Slides

ME 475/675 Introduction to Combustion Lecture 32 Midterm 2 review Announcements • Midterm 2 • Wednesday, November 12, 2014 • Extra Tutorials: • Monday 10-11 MS 227 • Monday 5-6 PE 113 • Tuesday 4-5 PE 105 • HW 13 Ch 8 (3, 12) • Due now • Term Project • Reduce to 3% of grade (originally 5%) • Move other 2% to HW, or to Midterm 2 and Final? • http://wolfweb.unr.edu/homepage/greiner/teaching/MECH.475.675.Combustion/TermProjectAssignment.pdf Midterm 2 • Test Format • 3-4 problems (with parts) • Test Rules • Open book (with bookmarks and notes in book) • One page of notes (if needed) • Test Coverage • All material since last exam • Chapters 4-7 plus the part of 8 that we covered in class and HW • HW 6-13, lecture notes and examples Chapter 5, Global Reaction Rate • One-step hydrocarbon combustion reaction • 𝐶𝑥 𝐻𝑦 + 𝑥 + 𝑦 4 𝑂2 𝑘𝐺 𝑦 𝑥𝐶𝑂2 + 𝐻2 𝑂 2 (𝐹 + 𝑎𝑂𝑥 → 𝑏𝑃𝑟) • For stoichiometric mixture with 𝑂2 , not air • Made up of many intermediate steps that are not seen • Overall reaction rate (empirical, black box, approximates observations) • 𝑑 𝐶𝑥 𝐻𝑦 𝑑𝑡 = −𝐴𝑒𝑥𝑝 • 𝑖 = 𝜒𝑖 𝑃 𝑅𝑢 𝑇 = 𝐸𝑎 𝑅𝑢 𝑇 𝑃𝑖 𝑅𝑢 𝑇 • Page 157, Table 5.1 = 𝜒𝑖 𝐶𝑥 𝐻𝑦 ρ 𝑀𝑊𝑀𝑖𝑥 𝑚 𝑂2 𝑛 = 𝑔𝑚𝑜𝑙𝑒 𝑐𝑚3 𝑠 • 𝐴, 𝐸𝑎 𝑅𝑢 , 𝑚 𝑎𝑛𝑑 𝑛 for different HC fuels • These values are based on flame speed data fit (Ch. 8) • Units: 𝐴 1 𝑘𝑚𝑜𝑙𝑒 1−𝑚−𝑛 𝑠 𝑚3 Usually Want These Units (add to table?) = 𝐴 𝑇𝑒𝑥𝑡𝑏𝑜𝑜𝑘 10001−𝑚−𝑛 Given in Table 5.1, p. 157 1 𝑔𝑚𝑜𝑙𝑒 1−𝑚−𝑛 𝑠 𝑐𝑚3 Chapter 4 Chemical Kinetics • Global (apparent) reaction (are what are observed) • 𝐹 + 𝑎𝑂𝑥 → 𝑏𝑃𝑟 • They are made up of many intermediate steps that are not seen • Bi-molecular, 𝐴 + 𝐵 → 𝐶 + 𝐷 • 𝑑𝐴 𝑑𝑡 = −𝑘𝑏𝑖𝑚𝑜𝑙𝑒𝑐 𝐴 1 𝐵 1 • Uni-molecular, 𝐴 → 𝐵, or 𝐴 → 𝐵 + 𝐶 • Low pressure: 𝑑𝐴 𝑑𝑡 = −𝑘𝑢𝑛𝑖 𝐴 ; High pressure: 𝑑𝐴 𝑑𝑡 = −𝑘𝑢𝑛𝑖 𝐴 𝑀 • Ter-molecular : 𝐴 + 𝐵 + 𝑀 → 𝐶 + 𝑀 (Recombination) • 𝑑𝐴 𝑑𝑡 = −𝑘𝑡𝑒𝑟 𝐴 𝐵 𝑀 • For bi-molecular reaction, collision theory can be used to predict • 𝑘𝑏𝑖𝑚𝑜𝑙𝑒𝑐 = 𝑘 𝑇 • Where 𝑝 = steric 1 2 𝐸 2 8𝜋𝑘𝐵 𝑇 = 𝑝𝑁𝐴𝑉 𝜎𝐴𝐵 𝑒𝑥𝑝 − 𝐴 ; 𝜇 𝑅𝑢 𝑇 factor, and 𝐸𝐴 = Activation Energy, but both = ? (data) • “Semi-empirical” three parameter form: • 𝑘 𝑇 = 𝐴𝑇 𝑏 𝑒𝑥𝑝 − 𝐸𝐴 𝑅𝑢 𝑇 , 𝐴, 𝑏 and 𝐸𝐴 values are tabulated p. 112 Multistep Mechanism Reaction Rates • A sequence of intermediate reactions leading from overall-reactants to overall-products • L steps, i = 1, 2,… L • N species, j = 1, 2,… N; some are intermediate (not in overall products or reactants) • Example 2𝐻2 + 𝑂2 → 2𝐻2 𝑂 • Forward and reverse intermediate steps • R1: • R2: • R3: 𝑘𝐹1 ,𝑘𝑅1 𝐻2 + 𝑂2 𝐻 + 𝑂2 𝑘𝐹2 ,𝑘𝑅2 𝑂𝐻 + 𝐻2 • R4: H + 𝑂2 + 𝑀 𝑘𝐹3 ,𝑘𝑅3 𝑘𝐹4 ,𝑘𝑅4 𝐻𝑂2 + 𝐻 𝑖=1 𝑂𝐻 + 𝑂 𝑖=2 𝐻2 𝑂 + 𝐻 𝑖=3 𝐻𝑂2 + 𝑀 𝑖=4 • Number of steps: L = 4 • Number of Species (𝐻2 , 𝑂2 , 𝐻𝑂2 , 𝐻, 𝑂𝐻, 𝑂, 𝐻2 𝑂, 𝑀): N = 8 • Number of unknowns: 8, 𝑖 𝑡 , 𝑖=1, 2, …,8 • Need 8 differential equations (constraints) General method for species net production rates •j=1 • 𝑑 𝑂2 𝑑𝑡 = 𝑘𝑅1 𝐻𝑂2 𝐻 + 𝑘𝑅2 𝑂𝐻 𝑂 + 𝑘𝐹3 𝑂𝐻 𝐻2 −𝑘𝐹1 𝐻2 𝑂2 − 𝑘𝐹2 𝐻 𝑂2 − 𝑘𝐹4 𝐻 𝑂2 𝑀 •j=2 𝑑𝐻 𝑑𝑡 = 𝑘𝐹1 𝐻2 𝑂2 + 𝑘𝑅2 𝑂𝐻 𝑂 + 𝑘𝐹3 𝑂𝐻 𝐻2 + 𝑘𝑅4 𝐻𝑂2 𝑀 −𝑘𝑅1 𝐻𝑂2 𝐻 − 𝑘𝐹2 𝐻 𝑂2 − 𝑘𝑅3 𝐻2 𝑂 𝐻 − 𝑘𝐹4 𝐻 𝑂2 𝑀 • i = 3, 4, …8 •… • What happens at equilibrium? • A general reaction • 𝑎𝐴 + 𝑏𝐵 𝑘𝑓 𝑘𝑟 𝑐𝐶 + 𝑑𝐷 • Consumption minus Generation of A • 𝑑𝐴 𝑑𝑡 = 𝑎 −𝑘𝑓 𝐴 • At equilibrium • 𝑘𝑓 𝐴 • 𝑘𝑓 𝑇 𝑘𝑟 𝑇 𝑎 𝐵 𝐵 𝑑𝐴 𝑑𝑡 • 𝐾𝐶 𝑇 = 𝑘𝑟 𝑇 𝐵 𝑏 + 𝑘𝑟 𝐶 𝑐 𝐷 𝑑 = 0, so = 𝑘𝑟 𝐶 = 𝐾𝐶 𝑇 = 𝑘𝑓 𝑇 𝑎 𝑐 𝐷 𝑑 𝐶 𝑐𝐷𝑑 𝐴𝑎𝐵𝑏 = Rate Coefficient • Looks like the Equilibrium Constant from Chapter 2 Relationship between Rate Coefficients and Equilibrium Constant (Chapter 2) • 𝑎𝐴 + 𝑏𝐵 𝑘𝑓 𝑐𝐶 + 𝑑𝐷 𝑘𝑟 • Equilibrium Constant: • 𝐾𝑃 𝑇 = 𝑃𝐶 𝑐 𝑃𝐷 𝑑 𝑃𝑜 𝑃𝑜 𝑃𝐴 𝑎 𝑃𝐵 𝑏 𝑃𝑜 𝑃𝑜 • Rate coefficient: • 𝐾𝐶 𝑇 = 𝑘𝐹 𝑇 𝑘𝑅 𝑇 = 𝑃𝑖 𝑢𝑇 𝑐 𝑑 𝐶 𝐷 𝐴𝑎𝐵𝑏 = 𝑃𝐶 𝑐 𝑃𝐷 𝑑 𝑃𝑜 𝑃𝑜 𝑃𝐴 𝑎 𝑃𝐵 𝑏 𝑃𝑜 𝑃𝑜 𝑃𝑜 𝑐+𝑑−(𝑎+𝑏) 𝑅𝑢 𝑇 = 𝐾𝑃 𝑇 • Using 𝑖 = 𝑅 • 𝐾𝑃 𝑇 = 𝐾𝐶 𝑇 𝑅𝑢 𝑇 𝑐+𝑑−(𝑎+𝑏) 𝑃𝑜 = exp −Δ𝐺𝑇𝑜 𝑅𝑢 𝑇 • Note: If 𝑁𝑅 = 𝑎 + 𝑏 = 𝑐 + 𝑑 = 𝑁𝑃 , then 𝐾𝑃 𝑇 = 𝐾𝐶 𝑇 𝑃𝑜 𝑐+𝑑−(𝑎+𝑏) 𝑅𝑢 𝑇 Steady State Approximation • In some reaction steps, the slow creation and rapid consumption of radicals cause the radical concentration to reach steady-state quickly • Makes some differential equation algebraic (simplifies solution) • Example: Zelovich two-step system 𝑘1 • 𝑂 + 𝑁2 → 𝑁𝑂 + 𝑁 • This is a relatively “slow” reaction, but N is highly reactive and is consumed as soon as it is created 𝑘2 • 𝑁 + 𝑂2 → 𝑁𝑂 + 𝑂 • Very fast consumption of N • Production minus Consumption of N radical • 𝑑𝑁 𝑑𝑡 = 𝑘1 𝑂 𝑁2 − 𝑘2 𝑁 𝑂2 • Since it is very fast and reaches steady state almost right away • 𝑁 𝑆𝑆 = 𝑘2 𝑂 𝑁2 𝑘1 𝑂2 𝑑𝑁 𝑑𝑡 ≈0 (algebraic equation, not differential) • This molar concentration is small and changes almost immediately when the other concentrations change • 𝑑 𝑁 𝑆𝑆 𝑑𝑡 𝑑 𝑘2 𝑂 𝑁2 𝑘1 𝑂2 = 𝑑𝑡 Uni-molecular Reaction Example • Apparent (global) • 𝐴 → 𝑝𝑟𝑜𝑑𝑢𝑐𝑡; 𝑑 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠 • Find = 𝑘𝑎𝑝𝑝 𝐴 𝑑𝑡 • Three-step mechanism 𝑘 𝑒 • 𝐴 + 𝑀 → 𝐴∗ + 𝑀 (𝐴∗ is an energized state of 𝐴 and highly reactive) 𝑘𝑑𝑒 ∗ • 𝐴 +𝑀 𝐴 + 𝑀 (De-energization of A) • 𝑘𝑢𝑛𝑖 ∗ 𝐴 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠 • 𝐴∗ will reach steady state conditions 𝑑 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠 • = 𝑘𝑢𝑛𝑖 𝐴∗ (need 𝐴∗ ) 𝑑𝑡 ∗ • Molecular Balance for 𝐴 ∗ • 𝑑𝐴 𝑑𝑡 ∗ • 𝐴 • = 𝑘𝑒 𝐴 𝑀 − 𝑘𝑑𝑒 𝐴∗ 𝑀 − 𝑘𝑒 𝐴∗ ≈ 0 (since 𝐴∗ is consumed as fast as it is produced) 𝑆𝑆 𝑑 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠 𝑑𝑡 ≈𝑘 = 𝑘𝑒 𝐴 𝑀 𝑑𝑒 𝑀 + 𝑘𝑒 𝑘𝑢𝑛𝑖 𝑘𝑒 𝑀 𝑘 𝑘𝑑𝑒 𝑀 + 𝑘𝑒 𝑘 𝑀 • 𝑘𝑎𝑝𝑝 = 𝑘 𝑢𝑛𝑖𝑀 𝑒+ 𝑘 𝑑𝑒 𝑒 𝐴 = 𝑘𝑎𝑝𝑝 𝐴 Partial Equilibrium • Some reaction steps of a mechanism are much faster in both forward and reverse directions than others • Usually chain propagating (or branching) reactions are bi-molecular and faster than ter-molecular recombination reactions • Treat fast reactions as if they are equilibrated • This allows them to be treated using algebraic equations and reduces the number of differential equations that must be solved. Example (fast bi-molecular, slow ter-molecular steps) • Reaction: 2𝐴2 + 𝐵2 → 2𝐴2 𝐵 • 𝐴 + 𝐵2 ↔ 𝐴𝐵 + 𝐵 • 𝑑𝐵 𝑑𝑡 = 𝑘1𝑓 𝐴 𝐵2 − 𝑘1𝑟 𝐴𝐵 𝐵 = 0; • 𝐵 + 𝐴2 ↔ 𝐴𝐵 + 𝐴 • 𝑘2𝑓 𝐵 𝐴2 = 𝑘2𝑟 𝐴 𝐴𝐵 ; • 𝐴𝐵 + 𝐴2 ↔ 𝐴2 𝐵 + 𝐴 • 𝑘3𝑓 𝐴𝐵 𝐴2 = 𝑘3𝑟 𝐴 𝐴2 𝐵 ; 𝐴𝐵 𝐵 𝐴 𝐵2 𝐴 𝐴𝐵 𝐵 𝐴2 𝐴 𝐴2 𝐵 𝐴𝐵 𝐴2 = = 𝑘1𝑓 𝑘1𝑟 𝑘2𝑓 𝑘2𝑟 = = 𝑘𝑃1 1 = 𝑘𝑃2 2 𝑘3𝑓 𝑘3𝑟 = 𝑘𝑃3 • 𝐴 + 𝐴𝐵 + 𝑀 → 𝐴2 𝐵 + 𝑀 • Slow ter-molecular recombination 𝑑 𝐴2 𝐵 𝑑𝑡 • = −𝑘𝑡𝑒𝑟 𝐴 𝐴𝐵 𝑀 • Get 𝐴 and 𝐴𝐵 = 𝑓𝑛( 𝐴2 , 𝐵2 , 𝐴2 𝐵 ) from 1, 2 and 3 3 Chemical Time Scales • How long does it take for the reactant with the smaller initial amount 𝐴 significantly decreases? • At time 𝑡 = 𝜏𝑐ℎ𝑒𝑚 , 𝐴 𝐴0 0 to = 𝑒 −1 = 0.368 • Uni-molecular Reaction, 𝐴 → 𝑃𝑟𝑜𝑑𝑢𝑐𝑡 • 𝑑𝐴 𝑑𝑡 = −𝑘𝑎𝑝𝑝 𝑇 𝐴 • 𝜏𝑐ℎ𝑒𝑚 = 𝑘 1 𝑎𝑝𝑝 • Assume T changes slowly, so that 𝑘𝑎𝑝𝑝 𝑇 ≈ 𝑘𝑎𝑝𝑝 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 • Bi-molecular Reaction 𝐴 + 𝐵 → 𝐶 + 𝐷 • 𝑑𝐴 𝑑𝑡 = −𝑘𝑏𝑖𝑚𝑜𝑙𝑒𝑐 𝐴 𝐵 • 𝜏𝑐ℎ𝑒𝑚,𝑏 = ln 𝑒− 𝑒−1 𝐴0 𝐵0 𝐵 0 − 𝐴 0 𝑘𝑏𝑖𝑚𝑜𝑙𝑒𝑐 ≈ 1 𝐵 0 𝑘𝑏𝑖𝑚𝑜𝑙𝑒𝑐 (can be 67% too small, but right order of magnitude) • Ter-molecular Reaction 𝐴 + 𝐵 + 𝑀 → 𝐶 + 𝑀 • 𝑑𝐴 𝑑𝑡 = − 𝑀 𝑘𝑡𝑒𝑟 𝐴 𝐵 • 𝜏𝑐ℎ𝑒𝑚,𝑡𝑒𝑟 ≈ 1 𝐵 0 𝑀 𝑘𝑡𝑒𝑟 Chapter 5 Some Important Chemical Mechanisms • Hydrocarbon (𝐶𝑥 𝐻𝑦 ) combustion has 2 steps • 𝐶𝑥 𝐻𝑦 + 𝑂2 → 𝐶𝑂 + ⋯ • 𝐶𝑂 + 𝑂2 → 𝐶𝑂2 + ⋯ • This second step • is “slow” unless 𝐻2 or 𝐻2 𝑂 are present (these molecules help produce 𝑂𝐻) • Produces most of the chemical heat release Chapter 6 Coupling Chemical and Thermal Analysis of Reacting systems • Identify four reactor systems, p 184 1. Constant pressure and fixed Mass • Time dependent, well mixed 2. Constant-volume fixed-mass • Time dependent, well mixed 3. Well-stirred reactor • Steady, different inlet and exit conditions 4. Plug-Flow • Steady, dependent on location • Coupled Energy, species production, and state constraints • For plug flow also need momentum since speeds and pressure vary with location • Assume we know “production rates” per unit volume • 1 𝑑𝑁𝑖 𝑉 𝑑𝑡 = 𝜔𝑖 = 𝑘(𝑇) 𝑀 𝑖=1 𝑖 𝑛𝑖 Constant pressure and fixed Mass Reactor • Constituents • reactants and products, 𝑖 = 1, 2, … 𝑀 (book uses 𝑁) • P and m constant • Find as a function of time, t • Temperature 𝑇 • To find use conservation of energy • Molar concentration 𝑖 • use species generation/consumption rates from chemical kinetics •𝑉= 𝑚 , 𝑛𝑒𝑒𝑑 𝜌 𝜌 • state, mixture 𝑄 𝑊 Constant pressure and fixed Mass Reactor • Initial Conditions, at t = 0 𝑊 • 𝑖 = 𝑖 0 , 𝑖 = 1, 2, … 𝑀, (specie molar concentrations) • 𝑇 = 𝑇0 • Assume we also know 𝜔𝑖 = 𝐴𝑒𝑥𝑝(− 𝐸𝐴 ) 𝑅𝑢 𝑇 𝑀 𝑖=1 𝑖 𝑄 𝑛𝑖 • Use the first order differentials to find 𝑖 and 𝑇 at time t + Δ𝑡 • • 𝑑𝑖 𝑑𝑡 𝑑𝑇 𝑑𝑡 = 𝜔𝑖 − 𝑖 = 𝑄 − 𝑉 𝜔𝑖 ℎ𝑖 𝑇 𝑖 𝑐𝑝,𝑖 𝑇 𝜔𝑖 𝑖 ≈ + 1 𝑑𝑇 𝑇 𝑑𝑡 ≈ Δ𝑖 Δ𝑡 ; 𝑖 Δ𝑇 ; Δ𝑡 𝑡+Δ𝑡 = 𝑖 𝑡+ 𝑑𝑖 𝑑𝑡 𝑑𝑇 𝑑𝑡 Δ𝑡 𝑇𝑡+Δ𝑡 = 𝑇𝑡 + Δ𝑡 • System Volume • 𝑉 𝑡 = t T 𝑚 𝜌 𝑇 = [1] [2] 𝑚 ; 𝑖 𝑀𝑊𝑖 𝜌= 𝑚 𝑉 = … [M] w1 w2 0 T0 [1]0 [2]0 … [M]0 Dt 2Dt 𝑚𝑖 𝑉 = … 𝑁𝑖 𝑀𝑊𝑖 𝑉 wM V = 𝑖 𝑀𝑊𝑖 Q d[1]/dt d[2]/dt … d[M]/dt dT/dt Constant-Volume V Fixed-Mass m Reactor • Find T, 𝑖 • 1st Law • 𝑑𝑇 𝑑𝑡 = 𝑄 − 𝑉 𝑓𝑜𝑟 𝑖 = 1,2, … , 𝑀 , and P versus time 𝑡 ℎ𝑖 𝜔𝑖 +𝑅𝑢 𝑇 𝜔𝑖 𝑖 𝑐𝑝,𝑖 −𝑅𝑢 (true and useful) • Species Production • 𝑑𝑖 𝑑𝑡 𝑁 = 𝑑 𝑉𝑖 𝑑𝑡 = 1 𝑑𝑁𝑖 𝑉 𝑑𝑡 = 𝜔𝑖 = 𝑘(𝑇) 𝑀 𝑖=1 𝑖 𝑛𝑖 • Initial Conditions: • At t = 0, 𝑇 = 𝑇0 , and 𝑖 = 𝑖 0 , 𝑖 = 1, 2, … 𝑀 • State Equation • 𝑃= 𝑖 𝑅𝑢 𝑇 • Pressure Rate of change (affects detonation) • 𝑑𝑃 𝑑𝑡 = 𝑅𝑢 𝑇 𝜔𝑖 + 𝑑𝑇 𝑑𝑡 𝑖 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 • On test could be asked to write the needed equations 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 Steady-State Well-Stirred Reactor • Exit condition same as system • Conservation of M species • 0 = 𝜔𝑖 𝑀𝑊𝑖 V + 𝑚 𝑌𝑖,𝑖𝑛 − 𝑌𝑖,𝑜𝑢𝑡 , 𝑖 = 1,2, … , 𝑀 • To find 𝜔𝑖 , need molar concentrations 𝑖 from mass fractions 𝑌𝑖 𝑌𝑖 𝑀𝑊𝑚𝑖𝑥 𝑃 𝑀𝑊𝑖 𝑅𝑢 𝑇 1 𝑀𝑊𝑚𝑖𝑥 = 𝑌𝑖 𝑀𝑊𝑖 • 𝑖 = • • Energy • 𝑄=𝑚 𝑌𝑖,𝑜𝑢𝑡 ℎ𝑖,𝑜𝑢𝑡 𝑇 − 𝑌𝑖,𝑖𝑛 ℎ𝑖,𝑖𝑛 𝑇𝑖𝑛 𝑜 • ℎ𝑖 = ℎ𝑓,𝑖 + ∆ℎ𝑠,𝑖 • All equations are algebraic (not differential) For our simple example • 0 = 𝜔𝑖 𝑀𝑊𝑖 𝑉 + 𝑚 𝑌𝑖,𝑖𝑛 − 𝑌𝑖,𝑜𝑢𝑡 , 𝑖 = 1,2, … , 𝑀 = 3 • Fuel • 0 = 𝜔𝐹 𝑀𝑊𝐹 𝑉 + 𝑚 𝑌𝐹,𝑖𝑛 − 𝑌𝐹 1 • Oxidizer • 0 = 𝜔𝑂𝑥 𝑀𝑊𝑂𝑥 𝑉 + 𝑚 𝑌𝑂𝑥,𝑖𝑛 − 𝑌𝑂𝑥 • 0= 𝐴 𝑉 𝜔𝐹𝑢𝑒𝑙 𝑀𝑊 𝑉 + 𝑚 𝑌𝑂𝑥,𝑖𝑛 − 𝑌𝑂𝑥 2 • Product • 0 = 𝜔𝑃 𝑀𝑊𝑃𝑟 𝑉 + 𝑚 𝑌𝑃𝑟,𝑖𝑛 − 𝑌𝑃𝑟 • 0=− 𝐴 𝑉 + 1 𝜔𝐹𝑢𝑒𝑙 𝑀𝑊 𝑉 + 𝑚 𝑌𝑃𝑟,𝑖𝑛 − 𝑌𝑃𝑟 • 𝑌𝑃𝑟 = 1 − 𝑌𝐹 − 𝑌𝑂𝑥 3 MathCAD Solution 0.01 .01 3 510 f ( T2 mdot) 0 0  .005 . Yoxin  f ( T mdot)  mdot  cp Hff 1 phi 1 16 3  510 3  0.941 298  15098   Yfin  cp  ( T  298)  ( T  298)  MW  Vol  6.19  10  exp     Hff  T    9 3 110 3 210 310 3000 T2 0.1 AF  cp     0.233  Yoxin   ( T  298) Hff    1.65 • On test, could be asked to find equation whose roots must be found  P     Ru  T  1.75 Plug-Flow Reactors • Assumptions • Quasi-one dimensional (quantities are ≠ 𝑓𝑛(𝑟, 𝜃)) 𝑑 • Steady state, =0 𝑑𝑡 • No-viscosity 𝜇 = 0 • Axial turbulent and molecular diffusion is small compared to advection (high enough axial velocity) • If velocity is “constant” then pressure is “constant” • Integrate to find 𝑇 𝑥 , 𝑌𝑖 𝑥 , 𝜌 𝑥 • At each location also need to calculate 𝑚 𝜌 𝑥 𝐴(𝑥) 𝜌 𝑥 𝑅𝑢 𝑇(𝑥) 𝑀𝑊𝑚𝑖𝑥 • 𝑣𝑥 𝑥 = • 𝑃 𝑥 = • Like the transient constant-pressure reactor, but varies with location instead of time. What do we expect? • Equations to be solved • Use • 𝜔𝑖 = 𝑓𝑛 𝑌𝑖 , 𝑇, 𝑃 • 𝑃 = 𝜌𝑅𝑇; 𝑅 = • 𝑣𝑥 = 𝑅𝑢 1 ; 𝑀𝑊𝑚𝑖𝑥 𝑀𝑊𝑚𝑖𝑥 𝑚 𝜌𝐴 𝑌𝑖 𝑀𝑊𝑖 = • Need 𝜌 𝑥 and 𝑇 𝑥 • Assume 𝑄 " 𝑥 , 𝐴 𝑥 and 𝑚 are given • Find … (page 209) • 𝑑𝑌𝑖 𝑑𝑥 • 𝑑𝑇 𝑑𝑥 • 𝑑𝜌 𝑑𝑥 = 𝜔𝑖 𝑀𝑊𝑖 𝐴 ,𝑖 𝜌𝑣𝑥 = 1,2, … , 𝑀 = 𝑣𝑥2 𝑑𝜌 𝜌𝑐𝑃 𝑑𝑥 𝑣𝑥2 𝑑𝐴 𝑐𝑃 𝐴 𝑑𝑥 = + 𝑅𝑢 1− 𝑐𝑃 𝑀𝑊𝑚𝑖𝑥 − 𝜔𝑖 𝑀𝑊𝑖 ℎ𝑖 𝜌𝑣𝑥 𝑐𝑃 1 𝑑𝐴 𝜌2 𝑣𝑥2 𝐴 𝑑𝑥 − 𝑄"𝒫 𝑚𝑐𝑃 𝜌𝑅𝑢 + 𝑣𝑥 𝑐𝑃 𝑀𝑊𝑚𝑖𝑥 𝑃 • On Test could ask 𝑣2 1+𝑐 𝑥𝑇 𝑃 𝑀𝑊𝑚𝑖𝑥 𝑀𝑊𝑖 𝜔𝑖 ℎ𝑖 − 𝑐 𝑇 𝑀𝑊𝑖 𝑃 𝜌𝑅𝑢 𝑄" 𝒫 + 𝑣𝑥 𝐴𝑐𝑃 𝑀𝑊𝑚𝑖𝑥 −𝜌𝑣𝑥2 • Apply and simplify these equations for a particular problem 𝑑𝐴 𝑣𝑥2 • Derived equations for = 0, ≪ ℎ and species have same temperature-independent properties (problem X3) 𝑑𝑥 2 Problem X3 (homework) dx 𝑚 m x h Yi h + (dh/dx)dx Yi + (dYi/dx)dx 𝑄" (𝑥) • Consider a constant-area A to the wall small). 𝑄" (𝑥), 𝑑𝐴 𝑑𝑥 = 0 plug flow reactor. It has an axially-varying heat flux mass flow rate 𝑚 𝑘𝑔 𝑠 𝑊 𝑚2 applied , and operates a constant pressure P (velocity variations are • The following mass-based reaction is taking place within the reactor with a stoichiometric air/fuel ratio of ν: • 1 kg F + ν 𝑘𝑔 𝑂𝑥 → 1 + ν 𝑘𝑔 𝑃𝑟; • Assume • • • • 𝜔𝐹 𝑥 = 𝑑𝐹 𝑑𝑡 = −𝐴𝐹 𝑒 𝐸 𝑅 − 𝑎 𝑢 𝑇 𝑂𝑥 𝑚 𝐹 𝑛 𝑣𝑥2 (2 The mass flow kinetic energy is much less than its enthalpy ≪ ℎ) The fuel F, Oxidizer Ox, and products Pr, have the same 𝑀𝑊 and 𝑐𝑝 (and 𝑐𝑝 ≠ 𝑓𝑛(𝑇)) 𝑜 The oxidizer and product heat of formation are zero, and that of the fuel is ℎ𝑓,𝐹 The inlet equivalence ratio and temperature are Φ𝑖𝑛 and 𝑇𝑖𝑛 • Use conservation of species and energy to find equations that can be used to find the axial variation of 𝑌𝑖 𝑥 , 𝑇 𝑥 , 𝜌 𝑥 , 𝑣𝑥 𝑥 Ch. 8 Laminar Premixed Flames 𝛼 𝑣𝑢 • • • • 𝑆𝐿 𝛼 Bunsen Burner Inner Cone angle, 𝛼 𝑆𝐿 is the laminar flame speed relative to the premixed reactants 𝑣𝑢 is the unburned reactant speed If 𝑣𝑢 > 𝑆𝐿 , then cone will adjust it’s angel 𝛼 so that 𝑆𝐿 = 𝑣𝑢 sin 𝛼 • The angel 𝛼 and its sine sin 𝛼 = 𝑆𝐿 𝑣𝑢 decrease as increases 𝑣𝑢 (inner cone length increases) • If 𝑣𝑢 < 𝑆𝐿 , then flame will flash back to air holes (unless quenched in tube). Tube filled with stationary premixed Oxidizer/Fuel Products shoot out as flame burns in 𝑆𝐿 Laminar Flame Speed Burned Products 𝛿 Unburned Fuel + Oxidizer • Flame reference frame: 𝑣𝑏 , 𝜌𝑏 𝑣𝑢 = 𝑆𝐿 , 𝜌𝑢 , 𝛿~1 𝑚𝑚 • 1 𝑘𝑔 𝐹𝑢 + 𝜈 𝑘𝑔 𝑂𝑥 → 1 + 𝜈 𝑘𝑔 𝑃𝑟 • Conservation of mass: 𝑚 = 𝜌𝑢 𝑣𝑢 = 𝑣𝑏 𝜌𝑏 𝑣𝑏 ; 𝑣𝑢 = 𝜌𝑢 𝜌𝑏 = 𝑃𝑢 𝑅𝑇𝑏 𝑅𝑇𝑢 𝑃𝑏 = • Diffusion of heat and species cause flame to propagate • How to estimate the laminar flame speed 𝑆𝐿 and thickness 𝛿? 𝑇𝑏 𝑇𝑢 ≈ • Depends on the pressure, fuel, equivalence ratio, heat and mass diffusion,… 2100𝐾 300𝐾 =7 Heat Flux with diffusion • Heat: Energy transfer at a boundary due to temperature difference • When there is a large species gradient, diffusion contributes to heat flux • 𝑄𝑥′′ = 𝑑𝑇 −𝑘 𝑑𝑥 • For 𝐿𝑒 = 𝛼 𝒟 = ′′ 𝑚𝑖,𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑜𝑛 ℎ𝑖 + 𝑘 𝒟𝜌𝑐𝑃 ≈ 𝑂(1), appropriate for most combustion mixtures • Shvab-Zeldovich form: 𝑄𝑥′′ = −𝜌𝒟 𝑑ℎ 𝑑𝑥 • Approximate Solution • 𝑆𝐿 = 2𝛼 1+𝜈 𝜌𝑢 • 𝛿= 2𝛼𝜌𝑢 ′′′ 1+𝜈 −𝑚𝐹 • 𝑚𝐹′′′ = 𝜔𝐹 𝑀𝑊𝐹 −𝑚𝐹′′′ = 2𝛼 𝑆𝐿 (Fast flames are thin) • 𝜔𝐹 at average 𝑇 that is closer to 𝑇𝑏 than 𝑇𝑢 , and average values of 𝐹𝑢𝑒𝑙 and 𝑂𝑥 Pressure and temperature dependence of SL and 𝛿 𝑆𝐿 𝑆𝐿 𝑆𝐿 2𝛼 1 + 𝜈 • For methane: Φ 𝑇𝑢 𝑃 • 𝑆𝐿 = ′′′ −𝑚𝐹 𝜌𝑢 ~𝑃0 𝑇𝑢 𝑇 0.375 𝑇𝑏−1 𝑒𝑥𝑝 • Actually decreases as P increases: 𝑆𝐿 cm 𝑆𝐿 s •𝛿= 𝛿 cm s = Φ 𝐸𝑎 𝑅𝑢 − 2𝑇𝑏 ≠ 𝑓𝑛 𝑃 (for HC fuels, N=2) 43 𝑃 [𝑎𝑡𝑚] • Increases with temperature: = 10 + 3.71 × 10−4 𝑇𝑢2 𝐾 • Decreases for Φ above or below 1.05 (because that decreases temperature) 2𝛼 𝐸 𝑅 ~𝑃−1 𝑇 0.375 𝑇𝑏1 𝑒𝑥𝑝 𝑎 𝑢 𝑆𝐿 2𝑇𝑏 • Fast (high temperature) flame are thin Dependence on Fuel Type 𝑆𝐿 𝑆𝐿,𝐶3 𝐻8 𝑇𝑓 • Table 8.2, P = 1 atm, Φ = 1, Tu = Room temperature • Figure 8.17 presents ratio flame speed some hydrocarbons [C2-C6 alkanes (single bonds), alkenes (double bonds), and alkynes (triple bonds)] to propane speed (C3H8) • C3-C6 follow same trend • Consistency of data for Methane with P = 1 atm, Tu = 298K • Table 8.2: 𝑆𝐿 = • 𝑆𝐿 cm s = 43 cm 40 ; s 𝑃 [𝑎𝑡𝑚] = 43 cm cm ; 𝑆𝐿 s s = 10 + 3.71 × 10−4 𝑇𝑢2 𝐾 = 43 cm s Flame Speed Correlations for Selected Fuels • Be aware of what fuels are in this and other tables so you’ll know the easiest way to find the results you need • 𝑆𝐿 = 𝑆𝐿,𝑟𝑒𝑓 𝑇𝑢 𝑇𝑢,𝑟𝑒𝑓 𝛾 𝑃 𝑃𝑟𝑒𝑓 𝛽 1 − 2.1𝑌𝑑𝑖𝑙 • 𝑇𝑢 > ~350𝐾, 𝑇𝑟𝑒𝑓 = 298 𝐾, 𝑃𝑟𝑒𝑓 = 1 𝑎𝑡𝑚 • 𝑆𝐿,𝑟𝑒𝑓 = 𝐵𝑀 + 𝐵2 Φ − Φ𝑚 2 • 𝛾 = 2.18 − 0.8 Φ − 1 • 𝛽 = −0.16 + 0.22 Φ − 1 • RMFD-303 is a research fuel that simulates gasalines

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