R.K.Yadav/Automobile Engg Dept/New Polytechnic Kolhapur. Page 1

Fuel : A fuel is defined as a substance (containing mostly carbon and hydrogen) which on burning with oxygen in atmospheric air, produces a large amount of heat. fuel. The amount of heat generated is known as calorific value of the Classification of Fuels : Types of gaseous fuels: (Any four) 1) Natural fuel 2) CNG 3) LPG 4) Water gas 5) Producer gas R.K.Yadav/Automobile Engg Dept/New Polytechnic Kolhapur. Page 1

6) Coal gas 7) Blast Furnace gas 8) Coke oven gas 9) Oil gas Requirement of good fuel : A good fuel should have; 1) Low ignition point. 2) High calorific value. 3) Freely burn with high efficiency. 4) Should not produce harmful gases. 5) Should be produce less quantity of smoke and gases. 6) Should be economical. 7) Easy to store. 8) Easy, safe and convenient for transportation. Calorific value of fuels : The calorific value (or) heating value of solid (or) liquid fuel may be defined as amount of heat given out by complete combustion of 1 Kg. of fuel. It is expressed in Kcal/Kg. of fuel at N.T.P. in S.I. system J/Kg. KJ/Kg (or) MJ/Kg. Types of calorific values of fuels : 1. Higher calorific values of fuel (H.C.V.) (or) Gross C.V. It is a total heat is liberated by complete burning of 1 kg or 1m 3 of fuel including heat of steam formed by combustion of Hydrogen in the fuel. OR The amount of heat obtained by complete combustion of fuel, when the products of its combustion are cooled down to temperature of surrounding supplied air (i.e. 15 o C) is called as higher calorific value. If C, H, O and S are the percentage of Carbon, Hydrogen, Oxygen and Sulphur by weight respectively present in a fuel then the higher calorific value of fuel can be calculated from the following formula known as Dulong s formula. H.C.V.= (33800 C +144000 (H 2 - O 2 /8) + 9270 S) KJ/Kg. R.K.Yadav/Automobile Engg Dept/New Polytechnic Kolhapur. Page 2

2. Lower calorific value of fuel (L.C.V.) (or) Net C.V. It is a total heat is liberated by complete burning of 1 kg or 1m 3 of fuel deducting heat of steam formed by combustion of Hydrogen in the fuel. OR When heat absorbed (or) carried away by the products of combustion is not recovered and steam is formed during combustion is not condensed then amount of heat obtained per Kg of fuel is known as net (or) lower calorific value. If H.C.V. is known then L.C.V. is obtained by L.C.V. = H.C.V. Heat of steam formed during combustion let. m S = Mass of steam formed in KJ/Kg. of fuel = 9 H 2 Since amount of heat per Kg. of steam is the latent heat of vaporization of water corresponding to a standard temperature of 25 o C is 2466 KJ/Kg. L.C.V. = H.C.V. m s x 2466 L.C.V. = H.C.V. 9 H 2 x 2466 KJ/Kg. Dulong s formula used to calculate the theoretical calorific value of fuel if ultimate analysis is available and the calorific value of elementary combustibles are known. Theoretical calorific Value of fuel =33800 C + 144500 ( - ) + 9300 S kj/kg Where C, H2 O2 & S repents the mass of carbon, hydrogen, oxygen and sulfur in kj/kg Analysis of solid fuels : We know that in today s life coal is used as solid fuel and which is generally found in nature which is not in a pure form. Also there is no uniform composition. A definite chemical formula is not available for coal. So following two methods are used to know composition of coal namely; i) Ultimate analysis. ii) Proximate analysis. R.K.Yadav/Automobile Engg Dept/New Polytechnic Kolhapur. Page 3

Ultimate analysis of solid fuels : In ultimate analysis a complete breakdown of coal into its chemical constituents is carried out by chemical process. This analysis is important for large scale trials i.e. boiler trial. This analysis useful for calculation of amount of air required for complete combustion of 1 Kg. of coal. This analysis gives percentage of carbon, hydrogen, oxygen, sulphur and ash on mass basis their sum is taken as equal to 100%. In this analysis moisture is consider as separate item. This analysis is also used to determine calorific value of the coal. Proximate analysis of solid fuels : In this analysis separation of coal into its physical components. This analysis made by means of chemical balance and temperature controlled furnace. In this analysis sample is heated into furnace. The components in analysis are fixed, carbon, volatile matter, moisture and ash. These components are expressed in percentage on mass basis and their sum is taken as 100% sulphur is determined separately. This analysis also used to determine heating value of the coal. Sr. No Ultimate analysis Proximate analysis Ultimate analysis is coal is Proximate analysis is coal is 01 complete breakdown of coal into chemical constituents complete breakdown of coal into physical constituents 02 This analysis gives percentage of carbon, hydrogen, oxygen, Sulpher and ash. This analysis gives percentage of moisture, volatile matter, fixed carbon and ash. R.K.Yadav/Automobile Engg Dept/New Polytechnic Kolhapur. Page 4

Advantages of liquid fuels over solid fuels used in boiler. Advantages : 1) Liquid fuel having higher calorific value. 2) Less space is required for storage. 3) Easy control of combustion by stopping supply of fuel. 4) It is very clean fuel, dust free. 5) Reduction in cost of handling. 6) Easily transported through pipes. 7) During burning it does not form ash. Disadvantages : 1) Cost of liquid fuel is high. 2) The storage tank specially designed. 3) It has higher cost. 4) Danger of explosion. 5) Liquid fuels mostly we import from other countries. So we depends on other countries. Advantages of gaseous fuels : 1. They are free from solid and liquid impurity. 2. Maximum complete combustion of gaseous fuel is possible. 3. The rate of combustion and temperature in the combustion chamber can be easily controlled. 4. For complete combustion less amount of excess air is required. 5. Do not produce ash and smoke. 6. Large amount of heat and temperature is obtained at a moderate cost. Disadvantages : 1) They are readily inflammable. 2) They require large storage capacity. 3) The cost of gaseous fuel are more. Advantages of solid fuel : 1. Solid fuel can be stored conveniently without any risk of explosion. R.K.Yadav/Automobile Engg Dept/New Polytechnic Kolhapur. Page 5

2. They can easily transported. 3. They have moderate ignition temperature. 4. Leakage problem is not takes place. Disadvantages : 1. Rate of combustion of solid fuel can t be easily controlled. 2. Large amount of heat is wasted. 3. The ash content of solid fuel is very high. 4. The cost of handling of solid fuel is high. 5. After burning it produce large quantity of smoke. Merits of liquid fuels over gaseous fuels: (Any four) 1. Required less space for storage. 2. Higher calorific value. 3. Easy control of consumption. 4. Easy handling & transportation. 5. Absences of danger from spontaneous combustion. Sr.No. Natural liquid fuel Artificial liquid fuel 01 It is obtained from It is obtained by distillation process of reservoirs in the earth. crude oil 02 Raw material of oil This is the final product of oil industries. industries 03 Impure form of fuel. Pure form of fuel. 04 It is cheap. It is costly. 05 Crude petroleum. Gasoline, Kerosene, Diesel, Lubricating oil, and grease. Sr.No. Solid Fuels Gaseous Fuel 01 Required Large space Required Large space 02 Low calorific value Low calorific value For combustion more air is 03 required Produce ash & smoke after 04 combustion 05 Low cost High cost 06 Impure form Pure form For combustion less air is required Do not Produce ash & smoke after combustion R.K.Yadav/Automobile Engg Dept/New Polytechnic Kolhapur. Page 6

Bomb Calorimeter : It is a apparatus used for finding the higher calorific value of solid and liquid fuels. In this calorimeter, as shown in Fig. the fuel is burnt at a constant volume and under a high pressure in a closed vessel called bomb. Construction : The bomb is made mainly of acid-resisting stainless steel, machined from the solid metal, which is capable of withstanding high pressure (up to 100 bar), heat and corrosion. The cover or head of the bomb carries the oxygen valve for admitting oxygen and a release valve for exhaust gases. A cradle or carrier ring, carried by the ignition rods, supports and Silica crucible, which in turn holds the sample of fuel under test. There is an ignition wire of Tungston, Platinum or Chrome which dips into the crucible. It is connected to a battery, kept outside, and can be sufficiently heated by passing current through it so as to ignite the fuel. The bomb is completely immersed in a measured quantity of water. The heat, liberated by the combustion of fuel, is absorbed by this water, the bomb and copper vessel. The rise in the temperature of water is measured by a precise thermometer, known as Beckmann thermometer which reads up to 0.01 o C. R.K.Yadav/Automobile Engg Dept/New Polytechnic Kolhapur. Page 7

Procedure : A carefully weighed sample of the fuel (usually one gram or so) is placed in the crucible. Pure oxygen is then admitted through the oxygen valve, till pressure inside the bomb rises to 30 atmosphere. The bomb is then completely submerged in a known quantity of water contained in a large copper vessel. This vessel is placed within a large insulated copper vessel shown in the figure to reduce loss of heat by radiation. When the bomb and its contents have reached steady temperature (this temperature being noted), fuse wire is heated up electrically. The fuel ignites, and continues to burn till whole of it is burnt. The heat released during combustion is absorbed by the surrounding water and the apparatus itself. The rise in temperature of water is noted. Let m f = Mass of fuel sample bunt in the bomb in Kg. H.C.V. = Higher calorific value of the fuel sample in KJ/Kg. m w = Mass of water filled in the calorimeter in Kg. m e = Water equivalent of apparatus in kg. t 1 = Initial temperature of water and apparatus in o C & t 2 = Final temperature of water and apparatus in o C. We know that heat liberated by fuel = m f x H.C.V. (i) and heat absorbed by water and apparatus = (m w + m e ) x C w x (t 2 t 1 ) (ii) Since the heat liberated is equal to the heat absorbed (neglecting losses), therefore equating equations (i) and (ii), m f x H.C.V. = (m w + m e ) x C w x (t 2 t 1 ) H.C.V. = (m w + m e ) x C w x (t 2 t 1 ) m f R.K.Yadav/Automobile Engg Dept/New Polytechnic Kolhapur. Page 8

Theoretical (or) Minimum mass of air required for complete combustion. We know proper supply of oxygen is very essential for the complete combustion of a fuel, for obtaining maximum amount of heat from a fuel. The theoretical or minimum mass (or) volume of oxygen required for complete combustion of 1 kg of fuel may be calculated from chemical analysis of the fuel.the mass of oxygen, required by each of the constituents of the fuel, may be calculated from the chemical equation. Now consider 1 kg of a fuel. Let; Mass of the carbon = C kg Mass of the hydrogen = H 2 Mass of sulphur = S kg we know that 1 kg of carbon requires 8/3 kg of oxygen for its complete combustion similarly; 1 kg of hydrogen requires 8 kg of oxygen and 1 kg of sulphur requires 1 kg of oxygen for its complete combustion Total oxygen required for complete combustion of 1 kg of fuel. = 8 C + 8 H 2 + 5 kg -------------------1 3 If some oxygen (say O 2 kg) is already present in the fuel, then total oxygen for complete combustion of 1 kg of fuel. = 8 C + 8 H 2 + S - O 2 kg -------------------2 3 Normally oxygen has to be obtained from atmospheric air which mainly consist of nitrogen & oxygen along with of rare gases like argon, neon and krypton etc. But for all calculations. the compositions of air is taken as; R.K.Yadav/Automobile Engg Dept/New Polytechnic Kolhapur. Page 9

Nitrogen (N 2 ) = 77% Oxygen (O 2 ) = 23% --------(By mass) and Nitrogen (N 2 ) = 79 % Oxygen (O 2 ) =21% ---------(By volume) for obtaining 1 kg of oxygen; amount of air required = 100 = 4.35 Kg ----------(By mass) 23 Theoretical (or) minimum air required for complete combustion of 1 kg of fuel, = 100 8 C + 8 H 2 + S - O 2 Kg. 23 3 Excess air supplied :- We know, minimum air required for complete combustion but many times for complete combustion and rapid combustion of fuel. some quantity of air in form of excess is supplied. If just minimum amount of air is supplied a part of the fuel may not burn properly. The amount of excess air supplied varies with the type of fuel and firing conditions. It may approach to a value of 100 percent; but in modern days it uses 25 to 50% excess air. Mass of excess air supplied :- The mass of excess air supplied may determined by the mass of unused oxygen found in a flue gases. we know that in order to supply one kg of oxygen we need 100/23 kg of air. Similarly; Mass of excess air supplied = 100 x Mass of excess oxygen. 23 Total Mass of air supplied = Mass of necessary air + Mass of excess air. R.K.Yadav/Automobile Engg Dept/New Polytechnic Kolhapur. Page 10

IMPORTANT FORMULAES : 1. Theoretical C.V. of the fuel =33800 C +144000 (H 2 O 2 /8) +9270 S KJ/Kg. 2. L.C.V. = H.C.V. 9 H 2 x 2466 KJ/Kg. 3. Mass of air required for complete combustion of 1 Kg. of fuel. = 100/23 (2.67 C + 8 H 2 +S O 2 ) Kg. 4. Mass of excess air supplied = 100/23 x Mass of excess Oxygen. Numerical on fuels and combustion : TYPE1 : Ex.1 A sample of coal has following composition on mass basis Carbon 82%, Hydrogen 8%, Sulphur 2%, Oxygen 4% and Ash 4%. Calculate using Dulong s formula higher and lower calorific value of fuel. (S-08,09,W-10,11) Sol n : Given Composition of coal on mass basis. Carbon (C) = 0.82 Hydrogen (H 2 ) = 0.08 Sulphur (S) = 0.02 Oxygen (O 2 ) = 0.04 Ash = 0.04 We know Dulong s formula. H.C.V. of Coal = 33800 C+144000 (H 2 - O 2 /8) + 9270 S KJ/Kg. Putting above values in formula. = 33800 x 0.82 + 144000 (0.08 0.04/8) + 9270 x 0.02 R.K.Yadav/Automobile Engg Dept/New Polytechnic Kolhapur. Page 11

H.C.V. of coal = 27716 + 10800 + 185.4 = 38701.4 KJ/Kg. L.C.V. = H.C.V. 9 H 2 x 2466 = 38701.4 9 x (0.08) x 2466 = 38701.4 1775.52 L.C.V. =36925.88 KJ/Kg. Ex.2 A sample of coal has the following composition by mass, carbon 76%, Hydrogen 5%, Oxygen 8.5%, Nitrogen 2%, Sulphur 1.5% and Ash 7% calculate higher and lower calorific value of fuel per Kg. (S-11) Sol n : Given Composition of coal by mass. Carbon (C) = 0.76 Hydrogen (H 2 ) = 0.05 Oxygen (O 2 ) = 0.085 Nitrogen (N 2 ) = 0.02 Sulphur (S) = 0.015 Ash = 0.07 Now We know Dulong s formula. 1) H.C.V. of Coal = 33800 C +144000 (H 2 - O 2 /8) + 9270 S KJ/Kg. = 33800 x 0.76 + 144000 (0.05 0.085/8) + 9270 x 0.015 = 25688 + 5670 + 139.05 = 31497.05 KJ/Kg. 2) L.C.V. = H.C.V. 9 H 2 x 2466 KJ/Kg. = 31497.05 9 x (0.05) x 2466 = 31497.05 1109.7 L.C.V. = 30387.35 KJ/Kg. R.K.Yadav/Automobile Engg Dept/New Polytechnic Kolhapur. Page 12

Ex.3 A sample of coal has the following composition by mass Carbon 75%, Hydrogen 6%, Oxygen 8%, Nitrogen 2.5%, Sulphur 1.5% and Ash 7% calculate higher and lower calorific value of per Kg. Sol n : Given Composition of coal by mass. Carbon (C) = 0.75 Hydrogen (H 2 ) = 0.06 Oxygen (O 2 ) = 0.08 Nitrogen (N 2 ) = 0.025 Sulphur (S) = 0.015 Ash = 0.07 We know Dulong s formula. 1) H.C.V. of Coal = 33800 C+144000 (H 2 - O 2 /8) +9270 S KJ/Kg. = 33800 x 0.75 + 144000 (0.06 0.085/8) + 9270 x 0.015 = 25350 + 7200 + 139.05 H.C.V. = 32689.05 KJ/Kg. 2) L.C.V. of Coal = H.C.V. 9 H 2 x 2466 KJ/Kg. = 32689.05 9 x (0.06) x 2466 = 32689.05 1331.64 L.C.V. = 31357.41 KJ/Kg. Ex.4 A sample of coal has the following composition by mass Carbon 60%, Hydrogen 10%, Oxygen 15%, Nitrogen 4.5%, Sulphur 3.5% and remaining is ash calculate H.C.V. and L.C.V. of per Kg. Sol n : Given Composition of coal by mass. Carbon (C) = 0.60 Hydrogen (H 2 ) = 0.10 R.K.Yadav/Automobile Engg Dept/New Polytechnic Kolhapur. Page 13

Oxygen (O 2 ) = 0.15 Nitrogen (N 2 ) = 0.045 Sulphur (S) = 0.035 We know Dulong s formula. 1) H.C.V. of Coal = 33800 C +144000 (H 2 - O 2 /8) + 9270 S KJ/Kg. = 33800 x 0.60 + 144000 (0.10 0.015/8) + 9270 x 0.035 = 20280 + 11700 + 324.45 = 32304.45 KJ/Kg. 2) L.C.V. of Coal = H.C.V. 9 H 2 x 2466 KJ/Kg. = 32304.45 9 x (0.10) x 2466 = 32304.45 2219.4 L.C.V. = 30085.05 KJ/Kg. Ex 5.(S-14) A coal has the following composition by mass Carbon 80%, Hydrogen 5%, Oxygen 6%, Nitrogen 2.5%, Sulphur 1.5% and Ash 5%. Calculate HCV and LCV per kg of coal. Data: Carbon = C = 80% = 0.8 Hydrogen = H 2 = 5% = 0.05 Oxygen = O 2 = 6% = 0.06 Nitrogen = N = 2.5% = 0.025 Sulphur = S =1.5% = 0.015 Ash = 5% = 0.05 Dulong s formula: H.C.V. of coal = 33800 C + 144000 ( H 2 - O 2 /8 ) + 9270 S KJ / Kg = 33800 x 0.8 + 144000 (0.05-0.06/8) + 9270 x 0.015 = 33299.05 KJ / Kg L.C.V. of coal = H.C.V.- 9 H 2 x 2466 KJ / Kg = 33299.05 9 x 0.05 x 2466 = 32189.35 KJ / Kg R.K.Yadav/Automobile Engg Dept/New Polytechnic Kolhapur. Page 14

TYPE 2 : Ex.1 Sol n : Given The following is the percentage composition of a sample of coal on mass basis. C = 85, H 2 = 4, O 2 = 10 and remaining is ash find minimum mass of air required for complete combustion of 1 Kg. of coal. Composition of coal on mass basis. Carbon (C) = 0.85 Hydrogen (H 2 ) = 0.04 Oxygen (O 2 ) = 0.10 Minimum mass of air required for complete combustion of 1 Kg. of fuel. = 100/23 (2.67 C + 8 H 2 + S O 2 ) Kg. = 100/23 (2.67 x 0.85 + 8 + 0.04 + 0 0.10) = 100/23 (2.2695 + 0.32 0.1) = 40.82 Kg. per Kg. of Coal burnt. Ex.2 The following is the percentage composition of coal on mass basis. C = 80, H 2 = 3.3, O 2 = 4 and S = 0.9 and remaining is ash. Calculated theoretical air required to 1 Kg. of coal completely. Sol n : Given Composition of coal on mass basis. Carbon (C) = 0.80 Hydrogen (H 2 ) = 0.033 Oxygen (O 2 ) = 0.004 Sulphur (S) = 0.009 Minimum mass of air required for complete combustion of 1 Kg. of fuel. = 100/23 (2.67 C + 8 H 2 + S O 2 ) Kg. = 100/23 (2.67 x 0.80 + 8 x 0.033 + 0.009 0.004) R.K.Yadav/Automobile Engg Dept/New Polytechnic Kolhapur. Page 15

= 100/23 (2.136 + 0.264+ 0.009 0.004) = 10.456 Kg. per Kg. of Coal burnt. Ex.3 During a boiler trial the coal analysis on mass basis was reported as C = 62.4%, H 2 = 4.2%, O 2 = 4.5%, Moisture = 15% and Ash 13.9%. Calculated minimum air required to burn 1 Kg. of coal also calculate H.C.V. & L.C.V. Sol n : Given Composition of coal on mass basis. Carbon (C) = 62.4 = 0.624 Hydrogen (H 2 ) = 4.2% = 0.042 Oxygen (O 2 ) = 4.2% = 0.045 Moisture = 15% = 0.15 Ash = 13.9 = 0.139 Now Minimum mass of air required for complete combustion of 1 Kg. of fuel. = 100/23 (2.67 C + 8 H 2 + S O 2 ) Kg. = 100/23 (2.67 x 0.624 + 8 x 0.042 + 0 0.044) = 100/23 (2.136 + 0.264 + 0.009 0.004) = 100/23 (1.666 + 0.336 0.045) = 8.613 Kg. per Kg. of Coal burnt. We know Dulong s formula. H.C.V. of Coal = 33800 C +144000 (H 2 - O 2 /8) + 9270 S KJ/Kg. = 33800 x 0.624 + 144000 (0.042 0.045/8) + 9270 X 0 H.C.V. = 26329.2 KJ/Kg. L.C.V. of Coal = H.C.V. 9 H 2 x 2466 KJ/Kg. = 263329.2 9 x 0.042 x 2466 = 26329.2 932.148 = 25397.052 KJ/Kg R.K.Yadav/Automobile Engg Dept/New Polytechnic Kolhapur. Page 16

Question bank : Ch.6 Fuel and Combustion 16 marks 1 2 How fuels are classified? Define Calorific value of the fuel. 2 m. 2 m. 3 Define L.C.V. & its unit 2 m. 4 What is calorific value of fuel. What is H.C.V. 2 m. 5 What is H.C.V. & L.C.V.? 2 m. 6 7 Define fuel and state the type of fuel. Enlist any four types of gaseous fuels 2 m. 2 m. 8 List the properties of fuel.(any 4 ) 2 m. 9 List out the merit of liquid fuel over gaseous fuels. 2 m. 10 11 State requirement of good fuel. Differentiate between Natural and Artificial liquid fuel. 4 m. 4 m. 12 State & explain Dulong s formula for theoretical 8 m. determination of calorific value of fuel. 13 Give the significance of ultimate analysis of fuel. How is % 8 m. 14 of carbon & hydrogen determined in this analysis. Define calorific value of fuel. Differentiate between H.C.V. and L.C.V. of the fuel. State which value is used in calculation and why? 15 Explain Ultimate analysis and proximate analysis of coal 8 m. Explain H.C.V. and L.C.V. of the fuels. 16 Describe with neat sketch construction and working of Bomb 8 m. 17 calorimeter. Write Dulong s formula and state it s use. Compare i) Solid fuel and Gaseous fuel ii) Ultimate analysis and proximate analysis 8 m. R.K.Yadav/Automobile Engg Dept/New Polytechnic Kolhapur. Page 17