Fuel and combustion is all about different fuel used in combustion process. Also energy produced by these fuels during the combustion process. This energy measured in terms of calorific value of fuel.
Fuels
Fuels are nothing but substances burnt with oxygen to supply energy to the process.
Classification of Fuels
Fuels are classified mainly into three categories, solid, liquid and gaseous fuels.
Solid Fuels
Principally coal which is a mixture of carbon, hydrogen, non-combustible ash, water, and sulfur is the solid fuel. Other solid fuels are coke, wood, bagasse, etc.
Liquid Fuels
Hydrocarbons are obtained by the distillation of crude oil (petroleum) are liquid fuels. Like Petrol, diesel, Kerosine, etc.
Gaseous Fuels
Natural gas is the gaseous fuel. Which usually contains 80 to 95% methane, the balance being ethane, propane, and small quantities of other gases.
Combustion
The rapid reaction of a fuel with oxygen is known as combustion. In combustion operation, the union of carbon, hydrogen, and sulfur of fuel takes place with oxygen. When a fuel is burned, the carbon in the fuel reacts to form either CO2 or CO. Hydrogen reacts to form H2O and sulfur reacts to form SO2.
The combustion is termed as complete combustion if the products of combustion are CO2, H2O, and SO2. A combustion reaction in which CO is formed is referred to as partial or incomplete combustion.
The significance of combustion reactions lies in the tremendous quantities of heat released during the combustion process. The heat released is used to produce steam, which is then used to drive turbines to produce electricity. In the process industry, the heat released by burning a fuel is used to supply thermal energy. For economic reasons, the air is the source of oxygen in most combustion operations.
The product gases that leave a combustion chamber [containing CO2, CO, H2O, O, SO2 (SO3), and NO] are referred to as the stack gases or flue gases.
In the analysis of flue gases, the term composition on a wet basis is used. Which denotes the component mole fractions of the flue gases that contain water. Also, the term composition on a dry basis is used. Which denotes the component mole fractions of the flue gases without water.
Calorific Values of Fuels
The calorific value of a fuel is also known as the heating value of the fuel. It is the negative of the standard heat of combustion, usually expressed per unit mass of the fuel. Since the standard heat of combustion is always negative, the calorific value (heating value) is positive.
The calorific value of a fuel is defined as “the total heat produced when a unit mass of fuel is completely burnt with pure oxygen.“
When a fuel is burnt, the hydrogen in the fuel reacts with oxygen to produce water. When water is present in the flue gases as vapour, the latent heat of vaporisation is lost (heat associated with water vapour), and hence this quantity of heat is not available for any useful purpose.
The Net Calorific Value (NCV)/ Net Heating Value (NHV) /Low HeatingValue(LHV) of a fuel is the calorific value of the fuel when water is present in the combustion products in the vapour state. The latent heat of vaporisation of water can be made available for useful purposes if water vapours are condensed.
The Gross-Calorific Value (GCV)/ Higher Heating Value (HHV) / GrossHeating Value (GHV) of a fuel is the calorific value of the fuel when water is present in the combustion products in the liquid state. It is equal to the net calorific value of a fuel to which the latent heat of water vapours is added.
The NCV and GCV of the fuels are usually reported at 298 K (25°C). These may be expressed in kJ/kg of fuel, kJ/mol of fuel, or kJ/m³ of gaseous fuel. To calculate the net calorific value of a fuel from the gross calorific value of the fuel or vice-versa. We must have to calculate the moles of water produced when a unit mass of the fuel is burned.
Let n be the mol of water produced.
GCV = NCV + n ΔHv [H2O, 298 K (25°C)] kJ/mol of fuel
Where ΔHv [H2O, 298 K (25°C)] is the heat of vaporisation of water at 298 K (25°C) in KJ/mol.
The GCV calculated will be kJ/mol. If ‘m’ is the quantity of water produced in kg when a unit mass of fuel is burned and ‘λ’ is the latent heat of water vapours at 298 K (25°C) in kJ/kg, then GCV and NCV of a fuel in kJ/kg of the fuel are related by
GCV = NCV + mλ
λ = Latent heat of water vapours, at 298 K (25°C) is 2442.5 kJ/kg.
𝐺𝐶𝑉 = 𝑁𝐶𝑉 + (𝑤𝑡 % ℎ𝑦𝑑𝑟𝑜𝑔𝑒𝑛 9 (𝜆)/100), kJ/kg
If the fuel under consideration is a mixture of combustible substances, its heating value (HV) is given by the following equation.
HV = Σ xi (HV)i
where (HV) is the heating value of the ith combustible substance.
Air Requirement
For any combustion process, i.e., for the chemical oxidation of a fuel, oxygen is a must ich will combine with carbon, hydrogen, and sulphur.
In normal practice, the air is used for burning fuel as it is an inexpensive source of oxygen. Air contains about 21% oxygen and 79% nitrogen by volume.
It is normal practice to feed a less expensive reactant in excess valuable one to increase the conversion of the valuable reactant. Hence, combustion reactions are invariably run with more air than needed to supply oxygen in the stoichiometric proportion to the fuel.
Theoretical Oxygen
It is the amount of oxygen needed for the complete combustion of all the fuel fed to the combustion chamber, assuming that all carbon in the fuel reacts with oxygen to form CO2, all hydrogen reacts to form H2O and all sulfur reacts to form SO2.
Theoretical Air
It is the quantity of air that contains the theoretical oxygen, i.e., it is the minimum air required to burn the fuel completely so that all carbon gets converted into CO2, all hydrogen into H2O, and all sulfur into SO2. The theoretical air requirement of any fuel is calculated as follows.
Theoretical air requirement in moles = Theoretical oxygen requirement in moles / 0.21
In actual combustion practice, air is used more than what is theoretically required to ensure complete combustion.
Excess air
It is the amount by which the air fed to the combustion chamber exceeds the theoretical air.
Percent excess air = ((actual air supply- theoretical air requirement) / theoretical air requirement) × 100
% excess oxygen = % excess air
The quantities of actual air supplied and theoretical air required can be used either in weight units or in molar units in the equation. The actual air supply is obtained knowing the % excess and theoretical air requirement by the relation as shown below.
Actual air supply = Theoretical air requirement × (1 + (% excess / 100))
If you know the feed rate of fuel and the stoichiometric equation for the complete combustion of the fuel, you can calculate the theoretical oxygen demand/requirement and air feed rate. The equation for the calculation of % excess air cited above applies to the calculation of % excess oxygen.
The % excess air and % excess oxygen are the same terms (whenever % excess air is given then treat it as % excess O2 = % excess air).
The theoretical air required to burn a given quantity of fuel does not depend on how much fuel is actually burned and the value of the percentage excess air depends only on the theoretical air and the air supply rate and not on the quantity of oxygen consumed or whether the combustion is complete or partial.
The theoretical as well as actual air requirements are expressed in different units for the sake of convenience, e.g. kg per kg fuel or m’ per kg fuel.
For converting the amount of air expressed in kmol into the amount of air in kg, the average molecular weight of air can be taken as 29. For converting the amount of air from kmol/mol into l/m³ (volume units), we have to use the ideal gas law.
The excess air requirement depends upon the type of fuel burnt. Gaseous fuels require very little excess air. Liquid fuels require more excess air than gaseous fuel and solid fuels require higher amounts of excess air than that required for liquid fuels. Gaseous fuels are burnt with 5 to 15% excess air, while liquid and solid fuels are burnt with 10 to 50% excess air.
Analysis of Solid Fuels
Coal is mostly used as solid fuel and is generally found in nature which is not in a pure form. Also, there is no uniform composition of the coal. A definite chemical formula is not available for coal. So the following two methods are used to determine the composition of coal.
1. Ultimate analysis.
2. Proximate analysis.
Ultimate Analysis of Solid Fuels
In the ultimate analysis, a complete breakdown of coal into its chemical constituents is carried out by the chemical process. This analysis is important for large-scale trials i.e. boiler trials.
It is very useful for the calculation of the amount of air required for the complete combustion of 1 Kg. of coal. Also, this analysis can give the percentages of carbon, hydrogen, oxygen, sulfur, and ash on a mass basis their sum is taken as equal to 100%.
In this analysis, moisture is considered as a separate item. Also, this technique is used to determine the calorific value of the coal.
Proximate Analysis of Solid Fuels
This analysis technique separates the coal into its physical components. This separation can be made using chemical balance and temperature-controlled furnaces. In this analysis, the sample is heated in the furnace.
The components in the analysis are fixed carbon, volatile matter, moisture, and ash. These components are expressed in percentage on a mass basis and their sum is taken as 100%. Sulfur is determined separately. This analysis is also used to determine the heating value of the coal.
Comparison of Ultimate Analysis and Proximate Analysis of Solid Fuels
| Ultimate Analysis | Proximate Analysis |
| The ultimate analysis is coal is a complete breakdown of coal into chemical constituents. | This analysis gives percentages of carbon, hydrogen, oxygen, Sulphur and ash. |
| This analysis gives percentage of carbon, hydrogen, oxygen, Sulphur and ash. | This analysis gives percentages of moisture, volatile matter, fixed carbon and ash |
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