Conservation of mass: 44 kg on LHS and RHS one kmol of CO₂ produced by burning 1 kmol of C with one kmol of O₂

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1 L20 Stoichiometry of Reactions Gore Combustion provides the heat source for many power cycles (cars, airplanes, power plants), for home heating, cooking applications, and for manufacturing from melting of ores to heat treatment of resulting metals and also for cement and other chemical process plants. The high temperature heat source source is often assumed to be at the high temperature defined by the combustion temperature. The source is often assumed to be large enough that does not decrease because of transfer of. The source is often combustion of fossil fuels like coal, natural gas, and other hydrocarbons with air * Coal (mostly carbon), oxygen combustion is represented by the chemical equation:. An atom balanced chemical equation always satisfies conservation of mass: Conservation of mass: 44 kg on LHS and RHS one kmol of CO₂ produced by burning kmol of C with one kmol of O₂ Methane + (Oxygen + Nitrogen = Air) = Products of complete combustion Mass on two sides is balanced. In case of methane combustion with air the # of moles are also balanced but this is not a general result. See ½ moles going to mole in both reactions below. ; " "# $% "# &'376 ( )(% % 376$ *"# & +, Mass balance is guaranteed by correct atom balance. Mole fraction in reactants. / ##*.+, ,6 6*.*,. *.+, 4 5 ##*.+, 5*.+, 6*.*, : :5

2 Mass fraction in reactants ; /0 2 ; ; ; : :5 Mass fractions in products ; / ; ; Mole fractions in products. / : LHS has moles, RHS has moles. Moles can increase (or decrease) as result of RXN Air to fuel Ratio: molar and mass based. Since just enough oxygen is provided for complete combustion this process is called combustion with theoretically correct air DE F% GFH DE ; ;ID GEE ; GFH GEE ; ;ID

3 J =>???? J MNBO > All of the above reactions are for complete combustion of theoretically correct mixtures. Often excess air is used. However, some applications do need fuel rich combustion. 50% excess air makes the heptane complete combustion reaction to be: +,.5P 3.76 Q Within our simplified treatment of combustion, excess air just enters and leaves the combustor. Equivalence ratio RSIFTGD%UD VGWF X > = Z [ \DG% H R]UDEE =FH > Practice Questions Combustion Partial pressures of CH₄, C₂H₆, C₃H₈ and C₈H₁₀ in a fuel mixture of these 4 paraffin compounds are 0.5 atm, 0.3 atm, 0.5 atm and 0.05 atm. Find the stoichiometric Air/Fuel ratio, mass and mole fractions of products resulting from complete combustion, mole fractions of CO₂, N₂ in the products if the water vapor is completely condensed out, mole fraction of H₂O, CO₂ and N₂ in the products if exhaust gas temperature is reduced to 50 C and the water condensation process reaches equilibrium. Solution: Partial pressures mole fractions if mixture is at atm. Assume mixture at atm. based on addition of partial pressures , 0.5 * ˆ ] ]3.76 G U UŠDWD U IEWF% C atom balance G G.75 H atom balance O atom balance 2] 2G ] /2 ] 3 25 N atom balance (N₂ is normally an inert) except for ppm levels of deadly NO, NO₂, etc. For energy analysis ppm levels do not hurt, for environment they do. 3

4 U c =.75 Reaction equation is , 0.5 * ˆ =/>.+6 ˆ#*. 6* 5.05 (exactly same as C₈H₁₈) P7.6,#7.**7#7.6#7.76ŽˆQ HC fuels have similar =/> which is good because natural fuel quality varies a lot and some devices need to be often operated alternatively on CH₄ (natural gas) and C₃H₈ (propane cylinders) and their unknown mixtures. The buzz word for this facility is called fuel flexibility. 2. Product mole fractions # #; DE / , , R] GIEW FE ŠHD F%G%W.. 3. Many combustion analyzers rely on dry analysis. The H₂O must be condensed out first. Dozens or so key patents inh₂o removal Gaseous steam mole fractions are:. / # / Dry based analysis 2.75 Liquid Note: Wet based composition can be obtained from dry based analysis if fuel composition is known. Eg. Dry base analysis yields:. / and , find wet based analysis if fuel composition is as before (and complete combustion occurs) , 0.5. * ˆ ] 3.76] ],.,G% are unknowns, use C, O, H, N, atom balances to solve for these. 4

5 Combustion ] 0.7 ] Partial condensation of H₂O x Š N GHE Š H Original Š N 0.67 but must reduce to Š N as the material cools down. Original reactants š O Now œ L 7.+ *#7.7ˆ# œ L , JDE ; U% D%ED JDE ; HFF%G F]WIHD \FSIF JDE JDE F] X Actual combustion process does not result in all CO₂. CO is the most common unburned pollutant. Orsat Analysis Gives moles % on a dry basis G 3.76G N balance 23.76G.2 G22.27 O balance 2G C balance ] H balance Reaction is: * *