Transcript ME 475/675 Introduction to Combustion

ME 475/675 Introduction to
Combustion
Lecture 17
Announcements
• HW 6 Due 10/8/14
• Today session from 11 to 12 in PE 113
• Return midterm Monday?
• Nevada Advanced Autonomous Systems Innovation Center (NAASIC)
• Media Showcase, Today, 10 a.m. to noon, Lawler
Chapter 4 Chemical Kinetics (from last lecture)
• Global (apparent) Reaction (this is what is observed)
• 𝐹 + 𝑎𝑂𝑥 → 𝑏𝑃𝑟
• Made up of many intermediate steps that are not seen
•
𝑑𝐹
𝑑𝑡
= −𝑘𝐺 𝑇 𝐹
𝑛
𝑂𝑥
𝑚,
𝑘𝐺 𝑇 global rate constant (black box)
• Elementary bimolecular reaction:
• 𝐴+𝐵 →𝐶+𝐷
𝑑𝐴
•
= −𝑘𝑏𝑖𝑚𝑜𝑙𝑒𝑐 𝐴 1 𝐵 1
𝑑𝑡
• Using collision theory we predicted
• 𝑘 𝑇 =
1 2
2 8𝜋𝑘𝐵 𝑇
𝑝𝑁𝐴𝑉 𝜎𝐴𝐵
𝑒𝑥𝑝
𝜇
• but 𝑝 and 𝐸𝐴 = ?
−
𝐸𝐴
𝑅𝑢 𝑇
;
• “Semi-empirical” three parameter form:
• 𝑘 𝑇 = 𝐴𝑇 𝑏 𝑒𝑥𝑝 −
𝐸𝐴
𝑅𝑢 𝑇
, 𝐴, 𝑏 and 𝐸𝐴 values are tabulated pp. 112
Multistep Mechanisms Reaction Rates
• A sequence of intermediate reactions leading from overall-reactants to overallproducts
• L steps, i = 1, 2,… L
• N species, j = 1, 2,… N; some are intermediate (not in overall products for reactants)
• Example (forward and reverse reactions)
• R1: 𝐻2 + 𝑂2
• R2: 𝐻 + 𝑂2
𝑘𝐹1 ,𝑘𝑅1
𝑘𝐹2 ,𝑘𝑅2
• R3: 𝑂𝐻 + 𝐻2
𝑘𝐹3 ,𝑘𝑅3
• R4: H + 𝑂2 + 𝑀
𝐻𝑂2 + 𝐻
𝑖=1
𝑂𝐻 + 𝑂
𝑖=2
𝐻2 𝑂 + 𝐻
𝑖=3
𝑘𝐹4 ,𝑘𝑅4
𝐻𝑂2 + 𝑀 𝑖 = 4
• Number of steps: L = 4
• Number of Species (j = 𝐻2 , 𝑂2 , 𝐻𝑂2 , 𝐻, 𝑂𝐻, 𝑂, 𝐻2 𝑂, 𝑀): N = 8
• 8 time-dependent unknowns molar concentration: 𝑗 𝑡
• Need 8 equations (constraints)
Species net reaction 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
•…
• Book describes compact notation
•
What happens at equilibrium?
• A general reaction
• 𝑎𝐴 + 𝑏𝐵
𝑘𝑓
𝑘𝑟
𝑐𝐶 + 𝑑𝐷
• Consumption – 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
𝑘𝑟
𝑐𝐶 + 𝑑𝐷
• Equilibrium Constant: 𝐾𝑃 𝑇 =
• Rate constant: 𝐾𝐶 𝑇 =
• Using 𝑖 =
𝑃𝑖
𝑅𝑢 𝑇
• 𝐾𝐶 𝑇 = 𝐾𝑃 𝑇
• 𝐾𝑃 𝑇 = 𝐾𝐶 𝑇
𝑘𝐹 𝑇
𝑘𝑅 𝑇
𝑃𝐶 𝑐 𝑃𝐷 𝑑
𝑃𝑜
𝑃𝑜
𝑃𝐴 𝑎 𝑃𝐵 𝑏
𝑃𝑜
𝑃𝑜
=
𝑃𝑜 𝑐+𝑑−(𝑎+𝑏)
,
𝑅𝑢 𝑇
𝑅 𝑇 𝑐+𝑑−(𝑎+𝑏)
𝑢
𝑃𝑜
𝐶 𝑐𝐷𝑑
𝐴𝑎𝐵𝑏
=
𝑃𝐶 𝑐 𝑃𝐷 𝑑
𝑅𝑢 𝑇
𝑅𝑢 𝑇
𝑃𝐴 𝑎 𝑃𝐵 𝑏
𝑅𝑢 𝑇
𝑅𝑢 𝑇
=
or
• Note: If 𝑁𝑅 = 𝑎 + 𝑏 = 𝑐 + 𝑑 = 𝑁𝑃 , then 𝐾𝑃 𝑇 = 𝐾𝐶 𝑇
𝑃𝐶 𝑐 𝑃𝐷 𝑑
𝑃𝑜
𝑃𝑜
𝑃 𝐴 𝑎 𝑃𝐵 𝑏
𝑃𝑜
𝑃𝑜
𝑃𝑜 𝑐+𝑑−(𝑎+𝑏)
𝑅𝑢 𝑇
Example 4.2 (page 120, turn in next time)
• In their survey of experimental determinations of rate coefficients for
the N-H-O system, Nanson and Salimian [reference 10 in book]
recommend the following rate coefficient for the reaction
• 𝑁𝑂 + 𝑂 → 𝑁 + 𝑂2 .
• 𝑘𝑓 = 2.80 ∗
−20,820
109 𝑇 1.0 𝑒𝑥𝑝
𝑇
=
𝑐𝑚3
𝑔𝑚𝑜𝑙∗𝑠
• These coefficients generally use [cgs] (centimeter, gram, seconds)] units, not [mks]
• Determine the rate coefficient at 2300 K for the reverse reaction, i.e.
• 𝑁 + 𝑂2 → 𝑁𝑂 + 𝑂
Reaction Mechanisms
• Describes concepts used to find global (apparent) rate
• Steady State Approximation
• Slow creation and rapid consumption of radicals cause them to reach steady-state quickly
• 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 – Consumption of N
•
𝑑𝑁
𝑑𝑡
= 𝑘1 𝑂 𝑁2 − 𝑘2 𝑁 𝑂2
• Since it is very fast and reaches steady state almost right away
• 𝑁
𝑆𝑆
=
𝑘1 𝑂 𝑁2
𝑘2 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 𝑘1 𝑂 𝑁2
𝑘2 𝑂2
= 𝑑𝑡
Uni-molecular Reaction
• Apparent (global)
• 𝐴 → 𝑝𝑟𝑜𝑑𝑢𝑐𝑡;
• Find
𝑑 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠
𝑑𝑡
= 𝑘𝑎𝑝𝑝 𝐴
• Measurements show that, for uni-molecular reactions
• At high pressures reaction rate is not a function of pressure
• At low pressure, reaction rate increases with pressure
• This can be explained using a “three-step” mechanism
𝑘𝑒
• 𝐴 + 𝑀 → 𝐴∗ + 𝑀 (𝐴∗ is an energized state of 𝐴 and highly reactive)
•
𝐴∗
+𝑀
𝑘
∗ 𝑢𝑛𝑖
• 𝐴
𝑘𝑑𝑒
𝐴 + 𝑀 (De-energization of A)
𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠
• 𝐴∗ will reach steady state conditions
Products production rate
𝑑 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠
𝑑𝑡
•
= 𝑘𝑢𝑛𝑖 𝐴∗
(need 𝐴∗ )
• Molecular Balance for 𝐴∗
•
𝑑 𝐴∗
𝑑𝑡
∗
• 𝐴
•
•
= 𝑘𝑒 𝐴 𝑀 − 𝑘𝑑𝑒 𝐴∗ 𝑀 − 𝑘𝑒 𝐴∗ ≈ 0 (since 𝐴∗ is consumed as fast as it is produced)
𝑆𝑆
≈
𝑘𝑒 𝐴 𝑀
𝑘𝑑𝑒 𝑀 + 𝑘𝑒
𝑘𝑢𝑛𝑖 𝑘𝑒 𝑀
𝑑 𝑃𝑟𝑜𝑑𝑢𝑐𝑡𝑠
=
𝑑𝑡
𝑘𝑑𝑒 𝑀 + 𝑘𝑒
𝑘
𝑘 𝑀
𝑘𝑎𝑝𝑝 = 𝑢𝑛𝑖 𝑒
𝑘𝑑𝑒 𝑀 + 𝑘𝑒
• Recall 𝑀 =
• 𝑘𝑎𝑝𝑝 =
• For 𝑃
• For 𝑃
𝑁𝑀
𝑉
= 𝜒𝑀
𝑃
𝑁
𝑉
𝐴 = 𝑘𝑎𝑝𝑝 𝐴
= 𝜒𝑀
𝑃
𝑅𝑢 𝑇
𝑘𝑢𝑛𝑖 𝑘𝑒 𝜒𝑀 𝑅 𝑇
𝑢
𝑃
𝑘𝑑𝑒 𝜒𝑀 𝑅 𝑇+ 𝑘𝑒
𝑢
𝑘𝑒 𝑅𝑢 𝑇
≫
,𝑘
𝑘𝑑𝑒 𝜒𝑀 𝑎𝑝𝑝
𝑘𝑒 𝑅𝑢 𝑇
≪
,𝑘
𝑘𝑑𝑒 𝜒𝑀 𝑎𝑝𝑝
→
→
𝑘𝑢𝑛𝑖 𝑘𝑒
≠ 𝑓𝑛 𝑃
𝑘𝑑𝑒
𝑘𝑢𝑛𝑖 𝜒𝑀 𝑃
= 𝑘𝑢𝑛𝑖
𝑅𝑢 𝑇
(consistent with observations)
𝑀 ~𝑃1 (also consistent with observations)
Hypothetical Chain reactions (example)
• Globally: 𝐴2 + 𝐵2 → 2𝐴𝐵 (𝐴 and 𝐵 are general atoms)
•
𝑑 𝐴𝐵
𝑑𝑡
= 2𝑘𝑎𝑝𝑝 𝐴2 𝐵2
• Proposed intermediate steps
• Slow creation of free radicals 𝐴 (and 𝐵)
𝑘1
• 𝐴2 + 𝑀 → 𝐴 + 𝐴 + 𝑀
• Fast consumption of 𝐴 and 𝐵 (neglect reverse because 𝐴 and 𝐵 are small)
𝑘2
• 𝐴 + 𝐵2 → 𝐴𝐵 + 𝐵
𝑘3
• 𝐵 + 𝐴2 → 𝐴𝐵 + 𝐴
• De-energization ter-molecular reaction is slow
𝑘4
• 𝐴 + 𝐵 + 𝑀 → 𝐴𝐵 + 𝑀
• Assume 𝑘2 and 𝑘3 are much greater than 𝑘1 and 𝑘4
Production - Consumption equations
• Reactants
•
•
𝑑 𝐴2
𝑑𝑡
𝑑 𝐵2
𝑑𝑡
= 0 − 𝑘1 𝐴2 𝑀 − 𝑘3 𝐵 𝐴2
= 0 − 𝑘1 𝐴 𝐵2
• Products
•
𝑑 𝐴𝐵
𝑑𝑡
= 𝑘2 𝐴 𝐵2 + 𝑘3 𝐵 𝐴2 + 𝑘4 𝐴 𝐵 𝑀
• Intermediates (fast)
•
•
𝑑𝐴
𝑑𝑡
𝑑𝐵
𝑑𝑡
= 2𝑘1 𝐴2 𝑀 − 𝑘2 𝐴 𝐵2 + 𝑘3 𝐵 𝐴2 − 𝑘4 𝐴 𝐵 𝑀 ≈ 0
= 𝑘2 𝐴 𝐵2 − 𝑘3 𝐵 𝐴2 − 𝑘4 𝐴 𝐵 𝑀 ≈ 0
• 𝐵 =
𝑘2 𝐴 𝐵2
𝑘3 𝐴2 −𝑘4 𝐴 𝑀
• 2𝑘1 𝐴2 𝑀 − 𝑘2 𝐴 𝐵2 +
• 2𝑘1 𝐴2 𝑀 − 𝑘2 𝐴 𝐵2
𝑘3 𝐴2 −𝑘4 𝐴 𝑀 𝑘2 𝐴 𝐵2
𝑘3 𝐴2 −𝑘4 𝐴 𝑀
𝑘3 𝐴2 − 𝑘4 𝐴 𝑀
=0
+ 𝑘3 𝐴2 − 𝑘4 𝐴 𝑀 𝑘2 𝐴 𝐵2 = 0