Transcript Slides
ME 475/675 Introduction to Combustion Lecture 35 Midterm 2 • Average = 84 Announcements • HW 14 Ch. 8 (1, 13, 15, 16) • Due Friday, November 20, 2015 • Integrated BS/MS Degree • http://www.unr.edu/engineering/academics/accelerated • Term Project • http://wolfweb.unr.edu/homepage/greiner/teaching/MECH.475.675.Combustion/TermProjectAssignment.pdf • Due December 4, 2015 • Email Hasib ([email protected]) to schedule times to complete stove construction and measurements in PE 113 • Bring one or two aluminum cans with you Laminar Flame Propagation Demonstration • Courtesy of Corey Trujillo (2015) • Slow motion • Full speed • How to Flame Quenching, Mixture Flammability, Ignition • What does it take to ignite a mixture? • What does it take to extinguish a flame? • “Williams Criteria” (a rough model, rule of thumb) • Ignition will occur if enough energy is added to a slab of thickness 𝛿 (laminar flame thickness) to raise it to the adiabatic flame temperature, Tad. • A flame will be sustained if its rate of chemical heat release insides a slab is roughly equal to heat loss by conduction out of the slab. Last time: Quenching Distance Data Table 8.4 page 291 • Smallest dimension that allows flame to pass • • Near stoichiometric conditions many hydrocarbon fuels require a mesh sizes around 2 mm to quench • For Ethylene and (C2H4) and hydrogen (H2), it must be smaller • For non-stoichiometric conditions, larger meshes will quench flame Flammability Limits 𝜈= 𝐴 𝐹 𝑠𝑡𝑜𝑖𝑐 • Flames only propagate within certain equivalence ratio ranges • Φ𝑚𝑖𝑛 < Φ < Φ𝑚𝑎𝑥 , • Φ𝑚𝑖𝑛 = Φ𝑙𝑜𝑤𝑒𝑟 = Φ𝑙𝑒𝑎𝑛 • Φ𝑚𝑎𝑥 = Φ𝑢𝑝𝑝𝑒𝑟 = Φ𝑟𝑖𝑐ℎ • See page 291, Table 8.4 for limits • Φ= 𝐴 𝐹 𝑠𝑡𝑜𝑖𝑐 𝐴 𝐹 • 𝜒𝐶𝐻4 ,𝐿𝑒𝑎𝑛 = 𝜈 =𝐴 = 𝐹 𝑁𝐶𝐻4 𝑚𝐹 𝑚𝐴,𝑆𝑡 𝑚𝐹,𝑆𝑡 𝑚𝐴 1 𝑁𝐶𝐻4 +𝑁𝐴𝑖𝑟 = 𝑁 1+𝑁 𝐴𝑖𝑟 𝐶𝐻4 = 1 𝑚 𝑀𝑊𝐶𝐻 1+𝑚 𝐴𝑖𝑟 𝑀𝑊 4 𝐶𝐻4 𝐴𝑖𝑟 Ignition 𝑑𝑇 𝑑𝑥 𝑅𝐶𝑟𝑖𝑡 • The minimum electrical spark energy capable of igniting a flammable mixture. • It is dependent on the temperature, pressure and equivalence ratio of the mixture • What is the critical (minimum) radius of a spark that will propagate • 𝑄′′′ 𝑉 > 𝑄𝑐𝑜𝑛𝑑 • −𝑚𝐹′′′ 2 • 𝑅𝑐𝑟𝑖𝑡 ≥ • 𝑆𝐿2 Δℎ𝑐 4 3 𝜋𝑅𝑐𝑟𝑖𝑡 3 3𝑘 𝑇𝑏 −𝑇𝑢 ′′′ Δℎ −𝑚𝐹 𝑐 = 2𝛼 1 + 𝜈 = ′′′ −𝑚𝐹 𝜌𝑢 𝑑𝑇 𝑇𝑏 −𝑇𝑢 2 >𝑘 ~𝑘 4𝜋𝑅𝑐𝑟𝑖𝑡 𝑑𝑥 𝑅𝐶𝑟𝑖𝑡 𝑅𝑐𝑟𝑖𝑡 2𝛼 1+𝜈 3𝑘 𝑇𝑏 −𝑇𝑢 6𝛼2 𝛼 6 = ; 𝑅 ≥ 6 = 𝑐𝑟𝑖𝑡 𝜌𝑢 𝑆𝐿2 𝑐𝑝 1+𝜈 𝑇𝑏 −𝑇𝑢 𝑆𝐿2 𝑆𝐿 2 ; 2 4𝜋𝑅𝑐𝑟𝑖𝑡 1 ′′′ −𝑚𝐹 = 2𝛼 1+𝜈 𝜌𝑢 𝑆𝐿2 ; Δℎ𝑐 = 𝑐𝑝 1 + 𝜈 𝑇𝑏 − 𝑇𝑢 ; 𝛼 = 𝑘 ; 𝜌𝑢 𝑐𝑝 𝛿 𝛿= 2𝛼 𝑆𝐿 Energy to bring critical volume from Tu to Tb • 𝐸𝑖𝑔𝑛 = 𝑚𝑐𝑟𝑖𝑡 𝑐𝑝 𝑇𝑏 − 𝑇𝑢 • 𝑚𝑐𝑟𝑖𝑡 = 4 3 𝜋𝑅𝑐𝑟𝑖𝑡 𝜌𝑏 3 • 𝐸𝑖𝑔𝑛 = 61.56𝑃 • • = 4 𝜋 3 𝛼 3 6 𝜌𝑏 𝑆𝐿 𝛼 3 𝑐𝑝 𝑇𝑏 −𝑇𝑢 𝑆𝐿 𝑅𝑏 𝑇𝑏 𝑇𝑢 𝑇 0.75 𝛼~ ; 𝑃 𝐸𝐴 0.375 0 𝑆𝐿 ~𝑇 𝑃 𝑒𝑥𝑝 2𝑅𝑢 𝑇𝑏 = 61.56 𝑃 ~ 3 ~𝑃−2 𝑃 • not normally considered reliable • Agrees with measurements at low pressure • Need lots of energy at low pressure • Hard to restart jet engines at high altitudes • 𝐸𝑖𝑔𝑛 decreases as Tu increases • Table 8.5 page 298 Different fuels 𝛼 3 𝑃 𝑆𝐿 𝑅𝑏 𝑇 𝑏 (why 𝜌𝑏 and not 𝜌𝑢 ?) Data • Acetylene (C2H2), hydrogen (H2) and ethylene (C2H4) require very little energy to ignite • Depends on fuel and equivalence ratio • And initial (unburned) temperature and pressure (next slide) 𝑚𝑒𝑡ℎ𝑎𝑛𝑒 • Chapter 16 Detonations • Flames (deflagrations) are subsonic • Detonations are supersonic • Flame speed 𝑣𝑥 > 𝑐 = • Mach Number: 𝑀𝑎 = 𝑣𝑥 𝑐 𝛾𝑅𝑇 = 𝛾 𝑅𝑢 𝑇, 𝑀𝑊𝑚𝑖𝑥 𝛾= 𝑐𝑃 𝑐𝑣 >1 >1 • Sound speed in room temperature air 𝑃𝑎 𝑚3 • 𝑐= 1.4 8315𝑘𝑚𝑜𝑙 𝐾 𝑘𝑔 𝑘𝑚𝑜𝑙 28.85 298𝐾 = 347 𝑚 3600𝑠 𝑠 ℎ𝑟 100𝑐𝑚 1𝑖𝑛 𝑓𝑡 1 𝑚𝑖𝑙𝑒 𝑚 2.54𝑐𝑚 12 𝑖𝑛 5280 𝑓𝑡 = 776 𝑚𝑖𝑙𝑒 ℎ𝑟 = 0.21 𝑚𝑖𝑙𝑒 𝑠 • 𝑐 increases with increasing T and decreasing 𝑀𝑊𝑚𝑖𝑥 • A detonation is a shockwave sustained by combustion energy release • 𝐶𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑠𝑖𝑜𝑛 𝑠ℎ𝑜𝑐𝑘 → ℎ𝑖𝑔ℎ 𝑡𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 → 𝑐𝑜𝑚𝑏𝑢𝑠𝑡𝑖𝑜𝑛 → 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑠ℎ𝑜𝑐𝑘 • Mass flux rates • ′′ 𝑚𝐷𝑒𝑓 ~ 𝑘𝑔 1 2 𝑚 𝑠 ′′ • 𝑚𝐷𝑒𝑡 ~ 2000 𝑘𝑔 𝑚2 𝑠 • https://www.youtube.com/watch?v=r17czTWHFmU&index=1&list=PL8ilW_a2YUUC3fWrYT06kq8x_LuEHGJPE Confined planar and spherical systems Hot Compression Shock Combustible Mixture 𝑣𝑥1 Combustion Zone Products of Combustion High Temperature T and Pressure P Low density, r Combustible Mixture 𝑣𝑥2 • Closed tube or expanding spark (both have confined hot products) • Expansion of hot gases behind shock may be much faster than laminar or turbulent flame speed • Flame Frame 2 1 Shock Detonation • Unknowns: 𝑃2 , 𝑇2 , 𝜌2 , 𝑐2 = 𝛾𝑅𝑇2 , 𝑎𝑛𝑑 𝑀𝑎2 (𝑣𝑥,2 ) Typical Properties • Analysis 1 2 Detonation • One-dimensional, const. area, 𝑃𝑉 = 𝑁𝑅𝑢 𝑇; 𝑐𝑝,1 = 𝑐𝑝,2 ≠ 𝑓𝑛 𝑇 ; 𝑄 = 0; no body forces 𝑚 • Mass Conservation: = 𝑚′′ = 𝜌1 𝑣𝑥,1 = 𝜌2 𝑣𝑥,2 (𝜌1 ≠ 𝜌2 compressible) • 𝑣𝑥,1 = 𝑚′′ ; 𝜌1 • Momentum: 𝑣𝑥,2 = 𝑚′′ 𝜌2 𝐴 𝐹 = Δ𝑀 (Rate of momentum flux change) 2 2 • 𝐴 𝑃1 − 𝑃2 = 𝐴 𝜌2 𝑣𝑥,2 − 𝜌1 𝑣𝑥,1 2 2 • 𝑃1 + 𝜌1 𝑣𝑥,1 = 𝑃2 + 𝜌2 𝑣𝑥,2 • Combine (find relation between 𝑃2 and 𝑣2 = • • • 1 𝜌2 2 2 𝑚′′ 𝑚′′ 1 1 𝑃1 + 𝜌1 = 𝑃2 + 𝜌2 ; 𝑚′′ 2 − 𝜌1 𝜌2 𝜌1 𝜌2 𝑃2 −𝑃1 𝑃 −𝑃 − 𝑚′′ 2 = = 2 1 < 0, where 𝑣 = 1 1 𝜌2 −1 𝜌1 𝑣2 −𝑣1 𝑃2 = 𝑃1 − 𝑚′′ 2 𝑣2 − 𝑣1 = − 𝑚′′ 2 𝑣2 + 𝑃1 + • As 𝑣2 increased 𝑃2 decreases. given 𝑃1 and 𝑣1 ) = 𝑃2 − 𝑃1 𝜌 𝑚′′ 2 𝑣1 Rayleigh Line 1 2 Detonation • Momentum and mass conservation require • − 𝑚′′ 2 = • 𝑃2 = 𝑃1 − 𝑃2 −𝑃1 𝑃2 −𝑃1 = < 0 (negative slope) 1 𝜌2 −1 𝜌1 𝑣2 −𝑣1 𝑚′′ 2 𝑣2 − 𝑣1 = − 𝑚′′ 2 𝑣2 + 𝑃1 + • No solutions are possible in quadrants A and B • If 𝑣2 > 𝑣1 𝑣𝑥,2 > 𝑣𝑥,1 , then 𝑃2 < 𝑃1 • If 𝑣2 < 𝑣1 𝑣𝑥,2 < 𝑣𝑥,1 , then 𝑃2 > 𝑃1 𝑚′′ 2 𝑣1 Energy Conservation 1 • Include kinetic energy • 𝑚 ℎ1 + 2 𝑣𝑥1 2 • ℎ 𝑇 = • 𝑜 𝑌 ℎ 1 𝑖 𝑓,𝑖 • 𝑜 𝑌 ℎ 1 𝑖 𝑓,𝑖 = 𝑚 ℎ2 + 𝑜 𝑌𝑖 ℎ𝑓,𝑖 2 𝑣𝑥2 2 + 𝑐𝑝 𝑇1 − 𝑇𝑟𝑒𝑓 + − 𝑜 𝑌 ℎ 2 𝑖 𝑓,𝑖 • 𝑞 + 𝑐𝑝 𝑇1 − 𝑇2 = Detonation 𝑇 𝑌𝑖 𝑇 𝑐𝑝,𝑖 𝑑𝑇 𝑟𝑒𝑓 2 𝑣𝑥1 + 2 𝑜 𝑌𝑖 ℎ𝑓,𝑖 = = 𝑜 𝑜 • 𝑞 = 1 𝑌𝑖 ℎ𝑓,𝑖 − 2 𝑌𝑖 ℎ𝑓,𝑖 = = heat of combustion • 𝑞 = 𝑓𝑛 𝑓𝑢𝑒𝑙, 𝑜𝑥𝑖𝑑𝑖𝑧𝑒𝑟, Φ 𝑇 𝑌𝑖 𝑐𝑝,𝑖 𝑇 𝑑𝑇 𝑟𝑒𝑓 + 𝑜 𝑌 ℎ 2 𝑖 𝑓,𝑖 + 𝑐𝑝 𝑇1 − 𝑇2 = 2 −𝑣 2 𝑣𝑥2 𝑥1 2 2 + 𝑐𝑝 𝑇2 − 𝑇𝑟𝑒𝑓 + 2 −𝑣 2 𝑣𝑥2 𝑥1 2 𝑜 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠 𝑌𝑖 ℎ𝑓,𝑖 − = 𝑜 𝑌 ℎ 𝑖 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 𝑓,𝑖 𝑜 𝑌𝑖 ℎ𝑓,𝑖 + 𝑐𝑝 𝑇 − 𝑇𝑟𝑒𝑓 2 𝑣𝑥2 2 Rankine-Hugoniot Curve • 𝑞 + 𝑐𝑝 𝑇1 − 𝑇2 = 2 −𝑣 2 𝑣𝑥2 𝑥1 2 • Eliminate T using ideal gas state, • and assuming 𝑀𝑊1 = 𝑀𝑊2 so 𝑅1 = 𝑅2 = 𝑅 • 𝑃1 = 𝜌1 𝑅1 𝑇1 ; 𝑅1 = 𝑅𝑢 𝑀𝑊1 • 𝑃2 = 𝜌2 𝑅2 𝑇2 ; 𝑅2 = 𝑅𝑢 𝑀𝑊2 • Also use 𝑐𝑃 𝑅 = 𝑐𝑃 𝑐𝑃 −𝑐𝑣 = 𝛾 ; 𝛾−1 𝛾= 𝑐𝑃 𝑐𝑣 >1 • Get… (Homework problem 16.1, p. 637) • 𝛾 𝑃2 𝛾−1 𝜌2 or • 𝛾 𝛾−1 − 𝑃1 𝜌1 − 1 2 𝑃2 𝑣2 − 𝑃1 𝑣1 − 𝑃2 − 𝑃1 𝑃2 −𝑃1 2 1 𝜌2 + 𝑣2 + 𝑣1 1 𝜌1 −𝑞 =0 − 𝑞=0 • 𝑃2 versus 𝑣2 is a hyperbola (for given 𝑃1 , 𝑣1 , 𝛾 and 𝑞) Rankine-Hougoniot Curve 𝑃 = 𝑃2 Rayleigh Lines • 𝛾 𝛾−1 𝑃2 𝑣2 − 𝑃1 𝑣1 − 𝑃2 −𝑃1 2 𝑣2 + 𝑣1 − 𝑞 = 0 • It can be shown (take may word for it) • • • • • Above D (tangent), 𝑣𝑥,2 < 𝑐: Strong detonation (small 𝑣2 ) Between D and B, 𝑣𝑥,2 > 𝑐: Weak detonation Between B and C, impossible (positive slope) Between C and E, 𝑣𝑥,2 < 𝑐: Weak deflagration (large 𝑣2 ) Below E, 𝑣𝑥,2 > 𝑐: Strong deflagration • Tangent with Rayleigh Lines • Upper and lower Chapman-Jouguet points • Real Detonation at point D (based on stability, c = 1) • Note 𝑘𝑔 ′′ 𝑚𝑑𝑒𝑓𝑙𝑎𝑔𝑟𝑎𝑡𝑖𝑜𝑛 ~1 2 ; 𝑚 𝑘𝑔 ′′ 𝑚𝑑𝑒𝑡𝑜𝑛𝑎𝑡𝑖𝑜𝑛 ~2000 2 𝑚 𝑣 = 𝑣2 End 2015 • Had a little extra time Example 16.1 • A combustion wave propagates with a mass flux of 3500 kg/(s m2) through a mixture initially at 298 K and 1 atm. The molecular weights and specific-heat ratio of the mixture (burned and unburned) are 29.0 kg/kmol and 1.3, respectively, and the heat release is 3.40x106 J/kg. • Determine the state (P2, v2) of the burned gas and determine in which region the state lies on the Rankine-Hougoniot curve. • Also, determine the Mach number of the burned gas.