Transcript Slides
ME 475/675 Introduction to Combustion Lecture 25 Finish Ex. 6.2 and its numerical solution Announcements • HW 9 Due Friday, Oct. 23, 2015 • Example 6.2 and its numerical solution • For Φ = 1, plot product mass fraction versus mass flow rate • Plot blow out Φ versus mass flow rate • At least 4 data points for each plot and compare to Fig. 6.8 of Textbook • Due Friday, Oct. 23, 2015 Steady-State Well-Stirred Reactor • Mass Conservation for M species, 𝑖 = 1, 2, … , 𝑀 • One inlet and one outlet: 𝑚𝑖𝑛 = 𝑚𝑖𝑛 = 𝑚 • 𝑑𝑚𝑖,𝐶𝑉 𝑑𝑡 = 𝜔𝑖 𝑀𝑊𝑖 𝑉 + 𝑚 𝑌𝑖,𝑖𝑛 − 𝑌𝑖,𝑜𝑢𝑡 , 𝑖 = 1,2, … , 𝑀 Accumulation = generation + net inflow • Since well-mixed properties leaving are same as this inside CV • Steady State 𝑑𝑚𝑖,𝐶𝑉 𝑑𝑡 =0 • 𝑚 𝑌𝑖 − 𝑌𝑖,𝑖𝑛 = 𝜔𝑖 𝑀𝑊𝑖 𝑉, 𝑖 = 1,2, … , 𝑀 • Find molar concentrations [i] from mass fractions Yi (needed to find 𝜔𝑖 ) • 𝑖 = 𝑁𝑖 𝑉 = 𝑌𝑖 𝑀𝑊𝑚𝑖𝑥 𝑃 𝑀𝑊𝑖 𝑅𝑢 𝑇 Energy Conservation (to find T) • Steady State: •𝑄=𝑚 𝑑𝑈 𝑑𝑡 =0 𝑌𝑖 ℎ𝑖 𝑇 − 𝑌𝑖,𝑖𝑛 ℎ𝑖,𝑖𝑛 𝑇𝑖𝑛 𝑜 𝑜 • ℎ𝑖 = ℎ𝑓,𝑖 + ∆ℎ𝑠,𝑖 ≈ ℎ𝑓,𝑖 + 𝑐𝑝,𝑖 𝑇 − 𝑇𝑟𝑒𝑓 • Residence Time • 𝑡𝑅 = 𝑚𝐶𝑉 𝑚 = 𝑉𝜌 𝑚 • 𝑃 = 𝜌𝑅𝑇 = 𝜌 = 𝑉 𝑃𝑀𝑊𝑚𝑖𝑥 𝑚 𝑇𝑅𝑢 𝑅𝑢 𝑇; 𝑀𝑊𝑚𝑖𝑥 𝜌= 𝑃𝑀𝑊𝑚𝑖𝑥 𝑇𝑅𝑢 • All these conditions (equations) are algebraic (not differential) • Find 𝑌𝑖 and 𝑇 (M + 1 unknowns) • Linear in 𝑌𝑖 , nonlinear in 𝑇 Example 6.2 • Develop a simplified model of a well-stirred reactor using the same simplified chemistry and thermodynamics used in Example 6.1 (equal and constant cp’s and MW’s, and one-step global kinetics). Use the model to determine the blowout characteristics of a spherical (80-mm-diameter) reactor with premixed reactants (C2H6-air) entering at 298 K. Plot the equivalence ratio at blowout as a function of mass flowrate for Φ ≤ 1.0. Assume the reactor is adiabatic. • MWF= MWOx= MWP= 29; 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, constant volume, or well stirred reactor? Last Lecture • Species Production • 𝜔𝐹𝑢𝑒𝑙 = −6.186 ∗ 109 𝑒𝑥𝑝 • 𝜔𝐹𝑢𝑒𝑙 = −6.186 ∗ 𝐴 𝐹 𝑆𝑇 • 𝜔𝑂𝑥 = −15098𝐾 109 𝑒𝑥𝑝 𝑇 𝜔𝐹𝑢𝑒𝑙 𝐴 𝐹 𝑆𝑇 • 𝜔𝑃𝑟𝑜𝑑 = − + 1 𝜔𝐹𝑢𝑒𝑙 • Inlet Conditions • 𝑌𝐹,𝑖𝑛 = 1 1+ • 𝑌𝑂𝑥,𝑖𝑛 = −15098𝐾 𝑇 𝐴 /Φ 𝐹 𝑆𝑇 1 𝐴 1+Φ/ 𝐹 𝑆𝑇 • 𝑌𝑃𝑟,𝑖𝑛 = 0; • 𝑇𝑖𝑛 = 298𝐾 = 𝑇𝑅𝑒𝑓 𝐹𝑢 𝑌𝐹𝑢 0.1 0.1 𝑂2 1.65 0.233𝑌𝑂𝑥 1.65 𝑃 1.75 𝑅𝑢 𝑇 Species Conservation • 𝑚 𝑌𝑖 − 𝑌𝑖,𝑖𝑛 = 𝜔𝑖 𝑀𝑊𝑖 𝑉, 𝑖 = 𝐹𝑢, 𝑂𝑥, 𝑃𝑟; 𝑀 = 3 • Fuel • 𝑚 𝑌𝐹 − 𝑌𝐹,𝑖𝑛 = 𝜔𝐹 𝑀𝑊𝐹 𝑉 T, 𝑌𝐹 , 𝑌𝑂𝑥 1 • Oxidizer • 𝑚 𝑌𝑂𝑥 − 𝑌𝑂𝑥,𝑖𝑛 = 𝜔𝑂𝑥 𝑀𝑊𝑂𝑥 𝑉 • 𝑚 𝑌𝑂𝑥 − 𝑌𝑂𝑥,𝑖𝑛 = 𝐴 𝐹 𝑆𝑡 𝜔𝐹𝑢𝑒𝑙 𝑀𝑊𝑂𝑥 𝑉 2 T, 𝑌𝐹 , 𝑌𝑂𝑥 • Product • 𝑚 𝑌𝑃𝑟 − 𝑌𝑃𝑟,𝑖𝑛 = 𝜔𝑃 𝑀𝑊𝑃𝑟 𝑉 • 𝑚 𝑌𝑃𝑟 − 𝑌𝑃𝑟,𝑖𝑛 = − • or • 𝑌𝑃𝑟 = 1 − 𝑌𝐹 − 𝑌𝑂𝑥 𝐴 𝐹 𝑆𝑡 + 1 𝜔𝐹𝑢𝑒𝑙 𝑀𝑊𝑃𝑟 𝑉 3 𝑌𝐹 , 𝑌𝑂𝑥 , 𝑌𝑃𝑟 Energy Conservation (to find T) •𝑄=𝑚 𝑌𝑖 ℎ𝑖 𝑇 − 𝑌𝑖,𝑖𝑛 ℎ𝑖,𝑖𝑛 𝑇𝑖𝑛 𝑜 𝑜 • ℎ𝑖 = ℎ𝑓,𝑖 + ∆ℎ𝑠,𝑖 ≈ ℎ𝑓,𝑖 + 𝑐𝑃,𝑖 𝑇 − 𝑇𝑅𝑒𝑓 𝑜 𝑜 • 0 = 𝑌𝑂𝑥,𝑜𝑢𝑡 ℎ𝑓,𝑂𝑥 + 𝑐𝑃 𝑇 − 𝑇𝑅𝑒𝑓 + 𝑌𝐹,𝑜𝑢𝑡 ℎ𝑓,𝐹𝑢 + 𝑐𝑃 𝑇 − 𝑇𝑅𝑒𝑓 + 𝑜𝑢𝑡 𝑜𝑢𝑡 𝑐𝑃 𝑇 − 𝑇𝑅𝑒𝑓 − 𝑌𝑂𝑥 𝑐𝑃 𝑇𝑖𝑛 − 𝑇𝑅𝑒𝑓 − 𝑌𝐹 𝑐𝑃 𝑇𝑖𝑛 − 𝑇𝑅𝑒𝑓 − 𝑜𝑢𝑡 𝑖𝑛 𝑖𝑛 𝑌𝑃𝑟 𝑐𝑃 𝑇𝑖𝑛 − 𝑇𝑅𝑒𝑓 𝑖𝑛 • − 𝑌𝑖,𝑖𝑛 ℎ𝑖,𝑖𝑛 𝑇𝑖𝑛 MathCAD Solution 0.01 .01 3 510 f ( T2 mdot) 0 0 .005 . 1 Yoxin 0.941 phi 1 16 3 510 3 3 110 298 mdot cp 9 15098 Yfin cp (T 298) f ( T mdot) (T 298) MW Vol 6.19 10 exp Hff Hff T 210 T2 0.1 AF cp 0.233 Yoxin ( T 298) Hff 3 310 3000 1.65 P Ru T 1.75 •