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ME 475/675 Introduction to Combustion Lecture 3 Thermodynamic Systems (reactors) 1π2 1π2 Inlet i ππ π + ππ£ Outlet o π Dm=DE=0 π0 π + ππ£ π m, E ππΆπ ππΆπ β’ Closed systems β’ 1π2 β 1π2 = π π’2 β π’1 + π£22 2 β π£12 2 + π π§2 β π§1 β’ Open Steady State, Steady Flow (SSSF) Systems β’ ππΆπ β ππΆπ = π βπ β βπ + π£π2 2 β π£π2 2 + π π§π β π§π β’ How to find changes, π’2 β π’1 and βπ β βπ , for mixtures when temperatures and composition change due to reactions (not covered in Thermodynamics I) Calorific Equations of State for a pure substance β’ π’ = π’ π, π£ = π’(π) β ππ(π£) β’ β = β π, π = β(π) β ππ(π) For ideal gases β’ Differentials (small changes) β’ ππ’ = β’ πβ = ππ’ ππ ππ π£ πβ ππ ππ π β’ For ideal gas β’ ππ’ = ππ£ π 0; + + ππ’ ππ£ π πβ ππ π ππ’ ππ π£ = ππ£ π πβ ππ π = ππ π β’ π π = ππ π» π π» β’ πβ = ππ π 0; β’ π π = ππ· π» π π» β’ Specific Heats, ππ£ and ππ [kJ/kg K] β’ Energy input to increase temperature of one kg of a substance by 1°C at constant volume or pressure β’ How are ππ£ π and ππ π measured? ππ£ ππ ππ ππ£ w Q m, T V = constant Q m, T P = wg/A = constant β’ Calculate ππ ππ π£ = π πΞπ π ππ π£ β’ Molar based β’ ππ = ππ β ππ; ππ£ = ππ£ β ππ T [K] Q [joules] Molar Specific Heat Dependence on Temperature ππ π ππ½ ππππ πΎ π [K] β’ Monatomic molecules: Nearly independent of temperature β’ Only possess translational kinetic energy β’ Multi-Atomic molecules: Increase with temperature and number of molecules β’ Also possess rotational and vibrational kinetic energy Specific Internal Energy and Enthalpy β’ Once ππ£ π and ππ π are known, specific enthalpy h(T) and internal energy u(T) can be calculated by integration β’ π’ π = π’πππ + β’ β π = βπππ + π π ππππ π£ π π ππππ π π ππ π ππ β’ Primarily interested in changes, i.e. β π2 β β π1 = π2 π π1 π π ππ, β’ When composition does not change ππππ and βπππ are not important β’ Tabulated: Appendix A, pp. 687-699, for combustion gases β’ bookmark (show tables) β’ Curve fits, Page 702, for Fuels β’ Use in spreadsheets β’ ππ£ = ππ β π π’ ; β’ ππ =ππ /ππ; ππ£ =ππ£ /ππ Mixture Properties β’ Extensive Enthalpy β’ π»πππ₯ = ππ βπ = π πππ‘ππ βπππ₯ β’ ππππ (π») = β’ π»πππ₯ = π π βπ ππππ‘ππ = ππ ππ (π») ππ βπ = ππππ‘ππ βπππ₯ β’ ππππ (π») = ππ βπ ππππ‘ππ = ππ ππ (π») β’ Specific Internal Energy β’ ππππ (π») = ππ ππ (π») β’ ππππ π» = ππ ππ π» β’ Use these relations to calculate mixture specific enthalpy and internal energy (per mass or mole) as functions of the properties of the individual components and their mass or molar fractions. β’ u and h depend on temperature, but not pressure Standardized Enthalpy and Enthalpy of Formation β’ Needed to find π’2 β π’1 and βπ β βπ for chemically-reacting systems because energy is required to form and break chemical bonds β’ Not considered in Thermodynamics I π β’ βπ π = βπ,π ππππ + Ξβπ ,π (π) β’ Standard Enthalpy at Temperature T = β’ Enthalpy of formation from βnormally occurring elemental compounds,β at standard reference state: Tref = 298 K and P° = 1 atm β’ Sensible enthalpy change in going from Tref to T = π π ππππ π π ππ β’ Normally-Occurring Elemental Compounds β’ Examples: O2, N2, C, He, H2 π β’ Their enthalpy of formation at ππππ = 298 K are defined to be βπ,π ππππ = 0 β’ Use these compounds as bases to tabulate the energy to form other compounds Standard Enthalpy of O atoms β’ To form 2O atoms from one O2 molecule requires 498,390 kJ/kmol of energy input to break O-O bond (initial and final T and P are same) β’ At 298K (1 mole) O2 + 498,390 kJ ο (2 mole) O 498,390 kJ ππ½ π β’ βπ,π ππππ = = + 249,195 2 πππππ πππππ π β’ βπ,π ππππ for other compounds are in Appendices A and B, pp 687-702 β’ To find enthalpy of O at other temperatures use π β’ β π2 π = βπ, π2 ππππ + Ξβπ , π2 (π) Example: β’ Problem 2.14, p 71: Consider a stoichiometric mixture of isooctane and air. Calculate the enthalpy of the mixture at the standard-state temperature (298.15 K) on a per-kmol-of-fuel basis (kJ/kmolfuel), on a per-kmol-of-mixture basis (kJ/kmolmix), and on a per-mass-of-mixture basis (kJ/kgmix). β’ Find enthalpy at 298.15 K of different bases β’ Problem 2.15: Repeat for T = 500 K Standard Enthalpy of Isooctane T [K] 298.15 theta 0.29815 h [kJ/Kmol] -224108.82 a1 a2 -0.55313 181.62 -0.16492 8.072412 a3 a4 a5 -97.787 20.402 -0.03095 -0.8639 0.040304 0.103807 a6 -60.751 -60.751 β’ Coefficients π1 to π8 from Page 702 π [πΎ] ; 1000 πΎ ππ½ π β = πππππ β’ π= β’ 4184(π1 π π2 + π2 2 π3 + π3 3 β’ Spreadsheet really helps this calculation + π4 π4 4 β π5 π + π6 ) a8 20.232 Enthalpy of Combustion (or reaction) Reactants 298.15 K, P = 1 atm Stoichiometric ππΌπ < 0 ππππ Products Complete Combustion Cο CO2 Hο H2O = 0 298.15 K, 1 atm β’ How much energy can be released if product temperature and pressure are the same as those of the reactant? β’ Steady Flow Reactor β’ ππΌπ β ππππ = π»π β π»π = π βπ β βπ β’ ππππ = π»π β π»π = π βπ β βπ