Slides

Report
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 1 to 8 from Page 702
 []
;
1000 


ℎ
=

• =
•
4184(1 
2
+ 2
2
3
+ 3
3
• Spreadsheet really helps this calculation
+
4
4
4
−
5

+ 6 )
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
 < 0
 = 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
•  −  =  −  =  ℎ − ℎ
•  = ∆ =  −  =  ℎ − ℎ = ∆ℎ
• ∆ and ∆ℎ Enthalpy of Reaction (< 0 for combustion)
• Dependent on T and P of reaction
• Heat of Combustion ∆ℎ = −∆ℎ = ℎ − ℎ > 0
Stoichiometric Methane Combustion, CH4
• CH4 + __ (O2 + 3.76 N2)  __ CO2 + __ H2O + ___ N2
• @ 25C and 1 kmol CH4
• ∆ =  −  =
Water Vapor
1 ℎ + ∆ℎ
+ 2 ℎ + ∆ℎ
+ 7.52 ℎ + ∆ℎ
2
− 1

ℎ
+ ∆ℎ
4
2 
+ 2

ℎ
+ ∆ℎ



= ℎ,
+
2
ℎ
−
1
ℎ
=
,

,
2
2
4
=

−393,546

p 688
= −802,405
+2


2
+ 7.52

 ℎ,

−241,845


−1

ℎ
−
2
+ ∆ℎ
2

 ℎ,


−74,831

p 692
p 701
(Heat in to system for TR = TP)
Other Bases
• Per kg fuel
• 4 = 16.043
• ∆ℎ = −





16.043

802,405
• Heat of Combustion
• ∆ℎ = −∆ℎ = 50,016
= −50,016




(Heat out for TR = TP)

• See page 701, LHV = Lower Heating Value = 50,016

• Corresponds to water vapor in the products



• ∆, = ℎ,
+
2
ℎ
−
1
ℎ
,2 ,
,4
2
•

ℎ,
2 ,
=

ℎ,
2 ,
− ℎ2, =

−241,845

p 692



• ∆,ℎ = ℎ,
+
2
ℎ
−
1
ℎ
,
,
,
2
2
4
• = −393,546 + 2 −241,845 − 1 −74,831 = −890,425

• ∆ℎ = −
−890,425


16.043
= 55,502




• p. 701: Higher Heating Value = HHV = 55,528
−

44,010

p 692


(slightly larger due to dissociation?)
=

−285,855

Per kg of reactant mixture
•


•


=
=

 +
 
 
=
• LHV = ∆ℎ, =
=
1

1+

=
1

1+
=
1
1+17.12
=
1 
18.12 
2∗ 3.76+1 ∗28.85

= 17.12
1∗16.043


1 

50,016
∗
= 2760
 18.12 

Adiabatic ( = 0) Flame Temperature
Complete Combustion Products
CCO2 HH2O
PP = PR, T = TAd
Stoichiometric
Reactants
TR PR
 = 0
 = 0
• 1st Law, Steady Flow Reactor
•  −  = 0 =  −  =  ℎ − ℎ
• All chemical energy goes into heating the
products
• To find adiabatic flame temperature use
• PP = PR and ℎ = ℎ
Adiabatic Methane Combustion TR = 25°C
• CH4 + 2 (O2 + 3.76 N2)  1 CO2 + 2 H2O + 7.52 N2
•  = 1 ℎ + ∆ℎ
4
• =  = 1 ℎ + ∆ℎ
+ 2 ℎ + ∆ℎ
2

= ℎ,
4
2
+ 2 ℎ + ∆ℎ
+ 7.52 ℎ + ∆ℎ
2 
2 
+ 7.52 ℎ + ∆ℎ



• ℎ,
−
1
ℎ
−
2
ℎ
,2
,2  = 1∆ℎ,2 + 2∆ℎ,2  + 7.52∆ℎ,2
4
• ∆ℎ,, =


 ,
  ≈ ,  − 
2 

Example (Turn in next time for Extra Credit)
• Find TAd for a 25°C Stoichiometric mixture of Acetylene and air

similar documents