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ME 475/675 Introduction to
Combustion
Lecture 4
Announcements
• Extra Credit example due now
• HW 1 Due Friday
• Tutorials
• Wednesday 1 pm PE 113
• Thursday 2 pm PE 113
• Please bring you textbook to class
• Please turn in HW on white or engineering paper
• Grading based on solution (not solely on answers)
Example:
• Last lecture (turned in today)
• Problem 2.14, p 71: Consider a stoichiometric mixture of isooctane and air.
Calculate the enthalpy of the mixture at the standard-state temperature (298.15
K) on a per-kmol-of-fuel basis (kJ/kmolfuel), on a per-kmol-of-mixture basis
(kJ/kmolmix), and on a per-mass-of-mixture basis (kJ/kgmix).
• Find enthalpy at 298.15 K of different bases
• This time (turn in next lecture)
• Problem 2.15: Repeat for T = 500 K
Standard Enthalpy of Isooctane
T [K]
298.15
theta
0.29815
h [kJ/Kmol]
-224108.82
a1
a2
-0.55313 181.62
-0.16492 8.072412
a3
a4
a5
-97.787 20.402 -0.03095
-0.8639 0.040304 0.103807
a6
-60.751
-60.751
• Coefficients π‘Ž1 to π‘Ž8 from Page 702
𝑇 [𝐾]
;
1000 𝐾
π‘˜π½
π‘œ
β„Ž
=
π‘˜π‘šπ‘œπ‘™π‘’
• πœƒ=
•
4184(π‘Ž1 πœƒ
πœƒ2
+ π‘Ž2
2
πœƒ3
+ π‘Ž3
3
• Spreadsheet really helps this calculation
+
πœƒ4
π‘Ž4
4
−
π‘Ž5
πœƒ
+ π‘Ž6 )
a8
20.232
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 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‘…
• 𝑄𝐼𝑁 = βˆ†π»π‘… = 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
Stoichiometric Methane Combustion, CH4
• CH4 + __ (O2 + 3.76 N2) οƒ  __ CO2 + __ H2O + ___ N2
• @ 25C 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
+ 7.52
π‘œ
𝑁𝑖 β„Žπ‘“,𝑖
π‘˜π½
−241,845
π‘˜π‘šπ‘œπ‘™
π‘ƒπ‘Ÿπ‘œπ‘‘
−1
π‘œ
β„Žπ‘“
−
𝑁2
+ 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 692
p 701
(Heat in to system for TR = TP)
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‘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‘™
π‘˜π‘”
𝐹𝑒𝑒𝑙
16.043π‘˜π‘šπ‘œπ‘™
= 55,502
π‘˜π½
π‘˜π‘”πΉπ‘’π‘’π‘™
π‘˜π½
π‘˜π‘”πΉπ‘’π‘’π‘™
• p. 701: Higher Heating Value = HHV = 55,528
−
π‘˜π½
44,010
π‘˜π‘šπ‘œπ‘™
p 692
π‘˜π½
π‘˜π‘”πΉπ‘’π‘’π‘™
(slightly larger due to dissociation?)
=
π‘˜π½
−285,855
π‘˜π‘šπ‘œπ‘™
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
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
• 𝑄𝐼𝑁 − π‘Šπ‘‚π‘ˆπ‘‡ = 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‘…
Adiabatic Methane Combustion TR = 25°C
• CH4 + 2 (O2 + 3.76 N2) οƒ  1 CO2 + 2 H2O + 7.52 N2
• π»π‘…π‘’π‘Žπ‘π‘‘ = 1 β„Žπ‘“π‘œ + βˆ†β„Žπ‘ 
𝐢𝐻4
• = π»π‘ƒπ‘Ÿπ‘œπ‘‘ = 1 β„Žπ‘“π‘œ + βˆ†β„Žπ‘ 
+ 2 β„Žπ‘“π‘œ + βˆ†β„Žπ‘ 
𝐢𝑂2
π‘œ
= β„Žπ‘“,𝐢𝐻
4
𝑂2
+ 2 β„Žπ‘“π‘œ + βˆ†β„Žπ‘ 
+ 7.52 β„Žπ‘“π‘œ + βˆ†β„Žπ‘ 
𝐻2 𝑂
𝑁2 π‘…π‘’π‘Žπ‘π‘‘
+ 7.52 β„Žπ‘“π‘œ + βˆ†β„Žπ‘ 
π‘œ
π‘œ
π‘œ
• β„Žπ‘“,𝐢𝐻
−
1
β„Ž
−
2
β„Ž
𝑓,𝐢𝑂2
𝑓,𝐻2 𝑂 = 1βˆ†β„Žπ‘ ,𝐢𝑂2 + 2βˆ†β„Žπ‘ ,𝐻2 𝑂 + 7.52βˆ†β„Žπ‘ ,𝑁2
4
• βˆ†β„Žπ‘ ,𝑖,𝑇𝐴𝑑 =
𝑇
𝑐
𝑇𝑅𝑒𝑓 𝑝,𝑖
𝑇 𝑑𝑇 ≈ 𝑐𝑝,𝑖 𝑇𝐴𝑑 − 𝑇𝑅𝑒𝑓
𝑁2 π‘ƒπ‘Ÿπ‘œπ‘‘
𝑇𝐴𝑑
Example (Turn in next time for Extra Credit)
• Find TAd for a 25°C Stoichiometric mixture of Acetylene and air
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