Transcript Slides
ME 475/675 Introduction to Combustion Lecture 23 Example 6.1 Excel programing, Effect of equivalence ratio Announcements • HW 8, Numerical Solution to Example 6.1 • Due Monday, Oct. 19, 2015 • Course Project • Construct a homemade camp stove fueled with absolute alcohol • See, for example: • http://www.backpacking.net/makegear/stove/ • https://www.google.com/search?q=homemade+backpacking+stove&biw=1461&bih=708&tb m=isch&tbo=u&source=univ&sa=X&sqi=2&ved=0CBwQsARqFQoTCN_ctrCmx8gCFck0iAodLok G7g • To bring 0.5 liters of room-temperature water to a boil, determine • How much time is required • How much fuel is consumed • Is soot deposited on the water container? • Work in groups of two with our teaching assistant • Hasibul Alam [email protected] • Due date and group assignments soon Example 6.1 • Create a simple constant-volume model of the autoignition process and determine the temperature and the fuel and product concentration histories. Also determine the dP/dt as a function of time. Assume initial conditions corresponding to compression of a fuel-air mixture from 300 K and 1 atm to top-dead-center for a compression ratio of 10:1. The initial volume before compression is 3.68*10-4 m3, which corresponds to an engine with both a bore and a stroke of 75 mm. Use ethane as fuel. Assume: • One-step global kinetics using the rate parameters for ethane C2H6 (Table 5.1) • Fuel, air, and products all have equal molecular weights: MWF= MWOx= MWP= 29 • The specific heats of the fuel, air and products are constants and equal: • cp,F = cp,Ox = cp,Pr = 1200 J/kgK • The enthalpy of formation of the air and products are zero, and that of the fuel is • 4*107 J/kg • The stoichiometric air-fuel ratio is 16.0 and restrict combustion to stoichiometric or lean conditions, Φ ≤ 1. • Constant Pressure, or constant Volume? Last Lecture found Numerical values • Species Production • Assuming 𝑂2 = 0.21 𝐴𝑖𝑟 = 0.21 𝑂𝑥 (only true initially since 𝑁2 is not consumed) • • • 𝑑 𝐹𝑢 𝑑𝑡 𝑑 𝑂𝑥 𝑑𝑡 𝑑 𝑃𝑟 𝑑𝑡 = 𝜔𝐹𝑢𝑒𝑙 = −6.186 ∗ 109 𝑒𝑥𝑝 −15098𝐾 𝑇 𝐹 0.1 0.21 𝑂𝑥 1.65 = 𝜔𝑂𝑥 = 𝐴 𝐹 16𝜔𝐹𝑢𝑒𝑙 = 16𝜔𝐹𝑢𝑒𝑙 = 𝜔𝑃𝑟𝑜𝑑 = − 𝐴 𝐹 + 1 𝜔𝐹𝑢𝑒𝑙 = −17𝜔𝐹𝑢𝑒𝑙 • 1st Law • 𝑑𝑇 𝑑𝑡 = 𝑄 − 𝑉 ℎ𝑗 𝜔𝑗 +𝑅𝑢 𝑇 𝑗 𝑐𝑝,𝑗 −𝑅𝑢 𝜔𝑗 = ℎ𝐹𝑢𝑒𝑙 𝜔𝐹𝑢𝑒𝑙 𝑃 𝑐𝑝 −𝑅𝑢 𝑅 𝑇 𝑢 = 𝑘𝑃𝑚3 𝜔𝐹𝑢𝑒𝑙 𝑇 −364,184 𝑘𝑚𝑜𝑙 𝑃 • Pressure Rate of change (affects detonation) • 𝑃 𝑇 = 𝑁𝑅𝑢 𝑉 = 𝑃0 𝑑𝑃 ; 𝑇0 𝑑𝑡 = 𝑃0 𝑑𝑇 𝑇0 𝑑𝑡 • Initial Conditions at t = 0 • 𝑇0 = • 𝑃0 = 𝑉𝐵𝐷𝐶 𝛾−1 𝑇𝐵𝐷𝐶 = 300𝐾 10 0.4 = 754 𝐾 𝑉𝑇𝐷𝐶 𝑉𝐵𝐷𝐶 𝛾 𝑃𝐵𝐷𝐶 = 101.3𝑘𝑃 10 1.4 = 2545 𝑘𝑃 𝑉𝑇𝐷𝐶 • 𝑃𝑟 • 𝑂𝑥 0 0 =0 = 1 𝑃 𝜙 +1 𝑅𝑢 𝑇 𝐴 𝐹 • 𝐹𝑢 0 = 1 𝑃 𝐴 𝐹 1+ 𝜙 𝑅𝑢 𝑇 Numerical Solution (Excel) dt [sec] phi 1.00E-07 t [sec] [Fuel] 1 0 0.023888 1.00E-07 2.39E-02 2.00E-07 2.39E-02 [Oxidizer] [Products] T P [kPa] d[F]/dt 0.382206 0 753.5659 2544.541 -0.13195 3.82E-01 2.24E-07 7.54E+02 2.54E+03 -0.13196 3.82E-01 4.49E-07 7.54E+02 2.54E+03 -0.13196 • Starting point • http://wolfweb.unr.edu/homepage/greiner/teaching/MECH.475.675.Combustion/Ex6.1.start.xlsx • Make sure [Fu] and [Ox] are ≥ 0 (use MAX(0,value) function in Excel) • Results, Φ = 1, Δ𝑡 = 10−7 𝑠𝑒𝑐, • Low temperature, slow start • Temperature and pressure spike and rapid reaction at t = 0.00307 sec = 3.07 ms • Extinction when both fuel and oxidizer are consumed • Realistic results, but not realistic reaction model, which should be multi-stepped d[Ox]/dt -2.11122 -2.11129 -2.11137 d[Pr]/dt 2.243168 2.243251 2.243333 dT/dt [K/s] dP/dT 14231.31 48054.4 14231.84 48056.17 14232.36 48057.94 Dependence on Φ, Δ𝑡 = −7 10 𝑠𝑒𝑐 Φ = 0.5 Φ = 0.1 • As before the reaction is “delayed” • Initial [Fuel] is decreases • Due to excess air, the mixture takes longer to heat up, and reaches lower temperatures and pressures • Lower temperatures delay reaction to 3.12 and 3.53 ms (3.07 ms for Φ = 1) •