department of mechanical engineering me 322 mechanical
TRANSCRIPT
Department of Mechanical Engineering
ME 322 – Mechanical Engineering
Thermodynamics
Lecture 36
Combustion Reactions
Combustion Processes
2
Why do mechanical engineers have to know about
combustion? Consider a combustion chamber in a gas
turbine cycle,
inQ
3 2in aQ m h h
2 3
Air from
compressor Air to
turbine
Our current model What really happens
2 3
Air from
compressor Combustion
products to
the turbine
Fuel input
How is this analyzed??
Combustion
• Fuels
– Stored chemical energy
• Combustion Reaction
– Transforms the chemical energy stored in the fuel to thermal energy (heat)
• Goals of this section of the course
– Understand combustion chemistry
– Use combustion chemistry to determine the heat released during a combustion process
• Heat of reaction
3
Combustion
The combustion of a fuel requires oxygen,
fuel + oxidant products
Fuel
In the most general sense, a fossil fuel makeup is,
C H S O N
4
The Greek letters signify the atomic composition of the fuel.
For example ... 8 18C H octane
2 5C H OH ethanol (ethyl alcohol)
Combustion
Oxidant
The oxidant must contain oxygen. The most abundant ‘free’
source is atmospheric air. By molar percent, atmospheric air
is considered to be ...
Nitrogen 79%
Oxygen 21%
For every mole of
oxygen involved in a
combustion reaction,
there are 79/21 = 3.76
moles of nitrogen.
5
fuel + oxidant products
Combustion
Products (for fuels with no sulfur content)
Complete Combustion: CO2, H2O, and N2
Incomplete Combustion: CO2, H2O, N2, CO, NOx
Combustion with Excess Oxygen: CO2, H2O, N2, and O2
NOTE: Fuels containing sulfur have the potential of
introducing sulfuric acid into the product stream.
6
fuel + oxidant products
Combustion Terminology
• Theoretical or Stoichiometric Air
– The amount of air required for complete combustion of the fuel
• Determined by balancing the combustion reaction
• Excess or Percent Theoretical Air
– The amount of air actually used in the combustion process relative to the stoichiometric value
• Can cause incomplete combustion or excess oxygen
7
Combustion Terminology
8
In many combustion processes, one of the parameters we
are interested in is how much air (or oxygen) is required per
unit quantity (moles or mass) of fuel.
Air-Fuel and Fuel-Air Ratios
1/ /
/
1/ / /
/
mol mol
mol
air
mass mol massfuel mass
A F F AA F
MA F A F F A
M A F
moles of air moles of fuel
moles of fuel moles of air
mass of fuel
mass of air
Equivalence Ratio
/
/
actual
stoichiometric
F A
F A
Equivalence Ratio and Products
• Stoichiometric (=1)
– CO2, H2O, N2
• Lean ( < 1 with T < 1800 R)
– CO2, H2O, N2, O2
• Rich ( > 1 with T < 1800 R)
– CO2, H2O, N2, O2, CO, H2
• Rich ( > 1 with T > 1800 R)
– CO2, H2O, N2, O2, CO, H2, H, O, OH, N, C(s), NO2, CH4
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ME 322
Advanced
courses
ME 422 & 433
Stoichiometric (Complete) Combustion
0 1 2 3 4 2 2 2 2 2 2C H S O N + (O + 3.76 N ) CO + H O + SO + N
1
2
3
0 1 2 3
0 4
2
2 2 2
2(3.76) 2
C:
H:
S:
O:
N:
5 equations
5 unknowns
(0 through 4)
Atomic Balances
Stoichiometric Combustion of a General Fuel in Air
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Lean Combustion
0 1 2 3 4 5 2 2 2 2 2 2 2C H S O N + (O + 3.76 N ) CO + H O + SO + O + N
0 0
1
2
3
0 1 2 3 4
0 5
2
2 2 2 2
2(3.76) 2
( )
C :
H:
S :
O:
N:
PTA PTA = Percent Theoretical Air
expressed as a decimal
6 equations
6 unknowns
(0 through 5)
Requires a stoichiometric
balance first (to get 0)
Lean Combustion of a General Fuel in Excess Air
11
Example – Octane Combustion
Given: Gasoline (modeled as octane - C8H18) burns
completely in 150% theoretical air (or 50% excess air).
Find:
(a) the A/F ratios (mass and molar)
(b) the equivalence ratio
(c) the dew point of the products of combustion at assuming
that the products are at 1 atm
12
Example – Octane Combustion
13
In order to calculate the air-fuel ratios and the equivalence
ratio, we need to know how much air is used in the
combustion reaction. This is determined by balancing the
combustion reaction.
In order to determine the dew point of the products, we need
to know the molar composition of the products. This is also
determined by balancing the combustion reaction.
Everything depends on the correct balance
of the combustion reaction!
Example – Octane Combustion
1. Balance the stoichiometric reaction to get 0
0 1 2 4 8 18 2 2 2 2 2C H + (O + 3.76 N ) CO + H O + N
0 1 2 4 5 8 18 2 2 2 2 2 2C H + 1.5 (O + 3.76 N ) CO + H O O + N
2. Balance the reaction with 150% theoretical air
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Solution strategy ...
3. Calculate the required (A/F) ratios, the equivalence ratio,
and the dew point temperature of the products
Example – Octane Combustion
0 1 2 4 8 18 2 2 2 2 2C H + (O + 3.76 N ) CO + H O + N
Stoichiometric Reaction
8 18 2 2 2 2 2C H + 12.5(O + 3.76 N ) 8CO + 9H O + 47N
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0 4 42(3.76) 2 3.76 12.5 47 N:
0 1 2 3 02 2 2 8 9 / 2 12.5 O:
30 S:
2 218 2 9 H:
18 C:
Example – Octane Combustion
Combustion in 150% theoretical air
0 1 2 4 5 8 18 2 2 2 2 2 2C H + 1.5 (O + 3.76 N ) CO + H O O + N
0 0 0 12.5 1.5 18.75 ( )PTA
8 18 2 2 2 2 2 2C H + 18.75(O + 3.76 N ) 8CO + 9H O + 6.25O + 70.5N
16
1
2 2
3
0 1 2 3 4 4
0 5 5
8
18 2 9
0
2 2 2 2 18.75 8 9 / 2 6.25
2(3.76) 2 3.76 18.75 70.5
C:
H:
S:
O:
N:
Example – Octane Combustion
The molar (A/F) ratio can now be found ...
17
8 18 2 2 2 2 2C H + 12.5(O + 3.76 N ) 8CO + 9H O + 47N
8 18 2 2 2 2 2 2C H + 18.75(O + 3.76 N ) 8CO + 9H O + 6.25O + 70.5N
0
,
1 3.76/
1mol actA F
0
,
1 3.76/
1mol stoichA F
/mol
A F 2 2moles of O + moles of Nmoles of air= =
moles of fuel moles of fuel
04.76 4.76 12.5 59.5
04.76 4.76 18.75 89.25
Example – Octane Combustion
/ / air
mass molfuel
MA F A F
M
The mass-based (A/F) ratio can be found knowing the
molecular masses of the air and the fuel,
18
Table C.13a
lbm28.97
lbmolairM
fuel C H S O NMW M M M M M
The molecular mass of the air is,
The molecular mass of the fuel is,
Table C.20 Table C.20
8 12.01115 lbm/lbmol 18 1.00797 lbm/lbmol 114.23 lbm/lbmolfuelMW
Example – Octane Combustion
Now, the mass-based (A/F) can be found ...
19
,
,
28.97/ 59.5 15.09
114.23
28.97/ 89.25 22.63
114.23
mass stoich
mass act
A F
A F
Once the (A/F) ratios are determined, the equivalence ratio
can be found,
/ 1/ /
/ 1/ /
actual act
stoich stoich
F A A F
F A A F
1/ 89.250.667
1/ 59.5
1/ 22.630.667
1/15.09
(molar based)
(mass based)
Example – Octane Combustion
The dew point of the products is the temperature where the
water vapor condenses,
Tdp = Tsat at Pw (partial pressure of the water vapor)
ww w w
Py P y P
P
20
8 18 2 2 2 2 2 2C H + 18.75(O + 3.76 N ) 8CO + 9H O + 6.25O + 70.5N
2
1 2 3 4 5
90.096
8 9 0 6.25 70.5vy
0.096 1 atm 0.096 atm 1.411 psiaw wP y P
113.4 Fdp sat wT T P
Example – Problem 15.42
21
Given: Combustion exhaust with 9.1% CO2, 8.9% CO,
82% N2, and no O2
Find:
a) fuel model CnHm
b) mass percent of carbon and hydrogen in fuel
c) molar air/fuel ratio and percent theoretical air (PTA)
d) dew point temperature at .106 MPa
Example – Problem 15.42
22
STEP 1: Write balance equation using ORSAT data
CnHm + a(O2 + 3.76 N2)
9.1 CO2 + 8.9 CO + bH2O + 82 N2
STEP 2: Solve for unknowns and write fuel model CnHm
n = ? m = ? a=? b=?
Example – Problem 15.42
23
STEP 3: Compute molar mass of fuel & mass composition
Mfuel = 18lbmolC/lbmolfuel*(12lbm/lbmolC)
+ 33lbmolH/lbmolfuel*(1lbm/lbmolH)
= 249 lbm/lbmolfuel
C: 18*(12)/249 87%
H: 33*(1)/249 13%
Example – Problem 15.42
24
STEP 4: Calculate molar air/fuel ratio
21.8 * (1 + 3.76) /1 = 103.8 moles air / mole fuel
STEP 5: Write equation for stoichiometric combustion
C18H33 + 26.25(O2 + 3.76 N2)
18 CO2 + 16.5 H2O + 98.7 N2
STEP 6: Find theoretical air
%TA = (21.8 / 26.75) * 100 = 83%