Transcript Slides

ME 475/675 Introduction to
Combustion
Lecture 29
Shvab-Zeldovich form of Energy Conservation
Announcements
• Midterm 2
• November 13, 2015
• HW 12 (problem 8.2)
• Due Wednesday, November 4, 2015
Broader Impact Assignment
• Two important ABET Student Learning Objectives:
• Students will demonstrate :
• A recognition of a need for, and an ability to engage in, life long learning (graduate school, continuing
education, short courses, technical training, self instruction by reading articles or textbook)
• A knowledge of contemporary issues
• Two choices, Both due Friday, November 6, 2015, two paragraph summaries
• Seminar: Used Nuclear Fuel: Storage, Transportation, and Disposal – Technical, Political and
Other Issues
• John Wagner, Director, Reactor & Nuclear Systems Division, Oak Ridge National Laboratory
• Today, Noon, November 2, 2015, DMS 102
• Hosted by UNR American Nuclear Society’s Student Chapter
• President: Kodi Summers [email protected]
• Or
• Article: Dependence of Fire Time of Concern on Location of a One-assembly Transport
Packages
Tube filled with stationary premixed Oxidizer/Fuel
Products shoot
out as flame
burns in
𝑆𝐿 Laminar Flame Speed
Burned
Products
𝛿
Unburned Fuel + Oxidizer
• Video: http://www.youtube.com/watch?v=CjGuHbsi3a8&feature=related (1.49; 3:19; 3:45; 4:05; 4:31; 5:05; 5:35; 6:17; 6:55)
• How to estimate the laminar flame speed 𝑆𝐿 and thickness 𝛿?
• Depends on the pressure, fuel, equivalence ratio,…
• 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
=
• For hydrocarbon fuels at P = 1 atm, 𝑇𝑢 ≈ 300𝐾, 𝑇𝑏 ≈ 2100𝐾, 𝑣𝑏 ≈ 7𝑣𝑢
• What happens within a premixed flame?
𝑇𝑏
𝑇𝑢
Simplified Analysis Assumptions
• Assumptions
•
•
•
•
•
•
•
One dimensional flow
Neglect kinetic energy, viscosity, radiation
Constant pressure
𝑐𝑃,𝑖 = 𝑐𝑃 ≠ 𝑓𝑛 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒 𝑜𝑟 𝑆𝑝𝑒𝑐𝑖𝑒
Single Step Kinetics
Φ < 1, Fuel Lean, so fuel is completely consumed
𝛼
𝑘
𝑘
Lewis Number, 𝐿𝑒 = =
≈ 1, so ≈ 𝒟𝜌
𝒟
𝒟𝜌𝑐𝑃
𝑐𝑃
• Calculate heat and species transport per unit area normal to flow
• 𝑞𝑥" = 𝑚′′ ℎ − 𝑘
𝑑𝑇
𝑑𝑥
and 𝑚𝑖" = 𝑚′′ 𝑌𝑖 − 𝜌𝒟
𝑑𝑌𝑖
𝑑𝑥
• Transport of heat and species ahead of reaction zone allows the flame to move
forward.
• Use this to calculate laminar flame speed and flame thickness
• Thinner flames have bigger gradients and so diffuse and move faster
Conservations Laws
Heat and Radical
• Mass Conservation
• 𝑚" = 𝜌𝑣𝑥 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡;
𝑑 𝜌𝑣𝑥
𝑑𝑥
=0
• Species Conservation (in + generation = out)
•
𝐴𝑚𝑖"
𝑚𝑖′′′
+ 𝐴 𝑑𝑥
𝑚𝑖′′′
"
=𝐴
𝑑𝑚
𝑚𝑖" + 𝑖 𝑑𝑥
𝑑𝑥
𝑑 𝑚𝑖"
•
=
𝑑𝑥
• Generation of 𝑖 𝑚𝑖′′′ causes mass flux of 𝑖 𝑚𝑖" to increase with 𝑥.
• Apply to stoichiometric combustion 1 𝑘𝑔 𝐹 + 𝜈 𝑘𝑔 𝑂𝑥 → 1 + 𝜈 𝑃𝑟
𝑘𝑔 𝑂𝑥
= Stoichiometric Air/Fuel
𝑘𝑔 𝐹
1 ′′′
1
′′′
• 𝑚𝐹′′′ = 𝑚𝑂𝑥
=−
𝑚𝑃𝑟
𝜈
𝜈+1
"
𝑑 𝑚𝐹𝑢
′′′
• 𝜈=
•
𝑑𝑥
ratio
= 𝑚𝐹𝑢 (< 0)
•
"
𝑑 𝑚𝑂𝑥
𝑑𝑥
′′′
′′′
= 𝑚𝑂𝑥
= 𝜈 𝑚𝐹𝑢
(< 0)
•
"
𝑑 𝑚𝑃𝑟
𝑑𝑥
′′′
= 𝑚𝑃𝑟
= − 𝜈 + 1 𝑚𝐹′′′ (> 0)
𝑑𝑥
𝑚𝑖′′′
𝐴𝑚𝑖"
𝐴
"
𝑑
𝑚
𝑖
𝐴 𝑚𝑖" +
𝑑𝑥
𝑑𝑥
Energy Conservation (Ch. 7, pp. 239-244)
𝐴𝑚′′ ℎ
𝐴𝑄𝑥′′
• 𝑄𝑄𝑉 − 𝑊𝑄𝑉 = 𝑚 ℎ𝑜𝑢𝑡 − ℎ𝑖𝑛
•𝐴
•
𝑄𝑥′′
𝑑 𝑄𝑥′′
−
𝑑𝑥
−
𝑄𝑥′′
=
𝑑ℎ
′′
𝑚
𝑑𝑥
+
𝑑 𝑄𝑥′′
𝑑𝑥
𝑑𝑥
=
𝑑𝑥
𝐴𝑚′′
ℎ+
𝑑ℎ
𝑑𝑥
𝑑𝑥
−ℎ
𝑊𝑄𝑉
• Decreasing heat flux in x-direction increases enthalpy in the +x-direction
𝑑𝑥
𝑄𝑥′′
ℎ
𝐴𝑚′′
𝑑ℎ
ℎ+
𝑑𝑥
𝑑𝑥
′′
𝑑
𝑄
𝑥
𝐴 𝑄𝑥′′ +
𝑑𝑥
𝑑𝑥
What is the Heat Flux?
• Heat: Energy transfer at a boundary due to temperature difference
• When there are species and temperature gradients, diffusion contributes to heat
flux
•
𝑄𝑥′′
=
• But
•
𝑑𝑇
−𝑘
𝑑𝑥
𝑑ℎ
𝑑𝑥
𝑑ℎ
−𝜌𝒟
𝑑𝑥
=
′′
𝑚𝑖,𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑜𝑛
ℎ𝑖
+
𝑑
𝑑𝑥
𝑌𝑖 ℎ𝑖 =
= −𝜌𝒟
𝑑𝑌𝑖
ℎ
𝑑𝑥 𝑖
𝑑𝑌𝑖
ℎ
𝑑𝑥 𝑖
−
+
=
𝑑𝑇
−𝑘
𝑑𝑥
−
𝑌𝑖
𝑑ℎ𝑖
𝑑𝑥
𝑑𝑌𝑖
ℎ
𝑑𝑥 𝑖
𝑑𝑇
𝜌𝒟𝑐𝑝
𝑑𝑥
=
= −𝜌𝒟
𝑑𝑌𝑖
𝜌𝒟𝑖
ℎ𝑖
𝑑𝑥
+
𝑌𝑖 𝑐𝑝,𝑖
𝑑𝑌𝑖
ℎ
𝑑𝑥 𝑖
−
=
𝑑𝑇
𝑑𝑥
𝑑𝑇
−𝑘
𝑑𝑥
=
𝑑𝑇
𝑘
𝑑𝑥
− 𝜌𝒟
𝑑𝑌𝑖
ℎ
𝑑𝑥 𝑖
= 𝑄𝑥′′
• Shvab-Zeldovich assumption: 𝐿𝑒 ≈ 𝑂 1 , valid for most combustion gases
• 𝐿𝑒 =
•
𝑄𝑥′′
=
𝛼
𝒟
=
𝑘
𝒟𝜌𝑐𝑃
𝑑ℎ
−𝜌𝒟
𝑑𝑥
≈ 1; 𝑘 ≈ 𝒟𝜌𝑐𝑃
• heat flux in x-direction due to both conduction and diffusion
+ 𝑐𝑝
𝑑𝑌𝑖
ℎ𝑖
𝑑𝑥
𝑑𝑇
𝑑𝑥
Shvab-Zeldovich form of Energy Conservation
•
𝑑 𝑄𝑥′′
−
𝑑𝑥
•ℎ=
•
•
𝑄𝑥′′
•
•
•
•
𝑑
𝑑𝑥
𝑑
𝑑𝑥
𝑑ℎ
𝑑𝑥
=
=
𝑑ℎ
′′
𝑚
𝑑𝑥
𝑜
𝑌𝑖 ℎ𝑓,𝑖
𝑌𝑖 ℎ𝑖 =
=
𝑑𝑌𝑖 𝑜
ℎ
𝑑𝑥 𝑓,𝑖
𝑑ℎ
−𝜌𝒟
𝑑𝑥
𝑑 𝑄𝑥′′
−
𝑑𝑥
=
𝑑
𝑑𝑥
(Energy Equation)
+
𝑜
𝑑ℎ𝑓,𝑖
𝑑 𝑇
𝑌𝑖
+
𝑐 𝑑𝑇
𝑑𝑥
𝑑𝑥 𝑇𝑟𝑒𝑓 𝑝
+
= −𝜌𝒟
𝑑𝑌𝑖 𝑜
ℎ
𝑑𝑥 𝑓,𝑖
𝜌𝒟
𝑑𝑌𝑖 𝑜
𝜌𝒟
ℎ𝑓,𝑖
𝑑𝑥
𝑑𝑌𝑖 𝑜
𝜌𝒟
ℎ
𝑑𝑥 𝑓,𝑖
𝑑
𝑜
−
ℎ𝑓,𝑖
𝑚′′ 𝑌𝑖
𝑑𝑥
𝑇
𝑌𝑖 𝑇 𝑐𝑝,𝑖 𝑑𝑇
𝑟𝑒𝑓
+
−
−
𝑚𝑖"
𝑑𝑇
𝑐𝑝
𝑑𝑥
𝑑𝑌𝑖 𝑜
ℎ
𝑑𝑥 𝑓,𝑖
+
+
=
=
𝑜
𝑌𝑖 ℎ𝑓,𝑖
𝑑𝑌𝑖 𝑜
ℎ
𝑑𝑥 𝑓,𝑖
+
+
𝑇
𝑐 𝑑𝑇;
𝑇𝑟𝑒𝑓 𝑝
𝑑𝑇
𝑐𝑝
𝑑𝑥
𝑑𝑇
𝑐𝑝
𝑑𝑥
𝑑𝑇
𝑐𝑝
𝑑𝑥
𝑑𝑌𝑖 𝑜
𝑑𝑇
=
ℎ𝑓,𝑖 + 𝑐𝑝
𝑑𝑥
𝑑𝑥
𝑑𝑇
𝑑
𝑑𝑇
𝑜
′′
′′
𝑚 𝑌𝑖 ℎ𝑓,𝑖 = 𝑚 𝑐𝑝 −
𝜌𝒟𝑐𝑝
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑑𝑌𝑖
𝑑𝑇
𝑑
𝑑𝑇
′′
𝜌𝒟
= 𝑚 𝑐𝑝 −
𝜌𝒟𝑐𝑝
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑚′′
𝑐𝑝 =
𝑌𝑖 𝑐𝑝,𝑖
Shvab-Zeldovich form of Energy Conservation
•
𝑑
−
𝑑𝑥
𝑜
ℎ𝑓,𝑖
𝑚′′ 𝑌𝑖
𝑑𝑌𝑖
𝜌𝒟
𝑑𝑥
−
𝑚𝑖"
• Left hand side=
𝑑
−
𝑑𝑥
• Species conservation:
=
𝑑𝑇
′′
𝑚 𝑐𝑝
𝑑𝑥
𝑜
ℎ𝑓,𝑖
𝑚𝑖" = −
𝑑𝑚𝑖"
𝑑𝑥
−
𝑑
𝑑𝑥
"
𝑑
𝑚
𝑜
𝑖
ℎ𝑓,𝑖
𝑑𝑥
𝑑𝑇
𝜌𝒟𝑐𝑝
𝑑𝑥
=−
𝑜
ℎ𝑓,𝑖
𝑚𝑖′′′
= 𝑚𝑖′′′
•
•
𝑜
𝑜
𝑜
𝑜
′′′
′′′
ℎ𝑓,𝑖
𝑚𝑖′′′ = ℎ𝑓,𝐹
𝑚𝐹′′′ + ℎ𝑓,𝑂𝑥
𝑚𝑂𝑥
+ ℎ𝑓,𝑃𝑟
𝑚𝑃𝑟
𝑜
𝑜
𝑜
= ℎ𝑓,𝐹
𝑚𝐹′′′ + ℎ𝑓,𝑂𝑥
𝜈𝑚𝐹′′′ − ℎ𝑓,𝑃𝑟
1 + 𝜈 𝑚𝐹′′′
•
𝑜
𝑜
𝑜
= 𝑚𝐹′′′ ℎ𝑓,𝐹
+ ℎ𝑓,𝑂𝑥
𝜈 − ℎ𝑓,𝑃𝑟
1+𝜈
= 𝑚𝐹′′′ Δℎ𝐶
𝑜
𝑜
𝑜
• Δℎ𝐶 = ℎ𝑓,𝐹
+ ℎ𝑓,𝑂𝑥
𝜈 − ℎ𝑓,𝑃𝑟
1+𝜈
• For
• Heat of combustion for 1 𝑘𝑔 𝐹 + 𝜈 𝑘𝑔 𝑂𝑥 → 1 + 𝜈 𝑃𝑟
𝛼
𝑘
𝐿𝑒 = =
≈ 1; 𝑘 ≈ 𝒟𝜌𝑐𝑃
𝒟
𝒟𝜌𝑐𝑃
𝑑𝑇
𝑑
𝑑𝑇
• −𝑚𝐹′′′ Δℎ𝐶 = 𝑚′′ 𝑐𝑝 −
𝑘
𝑑𝑥
𝑑𝑥
𝑑𝑥
• Generation = (net flow out) + (net conduction out)
nd
2
order differential
equation for T(x)
•
′′ 𝑑𝑇
𝑚
𝑑𝑥
−
1 𝑑
𝑐𝑝 𝑑𝑥
𝑑𝑇
𝑘
𝑑𝑥
=
′′′
𝑚𝐹
Δℎ𝐶
−
𝑐𝑝
• Where 𝑚′′ = 𝜌𝑢 𝑆𝐿
• Only accepts 2 boundary conditions,
• But we have 4 (Eigenvalue problem)
• For 𝑥 → −∞: 𝑇 → 𝑇𝑢 𝑎𝑛𝑑
• For 𝑥 → +∞: 𝑇 → 𝑇𝐵 𝑎𝑛𝑑
𝑑𝑇
𝑑𝑥
𝑑𝑇
𝑑𝑥
→0
→0
• For an approximate solution, assume a simple profile
• Find flame thickness 𝛿 and laminar flame speed 𝑆𝐿 =
conditions can be satisfied
𝑚′′
𝜌𝑢
so that all four boundary
Approximate Solution
•
𝑑𝑇
′′
𝑚
𝑑𝑥
−
1 𝑑
𝑐𝑝 𝑑𝑥
• Integrate
•
𝑚′′
𝑇
𝑇𝑏
𝑇𝑢
−
𝑑𝑇
𝑘
𝑑𝑥
+∞
−∞
1
𝑐𝑝
=
′′′
𝑚𝐹
Δℎ𝐶
−
𝑐𝑝
𝑑𝑥
𝑑𝑇 0
𝑘
𝑑𝑥 0
=
Δℎ𝐶 +∞ ′′′
−
𝑚𝐹 𝑑𝑥
𝑐𝑝 −∞
• 𝑚𝐹′′′ = 𝑓𝑛 𝑇 = 0 𝑓𝑜𝑟 𝑥 < 0 𝑜𝑟 𝑥 > 𝛿
• Inside 0 < 𝑥 < 𝛿,
′′
• 𝑚 𝑇𝑏 − 𝑇𝑢 =
•
𝑚′′
𝑇𝑏 − 𝑇𝑢 =
𝑑𝑇
𝑑𝑥
=
𝑇𝑏 −𝑇𝑢
,
𝛿
so 𝑑𝑥 =
Δℎ𝐶 +∞ ′′′
−
𝑚𝐹
𝑐𝑝 −∞
Δℎ𝐶 𝛿 ′′′
−
𝑚𝐹
𝑐𝑝
𝑇
𝛿
𝑑𝑇
𝑇𝑏 −𝑇𝑢
eqn. 1
• Two unknowns: 𝑚′′ = 𝑆𝐿 𝜌𝑢 and 𝛿
• Need another equation
𝛿
𝑑𝑇
𝑇𝑏 −𝑇𝑢
=
𝑇𝑏 ′′′
Δℎ𝐶 𝛿 1
−
𝑚𝐹
𝑐𝑝 𝑇𝑏 −𝑇𝑢 𝑇𝑢
𝑇 𝑑𝑇
Average over
temperature: 𝑚𝐹′′′
Approximate Solution
•
𝑑𝑇
′′
𝑚
𝑑𝑥
−
1 𝑑
𝑐𝑝 𝑑𝑥
• Integrate
•
•
•
•
𝑑𝑇
𝑘
𝑑𝑥
𝛿/2
−∞
=
′′′
𝑚𝐹
Δℎ𝐶
−
𝑐𝑝
𝑑𝑥
𝑇𝑏 −𝑇𝑢
𝛿
𝑇𝑏 +𝑇𝑢
2
𝑇𝑢
𝑘 𝑑𝑇
Δℎ𝐶 𝛿/2 ′′′
𝑚 𝑇
−
=−
𝑚𝐹 𝑑𝑥
𝑐𝑝 𝑑𝑥 0
𝑐𝑝 −∞
𝑇 +𝑇
𝑘 𝑇𝑏 −𝑇𝑢
𝑚′′ 𝑏 𝑢 − 𝑇𝑢 −
=0
2
𝑐𝑝
𝛿
′′
≈0
𝑇𝑏 −𝑇𝑢
𝑘 𝑇𝑏 −𝑇𝑢
′′
𝑚
=
2
𝑐𝑝
𝛿
2𝑘
𝛿 = ′′ eqn. 2
𝑚 𝑐𝑝
• From eqn. 1
• 𝑚′′ =
𝑚′′
𝑇𝑏 − 𝑇𝑢 =
2𝑘Δℎ𝐶
𝑐𝑝2 𝑇𝑏 −𝑇𝑢
Δℎ𝐶 𝛿 ′′′
−
𝑚𝐹
𝑐𝑝
−𝑚𝐹′′′ = 𝑆𝐿 𝜌𝑢 ; 𝑆𝐿 =
1
𝜌𝑢
=
Δℎ𝐶 ′′′ 2𝑘
−
𝑚𝐹 ′′
𝑐𝑝
𝑚 𝑐𝑝
2𝑘Δℎ𝐶
𝑐𝑝2 𝑇𝑏 −𝑇𝑢
−𝑚𝐹′′′
Approximate Solution
• 𝑆𝐿 =
1
𝜌𝑢
2𝑘Δℎ𝐶
𝑐𝑝2 𝑇𝑏 −𝑇𝑢
−𝑚𝐹′′′
• But Δℎ𝐶 = 1 + 𝜈 𝑐𝑃 𝑇𝑏 − 𝑇𝑢
• Show this in HW (problem 8.2)
• 𝑆𝐿 =
•
•
2𝑘 1+𝜈 𝑐𝑃 𝑇𝑏 −𝑇𝑢
𝜌𝑢 𝜌𝑢 𝑐𝑝2 𝑇𝑏 −𝑇𝑢
−𝑚𝐹′′′
2𝛼 1+𝜈
𝑆𝐿 =
−𝑚𝐹′′′
𝜌𝑢
2𝑘
2𝑘
eqn. 2: 𝛿 = ′′ =
𝑚 𝑐𝑝
𝑆𝐿 𝜌𝑢 𝑐𝑝
•𝛿=
2𝛼𝜌𝑢
′′′ 1+𝜈
−𝑚 𝐹
=
2𝛼
𝑆𝐿
=
; where 𝛼 =
2𝛼
𝑆𝐿
=
𝑘
𝜌𝑢 𝑐𝑃
2𝛼
2𝛼 1+𝜈
𝜌𝑢
′′′
−𝑚𝐹
(Fast flames are thin)
Example 8.2
• Estimate the laminar flame speed of a stoichiometric propane-air mixture using
the simplified theory results (Eqn. 8.20). Use the global one-step reaction
mechanism (Eqn. 5.2, Table 5.1, pp. 156-7) to estimate the mean reaction rate.
• Find: ___?
Conservations Laws
𝑑𝑥
𝐴𝑚𝑖"
𝐴
•
𝑚𝑖′′′
=
𝑑 𝑚𝑖"
𝑑𝑥
=
𝑑
𝑑𝑥
𝑚" 𝑌𝑖 − 𝜌𝒟
𝑑𝑌𝑖
𝑑𝑥
(using Flick’s Law)
• 𝑖 = 1, 2, … , 𝑀 = 3; Fuel, Oxidizer, Products
•
•
•
𝑑
𝑑𝑥
𝑑
𝑑𝑥
𝑑
𝑑𝑥
𝑑𝑌𝐹
𝑑𝑥
𝑑𝑌
𝜌𝒟 𝑂𝑥
𝑑𝑥
𝑑𝑌
𝜌𝒟 𝑃𝑟
𝑑𝑥
𝑚" 𝑌𝐹 − 𝜌𝒟
= 𝑚𝐹′′′
𝑚" 𝑌𝑂𝑥 −
′′′
= 𝑚𝑂𝑥
= 𝜈 𝑚𝐹′′′
𝑚" 𝑌𝑃𝑟 −
′′′
= 𝑚𝑃𝑟
= − 1 + 𝜈 𝑚𝐹′′′
𝑚𝑖′′′
"
𝑑
𝑚
𝑖
𝐴 𝑚𝑖" +
𝑑𝑥
𝑑𝑥
What is the Heat Flux?
• Heat: Energy transfer at a boundary due to temperature difference
• When there are species and temperature gradients, diffusion contributes to heat flux
•
𝑄𝑥′′
=
𝑑𝑇
−𝑘
𝑑𝑥
• Note:
•
𝑄𝑥′′
𝑑ℎ
𝑑𝑥
=
+
′′
𝑚𝑖,𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑜𝑛
ℎ𝑖
𝑑
𝑑𝑥
𝑌𝑖 ℎ𝑖 =
•
𝑑𝑌𝑖
ℎ
𝑑𝑥 𝑖
= 𝑑𝑥 − 𝑐𝑝 𝑑𝑥
𝑑ℎ
𝑑𝑇
=
𝑑𝑇
−𝑘
𝑑𝑥
− 𝜌𝒟
𝑑ℎ
𝑑𝑥
• Heat Flux =
𝑑𝑌𝑖
ℎ
𝑑𝑥 𝑖
−
𝑑𝑇
𝑐𝑝
𝑑𝑥
𝑑𝑇
−𝑘
𝑑𝑥
=
+
𝑌𝑖
=
𝑑ℎ𝑖
𝑑𝑥
=
𝑑𝑇
−𝑘
𝑑𝑥
−
𝑑𝑌𝑖
𝜌𝒟𝑖
ℎ
𝑑𝑥 𝑖
𝑑𝑌𝑖
ℎ
𝑑𝑥 𝑖
+
𝑑ℎ
− 𝜌𝒟
𝑑𝑥
=
𝑌𝑖 𝑐𝑝,𝑖
+
𝑑𝑇
𝑑𝑥
𝑑𝑇
−𝑘
𝑑𝑥
=
− 𝜌𝒟
𝑑𝑌𝑖
ℎ
𝑑𝑥 𝑖
𝑑𝑇
𝜌𝒟𝑐𝑝
𝑑𝑥
• Flux due to conduction +
• Flux of standardized enthalpy due to species diffusion +
• Flux of sensible enthalpy due to species diffusion
• For 𝐿𝑒 =
𝛼
𝒟
=
𝑘
𝒟𝜌𝑐𝑃
≈ 1; 𝑘 ≈ 𝒟𝜌𝑐𝑃
• Shvab-Zeldovich assumption (𝐿𝑒 ≈ 𝑂 1 , valid for most combustion gases)
𝑑ℎ
′′
• 𝑄𝑥 = −𝜌𝒟
(due to both conduction and diffusion)
𝑑𝑥
+ 𝑐𝑝
𝑑𝑌𝑖
ℎ
𝑑𝑥 𝑖
𝑑𝑇
𝑑𝑥