# Slides

```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
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1 to π8 from Page 702
π [πΎ]
;
1000 πΎ
ππ½
π
β
=
πππππ
• π=
•
4184(π1 π
π2
+ π2
2
π3
+ π3
3
• Spreadsheet really helps this calculation
+
π4
π4
4
−
π5
π
+ π6 )
a8
20.232
Enthalpy of Combustion (or reaction)
Products
Complete Combustion
Cο CO2 Hο H2O
298.15 K, 1 atm
Reactants
298.15 K, P = 1 atm
Stoichiometric
ππΌπ &lt; 0
ππππ = 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
• ππΌπ − ππππ = π»π − π»π = π βπ − βπ
• ππΌπ = βπ»π = π»π − π»π = π βπ − βπ = πββπ
• βπ»π and ββπ Enthalpy of Reaction (&lt; 0 for combustion)
• Dependent on T and P of reaction
• Heat of Combustion ββπΆ = −ββπ = βπ − βπ &gt; 0
Stoichiometric Methane Combustion, CH4
• CH4 + __ (O2 + 3.76 N2) ο  __ CO2 + __ H2O + ___ N2
• @ 25C and 1 kmol CH4
• βπ»π = π»π − π»π =
Water Vapor
1 βππ + ββπ
+ 2 βππ + ββπ
+ 7.52 βππ + ββπ
πΆπ2
− 1
π
βπ
+ ββπ
πΆπ»4
π»2 π
+ 2
π
βπ
+ ββπ
π
π
π
= βπ,πΆπ
+
2
β
−
1
β
=
π,π»
π
π,πΆπ»
2
2
4
=
ππ½
−393,546
ππππ
p 688
= −802,405
+2
ππ½
πππππΉπ’ππ
π2
+ 7.52
π
ππ βπ,π
ππ½
−241,845
ππππ
ππππ
−1
π
βπ
−
π2
+ ββπ
π2
π
ππ βπ,π
πππππ‘
ππ½
−74,831
ππππ
p 692
p 701
(Heat in to system for TR = TP)
Other Bases
• Per kg fuel
• πππΆπ»4 = 16.043
• ββπ = −
ππ
ππππ
ππ½
πππππΉπ’ππ
ππ
16.043
ππππ
802,405
• Heat of Combustion
• ββπ = −ββπ = 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
π
π
• βπ»π,πΏππ€ππ = βπ,πΆπ
+
2
β
−
1
β
π,π»2 π,π£ππππ
π,πΆπ»4
2
•
π
βπ,π»
2 π,πΏπππ’ππ
=
π
βπ,π»
2 π,π£ππππ
− βπ»2π,ππ =
ππ½
−241,845
ππππ
p 692
π
π
π
• βπ»π,π»ππβππ = βπ,πΆπ
+
2
β
−
1
β
π,π»
π,πΏπππ’ππ
π,πΆπ»
2
2
4
• = −393,546 + 2 −241,845 − 1 −74,831 = −890,425
ππ½
• ββπΆ = −
−890,425ππππ
ππ
πΉπ’ππ
16.043ππππ
= 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ππ
p 692
ππ½
πππΉπ’ππ
(slightly larger due to dissociation?)
=
ππ½
−285,855
ππππ
Per kg of reactant mixture
•
ππΉπ’ππ
ππππ₯
•
π΄
πΉ
=
=
ππΉπ’ππ
ππΉπ’ππ +ππ΄ππ
ππ΄ππ πππ΄ππ
ππΉπ’ππ πππΉπ’ππ
=
• LHV = ββπ,πΏππ€ππ =
=
1
ππ΄ππ
1+π
πΉπ’ππ
=
1
π΄
1+πΉ
=
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
= 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 (π = 0) Flame Temperature
Complete Combustion Products
Cο CO2 Hο H2O
PP = PR, T = TAd
Stoichiometric
Reactants
TR PR
ππΌπ = 0
ππππ = 0
• 1st Law, Steady Flow Reactor
• ππΌπ − ππππ = 0 = π»π − π»π = π βπ − βπ
• All chemical energy goes into heating the
products
• To find adiabatic flame temperature use
• PP = PR and βπ = βπ
Adiabatic Methane Combustion TR = 25&deg;C
• CH4 + 2 (O2 + 3.76 N2) ο  1 CO2 + 2 H2O + 7.52 N2
• π»πππππ‘ = 1 βππ + ββπ
πΆπ»4
• = π»ππππ = 1 βππ + ββπ
+ 2 βππ + ββπ
πΆπ2
π
= βπ,πΆπ»
4
π2
+ 2 βππ + ββπ
+ 7.52 βππ + ββπ
π»2 π
π2 πππππ‘
+ 7.52 βππ + ββπ
π
π
π
• βπ,πΆπ»
−
1
β
−
2
β
π,πΆπ2
π,π»2 π = 1ββπ ,πΆπ2 + 2ββπ ,π»2 π + 7.52ββπ ,π2
4
• ββπ ,π,ππ΄π =
π
π
ππππ π,π
π ππ ≈ ππ,π ππ΄π − ππππ
π2 ππππ
ππ΄π
Example (Turn in next time for Extra Credit)
• Find TAd for a 25&deg;C Stoichiometric mixture of Acetylene and air
```