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Fuels ppts 674816

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fuels

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Fuels ppts 674816

  1. 1. * CHEMISTRY OF FUELS Visit www.gyandharaonline.com for more notes n updates
  2. 2. * • Fuels: substances which undergo combustion in the presence of air to produce a large amount of heat that can be used economically for domestic and industrial purpose. • This definition does not include nuclear fuel because it cannot be used easily by a common man. • The various fuels used economically are wood, coal, kerosene, petrol, diesel gasoline, coal gas, producer gas, water gas, natural gas (LPG) etc. Visit www.gyandharaonline.com for more notes n updates
  3. 3. * Classifications Fuels can be broadly classified by origin as, (i)Primary or natural fuels: coal, wood etc (ii)Secondary or artificial or derived fuels: petrol, diesel On the basis of physical state, as : (i) Solid fuels (ii) Liquid fuels (iii) Gaseous fuelsV isit www.gyandharaonline.com for more notes n updates
  4. 4. * Basis Origin Physical State Source Natural or primary Artificial or Secondary or Derived Wood, peat, lignite, coal Semi coke, charcoal Solid fuels Crude oil, Vegetable oils Petrol, kerosene, gas oil, coal tar, alcohol Liquid fuels Natural gas Producer gas, coke-oven gas, water gas, blast furnace gas, compressed butane gas, LPG Gaseous fuels Visit www.gyandharaonline.com for more notes n updates
  5. 5. * Characteristics of Fuels The physical properties for which fuels are tested and their ideal requirements are listed below : (i)Calorific value or specific heat of combustion. - efficiency of fuel: how much heat it produces (ii) Ignition temperature (iii) Flame temperature (iv) Flash and Fire point. Visit www.gyandharaonline.com for more notes n updates
  6. 6. * (v) Aniline point (vi) Knocking. (vii) Specific gravity (viii) Cloud and Pour point (ix) Viscosity (x) Coke number. Visit www.gyandharaonline.com for more notes n updates
  7. 7. * The chemical properties include the compositional analysis of fuel. For solid and liquids fuels : (i) Percentage of various elements such as C, H, O, N, S, etc. (ii) Percentage of moisture (iii) Percentage of volatile matter Visit www.gyandharaonline.com for more notes n updates
  8. 8. * For gaseous fuels : (i) Percentage of combustible gases e.g. – CO, H2, CH4, C2H4, C2H6, C4H10, H2S etc. (ii) Percentage of non-combustible gases e.g. N2, CO2 etc. Visit www.gyandharaonline.com for more notes n updates
  9. 9. * Calorific Value • number of units of heat evolved during complete combustion of unit weight of the fuel. • A British Thermal Unit: the heat required to raise the temperature of one pound of water from 60° F to 61° F. • The Calorie: the heat required to raise the temperature of one kg of water from 15°C to 16°C. Visit www.gyandharaonline.com for more notes n updates
  10. 10. * High and Low Calorific Values Calorific values are of two types as, (i)High or Gross Calorific Value (H.C.V. or G.C.V.) (ii)Low or Net Calorific Value (L.C.V. or N.C.V.) High calorific value may be defined as, the total amount of heat produced when one unit of the fuel has been burnt completely and the combustion products have been cooled to 16°C or 60°F. Visit www.gyandharaonline.com for more notes n updates
  11. 11. * • LCV: is the net heat produced when unit mass or volume of fuel is completely burnt and products are allowed to escape. • Net or Low C.V.= Gross C.V. – loss due to water formed • Or Gross C.V – Mass of hydrogen ´ 9 ´ Latent heat of steam (587 cal/g) • (Because 1 part by weight of hydrogen produces 9 parts (1 + 8) by mass of water) Visit www.gyandharaonline.com for more notes n updates
  12. 12. * • The calorific value of fuels (e.g. Coal) is determined theoretically by Dulong formula, or I.A. Davies formula. • Dulong formula can be expressed as, HCV = 1/100 [8,080 C+ 34,500(H- O/8)+ 2240 S] Where C = % Carbon, H = % Hydrogen, O = % Oxygen, S = % Sulphur Visit www.gyandharaonline.com for more notes n updates
  13. 13. * • Oxygen in fuel (coal) is in combined state as water and hence it does not contribute to heating value of fuel. • LCV = [HCV – 0.09 H(%) × 587] cal/g Visit www.gyandharaonline.com for more notes n updates
  14. 14. * Bomb Calorimeter Visit www.gyandharaonline.com for more notes n updates
  15. 15. * • Let x = mass in g of fuel taken in crucible • W = mass of water in calorimeter • w = water equivalent in g of calorimeter, stirrer, thermometer, bomb etc. • t1 & t2 are initial & final temperatures of water in calorimeter • L = higher calorific value of fuel in cal/g • Then heat liberated by buring of fuel = xL • Heat absorbed by water & apparatus = (W+w)(t2-t1) • But heat liberated by fuel = heat absorbed by water, apparatus • so, xL = (W+w)(t2-t1) • L = (W+w)(t2-t1)/x cal/g or kcal/kg Visit www.gyandharaonline.com for more notes n updates
  16. 16. • If H = % of hydrogen in fuel • 9H/100 g = mass of water from 1 g of fuel= 0.09H g • So heat taken by water in forming steam = 0.09 H × 587 cal • LCV = HCV - 0.09 H × 587 cal/g • By considering fuse wire correction, acid correction & cooling corection • L = [{(W+w)(t2-t1+ cooling correction)}- {acid +fuse correction}]/x cal/g or kcal/kg * Visit www.gyandharaonline.com for more notes n updates
  17. 17. * Sr. No. Property Solid Fuels Liquid Fuels Gaseous Fuels 1. Calorific value Low Higher Highest 2. Specific gravity Highest Medium Lowest 3. Ignition point High Low Lowest 4. Efficiency Poor Good Best 5. Air required for combustion Large and excess of air Less excess of air Slight excess of air 6. Use in I.C. engine Cannot be used Already in use Can be used 7. Mode of supply Cannot be piped Can be piped Can be piped 8. Space for storage Large 50% less than solid fuel Very high space 9. Relative cost Cheaper Costly More costly than other two 10. Care in storage and transport Less care required Care is necessary Great care required Visit www.gyandharaonline.com for more notes n updates
  18. 18. * Combustion • Combustion is a process in which oxygen from the air reacts with the elements or compounds to give heat. • As the elements or compounds combine in indefinite proportions with oxygen, we need to calculate what is minimum oxygen or air required for the complete combustion of compounds. The commonly involved combustion reactions are : Visit www.gyandharaonline.com for more notes n updates
  19. 19. i) C + O2 ® CO2 ii) 2H2 + O2 ® 2H2O OR H2 + (O) ® H2O iii) S + O2 ® SO2 iv) 2CO + O2 ® 2CO2 OR CO + (O) ® * CO2 v) CH4 + 2O2 ® CO2 + 2H2O vi) 2C2H6 + 7O2 ® 4CO2 + 6H2O vii) C2H4 +3O2­® 2CO2 + 2H2O viii) 2C2H2 + 5O2 ® 4CO2 + 2H2O Visit www.gyandharaonline.com for more notes n updates
  20. 20. * Hint to Solve Problems on Calculation of Quantity of Air Required for Combustion of Fuel : • 1. First write the appropriate chemical reaction with oxygen and find their relation between the element or compound on weight or volume basis. e.g C + O2 ® CO2 by weight (12 gm) + (32 gm) (44gm) by volume 1 1 1 2H2 + O2 ® 2H2O by weight (4 gm) + (32 gm) (36gm) by volume 2 1 2 S + O2 ® SO2 by weight (32 gm) + (32 gm) (64gm) by volume 1 1 1
  21. 21. * 2) Calculate the oxygen required on the basis of unit quantity of fuel. 3) Calculate the total oxygen required for the combustion and subtract the oxygen which is present in the fuel. 4) The oxygen calculated should be converted into air by knowing that air contains 23 parts by weight of oxygen OR 21 parts by volume of oxygen. 5) The average molecular weight of air is 28.94 gm. Visit www.gyandharaonline.com for more notes n updates
  22. 22. * Calculate the weight and volume of air required for complete combustion of 5 kg. coal with following compositions, C = 85%; H = 10%; O = 5% Soln. : Combustion reactions : C + O2 ® CO2 12 + 32 ® 44 H2 + O2 ® H2O 2 + 16 ® 18 Visit www.gyandharaonline.com for more notes n updates
  23. 23. * Weight of elements per kg. of coal Weight of O2 required for complete combustion in kg. C = 0.85 0.85 ´32/12 = 2.26 kg. H = 0.1 0.1 ´ 8 = 0.8 kg. O = 0.05 – Total oxygen = 3.06 kg. Weight of oxygen required = Weight of oxygen needed – weight of oxygen present = 3.06 – 0.05 = 3.01 Visit www.gyandharaonline.com for more notes n updates
  24. 24. * Air required for complete combustion = 3.01 ´ 100/23 = 13.08 kg. per 1 kg. coal. Air required for 5 kg. of coal = 13.08 ´ 5 = 65.40 kg. Volume of Air 28.94 kg. of air = 22,400 ml volume at NTP 65.4 kg. of air =22400× 65.4/ 28.94 =50815.8 ml. Air =50.8158 litres of air Visit www.gyandharaonline.com for more notes n updates
  25. 25. * Sr No Types of Coal Classification of Coal Moisture of Air Dried At 40°C C% H% O% Ash % Calorific Value (kcal/kg) Uses 1. Peat 25 57 6 35 2 5400 Power generation and domestic purpose. 2. Lignite 20 67 5 20 8 6500 Manufacture of producer gas, thermal power plants. 3. Bitumi nous 4 83 5 15 7 8000 For metallurgical coke, coal gas, boiler, domestic purpose 4. Anthra cite 2 92 3 2 3 8600 Boilers, metallurgical fuel, domestic Visit www.gyandharaonline.com for more notes n updates
  26. 26. * Analysis of Coal The proximate analysis is easy and quicker and it gives a fair idea of the quality of coal. The ultimate analysis is essential for calculating heat balances in any process for which coal is employed as a fuel. Visit www.gyandharaonline.com for more notes n updates
  27. 27. * Moisture • It is determined by heating about one gm. of finely powdered coal at 105°C to 110°C for an hour in electric oven. The loss in weight is reported as due to moisture. • % Moisture = [loss in wt of sample × 100]/wt of coal taken • Decreases calorific value of coal • Takes away heat in the form of latent heat Visit www.gyandharaonline.com for more notes n updates
  28. 28. * Volatile matter • For determining volatile matter content, a known weight of dried sample is taken in a crucible with properly fitting lid. It is then heated at 950°C ± 20°C for exactly seven minutes in previously heated muffle furnace. The loss in weight is due to volatile matter which is calculated as • Volatile matter = [loss in wt at 9500C × 100]/wt of coal sample • Decreases calorific value • Forms smoke and pollutes air Visit www.gyandharaonline.com for more notes n updates
  29. 29. * Ash (non combustible matter) • A known weight of sample is taken in a crucible and the coal is burnt completely at 700°C – 750°C in muffle furnace until a constant weight is obtained. The residue left in the crucible is ash content in coal which is calculated as • % of Ash = [wt of residue left in crucible´ 100]/ wt of coal taken • Reduces calorific value as it is non burning part • Ash disposal is a problem Visit www.gyandharaonline.com for more notes n updates
  30. 30. * Fixed carbon % of Fixed carbon = 100 – (% of moisture + % of ash + % of volatile matter) • In any good sample of coal, the percentages of moisture, ash, volatile matter should be as low as possible and thus the percentage of fixed carbon should be as high as possible. Visit www.gyandharaonline.com for more notes n updates
  31. 31. * Determination of C & H • Accurately weighed coal sample is burnt in a current of oxygen in a combustion apparatus, which is heated to about 350°C. • Carbon and hydrogen of coal are converted into water vapour and carbon-dioxide. The products of combustion are absorbed in anhydrous CaCl2 and KOH tubes respectively of known weights. • After complete absorption of H2O and CO2, the tubes are again weighted. Visit www.gyandharaonline.com for more notes n updates
  32. 32. * C + O2 ® CO2 12 parts ® 44 parts 2H2 + O2 ® 2H2O 4 parts ® 36 parts • % of Carbon = [increase in wt of KOH tube × 12 × 100]/wt of coal taken × 44 • % of Hydrogen =[increase in wt of CaCl2 tube × 4 ´100]/wt of coal taken × 36 Visit www.gyandharaonline.com for more notes n updates
  33. 33. * Determination of Nitrogen Nitrogen is calculated by Kjehldals Method. The nitrogen is converted to NH3 and passed through a known volume of standard acid. On neutralization, the excess acid is back titrated with a base. 1000 ml of x N acid 14 gm of Nitrogen % N = [volume of acid consumed in neutralizing NH3 × N x 14 x 100 wt of coal taken x 1000 Visit www.gyandharaonline.com for more notes n updates
  34. 34. Determination of Sulphur % Sulphur: [wt of BaSO4 obtained × 32× 100]/wt of coal taken × 233 Determination of Oxygen: The oxygen is determined indirectly by calculation as % of Oxygen = 100 – (% of C + % of H + % of N + % of S + % of Ash) Visit www.gyandharaonline.com for more notes n updates *
  35. 35. Importance of Ultimate analysis: • Carbon: Greater the % carbon, better is the quality and calorific value of coal • Hydrogen: most of hydrogen is in form of moisture and volatile matter. Only a small % is combustible, hence it decreases C.V. Smaller the H% better is quality of coal • Nitrogen: does not burn, hence it has no C.V. Negligible N% is good coal • Sulphur: it increases C.V, but causes Sox pollution. Hence lower S% is better • Oxygen: most of oxygen is in form of moisture, hence it decreases C.V. Smaller the H% better is quality of coal *

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