ME 475/675 Introduction to Combustion Lecture 3 Thermodynamic properties of pure, mixtures and reacting systems

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

Thermodynamic Systems (reactors) m, E 1 𝑄 2 1 𝑊 2 Dm=DE=0 Inlet i Outlet o 𝑄 𝐶𝑉 𝑚 𝑖 𝑒+𝑃𝑣 𝑖 𝑚 0 𝑒+𝑃𝑣 𝑜 𝑊 𝐶𝑉 Closed systems 1 𝑄 2 − 1 𝑊 2 =𝑚 𝑢 2 − 𝑢 1 + 𝑣 2 2 2 − 𝑣 1 2 2 +𝑔 𝑧 2 − 𝑧 1 Open Steady State, Steady Flow (SSSF) Systems 𝑄 𝐶𝑉 − 𝑊 𝐶𝑉 = 𝑚 ℎ 𝑜 − ℎ 𝑖 + 𝑣 𝑜 2 2 − 𝑣 𝑖 2 2 +𝑔 𝑧 𝑜 − 𝑧 𝑖 How to find changes, 𝑢 2 − 𝑢 1 and ℎ 𝑜 − ℎ 𝑖 , for mixtures when temperatures and composition change due to reactions (not covered in Thermodynamics I)

Calorific Equations of State for a pure substance 𝑢=𝑢 𝑇,𝑣 =𝑢(𝑇)≠𝑓𝑛(𝑣) ℎ=ℎ 𝑇,𝑃 =ℎ(𝑇)≠𝑓𝑛(𝑃) For ideal gases Differentials (small changes) 𝑑𝑢= 𝜕𝑢 𝜕𝑇 𝑣 𝑑𝑇+ 𝜕𝑢 𝜕𝑣 𝑇 𝑑𝑣 𝑑ℎ= 𝜕ℎ 𝜕𝑇 𝑃 𝑑𝑇+ 𝜕ℎ 𝜕𝑃 𝑇 𝑑𝑃 For ideal gas 𝜕𝑢 𝜕𝑣 𝑇 = 0; 𝜕𝑢 𝜕𝑇 𝑣 = 𝑐 𝑣 𝑇 𝒅𝒖= 𝒄 𝒗 𝑻 𝒅𝑻 𝜕ℎ 𝜕𝑃 𝑇 = 0; 𝜕ℎ 𝜕𝑇 𝑃 = 𝑐 𝑃 𝑇 𝒅𝒉= 𝒄 𝑷 𝑻 𝒅𝑻 Specific Heats, 𝑐 𝑣 and 𝑐 𝑃 [kJ/kg K] Energy input Q to increase temperature of one kg of a substance by 1°C at constant volume or pressure How are 𝑐 𝑣 𝑇 and 𝑐 𝑃 𝑇 measured? Calculate 𝑐 𝑝 𝑜𝑟 𝑐 𝑝 = 𝑄 𝑚Δ𝑇 𝑝 𝑜𝑟 𝑣 𝑘𝐽 𝑘𝑔𝐾 Molar based 𝑐 𝑝 = 𝑐 𝑝 ∗𝑀𝑊; 𝑐 𝑣 = 𝑐 𝑣 ∗𝑀𝑊 𝑘𝐽 𝑘𝑚𝑜𝑙 𝐾 m, T 1 kg Q w P = wg/A = constant 𝑐 𝑝 m, T 1 kg Q 𝑐 𝑣 V = constant

Molar Specific Heat Dependence on Temperature 𝑐 𝑝 𝑇 𝑘𝐽 𝑘𝑚𝑜𝑙 𝐾 Monatomic molecules: Only possess translational kinetic energy All energy input increases kinetic energy (associated with temperature) 𝑐 𝑝 𝑇 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) 𝑐 𝑝 𝑇 increases with temperature as energy gets diverted to more modes 𝑇 [K]

Specific Internal Energy and Enthalpy Once 𝑐 𝑣 𝑇 and 𝑐 𝑝 𝑇 are known, specific enthalpy h(T) and internal energy u(T) can be calculated by integration 𝑢 𝑇 = 𝑢 𝑟𝑒𝑓 + 𝑇 𝑟𝑒𝑓 𝑇 𝑐 𝑣 𝑇 𝑑𝑇 ℎ 𝑇 = ℎ 𝑟𝑒𝑓 + 𝑇 𝑟𝑒𝑓 𝑇 𝑐 𝑝 𝑇 𝑑𝑇 Primarily interested in changes, i.e. ℎ 𝑇 2 − ℎ 𝑇 1 = 𝑇 1 𝑇 2 𝑐 𝑝 𝑇 𝑑𝑇 , When composition does not change 𝑇 𝑟𝑒𝑓 and ℎ 𝑟𝑒𝑓 are not important Tabulated and Curve Fit: Appendix A, pp. 687-699 For combustion gases: CO, CO2, H2, H, OH, H2O, O, N, NO, NO2, O2 bookmark (show tables) T, 𝑐 𝑝 , ℎ 𝑓 𝑜 𝑇 − ℎ 𝑓 𝑜 298 = ℎ 𝑇 − ℎ 298𝐾 ,… Superscript o ( ℎ 𝑓 𝑜 ) is for 1 atm, not important for ideal gas Curve fits for Fuels, Appendix B, Page 702 Use in spreadsheets 𝑐 𝑣 = 𝑐 𝑝 − 𝑅 𝑢 𝑐 𝑝 = 𝑐 𝑝 /𝑀𝑊; 𝑐 𝑣 = 𝑐 𝑣 /𝑀𝑊

Mixture Properties Extensive Enthalpy 𝐻 𝑚𝑖𝑥 = 𝑚 𝑖 ℎ 𝑖 = 𝑚 𝑇𝑜𝑡𝑎𝑙 ℎ 𝑚𝑖𝑥 𝐻 𝑚𝑖𝑥 = 𝑚 𝑖 ℎ 𝑖 = 𝑚 𝑇𝑜𝑡𝑎𝑙 ℎ 𝑚𝑖𝑥 𝒉 𝒎𝒊𝒙 (𝑻)= 𝑚 𝑖 ℎ 𝑖 𝑚 𝑇𝑜𝑡𝑎𝑙 = 𝒀 𝒊 𝒉 𝒊 (𝑻) 𝐻 𝑚𝑖𝑥 = 𝑁 𝑖 ℎ 𝑖 = 𝑁 𝑇𝑜𝑡𝑎𝑙 ℎ 𝑚𝑖𝑥 𝒉 𝒎𝒊𝒙 (𝑻)= 𝑁 𝑖 ℎ 𝑖 𝑁 𝑇𝑜𝑡𝑎𝑙 = 𝝌 𝒊 𝒉 𝒊 (𝑻) Specific Internal Energy 𝒖 𝒎𝒊𝒙 (𝑻)= 𝒀 𝒊 𝒖 𝒊 (𝑻) 𝒖 𝒎𝒊𝒙 𝑻 = 𝝌 𝒊 𝒖 𝒊 𝑻 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

Standardized Enthalpy and Enthalpy of Formation Needed to find 𝑢 2 − 𝑢 1 and ℎ 𝑜 − ℎ 𝑖 for chemically-reacting masses because energy is required to form and break chemical bonds Not considered in Thermodynamics I ℎ 𝑖 𝑇 = ℎ 𝑓,𝑖 𝑜 𝑇 𝑟𝑒𝑓 +Δ ℎ 𝑠,𝑖 (𝑇) Standard Enthalpy at Temperature T = Enthalpy of formation from “normally occurring elemental compounds,” at standard reference state: 𝑇 𝑟𝑒𝑓 = 298 K and P° = 1 atm Sensible enthalpy change in going from Tref to T = 𝑇 𝑟𝑒𝑓 𝑇 𝑐 𝑝 𝑇 𝑑𝑇 Normally-Occurring Elemental Compounds Examples: O2, N2, C, He, H2 Their enthalpy of formation at 𝑇 𝑟𝑒𝑓 =298 K are defined to be ℎ 𝑓,𝑖 𝑜 𝑇 𝑟𝑒𝑓 = 0 Use these Normally-Occurring compounds as bases to tabulate the energy to form other compounds

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) O2 + 498,390 kJ  (2 mole) O ℎ 𝑓,𝑂 𝑜 𝑇 𝑟𝑒𝑓 = 498,390 kJ 2 𝑘𝑚𝑜𝑙 𝑂 =+ 249,195 𝑘𝐽 𝑘𝑚𝑜𝑙 𝑂 energy input to form 1 mole O from 0.5 mole O2 (normally-occurring form of O) ℎ 𝑓,𝑖 𝑜 𝑇 𝑟𝑒𝑓 for other compounds are in Appendices A and B, pp 687-702 To find enthalpy of O at other temperatures use ℎ 𝑂 2 𝑇 = ℎ 𝑓, 𝑂 2 𝑜 𝑇 𝑟𝑒𝑓 +Δ ℎ 𝑠, 𝑂 2 (𝑇) (Appendix A)

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). ID: Find enthalpy at 298.15 K on different bases Problem 2.15: Repeat for T = 500 K do this in lecture 4

Standard Enthalpy of Isooctane Coefficients 𝑎 1 to 𝑎 8 from Page 702 𝜃= 𝑇 [𝐾] 1000 𝐾 ; ℎ 𝑜 𝑘𝐽 𝑘𝑚𝑜𝑙𝑒 =4184( 𝑎 1 𝜃+ 𝑎 2 𝜃 2 2 + 𝑎 3 𝜃 3 3 + 𝑎 4 𝜃 4 4 − 𝑎 5 𝜃 + 𝑎 6 ) Spreadsheet really helps this calculation