Transcript 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
510
f ( T2 mdot)
0
0
.005
.
Yoxin
f ( T mdot)
mdot cp
Hff
1
phi
1
16
3
510
3
0.941
298
15098 Yfin cp ( T 298)
( T 298) MW Vol 6.19 10 exp
Hff
T
9
3
110
3
210
310
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