Transcript Slides
ME 475/675 Introduction to Combustion Lecture 4 Announcements • Extra Credit example due now • HW 1 Due Friday • Tutorials • Wednesday 1 pm PE 113 • Thursday 2 pm PE 113 • Please bring you textbook to class • Please turn in HW on white or engineering paper • Grading based on solution (not solely on answers) Example: • Last lecture (turned in today) • 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 • This time (turn in next lecture) • Problem 2.15: Repeat for T = 500 K Standard Enthalpy of Isooctane T [K] 298.15 theta 0.29815 h [kJ/Kmol] -224108.82 a1 a2 -0.55313 181.62 -0.16492 8.072412 a3 a4 a5 -97.787 20.402 -0.03095 -0.8639 0.040304 0.103807 a6 -60.751 -60.751 • Coefficients 𝑎1 to 𝑎8 from Page 702 𝑇 [𝐾] ; 1000 𝐾 𝑘𝐽 𝑜 ℎ = 𝑘𝑚𝑜𝑙𝑒 •𝜃= • 4184(𝑎1 𝜃 𝜃2 + 𝑎2 2 𝜃3 + 𝑎3 3 • Spreadsheet really helps this calculation + 𝜃4 𝑎4 4 − 𝑎5 𝜃 + 𝑎6 ) a8 20.232 Enthalpy of Combustion (or reaction) Products Complete Combustion CCO2 HH2O 298.15 K, 1 atm Reactants 298.15 K, P = 1 atm Stoichiometric 𝑄𝐼𝑁 < 0 𝑊𝑂𝑈𝑇 = 0 • How much energy is released from a reaction if the product and reactant temperatures and pressures are the same? • 1st Law, Steady Flow Reactor • 𝑄𝐼𝑁 − 𝑊𝑂𝑈𝑇 = 𝐻𝑃 − 𝐻𝑅 = 𝑚 ℎ𝑃 − ℎ𝑅 • 𝑄𝐼𝑁 = ∆𝐻𝑅 = 𝐻𝑃 − 𝐻𝑅 = 𝑚 ℎ𝑃 − ℎ𝑅 = 𝑚∆ℎ𝑅 • ∆𝐻𝑅 and ∆ℎ𝑅 Enthalpy of Reaction (< 0 for combustion) • Dependent on T and P of reaction • Heat of Combustion ∆ℎ𝐶 = −∆ℎ𝑅 = ℎ𝑅 − ℎ𝑃 > 0 Stoichiometric Methane Combustion, CH4 • CH4 + __ (O2 + 3.76 N2) __ CO2 + __ H2O + ___ N2 • @ 25C and 1 kmol CH4 • ∆𝐻𝑅 = 𝐻𝑃 − 𝐻𝑅 = Water Vapor 1 ℎ𝑓𝑜 + ∆ℎ𝑠 + 2 ℎ𝑓𝑜 + ∆ℎ𝑠 + 7.52 ℎ𝑓𝑜 + ∆ℎ𝑠 𝐶𝑂2 − 1 𝑜 ℎ𝑓 + ∆ℎ𝑠 𝐶𝐻4 𝐻2 𝑂 + 2 𝑜 ℎ𝑓 + ∆ℎ𝑠 𝑜 𝑜 𝑜 = ℎ𝑓,𝐶𝑂 + 2 ℎ − 1 ℎ = 𝑓,𝐻 𝑂 𝑓,𝐶𝐻 2 2 4 = 𝑘𝐽 −393,546 𝑘𝑚𝑜𝑙 p 688 = −802,405 +2 𝑘𝐽 𝑘𝑚𝑜𝑙𝐹𝑢𝑒𝑙 𝑂2 + 7.52 𝑜 𝑁𝑖 ℎ𝑓,𝑖 𝑘𝐽 −241,845 𝑘𝑚𝑜𝑙 𝑃𝑟𝑜𝑑 −1 𝑜 ℎ𝑓 − 𝑁2 + ∆ℎ𝑠 𝑁2 𝑜 𝑁𝑖 ℎ𝑓,𝑖 𝑅𝑒𝑎𝑐𝑡 𝑘𝐽 −74,831 𝑘𝑚𝑜𝑙 p 692 p 701 (Heat in to system for TR = TP) Other Bases • Per kg fuel • 𝑀𝑊𝐶𝐻4 = 16.043 𝑘𝑔 𝑘𝑚𝑜𝑙 𝑘𝐽 • ∆ℎ𝑅 = − 802,405𝑘𝑚𝑜𝑙 𝐹𝑢𝑒𝑙 𝑘𝑔 16.043𝑘𝑚𝑜𝑙 • Heat of Combustion • ∆ℎ𝑐 = −∆ℎ𝑅 = 50,016 = −50,016 𝑘𝐽 𝑘𝑔𝐹𝑢𝑒𝑙 𝑘𝐽 𝑘𝑔𝐹𝑢𝑒𝑙 (Heat out for TR = TP) 𝑘𝐽 • See page 701, LHV = Lower Heating Value = 50,016 𝑘𝑔𝐹𝑢𝑒𝑙 • Corresponds to water vapor in the products 𝑜 𝑜 𝑜 • ∆𝐻𝑅,𝐿𝑜𝑤𝑒𝑟 = ℎ𝑓,𝐶𝑂 + 2 ℎ − 1 ℎ 𝑓,𝐻2 𝑂,𝑣𝑎𝑝𝑜𝑟 𝑓,𝐶𝐻4 2 • 𝑜 ℎ𝑓,𝐻 2 𝑂,𝐿𝑖𝑞𝑢𝑖𝑑 = 𝑜 ℎ𝑓,𝐻 2 𝑂,𝑣𝑎𝑝𝑜𝑟 − ℎ𝐻2𝑂,𝑓𝑔 = 𝑘𝐽 −241,845 𝑘𝑚𝑜𝑙 p 692 − 𝑘𝐽 44,010 𝑘𝑚𝑜𝑙 p 692 𝑜 𝑜 𝑜 • ∆𝐻𝑅,𝐻𝑖𝑔ℎ𝑒𝑟 = ℎ𝑓,𝐶𝑂 + 2 ℎ − 1 ℎ 𝑓,𝐻 𝑂,𝐿𝑖𝑞𝑢𝑖𝑑 𝑓,𝐶𝐻 2 2 4 • = −393,546 + 2 −241,845 − 1 −74,831 = −890,425 • ∆ℎ𝐶 = − 𝑘𝐽 𝑘𝑚𝑜𝑙𝐹𝑢𝑒𝑙 𝑘𝑔 16.043𝑘𝑚𝑜𝑙 −890,425 = 55,502 𝑘𝐽 𝑘𝑔𝐹𝑢𝑒𝑙 • p. 701: Higher Heating Value = HHV = 55,528 𝑘𝐽 𝑘𝑔𝐹𝑢𝑒𝑙 𝑘𝐽 𝑘𝑔𝐹𝑢𝑒𝑙 (slightly larger due to dissociation?) = 𝑘𝐽 −285,855 𝑘𝑚𝑜𝑙 Per kg of reactant mixture • 𝑚𝐹𝑢𝑒𝑙 𝑚𝑀𝑖𝑥 • 𝐴 𝐹 = = 𝑚𝐹𝑢𝑒𝑙 𝑚𝐹𝑢𝑒𝑙 +𝑚𝐴𝑖𝑟 𝑁𝐴𝑖𝑟 𝑀𝑊𝐴𝑖𝑟 𝑁𝐹𝑢𝑒𝑙 𝑀𝑊𝐹𝑢𝑒𝑙 = • LHV = ∆ℎ𝑐,𝐿𝑜𝑤𝑒𝑟 = = 1 𝑚𝐴𝑖𝑟 1+ 𝑚𝐹𝑢𝑒𝑙 = 1 𝐴 1+ 𝐹 = 1 1+17.12 = 1 𝑘𝑔𝐹𝑢𝑒𝑙 18.12 𝑘𝑔𝑀𝑖𝑥 2∗ 3.76+1 ∗28.85 𝑘𝑔𝐴𝑖𝑟 = 17.12 1∗16.043 𝑘𝑔𝐹𝑢𝑒𝑙 𝑘𝐽 1 𝑘𝑔𝐹𝑢𝑒𝑙 𝑘𝐽 50,016 ∗ = 2760 𝑘𝑔𝐹𝑢𝑒𝑙 18.12 𝑘𝑔𝑀𝑖𝑥 𝑘𝑔𝑀𝑖𝑥 Adiabatic (𝑄 = 0) Flame Temperature Complete Combustion Products CCO2 HH2O PP = PR, T = TAd Stoichiometric Reactants TR PR 𝑄𝐼𝑁 = 0 𝑊𝑂𝑈𝑇 = 0 • 1st Law, Steady Flow Reactor • 𝑄𝐼𝑁 − 𝑊𝑂𝑈𝑇 = 0 = 𝐻𝑃 − 𝐻𝑅 = 𝑚 ℎ𝑃 − ℎ𝑅 • All chemical energy goes into heating the products • To find adiabatic flame temperature use • PP = PR and ℎ𝑃 = ℎ𝑅 Adiabatic Methane Combustion TR = 25°C • CH4 + 2 (O2 + 3.76 N2) 1 CO2 + 2 H2O + 7.52 N2 • 𝐻𝑅𝑒𝑎𝑐𝑡 = 1 ℎ𝑓𝑜 + ∆ℎ𝑠 𝐶𝐻4 • = 𝐻𝑃𝑟𝑜𝑑 = 1 ℎ𝑓𝑜 + ∆ℎ𝑠 + 2 ℎ𝑓𝑜 + ∆ℎ𝑠 𝐶𝑂2 𝑜 = ℎ𝑓,𝐶𝐻 4 𝑂2 + 2 ℎ𝑓𝑜 + ∆ℎ𝑠 + 7.52 ℎ𝑓𝑜 + ∆ℎ𝑠 𝐻2 𝑂 𝑁2 𝑅𝑒𝑎𝑐𝑡 + 7.52 ℎ𝑓𝑜 + ∆ℎ𝑠 𝑜 𝑜 𝑜 • ℎ𝑓,𝐶𝐻 − 1 ℎ − 2 ℎ 𝑓,𝐶𝑂2 𝑓,𝐻2 𝑂 = 1∆ℎ𝑠,𝐶𝑂2 + 2∆ℎ𝑠,𝐻2 𝑂 + 7.52∆ℎ𝑠,𝑁2 4 • ∆ℎ𝑠,𝑖,𝑇𝐴𝑑 = 𝑇 𝑐 𝑇𝑅𝑒𝑓 𝑝,𝑖 𝑇 𝑑𝑇 ≈ 𝑐𝑝,𝑖 𝑇𝐴𝑑 − 𝑇𝑅𝑒𝑓 𝑁2 𝑃𝑟𝑜𝑑 𝑇𝐴𝑑 Example (Turn in next time for Extra Credit) • Find TAd for a 25°C Stoichiometric mixture of Acetylene and air