ME 475/675 Introduction to Combustion Lecture 32 Midterm 2 review

Announcements HW 13 Midterm 2 Project Due Now Review today (Wednesday is a holiday) Tutorial: Thursday, November 12, 2015, 5-7 pm, SEM 257 Will try to return all HW, and post all solutions Exam: Friday, November 13, 2015 PE 104, 8:00 AM to 9:30 (or 9:45?) AM Project Form your group by today Inform TA: Hasibul Alam hasibul.alam.bd@gmail.com http://wolfweb.unr.edu/homepage/greiner/teaching/MECH.475.675.Combustion/TermProjectAssignment.pdf

Midterm 2 Test Format Test Rules Test Coverage 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 Some Important Chemical Mechanisms Hydrocarbon ( 𝐶 𝑥 𝐻 𝑦 ) combustion has 2 “main” steps (really many more) 𝐶 𝑥 𝐻 𝑦 + 𝑂 2 →𝐶𝑂+… 𝐶𝑂+ 𝑂 2 →𝐶 𝑂 2 +… This second step is “slow” unless 𝐻 2 or 𝐻 2 𝑂 are present (these molecules help produce 𝑂𝐻) Produces most of the chemical heat release

Chapter 5, Global Reaction Rate One-step hydrocarbon combustion reaction 𝐶 𝑥 𝐻 𝑦 + 𝑥+ 𝑦 4 𝑂 2 𝑘 𝐺 𝑥𝐶 𝑂 2 + 𝑦 2 𝐻 2 𝑂 (𝐹+𝑎𝑂𝑥→𝑏𝑃𝑟) For stoichiometric mixture with 𝑂 2 , not air Made up of many intermediate steps that are not seen Overall reaction rate empirical, black box, approximates observations 𝑑 𝐶 𝑥 𝐻 𝑦 𝑑𝑡 =−𝐴𝑒𝑥𝑝 𝐸 𝑎 𝑅 𝑢 𝑇 𝐶 𝑥 𝐻 𝑦 𝑚 𝑂 2 𝑛 = 𝑔𝑚𝑜𝑙𝑒 𝑐𝑚 3 𝑠 𝑖 = 𝑁 𝑖 𝑉 = 𝜒 𝑖 𝑃 𝑅 𝑢 𝑇 = 𝑃 𝑖 𝑅 𝑢 𝑇 = 𝜒 𝑖 ρ 𝑀𝑊 𝑀𝑖𝑥 = 𝑌 𝑖 𝑀𝑊 𝑖 ρ= 𝑌 𝑖 𝑀𝑊 𝑀𝑖𝑥 𝑀𝑊 𝑖 𝑃 𝑅 𝑢 𝑇 Page 157, Table 5.1 𝐴, 𝐸 𝑎 𝑅 𝑢 , 𝑚 𝑎𝑛𝑑 𝑛 for different hydrocarbon fuels These values are based on flame speed data fit (Ch. 8) Units: 𝐴 1 𝑠 𝑘𝑚𝑜𝑙𝑒 𝑚 3 1−𝑚−𝑛 =𝐴 𝑇𝑒𝑥𝑡𝑏𝑜𝑜𝑘 1000 1−𝑚−𝑛 Given in Table 5.1, p. 157 1 𝑠 𝑔𝑚𝑜𝑙𝑒 𝑐𝑚 3 1−𝑚−𝑛 Usually Want These Units (add to table?)

Chapter 4 Chemical Kinetics (Use to find global rates) Global (apparent) reactions (what are observed) 𝐹+𝑎𝑂𝑥→𝑏𝑃𝑟 They are made up of many intermediate steps that are not seen Uni-molecular, 𝐴→𝐵, or 𝐴→𝐵+𝐶 Low pressure: 𝑑 𝐴 𝑑𝑡 =− 𝑘 𝑢𝑛𝑖 𝐴 Bi-molecular, 𝐴+𝐵→𝐶+𝐷 𝑑 𝐴 𝑑𝑡 =− 𝑘 𝑏𝑖𝑚𝑜𝑙𝑒𝑐 𝐴 1 𝐵 1 Ter-molecular : 𝐴+𝐵+𝑀→𝐶+𝑀 (Recombination) 𝑑 𝐴 𝑑𝑡 =− 𝑘 𝑡𝑒𝑟 𝐴 𝐵 𝑀 For bi-molecular reaction, collision theory can be used to predict 𝑘 𝑏𝑖𝑚𝑜𝑙𝑒𝑐 =𝑘 𝑇 =𝑝 𝑁 𝐴𝑉 𝜎 𝐴𝐵 2 8𝜋 𝑘 𝐵 𝑇 𝜇 1 2 𝑒𝑥𝑝 − 𝐸 𝐴 𝑅 𝑢 𝑇 ; Where 𝑝 = steric factor, and 𝐸 𝐴 = Activation Energy, both = ?(data) “Semi-empirical” three parameter form: 𝑘 𝑇 =𝐴 𝑇 𝑏 𝑒𝑥𝑝 − 𝐸 𝐴 𝑅 𝑢 𝑇 , 𝐴, 𝑏 and 𝐸 𝐴 values are tabulated p. 112

Multistep (chain) 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 intermediates, which are not in overall products or reactants Example 2 𝐻 2 + 𝑂 2 →2 𝐻 2 𝑂 In this example all intermediate steps include forward and reverse reactions R1: 𝐻 2 + 𝑂 2 𝑘 𝐹1 , 𝑘 𝑅1 𝐻 𝑂 2 +𝐻 𝑖=1 R2: 𝐻+ 𝑂 2 𝑘 𝐹2 , 𝑘 𝑅2 𝑂𝐻 +𝑂 𝑖=2 R3: 𝑂𝐻+ 𝐻 2 𝑘 𝐹3 , 𝑘 𝑅3 𝐻 2 𝑂+𝐻 𝑖=3 R4: H+ 𝑂 2 +𝑀 𝑘 𝐹4 , 𝑘 𝑅4 𝐻 𝑂 2 +𝑀 𝑖=4

What is the relationship between forward and reverse reactions What is the relationship between forward and reverse reactions? (think about equilibrium) A general reaction 𝑎𝐴+𝑏𝐵 𝑘 𝑓 𝑘 𝑟 𝑐𝐶+𝑑𝐷 Consumption minus Generation of A 𝑑 𝐴 𝑑𝑡 =𝑎 − 𝑘 𝑓 𝐴 𝑎 𝐵 𝑏 + 𝑘 𝑟 𝐶 𝑐 𝐷 𝑑 At equilibrium 𝑑 𝐴 𝑑𝑡 =0, so 𝑘 𝑓 𝐴 𝑎 𝐵 𝐵 = 𝑘 𝑟 𝐶 𝑐 𝐷 𝑑 𝑘 𝑓 𝑇 𝑘 𝑟 𝑇 = 𝐾 𝐶 𝑇 = 𝐶 𝑐 𝐷 𝑑 𝐴 𝑎 𝐵 𝑏 𝐾 𝐶 𝑇 = 𝑘 𝑓 𝑇 𝑘 𝑟 𝑇 = Rate Coefficient Looks like the Equilibrium Constant from Chapter 2

Relationship between Rate Coefficients and Equilibrium Constant (Chapter 2) 𝑎𝐴+𝑏𝐵 𝑘 𝑓 𝑘 𝑟 𝑐𝐶+𝑑𝐷 Equilibrium Constant: 𝐾 𝑃 𝑇 = 𝑃 𝐶 𝑃 𝑜 𝑐 𝑃 𝐷 𝑃 𝑜 𝑑 𝑃 𝐴 𝑃 𝑜 𝑎 𝑃 𝐵 𝑃 𝑜 𝑏 =exp −Δ 𝐺 𝑇 𝑜 𝑅 𝑢 𝑇 , ()where 𝑃 𝑜 =1𝑎𝑡𝑚) Rate coefficient: 𝐾 𝐶 𝑇 = 𝑘 𝐹 𝑇 𝑘 𝑅 𝑇 = 𝐶 𝑐 𝐷 𝑑 𝐴 𝑎 𝐵 𝑏 = 𝑃 𝐶 𝑃 𝑜 𝑐 𝑃 𝐷 𝑃 𝑜 𝑑 𝑃 𝐴 𝑃 𝑜 𝑎 𝑃 𝐵 𝑃 𝑜 𝑏 𝑃 𝑜 𝑅 𝑢 𝑇 𝑐+𝑑−(𝑎+𝑏) = 𝐾 𝑃 𝑇 𝑃 𝑜 𝑅 𝑢 𝑇 𝑐+𝑑−(𝑎+𝑏) Using 𝑖 = 𝑃 𝑖 𝑅 𝑢 𝑇 = 𝑃 𝑖 𝑃 𝑜 𝑃 𝑜 𝑅 𝑢 𝑇 𝑘 𝐹 𝑇 𝑘 𝑅 𝑇 =exp −Δ 𝐺 𝑇 𝑜 𝑅 𝑢 𝑇 𝑃 𝑜 𝑅 𝑢 𝑇 𝑐+𝑑−(𝑎+𝑏) , so If you know 𝑘 𝐹 𝑇 or 𝑘 𝑅 𝑇 , you can find the other! 𝐾 𝑃 𝑇 = 𝐾 𝐶 𝑇 𝑅 𝑢 𝑇 𝑃 𝑜 𝑐+𝑑−(𝑎+𝑏) =exp −Δ 𝐺 𝑇 𝑜 𝑅 𝑢 𝑇 Note: If 𝑁 𝑅 =𝑎+𝑏=𝑐+𝑑= 𝑁 𝑃 , then 𝐾 𝑃 𝑇 = 𝐾 𝐶 𝑇

Find overall reaction form Multistep Mechanism Rates Return to example 2 𝐻 2 + 𝑂 2 →2 𝐻 2 𝑂 In this example all intermediate steps include forward and reverse reactions R1: 𝐻 2 + 𝑂 2 𝑘 𝐹1 , 𝑘 𝑅1 𝐻 𝑂 2 +𝐻 𝑖=1 R2: 𝐻+ 𝑂 2 𝑘 𝐹2 , 𝑘 𝑅2 𝑂𝐻 +𝑂 𝑖=2 R3: 𝑂𝐻+ 𝐻 2 𝑘 𝐹3 , 𝑘 𝑅3 𝐻 2 𝑂+𝐻 𝑖=3 R4: H+ 𝑂 2 +𝑀 𝑘 𝐹4 , 𝑘 𝑅4 𝐻 𝑂 2 +𝑀 𝑖=4 Number of steps: L = 4 (or 8?) Number of Species ( 𝐻 2 , 𝑂 2 , 𝐻 𝑂 2 ,𝐻, 𝑂𝐻, 𝑂, 𝐻 2 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 𝑂 2 𝑑𝑡 = 𝑘 𝑅1 𝐻 𝑂 2 𝐻 + 𝑘 𝑅2 𝑂𝐻 𝑂 + 𝑘 𝐹3 𝑂𝐻 𝐻 2 − 𝑘 𝐹1 𝐻 2 𝑂 2 − 𝑘 𝐹2 𝐻 𝑂 2 − 𝑘 𝐹4 𝐻 𝑂 2 𝑀 j = 2 𝑑 𝐻 𝑑𝑡 = 𝑘 𝐹1 𝐻 2 𝑂 2 + 𝑘 𝑅2 𝑂𝐻 𝑂 + 𝑘 𝐹3 𝑂𝐻 𝐻 2 + 𝑘 𝑅4 𝐻 𝑂 2 𝑀 − 𝑘 𝑅1 𝐻𝑂 2 𝐻 − 𝑘 𝐹2 𝐻 𝑂 2 − 𝑘 𝑅3 𝐻 2 𝑂 𝐻 − 𝑘 𝐹4 𝐻 𝑂 2 𝑀 j = 3, 4, …8 … Could solve all of these coupled 1st order dif equns., but some may be simplified

Steady State Approximation (simplification) 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 𝑂+ 𝑁 2 𝑘 1 𝑁𝑂+𝑁 This is a relatively “slow” reaction, but N is highly reactive and is consumed as soon as it is created 𝑁+ 𝑂 2 𝑘 2 𝑁𝑂+𝑂 Very fast consumption of N Production minus Consumption of N radical 𝑑 𝑁 𝑑𝑡 = 𝑘 1 𝑂 𝑁 2 − 𝑘 2 𝑁 𝑂 2 Since it is very fast and reaches steady state almost right away 𝑑 𝑁 𝑑𝑡 ≈0 𝑁 𝑆𝑆 = 𝑘 2 𝑘 1 𝑂 𝑁 2 𝑂 2 (algebraic equation, not differential) This molar concentration is small and changes almost immediately when the other concentrations change 𝑑 𝑁 𝑆𝑆 𝑑𝑡 = 𝑑 𝑑𝑡 𝑘 2 𝑘 1 𝑂 𝑁 2 𝑂 2

Uni-molecular Reaction Example Apparent (global) 𝐴→𝑝𝑟𝑜𝑑𝑢𝑐𝑡; Find 𝑑 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠 𝑑𝑡 = 𝑘 𝑎𝑝𝑝 𝐴 Proposed three-step mechanism 𝐴+𝑀 𝑘 𝑒 𝐴 ∗ +𝑀 ( 𝐴 ∗ is an energized state of 𝐴 and highly reactive) 𝐴 ∗ +𝑀 𝑘 𝑑𝑒 𝐴+𝑀 (De-energization of A) 𝐴 ∗ 𝑘 𝑢𝑛𝑖 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠 𝐴 ∗ will reach steady state conditions 𝑑 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠 𝑑𝑡 = 𝑘 𝑢𝑛𝑖 𝐴 ∗ ( need 𝐴 ∗ ) Molecular Balance for 𝐴 ∗ 𝑑 𝐴 ∗ 𝑑𝑡 = 𝑘 𝑒 𝐴 𝑀 − 𝑘 𝑑𝑒 𝐴 ∗ 𝑀 − 𝑘 𝑢𝑛𝑖 𝐴 ∗ ≈0 (since 𝐴 ∗ is consumed as fast as it is produced) 𝐴 ∗ 𝑆𝑆 ≈ 𝑘 𝑒 𝐴 𝑀 𝑘 𝑑𝑒 𝑀 + 𝑘 𝑒 𝑑 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠 𝑑𝑡 = 𝑘 𝑢𝑛𝑖 𝑘 𝑒 𝑀 𝑘 𝑑𝑒 𝑀 + 𝑘 𝑢𝑛𝑖 𝐴 = 𝑘 𝑎𝑝𝑝 𝐴 𝑘 𝑎𝑝𝑝 = 𝑘 𝑢𝑛𝑖 𝑘 𝑒 𝑀 𝑘 𝑑𝑒 𝑀 + 𝑘 𝑢𝑛𝑖

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: 2 𝐴 2 + 𝐵 2 →2 𝐴 2 𝐵 𝐴+ 𝐵 2 ↔𝐴𝐵+𝐵 𝑑 𝐵 𝑑𝑡 = 𝑘 1𝑓 𝐴 𝐵 2 − 𝑘 1𝑟 𝐴𝐵 𝐵 =0; 𝐴𝐵 𝐵 𝐴 𝐵 2 = 𝑘 1𝑓 𝑘 1𝑟 = 𝑘 𝑃1 1 𝐵+ 𝐴 2 ↔𝐴𝐵+𝐴 𝑘 2𝑓 𝐵 𝐴 2 = 𝑘 2𝑟 𝐴 𝐴𝐵 ; 𝐴 𝐴𝐵 𝐵 𝐴 2 = 𝑘 2𝑓 𝑘 2𝑟 = 𝑘 𝑃2 2 𝐴𝐵+ 𝐴 2 ↔ 𝐴 2 𝐵+𝐴 𝑘 3𝑓 𝐴𝐵 𝐴 2 = 𝑘 3𝑟 𝐴 𝐴 2 𝐵 ; 𝐴 𝐴 2 𝐵 𝐴𝐵 𝐴 2 = 𝑘 3𝑓 𝑘 3𝑟 = 𝑘 𝑃3 3 𝐴+𝐴𝐵+𝑀→ 𝐴 2 𝐵+𝑀 Slow ter-molecular recombination 𝑑 𝐴 2 𝐵 𝑑𝑡 =− 𝑘 𝑡𝑒𝑟 𝐴 𝐴𝐵 𝑀 Get 𝐴 and 𝐴𝐵 = 𝑓𝑛( 𝐴 2 , 𝐵 2 , 𝐴 2 𝐵 ) from 1, 2 and 3

Chemical Time Scales (which reactions are relatively “fast”) How long does it take for the reactant with the smaller initial amount 𝐴 0 to significantly decreases? At time 𝑡= 𝜏 𝑐ℎ𝑒𝑚 , 𝐴 𝐴 0 = 𝑒 −1 =0.368 Uni-molecular Reaction, 𝐴→𝑃𝑟𝑜𝑑𝑢𝑐𝑡 𝑑 𝐴 𝑑𝑡 =− 𝑘 𝑎𝑝𝑝 𝑇 𝐴 𝜏 𝑐ℎ𝑒𝑚 = 1 𝑘 𝑎𝑝𝑝 Assume T changes slowly, so that 𝑘 𝑎𝑝𝑝 𝑇 ≈ 𝑘 𝑎𝑝𝑝 =𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 Bi-molecular Reaction 𝐴+𝐵→𝐶+𝐷 𝑑 𝐴 𝑑𝑡 =− 𝑘 𝑏𝑖𝑚𝑜𝑙𝑒𝑐 𝐴 𝐵 𝜏 𝑐ℎ𝑒𝑚,𝑏 = ln 𝑒− 𝑒−1 𝐴 0 𝐵 0 𝐵 0 − 𝐴 0 𝑘 𝑏𝑖𝑚𝑜𝑙𝑒𝑐 ≈ 1 𝐵 0 𝑘 𝑏𝑖𝑚𝑜𝑙𝑒𝑐 (can be 67% too small, but right order of magnitude) Ter-molecular Reaction 𝐴+𝐵+𝑀→𝐶+𝑀 𝑑 𝐴 𝑑𝑡 =− 𝑀 𝑘 𝑡𝑒𝑟 𝐴 𝐵 𝜏 𝑐ℎ𝑒𝑚,𝑡𝑒𝑟 ≈ 1 𝐵 0 𝑀 𝑘 𝑡𝑒𝑟

Chapter 6 Coupling Chemical and Thermal Analysis of Reacting systems Identify four reactor systems, p 184 Constant pressure and fixed Mass Time dependent, well mixed Constant-volume fixed-mass Well-stirred reactor Steady, different inlet and exit conditions 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 𝑉 𝑑 𝑁 𝑖 𝑑𝑡 = 𝜔 𝑖 =𝑘(𝑇) 𝑗=1 𝑀 𝑗 𝑛 𝑗

Constant pressure and fixed Mass Reactor 𝑊 𝑄 Initial Conditions, at t = 0 𝑖 = 𝑖 0 , 𝑖=1, 2, …𝑀, (specie molar concentrations) 𝑇= 𝑇 0 Assume we also know 𝜔 𝑖 =𝐴𝑒𝑥𝑝(− 𝐸 𝐴 𝑅 𝑢 𝑇 ) 𝑗=1 𝑀 𝑗 𝑛 𝑗 Use the first order differentials to find 𝑖 and 𝑇 at time t+Δ𝑡 𝑑 𝑖 𝑑𝑡 = 𝜔 𝑖 − 𝑖 𝜔 𝑖 𝑖 + 1 𝑇 𝑑𝑇 𝑑𝑡 ≈ Δ 𝑖 Δ𝑡 ; 𝑖 𝑡+Δ𝑡 = 𝑖 𝑡 + 𝑑 𝑖 𝑑𝑡 Δ𝑡 𝑑𝑇 𝑑𝑡 = 𝑄 𝑉 − 𝜔 𝑖 ℎ 𝑖 𝑇 𝑖 𝑐 𝑝,𝑖 𝑇 ≈ Δ𝑇 Δ𝑡 ; 𝑇 𝑡+Δ𝑡 = 𝑇 𝑡 + 𝑑𝑇 𝑑𝑡 Δ𝑡 System Volume 𝑉 𝑡 = 𝑚 𝜌 𝑇 = 𝑚 𝑖 𝑀𝑊 𝑖 ; 𝜌= 𝑚 𝑉 = 𝑚 𝑖 𝑉 = 𝑁 𝑖 𝑀𝑊 𝑖 𝑉 = 𝑖 𝑀𝑊 𝑖

Constant-Volume V Fixed-Mass m Reactor Find T, 𝑖 𝑓𝑜𝑟 𝑖=1,2,…,𝑀 , and P versus time 𝑡 1st Law 𝑑𝑇 𝑑𝑡 = 𝑄 𝑉 − ℎ 𝑖 𝜔 𝑖 + 𝑅 𝑢 𝑇 𝜔 𝑖 𝑖 𝑐 𝑝,𝑖 − 𝑅 𝑢 (true and useful) Species Production 𝑑 𝑖 𝑑𝑡 = 𝑑 𝑁 𝑖 𝑉 𝑑𝑡 = 1 𝑉 𝑑 𝑁 𝑖 𝑑𝑡 = 𝜔 𝑖 =𝑘(𝑇) 𝑗=1 𝑀 𝑗 𝑛 𝑗 Initial Conditions: At t = 0, 𝑇= 𝑇 0 , and 𝑖 = 𝑖 0 , 𝑖=1, 2, …𝑀 State Equation 𝑃= 𝑖 𝑅 𝑢 𝑇 Pressure Rate of change (affects detonation) 𝑑𝑃 𝑑𝑡 = 𝑅 𝑢 𝑇 𝜔 𝑖 + 𝑑𝑇 𝑑𝑡 𝑖

Numerical Solution (Excel) On test I could asked you to Write the needed equations and Perform one time-step calculation

Steady-State Well-Stirred Reactor Exit condition same as system Conservation of M species 𝜔 𝑖 𝑀𝑊 𝑖 V= 𝑚 𝑌 𝑖,𝑜𝑢𝑡 − 𝑌 𝑖,𝑖𝑛 , 𝑖=1,2,…,𝑀 To find 𝜔 𝑖 , need molar concentrations 𝑖 from mass fractions 𝑌 𝑖 𝑖 = 𝑌 𝑖 𝑀𝑊 𝑖 𝑀𝑊 𝑚𝑖𝑥 𝑃 𝑅 𝑢 𝑇 𝑀𝑊 𝑚𝑖𝑥 = 1 𝑌 𝑖 𝑀𝑊 𝑖 Energy 𝑄 = 𝑚 𝑌 𝑖,𝑜𝑢𝑡 ℎ 𝑖,𝑜𝑢𝑡 𝑇 − 𝑌 𝑖,𝑖𝑛 ℎ 𝑖,𝑖𝑛 𝑇 𝑖𝑛 ℎ 𝑖 = ℎ 𝑓,𝑖 𝑜 +∆ ℎ 𝑠,𝑖 All equations are algebraic (not differential)

For our simple example 0= 𝜔 𝑖 𝑀𝑊 𝑖 𝑉+ 𝑚 𝑌 𝑖,𝑖𝑛 − 𝑌 𝑖,𝑜𝑢𝑡 , 𝑖=1,2,…,𝑀=3 0= 𝜔 𝑖 𝑀𝑊 𝑖 𝑉+ 𝑚 𝑌 𝑖,𝑖𝑛 − 𝑌 𝑖,𝑜𝑢𝑡 , 𝑖=1,2,…,𝑀=3 Fuel 𝜔 𝐹 𝑀𝑊 𝐹 𝑉= 𝑚 𝑌 𝐹 − 𝑌 𝐹,𝑖𝑛 1 Oxidizer 0= 𝜔 𝑂𝑥 𝑀𝑊 𝑂𝑥 𝑉+ 𝑚 𝑌 𝑂𝑥,𝑖𝑛 − 𝑌 𝑂𝑥 𝐴 𝐹 𝜔 𝐹𝑢𝑒𝑙 𝑀𝑊 𝑉= 𝑚 𝑌 𝑂𝑥 − 𝑌 𝑂𝑥,𝑖𝑛 2 Product 0= 𝜔 𝑃 𝑀𝑊 𝑃𝑟 𝑉+ 𝑚 𝑌 𝑃𝑟,𝑖𝑛 − 𝑌 𝑃𝑟 − 𝐴 𝐹 +1 𝜔 𝐹𝑢𝑒𝑙 𝑀𝑊 𝑉= 𝑚 𝑌 𝑃𝑟 − 𝑌 𝑃𝑟,𝑖𝑛 or better 𝑌 𝑃𝑟 =1− 𝑌 𝐹 − 𝑌 𝑂𝑥 3

MathCAD Solution . On test, could be asked to find equation whose roots must be found Can you use your calculator to find roots?

Plug-Flow Reactors Assumptions Integrate to find 𝑇 𝑥 , 𝑌 𝑖 𝑥 ,𝜌 𝑥 Quasi-one dimensional (quantities are ≠𝑓𝑛(𝑟,𝜃)) Steady state, 𝑑 𝑑𝑡 =0 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 , 𝑌 𝑖 𝑥 ,𝜌 𝑥 At each location also need to calculate 𝑣 𝑥 𝑥 = 𝑚 𝜌 𝑥 𝐴(𝑥) 𝑃 𝑥 = 𝜌 𝑥 𝑅 𝑢 𝑇(𝑥) 𝑀𝑊 𝑚𝑖𝑥 Like the transient constant-pressure reactor, but varies with location instead of time.

What do we expect?

Equations to be solved Use Need 𝜌 𝑥 and 𝑇 𝑥 𝜔 𝑖 =𝑓𝑛 𝑌 𝑖 ,𝑇,𝑃 𝑃=𝜌𝑅𝑇; 𝑅= 𝑅 𝑢 𝑀𝑊 𝑚𝑖𝑥 ; 1 𝑀𝑊 𝑚𝑖𝑥 = 𝑌 𝑖 𝑀𝑊 𝑖 𝑣 𝑥 = 𝑚 𝜌𝐴 Need 𝜌 𝑥 and 𝑇 𝑥 Assume 𝑄 " 𝑥 , 𝐴 𝑥 and 𝑚 are given Find … (page 209) 𝑑 𝑌 𝑖 𝑑𝑥 = 𝜔 𝑖 𝑀𝑊 𝑖 𝐴 𝜌 𝑣 𝑥 , 𝑖=1,2,…, 𝑀 𝑑𝑇 𝑑𝑥 = 𝑣 𝑥 2 𝜌 𝑐 𝑃 𝑑𝜌 𝑑𝑥 + 𝑣 𝑥 2 𝑐 𝑃 𝐴 𝑑𝐴 𝑑𝑥 − 𝜔 𝑖 𝑀𝑊 𝑖 ℎ 𝑖 𝜌 𝑣 𝑥 𝑐 𝑃 − 𝑄 " 𝒫 𝑚 𝑐 𝑃 𝑑𝜌 𝑑𝑥 = 1− 𝑅 𝑢 𝑐 𝑃 𝑀𝑊 𝑚𝑖𝑥 𝜌 2 𝑣 𝑥 2 1 𝐴 𝑑𝐴 𝑑𝑥 + 𝜌 𝑅 𝑢 𝑣 𝑥 𝑐 𝑃 𝑀𝑊 𝑚𝑖𝑥 𝑀𝑊 𝑖 𝜔 𝑖 ℎ 𝑖 − 𝑀𝑊 𝑚𝑖𝑥 𝑀𝑊 𝑖 𝑐 𝑃 𝑇 + 𝜌 𝑅 𝑢 𝑄 " 𝒫 𝑣 𝑥 𝐴 𝑐 𝑃 𝑀𝑊 𝑚𝑖𝑥 𝑃 1+ 𝑣 𝑥 2 𝑐 𝑃 𝑇 −𝜌 𝑣 𝑥 2 On Test could ask Apply and simplify these equations for a particular problem Derived equations for 𝑑𝐴 𝑑𝑥 =0, 𝑣 𝑥 2 2 ≪ℎ and species have same temperature-independent properties (problem X5)

Problem X5 (homework) 𝑚 x 𝑄 " (𝑥) 𝑄 " (𝑥) 𝑚 Consider a constant-area A 𝑑𝐴 𝑑𝑥 =0 plug flow reactor. It has an axially-varying heat flux 𝑊 𝑚 2 applied to the wall 𝑄 " (𝑥), mass flow rate 𝑚 𝑘𝑔 𝑠 , and operates a constant pressure P (velocity variations are small). 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; 𝜔 𝐹 𝑥 = 𝑑 𝐹 𝑑𝑡 =− 𝐴 𝐹 𝑒 − 𝐸 𝑎 𝑅 𝑢 𝑇 𝑂𝑥 𝑚 𝐹 𝑛 Assume The mass flow kinetic energy is much less than its enthalpy ( 𝑣 𝑥 2 2 ≪ℎ) The fuel F, Oxidizer Ox, and products Pr, have the same 𝑀𝑊 and 𝑐 𝑝 (and 𝑐 𝑝 ≠𝑓𝑛(𝑇)) The oxidizer and product heat of formation are zero, and that of the fuel is ℎ 𝑓,𝐹 𝑜 The inlet equivalence ratio and temperature are Φ 𝑖𝑛 and 𝑇 𝑖𝑛 Use conservation of species and energy to find equations that can be used to find the axial variation of 𝑌 𝑖 𝑥 , 𝑇 𝑥 , 𝜌 𝑥 , 𝑣 𝑥 𝑥

Ch. 8 Laminar Premixed Flames 𝑣 𝑢 𝑆 𝐿 𝛼 Bunsen Burner Inner Cone angle, 𝛼 𝑆 𝐿 is the laminar flame speed relative to the premixed reactants 𝑣 𝑢 is the unburned reactant speed If 𝑣 𝑢 > 𝑆 𝐿 , then cone will adjust it’s angel 𝛼 so that 𝑆 𝐿 = 𝑣 𝑢 sin 𝛼 The angel 𝛼 and its sine sin 𝛼 = 𝑆 𝐿 𝑣 𝑢 decrease as increases 𝑣 𝑢 (inner cone length increases) If 𝑣 𝑢 < 𝑆 𝐿 , then flame will flash back to air holes (unless quenched in tube).

Tube filled with stationary premixed Oxidizer/Fuel Unburned Fuel + Oxidizer Burned Products 𝑆 𝐿 Laminar Flame Speed 𝛿 Products shoot out as flame burns in Flame reference frame: 1 𝑘𝑔 𝐹𝑢+ 𝜈 𝑘𝑔 𝑂𝑥→ 1+𝜈 𝑘𝑔 𝑃𝑟 Conservation of mass: 𝑚 = 𝜌 𝑢 𝑣 𝑢 = 𝜌 𝑏 𝑣 𝑏 ; 𝑣 𝑏 𝑣 𝑢 = 𝜌 𝑢 𝜌 𝑏 = 𝑃 𝑢 𝑅 𝑇 𝑢 𝑅 𝑇 𝑏 𝑃 𝑏 = 𝑇 𝑏 𝑇 𝑢 ≈ 2100𝐾 300𝐾 =7 Diffusion of heat and species cause flame to propagate How to estimate the laminar flame speed 𝑆 𝐿 and thickness 𝛿? Depends on the pressure, temperature, fuel, equivalence ratio, heat and mass diffusion,… 𝛿~1 𝑚𝑚 𝑣 𝑢 = 𝑆 𝐿 , 𝜌 𝑢 , 𝑣 𝑏 , 𝜌 𝑏

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 𝑄 𝑥 ′′ =−𝑘 𝑑𝑇 𝑑𝑥 + 𝑚 𝑖,𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑜𝑛 ′′ ℎ 𝑖 For 𝐿𝑒= 𝛼 𝒟 = 𝑘 𝒟𝜌 𝑐 𝑃 ≈𝑂(1), appropriate for most combustion mixtures Shvab-Zeldovich form: 𝑄 𝑥 ′′ =−𝜌𝒟 𝑑ℎ 𝑑𝑥 Approximate Solution 𝑆 𝐿 = 2𝛼 1+𝜈 𝜌 𝑢 − 𝑚 𝐹 ′′′ 𝛿= 2𝛼 𝜌 𝑢 − 𝑚 𝐹 ′′′ 1+𝜈 = 2𝛼 𝑆 𝐿 (Fast flames are thin) 𝑚 𝐹 ′′′ = 𝜔 𝐹 𝑀𝑊 𝐹 𝜔 𝐹 at average 𝑇 that is closer to 𝑇 𝑏 than 𝑇 𝑢 , and average values of 𝐹𝑢𝑒𝑙 and 𝑂𝑥

Pressure and temperature dependence of SL and 𝛿 𝑆 𝐿 𝑆 𝐿 𝑆 𝐿 𝛿 𝑃 𝑇 𝑢 Φ Φ 𝑆 𝐿 = 2𝛼 1+𝜈 − 𝑚 𝐹 ′′′ 𝜌 𝑢 ~ 𝑃 0 𝑇 𝑢 𝑇 0.375 𝑇 𝑏 −1 𝑒𝑥𝑝 − 𝐸 𝑎 𝑅 𝑢 2 𝑇 𝑏 ≠𝑓𝑛 𝑃 (for HC fuels, N=2) For methane: Actually decreases as P increases: 𝑆 𝐿 cm s = 43 𝑃 [𝑎𝑡𝑚] Increases with temperature: 𝑆 𝐿 cm s =10+3.71× 10 −4 𝑇 𝑢 2 𝐾 Decreases for Φ above or below 1.05 (because that decreases temperature) 𝛿= 2𝛼 𝑆 𝐿 ~ 𝑃 −1 𝑇 0.375 𝑇 𝑏 1 𝑒𝑥𝑝 𝐸 𝑎 𝑅 𝑢 2 𝑇 𝑏 Fast (high temperature) flame are thin

Dependence on Fuel Type 𝑆 𝐿 𝑆 𝐿, 𝐶 3 𝐻 8 𝑇 𝑓 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: 𝑆 𝐿 =40 cm s ; 𝑆 𝐿 cm s = 43 𝑃 [𝑎𝑡𝑚] =43 cm s ; 𝑆 𝐿 cm s =10+3.71× 10 −4 𝑇 𝑢 2 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 𝑆 𝐿 = 𝑆 𝐿,𝑟𝑒𝑓 𝑇 𝑢 𝑇 𝑢,𝑟𝑒𝑓 𝛾 𝑃 𝑃 𝑟𝑒𝑓 𝛽 1−2.1 𝑌 𝑑𝑖𝑙 𝑇 𝑢 >~350𝐾, 𝑇 𝑟𝑒𝑓 =298 𝐾, 𝑃 𝑟𝑒𝑓 =1 𝑎𝑡𝑚 𝑆 𝐿,𝑟𝑒𝑓 = 𝐵 𝑀 + 𝐵 2 Φ− Φ 𝑚 2 𝛾=2.18−0.8 Φ−1 𝛽=−0.16+0.22 Φ−1 RMFD-303 is a research fuel that simulates gasolines

Constant pressure and fixed Mass Reactor 𝑊 𝑄 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 𝜌 ,𝑛𝑒𝑒𝑑 𝜌 state, mixture