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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 • Closed systems • 1π2 − 1π2 = π π’2 − π’1 + π£22 2 − π£12 2 + π π§2 − π§1 • Open Steady State, Steady Flow (SSSF) Systems • ππΆπ − ππΆπ = π βπ − βπ + π£π2 2 − π£π2 2 + π π§π − π§π • How to find changes, π’2 − π’1 and βπ − βπ , for mixtures when temperatures and composition change due to reactions (not covered in Thermodynamics I) Calorific Equations of State for a pure substance • π’ = π’ π, π£ = π’(π) ≠ ππ(π£) • β = β π, π = β(π) ≠ ππ(π) For ideal gases • Differentials (small changes) • ππ’ = • πβ = ππ’ ππ ππ π£ πβ ππ ππ π • For ideal gas • ππ’ = ππ£ π 0; + + ππ’ ππ£ π πβ ππ π ππ’ ππ π£ = ππ£ π πβ ππ π = ππ π • π π = ππ π» π π» • πβ = ππ π 0; • π π = ππ· π» π π» • Specific Heats, ππ£ and ππ [kJ/kg K] • Energy input to increase temperature of one kg of a substance by 1°C at constant volume or pressure • 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 • Calculate ππ ππ π£ = π πΔπ π ππ π£ • Molar based • ππ = ππ ∗ ππ; ππ£ = ππ£ ∗ ππ T [K] Q [joules] Molar Specific Heat Dependence on Temperature ππ π ππ½ ππππ πΎ π [K] • Monatomic molecules: Nearly independent of temperature • Only possess translational kinetic energy • Multi-Atomic molecules: Increase with temperature and number of molecules • Also possess rotational and vibrational kinetic energy Specific Internal Energy and Enthalpy • Once ππ£ π and ππ π are known, specific enthalpy h(T) and internal energy u(T) can be calculated by integration • π’ π = π’πππ + • β π = βπππ + π π ππππ π£ π π ππππ π π ππ π ππ • Primarily interested in changes, i.e. β π2 − β π1 = π2 π π1 π π ππ, • When composition does not change ππππ and βπππ are not important • Tabulated: Appendix A, pp. 687-699, for combustion gases • bookmark (show tables) • Curve fits, Page 702, for Fuels • Use in spreadsheets • ππ£ = ππ − π π’ ; • ππ =ππ /ππ; ππ£ =ππ£ /ππ Mixture Properties • Extensive Enthalpy • π»πππ₯ = ππ βπ = π πππ‘ππ βπππ₯ • ππππ (π») = • π»πππ₯ = π π βπ ππππ‘ππ = ππ ππ (π») ππ βπ = ππππ‘ππ βπππ₯ • ππππ (π») = ππ βπ ππππ‘ππ = ππ ππ (π») • Specific Internal Energy • ππππ (π») = ππ ππ (π») • ππππ π» = ππ ππ π» • 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. • u and h depend on temperature, but not pressure Standardized Enthalpy and Enthalpy of Formation • Needed to find π’2 − π’1 and βπ − βπ for chemically-reacting systems because energy is required to form and break chemical bonds • Not considered in Thermodynamics I π • βπ π = βπ,π ππππ + Δβπ ,π (π) • Standard Enthalpy at Temperature T = • Enthalpy of formation from “normally occurring elemental compounds,” at standard reference state: Tref = 298 K and P° = 1 atm • Sensible enthalpy change in going from Tref to T = π π ππππ π π ππ • Normally-Occurring Elemental Compounds • Examples: O2, N2, C, He, H2 π • Their enthalpy of formation at ππππ = 298 K are defined to be βπ,π ππππ = 0 • Use these compounds as bases to tabulate the energy to form other compounds Standard Enthalpy of O atoms • 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) • At 298K (1 mole) O2 + 498,390 kJ ο (2 mole) O 498,390 kJ ππ½ π • βπ,π ππππ = = + 249,195 2 πππππ πππππ π • βπ,π ππππ for other compounds are in Appendices A and B, pp 687-702 • To find enthalpy of O at other temperatures use π • β π2 π = βπ, π2 ππππ + Δβπ , π2 (π) Example: • 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). • Find enthalpy at 298.15 K of different bases • 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 • Coefficients π1 to π8 from Page 702 π [πΎ] ; 1000 πΎ ππ½ π β = πππππ • π= • 4184(π1 π π2 + π2 2 π3 + π3 3 • Spreadsheet really helps this calculation + π4 π4 4 − 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 • How much energy can be released if product temperature and pressure are the same as those of the reactant? • Steady Flow Reactor • ππΌπ − ππππ = π»π − π»π = π βπ − βπ • ππππ = π»π − π»π = π βπ − βπ
