ME 475/675 Introduction to Combustion

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ME 475/675 Introduction to
Combustion
Lecture 4
Heat of reaction and combustion, constant pressure adiabatic flame
temperature
Announcements
• You may turn in extra credit problem 2.15 next time, if you wish
• HW 1 and Extra Credit example Problem 2.14 due now
• Problem X1;
• Consider the broad education necessary to understand the impact of engineering
solutions (Introduction to Combustion) in a global and societal context
• Read a news article that describes some issue related to combustion
• For example: Energy efficiency, pollution, range land fires, fire safety, nuclear safety
• In one paragraph, summarize the article, and indicate how it is related to your interest in
combustion and/or this class
• Do you need more time? I need everyone to complete this problem.
• 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).
• ID: Find enthalpy of mixture at 298.15 K on 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
-97.787
-0.8639
a4
a5
20.402 -0.03095
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 of Ideal
Complete Combustion
All CCO2 HH2O
298.15 K, 1 atm
Reactants
298.15 K, P = 1 atm
Stoichiometric
Find 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
• β„Žπ‘– = standardized enthalpies
Next year use acetylene on next sides instead
of CH4
Stoichiometric Methane Combustion, CH4
• CH4 + 2 (O2 + 3.76 N2) οƒ  1 CO2 + 2 H2O + 7.52 N2
• @ 25°C and 1 kmol CH4
• βˆ†π»π‘… = 𝐻𝑃 − 𝐻𝑅 =
Water Vapor
π‘œ
π‘œ
π‘œ
1 β„Žπ‘“ + βˆ†β„Žπ‘ 
+ 2 β„Žπ‘“ + βˆ†β„Žπ‘ 
+ 7.52 β„Žπ‘“ + βˆ†β„Žπ‘ 
𝐢𝑂2
− 1 β„Žπ‘“π‘œ + βˆ†β„Žπ‘ 
=
=
𝐢𝐻4
𝐻2 𝑂
+ 2 β„Žπ‘“π‘œ + βˆ†β„Žπ‘ 
=
+ 7.52 β„Žπ‘“π‘œ + βˆ†β„Žπ‘ 
𝑁2
π‘œ
π‘œ
π‘œ
π‘œ
π‘œ
β„Žπ‘“,𝐢𝑂
+
2
β„Ž
−
1
β„Ž
=
𝑁
β„Ž
−
𝑁
β„Ž
𝑖 𝑓,𝑖 π‘ƒπ‘Ÿπ‘œπ‘‘
𝑖 𝑓,𝑖 π‘…π‘’π‘Žπ‘π‘‘
𝑓,𝐻2 𝑂,𝑔
𝑓,𝐢𝐻4
2
π‘˜π½
π‘˜π½
π‘˜π½
−393,546
+ 2 −241,845
− 1 −74,831
π‘˜π‘šπ‘œπ‘™
π‘˜π‘šπ‘œπ‘™
π‘˜π‘šπ‘œπ‘™
p 688 (top of page)
π‘˜π½
−802,405
π‘˜π‘šπ‘œπ‘™πΉπ‘’π‘’π‘™
𝑂2
𝑁2
p 692
(Heat into system for TR = TP)
p 701
Other Bases
• Per kg fuel
π‘˜π‘”
π‘˜π‘šπ‘œπ‘™
−802,405
• π‘€π‘ŠπΆπ»4 = 16.043
• βˆ†β„Žπ‘… =
βˆ†π»π‘…
π‘€π‘ŠπΆπ»4
=
• Heat of Combustion
π‘˜π½
π‘˜π‘šπ‘œπ‘™πΉπ‘’π‘’π‘™
π‘˜π‘”
16.043π‘˜π‘šπ‘œπ‘™
• βˆ†β„Žπ‘ = −βˆ†β„Žπ‘… = 50,016
π‘˜π½
π‘˜π‘”πΉπ‘’π‘’π‘™
= −50,016
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,𝐻
𝑂,π‘£π‘Žπ‘π‘œπ‘Ÿ
𝑓,𝐢𝐻
2
2
4
π‘œ
π‘œ
• β„Žπ‘“,𝐻
=
β„Ž
𝑓,𝐻2 𝑂,π‘£π‘Žπ‘π‘œπ‘Ÿ − β„Žπ»2 𝑂,𝑓𝑔 = −241,845
2 𝑂,πΏπ‘–π‘žπ‘’π‘–π‘‘
p 692
π‘˜π½
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‘™
π‘˜π½
• p. 701: Higher Heating Value = HHV = 55,528π‘˜π‘”
𝐹𝑒𝑒𝑙
− 44,010
p 692
π‘˜π½
π‘˜π‘šπ‘œπ‘™
π‘˜π½
π‘˜π‘”πΉπ‘’π‘’π‘™
(slightly larger due to dissociation?)
= −285,855
π‘˜π½
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
2∗ 3.76+1 ∗28.85
1∗16.043
π‘˜π½
50,016
π‘˜π‘”πΉπ‘’π‘’π‘™
1
𝐴
1+𝐹
=
1
1+17.12
=
π‘˜π‘”π΄π‘–π‘Ÿ
17.12
π‘˜π‘”πΉπ‘’π‘’π‘™
∗
1 π‘˜π‘”πΉπ‘’π‘’π‘™
18.12 π‘˜π‘”π‘€π‘–π‘₯
=
=
1 π‘˜π‘”πΉπ‘’π‘’π‘™
18.12 π‘˜π‘”π‘€π‘–π‘₯
π‘˜π½
2760
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‘…
Example: Problem 2.30 page 72
• Next year use Acetylene
• Determine the adiabatic flame temperature for constantpressure combustion of a stoichiometric propane-air mixture
assuming reactants at 298 K, no dissociation of the products,
and constant specific heats evaluated at 298 K.
End 2015
Adiabatic Methane Combustion TR = 25°C
(next year do problem 2.30 here)
• 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
“Meaning” of Mixture Standardized Enthalpy
• Reactants and products are both at 27°C and 1 atm
• 8𝐢 + 9𝐻2 + 12.5𝑂2 + 47𝑁2 + β„Žπ‘– 𝑇 → 𝐢8 𝐻18 + 12.5𝑂2 + 47𝑁2
• Reactants are naturally-occurring elemental compounds
• All have zero standardized enthalpy at 27°C and 1 atm
• Need to add β„Žπ‘– 𝑇 to reactants to get product mixture at same temperature and
pressure
“Meaning” of Enthalpy (Heat) of Formation
• Reactants and products are both at 27°C and 1 atm
• 8𝐢 + 9𝐻2 + β„Žπ‘– 𝑇 → 𝐢8 𝐻18 (fuel heat of formation)
• Reactants are naturally-occurring elemental compounds
• All have zero standardized enthalpy at 27°C and 1 atm
• Describes how much energy needs to be added to C and 𝐻2 to create the fuel.
• See page 701
• For most hydrocarbon fuels β„Žπ‘“π‘œ < 0,
• 𝐢 + 𝐻2 mixture has more energy than fuel
• Exceptions
• Acetylene 𝐢2 𝐻2 , Ethene 𝐢2 𝐻4 , propene 𝐢2 𝐻4 , 1-Butene 𝐢4 𝐻8 , Benzene 𝐢6 𝐻6
• Small number of H per C (lots of double bonds between C atoms)
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