COMBUSTION THERMODYNAMICS Introduction: A fuel is a substance which releases energy in the form of heat while undergoing combustion. The fuels are classified.

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COMBUSTION THERMODYNAMICS Introduction:

A fuel is a substance which releases energy in the form of heat while undergoing combustion.

The fuels are classified as chemical fuels and nuclear fuels.

A chemical fuel is a substance which releases energy in the form of heat while undergoing combustion.

The main combustible elements in any chemical fuel are carbon and hydrogen. Some fuel contains small quantities of sulphur which is also combustible. However presence of sulphur in fuel is considered undesirable. Fuels are further classified as below;

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TYPE OF FUEL PRIMARY FUEL (WHICH OCCUR NATURALLY)

SECONDARY (PREPARED)

Solid Wood CokePeat CharcoalCoal Briquettes

Liquid Petroleum (Crude Oil)

Gasoline (Petrol)KeroseneDiesel oilFuel oilAlcoholBenzol

Shale oilGaseous Natural gas Petroleum gas

(LPG)Producer gas

Coal gasCoke-oven gas

Blast furnace gas

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Combustion or burning is the chemical process in which the inflammable matter in a substance combines with oxygen at a temperature above the spontaneous ignition temperature of that substance and results in the evolution of heat and light.

The combustion process involves the oxidation of constituents in the fuel, that are capable of being oxidized and it can be represented by a chemical equation.

These equations indicate the required amount of oxygen combined with required amount of fuel.In a chemical reaction the terms, reactants and the products are frequently used.

Reactants comprise of initial constituents which start the reaction while products comprise of final constituents which are formed by the chemical reaction.

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The smallest particle which can take part in a chemical change is called an atom.

It is rare to find elements to exist naturally as single atom.

Some elements have atoms which exist in pairs, each pair forming a molecule, and the atoms of each molecule are held together by stronger inter-atomic forces. The symbols and molecular weight of some important elements, compounds and gases are given in table below:

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Elements/Compounds/

Gases

Molecule Atom

Symbol Molecular weight

Symbol Atomic weight

HydrogenOxygenNitrogenCarbonSulphurWaterCarbon monoxideCarbon dioxideSulphur dioxideMethane (Marsh gas)EthyleneEthaneAcetylenePropane

H2

O2

N2

CS

H2O

COCO2

SO2

CH4

C2H4

C2H6

C2H2

C3H8

232281232182844641628302644

HONCS---------

116141232---------

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Generally combustion requires reaction with an oxidant or supporter of combustion.

Oxygen is the main oxidant, but the halogens (chlorine and fluorine), hydrogen peroxide and also nitric acid, may act as oxidants as in rocket propulsion.

Air is the commonest oxidant because it is cheap and readily available. The following analysis of air is used in combustion calculations:

Air components

By volume % By mass %

O2

N2

21.0079.00

23.0077.00

Total 100.00 100.00

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Composition of fuels:An accurate chemical analysis by mass of the different elements in the fuel is called the ultimate analysis.

The elements usually included being carbon, hydrogen, nitrogen, oxygen and sulphur.

The ultimate analysis is normally used for solid and liquid fuels and is also known as gravimetric analysis.Another analysis of solid fuels like coal is called proximate analysis, gives the percentage by mass of moisture, volatile matter, combustible solid (fixed carbon) and ash.Gaseous fuels are usually mixtures of several gases like hydrogen, methane, oxygen, carbon dioxide, carbon monoxide, nitrogen and other hydrocarbons.

The composition of gaseous fuels is normally given by percent volume known as volumetric analysis.

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Combustion equations:In a combustion chamber proportionate masses of air and fuel enter where the chemical reaction takes place, and then the combustion products pass to the exhaust.

By the conservation of mass the mass flow remains constant (i.e. total mass of products = total mass of reactants), but the reactants are chemically different from the products, and the products leave at a higher temperature.

The total number of atoms of each element concerned in the combustion remains constant, but the atoms are rearranged into groups having different chemical properties. This information is expressed in the chemical equation which shows (i) the reactants and the products of combustion, (ii) the relative quantities of the reactants and products. The two sides of the equation must be consistent, each having the same number of atoms of each element involved.Some important combustion equations are given below:

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1. Combustion of hydrogen:2H2 + O2 → 2H2O

This equation tells us that:2 volumes of Hydrogen + 1 volume of Oxygen → 2

volumes of H2O

This shows that there is volumetric contraction on combustion.The H2O may be liquid or a vapor depending on whether the product has been cooled sufficiently to cause condensation.The proportions by mass are obtained by using atomic weights as follows:

2H2 + O2 → 2H2O

2(2 X 1) + 2 X 16 → 2(2X1 + 16)4kg of H2 + 32kg of O2 → 36kg of H2O

1kg of H2 + 8kg of O2 → 9kg of H2O

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2. Combustion of carbon:(i) Complete combustion of carbon to carbon-dioxide C + O2 → CO2

1 volume Carbon + 1 volume Oxygen → 1 volume CO2

And 1kg C + 8/3 kg O2 → 11/3 kg CO2

(ii) The incomplete combustion of carbon. The incomplete combustion of carbon occurs when there is

an insufficient supply of oxygen to burn the carbon completely to carbon-dioxide.

2C + O2 → 2CO

1kg C + 4/3 kg O2 → 7/3 kg CO

If a further supply of oxygen is available then the combustion can continue to completion.

2CO + O2 → 2CO2

1kg CO + 4/7 kg O2 → 11/7 kg CO2

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3. Combustion of sulphur to sulphur-dioxide:S + O2 → SO2

1kg S + 1kg O2 → 2kg SO2

4. Combustion of methane:

CH4 + 2O2 → CO2 + 2H2O

1kg CH4 + 4kg O2 → 11/4 kg CO2 + 9/4 kg H2O

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Theoretical Air and Excess Air:The minimum amount of air that supplies sufficient oxygen for the complete combustion of all the carbon, hydrogen, and any other elements in the fuel that may oxidize is called the theoretical air.

When complete combustion is achieved with theoretical air, the products contain no oxygen.Complete combustion is not attained in practice unless more than the amount of theoretically required air is used.

The need of excess air is due to the difficulty of obtaining intimate contact between the air and the fuel.

It is also partly due to the need to complete the combustion within the combustion space.

The excess air is the amount of air supplied over and above theoretical air.

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The complete combustion of methane with minimum amount of theoretical air and 150% theoretical air respectively is written as:

Mass of excess air supplied may be determined by the mass of unused oxygen found in the flue gases. We know that in order to supply one kg of oxygen, we need 100/23 kg of air. Therefore, mass of air supplied = (100/23) X mass of excess oxygen.

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Stoichiometric Air-Fuel (A/F) Ratio: Stoichiometric (or chemically correct) mixture of air and fuel is one that contains just sufficient oxygen for complete combustion of fuel.

A weak mixture is one which has excess air. A rich mixture is one which has a deficiency of air. The percentage of excess air is given as under:

An alternate method of expressing the same is in terms of mixture strength;

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For a stoichiometric mixture, the mixture strength is 100%.

A weak mixture is one whose mixture strength is less than 100%, while a rich mixture has mixture strength of more than 100%.

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Theoretical or stoichiometric or minimum air required for complete combustion:We know that fuel mainly consists of constituents like carbon, hydrogen and sulphur and we also know that for combustion;Carbon requires 2.67 times its own mass of oxygenHydrogen requires 8 times its own mass of oxygenSulphur require its own mass of oxygen

Now consider 1kg of fuel whose ultimate analysis gives:Mass of carbon = C kgMass of hydrogen = H2 kg

Mass of sulphur = S kg Then oxygen required to burn

C kg of carbon = 2.67C kgH2 kg of hydrogen = 8 H2 kg

S kg of sulphur = S kg

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Therefore total oxygen required for complete combustion of 1 kg of fuel

= (2.67C + 8H2 + S) kg

As fuel already contains O2 kg of oxygen which we assume can be used for combustion.

Then total oxygen required for complete combustion of 1 kg of fuel is,

= (2.67C + 8H2 + S – O2) kg

Since air contains 23% of oxygen on mass basis, i.e. 1 kg of oxygen is associated with 100/23 = 4.35kg of air.

Therefore, Minimum or theoretical air required for complete combustion of 1kg of fuel,

= 100/23 (2.67C + 8H2 + S – O2) kg

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Theoretical or stoichiometric or minimum air required for complete combustion:Consider 1m3 of gaseous fuel whose ultimate volume analysis is:Volume of carbon monoxide = CO m3

Volume of hydrogen = H2 m3

Volume of methane = CH4 m3

Volume of ethylene = C2H4 m3

Then oxygen required to burn CO m3 of carbon = 0.5 CO m3

H2 m3 of hydrogen = 0.5 H2 m3

CH4 m3 of sulphur = 2CH4 m3

C2H4 m3 of hydrogen = 3C2H4 m3

Therefore total oxygen required for complete combustion of 1 m3 of fuel

= (0.5CO + 0.5H2 + 2CH4 + 3C2H4) m3

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As fuel already contains O2 m3 of oxygen which we assume can be used for combustion.

Then total oxygen required for complete combustion of 1 kg of fuel is,

= (0.5CO + 0.5H2 + 2CH4 + 3C2H4 – O2) m3

Since air contains 21% of oxygen on volume basis, i.e. 1 m3 of oxygen is associated with 100/21 = 3.76m3 of air.

Therefore, Minimum or theoretical air required for complete combustion of 1kg of fuel,= 100/21 (0.5CO + 0.5H2 + 2CH4 + 3C2H4 – O2) kg

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Air-fuel Ratio from analysis of products of combustion:By making analysis of the combustion products, the air-fuel ratio can be calculated by the following methods.

1. When fuel composition is knowni) Carbon Balance Method- Quite accurate when

combustion takes place with excess air and when no free (solid) carbon is present in the products.

ii) Hydrogen Balance Method- Suitable when solid carbon is suspected to be present in the products.

iii) Carbon-Hydrogen Balance Method- Used when there is some uncertainty about percentage of N2 present.

2. When fuel composition is unknownOnly Carbon-Hydrogen Balance Method can be employed.

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Analysis of exhaust (flue) gas: The combustion products are mainly gaseous.

When a sample is taken for analysis it is usually cooled down to a temperature which is below the saturation temperature of the steam present.

The steam content is therefore not included in the analysis, which is quoted as the analysis of the dry products.

Since the products are gaseous, it is usual to quote the analysis by volume.

An analysis which includes the steam in the exhaust is called a wet analysis.Practical analysis of combustion products:The most common means of analysis of the combustion products is the Orsat analysis using Orsat apparatus.

It gives an analysis of dry products of combustion.

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1. A burette- used to measure the volume of the exhaust gas sample inside the apparatus at any time.

2. A gas cleaner- to filter out solid particles in the exhaust gas sample.

3. A leveling bottle partly filled with water and connected to the bottom of the burette by tubing.

4. Three absorption pipettes- each containing different absorbent chemical solutions used to absorb different constituents of the gas sample.

The pipettes are interconnected by means of a manifold fitted with cocks with S1, S2, S3 and contain different chemicals to absorb CO2, CO and O2. Each pipette is also fitted with a number of small glass tubes which provide a greater amount of surface. These tubes are wetted by absorbing agents and are exposed to the gas under analysis.

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The pipette 1, 2, 3 contain the following chemicals:Pipette 1: caustic soda (KOH) to absorb CO2

Pipette 2: alkaline solution of pyrogallic acid to absorb O2

Pipette 3: acid solution of cuprous chloride to absorb CO Further-more the apparatus has a leveling bottle and a

three way cock to connect the apparatus either to gases or atmosphere. The procedure is as follows:

100cc of gas whose analysis is to be made is drawn into the bottle by lowering the leveling bottle. The stop cock S1 is then opened and the whole flue gas is forced to pipette 1. The gas remains in this pipette for some time and most of the CO2 is absorbed. The leveling bottle is then lowered to allow the chemical to come its original level.

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The volume of gas thus absorbed is read on the scale of the measuring bottle.

The flue gas is then forced through pipette 1 for a number of times to ensure that the whole of the CO2 is absorbed.

Further the remaining flue gas is then forced to pipette 2 which absorb whole of O2.

The reading on the measuring bottle will be the sum of volume of CO2 and O2.

The oxygen content then is found by subtraction. Finally, as before the sample of gas forced through the pipette 3 to absorb CO completely.The amount of N2 in the sample can be determined by subtracting from total volume of the gas the sum of CO2, CO and O2 contents.

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Enthalpy of formation (ΔHf):

A combustion reaction is a particular kind of reaction in which the products are formed from reactants with the release or absorption of energy as heat is transferred to and from the surroundings. In some substances like hydrocarbon fuels the heat of reaction or combustion may be calculated on the basis of known values of the enthalpy of formation, ΔHf of the constituent of the reactants and products at the temperature T0 (reference temperature). The enthalpy of formation (ΔHf) is the increase in enthalpy when a compound is formed from its constituent elements in their natural form and in a standard state. The standard state is 250C, and 1 atmospheric pressure.

Therefore the enthalpy of combustion at T0 or the constant pressure heat of combustion at T0, may be given as:

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Enthalpy of reaction:The enthalpy of reaction hR, is defined as the difference between the enthalpy of products at a specified state and the enthalpy of the reactants at the same state for a complete reaction.For combustion processes, the enthalpy of reaction is usually referred as the enthalpy of combustion hC, which represents the amount of heat released during a steady flow combustion process when 1 k-mol (or 1 kg) of fuel is burned completely at a specified temperature and pressure (Fig. below).

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It is expressed as

hR = hC = Hprod - Hreact

Which is -393.520 kJ/k-mol for carbon at the standard reference state.

The enthalpy of combustion of a particular fuel is different at different temperatures and pressures

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Internal energy of reaction:When the combustion process is carried out at constant volume, the non-flow energy equation, Q = (U2 – U1) + W, can be applied to give

Q = (UPo – URo)

Where, W = 0 for constant volume combustionU1 = URo

U2 = UPo

The internal energy change is independent of the path between the two states and depends only on the initial and final values and is given by the quantity Q. The heat transferred during this constant volume process is called the internal energy of combustion at T0 (or constant volume heat of combustion), and is denoted by . Thus

is a negative quantity since the internal energy of the reactants includes the potential chemical energy and heat is transferred from the system.

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Combustion efficiency:In evaluating the performance of an actual combustion process a number of different parameters can be defined depending on the nature of the process and the system considered. The combustion efficiency in a gas turbine can be defined as

Where, (F/A) = Fuel-Air ratio for adiabatic and complete combustion and in which the products would attain the adiabatic flame temperature.

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In case of steam generator or boiler

In case of an internal combustion engine

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Adiabatic flame temperature:In a given combustion process that takes place adiabatically and with no work or changes in kinetic or potential energy involved, the temperature of the products is referred to as adiabatic flame temperature.

With the assumption of no work and no changes in K.E or P.E, this is the maximum temperature that can be achieved for the given reactants because any heat transfer from the reacting substances and any incomplete would tend to lower the temperature of the products.

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The following points to be noted:1. The maximum temperature achieved through

adiabatic complete combustion varies with the type of reaction and percent of theoretical air supplied. An increase in the Air-Fuel ratio will affect a decrease in the maximum temperature.

2. For a given fuel and given pressure and temperature of the reactants, the maximum adiabatic flame temperature that can be achieved is with a stoichiometric mixture.

3. The adiabatic flame temperature can be controlled by the amount of excess air that is used.

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Heating values of the fuel: In case of a constant pressure process (steady

flow process), the negative of the enthalpy of combustion is called heating value at constant pressure.

This represents the heat evolved during the combustion at constant pressure.Similarly, the negative of the internal energy of combustion is called the heating value at constant volume and it represents the heat released during the constant volume combustion process.

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In general, the heating value also called calorific value of a fuel is the quantity of heat energy released by the combustion of the fuel.

For solid and liquid fuels it is expressed in kJ/kg of fuel, while for gaseous fuels the unit used is kJ/m3 of fuel at standard temperature and pressure.Heating values of fuel can be experimentally determined using calorimeters.

E.g., Bomb calorimeter, Boyce calorimeter and Junkers gas calorimeter.

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The Higher Heating Value (HHV) or Higher Calorific Value (HCV) is the heat released when H2O in the products of combustion is in the liquid state.

The Lower Heating Value (LHV) or Lower Calorific Value (LCV) is the heat released in the reaction when H2O in the products is in the vapor state.

Therefore, LCV = HCV – mH2O.hfg

Where, mH2O is the mass of water formed during the combustionhfg is the latent heat of vaporization.

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Example 1 Calculate the minimum volume of air required to burn one

Kg of coal having the following composition by weight C = 72.4%, H2 5.3%, N2 = 1.81, O2 = 8.5%, moisture 7.2%

S = 0.9% and ash 3.9% On weight basis:Taking 1kg coal as basis weight of oxygen required to burn

1kg of coal C + O2 → CO2

0.724 x 32/12 = 1.93 kg 0.53x 16/2 = 0.424 kg0.009x32/32 = 0.009 kg Total O2 = 2.363 kg per kg of coal

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But 0.085kg O2 is available in coal, therefore O2 required

= 2.363 – 0.085 = 2.278kg per Kg of coal.

Air contains 23% of oxygen by weight.

Therefore the weight of the air supplied is

2.278x 100/23 = 9.9 kg per kg of coal

Density of air required at NTP

P v = mRT

P = m/v RT = ρRT, ρ = Molecular weight

Volume = P/RT = 1.013x105/287x273 = 1.29 kg/m3

Therefore volume of air required = 9.9(kg)/1.29(kg) = 7.67 m3

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On mole basis Consider 100kg of coal C = 72.4/12= 6.03K mol, O2 = 8.5/32 = 0.265K mol

H2 = 5.3/2 = 2.65K mol, H2O = 7.2/18 = 0.4K mol

N2 = 1.8/28 = 0.064K mol, S = 0.9/32 = 0.028K mol

1K mol C + 1K mol O2 → 1Km CO2

Therefore 6.03 K mol of carbon requires 6.03 K mol of oxygen 1 K mol H2 + ½ K mol O2 → 1K mol H2O

H2 - 2.65 x ½ = 1.325K mol,S -0.028x1 = 0.028Total O2 required 6.03 + 1.325 + 0.028 = 7.383

The oxygen present in coal 0.265K mol Net O2 required = 7.383 – 0.265 = 7.118K mol

Air required 7.118x100/21= 33.89K mol / 100kg of coal = 0.3389K mol / 1kg coal Volume of air supplied

0.3389K mol/kg X 22.4m3= 7.59m3/kg of coal

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Example 2 Calculate the volumetric analysis of the flue gases when

coal burns with 20% excess air from the previous calculation the actual air required 33.89K mol/100kg coal.

Therefore the actual air is33.89 x 120/100=40.67K mol/ 100 kg coal

The amount of N2 associated with this

40.67 x 79/100 = 32.13K mol The amount of O2 present

40.67 x 21/100 = 8.54K mol The actual amount of O2 required was 7.118K mol excess O2

will appear in exhaust gas = 8.54 – 7.118 = 1.422K mol.

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Therefore:CO2 = 6.03K mol

SO2 = 0.028K mol

N2 = 32.13K mol (air) + 0.064 (fuel)

= 32.194K molO2 = 1.422K mol os excess oxygen.

Therefore the Total volume = (6.03 + 0.028 + 32.194 + 1.422)

= 39.674K mol The volumetric composition of the gas CO2= (6.03/39.674) x 100 = 15.12%

SO2= (0.028/39.674) x 100 = 0.07%

N2 = (32.13/39.674) x 100 = 81.15%

O2= (1.422/39.674) x 100 = 3.58%

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3. The composition of dry flue gases obtained by burning a liquid fuel containing only hydrogen and carbon is CO2 10.7%, O2 5.1%, N2 84.2%. Calculate the composition of fuel by weight and excess air used.

Solution: consider 100K mol of dry flue gases. They will contain

10.7K mol of O2 (from CO2) + 5.1K mole of (as max. oxygen) = 15.8K mol

Using nitrogen balance the actual air used 84.2 x 100/79 = 106.58K mol of dry flue gases and oxygen in the air supplied 106.58 x 21/100 = 22.38K mol. Therefore the amount of O2 present in the water produced by the combustion of H2 is 22.38 – 15.8 = 6.58K mol O2.

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We know that 1 K mole of H2 combines with ½ K mol O2 to produce water. Therefore the amount of hydrogen present is 6.58x2 = 13.16K mol/100K mol of dry flue gases, and the carbon present is 12X10.7 = 128.4kg/100K mol of dry flue gas.

Therefore the composition of fuel (by weight) is 128.4kgC and 26.32Kg H2 on the %age basis.

C = (128.4/(128.4+26.32) x 100 = 82.99% H = (26.32/(128.4+26.32) x 100 = 17.01% Excess air supplied The amount of O2 required to burn 10.7K mol C is 10.7K mol

and to burn 13.16K mol H2 is 13.16 X ½ = 6.58

Total O2 required = 10.7 + 6.58 = 17.28K mol/100K mol of dry flue gases

%age of excess air = (22.38 – 17.28)/(17.28) x 100 = 29.5%

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Example 4 A blast furnace gas has the following volumetric

analysis H2 CO-24%, CH4 – 2%, CO2-6%, O2-3% and N2-56%

Determine the Ultimate gravimetric analysis Given volumetric analysis, H2 – 9%, CO-24%,

CH4 – 2%, CO2-6%, O2-3% and N2-56%

Solution: The volumetric analysis may be converted into mass or granite metric analysis by completing the table as follows:

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Constituent

Volume in 1m3 of flue gas (a)

Molecular mass (b)

Proportional mass (c)=(a)x(b)

Mass in kg per kg of the gas (d)=(c)/ Σ ©

% by mass = (d)x100

CO 0.24 28 6.72 6.72/18.48 = 0.36

36%

CH4 0.02 16 0.32 0.32/18.48 = 0.0173

1.73%

CO2 0.06 44 2.64 264/18.48 = 0.142

14.2%

O2 0.03 32 0.96 0.96/18.48 = 0.0519

5.19%

N2 0.56 14 7.84 7.84/18.48 = 0.42

42%

Σc = 18.48 Σ (d) = 1 100

The volumetric analysis of flue gas components becomesCO-0.36, CH4 – 0.0173, CO2- 0.142, O2-0.0519 and N2-0.42

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5.Determine the fuel gas analysis and air fuel ratio by weight when fuel oil with 84.9% carbon, 11.4% hydrogen, 3.2% sulphur, 0.4% oxygen and 0.1% ash by weight is burnt with 20% excess air, assume complete combustion.

Solution: Consider 1kg of fuel Oxygen required / Kg of fuel For burning of 1kg C - 0.849 x32/12For burning of 1kg H - 0.114 x16/2For burning of 1kg S - 0.032 x32/32Total O2 required is 3.208 kg.

Amount of O2 contained in the fuel = 0.004Kg

Net O2 supplied / kg of fuel = 3.208 – 0.004

= 3.204 kg O2

Net air supplied = 3.204x100/23 = 13.93 kg/kg of fuel

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When 20% excess air supplied Total air supplied = 13.93 x 1.2 = 16.716 kg/kg of fuel.N2 actually supplied = 16.716 x 77/100 = 12.871 kg/kg of fuel

O2 actually supplied = 16.716 x 23/100 = 3.845 kg/kg of fuel

Total free O2 in fuel gas = 3.845 – 6.204

= 0.641 kg/kg of fuelTotal free N2 in fuel gas = 12.87 kg/kg of fuel

Flue gas analysis: C converted to CO2 = 0.849x44/12 = 3.113 kg CO2

H converted to H2O = 0.114x18/2 = 1.026 kg H2O

S converted to SO2= 0.032x64/32 = 0.064 kg SO2

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Flue gas / kg of fuel: = 3.113 + 1.26 + 0.064 + 0.641 + 12.871 CO2 H2O SO2 O2 N2

= 17.715kg.Therefore: CO2 = (3.113/17.715)x100 = 17.573%

SO2 = (0.064/17.715)x100 = 0.36%

O2 = (0.641/17.715)x100 = 3.618%

H2O = (1.026/17.715)x100 = 5.79%

N2 = (12.871/17.715)x100 = 72.656%

Air fuel mixture ratio is = 16.716 : 1

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6.A blast furnace gas has the following volumetric analysis. H2 = 9%, CO = 24%, CH4 = 2%, CO2 = 6%, O2 = 3% and N2 =

56 % Determine the ultimate gravimetric analysis. Solution: Total H2 in the blast furnace gas.% volumetric analysis = 9H2 + 2H4

Proportional mass = % volumetric analysis X mol. Mass of element = (9x2) + (2x4) = 18 + 8 = 26 kg.

Total ‘C’ in the blast furnace gas.% of volumetric analysis = 24C + 2C + 6CProportional mass = (24+2+6) x 12 = 384 kg

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Total O2 in the blast furnace gas

% of volumetric analysis = 24xO + 6O2 + 3O2

Proportional mass = (24+16) x 9 (32) = 672 kgTotal N2 in the blast furnace gas

% of volumetric analysis = 56 N2

Proportional mass of N2 = 56 x 28 = 1568 Kg.

Total weight of blast furnace gas:= 384kg C + 26kg H2 + 672kg O2 + 1568kg N = 2650kgs

Gravimetric %age compositon: C = (384/2650)x100 = 14.49% H2 = (26/2650)x100 = 0.98%

O2= (672/2650)x100 = 25.36%

N2 = (1568/2650)x100 = 59.17%

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7.The analysis of coal used in a boiler trial is as follows. 82% carbon, 6% hydrogen, 4% oxygen, 2% moisture and 8% ash. Determine the theoretical air required for complete combustion of 1kg of coal. If the actual air supplied is 18kg per kg of coal the hydrogen is completely burned & 80% carbon burned to CO2

,the reminder is CO, Determine the volumetric analysis of the dry products of combustion.

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Solution: For complete combustion. O2 required is

For carbon - 0.82 = 2.186 kg of O2

For hydrogen - 0.006 = 0.48 kg of O2

Total O2 required = 2.666kg.

Net O2 supplied = Total O2 required – O2 present in the fuel = 2.66 – 0.004 = 2.662 kg/kg of coal

Theoretical minimum air required for complete combustion [C burns to CO2 totally]

Air supplied = 2.626x100/23 = 11.417 kg/kg of coalFlue gas analysis:But actually only 80% carbon is burns to CO2

CO2 = 0.8 x 0.82 x44/12 = 2.405kg of CO2

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20% carbon is burnt to COCO = 0.2 x 0.82 x 28/12= 0.383 kg of CO

O2 actually required for 80% carbon burnt to CO2

= 0.8 x 0.82 x 3232/12 = 1.749 kg of O2

O2 actually required for 20% carbon burnt to CO

= 0.2 x 0.82 x 16/12 = 0.219 kg of O2

O2 required by Hydrogen:

= 0.06 x 8 = 0.48 kg of O2.

H2O produced = 0.06 x 9 = 0.54 kg of H2O

But actual air supplied = 18kgActually O2 supplied = 18 x23/100 = 4.14 kg of O2

Free O2 in the flue gas = 4.14 + 0.04 – 1.749 – 0.219 – 0.48 = 1.732 kg of O2/kg of coal

N2 in the flue gas = 18x77/100 = 13.86 kg/kg of coal

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Volumetric analysis of the dry products of combustion. CO2 = (2.405/44)x 100 = 0.0546m3/K. mol

CO = (0.383/28) x 100=0.0137 m3/K.mol O2= (1.732/32)x 100 = 0.0541m3/K.mol

N2 = (13.86/28)x 100 = 0.495m3/K.mol

In % of volume: CO2= (0.0546/0.6174)x 100 = 8.84%

CO = (0.0137/0.6174)x 100 = 2.22% O2 = (0.0541/0.6174)x 100 = 8.76%

N2 = (0.495/0.6174)x 100 = 80.70%

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8.Consider the following reaction, which occurs in a steady state, steady flow processes.

CH4 + 2O2 → CO2 + 2H2O (l)

The reactants and products are each at total pressure of 0.1Mpa and 25oC. Determine the heat transfer for per K mol of fuel entering the combustion chamber.

Solution : using the values of enthalpy of formation Q = hf = Σp nehf - ΣR nihf

ΣR nihf = (hf) CH4 = -74873KJ

Σp nehf = (hf) CO2 + 2 (hf) H2O (l)

= - 393522+2 (-2852830) = - 965182KJ Therefore Q = - 965182 – (-74873) = - 890309KJ

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9. A small gas turbine uses C8H18 as fuel and 400% theoretical air. The air and fuel enters at 25oC and the products of combustion leaves at 900K. The output of the engine and the fuel consumption are measured and it is found that the specific fuel consumption is 0.25kg/Sec of fuel per MW out put. Determine the heat transfer from the engine per K mol of fuel. Assume complete combustion

Solution:The combustion equation is C8H18 + 4 (12.5)O2 + 4 (12.5) ( 3.76)N2 - 8CO2 + 9H2O + 37.5O2 + 188N2

By first law Q + ΣR ni (hf + Δh); = W + Σp ne (hf + Δh)

ΣR ni (hf + Δh) = (hf) C8H18 = 250105KJ/K mol fuel at 25oC

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Considering the products Σp ne (hf + Δh) = nCO2 (hf + Δh) CO2+ nH2O (hf + Δh) H2O + nO2

(Δh)O2 + nN2 (Δh)N2

hf of O2, N2 = O Δh = Enthalpy of formation from 298oK to 900K

Therefore Σp ne (hf + Δh) = 8 (- 393522+288030)+ 9(-241826+21892) +37.5(19249)+188(18222)

= -755769KJ/K mol fuel.W = 1000(KW) X 114 Kg = 456920KJ/K mol 0.25 K mol

Therefore Q = -755769+456920- (-250105) = -48744KJ/K mol fuel