# 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 𝑎1 to 𝑎8 from Page 702
𝑇 [𝐾]
;
1000 𝐾
𝑘𝐽
𝑜
ℎ
=
𝑘𝑚𝑜𝑙𝑒
• 𝜃=
•
4184(𝑎1 𝜃
𝜃2
+ 𝑎2
2
𝜃3
+ 𝑎3
3
• Spreadsheet really helps this calculation
+
𝜃4
𝑎4
4
−
𝑎5
𝜃
+ 𝑎6 )
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 &lt; 0
𝑊𝑂𝑈𝑇 = 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 − 𝑊𝑂𝑈𝑇 = 𝐻𝑃 − 𝐻𝑅 = 𝑚 ℎ𝑃 − ℎ𝑅
• 𝑄𝐼𝑁 = ∆𝐻𝑅 = 𝐻𝑃 − 𝐻𝑅 = 𝑚 ℎ𝑃 − ℎ𝑅 = 𝑚∆ℎ𝑅
• ∆𝐻𝑅 and ∆ℎ𝑅 Enthalpy of Reaction (&lt; 0 for combustion)
• Dependent on T and P of reaction
• Heat of Combustion ∆ℎ𝐶 = −∆ℎ𝑅 = ℎ𝑅 − ℎ𝑃 &gt; 0
Stoichiometric Methane Combustion, CH4
• CH4 + __ (O2 + 3.76 N2)  __ CO2 + __ H2O + ___ N2
• @ 25C and 1 kmol CH4
• ∆𝐻𝑅 = 𝐻𝑃 − 𝐻𝑅 =
Water Vapor
1 ℎ𝑓𝑜 + ∆ℎ𝑠
+ 2 ℎ𝑓𝑜 + ∆ℎ𝑠
+ 7.52 ℎ𝑓𝑜 + ∆ℎ𝑠
𝐶𝑂2
− 1
𝑜
ℎ𝑓
+ ∆ℎ𝑠
𝐶𝐻4
𝐻2 𝑂
+ 2
𝑜
ℎ𝑓
+ ∆ℎ𝑠
𝑜
𝑜
𝑜
= ℎ𝑓,𝐶𝑂
+
2
ℎ
−
1
ℎ
=
𝑓,𝐻
𝑂
𝑓,𝐶𝐻
2
2
4
=
𝑘𝐽
−393,546
𝑘𝑚𝑜𝑙
p 688
= −802,405
+2
𝑘𝐽
𝑘𝑚𝑜𝑙𝐹𝑢𝑒𝑙
𝑂2
+ 7.52
𝑜
𝑁𝑖 ℎ𝑓,𝑖
𝑘𝐽
−241,845
𝑘𝑚𝑜𝑙
𝑃𝑟𝑜𝑑
−1
𝑜
ℎ𝑓
−
𝑁2
+ ∆ℎ𝑠
𝑁2
𝑜
𝑁𝑖 ℎ𝑓,𝑖
𝑅𝑒𝑎𝑐𝑡
𝑘𝐽
−74,831
𝑘𝑚𝑜𝑙
p 692
p 701
(Heat in to system for TR = TP)
Other Bases
• Per kg fuel
• 𝑀𝑊𝐶𝐻4 = 16.043
• ∆ℎ𝑅 = −
𝑘𝑔
𝑘𝑚𝑜𝑙
𝑘𝐽
𝑘𝑚𝑜𝑙𝐹𝑢𝑒𝑙
𝑘𝑔
16.043
𝑘𝑚𝑜𝑙
802,405
• Heat of Combustion
• ∆ℎ𝑐 = −∆ℎ𝑅 = 50,016
= −50,016
𝑘𝐽
𝑘𝑔𝐹𝑢𝑒𝑙
𝑘𝐽
𝑘𝑔𝐹𝑢𝑒𝑙
(Heat out for TR = TP)
𝑘𝐽
• See page 701, LHV = Lower Heating Value = 50,016
𝑘𝑔𝐹𝑢𝑒𝑙
• Corresponds to water vapor in the products
𝑜
𝑜
𝑜
• ∆𝐻𝑅,𝐿𝑜𝑤𝑒𝑟 = ℎ𝑓,𝐶𝑂
+
2
ℎ
−
1
ℎ
𝑓,𝐻2 𝑂,𝑣𝑎𝑝𝑜𝑟
𝑓,𝐶𝐻4
2
•
𝑜
ℎ𝑓,𝐻
2 𝑂,𝐿𝑖𝑞𝑢𝑖𝑑
=
𝑜
ℎ𝑓,𝐻
2 𝑂,𝑣𝑎𝑝𝑜𝑟
− ℎ𝐻2𝑂,𝑓𝑔 =
𝑘𝐽
−241,845
𝑘𝑚𝑜𝑙
p 692
𝑜
𝑜
𝑜
• ∆𝐻𝑅,𝐻𝑖𝑔ℎ𝑒𝑟 = ℎ𝑓,𝐶𝑂
+
2
ℎ
−
1
ℎ
𝑓,𝐻
𝑂,𝐿𝑖𝑞𝑢𝑖𝑑
𝑓,𝐶𝐻
2
2
4
• = −393,546 + 2 −241,845 − 1 −74,831 = −890,425
𝑘𝐽
• ∆ℎ𝐶 = −
−890,425𝑘𝑚𝑜𝑙
𝑘𝑔
𝐹𝑢𝑒𝑙
16.043𝑘𝑚𝑜𝑙
= 55,502
𝑘𝐽
𝑘𝑔𝐹𝑢𝑒𝑙
𝑘𝐽
𝑘𝑔𝐹𝑢𝑒𝑙
• p. 701: Higher Heating Value = HHV = 55,528
−
𝑘𝐽
44,010
𝑘𝑚𝑜𝑙
p 692
𝑘𝐽
𝑘𝑔𝐹𝑢𝑒𝑙
(slightly larger due to dissociation?)
=
𝑘𝐽
−285,855
𝑘𝑚𝑜𝑙
Per kg of reactant mixture
•
𝑚𝐹𝑢𝑒𝑙
𝑚𝑀𝑖𝑥
•
𝐴
𝐹
=
=
𝑚𝐹𝑢𝑒𝑙
𝑚𝐹𝑢𝑒𝑙 +𝑚𝐴𝑖𝑟
𝑁𝐴𝑖𝑟 𝑀𝑊𝐴𝑖𝑟
𝑁𝐹𝑢𝑒𝑙 𝑀𝑊𝐹𝑢𝑒𝑙
=
• LHV = ∆ℎ𝑐,𝐿𝑜𝑤𝑒𝑟 =
=
1
𝑚𝐴𝑖𝑟
1+𝑚
𝐹𝑢𝑒𝑙
=
1
𝐴
1+𝐹
=
1
1+17.12
=
1 𝑘𝑔𝐹𝑢𝑒𝑙
18.12 𝑘𝑔𝑀𝑖𝑥
2∗ 3.76+1 ∗28.85
𝑘𝑔𝐴𝑖𝑟
= 17.12
1∗16.043
𝑘𝑔𝐹𝑢𝑒𝑙
𝑘𝐽
1 𝑘𝑔𝐹𝑢𝑒𝑙
𝑘𝐽
50,016
∗
= 2760
𝑘𝑔𝐹𝑢𝑒𝑙 18.12 𝑘𝑔𝑀𝑖𝑥
𝑘𝑔𝑀𝑖𝑥
Adiabatic (𝑄 = 0) Flame Temperature
Complete Combustion Products
CCO2 HH2O
PP = PR, T = TAd
Stoichiometric
Reactants
TR PR
𝑄𝐼𝑁 = 0
𝑊𝑂𝑈𝑇 = 0
• 1st Law, Steady Flow Reactor
• 𝑄𝐼𝑁 − 𝑊𝑂𝑈𝑇 = 0 = 𝐻𝑃 − 𝐻𝑅 = 𝑚 ℎ𝑃 − ℎ𝑅
• All chemical energy goes into heating the
products
• To find adiabatic flame temperature use
• PP = PR and ℎ𝑃 = ℎ𝑅
Adiabatic Methane Combustion TR = 25&deg;C
• CH4 + 2 (O2 + 3.76 N2)  1 CO2 + 2 H2O + 7.52 N2
• 𝐻𝑅𝑒𝑎𝑐𝑡 = 1 ℎ𝑓𝑜 + ∆ℎ𝑠
𝐶𝐻4
• = 𝐻𝑃𝑟𝑜𝑑 = 1 ℎ𝑓𝑜 + ∆ℎ𝑠
+ 2 ℎ𝑓𝑜 + ∆ℎ𝑠
𝐶𝑂2
𝑜
= ℎ𝑓,𝐶𝐻
4
𝑂2
+ 2 ℎ𝑓𝑜 + ∆ℎ𝑠
+ 7.52 ℎ𝑓𝑜 + ∆ℎ𝑠
𝐻2 𝑂
𝑁2 𝑅𝑒𝑎𝑐𝑡
+ 7.52 ℎ𝑓𝑜 + ∆ℎ𝑠
𝑜
𝑜
𝑜
• ℎ𝑓,𝐶𝐻
−
1
ℎ
−
2
ℎ
𝑓,𝐶𝑂2
𝑓,𝐻2 𝑂 = 1∆ℎ𝑠,𝐶𝑂2 + 2∆ℎ𝑠,𝐻2 𝑂 + 7.52∆ℎ𝑠,𝑁2
4
• ∆ℎ𝑠,𝑖,𝑇𝐴𝑑 =
𝑇
𝑐
𝑇𝑅𝑒𝑓 𝑝,𝑖
𝑇 𝑑𝑇 ≈ 𝑐𝑝,𝑖 𝑇𝐴𝑑 − 𝑇𝑅𝑒𝑓
𝑁2 𝑃𝑟𝑜𝑑
𝑇𝐴𝑑
Example (Turn in next time for Extra Credit)
• Find TAd for a 25&deg;C Stoichiometric mixture of Acetylene and air
```