ME 475/675 Introduction to Combustion Lecture 2 Ideal stoichiometric hydrocarbon combustion, Mixture molecular weight, Air/Fuel mass ratio

Ideal Stoichiometric Hydrocarbon Combustion air CxHy + a(O2+3.76N2)  (x)CO2 + (y/2) H2O + 3.76a N2 a = number of oxygen molecules per fuel molecule, 𝑁 𝑂 2 𝑁 𝐹𝑢𝑒𝑙 Number of air molecules per fuel molecule is a(1+3.76) If a = aST = x + y/4, then the reaction is Stoichiometric No O2 or Fuel in products This mixture produces nearly the hottest flame temperature If a < x + y/4, then reaction is fuel-rich (oxygen-lean, “fuel” left over) If a > x + y/4, then reaction is fuel-lean (oxygen-rich, O2 left over) Equivalence Ratio (of fuel) Φ= 𝑎 𝑆𝑡 𝑎 𝐴𝑐𝑡𝑢𝑎𝑙 = 𝑁 𝑂 2 𝑁 𝐹𝑢𝑒𝑙 𝑆𝑡 𝑁 𝑂 2 𝑁 𝐹𝑢𝑒𝑙 𝐴𝑐𝑡𝑢𝑎𝑙 = 𝑁 𝐹𝑢𝑒𝑙,𝐴𝑐𝑡𝑢𝑎𝑙 𝑁 𝐹𝑢𝑒𝑙,𝑆𝑡 𝑁 𝑂 2 ,𝑆𝑡 𝑁 𝑂 2 ,𝐴𝑐𝑡𝑢𝑎𝑙 Φ=1→ Stiochiometric Φ>1→ Fuel Rich Φ<1→ Fuel Lean

Air to fuel mass ratio [kg air/kg fuel] of reactants 𝐴 𝐹 = 𝑚 𝐴𝑖𝑟 𝑚 𝐹𝑢𝑒𝑙 = 𝑁 𝐴𝑖𝑟 𝑀𝑊 𝐴𝑖𝑟 𝑁 𝐹𝑢𝑒𝑙 𝑀𝑊 𝐹𝑢𝑒𝑙 = 𝑁 𝑂 2 𝑁 𝐹𝑢𝑒𝑙 𝑁 𝐴𝑖𝑟 𝑁 𝑂 2 𝑀𝑊 𝐴𝑖𝑟 𝑀𝑊 𝐹𝑢𝑒𝑙 =𝑎 (1+3.76) 𝑀𝑊 𝐴𝑖𝑟 𝑀𝑊 𝐹𝑢𝑒𝑙 For a stoichiometric mixture 𝐴 𝐹 𝑆𝑡 =4.76 𝑎 𝑆𝑡 (1+3.76) 𝑀𝑊 𝐴𝑖𝑟 𝑀𝑊 𝐹𝑢𝑒𝑙 ; 𝑤ℎ𝑒𝑟𝑒 𝑎 𝑆𝑡 =𝑥+ 𝑦 4 Need to find molecular weights

Molecular Weight of a Pure Substance x Only one type of molecule: AxByCz… Molecular Weight MW = x(AWA) + y(AWB) + z(AWC) + … AWi = atomic weights Inside front cover of book Examples 𝑀𝑊 𝑂 2 = 2( AW 𝑂 ) = 2(15.9994) = 32.00 𝑘𝑔 𝑘𝑚𝑜𝑙 𝑀𝑊 𝐻 2 𝑂 = 2( AW 𝐻 ) + ( AW 𝑂 ) = 2(1.00794) + (15.9994) = 18.02 𝑘𝑔 𝑘𝑚𝑜𝑙 𝑀𝑊 𝑃𝑟𝑜𝑝𝑎𝑛𝑒 = 𝑀𝑊 𝐶 3 𝐻 8 =3 12.011 +8 1.00794 =44.097 𝑘𝑔 𝑘𝑚𝑜𝑙 See page 701 for fuels

Mixtures containing n components 𝑁 𝑖 = number of moles of species 𝑖 𝑖=1, 2,..𝑛 Total number of moles in system 𝑁 𝑇𝑜𝑡𝑎𝑙 = 𝑖=1 𝑛 𝑁 𝑖 Mole Fraction of species i 𝜒 𝑖 = 𝑁 𝑖 𝑁 𝑇𝑜𝑡𝑎𝑙 = 𝑁 𝑖 𝑖=1 𝑛 𝑁 𝑖 Mass of species 𝑖= 𝑚 𝑖 = 𝑁 𝑖 𝑀𝑊 𝑖 Total Mass 𝑚 𝑇𝑜𝑡𝑎𝑙 = 𝑖=1 𝑛 𝑚 𝑖 Mass Fraction of species i 𝑌 𝑖 = 𝑚 𝑖 𝑚 𝑇𝑜𝑡𝑎𝑙 = 𝑚 𝑖 𝑖=1 𝑛 𝑚 𝑖 Useful facts: 𝑖=1 𝑛 𝜒 𝑖 = 𝑖=1 𝑛 𝑌 𝑖 =1 but 𝜒 𝑖 ≠ 𝑌 𝑖 Mixture Molar Weight: 𝑀𝑊 𝑀𝑖𝑥 = 𝑚 𝑇𝑜𝑡𝑎𝑙 𝑁 𝑇𝑜𝑡𝑎𝑙 𝑀𝑊 𝑀𝑖𝑥 = 𝑚 𝑖 𝑁 𝑖 = 𝑁 𝑖 𝑀𝑊 𝑖 𝑁 𝑇𝑜𝑡𝑎𝑙 = 𝜒 𝑖 𝑀𝑊 𝑖 (weighted average) 𝑀𝑊 𝑀𝑖𝑥 = 𝑚 𝑖 𝑁 𝑖 = 𝑚 𝑇𝑜𝑡𝑎𝑙 𝑚 𝑖 / 𝑀𝑊 𝑖 = 1 𝑌 𝑖 / 𝑀𝑊 𝑖 Example 𝑀𝑊 𝐴𝑖𝑟 = 𝜒 𝑖 𝑀𝑊 𝑖 =0.21 𝑀𝑊 𝑂 2 +0.79 𝑀𝑊 𝑁 2 =0.21∗ 2∗15.9994 +0.79∗ 2∗14.0067 =0.21∗ 32.00 +0.79∗ 28.00 =28.85 𝑘𝑔 𝑘𝑚𝑜𝑙𝑒 Remember and/or write inside front cover of your book Relationship between 𝜒 𝑖 and 𝑌 𝑖 𝑌 𝑖 = 𝑚 𝑖 𝑚 𝑇𝑜𝑡𝑎𝑙 = 𝑁 𝑖 𝑀𝑊 𝑖 𝑁 𝑇𝑜𝑡𝑎𝑙 𝑀𝑊 𝑀𝑖𝑥 = 𝜒 𝑖 𝑀𝑊 𝑖 𝑀𝑊 𝑀𝑖𝑥 𝜒 𝑖 = 𝑌 𝑖 𝑀𝑊 𝑀𝑖𝑥 𝑀𝑊 𝑖

Stoichiometric Air/Fuel Mass Ratio For Hydrocarbon fuel CxHy 𝐴 𝐹 𝑆𝑡 = 𝑎 𝑆𝑡 (1+3.76) 𝑀𝑊 𝐴𝑖𝑟 𝑀𝑊 𝐹𝑢𝑒𝑙 𝑀𝑊 𝐴𝑖𝑟 = 𝜒 𝑖 𝑀𝑊 𝑖 = 28.85 𝑘𝑔 𝑘𝑚𝑜𝑙𝑒 𝑀𝑊 𝐹𝑢𝑒𝑙 = 𝑀𝑊 𝐶 𝑥 𝐻 𝑦 =𝑥 12.011 +𝑦(1.00794) aSt = x + y/4 𝐴 𝐹 𝑆𝑡 = 𝑥+ 𝑦 4 (4.76)28.85 𝑘𝑔 𝑘𝑚𝑜𝑙𝑒 𝑥 12.011 𝑘𝑔 𝑘𝑚𝑜𝑙𝑒 +𝑦(1.00794 𝑘𝑔 𝑘𝑚𝑜𝑙𝑒 ) 1 1.00794 𝑘𝑔 𝑘𝑚𝑜𝑙𝑒 𝑥 1 1.00794 𝑘𝑔 𝑘𝑚𝑜𝑙𝑒 𝑥 = 136.24 1+ 𝑦 𝑥 4 11.92+ 𝑦 𝑥 For 𝐶 𝑥 𝐻 𝑦 , 10< 𝐴 𝐹 𝑆𝑡 <35 Constraints on y/x later (see page 291 for some 𝐴 𝐹 𝑆𝑡 =𝜈)

Equivalence Ratio Φ Φ= 𝑎 𝑆𝑡 𝑎 𝐴𝑐𝑡𝑢𝑎𝑙 = 𝑎 𝑆𝑡 (1+3.76) 𝑀𝑊 𝐴𝑖𝑟 𝑀𝑊 𝐹𝑢𝑒𝑙 𝑎 𝐴𝑐𝑡𝑢𝑎𝑙 (1+3.76) 𝑀𝑊 𝐴𝑖𝑟 𝑀𝑊 𝐹𝑢𝑒𝑙 = 𝐴 𝐹 𝑆𝑡 𝐴 𝐹 𝐴𝑐𝑡𝑢𝑎𝑙 = 𝐹 𝐴𝑐𝑡𝑢𝑎𝑙 𝐹 𝑆𝑡 𝐴 𝑆𝑡 𝐴 𝐴𝑐𝑡𝑢𝑎𝑙 Φ=1→ Stiochiometric Φ>1→ Fuel Rich Φ<1→ Fuel Lean 𝑎= 𝑎 𝑆𝑡 Φ = 𝑥+ 𝑦 4 Φ ; 𝐴 𝐹 𝐴𝑐𝑡𝑢𝑎𝑙 = 𝐴 𝐹 𝑆𝑡 Φ = 𝑥+ 𝑦 4 4.76 𝑀𝑊 𝐴𝑖𝑟 𝑀𝑊 𝐹𝑢𝑒𝑙 Φ CxHy + a(O2+3.76N2) % Stoichiometric Air (%SA)= 100% Φ = 𝐹 𝑆𝑡 𝐹 𝐴𝑐𝑡𝑢𝑎𝑙 𝐴 𝐴𝑐𝑡𝑢𝑎𝑙 𝐴 𝑆𝑡 ∗100% % Excess Oxygen (%EO) = (%SA)-100%

Example For extra credit, this problem may be clearly reworked and turned in at the beginning of the next class period. Problem 2.11, Page 91: In a propane-fueled truck, 3 percent (by volume) oxygen is measure in the exhaust stream of the running engine. Assuming “complete” combustion without dissociation, determine the air-fuel ratio (mass) supplied to the engine. Also find Equivalence Ratio: Φ % Stoichiometric air: %SA % Excess Oxygen: %EA ID: Are reactants Fuel Rich, Fuel Lean, or Stoichiometric? Work on the board

Thermodynamic Systems (reactors) Closed systems Rigid tanks, piston/cylinders 1 = Initial state; 2 = Final state Mass: 𝑚 1 = 𝑚 2 =𝑚 Chemical composition inside can change But atoms are conserved 1st Law 1 𝑄 2 − 1 𝑊 2 =Δ𝐸=𝑚( 𝑒 2 − 𝑒 1 ) 𝑒=𝑢+ 𝑣 2 2 +𝑔𝑧 1 𝑄 2 − 1 𝑊 2 =𝑚 𝑢 2 − 𝑢 1 + 𝑣 2 2 2 − 𝑣 1 2 2 +𝑔 𝑧 2 − 𝑧 1 How to find internal energy 𝑢 for mixtures, and change 𝑢 2 − 𝑢 1 when composition changes due to reactions (not covered in Thermodynamics I) 1 𝑊 2 1 𝑄 2 m, E

Open Systems (control volume) Steady State, Steady Flow (SSSF) Fixed volume, no moving boundaries Properties are constant and uniform Inside CV (Dm = DE = 0), and At ports (ℎ=𝑢+𝑃𝑣) One inlet and one outlet: 𝑚 𝑖 = 𝑚 𝑜 = 𝑚 Atoms are conserved Energy: 𝑄 𝐶𝑉 − 𝑊 𝐶𝑉 = 𝑚 ℎ 𝑜 − ℎ 𝑖 + 𝑣 𝑜 2 2 − 𝑣 𝑖 2 2 +𝑔 𝑧 𝑜 − 𝑧 𝑖 Composition and temperature of inlet and outlet may not be the same due to reaction Need to find ℎ 𝑜 − ℎ 𝑖 (not covered in Thermodynamics I) Dm=DE=0 Inlet i Outlet o 𝑄 𝐶𝑉 𝑚 𝑖 𝑒+𝑃𝑣 𝑖 𝑚 0 𝑒+𝑃𝑣 𝑜 𝑊 𝐶𝑉

Combustion Thermochemistry We use thermodynamics to evaluate the internal energy, enthalpy and entropy of a systems at different states The difference in energy and enthalpy between the products and reactants is used with the first law to predict The heat of combustion (energy released during combustion, due to changes in chemical bonds) if the products are assumed to be at the same temperature as the reactants The adiabatic flame temperature (product temperature assuming all reaction heat release stays in the system). Determined by assuming the products have the same energy as the reactants The steady state product composition for a reaction may be determined from entropy considerations Need to find The internal energy, enthalpy and entropy of mixtures of gases, and How to account for effects of chemical bonds These are not covered in ME 311 Thermodynamics I, but we we’ll cover them now

Ideal Gas Equation of State 𝑃𝑉=𝑁 𝑅 𝑈 𝑇 Universal Gas Constant 𝑅 𝑈 =8.315 𝑘𝐽 𝑘𝑚𝑜𝑙𝑒 𝐾 =8315 𝐽 𝑘𝑚𝑜𝑙𝑒 𝐾 Inside book front cover kJ = kN*m= kPa*m3 𝑃𝑉= 𝑁∗𝑀𝑊 (𝑅 𝑈 /𝑀𝑊)𝑇=𝑚𝑅𝑇 Specific Gas Constant R = 𝑅 𝑈 /𝑀𝑊 MW = Molecular Weight of that gas 𝑃𝑣=𝑅𝑇;𝑣= 𝑉 𝑚 = 1 𝜌 𝑃=𝜌𝑅𝑇 Number of molecules N*NAV Avogadro's Number, 𝑁 𝐴𝑉 6.022∗ 10 26 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑘𝑚𝑜𝑙𝑒 6.022∗ 10 23 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑚𝑜𝑙𝑒 Number of molecules in 12 kg of C12

Partial Pressure 𝑃 𝑖 and volume fraction 𝑉 𝑖 𝑉 of a specie in a mixture at pressure 𝑃 and volume V Each specie acts as if it was the only component at the given V and T Specie 𝑖: 𝑃 𝑖 𝑉= 𝑁 𝑖 𝑅 𝑢 𝑇 Mixture: 𝑃 𝑉=𝑁 𝑅 𝑢 𝑇 Ratio: 𝑃 𝑖 𝑃 = 𝑁 𝑖 𝑁 = 𝜒 𝑖 𝑃 𝑖 = 𝜒 𝑖 𝑃 𝑃 𝑖 = 𝜒 𝑖 𝑃 =𝑃 𝜒 𝑖 =𝑃 Volume Fraction Each specie acts as if it was the only component at the given P and T Specie 𝑖: 𝑃 𝑉 𝑖 = 𝑁 𝑖 𝑅 𝑢 𝑇 Mixture: 𝑃 𝑉=𝑁 𝑅 𝑢 𝑇 Ratio: 𝑉 𝑖 𝑉 = 𝑁 𝑖 𝑁 = 𝜒 𝑖

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Extensive and Intensive System Properties Intensive Properties Independent of system size Examples Per unit mass (lower case) v = V/m [m3/kg] u = U/m [kJ/kg] h = H/m [kJ/kg] Denoted using lower-case letters Exceptions Temperature T [°C, K] Pressure P [Pa] Molar Basis (use bar ) V = vm = N 𝑣 U = um = N 𝑢 H = hm = N ℎ N number of moles in the system Useful because chemical equations deal with the number of moles, not mass Extensive thermodynamic properties depend on System Size (extent) Examples Volume V [m3] Internal Energy E [kJ] Enthalpy H = E + PV [kJ] Test: cut system in half Denoted with CAPITAL letters

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 For ideal gas 𝜕𝑢 𝜕𝑣 𝑇 = 0; 𝜕𝑢 𝜕𝑇 𝑣 = 𝑐 𝑣 𝑇 𝑑𝑢= 𝑐 𝑣 𝑇 𝑑𝑇 𝑑ℎ= 𝜕ℎ 𝜕𝑇 𝑃 𝑑𝑇+ 𝜕ℎ 𝜕𝑃 𝑇 𝑑𝑃 𝜕ℎ 𝜕𝑃 𝑇 = 0; 𝜕ℎ 𝜕𝑇 𝑃 = 𝑐 𝑃 𝑇 𝑑ℎ= 𝑐 𝑃 𝑇 𝑑𝑇 Specific Heat [kJ/kg C] 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 𝑘𝐽 𝑘𝑔𝐾 𝑐 𝑝 = 𝑐 𝑝 ∗𝑀𝑊; 𝑐 𝑣 = 𝑐 𝑣 ∗𝑀𝑊 𝑘𝐽 𝑘𝑚𝑜𝑙 𝐾 m, T Q w P = wg/A = constant 𝑐 𝑝 m, T Q 𝑐 𝑣 V = constant

Molar Specific Heat Dependence on Temperature Monatomic molecules: Nearly independent of temperature Only translational kinetic energy Multi-Atomic molecules: Increase with temperature and number of molecules Also possess rotational and vibrational kinetic energy

Internal Energy and Enthalpy Once cp(T) and cv(T) are known, internal energy change 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 Appendix A (pp. 687-699, bookmark) 𝑐 𝑝 𝑇 :𝑡𝑎𝑏𝑙𝑒𝑠 𝑎𝑛𝑑 𝑐𝑢𝑟𝑣𝑒 𝑓𝑖𝑡𝑠 Note 𝑐 𝑣 = 𝑐 𝑝 − 𝑅 𝑢 𝑐 𝑝 = 𝑐 𝑝 /𝑀𝑊

Mixture Properties Use these relations to calculate mixture enthalpy and internal energies (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 Individual gas properties are on pp. 687-699 as functions of gas and T Enthalpy 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) Internal Energy 𝒖 𝒎𝒊𝒙 (𝑻)= 𝒀 𝒊 𝒖 𝒊 (𝑻) 𝒖 𝒎𝒊𝒙 𝑻 = 𝝌 𝒊 𝒖 𝒊 𝑻

Standardized Enthalpy and Enthalpy of Formation Needed for chemically-reacting systems because energy is required to form and break chemical bonds Not considered in Thermodynamics I Needed to find 𝑢 2 − 𝑢 1 and ℎ 𝑜 − ℎ 𝑖 ℎ 𝑖 𝑇 = ℎ 𝑓,𝑖 𝑜 𝑇 𝑟𝑒𝑓 +Δ ℎ 𝑠,𝑖 (𝑇) Standard Enthalpy at Temperature T = Enthalpy of formation at standard reference state: Tref and P° + Sensible enthalpy change in going from Tref to T = 𝑇 𝑟𝑒𝑓 𝑇 𝑐 𝑃 𝑇 𝑑𝑇 Appendices A and B pp 687-702

Normally-Occurring Elemental Compounds For example: O2, N2, C, He, H2 ℎ 𝑓,𝑖 𝑜 𝑇 𝑟𝑒𝑓 = 0 𝑇 𝑟𝑒𝑓 =298K Use these as bases to tabulate the energy for form of more complex compounds Example: At 298K (1 mole) O2 + 498,390 kJ  (2 mole) O To form 2O atoms from one O2 molecule requires 498,390 kJ/kmol of energy input to break O-O bond 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