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ME 475/675 Introduction to Combustion
Lecture 3 Thermodynamic properties of pure, mixtures and reacting systems
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Announcement Dilesh will hold a problem session HW 1 due Wedensday
Tuesday 10 am or 4 pm. Which time? HW 1 due Wedensday Chapter 2 (2, 8, 12, 13, X1) Problem X1 Consider the broad education necessary to understand the impact of engineering solutions (Introduction to Combustion) in a global and societal context Read a news article that describes some issue related to combustion For example: Energy efficiency, pollution, range land fires, fire safety, nuclear safety In one paragraph, summarize the article, and indicate how it is related to your interest in combustion and/or this class
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Thermodynamic Systems (reactors)
m, E 1 π 2 1 π 2 Dm=DE=0 Inlet i Outlet o π πΆπ π π π+ππ£ π π 0 π+ππ£ π π πΆπ Closed systems 1 π 2 β 1 π 2 =π π’ 2 β π’ π£ β π£ π π§ 2 β π§ 1 Open Steady State, Steady Flow (SSSF) Systems π πΆπ β π πΆπ = π β π β β π + π£ π 2 2 β π£ π π π§ π β π§ π How to find changes, π’ 2 β π’ 1 and β π β β π , for mixtures when temperatures and composition change due to reactions (not covered in Thermodynamics I)
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Calorific Equations of State for a pure substance
π’=π’ π,π£ =π’(π)β ππ(π£) β=β π,π =β(π)β ππ(π) For ideal gases Differentials (small changes) ππ’= ππ’ ππ π£ ππ+ ππ’ ππ£ π ππ£ πβ= πβ ππ π ππ+ πβ ππ π ππ For ideal gas ππ’ ππ£ π = 0; ππ’ ππ π£ = π π£ π π
π= π π π» π
π» πβ ππ π = 0; πβ ππ π = π π π π
π= π π· π» π
π» Specific Heats, π π£ and π π [kJ/kg K] Energy input Q to increase temperature of one kg of a substance by 1Β°C at constant volume or pressure How are π π£ π and π π π measured? Calculate π π ππ π π = π πΞπ π ππ π£ ππ½ πππΎ Molar based π π = π π βππ; π π£ = π π£ βππ ππ½ ππππ πΎ m, T 1 kg Q w P = wg/A = constant π π m, T 1 kg Q π π£ V = constant
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Molar Specific Heat Dependence on Temperature
π π π ππ½ ππππ πΎ Monatomic molecules: Only possess translational kinetic energy All energy input increases kinetic energy (associated with temperature) π π π nearly independent of temperature Multi-Atomic molecules: Increase with temperature and number of molecules Also possess rotational and vibrational kinetic energy (only fraction of energy goes to kinetic) π π π increases with temperature as energy gets diverted to more modes π [K]
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Specific Internal Energy and Enthalpy
Once π π£ π and π π π are known, specific enthalpy h(T) and internal energy u(T) can be calculated by integration π’ π = π’ πππ + π πππ π π π£ π ππ β π = β πππ + π πππ π π π π ππ Primarily interested in changes, i.e. β π 2 β β π 1 = π 1 π 2 π π π ππ , When composition does not change π πππ and β πππ are not important Tabulated and Curve Fit: Appendix A, pp For combustion gases: CO, CO2, H2, H, OH, H2O, O, N, NO, NO2, O2 bookmark (show tables) T, π π , β π π π β β π π = β π β β 298πΎ ,β¦ Superscript o ( β π π ) is for 1 atm, not important for ideal gas Curve fits for Fuels, Appendix B, Page 702 Use in spreadsheets π π£ = π π β π
π’ π π = π π /ππ; π π£ = π π£ /ππ
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Mixture Properties Extensive Enthalpy π» πππ₯ = π π β π = π πππ‘ππ β πππ₯
π» πππ₯ = π π β π = π πππ‘ππ β πππ₯ π πππ (π»)= π π β π π πππ‘ππ = π π π π (π») π» πππ₯ = π π β π = π πππ‘ππ β πππ₯ π πππ (π»)= π π β π π πππ‘ππ = π π π π (π») Specific Internal Energy π πππ (π»)= π π π π (π») π πππ π» = π π π π 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. For ideal gases, u and h depend on temperature, but not pressure
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Standardized Enthalpy and Enthalpy of Formation
Needed to find π’ 2 β π’ 1 and β π β β π for chemically-reacting masses because energy is required to form and break chemical bonds Not considered in Thermodynamics I β π π = β π,π π π πππ +Ξ β π ,π (π) Standard Enthalpy at Temperature T = Enthalpy of formation from βnormally occurring elemental compounds,β at standard reference state: π πππ = 298 K and PΒ° = 1 atm Sensible enthalpy change in going from Tref to T = π πππ π π π π ππ Normally-Occurring Elemental Compounds Examples: O2, N2, C, He, H2 Their enthalpy of formation at π πππ =298 K are defined to be β π,π π π πππ = 0 Use these Normally-Occurring compounds as bases to tabulate the energy to form other compounds
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Standard Enthalpy of O atoms
298K Experiments show that 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= 298K and P= 1atm) At 298K: (1 mole) O ,390 kJ ο (2 mole) O β π,π π π πππ = 498,390 kJ 2 ππππ π =+ 249,195 ππ½ ππππ π energy input to form 1 mole O from 0.5 mole O2 (normally-occurring form of O) β π,π π π πππ for other compounds are in Appendices A and B, pp To find enthalpy of O at other temperatures use β π 2 π = β π, π 2 π π πππ +Ξ β π , π 2 (π) (Appendix A)
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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 ( 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). ID: Find enthalpy at K on different bases Problem 2.15: Repeat for T = 500 K do this in lecture 4
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Standard Enthalpy of Isooctane
Coefficients π 1 to π 8 from Page 702 π= π [πΎ] 1000 πΎ ; β π ππ½ πππππ =4184( π 1 π+ π 2 π π 3 π π 4 π 4 4 β π 5 π + π 6 ) Spreadsheet really helps this calculation
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