Lecture 3. Fuels And Combustion

Mechanical Engineering Department

Power Plant Engineering (MEng 5211)

Lecture 3. Fuels and Combustion

1

Prepared by: Addisu D.

May, 2018

Introduction  Any matter that is a source of heat is called fuel.  Fuel releases its energy either through a chemical reaction, such 

 

 2

as during combustion, or via nuclear fission or fusion. An important property of a useful fuel is that its energy can be stored and released only when needed, and that the release is controlled in such a way that the energy can be harnessed to produce the desired work. Fuels are broadly classified as fossil or organic or chemical fuel, nuclear fuel, and rocket 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. Fossil fuels generate substantial quantities of heat per unit of mass or volume by reacting with an oxidant in a combustion process.

Introduction  In most practical applications including combustion in a steam

generator, air is used as the oxidant, although in certain processes oxygen, oxygen-enriched air and other chemicals are used as the oxidant.  Energy generation by a nuclear fuel takes place either by the process of nuclear fission of heavy fissile elements in a nuclear reactor resulting in chain reactions or by the process of nuclear fusion, in which simple atomic nuclei are fused together to form complex nuclei, as in case of fusion of hydrogen isotopes to form helium.  The process of nuclear fusion is also known as a thermonuclear reaction, which is difficult to control even today. As a result, the main source of nuclear energy presently available is mainly from nuclear fission. 3

Introduction  The most common fissile radioactive heavy metals are naturally









4

occurring isotope of uranium, U235, artificial isotope of uranium, U233, and artificial element plutonium, P239. In a nuclear reactor plutonium, P239, is produced from naturally occurring isotope of uranium, U238, and U233 is produced from the naturally occurring element thorium, Th232. Nuclear fission of 1 kg of U235 generates about 85*106 MJ of heat, which is equivalent to the heat generated by combustion of about 5106 kg of coal with a high heating value (HHV) of 17 MJ/kg. In nuclear reactions, the product is either isotopes of the reactants or other nuclei. Unlike fossil fuel, rocket fuel does not depend on its surroundings for the oxidant. The oxidant is carried by the rocket itself. The propellant is the chemical mixture burned to produce thrust in rockets and consists of a fuel and an oxidizer.

Introduction  The increasing worldwide demand for energy has focused 



 

5

attention on fuels, their availability and environmental effects. The fuels available to utility industry are largely nuclear and fossil, both essentially nonrenewable. Nuclear fuels originated with the universe, and it takes nature millions of years to manufacture fossil fuels. Fossil fuels originate from the earth as a result of the slow decomposition and chemical conversion of organic material. They come in three basic forms: solid (coal). liquid (oil), and natural gas. Coal represents the largest fossil-fuel energy resource in the world. According to IEA report in 2015, coal is responsible for about 40 percent of electric-power generation in the world. Oil and natural gas are responsible for another 27 percent. The remaining percentage is mostly due to nuclear and hydraulic generation.

Introduction  Most widely used fossil fuels for the production of steam in

relation to power are available in all the states of matter.  Gaseous ✓ natural gas, blast furnace gas, coke-oven gas, refinery gas, liquefied petroleum gas or (LPG), etc.  Liquid ✓ high-speed diesel (HSD), light diesel oil (LDO), heavy fuel oil (HFO), furnace oil (FO), low sulfur heavy stock (LSHS), naphtha, liquefied natural gas (LNG), etc.  Solid ✓ bituminous coal, anthracite, lignite, peat, oil shale, biomass, rice husk, bagasse etc.

6

Introduction  Fossil fuels consist of a large number of complex compounds of



 

 7

five elements: carbon (C), hydrogen (H), oxygen (O), sulfur (S), and nitrogen (N). Besides these elements, all fuels contain mineral matter (A) and moisture (M) to some extent. However, there are just three combustible elements of significance in a fuel, e.g., carbon, hydrogen, and sulfur, of which carbon is the principal combustible element with a HHV of 32.780 MJ/kg. Hydrogen has a very high HHV of 141.953 MJ/kg, but its content in solid fuel is quite low, about 2 - 4%. The HHV of sulfur is only 9.257 MJ/kg, hence as a source of heat its presence is insignificant, although it is more so since its presence in coal is small in quantity. Major concern regarding sulfur is that it promotes corrosion and creates atmospheric pollution problems.

Introduction  Contrary to solid and liquid fuels, gaseous fuels are mixtures of



 



8

combustible and non-combustible gases. Natural gas is gaseous fuel occurring in nature and consisting mostly of organic compounds, normally methane (CH4), ethane (C2H6), propane (C3H8), and butane (C4H10). The calorific value of natural gas varies between 26.1 and 55.9 MJ/m3, the majority averaging 37.3 MJ/m3. Natural gas on average contains 80 - 90% methane, 6 - 9% ethane, and 2 - 5% propane. Noncombustible gases present in minor quantities in natural gases are nitrogen (0.5 - 2.0%) and carbon dioxide (0.1 - 1.0%). LPG is a material composed predominantly of the following hydrocarbons or mixtures of them: propane, propylene, n-butane, isobutene, and butylenes. The mixture is liquefied at room temperature at very high pressure. An average mixture of LPG is comprised of about 80% butane and about 20% propane.

Coal  Fossil solid fuels are available in almost all countries but are distributed

 





9

unevenly. The majority of countries is not rich in coal production. There are around 70 countries that contain recoverable coal reserves. The world currently consumes over 7,700 Mt of coal which is used by a variety of sectors including power generation, iron and steel production, cement manufacturing and as a liquid fuel. Coal is the major fuel used to generate over 40% of the world’s electricity demand and provides 30% of global primary energy needs. Coal is also used in the production of 70% of the world’s steel. According to World Energy Council Report in 2016 the largest producer of coal is the People’s Republic of China (about 3747 million tons of coal was produced in the year 2015) followed by the United States (813 million tons) , India (677 million tons), Australia (485 million tons), Indonesia (392 million tons). Russia, South Africa, Germany, Poland, and Kazakhstan also contribute to global coal production to some extent.

Coal  According to geological order of formation, coal may be of the





 

10

following types: (1) Peat, (2) Lignite, (3) Subbitumioous, (4) Bituminous, (5) Subanthracite, and (6) Anthracite, with increasing percentages of carbon. After anthracite, graphite is formed. Anthracite contains more than 86% fixed carbon (in amorphous form) and less volatile matter. Volatile matter helps in the ignition of coal. So, it is often difficult to bum anthracite. Bituminous coal is the largest group containing 46-86% of fixed carbon and 20-40% of volatile matter. It can be low-volatile, medium-volatile and high-volatile. The lower the volatility, the higher the heating value. Lignite is the lowest grade of coal containing moisture as high as 30% and high volatile matter. According to ASTM (American Society of Testing and Materials), peat is not regarded as a rank of coal. Peat contains up to 90% moisture and is not attractive as a utility fuel.

Coal Analysis  There are two methods ➢ ultimate analysis and ➢ proximate analysis.  The ultimate analysis determines all coal component elements,

solid or gaseous and the proximate analysis determines only the fixed carbon, volatile matter, moisture and ash percentages.  The ultimate analysis is determined in a properly equipped laboratory by a skilled chemist, while proximate analysis can be determined with a simple apparatus.  It may be noted that proximate has no connection with the word “approximate”.

11

Coal Analysis: Proximate Analysis Measurement of Moisture  When 1 g sample of coal is subjected to a temperature of about 105oC for a period of 1 hour, the loss in weight of the sample gives the moisture content of the coal.

Significance of Moisture:  High moisture content of the coal is undesirable for the following reasons: 1) Reduces the calorific value of coal 2) Increases the consumption of coal for heating purpose. 3) Lengthens the time of heating. 12

Coal Analysis: Proximate Analysis Measurement of Volatile Matter  When 1 g sample of coal is placed in a covered platinum crucible and heated to 950 oC and maintained at that temperature for about 7 min, there is a loss in weight due to the elimination of moisture and volatile matter. The latter may now be determined since moisture has been calculated from the previous test.

Significance of volatile matter:  During burning of coal, gases like CO, CO2, CH4, N2, O2, hydrocarbons etc. that come out are called volatile matter of the coal.  It has been found that the coal with higher volatile matter content ignites easily has lower calorific value burns with long yellow smoky flame will evolve more coal gas when heated in the absence of air. 13

Coal Analysis: Proximate Analysis

Measurement ash in coal  By subjecting 1 g sample of coal in an uncovered crucible to a temperature of about 720 oC until the coal is completely burned, a constant weight is reached, which indicates that there is only ash remaining in the crucible. Complete combustion of coal is determined by repeated weighing of the sample.

 High ash content in coal is undesirable because it i. ii. iii.

14

increases transporting, handling and storage costs is harder and stronger has lower calorific value

Coal Analysis: Proximate Analysis

Measurement of fixed carbon  Fixed carbon is the difference between 100% (original sample) and

the sum of the percentages of moisture, ash and volatile matter. However, this difference does not represent all the carbon that was in the coal. Some of the carbon may have been in the form of hydrocarbons which may have been distilled off while determining the volatile matter. It is also possible that some of this fixed carbon may include Sulphur, nitrogen and oxygen. % of fixed carbon in coal = 100 - % (moisture + volatile matter + ash)  Fixed carbon is the pure carbon present in coal. Higher the fixed carbon content of the coal, higher will be the calorific value of the sample.  So, the proximate analysis of coal gives FC + VM + M+ A = 100% by mass 15

Coal Analysis Ultimate Analysis  The ultimate analysis indicates the various elemental chemical constituents such as Carbon, Hydrogen, Oxygen, Nitrogen, Sulphur of pure dry coal. This analysis gives the elementary, ultimate constituents of coal.  It is useful in determining the quantity of air required for combustion and the volume and composition of the combustion gases.  This analysis is essential for calculating heat balances in any process for which coal is employed as a fuel.  It is useful to the designing of coal burning equipment and auxiliaries. This information is required for the calculation of flame temperature and the flue duct design etc. 16

Coal Analysis: Ultimate Analysis a) Determination of carbon and hydrogen in coal:  A known amount of coal is burnt in presence of oxygen thereby converting carbon and hydrogen of coal into(i) CO2 (C + O2 → CO2) and (ii) H2O (H2 + ½ O2 → H2O) respectively. The products of combustion CO2 and H2O are passing over weighed tubes of anhydrous CaCl2 and KOH which absorb H2O and CO2 respectively.  The increase in the weight of CaCl2 tube represents the weight of water formed while the increase in the weight of KOH tube represents the weight of CO2 formed.  The percentage of carbon and hydrogen in coal can be calculated in the following way17

Coal Analysis: Ultimate Analysis

18

Coal Analysis: Ultimate Analysis (b) Determination of hydrogen:

19

Coal Analysis: Ultimate Analysis (c) Determination of nitrogen: This is done by Kjeldhal’s method.  A known amount of powdered coal is heated with concentrated sulphuric acid in the presence of potassium sulfate (K2SO4) and copper sulfate (CuSO4) in a long necked Kjeldhal’s flask.  This converts nitrogen of coal to ammonium sulphate. When the clear solution is obtained (i.e., the whole of nitrogen is converted into ammonium sulphate), it is heated with 50 % sodium hydroxide (NaOH) solution and the following reaction occurs:  The ammonia thus formed is distilled over and is absorbed in a

20

known quantity of standard 0.1 N H2SO4 solution. The volume of unused 0.1 N H2SO4 is then determined by titrating against standard NaOH solution. Thus, the amount of acid neutralized by liberated ammonia from coal is determined using the formula.

Coal Analysis: Ultimate Analysis

(d) Determination of sulphur in coal:  A known amount of coal is burnt completely in Bomb calorimeter

in presence of oxygen. Ash thus obtained contains sulphur of coal as sulphate which is extracted with dil. HCl.  The acid extract is then treated with BaCl2 solution to precipitate sulphate as BaSO4. The precipitate is filtered, washed, dried and weighed. From the weight of BaSO4, the percentage of sulphur in coal is calculated in the following way. 21

Coal Analysis: Ultimate Analysis

22

Coal Analysis: Ultimate Analysis

e) Determination of oxygen in coal:  It is calculated indirectly in the following way% of oxygen in coal = 100 - % (C + H + N + S + ash) Significance:  The less the oxygen content, the better is the coal. As the oxygen content increases,  its moisture holding capacity also increases.

23

Coal Analysis Table : Proximate and ultimate analysis of some U.S. coals

24

HEATING VALUE  The heating value, J/kg of fuel, is the heat transferred when the

products of complete combustion of a sample of coal or other fuel are cooled to the initial temperature of air and fuel.  It is determined in a standard test in a bomb calorimeter.  There are two determinations:  The higher heating value (HHV) or gross calorific value (GCV), assumes that the water vapor in the products condenses and thus includes the latent heat of vaporization of the water vapor in the products;  When the latent heat of vaporization of water vapor contained in the combustion products is subtracted from the HHV we get the lower heating value (LHV) or net calorific value (NCV) of fuel.  Therefore, 25

HEATING VALUE LHV = HHV − 𝑚𝑤 ℎ𝑓𝑔

where mw is the mass of water vapour formed given by

𝑚𝑤 = 𝑀 + 9𝐻 + 𝛾𝐴 𝑤𝐴  where M and H are the mass fractions of moisture and hydrogen

in the coal, 𝛾𝐴 is the specific humidity of atmospheric air and 𝑤𝐴 is the actual amount of air supplied per kg of coal.\  If the ultimate analysis is known, the HHV of anthracite and bituminous coals can be determined approximately by using Dulong and Petit formula as given below:

HHV = 33.83 C + 144.45

H −

O 8

+ 9.38 S, in MJ/kg

 where C, H, 0 and S are mass fractions of carbon, hydrogen,

oxygen and sulphur in coal. 26

HEATING VALUE  Assuming the latent heat of vaporization hfg at the partial pressure

of water vapour in the combustion products as 2.395 MJ/kg, the lower heating value of coal is given by LHV = HHV − 2.395 𝑚𝑤 in MJ/kg  For lower-rank fuels the Dulong and Petit formula usually underestimates the HHV.

27

COMBUSTION REACTIONS  Combustion is the high temperature oxidation of the combustible

elements of a fuel with heat release.  In combustion reactions, rapid oxidation of combustible elements of the fuel results in energy release as combustion products are formed.  The three major combustible chemical elements in most common fuels are carbon, hydrogen, and sulfur. Sulfur is usually a relatively unimportant contributor to the energy released, but it can be a significant cause of pollution and corrosion problems.  Combustion is complete when all the carbon present in the fuel is burned to carbon dioxide, all the hydrogen is burned to water, all the sulfur is burned to sulfur dioxide, and all other combustible elements are fully oxidized. When these conditions are not fulfilled, combustion is incomplete. 28

COMBUSTION REACTIONS  The combustible elements in coal and fuel oil are carbon, hydrogen and

sulphur. The basic chemical equations for co1nplete combustion are C + O2 → CO2 2H2 + O2 →2H2O S + O2 → SO2  When insufficient oxygen is present, the carbon will be burned incompletely

with the formation of carbon monoxide. 2C + O2 → 2CO  When dealing with chemical reactions, it is necessary to remember that mass

is conserved, so the mass of the products equals the mass of the reactants.  The total mass of each chemical element must be the same on both sides of

the equation, even though the elements exist in different chemical compounds in the reactants and products.  However, the number of moles of products may differ from the number of

moles of reactants. for example. . . consider the complete combustion of hydrogen with oxygen 29

1H2

1 O + 2 2

→ 1H2 O

COMBUSTION REACTIONS  In this case, the reactants are hydrogen and oxygen. Hydrogen is

30

the fuel and oxygen is the oxidizer. Water is the only product of the reaction.  The numerical coefficients in the equation, which precede the chemical symbols to give equal amounts of each chemical element on both sides of the equation, are called stoichiometric coefficients. In words, 1 1 kmol H2 + kmol O2 → 1 kmol H2 O 2  Note that the total numbers of moles on the left and right sides of are not equal. However, because mass is conserved, the total mass of reactants must equal the total mass of products.  Since 1 kmol of H2 equals 2 kg, 1 kmol of O2 equals 16 kg, and 1/2 kmol of H2O equals 18 kg, the above equation can be interpreted as stating 2 kg H2 + 16 kg O2 → 18 kg H2 O

COMBUSTION REACTIONS  ln order to bum a fuel completely, four basic conditions must be

31

fulfilled: 1. Supply enough air for complete combustion of fuel. 2. Secure enough turbulence for thorough mixing of fuel and air. 3. Maintain a furnace temperature high enough to ignite the incoming fuel air mixture. 4. Provide a furnace volume large enough to allow time for combustion to be completed.  Apart from adequate air supply, the three T's, viz., time, temperature and turbulence have to be kept in mind while designing a furnace.  Since the complete mixing of the fuel and air is virtually impossible, excess air must be supplied to ensure complete combustion. The greater is the rate of mixing or turbulence, the lower would be the excess air required.

FUELS  A fuel is simply a combustible substance.

In this chapter emphasis is on hydrocarbon fuels, which contain hydrogen and carbon. Sulfur and other chemical substances also may be present.  Hydrocarbon fuels are denoted by the

general formula CnHm.  Hydrocarbon fuels exist in all phases, some examples being coal, gasoline, and natural gas  Liquid hydrocarbon fuels are commonly derived from crude oil through distillation and cracking processes. Examples are gasoline, diesel fuel, kerosene, and other types of fuel oils. 32

FUELS  Most liquid fuels are mixtures of hydrocarbons for which

compositions are usually given in terms of mass fractions. For simplicity in combustion calculations, gasoline is often modeled as octane, C8H18, and diesel fuel as dodecane, C12H26.  Gaseous hydrocarbon fuels are obtained from natural gas wells or are produced in certain chemical processes. Natural gas normally consists of several different hydrocarbons, with the major constituent being methane, CH4. The compositions of gaseous fuels are usually given in terms of mole fractions.  The main constituent of coal is carbon. Coal also contains varying amounts of oxygen, hydrogen, nitrogen, sulfur, moisture, and ash. It is difficult to give an exact mass analysis for coal since its composition varies considerably from one geographical area to the next and even within the same geographical location. 33

FUELS  For combustion calculations, the composition of coal is usually 

 



34

expressed as an ultimate analysis. The ultimate analysis gives the composition on a mass basis in terms of the relative amounts of chemical elements (carbon, sulfur, hydrogen, nitrogen, oxygen) and ash. Most liquid hydrocarbon fuels are a mixture of numerous hydrocarbons and are obtained from crude oil by distillation. The most volatile hydrocarbons vaporize first, forming what we know as gasoline. The less volatile fuels obtained during distillation are kerosene, diesel fuel, and fuel oil. The composition of a particular fuel depends on the source of the crude oil as well as on the refinery.

FUELS  Although liquid hydrocarbon fuels are mixtures of many different

hydrocarbons, they are usually considered to be a single hydrocarbon for convenience.  For example, gasoline is treated as octane, C8H18, and the diesel fuel as dodecane, C12H26. Another common liquid hydrocarbon fuel is methyl alcohol, CH3OH, which is also called methanol and is used in some gasoline blends. The gaseous hydrocarbon fuel natural gas, which is a mixture of methane and smaller amounts of other gases, is often treated as methane, CH4, for simplicity.  Natural gas is produced from gas wells or oil wells rich in natural gas. It is composed mainly of methane, but it also contains small amounts of ethane, propane, hydrogen, helium, carbon dioxide, nitrogen, hydrogen sulfate, and water vapor.

35

COMBUSTION AIR  Oxygen is required in every combustion reaction. Pure oxygen is  





used only in special applications such as cutting and welding. In most combustion applications, air provides the needed oxygen. On a mole or a volume basis, dry air is composed of 20.9 percent oxygen, 78.1 percent nitrogen, 0.9 percent argon, and small amounts of carbon dioxide, helium, neon, and hydrogen. In the analysis of combustion processes, the argon in the air is treated as nitrogen, and the gases that exist in trace amounts are disregarded. Then dry air can be approximated as 21 percent oxygen and 79 percent nitrogen by mole numbers. Therefore, each mole of oxygen entering a combustion chamber is accompanied by 0.79/0.21 = 3.76 mol of nitrogen. 1 kmol O2 + 3.76 kmol N2 = 4.76 kmol air

36

COMBUSTION AIR  For the combustion calculations, air is considered to be 21%

37

oxygen and 79% nitrogen on a molar basis. With this idealization the molar ratio of the nitrogen to the oxygen is 0.79/0.21 = 3.76. When air supplies the oxygen in a combustion reaction, therefore, every mole of oxygen is accompanied by 3.76 moles of nitrogen.  We also assume that the nitrogen present in the combustion air does not undergo chemical reaction. That is, nitrogen is regarded as inert.  The nitrogen in the products is at the same temperature as the other products, however, so the nitrogen undergoes a change of state if the products are at a temperature other than the air temperature before combustion.  At very high temperatures, such as those encountered in internal combustion engines, nitrogen can form compounds such as nitric oxide and nitrogen dioxide. Even trace amounts of oxides of nitrogen appearing in the exhaust of internal combustion engines can be a source of air pollution.

AIR–FUEL RATIO  Two parameters that are frequently used to quantify the amounts of fuel

and air in a particular combustion process are the air–fuel ratio and its reciprocal, the fuel–air ratio.  The air–fuel ratio is simply the ratio of the amount of air in a reaction to the amount of fuel. The ratio can be written on a molar basis (moles of air divided by moles of fuel) or on a mass basis (mass of air divided by mass of fuel). Conversion between these values is accomplished using the molecular weights of the air, Mair, and fuel, Mfuel,

 where 𝐴𝐹 is the air–fuel ratio on a molar basis and AF is the ratio on a

mass basis. For the combustion calculations the molecular weight of air is taken as 28.97. 38

THEORETICAL AIR  The minimum amount of air that supplies sufficient oxygen for the

complete combustion of all the carbon, hydrogen, and sulfur present in the fuel is called the theoretical (stoichiometric) amount of air.  For complete combustion with the theoretical amount of air, the products would consist of carbon dioxide, water, sulfur dioxide, the nitrogen accompanying the oxygen in the air, and any nitrogen contained in the fuel. No free oxygen would appear in the products.

 for example.. . let us determine the theoretical amount of air for the complete combustion of methane. For this reaction, the products contain only carbon dioxide, water, and nitrogen. The reaction is  where a, b, c, and d represent the numbers of moles of oxygen,

carbon dioxide, water, and nitrogen. In the reaction equation above 3.76 moles of nitrogen are considered to accompany each mole of oxygen. 39

THEORETICAL AIR  Applying the conservation of mass principle to the carbon, hydrogen,

oxygen, and nitrogen, respectively, results in four equations among the four unknowns

 Solving these equations, the balanced chemical equation is  The coefficient 2 before the term (O2 + 3.76N2) is the number of moles

of oxygen in the combustion air, per mole of fuel, and not the amount of air. The amount of combustion air is 2 moles of oxygen plus 2 x 3.76 moles of nitrogen, giving a total of 9.52 moles of air per mole of fuel.  Thus, for the reaction given, the air–fuel ratio on a molar basis is 9.52. 40

THEORETICAL AIR  The air–fuel ratio on a mass basis can be calculated as

 Normally the amount of air supplied is either greater or less than the

theoretical amount.  The amount of air actually supplied is commonly expressed in terms of the percent of theoretical air. For example, 150% of theoretical air means that the air actually supplied is 1.5 times the theoretical amount of air.  The amount of air supplied can be expressed alternatively as a percent excess or a percent deficiency of air. Thus, 150% of theoretical air is equivalent to 50% excess air, and 80% of theoretical air is the same as a 20% deficiency of air.

41

THEORETICAL AIR  for example.. . consider the complete combustion of methane with

150% theoretical air (50% excess air). The balanced chemical reaction equation is  In this equation, the amount of air per mole of fuel is 1.5 times the

theoretical amount determined previously. Accordingly, the air–fuel ratio is 1.5 times higher.  Since complete combustion is assumed, the products contain only carbon dioxide, water, nitrogen, and oxygen. The excess air supplied appears in the products as uncombined oxygen and a greater amount of nitrogen than the equation based on the theoretical amount of air.  The equivalence ratio is the ratio of the actual fuel–air ratio to the

fuel–air ratio for complete combustion with the theoretical amount of air. The reactants are said to form a lean mixture when the equivalence ratio is less than unity. When the ratio is greater than unity, the reactants are said to form a rich mixture. 42

E X A M P L E 1. Determining the Air–Fuel Ratio  Determine the air–fuel ratio on both a molar and mass basis for the

complete combustion of octane, C8H18, with (a) the theoretical amount of air, (b) 150% theoretical air (50% excess air).

Analysis:  (a) For complete combustion of C8H18 with the theoretical amount of

air, the products contain carbon dioxide, water, and nitrogen only. That is  Applying the conservation of mass principle to the carbon, hydrogen,

oxygen, and nitrogen, respectively, gives

43

E X A M P L E 1. Determining the Air–Fuel Ratio  Solving these equations, a = 12.5, b = 8, c = 9, d = 47. The

balanced chemical equation is  The air–fuel ratio on a molar basis is

 The air–fuel ratio expressed on a mass basis is

 For 150% theoretical air, the chemical equation for complete

combustion takes the form 44

E X A M P L E 1. Determining the Air–Fuel Ratio  Applying conservation of mass

 Solving this set of equations, b = 8, c = 9, d = 70.5, e = 6.25, giving a

balanced chemical equation  The air–fuel ratio on a molar basis is

 On a mass basis, the air–fuel ratio is 22.6 kg (air) / kg (fuel), as can be

45

verified.  Note: When complete combustion occurs with excess air, oxygen appears in the products, in addition to carbon dioxide, water, and nitrogen.

Example 2  A certain coal has the following analysis on a mass basis: 82

percent C, 5 percent H2O, 2 percent H2, 1 percent O2, and 10 percent ash. The coal is burned with 50 percent excess air. Determine the air–fuel ratio.  Analysis: The composition of the coal is given on a mass basis, but we need to know the composition on a mole basis to balance the combustion equation.  Considering 1 kg of coal, the numbers of mole of the each component are determined to be

46

Example 2

47

Dew Point Temperature  The saturation temperature at the partial pressure of water vapour is

called the dew point temperature (d.p.t.).  The partial pressure of water vapour in the mixture of gases constituting the flue gas is given by 𝑃𝐻2 𝑂 = 𝑥𝐻2 𝑂 𝑃  where P is the total pressure of the exhaust gas mixture and is 𝑥𝐻2 𝑂 is

the mole fraction of water vapour in the exhaust gas mixture 𝑛𝐻2 𝑂 𝑥𝐻2 𝑂 = 𝑛𝑡𝑜𝑡𝑎𝑙  The flue gases arc cooled in heat exchangers like economizer and air

48

preheater so as to minimize the exhaust losses through chimney. These gases, however, should never be cooled below the dew point temperature. If cooled below the d.p.t., the water vapour condenses into liquid droplets which react with S02 or S03 to form acid. This acid corrodes the metal surfaces of the ducts through which the flue gas flows

Example 3  Ethane (C2H6) is burned with 20 percent excess air during a

combustion process, as shown in Fig. below. Assuming complete combustion and a total pressure of 100 kPa, determine (a) the air– fuel ratio and (b) the dew-point temperature of the products

 Solution The fuel is burned completely with excess air. The AF

49

and the dew point of the products are to be determined.  Assumptions 1 Combustion is complete. 2 Combustion gases are ideal gases.  Analysis The combustion products contain CO2, H2O, N2, and some excess O2 only. Then the combustion equation can be written as

Example 3

 where ath is the stoichiometric coefficient for air. We have

automatically accounted for the 20 percent excess air by using the factor 1.2ath instead of ath for air.  The stoichiometric amount of oxygen (athO2) is used to oxidize the fuel, and the remaining excess amount (0.2athO2) appears in the products as unused oxygen. Notice that the coefficient of N2 is the same on both sides of the equation, and that we wrote the C and H balances directly since they are so obvious.  The coefficient ath is determined from the O2 balance to be

50

Example 3  (a) The air–fuel ratio is determined by taking the ratio of the mass

of the air to the mass of the fuel,

 That is, 19.3 kg of air is supplied for each kilogram of fuel during

51

this combustion process.  (b) The dew-point temperature of the products is the temperature at which the water vapor in the products starts to condense as the products are cooled at constant pressure.  The dew-point temperature of a gas–vapor mixture is the saturation temperature of the water vapor corresponding to its partial pressure. Therefore, we need to determine the partial pressure of the water vapor Pv in the products first. Assuming ideal-gas behavior for the combustion gases, we have

Example 3

52

EXAMPLE 4. Combustion of a Gaseous Fuel with Moist Air  A certain natural gas has the following volumetric analysis: 72

percent CH4, 9 percent H2, 14 percent N2, 2 percent O2, and 3 percent CO2. This gas is now burned with the stoichiometric amount of air that enters the combustion chamber at 20°C, 1 atm, and 80 percent relative humidity. Assuming complete combustion and a total pressure of 1 atm, determine the dewpoint temperature of the products.

53

EXAMPLE 4. Combustion of a Gaseous Fuel with Moist Air  Assumptions 1 The fuel is burned completely and thus all the carbon in

the fuel burns to CO2 and all the hydrogen to H2O. 2 The fuel is burned with the stoichiometric amount of air and thus there is no free O2 in the product gases. 3 Combustion gases are ideal gases.  Properties The saturation pressure of water at 20oC is 2.3392 kPa  Analysis We note that the moisture in the air does not react with anything; it simply shows up as additional H2O in the products. Therefore, for simplicity, we balance the combustion equation by using dry air and then add the moisture later to both sides of the equation.

54

EXAMPLE 4. Combustion of a Gaseous Fuel with Moist Air

55

EXAMPLE 4. Combustion of a Gaseous Fuel with Moist Air

56

EXAMPLE 4. Combustion of a Gaseous Fuel with Moist Air

57

DETERMINING PRODUCTS OF COMBUSTION  Combustion is the result of a series of very complicated and rapid 







58

chemical reactions, and the products formed depend on many factors. When fuel is burned in the cylinder of an internal combustion engine, the products of the reaction vary with the temperature and pressure in the cylinder. In combustion equipment of all kinds, the degree of mixing of the fuel and air is a controlling factor in the reactions that occur once the fuel and air mixture is ignited. Although the amount of air supplied in an actual combustion process may exceed the theoretical amount, it is not uncommon for some carbon monoxide and unburned oxygen to appear in the products. This can be due to incomplete mixing, insufficient time for complete combustion, and other factors. When the amount of air supplied is less than the theoretical amount of air, the products may include both CO2 and CO, and there also may be unburned fuel in the products.

DETERMINING PRODUCTS OF COMBUSTION  The products of combustion of an actual combustion process and

their relative amounts can be determined only by measurement.  Among several devices for measuring the composition of products of combustion are the Orsat analyzer, gas chromatograph, infrared analyzer, and flame ionization detector. Data from these devices can be used to determine the mole fractions of the gaseous products of combustion.  The analyses from these devices are often reported on a “dry” basis. In a dry product analysis, the mole fractions are given for all gaseous products except the water vapor.  If the gaseous products of combustion are cooled at constant mixture pressure, the dew point temperature is reached when water vapor begins to condense. Since water deposited on duct work, mufflers, and other metal parts can cause corrosion, knowledge of the dew point temperature is important 59

Example 5  A natural gas has the following molar analysis:

CH4, 80.62%; C2H6, 5.41%; C3H8, 1.87%; C4H10, 1.60%; N2, 10.50%. The gas is burned with dry air, giving products having a molar analysis on a dry basis: CO2, 7.8%; CO, 0.2%; O2, 7%; N2, 85%. (a) Determine the air–fuel ratio on a molar basis. (b) Assuming ideal gas behavior for the fuel mixture, determine the amount of products in kmol that would be formed from 100 m3 of fuel mixture at 300 K and 1 bar. (c) Determine the percent of theoretical air.

60

Example 5 Analysis:  (a) The solution can be conducted on the basis of an assumed amount

of fuel mixture or on the basis of an assumed amount of dry products. Let us illustrate the first procedure, basing the solution on 1 kmol of fuel mixture. The chemical equation then takes the form

 The products consist of b kmol of dry products and c kmol of water

vapor, each per kmol of fuel mixture.  Applying conservation of mass to carbon Solving gives b = 12.931.  Conservation of mass for hydrogen results in 61

which gives c = 1.93.

Example 5  The unknown coefficient a can be found from either an oxygen

balance or a nitrogen balance. Applying conservation of mass to oxygen giving a = 2.892.  The balanced chemical equation is then

 The air–fuel ratio on a molar basis is

62

Example 5  (b) By inspection of the chemical reaction equation, the total amount

of products is b + c = 12.931 + 1.93 = 14.861 kmol of products per kmol of fuel. The amount of fuel in kmol, nF, present in 100 m3 of fuel mixture at 300 K and 1 bar can be determined from the ideal gas equation of state as

 Accordingly, the amount of product mixture that would be formed

from 100 m3 of fuel mixture is (14.861)(4.01) = 59.59 kmol of product gas  (c) The balanced chemical equation for the complete combustion of the fuel mixture with the theoretical amount of air is 63

Example 5  The theoretical air–fuel ratio on a molar basis is

 The percent theoretical air is then

64

Example 6  Octane (C8H18) is burned with dry air. The volumetric analysis of

the products on a dry basis is

 Determine (a) the air–fuel ratio, (b) the percentage of theoretical

air used, and (c) the amount of H2O that condenses as the products are cooled to 25°C at 100 kPa.  Solution Combustion products whose composition is given are

cooled to 25°C. The AF, the percent theoretical air used, and the fraction of water vapor that condenses are to be determined.  Assumptions Combustion gases are ideal gases.  Properties The saturation pressure of water at 25°C is 3.1698 kPa 65

Example 6  Analysis: Note that we know the relative composition of the

products, but we do not know how much fuel or air is used during the combustion process. However, they can be determined from mass balances. The H2O in the combustion gases will start condensing when the temperature drops to the dewpoint temperature.  For ideal gases, the volume fractions are equivalent to the mole fractions. Considering 100 kmol of dry products for convenience, the combustion equation can be written as

66

Example 6  The O2 balance is not necessary, but it can be used to check the

values obtained from the other mass balances, as we did previously. Substituting, we get

67

Example 6

68

Example 6

69

Enthalpy of Combustion and Enthalpy of Formation  Enthalpy of combustion hC, which represents the amount of heat

released during a steady-flow combustion process when 1 kmol (or 1 kg) of fuel is burned completely at a specified temperature and pressure . It is expressed as  The enthalpy of combustion of a particular fuel is different at

70

different temperatures and pressures.  Since there are so many different fuels and fuel mixtures it is not practical to list hC values for all possible cases. Besides, the enthalpy of combustion is not of much use when the combustion is incomplete.  Therefore a more practical approach would be to have a more fundamental property to represent the chemical energy of an element or a compound at some reference state. This property is the enthalpy of formation ℎത𝑓 , which can be viewed as the enthalpy of a substance at a specified state due to its chemical composition.

Enthalpy of Combustion and Enthalpy of Formation  The enthalpy of formation of all stable

elements (such as O2, N2, H2, and C) is taken as zero at the standard reference state of 25°C and 1 atm. That is, ℎത𝑓 = 0 for all stable elements.  Consider the formation of CO2 (a compound) from its elements C and O2 at 25°C and 1 atm during a steady-flow process. The enthalpy change during this process is -393,520 kJ/kmol. However, Hreact = 0 since both reactants are elements at the standard reference state, and the products consist of 1 kmol of CO2 at the same state. Therefore, the enthalpy of formation of CO2 at the standard reference state is - 393,520 kJ/kmol. That is, 71

Fig. The enthalpy of formation of a compound represents the amount of energy absorbed or released as the component is formed from its stable elements during a steady-flow process at a specified state.

Enthalpy of Combustion and Enthalpy of Formation  The negative sign is due to the fact that the enthalpy of 1 kmol of CO2

at 25°C and 1 atm is 393,520 kJ less than the enthalpy of 1 kmol of C and 1 kmol of O2 at the same state. In other words, 393,520 kJ of chemical energy is released (leaving the system as heat) when C and O2 combine to form 1 kmol of CO2.  Therefore, a negative enthalpy of formation for a compound indicates that heat is released during the formation of that compound from its stable elements. A positive value indicates heat is absorbed.  Another term commonly used in conjunction with the combustion of fuels is the heating value of the fuel, which is defined as the amount of heat released when a fuel is burned completely in a steady-flow process and the products are returned to the state of the reactants. In other words, the heating value of a fuel is equal to the absolute value of the enthalpy of combustion of the fuel. That is,

72

Enthalpy of Combustion and Enthalpy of Formation  The heating value depends on the phase of

the H2O in the products.  The heating value is called the higher

heating value (HHV) when the H2O in the products is in the liquid form, and it is called the lower heating value (LHV) when the H2O in the products is in the vapor form. The two heating values are related by  where m is the mass of H2O in the

products per unit mass of fuel and hfg is the enthalpy of vaporization of water at the specified temperature.  The heating value or enthalpy of

73

combustion of a fuel can be determined from a knowledge of the enthalpy of formation for the compounds involved.

Example 7: Evaluation of the Enthalpy of Combustion  Determine the enthalpy of combustion of liquid octane

(C8H18) at 25°C and 1 atm, using enthalpy-of-formation data from Enthalpy of formation Table. Assume the water in the products is in the liquid form.  Properties The enthalpy of formation at 25°C and 1 atm is 393,520 kJ/kmol for CO2, -285,830 kJ/kmol for H2O(l), and 249,950 kJ/kmol for C8H18 (l)  Analysis The stoichiometric equation for combustion of C8H18

is  Both the reactants and the products are at the standard

reference state of 25°C and 1 atm. Also, N2 and O2 are stable elements, and thus their enthalpy of formation is zero. Then the enthalpy of combustion of C8H18 becomes 74

Example 7: Evaluation of the Enthalpy of Combustion

 Since the water in the products is assumed to be in the liquid phase,

this hC value corresponds to the HHV of liquid C8H18.  It can be shown that the result for gaseous octane is -5,512,200 kJ/kmol or -48,255 kJ/kg.  For fuels that exhibit considerable variations in composition depending on the source, such as coal, natural gas, and fuel oil, it is more practical to determine their enthalpy of combustion experimentally by burning them directly in a bomb calorimeter at constant volume or in a steady75 flow device.

FIRST-LAW ANALYSIS OF REACTING SYSTEMS

Steady-Flow Systems  The specific enthalpy of a compound at a state other than the

standard state is found by adding the specific enthalpy change ∆ℎത between the standard state and the state of interest to the enthalpy of formation  That is, the enthalpy of a compound is composed of

associated with the formation of the compound from its elements, and ∆ℎത associated with a change of state at constant composition. ത since it  An arbitrary choice of datum can be used to determine ∆ℎ is a difference at constant composition. Accordingly, ∆ℎത can be evaluated from tabular sources such as the steam tables, the ideal gas tables when appropriate, and so on. 76

FIRST-LAW ANALYSIS OF REACTING SYSTEMS  When the changes in kinetic and potential energies are negligible,

the steady-flow energy balance relation 𝐸ሶ 𝑖𝑛 = 𝐸ሶ 𝑜𝑢𝑡 can be expressed for a chemically reacting steady-flow system more explicitly as

 In combustion analysis, it is more convenient to work with

quantities expressed per mole of fuel.

77

FIRST-LAW ANALYSIS OF REACTING SYSTEMS  where Nr and Np represent the number of moles of the reactant r

and the product p, respectively, per mole of fuel. Note that Nr = 1 for the fuel, and the other Nr and Np values can be picked directly from the balanced combustion equation.  Taking heat transfer to the system and work done by the system to be positive quantities, the energy balance relation just discussed can be expressed more compactly as

78

FIRST-LAW ANALYSIS OF REACTING SYSTEMS  A combustion chamber normally involves heat output but no heat

input. Then the energy balance for a typical steady-flow combustion process becomes

79

EXAMPLE 8. First-Law Analysis of Steady-Flow Combustion  Liquid propane (C3H8) enters a combustion chamber at 25°C at a

rate of 0.05 kg/min where it is mixed and burned with 50 percent excess air that enters the combustion chamber at 7°C, as shown in the Fig. An analysis of the combustion gases reveals that all the hydrogen in the fuel burns to H2O but only 90 percent of the carbon burns to CO2, with the remaining 10 percent forming CO. If the exit temperature of the combustion gases is 1500 K, determine (a) the mass flow rate of air and (b) the rate of heat transfer from the combustion chamber.

80

EXAMPLE 8. First-Law Analysis of Steady-Flow Combustion

Analysis  We note that all the hydrogen in the fuel burns to H2O but 10

percent of the carbon burns incompletely and forms CO. Also, the fuel is burned with excess air and thus there is some free O2 in the product gases.  The theoretical amount of air is determined from the stoichiometric reaction to be

81

EXAMPLE 8. First-Law Analysis of Steady-Flow Combustion

82

EXAMPLE 8. First-Law Analysis of Steady-Flow Combustion

83

EXAMPLE 8. First-Law Analysis of Steady-Flow Combustion

84

ADIABATIC FLAME TEMPERATURE  In the absence of any work interactions and any changes in kinetic

or potential energies, the chemical energy released during a combustion process either is lost as heat to the surroundings or is used internally to raise the temperature of the combustion products. The smaller the heat loss, the larger the temperature rise.  In the limiting case of no heat loss to the surroundings (Q = 0), the temperature of the products reaches a maximum, which is called the adiabatic flame or adiabatic combustion temperature of the reaction Fig. The temperature of a combustion chamber becomes maximum when combustion is complete and no heat is lost to the surroundings (Q = 0). 85

ADIABATIC FLAME TEMPERATURE  The adiabatic flame temperature of a steady-flow combustion

process is determined by setting Q = 0 and W = 0. It yields

 Once the reactants and their states are specified, the enthalpy of

the reactants Hreact can be easily determined. The calculation of the enthalpy of the products Hprod is not so straightforward, however, because the temperature of the products is not known prior to the calculations.  Therefore, the determination of the adiabatic flame temperature requires the use of an iterative technique unless equations for the sensible enthalpy changes of the combustion products are available. 86

ADIABATIC FLAME TEMPERATURE  A temperature is assumed for the product gases, and the Hprod is







 87

determined for this temperature. If it is not equal to Hreact, calculations are repeated with another temperature. The adiabatic flame temperature is then determined from these two results by interpolation. When the oxidant is air, the product gases mostly consist of N2, and a good first guess for the adiabatic flame temperature is obtained by treating the entire product gases as N2. In combustion chambers, the highest temperature to which a material can be exposed is limited by metallurgical considerations. Therefore, the adiabatic flame temperature is an important consideration in the design of combustion chambers, gas turbines, and nozzles. The maximum temperatures that occur in these devices are considerably lower than the adiabatic flame temperature, however, since the combustion is usually incomplete, some heat loss takes place, and some combustion gases dissociate at high temperatures The maximum temperature in a combustion chamber can be controlled by adjusting the amount of excess air, which serves as a coolant.

Example 9. Adiabatic Flame Temperature in Steady Combustion  Liquid octane (C8H18) enters the combustion chamber of a gas

turbine steadily at 1 atm and 25°C, and it is burned with air that enters the combustion chamber at the same state, as shown in Fig. below. Determine the adiabatic flame temperature for (a) complete combustion with 100 percent theoretical air, (b) complete combustion with 400 percent theoretical air, and (c) incomplete combustion (some CO in the products) with 90 percent theoretical air.

88

Example 9. Adiabatic Flame Temperature in Steady Combustion

89

Example 9. Adiabatic Flame Temperature in Steady Combustion

 It appears that we have one equation with three unknowns. Actually

we have only one unknown—the temperature of the products Tprod since h = h(T ) for ideal gases. Therefore, we have to use an equation solver programs or a trial-and-error approach to determine the temperature of the products 90

Example 9. Adiabatic Flame Temperature in Steady Combustion  A first guess is obtained by dividing the right-hand side of the

equation by the total number of moles, which yields 5,646,081/(8 + 9 + 47) = 88,220 kJ/kmol. This enthalpy value corresponds to about 2650 K for N2, 2100 K for H2O, and 1800 K for CO2.  Noting that the majority of the moles are N2, we see that Tprod should be close to 2650 K, but somewhat under it. Therefore, a good first guess is 2400 K. At this temperature,

 This value is higher than 5,646,081 kJ. Therefore, the actual

temperature is slightly under 2400 K. Next we choose 2350 K. It yields  which is lower than 5,646,081 kJ. Therefore, the actual temperature 91

of the products is between 2350 and 2400 K. By interpolation, it is found to be Tprod = 2395 K.

Example 9. Adiabatic Flame Temperature in Steady Combustion

92

Exercises 1.

2.

3.

93

Propylene (C3H6) is burned with 50 percent excess air during a combustion process. Assuming complete combustion and a total pressure of 105 kPa, determine (a) the air–fuel ratio and (b) the temperature at which the water vapor in the products will start condensing. One kilogram of butane (C4H10) is burned with 25 kg of air that is at 30°C and 90 kPa. Assuming that the combustion is complete and the pressure of the products is 90 kPa, determine (a) the percentage of theoretical air used and (b) the dew-point temperature of the products. A certain natural gas has the following volumetric analysis: 65 percent CH4, 8 percent H2, 18 percent N2, 3 percent O2, and 6 percent CO2. This gas is now burned completely with the stoichiometric amount of dry air. What is the air–fuel ratio for this combustion process?

Exercises 4.

5.

6.

94

A certain coal has the following analysis on a mass basis: 82 percent C, 5 percent H2O, 2 percent H2, 1 percent O2, and 10 percent ash. The coal is burned with 50 percent excess air. Determine the air–fuel ratio. Methane (CH4) is burned with dry air. The volumetric analysis of the products on a dry basis is 5.20 percent CO2, 0.33 percent CO, 11.24 percent O2, and 83.23 percent N2. Determine (a) the air–fuel ratio and (b) the percentage of theoretical air used. Liquid propane (C3H8) enters a combustion chamber at 25°C at a rate of 1.2 kg/min where it is mixed and burned with 150 percent excess air that enters the combustion chamber at 12°C. If the combustion is complete and the exit temperature of the combustion gases is 1200 K, determine (a) the mass flow rate of air and (b) the rate of heat transfer from the combustion chamber.

Exercises 7.

8.

95

Ethane gas (C2H6) at 25°C is burned in a steady-flow combustion chamber at a rate of 5 kg/h with the stoichiometric amount of air, which is preheated to 500 K before entering the combustion chamber. An analysis of the combustion gases reveals that all the hydrogen in the fuel burns to H2O but only 95 percent of the carbon burns to CO2, the remaining 5 percent forming CO. If the products leave the combustion chamber at 800 K, determine the rate of heat transfer from the combustion chamber. Diesel fuel (C12H26) at 25°C is burned in a steady flow combustion chamber with 20 percent excess air that also enters at 25°C. The products leave the combustion chamber at 500 K. Assuming combustion is complete, determine the required mass flow rate of the diesel fuel to supply heat at a rate of 2000 kJ/s.

Exercises Acetylene gas (C2H2) at 25°C is burned during a steady-flow combustion process with 30 percent excess air at 27°C. It is observed that 75,000 kJ of heat is being lost from the combustion chamber to the surroundings per kmol of acetylene. Assuming combustion is complete, determine the exit temperature of the product gases. 10. Hydrogen (H2) at 7°C is burned with 20 percent excess air that is also at 7°C during an adiabatic steady-flow combustion process. Assuming complete combustion, determine the exit temperature of the product gases. 9.

96

End of Lecture 3 Next Lecture Lecture 4: Steam Generators (Boilers)

97