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Fuels and Combustion

Fuels and Combustion. Lecture 1. Introduction Definition Classification of fuels Calorific Value and its units. 2. Introduction.

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Fuels and Combustion

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  1. Fuels and Combustion

  2. Lecture 1 • Introduction • Definition • Classification of fuels • Calorific Value and its units 2

  3. Introduction • A chemical fuel is defined as a combustible substance containing carbon as the main constituent, which on complete combustion gives large amount of heat that can be used economically for domestic and industrial purposes. Different Energy Sources • Non-renewable Fossil Fuels • Renewable Sources such as solar energy, wind energy, biofuels etc. 3

  4. Characteristics of a good fuel • High Calorific value • Moderate ignition temperature • Low moisture content • Low ash content • Moderate velocity of combustion • Less hazardous combustion products • Low cost • Easy Storage 4

  5. Classification of fuels 5

  6. Calorific value • To grade a fuel and assess its quality, it is important to find the calorific value of the fuel. • Calorific value of a fuel is "the total quantity of heat liberated, when a unit mass (or volume) of the fuel is burnt completely." Units (1) CGS: Calorie/g (2) SI unit: Joule/g (3) British Thermal unit/pound (B.T.U./lb) 1 Calorie = 4.187 J 1 B.T.U. = 252 cal = 0.252 kcal 1 kcal = 3.968 6

  7. Gross Calorific Value & Net Calorific Value • Gross Calorific Value (GCV) GCVof a fuel can be defined as the total amount of heat obtained on complete combustion of unit mass of a solid or liquid fuel or unit volume of a gaseous fuel (STP) and on cooling the products of combustion to 15°C. The gross calorific value is also called as higher calorific value. • Net Calorific Value (NCV) NCV is defined as the amount of heat obtained practically on complete combustion of unit mass of solid or liquid fuel or unit volume of a gaseous fuel at STP and the products of combustion are allowed to escape with some heat. NCV is also called as lower calorific value. NCV = GCV- [mass of steam x specific latent heat of condensation of water] . 7

  8. Lecture 2 • General Principle of Calorimetry • Methods of determination of calorific value • Bomb calorimeter Principle Construction Working Formula Corrections Numericals 8

  9. Bomb Calorimeter • Principle of Bomb Calorimeter A known weight of solid / non-volatile liquid fuel is burnt in the presence of excess oxygen in the closed pot, and the products of combustion are cooled, to get GCV of the fuel. https://www.youtube.com/watch?v=nJOH29SGcCk 10

  10. Diagram of Bomb Calorimeter 11

  11. Calculations of Bomb Calorimeter Mass of fuel = x g      Mass of water in calorimeter = W g   Water equivalent of calorimeter set = w g Gross Calorific value of fuel = L cal/g    Rise in temperature of water = (t2 – t1) oC Heat liberated by burning fuel = Heat absorbed by water and calorimeter 12

  12. Corrections (i) Fuse wire correction (f) The heat liberated includes the heat given out by ignition of the fuse wire used. (ii) Acid correction (a) S + O2 → SO2 2SO2 + O2 + 2H2O → 2H2SO4 ∆H = -6.03 x 105 J         4N + 2H2O + 5O2 → 4HNO3 ∆H = - 2.39 x 105 J (iii) Cooling correction (tc) Corrected formula for GCV 13

  13. Numericals 1) The coal containing 5% hydrogen (dry / moisture free basis) and 10% moisture has gross calorific value of 33.5 MJ/kg. Calculate Net Calorific Value of Coal. Latent heat of water vapour is 2.45 MJ/kg. 2) 0.72 g of a fuel containing 80% Carbon, when burnt in a bomb calorimeter, increased the temperature of water from 27.3°C to 29.1°C. If the calorimeter contains 250 g of water and its water equivalent is 150 g, calculate GCV of the fuel. 14

  14. Numericals 3) The temperature of 950.0 g of water increased from 25.5 0C to 28.5 0C. On burning, 0.75 g of solid fuel in a Bomb calorimeter. Water equivalent of copper calorimeter and latent heat of steam are 400 g and 587 cal/g respectively. If the fuel contains 0.65% of hydrogen, calculate net calorific value. 4) A sample of coal containing 5% H when allowed to undergo combustion in Bomb Calorimeter, the following data were obtained weight of coal burnt = 0.95 gweight of water taken = 700 gwater equivalent of bomb calorimeter = 2000 grise in temperature = 2.48 °Ccooling correction = 0.02 °Cfuse wire correction = 10 calacid correction = 60 calCalculate Gross and Net Calorific Value of Coal. 15

  15. Lecture 3 • Boy’s Calorimeter Principle Construction Working Formula Numericals • Solid fuel - Coal Methods of Analysis Proximate Analysis and its numericals 16

  16. Boy’s calorimeter Principle of Boy’s Calorimeter A known volume of volatile liquid / gaseous fuel is burnt and the heat is transferred from burning the fuel for heating the water that circulates in a coil heat exchanger. 17

  17. Diagram 18

  18. Calculations Let, V = Volume in m3 W = mass of water used in kg m = mass of water condensate in kg L = GCV of the fuel Heat produced by combustion of fuel = heat absorbed by cooling water 19

  19. Numericals 1) Observations in the Boy’s gas calorimeter experiment on a gaseous fuel are given below; Find the G.C.V. and N.C.V. of the fuel, Volume of gas burnt (STP) = 0.08 m3Mass of cooling water used = 29.5 kgRise in temperature of circulating water = 9.1°CMass of steam condensed = 0.04 kg 2) In Boy’s gas calorimeter’s experiment when 0.1m3 of a fuel gas is burnt during which 25 kg of water is circulated. Temperature of incoming water and outgoing water is 20°C and 33°C respectively. Weight of steam condensed is 250 gm. Calculate gross calorific value and net calorific value, if heat liberated in condensing water vapour and cooling the condensate is 586 kcal/kg. 20

  20. COAL https://www.youtube.com/watch?v=nIYz4Ck3w-k 21

  21. Analysis of Coal Proximate analysis • Moisture • Volatile Matter • Ash • % Fixed carbon • Ultimate analysis • 1. Hydrogen • 2. Nitrogen • 3. Sulphur • 4. Oxygen 22

  22. Proximate analysis 23

  23. Proximate analysis • Volatile Matter (V.M.) % • Ash % • Fixed carbon % Fixed carbon % = 100 – (% Moisture + % V. M. + % ash) 24

  24. Significance of Proximate analysis • Mositure: Lower moisture content implies that the quality of coal is good • Volatile matter: Lower VM better is the quality of coal • Ash: Low ash content implies good quality coal • Carbon: Higher the carbon content better the quality of coal. 25

  25. Numericals • 1.508 g of coal sample was heated at 110 oC for 1h. On cooling the weight of the sample was found as 1.478 g. Strong heating of the sample at 950 oC for 7 min. carried out in a closed crucible. The sample on cooling weighed 1.058g. Calculate % moisture and % volatile matter present in the sample. • 1.2 g of coal sample heated at 105-110 oC for 1 h, after heating the sample weighed 1.16 g. This remaining sample of coal ignited at constant weight of 0.09 g. In another experiment, 1.2 g of sample was heated in a crucible at 950 oC for 7 min. After cooling the residue weighed 0.8 g. Calculate % of fixed carbon 26

  26. Lecture 4 Ultimate Analysis • Estimation of Carbon and hydrogen • Estimation of nitrogen • Estimation of Sulphur Numericals 27

  27. Ultimate Analysis of Coal The analysis of coal in which percentages of C, H, N, S and O elements are found out, is known as ultimate analysis. • Estimation of Carbon and Hydrogen 28

  28. 29

  29. Estimation of Nitrogen 31

  30. Estimation of Nitrogen Estimation of Sulphur 30

  31. Numericals 1) 0.25 g of a coal sample on burning in a combustion chamber in the current of pure oxygen was found to increase weight of U-tube with anhydrous CaCl2 by 0.075 g and of KOH U-tube by 0.52gm. Find C and H percentages in coal. 2) Find the % of C and H in coal sample from the following data- 0.20 g of coal on burning in a combustion tube in presence of pure oxygen was found to increase in the weight of CaCl2 tube by 0.08 g and KOH tube by 0.12 g. 32

  32. Lecture 5 • Petroleum • Classification of petroleum • Refining of petroleum • Knocking • Octane Number of petrol • Cetane number of diesel 33

  33. Liquid Fuels -Petroleum Petroleum is a naturally occurring substance consisting of organic compounds in the form of gas, liquid, or semisolid. Organic compounds are carbon molecules that are bound to hydrogen (e.g., hydrocarbons) and to a lesser extent sulphur, oxygen, or nitrogen. 34

  34. Classification of Petroleum • There are three types 1) Paraffinic base type: Saturated hydrocarbons from CH4 to C35H72 2) Asphaltic base type: Cycloparaffins with little amount of paraffins and aromatic hydrocarbons 3) Mixed base type: Both paraffinic and asphaltic hydrocarbons 35

  35. Refining of Petroleum The refining of petroleum consists of (i) Removal of water (ii) Removal of sulphur (iii) Fractional Distillation • https://www.youtube.com/watch?v=J2-tDV8KYEA&t=30s 36

  36. Fractional Distillation of Petroleum Principle of Fractional Distillation Separation of a liquid mixture into fractions differing in boiling point (and hence chemical composition) by means of distillation, typically using a fractionating column. Portion of fractionating tower Fractionating tower 37

  37. Various fractions of crude oil 38

  38. Octane Number of Petrol • Knocking of Petrol engine: The intense vibrations, strong rattling metal noises produced in petrol engine due to pre-ignition of petrol, is called knocking. • Octane Number • Definition: Octane number of a petrol sample is defined as the % of iso-octane • in the mixture of iso-octane and n-heptane, which has similar knocking to the petrol sample. • Aromatics > Cycloalkanes > alkenes > Br. alkane > n-alkane https://www.youtube.com/watch?v=uWx1cXR7x_M 39

  39. Anti-knocking Agents • Tetraethyl lead Pb(C2H5)4 0.5ml/litre • Branched chain alkane of higher molecular weight • Benzene, toluene, xylene • Methyl t-butyl ether (MTBE) or ethyl t-butyl ether. 40

  40. Cetane Number of Diesel • Knocking of Diesel engine: The intense vibrations, strong rattling metal noises produced in diesel engine due to time lag or delay in ignition of diesel, is called knocking. • Cetane number • Definition: Cetane number of a diesel is defined as the % of n-hexadecane in the mixture of n-hexadecane and 2-methyl naphthalene which has same ignition character like the ignition character of the diesel under test • 2-Methyl napthalene n-Hexadecane / Cetane • Cetane no. 0 Cetane no. 100 • CH3-(CH2)14-CH3 • Relative order of cetane no. values • n-alkanes > cycloalkanes > alkenes > branched alkanes > aromatics 41

  41. How to increase cetane number • Alkyl nitrates (principally 2-ethylhexyl nitrate) and di-tert-butyl peroxide are used as additives to raise the cetane number. • Antioxidant to improve oxidation resistance during storage. • Lubricity additives for lubrication of fuel injection system. 42

  42. Lecture 6 • Combustion • Calculations • Numericals 43

  43. Combustion • Combustion is a high-temperature exothermic redox chemical reaction between a fuel (the reductant) and an oxidant, usually atmospheric oxygen. • Substances always combine in definite proportions. These proportions are determined by their molecular masses. C + O2 → CO2 (12:32:44) • 22.4 L of any gas at 0°C and 760mm Hg pressure (STP) has a mass equal to its 1 mol. • Air contain 21% of oxygen by volume and 23% of oxygen by mass. • 28.94 g/mol is taken as molar mass of air. • O2 required for combustion = theoretical O2 required - O2 present in the fuel. 44

  44. Combustion • For solid or Liquid fuels: Where C, H, S and O are amounts of elements in Kg. Air quantity = For Gaseous fuels O2 volume required = volume gas component in m3 x volume of O2per volume of gas. 45

  45. Numericals on combustion 1) A petrol sample contains 14% H and 86% carbon. Calculate the quantity of air required for complete combustion of 1 kg petrol. 2) Volumetric analysis of producer gas is, H2 = 20 % CO = 22 % N2 = 50 %, CH4 = 2 % and CO2 = 6 %. Find volume of air required for complete combustion of 1 m3 of the gas. 3) A gas has following composition by volume, H2 = 20 %, CH4 = 6%, CO = 18%, O2 = 5%, N2 = 43 %. If 25 % excess air is used. Find volume of air actually supplied per m3 of the gas. 46

  46. Numericals on Combustion 4) A gas has following composition by volume: H2 = 20 %; CH4 = 6% , CO = 22%, CO2 = 44% O2= 4% N2 = 4%. Calculate the volume of air required for complete combustion of 1 m3 of the fuel. 5) A sample of coal requires 20% excess air for complete combustion. Calculate weight of air for 250 g of the coal, if its composition is C = 81%, H = 4%, N = 1.5%, S = 1.2%, O = 3%, ash = 9.35. 6) A gas used in internal combustion engine contains, H2 = 45 %, CO = 15%, CH4 =35%, N2= 5 % Find the minimum quantity (volume) of air required per 1 m3 of the gas for complete combustion. 47

  47. Lecture 7 • Carbon Credit • Carbon Emission 48

  48. Carbon Credit A carbon credit is a generic term for any tradable certificate or permit representing the right to emit one tonne of carbon dioxide or the mass of another greenhouse gas with a carbon dioxide equivalent (tCO2e) equivalent to one tonne of carbon dioxide. 49

  49. Greenhouse Effect • The greenhouse effect is the process by which radiation from a planet's atmosphere warms the planet's surface to a temperature above what it would be without its atmosphere. Greenhouse Gases - • Greenhouse gases are a group of compounds that are able to trap heat (longwave radiation) in the atmosphere, keeping the Earth's surface warmer than it would be if they were not present. Following are various greenhouse gases. • Carbon dioxide • Nitrous oxide • Chloro fluorocarbons • Methane • Hydrofluorocarbon 50

  50. Global warming • Global warming, also referred to as climate change, is the observed century-scale rise in the average temperature of the Earth's climate system and its related effects. • Anticipated effects include warming global temperature, rising sea levels, changing precipitation, and expansion of deserts in the subtropics. 51

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