ME 475/675 Introduction to Combustion
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ME 475/675 Introduction to Combustion Lecture 4 Heat of reaction and combustion, constant pressure adiabatic flame temperature Announcements • You may turn in extra credit problem 2.15 next time, if you wish • HW 1 and Extra Credit example Problem 2.14 due now • Problem X1; • Consider the broad education necessary to understand the impact of engineering solutions (Introduction to Combustion) in a global and societal context • Read a news article that describes some issue related to combustion • For example: Energy efficiency, pollution, range land fires, fire safety, nuclear safety • In one paragraph, summarize the article, and indicate how it is related to your interest in combustion and/or this class • Do you need more time? I need everyone to complete this problem. • 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). • ID: Find enthalpy of mixture at 298.15 K on 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 -97.787 -0.8639 a4 a5 20.402 -0.03095 0.040304 0.103807 a6 -60.751 -60.751 • Coefficients π1 to π8 from Page 702 π [πΎ] ; 1000 πΎ ππ½ π β = πππππ • π= • 4184(π1 π π2 + π2 2 π3 + π3 3 • Spreadsheet really helps this calculation + π4 π4 4 − π5 π + π6 ) a8 20.232 Enthalpy of Combustion (or reaction) Products of Ideal Complete Combustion All Cο CO2 Hο H2O 298.15 K, 1 atm Reactants 298.15 K, P = 1 atm Stoichiometric Find ππΌπ < 0 ππππ = 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 • ππΌπ − ππππ = π»π − π»π = π βπ − βπ • ππΌπ = βπ»π = π»π − π»π = π βπ − βπ = πββπ • βπ»π and ββπ Enthalpy of Reaction (< 0 for combustion) • Dependent on T and P of reaction • Heat of Combustion ββπΆ = −ββπ = βπ − βπ > 0 • βπ = standardized enthalpies Next year use acetylene on next sides instead of CH4 Stoichiometric Methane Combustion, CH4 • CH4 + 2 (O2 + 3.76 N2) ο 1 CO2 + 2 H2O + 7.52 N2 • @ 25°C and 1 kmol CH4 • βπ»π = π»π − π»π = Water Vapor π π π 1 βπ + ββπ + 2 βπ + ββπ + 7.52 βπ + ββπ πΆπ2 − 1 βππ + ββπ = = πΆπ»4 π»2 π + 2 βππ + ββπ = + 7.52 βππ + ββπ π2 π π π π π βπ,πΆπ + 2 β − 1 β = π β − π β π π,π ππππ π π,π π ππππ‘ π,π»2 π,π π,πΆπ»4 2 ππ½ ππ½ ππ½ −393,546 + 2 −241,845 − 1 −74,831 ππππ ππππ ππππ p 688 (top of page) ππ½ −802,405 πππππΉπ’ππ π2 π2 p 692 (Heat into system for TR = TP) p 701 Other Bases • Per kg fuel ππ ππππ −802,405 • πππΆπ»4 = 16.043 • ββπ = βπ»π πππΆπ»4 = • Heat of Combustion ππ½ πππππΉπ’ππ ππ 16.043ππππ • ββπ = −ββπ = 50,016 ππ½ πππΉπ’ππ = −50,016 ππ½ πππΉπ’ππ (Heat out for TR = TP) ππ½ • See page 701, LHV = Lower Heating Value = 50,016 πππΉπ’ππ • Corresponds to water vapor in the products π π π • βπ»π ,πΏππ€ππ = βπ,πΆπ + 2 β − 1 β π,π» π,π£ππππ π,πΆπ» 2 2 4 π π • βπ,π» = β π,π»2 π,π£ππππ − βπ»2 π,ππ = −241,845 2 π,πΏπππ’ππ p 692 ππ½ ππππ π π π • βπ»π ,π»ππβππ = βπ,πΆπ + 2 β − 1 β π,π» π,πΏπππ’ππ π,πΆπ» 2 2 4 • = −393,546 + 2 −241,845 − 1 −74,831 = −890,425 ππ½ • ββπΆ = − −890,425ππππ πΉπ’ππ ππ 16.043ππππ = 55,502 ππ½ πππΉπ’ππ ππ½ • p. 701: Higher Heating Value = HHV = 55,528ππ πΉπ’ππ − 44,010 p 692 ππ½ ππππ ππ½ πππΉπ’ππ (slightly larger due to dissociation?) = −285,855 ππ½ ππππ Per kg of reactant mixture • ππΉπ’ππ ππππ₯ • π΄ πΉ = = ππΉπ’ππ ππΉπ’ππ +ππ΄ππ ππ΄ππ πππ΄ππ ππΉπ’ππ πππΉπ’ππ = • LHV = ββπ,πΏππ€ππ = = 1 = ππ΄ππ 1+π πΉπ’ππ 2∗ 3.76+1 ∗28.85 1∗16.043 ππ½ 50,016 πππΉπ’ππ 1 π΄ 1+πΉ = 1 1+17.12 = πππ΄ππ 17.12 πππΉπ’ππ ∗ 1 πππΉπ’ππ 18.12 πππππ₯ = = 1 πππΉπ’ππ 18.12 πππππ₯ ππ½ 2760 πππππ₯ Adiabatic (π = 0) Flame Temperature Complete Combustion Products Cο CO2 Hο H2O PP = PR, T = TAd Stoichiometric Reactants TR PR ππΌπ = 0 ππππ = 0 • 1st Law, Steady Flow Reactor • ππΌπ − ππππ = 0 = π»π − π»π = π βπ − βπ • All chemical energy goes into heating the products • To find adiabatic flame temperature use • PP = PR and βπ = βπ Example: Problem 2.30 page 72 • Next year use Acetylene • Determine the adiabatic flame temperature for constantpressure combustion of a stoichiometric propane-air mixture assuming reactants at 298 K, no dissociation of the products, and constant specific heats evaluated at 298 K. End 2015 Adiabatic Methane Combustion TR = 25°C (next year do problem 2.30 here) • CH4 + 2 (O2 + 3.76 N2) ο 1 CO2 + 2 H2O + 7.52 N2 • π»π ππππ‘ = 1 βππ + ββπ πΆπ»4 • = π»ππππ = 1 βππ + ββπ + 2 βππ + ββπ πΆπ2 π = βπ,πΆπ» 4 π2 + 2 βππ + ββπ + 7.52 βππ + ββπ π»2 π π2 π ππππ‘ + 7.52 βππ + ββπ π π π • βπ,πΆπ» − 1 β − 2 β π,πΆπ2 π,π»2 π = 1ββπ ,πΆπ2 + 2ββπ ,π»2 π + 7.52ββπ ,π2 4 • ββπ ,π,ππ΄π = π π ππ ππ π,π π ππ ≈ ππ,π ππ΄π − ππ ππ π2 ππππ ππ΄π Example (Turn in next time for Extra Credit) • Find TAd for a 25°C Stoichiometric mixture of Acetylene and air “Meaning” of Mixture Standardized Enthalpy • Reactants and products are both at 27°C and 1 atm • 8πΆ + 9π»2 + 12.5π2 + 47π2 + βπ π → πΆ8 π»18 + 12.5π2 + 47π2 • Reactants are naturally-occurring elemental compounds • All have zero standardized enthalpy at 27°C and 1 atm • Need to add βπ π to reactants to get product mixture at same temperature and pressure “Meaning” of Enthalpy (Heat) of Formation • Reactants and products are both at 27°C and 1 atm • 8πΆ + 9π»2 + βπ π → πΆ8 π»18 (fuel heat of formation) • Reactants are naturally-occurring elemental compounds • All have zero standardized enthalpy at 27°C and 1 atm • Describes how much energy needs to be added to C and π»2 to create the fuel. • See page 701 • For most hydrocarbon fuels βππ < 0, • πΆ + π»2 mixture has more energy than fuel • Exceptions • Acetylene πΆ2 π»2 , Ethene πΆ2 π»4 , propene πΆ2 π»4 , 1-Butene πΆ4 π»8 , Benzene πΆ6 π»6 • Small number of H per C (lots of double bonds between C atoms)
