O and Sulphur Dioxide SO 2 which pass through the chimney along with the

O and sulphur dioxide so 2 which pass through the

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O) and Sulphur Dioxide (SO 2 ), which pass through the chimney along with the Nitrogen (N 2 ) in the air,. After surrendering useful heat in the heat absorption area of a furnace or boiler, the combustion products or fuel gases leave the system through the chimney, carrying away a significant quantity of heat with them. Rules for combustion of oil 1. Atomize the oil completely to produce a fine uniform spray 2. Mix the air and fuel thoroughly 3. Introduce enough air for combustion, but limit the excess air to a maximum of 15% 4. Keep the burners in good condition
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1. Fuels and Combustion 17 Bureau of Energy Efficiency Constituents % By weight Carbon 85.9 Hydrogen 12 Oxygen 0.7 Nitrogen 0.5 Sulphur 0.5 H 2 O` 0.35 Ash 0.05 GCV of fuel: 10880 kCal/kg The specifications of furnace oil from lab analysis is given below: Calculation of Stoichiometric Air Calculation for Requirement of Theoretical Amount of Air Considering a sample of 100 kg of furnace oil. The chemical reactions are: Element Molecular Weight kg / kg mole C 12 O 2 32 H 2 2 S 32 N 2 28 CO 2 44 SO 2 64 H 2 O 18 C + O 2 CO 2 H 2 + 1/2O 2 H 2 O S + O 2 SO 2 Constituents of fuel C + O 2 CO 2 12 + 32 44 12 kg of carbon requires 32 kg of oxygen to form 44 kg of carbon dioxide therefore 1 kg of carbon requires 32/12 kg i.e 2.67 kg of oxygen (85.9) C + (85.9 × 2.67) O2 315.25 CO 2 2H 2 + O 2 2H 2 O 4 + 32 36
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1. Fuels and Combustion 18 Bureau of Energy Efficiency 4 kg of hydrogen requires 32 kg of oxygen to form 36 kg of water, therefore 1 kg of hydro- gen requires 32/4 kg i.e 8 kg of oxygen (12) H 2 + (12 × 8) O2 (12 x 9 ) H 2 O S + O 2 SO 2 32 + 32 64 32 kg of sulphur requires 32 kg of oxygen to form 64 kg of sulphur dioxide, therefore 1 kg of sulphur requires 32/32 kg i.e 1 kg of oxygen (0.5) S + (0.5 × 1) O 2 1.0 SO 2 Total Oxygen required = 325.57 kg (229.07+96+0.5) Oxygen already present in 100 kg fuel (given) = 0.7 kg Additional Oxygen Required = 325.57 – 0.7 = 324.87 kg Therefore quantity of dry air reqd. = (324.87) / 0.23 (air contains 23% oxygen by wt.) = 1412.45 kg of air Theoretical Air required = (1412.45) / 100 = 14.12 kg of air / kg of fuel Calculation of theoretical CO 2 content in flue gases Nitrogen in flue gas = 1412.45 - 324.87 = 1087.58 kg Theoretical CO 2 % in dry flue gas by volume is calculated as below : Moles of CO 2 in flue gas = (314.97) / 44 = 7.16 Moles of N 2 in flue gas = (1087.58) / 28 = 38.84 Moles of SO 2 in flue gas = 1/64 = 0.016 = 15.5 % 2 2 % 100 ( ) Moles of CO Theoritical CO byvolume Total moles dry = × 7.16 100 7.16 38.84 0.016 = × + +
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1. Fuels and Combustion 19 Bureau of Energy Efficiency Calculation of constituents of flue gas with excess air % CO 2 measured in flue gas = 10% (measured) = 55% Theoretical air required for 100 kg of fuel burnt = 1412.45 kg Total quantity. of air supply required with 55% excess air = 1412.45 X 1.55 = 2189.30 kg Excess air quantity = 2189.30 – 1412.45 = 776.85 kg. O 2 = 776.85 X 0.23 = 178.68 N 2 = 776.85 – 178.68 = 598.17 kg The final constitution of flue gas with 55% excess air for every 100 kg fuel. CO 2 = 314.97 kg H 2 O = 108.00 kg SO 2 = 1 kg O 2 = 178.68 kg N 2 = 1087.58 + 598.17 = 1685.75 kg Calculation of Theoretical CO 2 % in Dry Flue Gas By Volume Moles of CO 2 in flue gas = 314.97/44 = 7.16 Moles of SO 2 in flue gas = 1/64 = 0.016 Moles of O 2 in flue gas = 178.68 / 32 = 5.58 Moles of N 2 in flue gas = 1685.75 / 28 = 60.20 2 2 % % 1 100 % Theoritical CO Excess air Actual CO = × 2 2 % 100 ( ) Moles of CO Theoritical CO byvolume Total moles dry =
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  • Spring '17
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