Combustion

Fuels and Combustion

FUELS AND COMBUSTION

1

1.1 Fuel

Fuel is any material that is burnt or altered in order to obtain energy. Fuel releases its energy either through chemical means, such as combustion, or nuclear means, such as nuclear fission or nuclear fusion. Wood was one of the first fuels used by humans and is still the primary energy source in much of the world. Fossil fuel

Fossil fuels are hydrocarbons, primarily coal and petroleum (liquid petroleum or natural gas), formed from the fossilized remains of dead plants and animals by exposure to heat and pressure in the Earth's crust over hundreds of millions of years. The burning of fossil fuels by humans is the largest source of emissions of carbon dioxide, which is one of the greenhouse gases (GHG) that enhances radiative forcing and contributes to global warming. The atmospheric concentration of CO2, a greenhouse gas, is increasing, raising concerns that solar heat will be trapped and the average surface temperature of the Earth will rise in response. A

1

A Course in Energy Conversion

global movement toward the generation of renewable energy is therefore under way to help meet the increased global energy needs.

1.2 Types of FuelsThe various types of fuels (like liquid, solid and gaseous fuels) that are available depend on various factors such as costs, availability, storage, handling, pollution and landed boilers, furnaces and other combustion equipments. The knowledge of the fuel properties helps in selecting the right fuel for the right purpose and for the efficient use of the fuel. (i) Solid Fuel (Coal)

Coal is a hard, black colored rock-like substance. It is made up of carbon, hydrogen, oxygen, nitrogen and varying amounts of sulphur. There are three main types of coal - anthracite, bituminous and lignite. Anthracite coal is the hardest and has more carbon, which gives it higher energy content. Lignite is the softest and is low in carbon but high in hydrogen and oxygen content. Bituminous is in between. Today, the precursor to coal - peat - is still found in many countries and is also used as an energy source.

Coal was first used as a fuel around 1000 BCE in China. With the development of the steam engine in 1769, coal came into more common use as a power source. Coal was later used to drive ships and locomotives. By the 19th century, gas extracted from coal was being used for street lighting in London. In the 20th century, the primary use of coal is for the generation of electricity, providing 40% of the world's electrical power supply in 2005.

Emissions : Coal contains an average of 1.5% sulphur (S) by weight, but this level may be as high as 3% depending upon where the coal was mined. During the combustion process: Sulphur will combine with oxygen (O2) from the air to form SO2 or SO3. Hydrogen (H) from the fuel will combine with oxygen (O2) from the air to form water (H2O). After the combustion process is completed, the SO3 will

2

Fuels and Combustion

combine with the water (H2O) to produce sulphuric acid (H2SO4), which can condense in the flue causing corrosion if the correct flue temperatures are not maintained. Alternatively, it is carried over into the atmosphere with the flue gases. This sulphuric acid is brought back to earth with rain, causing damage to the fabric of buildings, distress and damage to plants and vegetation. The ash produced by coal is light, and a proportion will inevitably be carried over with the exhaust gases, into the stack and expelled as particulate matter to the environment. Coal, however, is still used to fire many of the very large water-tube boilers found in power stations. The coal used in power stations is milled to a very fine powder, generally referred to as "pulverised fuel".

With regard to the quality of the gases released into the atmosphere. The boiler gases will be directed through an electrostatic precipitator where electrically charged plates attract ash and other particles, removing them from the gas stream. The sulphurous material will be removed in a gas scrubber. The final emission to the environment is of a high quality. Approximately 8 kg of steam can be produced from burning 1 kg of coal.

)ii (Liquid Fuel (Oil)

Oil is another fossil fuel. It was also formed more than 300 million years ago. Oil fuel is created from the residue produced from crude petroleum after it has been distilled to produce lighter oils like gasoline, paraffin, kerosene, diesel or gas oil. Various grades are available, each being suitable for different boiler ratings; the grades are as follows:

Class D - Diesel or gas oil.← Class E - Light fuel oil.← Class F - Medium fuel oil.← Class G - Heavy fuel oil.

The advantages of oil over coal:

3

A Course in Energy Conversion

a) A shorter response time between demand and the required amount of steam being generated.

b) Less energy had to be stored in the boiler water. The boiler could therefore be smaller, radiating less heat to the environment, with a consequent improvement in efficiency and less production space.

c) Mechanical stokers were eliminated, reducing maintenance workload.

d) Eliminating the problem of ash handling and disposal of the coal.Approximately 15 kg of steam can be produced from 1 kg of oil, or 14 kg of steam from 1 litre of oil. (iii) Gaseous Fuel

Gas fuels are the most convenient because they require the least amount of handling and are used in the simplest and most maintenance-free burner systems. Gas is delivered "on tap" via a distribution network and so is suited for areas with a high population or industrial density. However, large individual consumers do have gasholders and some produce their own gas.

Types of gaseous fuel

The following is a list of the types of gaseous fuel:(A)Fuels naturally found in nature:Natural gasMethane from coal mines(B)Fuel gases made from solid fuel:Gases derived from coalGases derived from waste and biomassFrom other industrial processes (blast furnace gas)(C)Gases made from petroleum:Liquefied Petroleum gas (LPG)Refinery gasesGases from oil gasification(D)Gases from some fermentation processGaseous fuels in common use are liquefied petroleum gases (LPG), Natural gas, producer gas, blast furnace gas, coke oven gas etc.

4

Fuels and Combustion

Typical physical and chemical properties of various gaseous fuels are given in the Table. 1.1.

Table 1.1 Typical physical and chemical properties of various gaseous fuels

Fuel GasRelative Density

Air/Fuel ratio m3 of air to m3

of Fuel

Flame Temp. oC

Flame Speed m/s

Natural Gas 0.6 10 1954 0.290Propane 1.52 25 1967 0.460Butane 1.96 32 1973 0.870

←← Natural gas←

Natural gas is lighter than air. Natural gas is mostly made up of a gas called methane. Methane is a simple chemical compound that is made up of carbon and hydrogen atoms. It's chemical formula is CH4 - one atom of carbon along with four atoms hydrogen. This gas is highly flammable. Natural gas is usually found near petroleum underground. It is pumped from below ground and travels in pipelines to storage areas. Natural gas usually has no odor and you can't see it. Before it is sent to the pipelines and storage tanks, it is mixed with a chemical that gives a strong odor. The odor smells almost like rotten eggs. The odor makes it easy to smell if there is a leak. Only a hint of sulphur is present in natural gas, meaning that the amount of sulphuric acid in the flue gas is virtually zero.

← Liquefied petroleum gases (LPG)← ← These are gases that are produced from petroleum refining and are

then stored under pressure in a liquid state until used. The most common forms of LPG are propane and butane.LPG may be defined as those hydrocarbons, which are gaseous at normal atmospheric pressure, but may be condensed to the liquid state at normal temperature, by the application of moderate pressures. Although they are normally used as gases, they are stored and transported as liquids under pressure for convenience and ease of

5

A Course in Energy Conversion

handling. Liquid LPG evaporates to produce about 250 times volume of gas.LPG vapour is denser than air: butane is about twice as heavy as air and propane about one and a half time as heavy as air. Consequently, the vapour may flow along the ground and into drains sinking to the lowest level of the surroundings and be ignited at a considerable distance from the source of leakage. In still air vapour will disperse slowly. Escape of even small quantities of the liquefied gas can give rise to large volumes of vapour / air mixture and thus cause considerable hazard. To aid in the detection of atmospheric leaks, all LPG’s are required to be odorized. There should be adequate ground level ventilation where LPG is stored. For this very reason LPG cylinders should not be stored in cellars or basements, which have no ventilation at ground level.

Approximately 42 kg of steam can be produced from 1 Therm of gas (equivalent to 105.5 MJ) for a 10 bar g boiler, with an overall operating efficiency of 80%.

(iv) Nuclear Fuel

The most common type of nuclear fuel used by humans is heavy fissile elements that can be made to undergo nuclear fission chain reactions in a nuclear fission reactor; nuclear fuel can refer to the material or to physical objects (for example fuel bundles composed of fuel rods) composed of the fuel material, perhaps mixed with structural, neutron moderating, or neutron reflecting materials. The most common fissile nuclear fuels are 235U and 239Pu, and the actions of mining, refining, purifying, using, and ultimately disposing of these elements together make up the nuclear fuel cycle, which is important for its relevance to nuclear power generation and nuclear weapons.

←Fuels that produce energy by the process of nuclear fusion are currently not utilized by man but are the main source of fuel for stars, the most powerful energy sources in nature. Fusion fuels tend to be light elements such as hydrogen which will combine easily.

6

Fuels and Combustion

1.3 Properties of fuels

Calorific Value

The calorific value is the measurement of heat or energy produced. This value may be expressed in two ways 'Gross or High ' and ‘Net or Low' calorific value. The difference is determined by the latent heat of condensation of the water vapour produced during the combustion process.

Gross calorific value (GCV): assumes all vapour produced during the combustion process is fully condensed. The gross calorific value of the fuel includes the energy used in evaporating this water.

Net calorific value (NCV): assumes the water leaves with the combustion products without fully being condensed. The net calorific value of the fuel excludes the energy in the steam discharged to the stack, and is the figure generally used to calculate boiler efficiencies. Fuels should be compared based on the net calorific value.

Since most gas combustion appliances cannot utilize the heat content of the water vapour, gross calorific value is of little interest. Fuel should be compared based on the net calorific value. This is especially true for natural gas, since increased hydrogen content results in high water formation during combustion.

The calorific value of coal varies considerably depending on the ash, moisture content and the type of coal while calorific values of fuel oils are much more consistent. Table 1.2 shows the Gross calorific values of fuel oil and some gases.

Density

Density is defined as the ratio of the mass of the fuel to the volume of the fuel at a reference temperature of 15°C. Density is measured by an instrument called a hydrometer. The knowledge of density is useful

7

A Course in Energy Conversion

for quantitative calculations and assessing ignition qualities. The unit of density is kg/m3.

Table 1.2 Gross Calorific Values for some fuel oil and gases

Oil FuelGross Calorific Value (MJ/lit.)

Class D - Diesel or gas oil. 40.1←Class E - Light fuel oil. 40.6

←Class F - Medium fuel oil 41.1←Class G - Heavy fuel oil. 41.8

Gas FuelGross Calorific Value (MJ/m3) at NTP

Natural 38.0Propane 93.0Butane 122.0

Specific gravity

This is defined as the ratio of the weight of a given volume of oil to the weight of the same volume of water at a given temperature. The density of fuel, relative to water, is called specific gravity. The specific gravity of water is defined as 1. The measurement of specific gravity is generally made by a hydrometer. Specific gravity is used in calculations involving weights and volumes.

Viscosity

The viscosity of a fluid is a measure of its internal resistance to flow. Viscosity depends on the temperature and decreases as the temperature increases. Any numerical value for viscosity has no meaning unless the temperature is also specified. Viscosity is measured in Stokes / Centistokes. Each type of oil has its own temperature - viscosity relationship. The measurement of viscosity is

8

Fuels and Combustion

made with an instrument called a Viscometer. Viscosity is the most important characteristic in the storage and use of fuel oil. It influences the degree of pre- heating required for handling, storage and satisfactory atomization. If the oil is too viscous, it may become difficult to pump, hard to light the burner, and difficult to handle.

Flash Point

The flash point of a fuel is the lowest temperature at which the fuel can be heated so that the vapour gives off flashes momentarily when an open flame is passed over it. The flash point for furnace oil is 66 oC.

Pour Point

The pour point of a fuel is the lowest temperature at which it will pour or flow when cooled under prescribed conditions. It is a very rough indication of the lowest temperature at which fuel oil is ready to be pumped.

Specific Heat

Specific heat is the amount of heat in(kJ or kCal) needed to raise the temperature of 1 kg of oil by 1 oC. The unit of specific heat is kJ/kg oC or kcal/kg oC. It varies from 0.92 to 1.17 kJ/kg oC depending on the oil specific gravity. The specific heat determines how much steam or electrical energy it takes to heat oil to a desired temperature. Light oils have a low specific heat, whereas heavier oils have a higher specific heat.

Sulphur

The amount of sulphur in the fuel oil depends mainly on the source of the crude oil and to a lesser extent on the refining process. The main disadvantage of sulphur is the risk of corrosion by sulphuric acid formed during and after combustion, and condensation in cool parts of the chimney or stack, air pre-heater and economizer.

9

A Course in Energy Conversion

The normal sulfur content for the residual fuel oil (furnace oil) is in the order of 2 - 4 % and for diesel oil is in order of 0.05 - 0.25 % , while for Kerosene is between 0.05 – 0.2 %.

Ash Content

The ash value is related to the inorganic material or salts in the fuel oil. The ash levels in distillate fuels are negligible. Residual fuels have higher ash levels. These salts may be compounds of sodium, vanadium, calcium, magnesium, silicon, iron, aluminum, nickel, etc. Typically, the ash value is in the range 0.03 - 0.07 %. Excessive ash in liquid fuels can cause fouling deposits in the combustion equipment. Ash has an erosive effect on the burner tips, causes damage to the refractories at high temperatures and gives rise to high temperature corrosion and fouling of equipments.

Carbon Residue

Carbon residue indicates the tendency of oil to deposit a carbonaceous solid residue on a hot surface, such as a burner or injection nozzle, when its vaporizable constituents evaporate. Residual oil contains carbon residue of 1 percent or more.

Water Content

The water content of furnace oil when it is supplied is normally very low because the product at refinery site is handled hot. An upper limit of 1% is specified as a standard.Water may be present in free or emulsified form and can cause damage to the inside surfaces of the furnace during combustion especially if it contains dissolved salts. It can also cause spluttering of the flame at the burner tip, possibly extinguishing the flame, reducing the flame temperature or lengthening the flame.

1.4 Combustion

10

Fuels and Combustion

Combustion or burning is a complex sequence of exothermic chemical reactions between a fuel and an oxidant accompanied by the production of heat or both heat and light in the form of either a glow or flames. Oxygen (O2) is one of the most common elements on earth making up 20.9% of air. Rapid fuel oxidation results in large amounts of heat. Solid or liquid fuels must be changed to a gas before they will burn. Usually heat is required to change liquids or solids into gases. Fuel gases will burn in their normal state if enough air is present. Most of the 79% of air is nitrogen. Nitrogen reduces combustion efficiency by absorbing heat from the combustion of fuels and diluting the flue gases. This reduces the heat available for transfer through the heat exchange surfaces. It also increases the volume of combustion by-products, which then have to travel through the heat exchanger and up the stack faster to allow the introduction of additional fuel-air mixture.This nitrogen also can combine with oxygen (particularly at high flame temperatures) to produce oxides of nitrogen (NOx), which are toxic pollutants. Carbon, hydrogen and sulphur in the fuel combine with oxygen in the air to form carbon dioxide, water vapour and sulphur dioxide, releasing 8,084 kcals, 28,922 kcals and 2,224 kcals of heat respectively.

The objective of good combustion is to release all of the heat in the fuel. This is accomplished by controlling: (1) Temperature high enough to ignite and maintain ignition of the fuel, (2) Turbulence or intimate mixing of the fuel and oxygen, and (3) Time, sufficient for complete combustion.

1.5 Complete combustion

In a complete combustion reaction, a compound reacts with an oxidizing element, such as oxygen , and the products are compounds of each element in the fuel with the oxidizing element. Complete combustion can be defined as stiochiometric combustion of the fuel .

A simpler example can be seen in the combustion of hydrogen and oxygen, which is a commonly used reaction in rocket engines:

11

A Course in Energy Conversion

2H2 + O2 = 2H2O + heat. The result is simply water vapor.

In the large majority of the real world uses of combustion, the oxygen (O2) oxidant is obtained from the ambient air and the resultant flue gas from the combustion will contain nitrogen.

For Methane:

CH4 + 2O2 = CO2 + 2H2O

CH4 + 2O2 + 7.52 N2 = CO2 + 2H2O + 7.52N2 + heat

As can be seen, when air is the source of the oxygen, nitrogen is by far the largest part of the resultant flue gas.

For example, the burning of propane is

C3H8 + 5 O2 = 3 CO2 + 4 H2O + heat

When not enough oxygen is present for complete combustion, propane burns to form water and carbon monoxide.

2 C3H8 + 7 O2 = 6CO + 8 H2O + heat

In complete combustion, the reactant will burn in oxygen, producing a limited number of products. It should be noted that complete combustion is almost impossible to achieve. In reality, as actual combustion reactions come to equilibrium, a wide variety of major and minor species will be present. For example, the combustion of methane in air will yield, in addition to the major products of carbon dioxide and water, the minor product carbon monoxide and nitrogen oxides, which are products of side reaction (oxidation of nitrogen).

Forms of combustions

(i) Rapid: Rapid combustion is a form of combustion in which large amounts of heat and light energy are released, which often results in a fire. This is used in a form of machinery such as internal combustion engines and in thermobaric weapons.

12

Fuels and Combustion

(ii) Slow: Slow combustion is a form of combustion which takes place at low temperatures. Respiration is an example of slow combustion. (iii) Turbulent: Turbulent combustion is a combustion characterized by turbulent flows. It is the most used for industrial application (e.g. gas turbines, diesel engines, etc.) because the turbulence helps the mixing process between the fuel and oxidizer.

1.6 Incomplete combustion

Incomplete combustion occurs when there isn't enough oxygen to allow the fuel (usually a hydrocarbon) to react completely with the oxygen to produce carbon dioxide and water.The quality of combustion can be improved by design of combustion devices, such as burners and internal combustion engines. Further improvements are achievable by catalytic after-burning devices (such as catalytic converters) or by the simple partial return of the exhaust gases into the combustion process. Such devices are required by environmental legislation for cars in most countries, and may be necessary in large combustion devices, such as thermal power plants, to reach legal emission standards.

Adiabatic Temperature

Assuming perfect combustion conditions, such as complete combustion under adiabatic conditions (i.e., no heat loss or gain), the adiabatic combustion temperature can be determined. The formula that yields this temperature is based on the first law of thermodynamics and takes note of the fact that the heat of combustion is used entirely for heating the fuel, the combustion air or oxygen, and the combustion product gases (flue gas). In the case of fossil fuels burnt in air, the combustion temperature depends on the heating value, the stoichiometric air to fuel ratio, the specific heat capacity of fuel and air and the air and fuel inlet temperatures. The adiabatic combustion temperature (also known as the adiabatic flame temperature) increases for higher heating values and inlet air

13

A Course in Energy Conversion

and fuel temperatures and for stoichiometric air ratios approaching one.Most commonly, the adiabatic combustion temperatures for coals are around 2200 °C (for inlet air and fuel at ambient temperatures and for the stoichiometric air to fuel ratio λ = 1.0), around 2150 °C for oil and 2000 °C for natural gas.

Air – Fuel Ratio

Air-fuel ratio (AFR): It is the ratio between the mass of air and the mass of fuel in the fuel-air mix at any given moment. When all the fuel is combined with all the free oxygen, typically within a vehicle's combustion chamber, the mixture is chemically balanced and this AFR is called the stoichiometric mixture (often abbreviated to stoich). AFR is an important measure for anti-pollution and performance tuning reasons.

1.7 Combustion analysis by mass and volume

(a) Gravimetric analysis – mass analysis :

For example to analysis by mass of combustion of H2 follow these steps:

2H2 +O2 = 2H2O this is complete combustion equationMolecular weight of H2 is 2Molecular weight of O2 is 36

So that, The relative mass of H2 is 2 × 2 = 4

The relative mass of O2 = 2 × 16 = 32Then, for H2O, the relative mass = 2 (2+ 16) = 36 Therefore, we can write the combustion equation as following:

4 masses H2 combined with 32 masses O2 = 36 masses H2ODividing by 4:

1mass H2 + 8 masses O2 = 9 masses H2O Take the mass in kilogrammes, then

14

Fuels and Combustion

1 kg H2 + 8 kg O2 = 9 kg H2OAir contains 0.23 % of oxygen and 0.77 % of nitrogen by mass.Hence,

Mass of air required to give 8 kg of O2 is kg of air.

The N2 which presented in 34.5 kg of air is 34.5 – 8 = 26.5 kgSo, 1 kg H2 is burnt in 34.5 kg of air the products of combustion is 9 kg H2O with 26.5 kg N2.

Similarly, we can find the combustion analysis by mass for any fuel composition such as : C , Co , S , Ch4 , etc.

(b) Volumetric analysis – volume analysis:

Again take the case of the complete combustion of H2. The combustion equation is

2H2 +O2 = 2H2OThus, 2 molecules of H2 + 1 molecule of O2 = 2 molecules of H2O

By Avogadro’s Hypothesis: Proportions by molecules are also proportions by volume. Hence,

2 volumes H2 + 1 volume O2 = 1 volume H2ODividing by 2.

1 vol. H2 + 0.5 vol. O2 = 1 vol. H2O

Assume that volume is in m3, then,

1 m3 H2 + 0.5 m3 O2 = 1 m3 H2O

Air contains 0.21% of oxygen and 0.79% of Nitrogen by volume.

Hence volume of air required to give 0.5 of O2 is m3 of

air. The N2 which presented in 2.38 m3 of air is 2.38 – 0.5 = 1.88 m3. So,

1 m3 H2 + 2.38 m3 of air = 1 m3 H2O + 1.88 m3 of air

15

A Course in Energy Conversion

The combustion analysis of fuel constituents by volume can determine by same method. Table 1.3 shows the oxygen, air required and the product of combustions for complete combustion of important constituents of fuels by mass. While Table 1.4 shows the complete combustion analysis of the constituents by volume.

Table 1.3 complete combustion analysis of some constituents by mass

One kg

Complete Combustion analysis by mass (kg)O2 Air H2O CO2 CO SO2 N2

H2 8 34.5 9 - - - 26.5C 2.67 11.61 - 3.67 - - 8.94

CO 0.57 2.49 - 1.57 - - 1.92

CH4 4 17.39 2.25 2.75 - -13.3

9S 1 4.35 - - - 2 3.35

Table 1.4 complete combustion analysis of some constituents by volume

One m3

Complete Combustion analysis by Volume (m3)O2 Air H2O CO2 CO SO2 N2

H2 0.5 2.38 1 - - - 1.88C 1 4.76 - 1 - - 3.76

CO 0.5 2.38 - 1 - - 1.88CH4 2 9.52 2 1 - - 7.52

S 1 4.76 - - - 1 3.76

1.8 Combustion analysis by the mole methodThe Mole: The mole is that quantity of a substance which is equal in mass to the molecular weight of the substance. For example one kilogram mole of hydrogen has a mass of 2 kilograms, and 1 kg mole of Nitrogen has a mass of 28 kg, etc. The mole may also consider as a unit of volume.

16

Fuels and Combustion

At standard temperature and pressure, the volume occupied by 1 mole of any ideal gas is the same and is given by:

= 22.48 m3/ kg mol.

The Solution of combustion equation by mole method:Oxygen needed for oxidation processes can be calculated as follows:

C + O2→ Co2

1 mole C + 1 mole O2= 1 mole Co2 1 volume C + 1 Volume O2 =1 volume Co2

Same thing; 2 H2 + O2 = 2 H2O

1 mole H2 + 0.5 mole O2 = 1 mole H2O1 mole Co + 0.5 mole O2 = 1 mole Co2

1 mole S + 1 mole O2 = 1 mole So2

In general, 1 mole Cm Hn + (m + n/4) mole O2 = m mole Co2 + n/2 mole H2O.

Example1.1

A fuel oil consists of the following analysis:

Constt C H2 O2 N2 S H2O Ash

% by mass

85.9 12 0.7 0.5 0.5 0.35 0.05

Determine the stoichiometric mass of air required to completely burn 1 kg of this fuel and also determine the percentage products of the combustion by mass.

Solution

It is better to solve such example by making a table. The calculations are shown in Table 1.5 for 1 kg of the fuel.

17

A Course in Energy Conversion

From the table : Total oxygen required =2.29+0.96+0.005 = 3.25 kg But, oxygen already present in 1 kg fuel I = 0.007Therefore, Supplied oxygen required = 3.255-0.007=3.248 kg/kg of fuelTherefore,

The theoretical air needed =

18

Fuels and Combustion 19

A Course in Energy Conversion

The products of combustion

=3.15+1.0835+.01+10.875 = 15.12 kg /kg of fuel

Hence, percentage of products by mass:

CO2 = (3.15/15.12) ×100 =20.83%

H2O = (1.0835/15.12) ×100 = 7.1%

SO2 = (0.01/15.12) ×100 = 0.07 %

N2 = (10.875/15.12) ×100 = 72%

1.9 Combustion analysis of some typical fuels

For solid and liquid type fuels, the fuel composition is given on weight fraction basis. In this analysis, CH4 is the only gas fuel considered. In order to keep the combustion analysis simple and straightforward, the CH4 composition is also provided on the weight fraction basis. Oxidant composition is usually given on the mole or volume basis. Table 1.6 provides some fuel compositions on a weight fraction basis. Again, in this combustion analysis, only the stoichiometric combustion is analyzed. Results of such analysis are provided, including the composition of the combustion gas products on a weight and mole/volume basis, the adiabatic flame temperature, the stoichiometric ratio and the fuel's higher heating value (HHV).

Table 1.6 Fuel Composition (weight fraction basis)

Fuel C H S N O H2O CH4

Coal 0.780 0.050 0.030 0.040 0.080 0.020 -Oil 0.860 0.140 0.000 0.000 0.000 0.000 -Fuel Gas

- - - - - - 1.00

Table 1.7 provides the combustion adiabatic flame temperature, stoichiometric ratio and the fuel's higher heating value. Note that, Stoichiometric ratio is the weight of air required for complete combustion of a unit weight of fuel. Thus, 1 kg of carbon fuel requires

20

Fuels and Combustion

11.444 kg of air for complete, ideal combustion. Table 1.8 provides stiochiometric air-fuel ratios of common fuels by weight and volume.

Table 1.7 Fuel Characteristics

FuelAdiab.Flame Temp. (K)

Stoichiometric Ratio (by wt.)

HHV (kJ/kg)

Carbon 2,460 11.444 32,779.8Hydrogen 2,525 34.333 141,866.8Sulfur (solid) 1,972 4.292 9,261.3Coal 2,484 10.487 32,937.9Oil 2,484 14.580 47,630.0Fuel Gas 2,327 17.167 50,151.2

Table 1.8 Stoichiometric air-fuel ratios of common fuels

Fuel By weight By volume Gasoline 14.7 : 1 -Natural Gas 17.2 : 1 9.7  : 1Propane (LP) 15.5 : 1 23.9 : 1Ethanol 9 : 1 -Methanol 6.4 : 1 -Hydrogen 34 : 1 2.39 : 1Diesel 14.6 : 1 -

1.10 Excess air

Excess air is the quantity of the air which supplied to burn the fuel over that required for complete combustion. In industrial fired heaters, power plant steam generators, and large gas-fired turbines, the more common term is percent excess combustion air. For example, excess combustion air of 15 percent means that 15 percent more than the required stoichiometric air is being used. Some instruments meter air and gas flow directly to aid control of air supply. The excess air percentages of oil fuel (in oil burners) and

21

RichWeak

Carbon Monoxide

Carbon Dioxide

Stochiometric Fuel-Air ratio

Oxygen

%

Fuel-Air ratio

A Course in Energy Conversion

natural gas (in gas burners) can be taken in the ranges 5 – 20 and 5 – 12 respectively.

Rich and Lean Mixture

For gasoline fuel, the stoichiometric air/fuel mixture is approximately 14.7 times the mass of air to fuel. Any mixture less than 14.7 to 1 is considered to be a rich mixture, any more than 14.7 to 1 is a lean mixture.

If a percentage analysis of dry flue gas is made from very weak to very rich mixtures of fuel and air then, Figure 1.1 shows the general shape of a graph of the results obtained. This shows that from a weak mixture the percentage CO2 increases while the O2 decreases. The CO2

is at a maximum at the stiochiometric fuel-air ratio and at this ratio, the O2 has theoretically reduced to zero. Over to the rich side shows the CO2 percentage decreasing while there is now an appearance of CO which increases as the richness increase. The O2 and CO lines generally overlap slightly practice.

Lean or weak mixtures produce cooler combustion gases than does a stoichiometric mixture, primarily due to the excessive dilution by unconsumed oxygen and its associated nitrogen. Rich mixtures also produce cooler combustion gases than does a stoichiometric mixture, primarily due to the excessive amount of carbon which oxidises to form carbon monoxide, rather than carbon dioxide. The chemical reaction oxidizing carbon to form carbon monoxide releases significantly less heat than the similar reaction to form carbon dioxide. Lean mixtures, when consumed in an internal combustion engine, produce less power than does the stoichiometric mixture. Similarly, rich mixtures return poorer fuel efficiency than the stoichiometric mixture.

22

Fuels and Combustion

Figure 1.1 Weak and Rich mixtures of fuel and air

Figure 1.2 and Figure 1.3 show a faster way to calculate the percentage of excess air, provided the percentage of CO2 or O2 in the flue gases.For optimum combustion of fuel oil the CO2 or O2 in flue gases should be maintained as follows: CO2 = 14.5–15 % O2 = 2–3 %

Figure 1.2 Relations between CO2 and Excess Air

23

A Course in Energy Conversion

Figure 1.3 Relations between residual oxygen and excess air

1.11 Flue gas

Exhaust gas is flue gas which occurs as a result of the combustion of fuels such as natural gas, gasoline/petrol, diesel, fuel oil or coal. It is discharged into the atmosphere through an exhaust pipe or flue gas stack. Its composition depends on what is being burned, but it will usually consist of mostly nitrogen (typically more than two-thirds) derived from the combustion air, carbon dioxide (CO2) and water vapor as well as excess oxygen (also derived from the combustion air). It further contains a small percentage of pollutants such as particulate matter, carbon monoxide, nitrogen oxides and sulfur oxides. The analysis of combustion products is necessary to determine, if the fuel burnt efficiently or not. It depends upon the amount of air and type of fuel. The steam generators in large power plants and the process furnaces in large refineries, petrochemical and chemical plants burn very

24

Fuels and Combustion

considerable amounts of fossil fuels and therefore emit large amounts of flue gas to the ambient atmosphere. Table 1.9 presents the total amounts of flue gas typically generated by the burning of fossil fuels such as natural gas, fuel oil and coal. The data in the table were obtained by stoichiometric calculations.

It is of interest to note that the total amount of flue gas generated by coal combustion is only 10 percent higher than the flue gas generated by natural gas combustion.

Table 1.9 Exhaust flue gas generated by combustion of fossil fuels. (In SI metric units and  Nm³ at 0 °C and 101.325 kPa,)

1.12 Determination of the mass of excess air supplied from the dry flue gas

Combustion Data Fuel Gas

Fuel Oil Coal

Gross caloric value 43.01 MJ/m³

43.50 MJ/kg

25.92 MJ/kg

Excess combustion air, % 12 15 20CO2 in wet exhaust gas, volume %

8.8 12.4 13.7

O2 in wet exhaust gas, volume %

2.0 2.6 3.4

Molecular weight of wet exhaust gas

27.7 29.0 29.5

CO2 in dry exhaust gas, volume %

10.8 14.0 15.0

O2 in dry exhaust gas, volume %

2.5 2.9 3.7

Molecular weight of dry exhaust gas

29.9 30.4 30.7

25

A Course in Energy Conversion

The Dry flue gas is exhaust gas when there is no water vapor (H2O). The procedures of determining the excess mass of air from given dry flue gas are as following:1. convert the volumetric analysis of the dry flue gas to a mass

analysis (if its given volumetric analysis)

2. determine the mass of dry flue gas/kg of fuele.g. in one kg CO2 there is 12/44 carbon

in one kg CO there is 12 / 28 CarbonTherefore, Mass of carbon, C/kg dry flue gas= (12/44 CO2 +12/28 CO) kg

3. compare the mass of C / kg dry flue gas with mass of C/kg of fuel Let m = mass of C / kg fuel

Therefore,

The mass of dry flue gas / kg fuel = kg

Mass of excess O2 / kg of fuel = mass of O2 / kg dry flue gas × mass of dry flue gas/ kg of fuel

Mass of excess air =

4. determine the mass of stiochiometric air required for the fuel using given composition of the fuel.

Then, Total mass of air = stiochiometric air + excess mass of air

Alternatively, the excess air can be determined using the following relation:

% excess air=(O2 – 0.5 CO2) / (0.264 N2 – (O2–0.5 CO)) × 100

Example 1.2

The dry exhaust gas from an oil engine and the fuel oil had the following percentage composition by volume and mass respectively as

26

Fuels and Combustion

shown in table below. Determine the mass of air supplied / kg fuel burnt.

CO2 CO O2 N2 C H2 O2

% Dry exhaust gas composition by vol.

8.85 1.2 6.8 83.15 - - -

% Fuel oil composition by mass

- - - - 84 14 2

Solution

1. Convert the volumetric analysis of flue gas into mass analysis:

Constt. % by Volume

% by Volume×Molecular

Weight

% by mass

CO2 8.85 8.85×44=389.5=13.1

Co 1.2 1.2×28=33.6=1.13

O2 6.8 6.8×32=217.6=7.32

N2 83.15 83.15×28=2330=78.45

Total 100 2970.7 100

2. Calculate the stiochiometric mass of oxygen and air:

Constituents Mass/kg fuel Oxygen required

27

A Course in Energy Conversion

C 0.84 0.84×2.67=2.24H2 0.14 0.14×8=1.12O2 0.02 -0.02

Total 1.0 3.34

Therefore, theoretical oxygen required to burn 1 kg of this fuel= 3.34 kg/kg fuelTheoretical air required= 3.34/0.23 = 14.52 kg/kg fuel

3. Determine the mass of dry flue gas/kg of fuel:Mass of C/kg of dry flue gas

=

4. Compare the mass of C / kg dry flue gas with mass of C /kg of fuel

Mass of C/kg fuel 0.84 kg (given)

Therefore, mass of dry flue gas / kg fuel =

5. Determine the total mass of air required :

Hence, mass of excess O2 / kg fuel = 20.65 × 0.0732 = 1.513 kg Mass of excess air = 1.513/0.23 = 6.58 kg/ kg fuelTherefore, mass of air supplied / kg fuel = theoretical air + excess air

= 14.54 + 6.58 = 21.08 kg/kg fuel

Alternatively:

Mass of N2 /kg fuel = 20.65 × 0.7845 = 16.2 kg/kg fuel

Therefore, Mass of air supplied / kg fuel =

Example 1.3 ( Solve by Mole method )

A gas has the following composition by volume:

28

Fuels and Combustion

H2 CO2 CH4 N2 CO O2

% by volume 28 16 2 42 10 2

The gas burned with 50 % excess air. Determine:1. Volume of air / m3 of gas.2. Volumetric analysis of dry products.3. Mass of products / kg fuel.4. Mass of dry products / kg fuel.5. Mass of air supplied / kg fuel.

Solution

The calculations are given in the Table 1.10 on the basis of 1 mole of gas fuel.

(a) Moles of oxygen required to burn 1 mole fuel = 0.315 Moles of air/ mole of fuel gas = 0.315/0.21 =1.5 OR Volume of air / m3 of fuel gas= 1.5 m3

(b) Moles of dry products per mole of fuel gas = 0.28+1.604+0.105=1.989 moles

Hence, CO2 = 0.28/1.989 × 100 = 14.1% N2 = 1.604/1.989 × 100 = 80.6 % O2 = 0.105/1.989 × 100 = 5.3 %(c) Mass of products per mole of fuel gas is given by:

= 0.28×44 + 0.32× 18 + 1.604 × 28 +0.105 × 32 = 66.34 kgMass of one mole of fuel gas is given by: = 0.28 × 2 + 0.02×16 +0.1×28 +0.16×44 + 0.42×28 +0.02×32 = 23.12 kgOR Mass of products per kg of fuel gas = 66.34/23.12 = 2.87 kg

29

A Course in Energy Conversion 30

Fuels and Combustion

(d) Mass of dry products per kg fuel =

(e) Mass of air per kg fuel = 2.87 – 1 =1.87 kg

The mass of air can be checked by considering the moles of air supplied per mole of fuel:

Mass of air = moles of air × Molecular weight of air = 1.5 × 29= 43.5 kgTherefore,Mass of air/ mass of fuel = 43.4/23.12 = 1.88 kg /kg fuel

QUESTIONS 1(1) A fuel oil has the analysis as shown in table below. Determine the

mole formula, and write the complete reaction equation with oxygen

C O2 S H2 N2

% by weight 83 3.0 1.0 11.0 2.0(2) Propane gas is reacted with air in such a ratio that an analysis of

the dry products of combustion gives: CO2 = 11.5%, O2 = 2.75%, and CO = 0.7%. What is the percentage excess air used?. Find the number of moles of the gas burnt.

(3) An unknown hydrocarbon fuel, Cx Hy, was allowed to react with air. An Orsat analysis was made of a reprehensive sample of the product gases with the following result: CO2 = 12.1%, O2 = 3.8%, and CO = 0.9%. Determine:

i) the composition of the fuel;ii) the chemical equation for the actual reaction;iii) the excess air used percentage;iv) air- fuel ratio.

(4) Find the stoichiometric value of oxygen required to burn a heptane C7 H16.

(5) A coal has the following analysis:

31

A Course in Energy Conversion

C O2 S H2 N2 Ash% gravimetric

analysis62 15 4 6 1

12

The coal is burned with 40% excess air. Calculate:

(a) Mass of air /kg fuel;(b) Mass of products / kg fuel;(c) Mass of H2O in products /kg fuel;(d) Volumetric analysis of dry products;(e) Specific volume of products at 1 atm. And 187 oC.

(6) The analysis, by mass of the coal fired to a boiler is

Carbon = 82%; Hydrogen = 8%; Oxygen = 3%; Ash = 7%The boiler uses 0.19 kg/s and air is supplied to the furnace by a fan, the air supply being 30% excess of that required for theoretically correct combustion. Calculate:(i) the volume of air taken in by the fan/s when the conditions at

the fan intake are 100 kN/m2 and 18 oC (R for air = 0.287 kJ/kg K)

(ii) the percentage composition by mass of the dry flue gases. (7) A single – cylinder, four-stroke, compression-ignition oil engine

gives 15 kW at 5 r.p.s. and uses fuel having the composition by mass: Carbon = 84%; Hydrogen=16% . The air supply is 100% in excess of that required for perfect combustion. The fuel has a calorific value of 45 000 kJ/kg and the brake thermal efficiency of the engine is 30 %. Calculate:(i) The mass of fuel used/cycle;(ii) The actual mass of air taken in / cycle;(iii) The volume of air taken in/cycle at 100 kN/m2 and 15 oC. Take R=0.29 kJ/kg K. Ans [(a) 0.445 g; (b) 30.7 kg/kg fuel; (c) 0.011 m3]

(8) A fuel oil consists of 86 % Carbon and 14 % Hydrogen, by mass. During a test on an engine using this fuel, the dry exhaust gas analysis by volume was :

32

Fuels and Combustion

CO2 = 11.25 %, O2 =1,2 %, CO = 2.8 % and the remainder N2. Estimate the air-fuel ratio by mass being supplied to the engine. Ans. [15.7:1].

(9) A gaseous fuel contains the following components on a volumetric or mole basis: Hydrogen = 2%; Methane=64%; and Ethane=34%. Calculate:

i) the air fuel ratio required, kg air/kg fuel,ii) the volume of air required per kg and per kg-mole of

fuel, if 20% excess air is used and the air contains are 27 0C and 0.98 bar.

Ans. [16.7, 17.6 m3/kg fuel, 360 m3/kgmol fuel].

(10) A natural gas delivered from Marib (Yemen) has the following Mole fraction analysis:

Constituents % by Mole Constituents % by MoleMethane 96.59 Iso Pentane 0.02Ethane 2.49 Pentane 0.01Propane 0.27 Hexane 0.01Iso Butane 0.05 Nitrogen 0.04Butane 0.06 Carbon Dioxide 0.46

Determine: the theoretical air required to burn 1m3 of the natural gas. What is the Molecular Weight of the natural gas and the Molecular Weight of the combustion products of this gas.

33