 38 28 23 62 0 4505,mf3 0 55 18 23 62 0 4191, As a check we sum the mass fractions 0 1304 0 4505 0 4191 1 0000. For a mixture of gases at a given temperature and pressure the ideal gas law. shows that pVi ni T holds for any component and pV n T for the mixture as a. whole Forming the ratio of the two equations we observe that the mole fractions have. the same values as the volume fraction,xi Vi V ni n dl. Similarly for a given volume of a mixture of gases at a given temperature piV ni T. for each component and pV n T for the mixture The ratio of the two equations. shows that the partial pressure of any component i is the product of the mole fraction. of i and the pressure of the mixture,pi pni n pxi,EXAMPLE 3 2.
What is the partial pressure of water vapor in Example 3 1 if the mixture pressure is. two atmospheres, The mole fraction of water vapor in the mixture of Example 3 1 is 0 55 The partial. pressure of the water vapor is therefore 0 55 2 1 1 atm. Characterizing Air for Combustion Calculations, Air is a mixture of about 21 oxygen 78 nitrogen and 1 other constituents by. volume For combustion calculations it is usually satisfactory to represent air as a 21. oxygen 79 nitrogen mixture by volume Thus for every 21 moles of oxygen that. react when air oxidizes a fuel there are also 79 moles of nitrogen involved Therefore. 79 21 3 76 moles of nitrogen are present for every mole of oxygen in the air. At room temperature both oxygen and nitrogen exist as diatomic molecules O2. and N2 respectively It is usually assumed that the nitrogen in the air is nonreacting at. combustion temperatures that is there are as many moles of pure nitrogen in the. products as there were in the reactants At very high temperatures small amounts of. nitrogen react with oxygen to form oxides of nitrogen usually termed NOx These small. quantities are important in pollution analysis because of the major role of even small. traces of NOx in the formation of smog However since these NOx levels are. insignificant in energy analysis applications nitrogen is treated as inert here. The molecular weight of a compound or mixture is the mass of 1 mole of the. substance The average molecular weight M of a mixture as seen earlier is the linear. combination of the products of the mole fractions of the components and their. respective molecular weights Thus the molecular weight for air Mair is given by the. sum of the products of the molecular weights of oxygen and nitrogen and their. respective mole fractions in air Expressed in words. Mair Mass of air Mole of air Moles of N2 Mole of air Mass of N2 Mole of N2. Moles of O2 Mole of air Mass of O2 Mole of O2,Mair 0 79 Mnitrogen 0 21 Moxygen. 0 79 28 0 21 32 28 84, The mass fractions of oxygen and nitrogen in air are then. mfoxygen 0 21 32 28 84 0 233 or 23 3,mfnitrogen 0 79 28 28 84 0 767 or 76 7.
3 2 Combustion Chemistry of a Simple Fuel, Methane CH4 is a common fuel that is a major constituent of most natural gases. Consider the complete combustion of methane in pure oxygen The chemical reaction. equation for the complete combustion of methane in oxygen may be written as. CH4 2O2 Y CO2 2H2O 3 1, Because atoms are neither created nor destroyed Equation 3 1 states that methane. consisting of one atom of carbon and four atoms of hydrogen reacts with four atoms. of oxygen to yield carbon dioxide and water products with the same number of atoms. of each element as in the reactants This is the basic principle involved in balancing all. chemical reaction equations, Carbon dioxide is the product formed by complete combustion of carbon through. the reaction C O2 Y CO2 Carbon dioxide has only one carbon atom per molecule. Since in Equation 3 1 there is only one carbon atom on the left side of the equation. there can be only one carbon atom and therefore one CO2 molecule on the right. Similarly water is the product of the complete combustion of hydrogen It has two. atoms of hydrogen per molecule Because there are four hydrogen atoms in the. reactants of Equation 3 1 there must be four in the products implying that two. molecules of water formed These observations require four atoms of oxygen on the. right which implies the presence of two molecules four atoms of oxygen on the left. The coefficients in chemical equations such as Equation 3 1 may be interpreted as. the number of moles of the substance required for the reaction to occur as written. Thus another way of interpreting Equation 3 1 is that one mole of methane reacts. with two moles of oxygen to form one mole of carbon dioxide and two moles of water. While not evident in this case it is not necessary that there be the same number of. moles of products as reactants It will be seen in numerous other cases that a different. number of moles of products is produced from a given number of moles of reactants. Thus although the numbers of atoms of each element must be conserved during a. reaction the total number of moles need not Because the number of atoms of each. element cannot change it follows that the mass of each element and the total mass must. be conserved during the reaction Thus using the atomic weights masses of each. element the sums of the masses of the reactants and products in Equation 3 1 are. CH4 2O2 Y CO2 2H2O,12 4 1 4 16 Y 12 2 16 2 2 1 16 80. Other observations may be made with respect to Equation 3 1 There are 2 moles of. water in the 3 moles of combustion products and therefore a mole fraction of water in. the combustion products of xwater 2 3 0 667 Similarly xCarbon dioxide 1 3 0 333. moles of CO2 in the products, There are 44 mass units of CO2 in the 80 mass units of products for a mass.
fraction of CO2 in the products,mfcarbon dioxide 44 80 0 55. Likewise the mass fraction of water in the products is 2 18 80 0 45. We also observe that there are 12 mass units of carbon in the products and. therefore a carbon mass fraction of 12 80 0 15 Note that because the mass of any. element and the total mass are conserved in a chemical reaction the mass fraction of. any element is also conserved in the reaction Thus the mass fraction of carbon in the. reactants is 0 15 as in the products,Combustion in Air. Let us now consider the complete combustion of methane in air The same combustion. products are expected as with combustion in oxygen the only additional reactant. present is nitrogen and it is considered inert Moreover because we know that in air. every mole of oxygen is accompanied by 3 76 moles of nitrogen the reaction equation. can be written as,CH4 2O2 2 3 76 N2 Y CO2 2H2O 2 3 76 N2 3 2. It is seen that the reaction equation for combustion in air may be obtained from the. combustion equation for the reaction in oxygen by adding the appropriate number of. moles of nitrogen to both sides of the equation, Note that both Equations 3 1 and 3 2 describe reactions of one mole of. methane fuel Because the same amount of fuel is present in both cases both reactions. release the same amount of energy We can therefore compare combustion reactions in. air and in oxygen It will be seen that the presence of nitrogen acts to dilute the. reaction both chemically and thermally With air as oxidizer there are 2 moles of water. vapor per 10 52 moles of combustion products compared with 2 moles of water per 3. moles of products for combustion in oxygen Similarly with air there is a mass fraction. of CO2 of 0 1514 and a carbon mass fraction of 0 0413 in the combustion products. compared with 0 55 and 0 15 respectively for combustion in oxygen. The diluting energetic effect of nitrogen when combustion is in air may be reasoned. as follows The same amount of energy is released in both reactions because the same. amount of fuel is completely consumed However the nonreacting nitrogen molecules. in the air have heat capacity This added heat capacity of the additional nitrogen. molecules absorbs much of the energy released resulting in a lower internal energy per. unit mass of products and hence a lower temperature of the products Thus the energy. released by the reaction is shared by a greater mass of combustion products when the. combustion is in air, Often products of combustion are released to the atmosphere through a chimney.
stack or flue These are therefore sometimes referred to as flue gases The flue gas. composition may be stated in terms of wet flue gas wfg or dry flue gas dfg because. under some circumstances the water vapor in the gas condenses and then escapes as a. liquid rather than remaining as a gaseous component of the flue gas When liquid water. is present in combustion products the combustion product gaseous mass fractions may. be taken with respect to the mass of flue gas products with the product water present. or omitted Thus for Equation 3 2 the mass of dry combustion products is 254 56. Hence the mass fraction of carbon dioxide is 44 254 56 0 1728 with respect to dry. flue gas and 44 290 56 0 1514 with respect to wet flue gas. In combustion discussions reference is frequently made to higher and lower heating. values The term higher heating value HHV refers to a heating value measurement in. which the product water vapor is allowed to condense As a consequence the heat of. vaporization of the water is released and becomes part of the heating value The lower. heating value LHV corresponds to a heating value in which the water remains a. vapor and does not yield its heat of vaporization Thus the energy difference between. the two values is due to the heat of vaporization of water and. HHV LHV mwater mfuel hfg Btu lbm kJ kg, where mwater is the mass of liquid water in the combustion products and hfg is the latent. heat of vaporization of water,Air Fuel Ratio, It is important to know how much oxygen or air must be supplied for complete. combustion of a given quantity of fuel This information is required in sizing fans and. ducts that supply oxidizer to combustion chambers or burners and for numerous . 85 C H A P T E R 3 FUELS AND COMBUSTION 3 1 Introduction to Combustion Combustion Basics The last chapter set forth the basics of the Rankine cycle and the principles