ME 475/675 Introduction to Combustion Lecture 3

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 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 𝑣 = 𝑄 𝑚Δ𝑇 𝑝 𝑜𝑟 𝑣 Molar based 𝑐 𝑝 = 𝑐 𝑝 ∗𝑀𝑊; 𝑐 𝑣 = 𝑐 𝑣 ∗𝑀𝑊 m, T Q w P = wg/A = constant 𝑐 𝑝 m, T Q 𝑐 𝑣 V = constant

Molar Specific Heat Dependence on Temperature 𝑐 𝑝 𝑇 𝑘𝐽 𝑘𝑚𝑜𝑙 𝐾 𝑇 [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 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: Appendix A, pp. 687-699, for combustion gases bookmark (show tables) Curve fits, Page 702, for Fuels 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. 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 systems 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: Tref = 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 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  (2 mole) O ℎ 𝑓,𝑂 𝑜 𝑇 𝑟𝑒𝑓 = 498,390 kJ 2 𝑘𝑚𝑜𝑙 𝑂 =+ 249,195 𝑘𝐽 𝑘𝑚𝑜𝑙 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)

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 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

Enthalpy of Combustion (or reaction) Reactants 298.15 K, P = 1 atm Stoichiometric Products Complete Combustion CCO2 HH2O 298.15 K, 1 atm 𝑄 𝐼𝑁 <0 𝑊 𝑂𝑈𝑇 =0 How much energy can be released if product temperature and pressure are the same as those of the reactant? Steady Flow Reactor 𝑄 𝐼𝑁 − 𝑊 𝑂𝑈𝑇 = 𝐻 𝑃 − 𝐻 𝑅 = 𝑚 ℎ 𝑃 − ℎ 𝑅 𝑄 𝑂𝑈𝑇 = 𝐻 𝑅 − 𝐻 𝑃 = 𝑚 ℎ 𝑅 − ℎ 𝑃