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UNIT 10 STOICHIOMETRY

Structure

10.1 Introduction

Objectives

10.2 Fundamental Physical Laws used for Stoichiometric Analysis

10.2.1 Conservation of Mass 10.2.2 Conservation of Energy 10.2.3 Law of Combination Weights 10.2.4 Ideal Gas Law

10.2.5 Avogadro’s Law 10.2.6 Dalton’s Law 10.2.7 Amagat’s Law

10.3 Application of Fundamental Laws for Stoichiometric Analysis

10.3.1 Stoichiometric Analysis for Combination with Oxygen 10.3.2 Stoichiometric Analysis for Combination with Air 10.3.3 Stoichiometric Analysis for Combination of Liquid Fuels 10.3.4 Stoichiometric Analysis for Combination of Gaseous Fuels 10.3.5 Stoichiometric Analysis for Combination of Coal and Solid Fuels 10.3.6 Stoichiometric Analysis for Combination with Insufficient Air

10.4 Theoretical Air Requirement for the Combustion of a Fuel 10.5 Summary

10.6 Key Words 10.7 Answers to SAQs

10.1 INTRODUCTION

Stoichiometry deals with the laws of conservation of mass and energy. For stoichiometric analysis of a system, stoichiometric calculations are done and these calculations deal with the weights of materials and quantities of energy entering and leaving the chemical reactions. Fundamental stoichiometric calculations dealing with combustion are idealized chemical reactions

wherein fuel and air combine chemically to form the products of combustion.

Actually, burning process is extremely complicated and no theory completely explains all the phenomena of combustion.

When the fuel is burned very rapidly under high pressure as in the cylinder of an internal-combustion engine, the products of reaction vary with the physical conditions of temperature and pressure in the cylinder. In other type of combustion equipment, the degree of mixing of the fuel and air is a

controlling factor in the reactions which occur, once the fuel and air mixture is ignited. What is more, it is thought that all the burning is the result of a series of very complicated and rapid chemical reactions.

In this unit, however, it is assumed that the fuel is completely mixed with air or oxygen, and that combustion proceeds directly to the simple end products of the complete chemical reaction, namely, carbon dioxide and water. Later, the effects of incomplete mixing of fuel and air, high temperatures and pressures of combustion gases, and rapid cooling of these gases have been discussed. The simplified reactions used in the stoichiometric calculations which follow, are found to yield results which are sufficiently accurate for most types of design or efficiency computations.

Objectives

After studying this unit, you should be able to understand the

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 function of fundamental laws in stoichiometric analysis of the combustion,

 stoichiometric calculations with oxygen or air and

 methods of finding the composition of combustion products, air-fuel ratio, etc., for combustion of different types of fuel.

10.2 FUNDAMENTAL PHYSICAL LAWS USER FOR STOICHIOMETRIC ANALYSIS

The calculations used involved34 combustion processes are comparatively simple, once the physical principles involved are thoroughly understood. The need for a thoroughly understanding of the seven laws enumerated below cannot be emphasized too strongly. These laws are the promises upon which all calculations that follow are based. In their simplest form, the seven fundamental physical laws of immediate concern are:

(i) Conservation of Mass: Matter can neither be destroyed not created.

(ii) Conservation of Energy: Energy can neither be destroyed nor created.

(iii) Law of Combining Weights: All substances combine in accordance with sample, definite weight relationship.

(iv) Ideal Gas Law: The volume of an ideal gas is directly proportional to its absolute temperature and inversely proportional to the absolute pressure.

(v) Avogadro’s Law: Equal volumes of perfect gases under identical conditions of temperature and pressure have the same number of molecules.

(vi) Dalton’s Law: The total pressure of a mixture of gases is the sum of the partial pressures which would be exerted by each of the

constituents if each gas were to occupy alone the same volume as that of the mixture.

(vii) Amagat’s Law: The total volume occupied by a mixture of gases is equal to the sum of the volume which would be occupied by each constituents when at the same temperature and the pressure as the mixture.

10.2.1 Conservation of Mass

The total mass of material that enters into a process remain unchanged, even though there may be some rearrangement of atom or molecules. In the normal thermodynamic processes associated with the best power field, this concept is easily understood because the molecules remain unchanged.

In the combustion process, the atoms rearrange themselves to form new molecules and liberate chemical energy in doing so. However, every atom in the material that leaves, there is no change of weight during the process.

The concept of the conservation of mass may be illustrated by an internal combustion engine operating on an air/fuel ratio of 15 to 1 by weight. In burning 1 kg of fuel, 15 kg of air is used and 16 kg of exhaust products are formed. The chemical energy in 1 kg of fuel is released as heat energy which is then converted to mechanical work or dissipated as heat.

Hydrogen burns to form water according to the following chemical equation.

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2H2 + O2 = 2H2O ...(10.1)

Two molecules of hydrogen plus one molecule of oxygen react to give the end product two molecules of water. A 'material balance' of the equation indicates that four atoms of hydrogen and two atoms of oxygen enter into reaction, although their grouping or combination is different. Likewise, the sum of the weights of hydrogen and oxygen entering the combustion process must equal the weight of water formed. As long as there are no change in the atomic structures of the elements involved, the material balance must exist.

10.2.2 Conservation of Energy

The sum of all energy entering a process is exactly equal to the sum of energy leaving the process. The total of the potential, kinetic, heat, chemical and electrical energy in the material entering must appear in the material leaving, although the proportionate amounts of each form of energy may change. The combustion process is characterized by the change of chemical energy into heat energy.

10.2.3 Law of Combining Weights

It has been found experimentally that elements combine in simple and constant proportions to form definite compounds. These proportions by weight are always exact ratios of molecular weights of the constituents entering into the reaction. Because of this method of combination, it is convenient to state chemical equations in terms of the number of molecular weights, or moles, of the elements or compounds in the reaction.

The molecular weights are based on an atomic weight of oxygen i.e., 16.00, so chosen because the atomic weight of lightest element, hydrogen, is then approximately 1. The atomic weight of carbon is 12, because the carbon atom is 3/4 as heavy as the atom. The atomic weights of some common elements used in combustion calculations are listed in Table 10.1.

Table 10.1: Atomic Weights of Common Elements Element Exact Atomic

Weight

Approximate Atomic Weight

Hydrogen 1.0080 1

Carbon 12.01 12

Nitrogen 14.008 14

Oxygen 16.00 16

Sulphur 32.06 32

It is seen from the values listed in the table that the approximate atomic weights are close to the exact values and hence can be used in most cases with sufficient accuracy.

A term called the gram-molecular weight, or gram-mole, is applied to the mass of a substance which has a molecular weight in gm equal to its molecular weight. Since the hydrogen has a molecular weight of 2, a gm- mole of hydrogen weights 2 gm. The gm-mole is a very convenient term to use in combustion calculations because the elements combine in simple and constant proportions. A better understanding of its advantages and use will be made clear in the calculations which follow later in the unit.

10.2.4 Ideal Gas Law

Ideal gas law, in equation form, is

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PV = NRT ... (10.2) where, P = Absolute pressure,

V = Volume,

N = No. of moles of gas, R = Universal gas constant, and T = Absolute temperature.

This law is the statement of the physical phenomena, first observed by Boyle and Charles. For I mole of gas, Eq. (10.2) becomes

= 𝑅…(10.3)

The universal gas constant R has been determined very accurately for many gases and approximately 1.987 cal. mole per °C. It should be noted that the specific gas constant, whose units are cal-mole/gm °C, is equal to the universal gas constant divided by the molecular weight of gas. Thus, for oxygen which has a molecular weight of 32 gm per

mole, the specific gas constant 𝑅, is

…(10.4)

= . = 0.062 cal/gm °C

= 0.08206 Litre-atm/degree-mole

While the universal gas constant does not vary appreciably, the specific gas constant for each gas is different since the gases have different weights.

Oxygen has a molecular weight of 32. Therefore, 1 mole of oxygen weights 32 gm, and the volume of 1 mole of oxygen at the scientific standard temperature and pressure (273 K and 1 atm) would be

V = = × . ×

.

= 22.4 litres

A gm-mole of any ideal gas must occupy the same volume at standard conditions since all terms in the ideal-gas equation are constant at the same temperature and pressure.

Appreciable deviations from the ideal gas law will occur with real gases under conditions of high pressure or low temperature, that is, in the vaporous region near saturation conditions for the gas. For highly superheated gases at normal pressures, the deviations are usually insignificant, and the ideal gas law can be applied with sufficient accuracy for combustion calculations.

10.2.5 Avogadro's Law

From the preceding discussion it has been seen that the volume occupied by 1 mole of any ideal gas is same under like conditions of temperature and pressure. Also, the weight of a mole of a gas is directly dependent upon the weight of the molecules making up the gas. It therefore, follows that a mole of any gas always has a fixed number of molecules.

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Avogadro stated this relationship. It has been further proved that a mole of gas contains 2.7 x 1026 molecules. The full importance of Avogadro's law is not readily recognized. However, in the calculations which follow, the volume and mole relationship should become evident as the law is applied.

10.2.6 Dalton's Law

In a mixture of gases, each gas fills the entire volume. In Figure 10.1, three gases considered - first individually, and then as a mixture.

Gas A + Gas B + Gas C = Mixture P = 1 atm P = 1 atm P = 1atm P = 3 atm

V = 22.4

litre V = 22.4 litre V = 22.4

litre V = 22.4 litre N = 1 mole N = 1 mole N = 1 mole N = 3 moles

T = 273 K T = 273 K T = 273 K T = 273 K Figure 10.1 : Partial Pressure of a Gaseous Mixture

If gases A, B and C are 1 mole each of CO2, N2 and O2, respectively, their weights will be 44, 28, and 32 gm in the order given. The 3 moles of gases, when combined, will have a total weight of 44 + 28 +32 = 104 gm. If each gas exerts a pressure of 1 atm. alone, it will exert the same pressure on the vessel containing the mixture if the final volume and temperature are the same. Thus the total pressure (3 atm) of the mixture is equal to the sum of the pressures created by each constituent.

The above is an example of Dalton's law of partial pressure which states that the total pressure of a gaseous mixture is the sum of the partial pressures which would be exerted by each of the constituent if each gas were to occupy alone the volume as the mixture. Mathematically it can be represented as follows:

PT = PA + PB + PC …(10.5)

where, PT is the total pressure, and PA, PB and PC are the partial pressures exerted by gases A, B and C.

If the ideal gas law is applied to gas-A, then PAV = NART ... (10.6)

For the mixture, the same law states, PT V = NT R T ...(10.7)

By dividing Eq. (10.6) by Eq. (10.7) we get =

or, = …(10.8)

Solving for the partial pressure of Gas A and substituting the numerical values, we get the following:

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PA = × PT = × 3 = 1 atm

The fraction NA/NT is often called the mole fraction. The partial pressure of a gas present in the gaseous mixture is equal to the mole fraction of that gas times the total pressure of the mixture. Although Dalton's law of partial pressure can be applied in principle to real gases, the expression in Eq. (10.7) is precise only for ideal gases because it was derived from the ideal gas law.

However, Eq. (10.7) may be used for all gases within the limits of accuracy required for combustion calculations.

Example 10.1

A tank having a volume of 100 litres contains 44 gm of carbon dioxide, 120 gm of nitrogen and 25 gm of oxygen. What pressure will be indicated by a gauge attached to the tank if the gas temperature is 55°C?

Solution

For CO2:

N = . = = 1 mole

𝑃 = = × . × = 0.269 atm

For N2:

N = = 4.29 moles

𝑃 = . × . × = 1.1546

For O2:

N = = 0.78 moles

𝑃 = . × . × = 0.2098

Therefore, total pressure is

PT = 1.633 atm

and the gas will indicate the pressure equivalent to PT - 1

= 0.633 atm.

Alternative Method

NT = 𝑁 + 𝑁 + 𝑁 = + + = 6.07 moles Now, PT V = NT RT or, PT = . × . × = 1.633 atm.

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10.2.7 Amagat's Law

Amagat's law is similar to Dalton's in that it concerns gaseous mixtures.

However, instead of considering the additive effect of partial pressures, it states a similar principle of additive partial volumes. Amagat's law states:

"The total volume occupied by a mixture of gases is equal to the sum of the volumes which would be occupied by each constituent when at the same temperature and pressure as the mixture".

The physical significance of the law may be visualized by referring to the gaseous mixture of Figure 10.2.

= + + +

P = 3 atm P = 3 atm P = 3 atm

V = 22.4 litre V = 22.4 litre V = 22.4 litre

P = 3 atm N = 1 N = 1 N = 1

V = 67.2 litre T = 273 K T = 273 K T = 273 K

N = 3 T = 273 K

Figure 10.2: Partial Volume of a Gaseous Mixture

If the gases making up the mixture were separated into individual containers so that each gas was at the same temperature and pressure as the mixture, the resulting volumes would be as shown in Figure 10.2. From this figure it is evident that the total volume V, is equal to the sum of the partial volumes V, V and V The following relation similar to that for Eq. (10.8). gives

VA = X VT = X 67.2 = 22.4 …(10.9)

Thus, the partial volume of a gas present in a gaseous mixture is equal to the mole fraction of that gas times the total volume of the mixture. Eq. (10.9) applies to ideal be used with sufficient accuracy for combustion calculations for any gas.

Example 10.2

What is the volumetric composition of gaseous mixture of Example 10.1?

Solution 𝑉 = .

. X 100 = 16.47 litre = 16.47%

𝑉 = .

. X 100 = 70.68 litre = 70.68%

𝑉 = .

. X 100 = 12.85 litre = 12.85%

Mixture Gas A Gas B Gas C

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10.3 APPLICATION OF FUNDAMENTAL LAWS FOR STOICHIOMETRIC ANALYSIS

The application of the above physical laws to the solution of problems dealing with the stoichiometric relationships of fuels, air, and combustion products is presented in the following sections.

10.3.1 Stoichiometric Analysis for Combustion with Oxygen

Since the combustible matter in fuels is composed mainly of carbon and hydrogen, combustion calculations deal mostly with the different

relationships between carbon, hydrogen, and oxygen and the combustion products which appear as a result of the reactions.

The reaction between hydrogen and oxygen may be stated in equation form as

2H2 + O2 = 2H2O ... (10.10)

This is the same as saying that two molecules of hydrogen plus one molecule of oxygen will unite to form two molecules of water. While it must be remembered that a chemical reaction between two substances is actually a combination of the individual molecules of those substances, it is usually a more convenient to work with molar units. The equation also states that 2 gm moles of hydrogen plus 1 gm mole of oxygen form 2 gm moles of water.

Also from Avogadro's law it can be concluded that 2 vol. of hydrogen plus 1 vol. of oxygen will form 2 vol. of water vapour. Since each mole of hydrogen weighs 2 gm, a .mole of oxygen weighs 32 gm and each mole of water weighs 18gm, it follows that 4 gm of hydrogen plus 32 gm of oxygen gives 36 gm of water.

In summary, the combustion equations for hydrogen may be written as follows:

2 H2 + O2 → 2H2O

2 molecules H2 + 1 molecule O2 →2 molecules H2O 2 moles. H2 + 1 mole. O2 → 2 mole. H2O

2 c.c H2 + 1 c.c O2 → 2 c.c H2O

2 gm H2 + 1 gm O2 → 2 gm H2O These relations must be thoroughly understood as they are the basis for all combustion calculations. Other fundamental combustion equations are C + O2 → CO2 ... (10.11) CO + O2→ CO2... (10.12)

C + O2→ CO... (10.13)

S + O2 → SO2 ... (10.14)

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The same reactions of the elements with oxygen will occur even if the elements are combined in a chemical compound; thus the combustion equation for the hydro-carbons methane and octane can be written, CH4 + 2O2 → CO2 + 2H2O ... (10.15)

C8H18 + 12.5 O2 → 8CO2 + 9H2O ... (10.16 )

Note that, it is not always necessary to balance the equation in such a way that all the coefficients are whole numbers. It is just as true that 1 mole of CO unites with 1/2 mole of O2 as if it were written 2 moles of CO units with 1 mole of O2. This simplifies balancing the equations.

10.3.2 Stoichiometric Analysis for Combustion with Air

In most applications of combustion, fuel is burned in presence of air rather than oxygen. This means that enough nitrogen from the air must accompany the oxygen which the chemical reaction. However, nitrogen being inert, does not enter into the reaction. Table 10.2 shows the average composition of dry air by volume.

Table 10.2: Composition of Dry Air Component % by

Volume

Component % by Volume

Nitrogen 78.03 Hydrogen 0.01

Oxygen 20.99 Neon 0.00123

Argon 0.94 Helium 0.0004

Krypton 0.00005

Carbon Dioxide

0.03 Xenon 0.000006

This analysis of the atmosphere remains rather constant throughout the world at the surface of the earth and upto elevations yet attainable by man. The small percentages of rare gases present in the atmosphere are also inert. They act very much like nitrogen and do not enter the combustion reaction.

Therefore, the composition of air by volume can be taken as 79 percent nitrogen and 21 percent oxygen with sufficient accuracy for most engineering calculations. The ratio of nitrogen to oxygen in air, then 79/21 = 3.76, which means 3.76 moles of nitrogen accompany each mole of oxygen when a fuel is burned in air. The combustion equation for methane in air could then be written as follows:

CH4 + 202 + (2 x 3.76 N2) → CO2 + 2H20 + 7.52 N2... (10.17)

The molecular weight for a mixture of gases can be computed as the sum of the mole fractions of the gases times their respective molecular weight. In equation form,

Mav = MA + MB + MC + … … (10.18) where, Mav = Average molecular weight of mixture,

MA, MB, MC = Molecular weights of gases A, B and C respectively, NA, NB, NC = Number of moles of gases A, B and C respectively, and

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NT = Total number of moles of gas.

Since the temperature and pressure of each constituent gas in the mixture are the same, the molefraction is the percent by volume of each constituent.

The molecular weight of mixture gases such as air, for example, can be computed as

Mav = % . MA + % . MB + % . X MC + … … (10.19)

= (0.21 X 32) + (0.79 X 28) = 28.9

Because the argon and the small amount of carbon dioxide in the air are somewhat heavier than nitrogen, the actual average molecular weight of air is closer to 29.0 than the 28.9 obtained above.

10.3.3 Stoichiometric Analysis for Combustion of Liquid Fuels If a liquid fuel can be represented by a chemical formula, one combustion equation for the chemical reaction can be written containing the quantity of oxygen required to burn the fuel, the combustion products formed, and the nitrogen

from the air which must accompany the required oxygen. For combustion of the hydrocarbon octane, C8H18 with air, the equation would be

C8H18 + 12.5 O2 + 12.5 x 3.76 N2 → 8CO2 + 9H20 + 47N2...(10.20) The chemical equation is balanced by noting that eight oxygen molecules (8O2) will be needed to burn the eight carbon atoms in the octane molecule to eight carbon dioxide molecules (8CO2). In equation form,

8C + 802→ 8CO2 ... (10.21)

Likewise, nine oxygen atoms, or four and one-half oxygen molecules (4.5 O) would be required to burn the 18 hydrogen atoms in the original fuel

molecule to nine water molecules

18H + 4.5 O2 → 9H20 ... (10.22)

Thus (8 + 4.5) or 12.5 oxygen molecules (12.5 O2) must be taken from the air for complete combustion of octane. This oxygen would be accompanied by (3.76 x 12.5) nitrogen molecules, and the nitrogen would appear unchanged in the combustion products.

A mixture of 1 mole of octane, 12.5 moles of oxygen, and 47 moles of nitrogen, as stated in the combustion equation for octane above, would be a chemically correct.

The ratio of air (A) to fuel (F) in any mixture may be stated as the volume (or moles) of air present per volume (or mole) of fuel, or as the weight of air per weight of fuel. For the volume (or mole) relationship, the theoretical air/fuel ratio for octane is:

= .

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= .

= 59.5 . .

. .

… (10.23)

It is important to note that the volume relationship hold only if the fuel is in the vapour state, as a mole of liquid fuel occupy considerably less than 22.4 litre under standard conditions.

On a weight basis, the air/fuel ratio is:

= . ×

×( × × )

= 15.1 … (10.24)

The liquid fuels used most commonly, namely gasoline, fuel oil, kerosine, etc. are not made up of a simple compound such as octane. Instead, these fuels are composed of hundreds of different hydrocarbon compounds, each having a different molecular formula, molecular weight, and physical and combustion characteristics. For the purpose of these calculations, a liquid hydrocarbon fuel may be represented by one simple formula which gives the relationship between the number of hydrogen atoms, and the number of carbon atoms.

Thus, while a gasoline may be composed of several hundred different compounds, a formula Cx Hy can be used to write a combustion equation for the fuel and determine the relationship of air, fuel and combustion products.

The composition of a liquid fuel may be given in several ways; two common methods are

(i) to give the ratio of hydrogen to carbon by weight, and

(ii) to list the percentage of hydrogen and the percentage of carbon by weight in the fuel.

The ratio of hydrogen to carbon by weight may be changed to the ratio of hydrogen atoms to carbon atoms by dividing each element by its atomic weight. For example, if a liquid fuel is composed of 8.72 gm of hydrogen per 18 gm of carbon,

. /

/ = . X 12

= .

A representative chemical formula for the fuel can then be written. Such a formula cannot be used to find the average molecular weight of the liquid fuel unless the average number of carbon atoms in the molecule has been

determined chemically, but it is useful in predicting the air-fuel-combustion products relationships.

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Many times the ratio of hydrogen atoms to carbon atoms is such that the representative formula cannot be expressed easily with integer subscripts.

This should cause little trouble, however, since it is no more difficult to balance combustion equations for C4H8.7 or CH2.18 than for C8H18, once it is realized that such a thing can be done.

Example 10.3

Determine the products of combustion and the air/fuel ratio by weight when a liquid fuel of 16 percent hydrogen and 84 percent carbon by weight is burned with 20 percent excess air, that is, 20 percent more air than is theoretically required.

Solution

To solve this problem it is advantageous first to find a representative formula for the fuel burned. In this case,

The ratio of H/C by weight is 16gm H/84 gm C. By dividing the weight of each element by its atomic weight, an

atomic ratio of H/C can be found. Thus, /

/ or shows that the fuel could be represented by the

chemical formula, C7H16. The theoretical combustion equation is,

C7H16 + 11 O2 + 3.76 X 11 N2 → 7 CO2 + 8 H2O + 41.36 N2

…(10.25)

Percentage of Excess Air

It is the ratio of the excess air supplied to the air theoretically required times 100.

Eq. (10.24) is balanced for the theoretical air, but since the problem states 20% excess air is supplied, the

coefficient of O2 and N2 on left side of the equation should be multiplied by 1.20. Again N2 is unchanged

during reaction, but since more O2 (20 percent of theoretical O2) appears in the products of combustion.

The combustion equation with 20 percent excess sir is C7H16 + (11 X 1.20) O2 + (12 X 3.76 X 1.20) N2

→ 7CO2 + 8H2O + 2.2O2 + 49.6N2…(10.26) (by weight) = ( . . ) . /

( × × ) /

= 18.2 gm air/gm fuel

10.3.4 Stoichiometric Analysis for Combustion of Gaseous Fuel

Gaseous fuels are usually mixtures of several combustible components. The composition of these fuels is always given in percent by volume of the each constituent of the mixture. The principal constituents of gaseous fuels are the higher hydrocarbons, hydrogen and carbon monoxide. The latter two

constituents are present in varying amounts in all manufactured gases, but not in petroleum gases.

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Since the composition of gases are given in percent by volume, the number of moles of each constituent in 100 moles of gaseous fuel is the same as percent by volume. It is convenient to choose 100 moles of fuel as the basis for computation, and find the amount of air required to burn 100 moles of gas and the number of moles of combustion products formed.

Example 10.4

A gaseous fuel of following composition by volume burns with 10%

excess air.

CO2 = 5.5%, CO = 38.3%, CH4 = 0.4%

O2 = 0.1%, H2 = 52.8%, N2= 2.9%

Flue gas leaves the heater at 400°C and 1 atm. Determine the composition of flue gases, the air/fuel ratio by volume.

Solution

For this type of problem, one should arrange the calculations in tabular form. From the basic combustion equations, the moles of oxygen theoretically required to burn each constituent and the moles of combustion products which result are determined.

Thus, from the basic equation, CO + O2 → CO2, it is seen that 19.15 moles of oxygen are theoretically required to burn 38.3 moles of carbon monoxide to 38.3 moles of carbon dioxide. The 0.1 mole of oxygen present in the fuel will enter into the reaction once the fuel is heated to combustion temperature. Table 10.3 is generated from the given data.

Table 10.3: Basic 100 Moles of Fuel Gas Component Moles

of Fuel

Moles of O2

Required

Moles of Combustion Products

CO2 O2 N2 H2O

CO2 5.5 - - - -

O2 0.1 -0.1 5.5 - -

CO 38.3 19.15 38.3 - -

H2 52.8 26.4 - - 52.8

CH4 0.4 0.8 0.4 - 0.8

N2 2.9 - - - 2.9 -

100.00 46.25 the or.

+ 4.63 (10%

excess)

44.2 4.63 2.9

191.3* 53.6

50.88 supplied

194.2

* Excess oxygen resulting from the 10% excess air + N2 with the air = 50.88 x 3.76 = 191.3 moles.

The 0.1 mole of oxygen is, therefore, entered in the "Moles of oxygen required" column as a negative value, indicating that it is not required from the atmosphere. The carbon dioxide and nitrogen are inert constituents and are carried over to the combustion products unchanged.

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(by volume) = . ×

= 2.42

= 2.42

10.3.5 Stoichiometric Analysis for Combustion of Coal (Solid Fuel) There are several methods of reporting the composition and burning characteristics of coal. However, to make stoichiometric calculations, the ultimate analysis is all that need to be known. The ultimate analysis is an accurate chemical analysis which lists the percent by weight of the five principal elements – carbon, hydrogen, oxygen, nitrogen and sulphur and the ash in the fuel. Since this analysis is given on the weight basis, it is

convenient to choose 100 kg of coal as the basis for the combustion

calculations. However, to determine the oxygen required for combustion and the moles of combustion products, it is necessary to know the number of moles of combustible elements in the 100 kg of coal. Therefore, the weight of the constituents are divided by their respective molecular weight to find the number of moles of the elements present.

Example 10.5

Determine the flue gas analysis, the air/fuel ratio by weight and the volume of combustion product at 400°C when coal of the following composition burns with 50 percent excess air. Assume that all fuel is burned and that dry air is used for combustion.

C = 72%, N2 = 1.4%, Ash = 11.0%

H = 6%, O2 = 8%, S = 1.6%

Solution

We take 100 kg of coal as the basis to generate the following Table.

Componen t

kg/10 0 kg Coal

Moles / 100

kg

Moles O2

(Theor.) Reqd./100k

g of coal

Moles of Products/100 kg of Coal

Carbon 72 ÷

12

6.0 6.0 6.

0

Hydrogen 6 ÷ 2 3.0 1.5 3.

0

Oxygen 8 ÷

32 0.25 -0.25

Nitrogen 1.4 ÷

28 0.05 - 0.05

Sulphur 1.6 ÷ 32

0.05 0.05 0.0

5

Ash 11

100.00

7.30 6.

0 3.

0 0.0

5 0.05 41.17

*

0.0 3.65*

*

41.22 3.65

* (7.30 x 1.50) × 3.76 = 41.17 molęs N2 with air

** Excess O2 = 0.50 x 7.30 = 3.65 moles Moles of Product (Wet) = 53.92

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Moles of Product (Dry) = 50.92 Analysis of Wet Flue-Gas

Moles(Wet) Volume %

CO2 = 6.0 + 53.92 11.13

H2O = 3.0 + 53.92 5.56

SO2 = 0.05 + 53.92 0.09 N2 = 41.22 + 53.92 76.45

O2 = 3.65 + 53.92 6.77

53.92 100.00

Analysis of Dry Flue-Gas

Moles(Wet) Volume %

CO2 + 5O2= 6.05 + 50.92

11.88

O2 = 3.65 + 50.92 7.17

CO = 0.0 + 50.92 0.00

N2 = 41.22 + 50.92 80.95

50.92 100.00

Air/Fuel Ratio

= ( . × . × )

= 15.12 kg air/kg coal

Volume of combustion products =

= 53.92 moles ꓫ . X 673

= 29,77.8 litres flue gas / 100 kg coal

10.3.6 Stoichiometric Analysis for the Combustion with Insufficient Air In certain applications of combustion, and particularly in spark-ignition engines, it is sometimes desirable to burn slightly rich mixtures. For example, the air/fuel mixture which gives maximum power in gasoline engines

provides only about 85 to 90 percent of theoretical air. Thus, with insufficient air, the fuel is not oxidized completely, and the products of combustion passing out the exhaust manifold always contain carbon monoxide. Example 10.6 shows an approximate method for computing the combustion products from rich mixture.

When making this type of calculation, we may assume, first, that all of the hydrogen will burn to water, next that the carbon will all burn to carbon monoxide, and finally that any oxygen remaining will unite with some of the carbon monoxide to form carbon dioxide. This is a logical assumption since the relative affinities of hydrogen, carbon and carbon monoxide for oxygen are in that respective order. As shown in the example below, it is helpful to find the number of oxygen atoms actually available for the reactions, and then

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to run an oxygen balance to determine the number of oxygen atoms used for each step of the reaction.

Example 10.6

Compute the composition of the exhaust gases resulting from the combustion of C8H18 with 85

percent theoretical air.

Solution

The first step is to find the amount of air theoretically required for combustion. The theoretical

equation for the combustion of C8H18 may be written as follows:

C8H18 + 12.5 O2 + 12.5 x 3.76 N2

→ 8 CO2 + 9 H2O + 47 N2 …(10.27) It is observed that 12.5 oxygen molecules (or 25 oxygen atoms) are required to complete combustion of one molecule of the fuel. Here only 85 percent of that number, or 0.85 x 25 = 21.25, is available and since, this is not enough to complete the reaction, it is necessary to find how far the

combustion will be carried. Any oxygen balance is established to determine where :

Step 1: 21.25 atoms available -9.00 to burn 18 H to 9 H2O

Step 2:

12.25 left over

- 8.00 to burn 8 C to 8 CO Step 3: 4.25 left over

-4.25 to burn 4.25 CO + CO2

0.00 left

Assuming the hydrogen gets its share of oxygen atoms first, the number needed to burn the hydrogen to water vapour is subtracted from the total number of oxygen atoms available. The number needed to burn carbon to carbon monoxide is then deducted. This leaves only 4.25 oxygen atoms to complete the combustion of the 8 CO formed. Therefore, all of the CO cannot be burned to CO2, but the reaction will proceed as long as oxygen is

available. As shown, 4.25 oxygen atoms left will burn 4.25 CO to 4.25 CO2. The rest of the 8 CO formed in the second problem remains as CO in the exhaust. The molecules of CO2 in the exhaust would be 4.25 and the CO, (8 – 4.25) or 3.75 The combustion equation for C8H18 with 85 percent theoretical air could then be written as follows:

C8H18 + (0.85 x 12.5) O2 + 39.9 N2 → 4.25 CO2 + 3.75 CO + 9 H20 + 39.9 N2

The dry analysis of the exhaust gases would be

Component Moles % by volume

CO2 4.25 8.9

CO 3.75 7.8

N2 39.90 83.8

47.90 100.0

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10.4 THEORETICAL AIR REQUIREMENT FOR THE COMBUSTION OF A FUEL

Carbon and hydrogen, in a fuel, are the principle elements that contribute towards combustion. Let us assume that one mole kg. of any fuel has the composition Cl Hm On where I, m, n are the number of atoms of carbon, hydrogen and oxygen respectively.

The number of moles, 𝜙, of oxygen for the stoichiometric combustion of the fuel is given by (Murgai and Ram Chandra, 2000)

𝜙 = l + - … (10.28)

The amount of air,𝜱, for the complete combustion of the fuel is 𝚽 = 128.5 𝜙 kg of air per k mole of the fuel ...(10.29)

Example 10.7

Determine the theoretical amount of air required for the complete combustion of coal with the

following composition:

C = 60%, H = 5%, O = 4.8%, S = 0.2%, Nitrogen = 2%, Moisture = 10% and Ash = 18%.

Solution

The number of atoms I, m and n are:

l = = 5.0 m = = 5.0 n = . = 0.3

From Eq. (10.28), we get 𝜙 = l + -

= 5 + 1.25 – 0.15

= 6.10 k moles of O2 per 100 kg of coal From Eq. (10.29), the theoretical air for complete combination of coal is

𝚽 = 128.5 X 6.10

= 7.83 kg of air per kg of coal

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SAQ 1

(a) What is stoichiometric calculation?

(b) What is the role of fundamental laws for stoichiometric calculation?

(c) What is the difference between laws of conservation of mass and energy?

(d) What is gm-molecular weight?

(e) What is the main difference between Avogadro's law and Amagat's law?

(f) List all relations, with examples, which are the basis for all combustion calculation?

(g) Compare the combustion equations with oxygen and that of with air.

(h) For stoichiometric calculations of combustion of coal which analysis of coal is required to be known?

(i) How are the compositions of liquid fuels are expressed?

(j) What is the effect of insufficient air on the combustion of fuel?

(k) Determine the air/fuel ratio by weight used in an engine if the

products of combustion show CO2 = 6.0%, O2 = 0.2%, CO = 13.4%, H2 = 8.0% and CH4 = 0.5% by volume on a dry basis. Since the constituents listed in the exhaust gas analysis do not total 100 percent, the remainder of the gas may be assumed to be N2.

10.5 SUMMARY

For stoichiometric analysis of a system, stoichiometric calculations are done and these calculations deal with the weights of materials and quantities of energy entering and leaving the chemical reactions. Fundamental

stoichiometric calculations dealing with combustion are idealized chemical reactions wherein fuel and air combine chemically to form products of combustion.

The fundamental laws, upon which the stoichiometric calculations are based are conservation of mass, conservation of energy, Law of combining weights, Ideal Gas Law, Avogadro's law, Dalton's law and Amagat's law.

The ratio of nitrogen to oxygen in air is 79/21 = 3.76, which means 3.76 moles of nitrogen accompany each mole of oxygen when a fuel is burned in air.

Air/fuel ratio for solid or liquid fuel is expressed on weight basis and for gaseous fuel is expressed on volume basis.

For the stoichiometric calculation for combustion with insufficient air, it may be assumed, first that all of the hydrogen will burn to water, next that all carbon will burn to carbon monoxide and finally that any oxygen remaining will unite with some of the carbon monoxide to form carbon dioxide.

10.6 KEY WORDS

Stoichiometric Calculation : It deals with the weights of materials and quantities of energy entering and leaving the chemical reaction.

Gram-molecular Weight : Molecular weight expressed in gram (in C. G.

S. unit).

Air-fuel Ratio : It serves as the reference condition for combustion reaction.

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Mole Fraction: Ratio of the number of moles of the component to the total number of moles of the mixture.

Percentage of Excess Air : It is the ratio of the excess air supplied to the air theoretically required times 100.

Theoretical Air : Air required for the stoichiometric combustion.

Combustion reaction : It is indicated by chemical equations which are the basis of combustion calculations.

10.7 ANSWERS TO SAQs

SAQ 1

(a) Stoichiometric calculations deal with the weights of material and quantities of energy entering and leaving the chemical reactions.

(b) The fundamental laws are the bases upon which all combustion calculations are based.

(c) Conservation of mass is related to the mass of the matter, whereas conservation of energy is related to energy of the materials.

(d) It is the molecular weight expressed in gm (C G. S. Unit).

(e) Avogadro's law is related to partial pressure, whereas Amagat's law is related to partial volumes of gaseous components.

(f) The relevant combustion equation is 2H2 – O2 → 2H2O

(i) Two molecules H2 + one molecule O2 - two molecules H2O (ii) Two mole H2 + one mole O2 → two moles H2O

(iii)Two c. c H2 + one cc O2 - two c.c. H2O (iv) Two gm H2 + one gm O2 - two gm H2O (g) Combustion equation with O2

CH4 + 2O2 = CO2 + 2 H2O Combustion equation with air

CH4 + 2O2 + 2 (3.76) N2 → CO2 + 2H2O + 7 52 N2

(h) For stoichiometric calcuiation for combustion of coal, the ultimate analysis is required to be known.

(i) Composition of a liquid fuel may be given by several ways; two common methods are :

(i) to give ratio of hydrogen to carbon by weight, and (ii) to list the percentage of carbon by weight in the fuel.

(j) With insufficient air, fuel is not oxidized completely, and the products of combustion passing out the exhaust always contain carbon

monoxide.

(k) = 18.8

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References

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