Richard Kelly, D.I.T. 1

Facilities Management and the Environment

BSc in Electrical Services and Energy Management

FUEL COMBUSTION AND BOILER OPERATION

What is a Boiler?

A boiler is an enclosed vessel in which water is heated and circulated, either as hot water, steam, or superheated steam for the purpose of heating, powering, and/or producing electricity.

The furnace of the boiler is where the fuel and air are introduced to combust;

Fuel/air mixtures are normally introduced into the furnace by using burners, where the flames are formed.

The resulting hot gases travel through a series of heat exchangers, where heat is transferred to the water flowing though them.

The combustion gases are finally released to the atmosphere via the stack of exhaust section of the boiler.

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Richard Kelly, D.I.T. 8

Good boiler design will:

• Seek to maximise the transfer of the energy released to heating of water,

while at the same time

• Attempt to minimise the inevitable pollution of the environment

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The most significant loss of energy fromthe overall process is the stack heat-loss due

to the exit of warm combustiongases through the boiler flue.

These losses can be estimated using combustion theory.

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•Air is mixed with the fuel in the boiler to provide oxygen for the combustion process.

•Excess air above the theoretical amount is needed in boiler operation to compensate for incomplete mixing of fuel and combustion air.

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Complete combustion occurs when all of the fuel is converted to Carbon

Dioxide and water.

This is also called stoichiometric combustion.

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The quantity of excess air provided is important.

Too much excess air leads to unnecessary loss of heat from the boiler.

A deficiency of combustion air, leads to fuel-rich combustion.This is dangerous because highly toxic carbon monoxide (CO) may be present in the flue gases in appreciable amounts.

The combustion efficiency

will increase with increased

excess air, until the heat

loss in the excess air is

larger than the heat

provided by more efficient

combustion.

Typical excess air quantities to achieve highest efficiency for

burning fuels are:

•Natural Gas 5%-10%•Fuel Oil 5%-20%

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Large, unnecessary amounts of excess air can occur because of: ♦ Burner/control system imperfections

♦ Variations in boiler room temperature, pressure, and relative humidity

♦ Need for burner maintenance

♦ Changes in fuel composition

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Air is comprised of approximately 21% oxygen and 79% nitrogen

When air is delivered for combustion,the nitrogen absorbs heat and is carried up the stack, resulting in energy losses.

If there is excess air, the result is unused oxygen as well as even more nitrogen to absorb heat that is carriedup the stack.

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Boiler efficiency can be improved by incorporating an excess air trim loop into the

boiler controls.

A stack gas oxygen analyzer can be installed to continuously monitor excess air and adjust or “trim” the boiler fuel-to-air ratio for optimum

efficiency.

A carbon monoxidetrim loop, used in conjunction with the oxygen analyzer, assures that incomplete combustion

cannot occur due to a deficient air supply.

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For combustion calculations we deal with quantities of materials in terms of

kilomoles of a substance

1 kilomole of a substance contains 6.022 x 1026 atoms or molecules of that

substance

6.022 x 1026 is known as Avogadro's Number

Formal definition of the kilomole

The amount of a substance which contains as many atoms or molecules as there are

atoms in 12 kg of Carbon 12

(i.e. 12 kg of Carbon 12 contains 1 kilomole of carbon 12).

Kilomolar Masses of some Molecules

SYMBOLMolar Mass (kg/kmol)

Hydrogen H 1

Carbon C 12

Sulphur S 32

Oxygen (in molecular form) O2 32

Nitrogen (in molecular form) N2 28

Air (approximate) - 29

Water H2O 18

Carbon Dioxide CO2 44

Carbon Monoxide CO 28

3.293.76Ratio

Nitrogen/Oxygen

23.3%21%Oxygen

76.7%79%Nitrogen

By MassBy Volume 

Approximate composition of Air

Calorific Value of Some Fuels

Gross Calorific Value(MJ/m3)

Net CalorificValue (MJ/

m3)Hydrogen (H2) 12.10 10.22Methane (CH4) 37.71 33.95

Natural Gas 38.6 34.8Propane(C3H8)

50.0 46.3

● Natural Gas is composed of several gases ofvarying ratios, typically 70-90% Methane, 5-15%Ethane (C2H6)and less than 5% Butane (C4H10)and Propane

● LPG (Liquid Petroleum Gas) is a Propane (60%)Butane (40%) mix.

Chemical reactions in symbolic form.

C + 02 → C0

2 + Heat

• Solid carbon reacts with oxygen gas to form carbon dioxide gas with the release of heat.

• A single molecule of carbon (one atom) combines with a single molecule of oxygen (containing two atoms).

• 1 kmol of C + 1 kmol of 02 = 1 kmol of C02

• Mass is conserved: 12kg Carbon + 32kg Oxygen = 44kg CO2

2H2 +O

2 → 2H

2O+ Heat

2 Hydrogen molecules combine with one one Oxygen molecule to

produce water and heat

Air Needed for Complete Combustion

“Write the combustion equation for Octane (C8H18) burning with the stoichiometric

quantity of air.”

- An idealised reaction is considered, i.e. additional products such as CO,

NO and NO2 are ignored.

• The reaction will involve the Octane reacting with Oxygen and Nitrogen to produce CO2, water and Nitrogen.

• The kilomolar quantities of Nitrogen and Oxygen are set equal to their approx volumetric proportions in air.

• 8 kmol of O2 will combine with the 8 kmol of Carbon to produce 8 kmol of CO2

• The 18 Hydrogen atoms combine with 9 Oxygen atoms (4.5 kmol of O2) to produce 9 kmol of water

• Total O2 required = 8+4.5=12.5• There will be 3.76 times more N2 than there is O2, so the

total N2 is 12.5*3.76=47• The stoichiometric equation is then:

C8H

18+ 12.50

2 + 47N

2 → 8C0

2 + 9H

20 + 47N

2

C8H

18+ x0

2 + 3.76xN

2 → yC0

2 + zH

20 + 3.76xN

2

Calculating the Volumetric Composition of Flue Gases

A fuel oil, C12

H26

, is burned with 25% excess air. Determine the volume

percentage of the products of combustion.

First, determine the stoichiometric combustion equation:

C12

H26

+x02 +3.76xN

2 →12C0

2 + 13H

20+69.56N

2

C12

H26

+x02 +3.76xN

2 →12C0

2 + 13H

20+69.56N

2

Stoichiometric combustion required 18.5 Oxygen molecules.

Next, increase the quantity of Oxygen and Nitrogen by a factor of 1.25 (25% excess air)

The final combustion equation becomes:

C12

H26

+x02 +3.76xN

2 →12C0

2 + 13H

20+4.63O

2 + 86.95N

2

C12

H26

+x02 +3.76xN

2 →12C0

2 + 13H

20+4.63O

2 + 86.95N

2

Adding the products on the right we get:116.58 kmol

We can now express each of the products as a percentage of 116.58:

C02 : 12/116.58 =10.29%

H20: 13/116.58 = 11.15%

O2 : 4.63/116.58 = 3.97%

N2 : 86.95/116.58 = 74.58%

Total 99.9%