Transcript Slides

ME 475/675 Introduction to
Combustion
Lecture 18
Hypothetical chain reaction steps, Global rate calculation, Example 4.3
Announcements
• HW 7, Due 10/12/15 (problem list next lecture)
• Midterm I, Average 74
Hypothetical Chain reactions (example)
• Globally: 𝐴2 + 𝐵2 → 2𝐴𝐵 (𝐴 and 𝐵 are general atoms)
• Find rate of product production in terms of reactant molar concentrations
•
𝑑 𝐴𝐵
𝑑𝑡
= 2𝑘𝑎𝑝𝑝 𝐴2
𝑚
𝐵2
𝑛
• Find 𝑘𝑎𝑝𝑝 , 𝑚 and 𝑛 for a proposed mechanism
• Until you have more experience, you can’t yet know how to propose these mechanisms
• Based on global (observed) reaction
•
1 𝑑 𝐴𝐵
2 𝑑𝑡
=−
𝑑 𝐴2
𝑑𝑡
=−
𝑑 𝐵2
𝑑𝑡
• So if we can any one of these three time-derivatives, then we can find the other two
Proposed intermediate steps
• For the Global (observed) reaction 𝐴2 + 𝐵2 → 2𝐴𝐵
• Someone has postulated four steps:
• Slow creation of free radicals 𝐴 (and 𝐵),
𝑘1
• 𝐴2 + 𝑀 → 𝐴 + 𝐴 + 𝑀
• Chain-initiating step. creates radicals
• Fast consumption of 𝐴 and 𝐵 (neglect reverse because 𝐴 and 𝐵 are small)
𝑘2
• 𝐴 + 𝐵2 → 𝐴𝐵 + 𝐵
𝑘3
• 𝐵 + 𝐴2 → 𝐴𝐵 + 𝐴
• Chain-propagating step (consumes and creates radical, net zero)
• De-energization ter-molecular reaction is slow
𝑘4
• 𝐴 + 𝐵 + 𝑀 → 𝐴𝐵 + 𝑀
• Chain-terminating step (consumes radicals)
• Assume 𝑘2 and 𝑘3 are much greater than 𝑘1 and 𝑘4
• Number of species (𝐴2 , 𝐵2 , 𝐴𝐵, 𝐴, 𝐵, 𝑀), N = 6
• 5 differential equations, 1 algebraic equation (M)
Product Production
• Products: 𝐴𝐵 (eventually want this in terms of 𝐴2 and 𝐵2 , alone)
•
𝑑 𝐴𝐵
𝑑𝑡
= 𝑘2 𝐴 𝐵2 + 𝑘3 𝐵 𝐴2 + 𝑘4 𝐴 𝐵 𝑀 (need to eliminate 𝐴 and 𝐵 )
• But remember
•
𝑑 𝐴𝐵
𝑑𝑡
=
𝑑 𝐴2
−2
𝑑𝑡
• So if it’s easier
𝑑 𝐵2
= −2
𝑑𝑡
𝑑 𝐴2
𝑑 𝐵2
to find
or
𝑑𝑡
𝑑𝑡
in terms of 𝐴2 and 𝐵2 , then that works also
• Reactant consumption, 𝐴2 , 𝐵2
•
•
𝑑 𝐴2
𝑑𝑡
𝑑 𝐵2
𝑑𝑡
= 0 − 𝑘1 𝐴2 𝑀 − 𝑘3 𝐵 𝐴2
= 0 − 𝑘2 𝐴 𝐵2
• So if we can in find 𝐴 terms of 𝐴2 and 𝐵2 (and 𝑀 , which is constant), then we will be successful
Production - Consumption equations
• Intermediates 𝐴, 𝐵 (fast, so becomes algebraic, not differential)
• Steady state approximation
•
•
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
Solve for [A]
• 2𝑘1 𝐴2 𝑀 − 𝑘2 𝐴 𝐵2
𝑘3 𝐴2 + 𝑘4 𝐴 𝑀
+ 𝑘3 𝐴2 − 𝑘4 𝐴 𝑀 𝑘2 𝐴 𝐵2 = 0
• 2𝑘1 𝑘3 𝐴2 2 𝑀 −𝑘2 𝑘3 𝐴2 𝐴 𝐵2 + 2𝑘1 𝑘4 𝐴 𝐴2 𝑀 2 − 𝑘2 𝑘4 𝐴
𝑘2 𝑘3 𝐴2 𝐴 𝐵2 − 𝑘2 𝑘4 𝐴 2 𝑀 𝐵2 = 0
• 2𝑘1 𝑘3 𝐴2 2 𝑀 + 2𝑘1 𝑘4 𝐴 𝐴2 𝑀 2 − 2𝑘2 𝑘4 𝐴 2 𝑀 𝐵2 = 0
2
𝑀 𝐵2 +
• Divide by -2 𝑀
• 𝑘2 𝑘4 𝐵2 𝐴
• 𝐴 =
2
− 𝑘1 𝑘4 𝐴2 𝑀 𝐴 − 𝑘1 𝑘3 𝐴2
−𝑏± 𝑏2 −4𝑎𝑐
2𝑎
=
−𝑏
2𝑎
1∓ 1−
4𝑎𝑐
𝑏2
=
2
= 0 (quadratic in 𝐴 )
𝑘1 𝑘4 𝐴2 𝑀
2𝑘2 𝑘4 𝐵2
1∓ 1+
4𝑘2 𝑘4 𝐵2 𝑘1 𝑘3 𝐴2 2
𝑘1 𝑘4 𝐴2 𝑀 2
• Use + since 𝐴 > 0; also simplify knowing 𝑘2 and 𝑘3 ≫ 𝑘1 and 𝑘4
• 𝐴 =
𝑘1 𝐴2 𝑀
2𝑘2 𝐵2
• Plug into 𝐵 =
1+ 1+
4𝑘2 𝑘3 𝐵2
𝑘1 𝑘4 𝑀 2
𝑘2 𝐴 𝐵2
𝑘3 𝐴2 +𝑘4 𝐴 𝑀
≈
𝑘1 𝐴2 𝑀
2𝑘2 𝐵2
2
𝑀
𝑘2 𝑘3 𝐵2
𝑘1 𝑘4
Production
= 𝐴2
𝑘1 𝑘3
𝐵2 𝑘2 𝑘4
Consumption
𝐵2 Production
•
𝑑 𝐵2
𝑑𝑡
•
𝑑 𝐵2
𝑑𝑡
= −𝑘2 𝐴 𝐵2 = −𝑘2 𝐴2
=
𝑑 𝐴2
𝑑𝑡
=
𝑑 𝐴𝐵
−2
𝑑𝑡
=−
𝑘1 𝑘 3
𝐵2 𝑘2 𝑘4
𝑘1 𝑘2 𝑘3
𝑘4
𝐴2 𝐵2
• We were trying to find 𝑘𝑎𝑝𝑝 , 𝑚 and 𝑛 for
• So 𝑘𝑎𝑝𝑝 = 4
𝑘1 𝑘2 𝑘3
;𝑚
𝑘4
𝐵2 = −
= 1 and 𝑛 = 0.5
𝑑 𝐴𝐵
𝑑𝑡
𝑘1 𝑘 2 𝑘3
𝑘4
𝐴2 𝐵2
0.5
= 2𝑘𝑎𝑝𝑝 𝐴2
𝑚
𝐵2
𝑛
0.5
Example 4.3 page 125
• As mentioned previously, a famous chain mechanism is the Zeldovich, or thermal,
mechanism for the formation of nitric oxide from atmospheric nitrogen:
• 𝑂 + 𝑁2
𝑘1𝑓
𝑁𝑂 + 𝑁
𝑘2
• 𝑁 + 𝑂2 → 𝑁𝑂 + 𝑂
• Because the second reaction is much faster than the first, the steady-state approximation can be used
to evaluate the N-atom concentration. Furthermore, in high-temperature systems, the NO formation
reaction is typically much slower than other reactions involving 𝑂2 and 𝑂. Thus 𝑂2 and 𝑂 can be
assumed to be in (partial) equilibrium:
• 𝑂2
𝐾𝑃
2𝑂.
• Construct a global mechanism
• 𝑁2 + 𝑂2
𝑘𝐺
2𝑁𝑂
• Represented as
•
𝑑 𝑁𝑂
𝑑𝑡
= 𝑘𝐺 𝑁2
𝑚
𝑂2 𝑛 ,
• i.e. determine 𝑘𝐺 , 𝑚, and 𝑛. Using the elementary rate coefficients, etc., from the detailed
mechanisms.
End 2015
Example 4.4 page 127
• Consider the shock-heating of air to 2500 K atm 3 atm. Use the results of
Example 4.3 to determine:
• A. The initial nitric oxide formation rate in ppm/s
• B. The amount of nitric oxide form (in ppm) in 0.25 ms.
• The rate coefficient, is [reference 10 from book]
• 𝑘1𝑓 = 1,82 ∗
1015 𝑒𝑥𝑝
−38,370/𝑇(𝐾) =
𝑐𝑚3
𝑔𝑚𝑜𝑙∗𝑠