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ME 475/675 Introduction to Combustion Lecture 14 Midterm I Review

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ME 475/675 Introduction to
Combustion
Lecture 14
Midterm I Review
Announcements
• HW 5 Due Now
• Solutions will be posted today
• Midterm 1
•
•
•
•
Monday, September 28, 2015
8-10 AM, PE 104
In-class review Today
Hasib Tutorial: Saturday, Sept. 26, 5 PM, in PE 101 (who will come?)
Midterm I
• See me after class today for any special accommodations
• Confirm by email (greiner@unr.edu)
• Open book with bookmarks, 1 page of notes
• 3-4 HW-like problems
• Coverage
• Chapter 1-3, HW 1-5
• All examples in class and book
• Handout
• Last year’s Midterm 1
• Intended to show problem style (like HW)
• Please do not ask Hasib or me to work with you on them or show you how
they are done.
Likely Problem Types
• Mixtures and their properties, air/fuel mass ratio
• Heat of Combustion
• for a specified temperature, TProd = TReact
• Adiabatic Flame Temperature
• no dissociation
• constant Pressure or Volume
• Chemical Equilibrium
• for specified T and P
• Simple Combustion Equilibrium Products
• water/shift reaction
• Diffusion
•
•
•
•
•
Stefan (planar) or radial (droplet)
mass flow rates and mass fraction profiles
Boundary mass fraction
Dependence of diffusion coefficient on T and P
Average density, dependence on temperature, pressure and composition
Ideal Stoichiometric Hydrocarbon Combustion
air
• CxHy + a(O2+3.76N2) οƒ  (x)CO2 + (y/2) H2O + 3.76a N2
• a = number of oxygen molecules per fuel molecule
(no dissociation)
• Number of air molecules per fuel molecule is a(1+3.76)
• If a = aST = x + y/4, then the reaction is Stoichiometric
• No O2 or Fuel in products
• This mixture produces nearly the hottest flame temperature
• If a < x + y/4, then reaction is fuel-rich (oxygen-lean)
• If a > x + y/4, then reaction is fuel-lean (oxygen-rich)
• Air to fuel mass ratio [kg air/kg fuel] of reactants
•
•
𝐴
𝐹
=
π‘šπ΄π‘–π‘Ÿ
π‘šπΉπ‘’π‘’π‘™
𝐴
𝐹 𝑆𝑑
=
𝑁𝑂2
π‘π΄π‘–π‘Ÿ
𝑁𝑂
2
π‘€π‘Šπ΄π‘–π‘Ÿ
π‘π΄π‘–π‘Ÿ π‘€π‘Šπ΄π‘–π‘Ÿ
=
=
𝑁𝐹𝑒𝑒𝑙 π‘€π‘ŠπΉπ‘’π‘’π‘™
𝑁𝐹𝑒𝑒𝑙
π‘€π‘ŠπΉπ‘’π‘’π‘™
(1+3.76)π‘€π‘Šπ΄π‘–π‘Ÿ
π‘Žπ‘†π‘‘
> 1 (generally ~20)
1∗π‘€π‘ŠπΉπ‘’π‘’π‘™
=
• Need to find molecular weights
π‘Ž
(1+3.76)π‘€π‘Šπ΄π‘–π‘Ÿ
1∗π‘€π‘ŠπΉπ‘’π‘’π‘™
Equivalence Ratio Φ
•Φ=
𝐴
𝐴
𝐹 𝑆𝑑
=
𝐹 π΄π‘π‘‘π‘’π‘Žπ‘™
πΉπ΄π‘π‘‘π‘’π‘Žπ‘™ 𝐴𝑆𝑑
𝐹𝑆𝑑 π΄π΄π‘π‘‘π‘’π‘Žπ‘™
(compared to stoichiometric fuel)
• Φ = 1 → Stiochiometric
• Φ > 1 → Fuel Rich
• Φ < 1 → Fuel Lean
•π‘Ž=
π‘Žπ‘†π‘‘
Φ
𝑦
=
π‘₯+ 4
Φ
• CxHy + a(O2+3.76N2)
•%
•%
100%
Φ
𝐹𝑆𝑑
π΄π΄π‘π‘‘π‘’π‘Žπ‘™
Stoichiometric Air (%SA)=
=
∗ 100%
πΉπ΄π‘π‘‘π‘’π‘Žπ‘™ 𝐴𝑆𝑑
1
Excess Oxygen (%EO) = (%SA)-100% =
− 1 100%
Φ
Molecular Weight of a Pure Substance
x
• Only one type of molecule:
• AxByCz…
• MW = x(AWA) + y(AWB) + z(AWC) + …
• AWi = atomic weights
• Inside front cover of book
• Hint: put in front cover of book
• π‘€π‘Šπ‘‚2 = 2(AWO) = 2(15.9994) = 32.00
π‘˜π‘”
π‘˜π‘šπ‘œπ‘™
• π‘€π‘Šπ»2𝑂 = 2(AWH) + (AWO) = 2(1.00794) + (15.9994) =
• Fuels
• Bookmark page 701 for fuels
π‘˜π‘”
18.02
π‘˜π‘šπ‘œπ‘™
x
x
x xx
x
x
x
Mixtures containing n components
• Total number of moles in system
• π‘π‘‡π‘œπ‘‘π‘Žπ‘™ =
𝑛
𝑖=1 𝑁𝑖
• Mole Fraction of species i
• πœ’π‘– = 𝑁
𝑁𝑖
π‘‡π‘œπ‘‘π‘Žπ‘™
=
𝑁𝑖
𝑛
𝑖=1 π‘šπ‘–
• Mass Fraction of species i
• π‘Œπ‘– = π‘š
π‘šπ‘–
π‘‡π‘œπ‘‘π‘Žπ‘™
=
• π‘€π‘Šπ‘€π‘–π‘₯ =
𝑛
𝑖=1 𝑁𝑖
• π‘šπ‘– = 𝑁𝑖 π‘€π‘Šπ‘– = mass of species 𝑖
• Total Mass
• π‘š π‘‡π‘œπ‘‘π‘Žπ‘™ =
• Mixture Molar Weight: π‘€π‘Šπ‘€π‘–π‘₯ =
• π‘€π‘Šπ‘€π‘–π‘₯ =
π‘šπ‘–
𝑛
𝑖=1 π‘šπ‘–
• Useful facts:
• 𝑛𝑖=1 πœ’π‘– = 𝑛𝑖=1 π‘Œπ‘– = 1
• but πœ’π‘– ≠ π‘Œπ‘–
π‘šπ‘–
𝑁𝑖
π‘šπ‘–
𝑁𝑖
=
=
x
x xx
x
x
x
o o
o x
x
• 𝑁𝑖 = number of moles of species 𝑖
• 𝑖 = 1, 2, . . 𝑛
o
𝑁𝑖 π‘€π‘Šπ‘–
=
π‘π‘‡π‘œπ‘‘π‘Žπ‘™
π‘šπ‘‡π‘œπ‘‘π‘Žπ‘™
=
π‘šπ‘– /π‘€π‘Šπ‘–
π‘šπ‘‡π‘œπ‘‘π‘Žπ‘™
π‘π‘‡π‘œπ‘‘π‘Žπ‘™
πœ’π‘– π‘€π‘Šπ‘–
1
π‘Œπ‘– /π‘€π‘Šπ‘–
• Hint: Put inside front cover of book
• π‘€π‘Šπ΄π‘–π‘Ÿ =
π‘˜π‘”
πœ’π‘– π‘€π‘Šπ‘– = 0.21π‘€π‘Šπ‘‚2 + 0.79π‘€π‘Šπ‘2 = 28.85 π‘˜π‘šπ‘œπ‘™π‘’
• Relationship between πœ’π‘– and π‘Œπ‘–
• π‘Œπ‘– = π‘š
• πœ’π‘– =
π‘šπ‘–
π‘‡π‘œπ‘‘π‘Žπ‘™
=𝑁
π‘€π‘Š
π‘Œπ‘– π‘€π‘Šπ‘€π‘–π‘₯
𝑖
𝑁𝑖 π‘€π‘Šπ‘–
π‘‡π‘œπ‘‘π‘Žπ‘™ π‘€π‘Šπ‘€π‘–π‘₯
π‘€π‘Šπ‘–
= πœ’π‘– π‘€π‘Š
𝑀𝑖π‘₯
Ideal Gas Equation of State
• 𝑃𝑉 = π‘π‘…π‘ˆ 𝑇
• Universal Gas Constant
• π‘…π‘ˆ = 8.315
8315
π‘˜π½
π‘˜π‘šπ‘œπ‘™π‘’ 𝐾
= 8.315
𝐽
π‘˜π‘šπ‘œπ‘™π‘’ 𝐾
π‘˜π‘ƒπ‘Ž π‘š3
π‘˜π‘šπ‘œπ‘™π‘’ 𝐾
• Number of molecules
=
• Inside book front cover
• kJ = kPa*m3
• 𝑃𝑉 = π‘šπ‘…π‘‡ = 𝑁 ∗ π‘€π‘Š (π‘…π‘ˆ /π‘€π‘Š)𝑇
• Specific Gas Constant
• R =π‘…π‘ˆ /π‘€π‘Š
• MW = Molecular Weight of that gas
• 𝑃𝑣 = 𝑅𝑇; 𝑣 =
• 𝑃 = πœŒπ‘…π‘‡
𝑉
π‘š
=
1
𝜌
• N*NAV
• Avogadro's Number, 𝑁𝐴𝑉
π‘šπ‘œπ‘™π‘’π‘π‘’π‘™π‘’π‘ 
π‘˜π‘šπ‘œπ‘™π‘’
π‘šπ‘œπ‘™π‘’π‘π‘’π‘™π‘’π‘ 
1023
π‘šπ‘œπ‘™π‘’
• 6.022 ∗ 1026
• 6.022 ∗
Thermodynamic Systems (reactors)
1π‘Š2
1𝑄2
Inlet i
π‘šπ‘– 𝑒 + 𝑃𝑣
Outlet o
𝑖
m, E
𝑄𝐢𝑉
• Closed systems
•
1𝑄2
Dm=DE=0
− 1π‘Š2 = π‘š 𝑒2 − 𝑒1 +
𝑣22
2
−
𝑣12
2
π‘š0 𝑒 + 𝑃𝑣
π‘œ
π‘ŠπΆπ‘‰
+ 𝑔 𝑧2 − 𝑧1
• Open Steady State, Steady Flow (SSSF) Systems
• 𝑄𝐢𝑉 − π‘ŠπΆπ‘‰ = π‘š β„Žπ‘œ − β„Žπ‘– +
π‘£π‘œ2
2
−
𝑣𝑖2
2
+ 𝑔 π‘§π‘œ − 𝑧𝑖
• How to find changes, 𝑒2 − 𝑒1 and β„Žπ‘œ − β„Žπ‘– , for mixture when temperatures and
composition change due to reactions (not covered in Thermodynamics I)
For a pure substances
• Mass and Molar ( ) Bases
• π‘ˆ = π‘šπ‘’ = 𝑁𝑒
• 𝐻 = π‘šβ„Ž = π‘β„Ž
•
• N number of moles in the system
𝑇
𝑐𝑝 (𝑇) and β„Ž 𝑇 = β„Žπ‘Ÿπ‘’π‘“ + 𝑇 𝑐𝑝
π‘Ÿπ‘’π‘“
𝑇 𝑑𝑇
• Appendix A, pp. 687-699, for combustion gases
• bookmark
• 𝑐𝑣 = 𝑐𝑝 − 𝑅𝑒
• 𝑒 𝑇 = π‘’π‘Ÿπ‘’π‘“ +
𝑇
𝑐
π‘‡π‘Ÿπ‘’π‘“ 𝑣
𝑇 𝑑𝑇
• 𝑐𝑝 =𝑐𝑝 /π‘€π‘Š; 𝑐𝑣 =𝑐𝑣 /π‘€π‘Š
• For mixtures
• β„Žπ‘šπ‘–π‘₯ (𝑇) =
• π‘’π‘šπ‘–π‘₯ 𝑇 =
πœ’π‘– β„Žπ‘– (𝑇), β„Žπ‘šπ‘–π‘₯ (𝑇) =
πœ’π‘– 𝑒𝑖 𝑇 , π‘’π‘šπ‘–π‘₯ (𝑇) =
π‘Œπ‘– β„Žπ‘– (𝑇)
π‘Œπ‘– 𝑒𝑖 (𝑇)
Standardized Enthalpy and Enthalpy of Formation
• Needed to find 𝑒2 − 𝑒1 and β„Žπ‘œ − β„Žπ‘– for chemically-reacting matter because
energy is required to form and break chemical bonds
π‘œ
• β„Žπ‘– 𝑇 = β„Žπ‘“,𝑖
π‘‡π‘Ÿπ‘’π‘“ + Δβ„Žπ‘ ,𝑖 (𝑇)
• Standard Enthalpy of substance 𝑖 at Temperature T =
• Enthalpy of formation from “normally occurring elemental compounds,” at standard
reference state: Tref = 298 K and P° = 1 atm
• Sensible enthalpy change in going from Tref to T =
𝑇
𝑐
π‘‡π‘Ÿπ‘’π‘“ 𝑝
𝑇 𝑑𝑇
• Normally-Occurring Elemental Compounds
• Examples: O2, N2, C, He, H2
π‘œ
• Their enthalpy of formation at π‘‡π‘Ÿπ‘’π‘“ = 298 K are defined to be β„Žπ‘“,𝑖
π‘‡π‘Ÿπ‘’π‘“ = 0
• Use these compounds as bases to tabulate the energy to form other compounds
Mixture Example: Stoichiometric Acetylene Combustion
• C2H2 + 2.5 (O2 + 3.76 N2) οƒ  2 CO2 + 1 H2O + 9.4N2
• Reactant standard enthalpy
• 𝐻𝑅 = 1 β„Žπ‘“π‘œ + βˆ†β„Žπ‘ 
𝐢2 𝐻2
+ 2.5 β„Žπ‘“π‘œ + βˆ†β„Žπ‘ 
• At 𝑇 = 𝑇𝑅 = 298𝐾, βˆ†β„Žπ‘– = 0
π‘œ
• 𝐻𝑅 = β„Žπ‘“,𝐢
= 226,748
2 𝐻2
π‘˜π½
π‘˜π‘šπ‘œπ‘™πΉπ‘’π‘’π‘™
𝑂2
+ 9.4 β„Žπ‘“π‘œ + βˆ†β„Žπ‘ 
𝑁2
=
π‘œ
𝑁
β„Ž
𝑅 𝑖 𝑓,𝑖
(page 701, >0, Add energy to Standard Elemental Compounds)
• Energy that must be added to 2C + H2 + 2.5 O2 + 9.4 N2 at 𝑇 = 𝑇𝑅 = 298𝐾 to form Reactants
• Product standard enthalpy
• 𝐻𝑃 = 2 β„Žπ‘“π‘œ + βˆ†β„Žπ‘ 
𝐢𝑂2
+ 1 β„Žπ‘“π‘œ + βˆ†β„Žπ‘ 
• At 𝑇 = 𝑇𝑅 = 298𝐾, βˆ†β„Žπ‘– = 0
𝐻2 𝑂
+ 9.4 β„Žπ‘“π‘œ + βˆ†β„Žπ‘ 
π‘˜π½
𝑁2
=
π‘œ
𝑃 𝑁𝑖 β„Žπ‘“,𝑖
π‘˜π½
π‘˜π½
π‘œ
π‘œ
• 𝐻𝑃 = 2β„Žπ‘“,𝐢𝑂
+
β„Ž
= 2 −393,546 π‘˜π‘šπ‘œπ‘™ + −241,845 π‘˜π‘šπ‘œπ‘™ = −1,028,937 π‘˜π‘šπ‘œπ‘™
𝑓,𝐻
2
2𝑂
• Energy added to 2C + H2 + 2.5 O2 + 9.4 N2 at 𝑇 = 𝑇𝑅 = 298𝐾 to form Products
𝐹𝑒𝑒𝑙
• Which one has more energy? Reactants or Products?
• Constant pressure heat of combustion at 𝑇 = 𝑇𝑅 = 298𝐾
• 𝐻𝑐 =
π‘œ
𝑁
β„Ž
𝑖
𝑅
𝑓
𝑖
−
π‘œ
𝑁
β„Ž
𝑖
𝑃
𝑓
𝑖
= 1,255,685
π‘˜π½
π‘˜π‘šπ‘œπ‘™πΉπ‘’π‘’π‘™
Energy is released (Exothermic)
Enthalpy of Combustion (or reaction)
Products
Complete Combustion
CCO2 HH2O
298.15 K, 1 atm
Reactants
298.15 K, P = 1 atm
Stoichiometric
𝑄𝐼𝑁 < 0
π‘Šπ‘‚π‘ˆπ‘‡ = 0
• How much energy is released π‘„π‘œπ‘’π‘‘ from a
reaction if the product and reactant
temperatures and pressures are the same?
• 1st Law, Steady Flow Reactor
• 𝑄𝐼𝑁 − π‘Šπ‘‚π‘ˆπ‘‡ = 𝐻𝑃 − 𝐻𝑅 = π‘š β„Žπ‘ƒ − β„Žπ‘… = π‘šβˆ†β„Žπ‘…
• βˆ†π»π‘… and βˆ†β„Žπ‘… Enthalpy of Reaction (< 0 for combustion)
• Dependent on T and P of reaction
• Heat of Combustion βˆ†β„ŽπΆ = −βˆ†β„Žπ‘… = β„Žπ‘… − β„Žπ‘ƒ > 0
• For exothermic reactions
Example: Stoichiometric Methane Combustion
• CH4 + 2 (O2 + 3.76 N2) οƒ  1 CO2 + 2 H2O + 7.52 N2
• Heat of reaction into system for TR = TP = 25°C and 1 kmol CH4
• βˆ†π»π‘… = 𝐻𝑃 − 𝐻𝑅 =
Water Vapor
1 β„Žπ‘“π‘œ + βˆ†β„Žπ‘ 
+ 2 β„Žπ‘“π‘œ + βˆ†β„Žπ‘ 
+ 7.52 β„Žπ‘“π‘œ + βˆ†β„Žπ‘ 
𝐢𝑂2
− 1
π‘œ
β„Žπ‘“
+ βˆ†β„Žπ‘ 
𝐢𝐻4
𝐻2 𝑂
+ 2
π‘œ
β„Žπ‘“
+ βˆ†β„Žπ‘ 
π‘œ
π‘œ
π‘œ
= β„Žπ‘“,𝐢𝑂
+
2
β„Ž
−
1
β„Ž
=
𝑓,𝐻
𝑂
𝑓,𝐢𝐻
2
2
4
=
π‘˜π½
−393,546
π‘˜π‘šπ‘œπ‘™
p 688
= −802,405
+2
π‘˜π½
π‘˜π‘šπ‘œπ‘™πΉπ‘’π‘’π‘™
𝑂2
𝑁2
+ 7.52
π‘œ
𝑁𝑖 β„Žπ‘“,𝑖
π‘˜π½
−241,845
π‘˜π‘šπ‘œπ‘™
p 692
(< 0, exothermic)
π‘ƒπ‘Ÿπ‘œπ‘‘
−1
π‘œ
β„Žπ‘“
−
+ βˆ†β„Žπ‘ 
𝑁2
π‘œ
𝑁𝑖 β„Žπ‘“,𝑖
π‘…π‘’π‘Žπ‘π‘‘
π‘˜π½
−74,831
π‘˜π‘šπ‘œπ‘™
p 701
Other Bases
• Per kg fuel
• π‘€π‘ŠπΆπ»4 = 16.043
• βˆ†β„Žπ‘… = −
π‘˜π‘”
π‘˜π‘šπ‘œπ‘™
π‘˜π½
π‘˜π‘šπ‘œπ‘™πΉπ‘’π‘’π‘™
π‘˜π‘”
16.043
π‘˜π‘šπ‘œπ‘™
802,405
• Heat of Combustion
• βˆ†β„Žπ‘ = −βˆ†β„Žπ‘… = 50,016
= −50,016
π‘˜π½
π‘˜π‘”πΉπ‘’π‘’π‘™
π‘˜π½
π‘˜π‘”πΉπ‘’π‘’π‘™
(Heat out for TR = TP)
π‘˜π½
• See page 701, LHV = Lower Heating Value = 50,016
π‘˜π‘”πΉπ‘’π‘’π‘™
• Corresponds to water vapor in the products
π‘œ
π‘œ
π‘œ
• βˆ†π»π‘…,πΏπ‘œπ‘€π‘’π‘Ÿ = β„Žπ‘“,𝐢𝑂
+
2
β„Ž
−
1
β„Ž
𝑓,𝐻2 𝑂,π‘£π‘Žπ‘π‘œπ‘Ÿ
𝑓,𝐢𝐻4
2
•
π‘œ
β„Žπ‘“,𝐻
2 𝑂,πΏπ‘–π‘žπ‘’π‘–π‘‘
=
π‘œ
β„Žπ‘“,𝐻
2 𝑂,π‘£π‘Žπ‘π‘œπ‘Ÿ
− β„Žπ»2𝑂,𝑓𝑔 =
π‘˜π½
−241,845
π‘˜π‘šπ‘œπ‘™
p 692
−
π‘˜π½
44,010
π‘˜π‘šπ‘œπ‘™
p 692
=
π‘˜π½
−285,855
π‘˜π‘šπ‘œπ‘™
π‘œ
π‘œ
π‘œ
• βˆ†π»π‘…,π»π‘–π‘”β„Žπ‘’π‘Ÿ = β„Žπ‘“,𝐢𝑂
+
2
β„Ž
−
1
β„Ž
𝑓,𝐻
𝑂,πΏπ‘–π‘žπ‘’π‘–π‘‘
𝑓,𝐢𝐻
2
2
4
• = −393,546 + 2 −241,845 − 1 −74,831 = −890,425
π‘˜π½
• βˆ†β„ŽπΆ = −
−890,425π‘˜π‘šπ‘œπ‘™
𝐹𝑒𝑒𝑙
π‘˜π‘”
16.043π‘˜π‘šπ‘œπ‘™
= 55,502
π‘˜π½
π‘˜π‘”πΉπ‘’π‘’π‘™
• p. 701: Higher Heating Value = HHV = 55,528
π‘˜π½
π‘˜π‘”πΉπ‘’π‘’π‘™
π‘˜π½
π‘˜π‘”πΉπ‘’π‘’π‘™
(slightly larger than book, not due to dissociation since temp is low)
Per kg of reactant mixture
•
π‘šπΉπ‘’π‘’π‘™
π‘šπ‘€π‘–π‘₯
•
𝐴
𝐹
=
=
π‘šπΉπ‘’π‘’π‘™
π‘šπΉπ‘’π‘’π‘™ +π‘šπ΄π‘–π‘Ÿ
π‘π΄π‘–π‘Ÿ π‘€π‘Šπ΄π‘–π‘Ÿ
𝑁𝐹𝑒𝑒𝑙 π‘€π‘ŠπΉπ‘’π‘’π‘™
=
• LHV = βˆ†β„Žπ‘,πΏπ‘œπ‘€π‘’π‘Ÿ =
=
1
π‘šπ΄π‘–π‘Ÿ
1+π‘š
𝐹𝑒𝑒𝑙
=
1
𝐴
1+𝐹
=
1
1+17.12
=
1 π‘˜π‘”πΉπ‘’π‘’π‘™
18.12 π‘˜π‘”π‘€π‘–π‘₯
2∗ 3.76+1 ∗28.85
π‘˜π‘”π΄π‘–π‘Ÿ
= 17.12
1∗16.043
π‘˜π‘”πΉπ‘’π‘’π‘™
π‘˜π½
1 π‘˜π‘”πΉπ‘’π‘’π‘™
π‘˜π½
50,016
∗
= 2760
π‘˜π‘”πΉπ‘’π‘’π‘™ 18.12 π‘˜π‘”π‘€π‘–π‘₯
π‘˜π‘”π‘€π‘–π‘₯
Adiabatic (𝑄 = 0) Flame Temperature, 𝑇𝐴𝑑
Complete Combustion Products
CCO2 HH2O
PP = PR, T = TAd
Stoichiometric
Reactants
TR PR
𝑄𝐼𝑁 = 0
π‘Šπ‘‚π‘ˆπ‘‡ = 0
• 1st Law, Steady Flow Reactor
• 0 = 𝐻𝑃 − 𝐻𝑅 = π‘š β„Žπ‘ƒ − β„Žπ‘…
• All chemical energy goes into heating the
products
• To find adiabatic flame temperature use
• PP = PR and β„Žπ‘ƒ = β„Žπ‘…
• To evaluate using specific heats
• 𝑇𝐴𝑑,𝑃 − 𝑇𝑅 =
𝐻𝑐
𝑃 𝑁𝑖 𝑐𝑝,𝑖,π‘Žπ‘£π‘”
• 𝑇𝐴𝑑 will be lower if we include dissociation
Constant Volume Adiabatic Flame Temperature
• 𝑉𝑃 = 𝑉𝑅 = 𝑉
• π‘ˆπ‘… = π‘ˆπ‘ƒ
𝑄=0
V, m
π‘Š=0
• Use definition: π‘ˆ = 𝐻 − 𝑃𝑉 (since standard internal energy U is not tabulated)
• 𝐻𝑅 − 𝑃𝑅 𝑉 = 𝐻𝑃 − 𝑃𝑃 𝑉
• Idea gas: 𝑃𝑅 𝑉 = 𝑁𝑅 𝑅𝑒 𝑇𝑅 ; 𝑃𝑃 𝑉 = 𝑁𝑃 𝑅𝑒 𝑇𝑃 = 𝑁𝑃 𝑅𝑒 𝑇𝐴𝑑,𝑣 ,
• 𝐻𝑅 𝑇𝑅 − 𝑁𝑅 𝑅𝑒 𝑇𝑅 = 𝐻𝑃 𝑇𝐴𝑑,𝑣 − 𝑁𝑃 𝑅𝑒 𝑇𝐴𝑑.𝑣
• Only 𝑇𝐴𝑑,𝑣 and 𝐻𝑅 𝑇𝐴𝑑,𝑣 are unknown
• To evaluate using specific heats: 𝑇𝐴𝑑,𝑉 − 𝑇𝑅 =
• 𝑇𝑅 = 298K
𝐻𝑐 + 𝑁𝑃 −𝑁𝑅 𝑅𝑒 𝑇𝑅
𝑃 𝑁𝑖 𝑐𝑝,𝑖,π‘Žπ‘£π‘” −𝑁𝑃 𝑅𝑒
Find equilibrium composition (reactants and products)
for a given Temperature, Pressure & Mass
• aA + bB + … οƒ  eE + fF + …
• Use Q and boundary work π‘Š =
• 𝐾𝑃 = exp
π‘Š=
𝑃𝑑𝑉
Q
𝑃𝑑𝑉 to achieve P and T
T,P
−ΔπΊπ‘‡π‘œ
𝑅𝑒 𝑇
• Equilibrium𝑒 Constant
𝑓
𝑃𝐸
𝑃𝐹
…
π‘œ
𝑃
π‘ƒπ‘œ
𝑃𝐴 π‘Ž 𝑃𝐡 𝑏
…
π‘ƒπ‘œ
π‘ƒπ‘œ
since 𝑃𝑖 = πœ’π‘– 𝑃
• 𝐾𝑃 =
•
=
πœ’πΈ 𝑒 πœ’πΉ 𝑓 … 𝑃 𝑒+𝑓−π‘Ž−𝑏…
πœ’ 𝐴 π‘Ž πœ’π΅ 𝑏 … 𝑃 π‘œ
=
π‘ƒπ‘Ÿπ‘œπ‘‘π‘’π‘π‘‘π‘ 
π‘…π‘’π‘Žπ‘π‘‘π‘Žπ‘›π‘‘π‘ 
,
• Standard State Gibbs Function Change
π‘œ
π‘œ
π‘œ
π‘œ
• ΔπΊπ‘‡π‘œ = 𝑒𝑔𝑓,𝐸
+ 𝑓𝑔𝑓,𝐹
+ β‹― − π‘Žπ‘”π‘“,𝐴
− 𝑏𝑔𝑓,𝐡
− β‹― = π‘ƒπ‘Ÿπ‘œπ‘‘π‘’π‘π‘‘π‘  − π‘…π‘’π‘Žπ‘π‘‘π‘Žπ‘›π‘‘π‘  = 𝑓𝑛(𝑇)
π‘œ
• In terms of Gibbs functions of formation 𝑔𝑓,𝑖
= 𝑓𝑛(𝑇) (tabulated in App. A and B)
• Products are “favored”
• As 𝐾𝑃 increases, which happens when ΔπΊπ‘‡π‘œ decreases
• If N > N , as P decreases
Number of unknowns is the number of species in
equilibrium
• May need additional Equations
• Simple reactions of few species have fewer unknowns
• πœ’π‘– = 1
• Use atomic balances if necessary
• May need to solve system of equations
• May be quadratic, or learn to use calculator to solve non-linear
Equilibrium Products of Combustion
• Combine Chemical Equilibrium (2nd law) & Adiabatic Flame
Temperature (1st law)
• For Example: Propane and air combustion
• Ideal Stoichiomectric
• 𝐢3 𝐻8 + 5 𝑂2 + 3.76𝑁2 → 3𝐢𝑂2 + 4𝐻2 𝑂 + 18.8𝑁2 + (0)𝑂2
• Four products for a range of air/fuel ratios: 𝐢𝑂2 , 𝐻2 𝑂, 𝑁2 , 𝑂2
• Now consider seven more possible dissociation products:
• 𝐢𝑂, 𝐻2 , 𝐻, 𝑂𝐻, 𝑂, 𝑁𝑂, 𝑁
minor
• What happens as air/fuel (equivalence) ratio changes
•Φ =
𝐴/𝐹 π‘†π‘‘π‘œπ‘–π‘β„Žπ‘–π‘œ
𝐴/𝐹 π΄π‘π‘‘π‘’π‘Žπ‘™
=
𝐹 𝐴 π΄π‘π‘‘π‘’π‘Žπ‘™
𝐹 𝐴 π‘†π‘‘π‘œπ‘–π‘β„Žπ‘–π‘œ
Simple Product Calculation method
• 𝐢π‘₯ 𝐻𝑦 + π‘Ž 𝑂2 + 3.76𝑁2 → 𝑏𝐢𝑂2 + 𝑐𝐢𝑂 + 𝑑𝐻2 𝑂 + 𝑒𝐻2 + 𝑓𝑂2 + (3.76π‘Ž)𝑁2
• No minor species
•π‘Ž=
π‘šπ‘œπ‘™π‘’π‘  π΄π‘–π‘Ÿ
π‘šπ‘œπ‘™π‘’π‘  𝐹𝑒𝑒𝑙
• Φ=
=
𝐴/𝐹 π‘†π‘‘π‘œπ‘–π‘β„Žπ‘–π‘œ
𝐴/𝐹 π΄π‘π‘‘π‘’π‘Žπ‘™
π‘₯+𝑦 4
Φ
• Assume π‘₯, 𝑦 and Φ are known
• What is a good assumption for lean or stoichiometric mixtures Φ ≤ 1?
• 𝐢π‘₯ 𝐻𝑦 + π‘Ž 𝑂2 + 3.76𝑁2 → 𝑏𝐢𝑂2 + 𝑑𝐻2 𝑂 + 𝑓𝑂2 + (3.76π‘Ž)𝑁2
• c = e = 0 (no CO or H2), but now include 𝑂2
• Mole Fractions
• πœ’πΆπ‘‚2 =
π‘₯
π‘π‘‡π‘œπ‘‘
• π‘π‘‡π‘œπ‘‘ = π‘₯
𝑦
; πœ’π»2𝑂 =
𝑦
+
2
𝑦
+
𝑦
2
π‘π‘‡π‘œπ‘‘
π‘₯+ 4
Φ
; πœ’π‘‚2 =
π‘₯+ 4
1 − Φ + 3.76
1−Φ
π‘π‘‡π‘œπ‘‘
Φ
π‘₯+
; πœ’π‘2 = 3.76
𝑦
4
π‘π‘‡π‘œπ‘‘
Φ
For Rich combustion Φ > 1
• 𝐢π‘₯ 𝐻𝑦 + π‘Ž 𝑂2 + 3.76𝑁2 → 𝑏𝐢𝑂2 + 𝑐𝐢𝑂 + 𝑑𝐻2 𝑂 + 𝑒𝐻2 + (3.76π‘Ž)𝑁2
• 𝑓 = 0; no 𝑂2 (or fuel)
1.0
0.9
• 4 unknowns: b, c, d and e
• 3 Atom balances: C, H, O
0.8
0.7
Kp
0.6
0.4
• Need one more constraint
0.3
0.2
• Consider “Water-Gas Shift Reaction” equilibrium
• 𝐢𝑂 + 𝐻2 𝑂 ↔ 𝐢𝑂2 + 𝐻2
• 𝐾𝑃 =
𝑃𝐢𝑂 1 𝑃𝐻 1
2
2
π‘œ
π‘œ
𝑃
𝑃
1
𝑃𝐢𝑂 1 𝑃𝐻2 𝑂
π‘ƒπ‘œ
π‘ƒπ‘œ
1
=
0.5
𝑏
1
πœ’πΆπ‘‚2
πœ’π»2
𝑃 0
1
π‘ƒπ‘œ
πœ’πΆπ‘‚ 1 πœ’π»2 𝑂
=
π‘π‘‡π‘œπ‘‘
𝑐
π‘π‘‡π‘œπ‘‘
1
1
0.1
0.0
1000
𝑒
π‘π‘‡π‘œπ‘‘
𝑑
1500
2000
3000
T [K]
1
1
2500
=
𝑏𝑒
𝑐𝑑
= 𝐾𝑃
π‘π‘‡π‘œπ‘‘
• Not dependent on P since number of moles of products and reactants are the same
•
ΔπΊπ‘‡π‘œ
=
π‘œ
1𝑔𝑓,𝐢𝑂
2
π‘œ
+ 1𝑔𝑓,𝐻
2
−
π‘œ
1𝑔𝑓,𝐢𝑂
π‘œ
− 1𝑔𝑓,𝐻
2𝑂
= 𝑓𝑛 π‘‡π‘’π‘šπ‘π‘’π‘Ÿπ‘Žπ‘‘π‘’π‘Ÿπ‘’ ; 𝐾𝑃 = exp
−ΔπΊπ‘‡π‘œ
𝑅𝑒 𝑇
• See plot from data on page 51
• KP = 0.22 to 0.1635 for T = 2000 to 3500 K (on a test you may need to evaluate at temperature T)
3500
Solution
•0=
•π‘=
𝑏2
𝐾𝑃 − 1 + b 2π‘Ž 1 − 𝐾𝑃 − π‘₯ −
− 2π‘Ž
𝑦
1−𝐾𝑃 −π‘₯− 2
−
2π‘Ž
𝑦 2
1−𝐾𝑃 −π‘₯− 2 −4
𝑦
2
+ 𝐾𝑃 π‘₯ 2π‘Ž − π‘₯
𝐾𝑃 −1 𝐾𝑃 π‘₯ 2π‘Ž−π‘₯
2 𝐾𝑃 −1
• Since 𝐾𝑃 − 1 < 0, use “-” root
• 𝑐 =π‘₯−𝑏
• 𝑑 = 2π‘Ž − 𝑏 − π‘₯
𝑦
2
• 𝑒 = − 2π‘Ž + 𝑏 + π‘₯
• Products: 𝑏𝐢𝑂2 + 𝑐𝐢𝑂 + 𝑑𝐻2 𝑂 + 𝑒𝐻2 + (3.76π‘Ž)𝑁2
• π‘π‘‡π‘œπ‘‘ = 𝑏 + 𝑐 + 𝑑 + 𝑒 + 3.76π‘Ž
• Mole Fractions
• πœ’πΆπ‘‚2 =
𝑏
π‘π‘‡π‘œπ‘‘
; πœ’πΆπ‘‚ =
𝑐
;
π‘π‘‡π‘œπ‘‘
πœ’π»2𝑂 =
𝑑
π‘π‘‡π‘œπ‘‘
; πœ’π»2 =
𝑒
;
π‘π‘‡π‘œπ‘‘
πœ’π‘2 = 3.76
π‘Ž
π‘π‘‡π‘œπ‘‘
Stefan Problem (no reaction)
x
L-
• One dimensional tube (Cartesian)
π‘Œπ΄,∞
• Gas B is stationary: π‘šπ΅" = 0
• Gas A moves upward π‘šπ΄" > 0
YB
• Want to find this
YA
Y
π‘Œπ΄,𝑖
B+A
•
π‘šπ΄"
=
π‘Œπ΄ π‘šπ΄"
+
π‘šπ΅"
+
π‘‘π‘Œπ΄
−πœŒπ’Ÿπ΄π΅
𝑑π‘₯
• πœŒπ’Ÿπ΄π΅ = 𝑓𝑛 π‘₯ but treat as constant
A
Liquid-Vapor Interface Boundary Condition
• At interface need π‘Œπ΄,𝑖 =
A+B
Vapor
• πœ’π΄,𝑖 =
𝑃𝐴,𝑖
π‘ƒπ‘‡π‘œπ‘‘π‘Žπ‘™
=
1
1
π‘€π‘Š
1+ πœ’ −1 π‘€π‘Šπ΅
𝐴
𝐴,𝑖
𝑃𝐴,π‘†π‘Žπ‘‘
𝑃
π‘Œπ΄,𝑖
• 𝑃𝐴,π‘†π‘Žπ‘‘ = 𝑓𝑛(𝑇) Saturation pressure at temperature T
Liquid
A
• For water, tables in thermodynamics textbook
• Or use Clausius-Slapeyron Equation (page 18 eqn. 2.19)
• πœ’π΄,𝑖 =
𝑃𝐴,π‘†π‘Žπ‘‘
π‘ƒπ‘‡π‘œπ‘‘π‘Žπ‘™
=
π‘ƒπ΅π‘œπ‘–π‘™
𝑒π‘₯𝑝
π‘ƒπ‘‡π‘œπ‘‘π‘Žπ‘™
β„Žπ‘“π‘”
𝑅
1
1
−
π‘‡π΅π‘œπ‘–π‘™
𝑇
• If given π‘‡π΅π‘œπ‘–π‘™ , π‘ƒπ΅π‘œπ‘–π‘™ , β„Žπ‘“π‘” π‘Žπ‘›π‘‘ 𝑇, we can use this to find π‘ƒπ‘†π‘Žπ‘‘
• Page 701, Table B: β„Žπ‘“π‘” , π‘‡π΅π‘œπ‘–π‘™ at π‘ƒπ΅π‘œπ‘–π‘™ = 1 π‘Žπ‘‘π‘š
Mass Flux and fraction of evaporating liquid A
For π‘Œπ΄,∞ = 0
• π‘šπ΄" =
1−π‘Œπ΄,∞
πœŒπ’Ÿπ΄π΅
ln
𝐿
1−π‘Œπ΄,𝑖
π‘Œπ΄ π‘₯ = 1 − 1 − π‘Œπ΄,𝑖
1
1 − π‘Œπ΄,𝑖
π‘₯
𝐿
8
6.908
1
0.99
π‘Œπ΄,𝑖 =0.99
6
π‘Œπ΄,𝑖 =0.9
0.8
π‘šπ΄"
πœŒπ’Ÿπ΄π΅m(Y) 4
𝐿
YA ( x .05)
YA ( x .1)
0.6
π‘Œπ΄YAπ‘₯( x .5)
π‘Œπ΄,𝑖 =0.5
YA ( x .9)
YA ( x .99)
2
0.4
0.2
0
0
0
0
0.2
0.4
0.6
Y
π‘Œπ΄,𝑖
π‘Œπ΄,𝑖 =0.1
π‘Œπ΄,𝑖 =0.05
0.8
1
0
0
0
0
0.2
0.4
π‘₯x
0.6
0.8
1
Possible Questions
• How long will it take for column to recede by given (measurable)
amount?
• For a given π‘šπ΄" , temperature, pressure and dimension, find π‘Œπ΄,∞
• Find variation with temperature
Spherical Droplet Evaporation
• A is evaporating, find π‘šπ΄
• B is stagnant π‘šπ΅" = 0
π‘Ÿπ‘  =
•
𝐷
2
π‘Œπ΄,∞
π‘šπ΄"
=
π‘šπ΄
𝐴
=
π‘Œπ΄ π‘šπ΄"
+
π‘˜π‘”
𝑠
and π‘Œπ΄ (π‘Ÿ)
+
π‘‘π‘Œπ΄
−πœŒπ’Ÿπ΄π΅
π‘‘π‘Ÿ
π‘šπ΅"
• π‘šπ΄ = 4πœ‹π‘Ÿπ‘  πœŒπ’Ÿπ΄π΅ ln 1 + π΅π‘Œ
• π΅π‘Œ =
4
3.932
π‘Œπ΄,𝑠 −π‘Œπ΄,∞
3
π‘šπ΄
4πœ‹π‘Ÿπ‘  πœŒπ’Ÿπ΄π΅
1−π‘Œπ΄,𝑠
m1 ( By) 2
• If π΅π‘Œ and πœŒπ’Ÿπ΄π΅ do not change with time
• Then π‘šπ΄ decreases as π‘Ÿπ‘  =
π‘Œπ΄,𝑠
• π‘Œπ΄ = 1 − 1 − π‘Œπ΄,𝑠
π‘Œπ΄,∞
𝐷
2
1−π‘Œπ΄,∞
decreases
0
0
0
0
10
20
30
40
50
By
( 1 ο€­ YAs)
YA1 ( r YAs) ο€Ίο€½ 1 ο€­
 1ο€­ 1 οƒΆ
 rοƒ·
οƒΈ
( 1 ο€­ YAs) 
π‘Ÿ
1− π‘Ÿπ‘ 
50
π΅π‘Œ
1
0.8
YA1 ( r .05)
1−π‘Œπ΄,𝑠
• For π‘Œπ΄,∞ = 0: π‘Œπ΄ = 1 −
1
YA1 ( r .1) 0.6
YA1 ( r .5)
1−π‘Œπ΄,𝑠
1−π‘Œπ΄,𝑠
π‘Ÿ
1− π‘Ÿπ‘ 
YA1 ( r .9)
YA1 ( r .95) 0.4
0.2
ο€­3
4.652ο‚΄10
0
2
1
4
6
r
8
10
11
Droplet Diameter 𝐷 = 2π‘Ÿπ‘  versus time 𝑑
• Mass Conservation
•
•
π‘‘π‘šπ·π‘Ÿπ‘œπ‘
𝑑𝐷2
𝑑𝑑
𝑑𝑑
= −π‘š
= −𝐾,
• 𝐾 = 8πœ‹
• π΅π‘Œ =
𝜌
π’Ÿ
πœŒπ‘™ 𝐴𝐡
ln 1 + π΅π‘Œ Evaporation Const.
π‘Œπ΄,𝑠 −π‘Œπ΄,∞
1−π‘Œπ΄,𝑠
• Constant slope for 𝐷2 versus 𝑑
• Confirmed by experiment
• Droplet life 𝑑𝐷 =
𝐷02
𝐾
Dependence of π’Ÿπ΄π΅ on Temperature and Pressure
•
•
1 2
3
2 π‘˜π΅
𝑇
𝑇
3 2 𝑃 −1
π’Ÿπ΄π΅ =
~𝑇
3 πœ‹ 3 π‘šπ΄
𝜎2 𝑃
π‘€π‘Š∗𝑃
πœŒπ’Ÿπ΄π΅ =
π’Ÿπ΄π΅ ~𝑇 1 2
𝑅𝑒 𝑇
• Fairly independent of T and P
Extra Slides
Flame temperature and major mole-fractions vs Φ
• Equivalence Ratio Φ =
Tad[K]
πœ’π‘–
%
• At Φ = 1, O2, CO, H2 all present due to
dissociation. Not present in “ideal”
combustion
• πœ’π»2𝑂
“Old”
• 𝑇𝐴𝑑
π‘€π‘Žπ‘₯
π‘€π‘Žπ‘₯
at Φ = 1.15
at Φ = 1.05
• 𝑇𝐴𝑑 − 𝑇𝑅𝑒𝑓 =
“New”
Fuel Lean
O2
Fuel Rich
Get CO, H2
𝐹 𝐴 π΄π‘π‘‘π‘’π‘Žπ‘™
𝐹 𝐴 π‘†π‘‘π‘œπ‘–π‘β„Žπ‘–π‘œ
𝐻𝐢
π‘π‘‡π‘œπ‘‘ 𝑐𝑝,π‘šπ‘–π‘₯
• 𝐻𝐢 and π‘π‘‡π‘œπ‘‘ 𝑐𝑝,π‘šπ‘–π‘₯ decrease for Φ > 1
• For Φ < 1.05 π‘π‘‡π‘œπ‘‘ 𝑐𝑝,π‘šπ‘–π‘₯ decreases faster
• For Φ > 1.05 𝐻𝐢 decreases faster
Minor-Specie Mole Fractions
1%
πœ’π‘–
ppm
• NO, OH, H, O
• πœ’π‘– < 4000 ppm = 0.4%
• Peak near Φ = 1
• πœ’π‘‚π» > 10πœ’π‘‚
• πœ’π‘(π‘›π‘œπ‘‘ π‘ β„Žπ‘œπ‘€π‘›) β‰ͺ πœ’π‘‚
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