Combustion Lecture Note Simple
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Transcript of Combustion Lecture Note Simple
Lecture 3 Combustion Chemistry
Fuels burn with oxygen,
CH4 + 2O2 CO2 + 2H2O
C3H8 + 5O2 3CO2 + 4H2O
C10H22 + 31/2O2 10CO2 + 11H2O
In general,
CmHn + (4m+ n) O2 mCO2 + n H2O
4 2
Lecture 3 Combustion Chemistry
If air is used, assume air is dry with composition:
21 mol% O2
79 mol% N2
Ratio of N2 to O2 in air:
3.764 moles N2: 1 mole O2
Theoretical air for some types of fuel
Anthracite Coal 7.83 1b / 10,000 Btu
Coke 7.73 1b / 10,000 Btu
Fuel Oil 7.58 1b / 10,000 Btu
Natural Gas 7.37 1b / 10,000 Btu
LPG 7.25 1b / 10,000 Btu
Biomass 6.60 1b / 10,000 Btu (oxygenated hydrocarbon CmHnOp)
Lecture 3 Combustion Chemistry
Example :
One mole propane is burned in air. What is the mass of air required for
the complete combustion of propane?
C3H8 + 5 O2 3CO2 + 4H2O
Stoichiometry:
32 g Mass O2 = 5 mol* = 160 g
Therefore mass of air (O2 and N2) = 0.69 kg
mol
Mass N2 = 3.764*5 mol* = 527 g 28 g
mol
N.B.
Alternately, the calculation is straightforward if molecular weight of air is known.
Lecture 3 Combustion Chemistry
Air-to-fuel ratio (AFR)
The standard measure of the amount of air used in a combustion process.
mair AFR =
mfuel
C3H8 + 5 O2 3CO2 + 4H2O
Thus, for the combustion of propane in air
mair AFR = = = 15.6 kg/kg
mfuel
687
1*44
Lecture 3 Combustion Chemistry
Combustion System
CmHn (fuel)
Air
O2
N2
CO2
H2O
N2
Combustion
System
• Excess air – products: CO2, H2O, N2, O2
• Incomplete (partial) combustion – products: CmHn, C, CO, CO2, H2O, N2
• Contaminated fuel, e.g. S – products: SO2
Lecture 3 Combustion Chemistry
Example :
Octane is burned in 1.4 excess air. Calculate the molar amount of
air fed into the combustion system.
C8H18 + (25/2) O2 8CO2 + 9H2O
C8H18 + 17.5O2 8CO2 + 9H2O + 5O2
Mass balance:
With 40% excess air::
Excess air feed :
(17.5 x 100/21) = 83.3 kmol air
Lecture 3 Combustion Chemistry
Example:
A fuel mixture 60 mol% ethane, and 40 mol% propane is
burned in stoichiometric air. Calculate the mass flow rate of
air required if the fuel mass is 12 kg/h.
Basis 1 kmol/h fuel,
0.6C2H6 + 2.1O2 1.2CO2 + 1.8H2O
0.4C3H8 + 2O2 1.2CO2 + 1.6H2O
Mass balance:
0.6 kmol/h ethane = 0.6x30 = 18 kg/h
0.4 kmol/h propane = 0.4x44 = 17.6 kg/h 35.6 kg/h
Stoichiometric air required :
35.6 kg/h fuel requires 4.1 kmol O2 ;
12 kg/h fuel requires 1.38 kmol O2 = 6.58 kmol/h air = ??? kg/h air
Lecture 3 Combustion Chemistry
Example :
Butane burns in incomplete combustion which produces CO in addition
to CO2 and H2O. Write a balanced chemical equation if the combustion
is 80% complete.
Basis 1 kmol/h fuel,
0.8 C4H10 + 5.2O2 3.2CO2 + 4H2O
Mass balance:
0.2 C4H10 + 0.9O2 0.8CO + H2O
C4H10 + 6.1O2 0.8CO + 3.2CO2 + 5H2O
Example :
Ethyl mercaptan, C2H6S, is intentionally added into domestic LPG
bottle to alert consumers of gas leakage. Write the stoichiometry
for the complete combustion of ethyl mercaptan in air.
Basis 1 kmol/h fuel,
C2H6 + 7/2O2 2CO2 + 3H2O
Mass balance:
S + O2 SO2
C2H6S + 9/2O2 2CO2 + SO2 + 3H2O
A furnace system
Tstack
TFT
To Tdew
T
DH
Qprocess Qlost
Qfuel = Qprocess + Qlost
hfurnace =
Qprocess
Qfuel
Lecture 4 Theoretical Flame Temperature
For adiabatic condition, DHc = 0
Then, use Hess’s Law to calculate Theoretical Flame Temperature (TFT)
• Enthalpy is a state property
• Change of enthalpy is independent of path
Lecture 4 Theoretical Flame Temperature
Fuel + Air
Tin
Combustion Products
(CO2, H2O, etc)
Tmax = TFT (adiabatic)
Fuel + Air
T0
∆H1 = Cp dT ∫ Tin
T0
∆H2 = ∆HoC
Combustion Products
T0
∆H3 = Cp dT ∫ To
TFT
∆HC
Theoretical Flame Tempeature
From Hess’s Law, DHc = ∆H1 + ∆HoC + ∆H3 = 0
Standards Heats of Combustion (∆HoC) are available from literatures
Lecture 4 Theoretical Flame Temperature
∆HoC
Cp (T) can be assumed a linear function between T and To.
An average value then can be used such that,
Average Cp values are available from literatures
∆H1 = Cp dT = [Cp]xDT
∫ Tin
T0
Lecture 4 Theoretical Flame Temperature
Cp
Example :
Methane at 25oC is burned in stoichiometric amount of air, also at
25oC. Calculate the theoretical flame temperature.
Basis 1 kmol/h fuel,
CH4 + 2O2 CO2 + 2H2O
Inlet Fuel + Air:
CH4 = 1 kmol
O2 = 2 kmol
N2 = 7.52 kmol
Combustion Products
CO2 = 1 kmol
H2O = 2 kmol
N2 = 7.52 kmol (inert)
Lecture 4 Theoretical Flame Temperature
Example (contd.):
∆H1 = Cp dT = 0
∫ Tin = T0
T0
∆H2 = ∆HoC = -802,310 kJ/kmol
∆H3= Cp dT
∆H3 = [(1x54.85) + (2x43.67) + (7.52x33.47)] x
(TFT-298)
= 393.88x(TFT – 298)
∫ To
TFT
(Guess TFT = 2000oC)
Lecture 4 Theoretical Flame Temperature
Example (contd.):
Energy Balance DHc = 0;
-802,310 + 393.88x(TFT – 298) = 0
TFT = 2335 K
= 2062oC
Answer is within +/- 5% deviation from Guess TFT.
Therefore TFT is acceptable.
Lecture 4 Theoretical Flame Temperature
Example :
Liquefied Petroleum Gas (LPG) is used as fuel to generate steam in
a boiler. The composition of LPG is 30 mol% propane and 70 mol%
butane. The air fed is 20% excess. The combustion is complete and
the system is adiabatic.
Calculate the standard heat of combustion for one kmole LPG. Then
calculate the theoretical flame temperature if air is fed at 100oC
while the fuel is at 25oC.
Tables for the standard heat of combustion and the mean molal heat
capacities are given.
Lecture 4 Theoretical Flame Temperature
Lecture 4 Theoretical Flame Temperature
For one kmol LPG, the fuel contains 0.3 mole C3H8 and 0.7 mole
C4H10. The individual DHoC are -2,220,000 kJ/kmol and – 2877000
kJ/kmol, respectively. Then,
DHoC,LPG = (0.3)(–2,220,000) + (0.7)(–2877000)
= – 2,679,900 kJ/kmol
Individual species stoichiometric balance:
C3H8 + 5O2 3CO2 + 4H2O
C4H10 + 6.5O2 4CO2 + 5H2O
One kmol LPG fuel stoichiometric balance:
0.3C3H8 + 1.5O2 0.9CO2 + 1.2H2O
0.7C4H10 + 4.55O2 2.8CO2 + 3.5H2O
Lecture 4 Theoretical Flame Temperature
Chemical equation with 20% excess air:
0.3C3H8 + 1.8O2 + (3.76)1.8N2
0.9CO2 + 1.2H2O + 0.3O2 + (3.76)1.8N2
0.7C4H10 + 5.46O2 + (3.76)5.46N2
2.8CO2 + 3.5H2O + 0.91O2 + (3.76)1.8N2
0.3C3H8 + 0.7C4H10 + 7.26O2 + 27.3N2
3.7CO2 + 4.7H2O + 1.21O2 + 27.3N2
Lecture 4 Theoretical Flame Temperature
Chemical equation with 20% excess air:
Inlet Fuel (25oC):
C3H8 0.3 kmol
C4H10 0.7 kmol
Inlet Air (100oC):
O2 7.26 kmol
N2 27.3 kmol
Combustion
Products (TFT):
CO2 3.7 kmol
H2O 4.7 kmol
O2 1.21 kmol
N2 27.3 kmol
Combustion
Process
∆H1 = (7.26x29.66 + 27.3x29.19) dT + ∫ Tin = 100oC
To = 25oC
CpLPG dT
∫ Tin = T0
T0 zero
= – 99,044 kJ
Example (contd.):
∆H2 = (1 kmol LPG) ∆HoC,LPG = – 2,679,900 kJ
∆H3= Cp dT
∆H3 = [(3.7x53.13) + (4.7x41.41) + (1.21x34.43) + (27.3x32.60)]
x (TFT-298)
= 1322.85x(TFT – 298)
∫ To
TFT
(Guess TFT = 1500oC)
Lecture 4 Theoretical Flame Temperature
Example (contd.):
Lecture 4 Theoretical Flame Temperature
Energy Balance DHc = 0;
= (– 99,044) + (– 2,679,900) + (1322.85x(TFT – 298)) = 0
TFT = 2398.7 K
= 2126oC
Answer deviates by more than 5% from Guess TFT.
Therefore TFT not acceptable.
Example (contd.):
Iteration
∆H3 = [(3.7x55.14) + (4.7x44.05) + (1.21x35.42) + (27.3x33.61)]
x (TFT-298)
= 1371.5x(TFT – 298)
(Guess TFT = 2100oC)
Lecture 4 Theoretical Flame Temperature
Energy Balance
(– 99,044) + (– 2,679,900) + (2460.5x(TFT – 298)) = 0
TFT = 2324.2 K
= 2026oC
Answer is within +/- 5% deviation from Guess TFT.
Therefore TFT is acceptable.