Combustion Lecture Note Simple

Lecture 3: Combustion Chemistry

CBB 4323 : Plant Utility Systems

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


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)


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.


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


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


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


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


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.


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.


C2H6 + 7/2O2 2CO2 + 3H2O

Mass balance:

S + O2 SO2

C2H6S + 9/2O2 2CO2 + SO2 + 3H2O

Lecture 4: Theoretical Flame Temperature

CBB 4323 : Plant Utility Systems

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


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


∆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


Cp

Example :

Methane at 25oC is burned in stoichiometric amount of air, also at

25oC. Calculate the theoretical flame temperature.


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)


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)


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.


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.





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


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


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



Example (contd.):



= (– 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)



Energy Balance

(– 99,044) + (– 2,679,900) + (2460.5x(TFT – 298)) = 0

TFT = 2324.2 K

= 2026oC



Example (contd.):



= (– 99,044) + (– 2,679,900) + (1364.6x(TFT – 298)) = 0

TFT = 2334.4 K

= 2061oC



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Combustion Lecture Note Simple