1 / 33

MSW (Municipal Solid Waste Incinerator) Definition :

MSW (Municipal Solid Waste Incinerator) Definition : Combustion - a process of burning, resulting from the rapid oxidation of substances Used for municipal solid wastes, industrial (hazardous) waste, sludges, fossil fuel

Download Presentation

MSW (Municipal Solid Waste Incinerator) Definition :

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. MSW (Municipal Solid Waste Incinerator) • Definition : • Combustion - a process of burning, resulting from the rapid oxidation of substances • Used for municipal solid wastes, industrial (hazardous) waste, sludges, fossil fuel • Typically operated with excess air (oxygen supply > combustion stoichiometry) • Advantages : • Volume and weight reduced (approx. 90% volume and 75% weight reduction). • Waste reduction is immediate, no long term residency required. • Destruction in seconds. • Incineration can be done at generation site (ex. medical waste incinerators). • Air discharges can be controlled (low health risk). • Small disposal area required. • Cost can be offset by heat recovery/ sale of energy. • Disadvantages : • High capital cost. • Skilled operators are required (particularly for boiler operations). • Not all material are incinerable (noncombustable solids). • Some material require supplemental fue.l • Public disapproval. • Risk imposed rather than voluntary. • Incineration will decrease property value. • Distrust of government/industry ability to regulate.

  2. Energy Consumption

  3. Gasoline • Petroleum (crude oil) • Thousands of organic compounds, mostly alkanes • Tb / oC • CnH2n + 2 methane CH4 -164 • ethane C2H6 H3C-CH3 -89 • propane C3H8 H3C-CH2-CH3 -42 • butane C4H10 H3C-(CH2)2-CH3 -0.5 • pentane C5H12 H3C-(CH2)3-CH3 36 • hexane C6H14 H3C-(CH2)4-CH3 69 • heptane C7H16 H3C-(CH2)5-CH3 98 • Octane C8H18 H3C-(CH2)6-CH3 126

  4. Other compounds in petroleum (and gasoline) Cyclic hydrocarbons (cycloalkanes) Cyclohexane (C6H12) Aromatic hydrocarbons Benzene Toluene p-Xylene BTX (or BTEX)

  5. Total moles of products = 1+2+7.52 moles = 10.52 moles Mole % of CO2 = 1/10.52x100(%) = 9.5%-vol CO2 =9.5x104 ppm CO2 Mole % of H2O = 2/10.52x100(%) = 19%-vol H2O Mole % of N2 = 7.52/10.52x100(%) = 71.5%-vol N2

  6. Air-Fuel Ratio (공연비) • Composition of air-fuel mixture may also be expressed in terms of % excess air, % theoretical air,γand equivalence ratio, φ. • Air-Fuel Ratio (AF) = mair/mfuel • Theoretical or stoichiometric AF ratio : [mair/mfuel]stoich • % theoretical air = Air-Fuel Ratio (AF)/ stoichiometric AF ratiox100 (%) • 공기비(과잉공기계수)= Air-Fuel Ratio (AF)/ stoichiometric AF ratio • % excess air = % theoretical air -100% • % theoretical air,γ=1/φ • mfuel/mair • Equivalence ratio, φ = --------------- • [mfuel/mair]stoich • φ<1 indicates that excess air is present. : Fuel-lean combustion • φ>1 indicates that too little air is present for complete combustion. : Fuel-rich combustion • φ=1 indicates that the stoichiometric amount of air is present. : stoichiometric combustion

  7. Balancing the stoichiometric equation, Air-fuel ratio : example • C8H18 + 20(O2 + 3.76N2) → aCO2+bH2O+cO2+dN2 • Determine the AF ratio, the %theoretical air, the %excess air, and equivalence ratio. • AF= [(20*4.76 kmol Air)(28.97 kg/kmol Air)] /[(1 kmol C8H18)(114.24 kg/kmol C8H18)] = 24.1

  8. C8H18 + 12.5(O2 + 3.76N2) → 8CO2+9H2O+cO2+94N2 • stoichiometric AF= [(12.5*4.76 kmol Air)(28.97 kg/kmol Air)] /[(1 kmol C8H18)(114.24 kg/kmol C8H18)] = 15.1 • % theoretical air = 24.1/15.1x100=160% • % excess air = 160-100=60% • Equivalence ratio = (1/24.1)/(1/15.1)=0.63 (Fuel-lean condition)

  9. Example: 2-5(p. 34) Excess air=30%, (70%의 상대습도, air)에서 75kg/hr의 waste를 소각하고 Waste 성분 분석은 다음과 같다. Waste 성분 분석(Ultimate Analysis:원소분석) : (100kg 기준시) C : 71%  71kg  5.912kgmoles H : 9.2%  9.2kg  9.2kgmoles O : 2.1%  2.1kg  0.132kgmoles S : 3.4%  3.4kg  0.106kgmoles N : 0.6%  0.6kg  0.042kgmoles H2O : 12.2%  12.2kg Ash : 1.5%  1.5kg 그러므로 waste의 분자식은 C5.912H9.2O0.132S0.106N0.042 이론적 공기량을x(O2+3.78N2) 라 두면, C5.912H9.2O0.132S0.106N0.042+(1+ 0.3)x(O2+3.78N2)  5.912CO2+4.6H2O+0.106SO2+0.3xO2+[0.021+(1.3x)3.78]N2 O2를 기준 : + 1.3x=5.912+2.3+0.106+0.3x  x=8.234 C5.912H9.2O0.132S0.106N0.042+(1+0.3)(8.234)(O2+3.78N2)  5.912CO2+4.6H2O+0.106SO2+2.47O2+40.483N2

  10. (9) Moles nitrogen in stoichiometric air 8.234  3.78 = 30.975 (10) Moles nitrogen in excess air = (0.3)(8.234x3.78)=(0.3)(30.975) = 9.293 (11) Moles oxygen in excess air = (0.3)(8.234) = 2.470 (12) Moles moistures in combustion air (psychrometric chart를 이용) 70%의 상대습도  0.008kg H2O/kg of dry air  (10.704  32 + 10.704  3.78 28)kg air = 0.653kgmoles (13) Total moles : CO2 = 5.912, H2O = 4.6 + 0.653 = 5.894, SO2 = 0.106 O2 = 2.47, N2 = 40.483, Total = 54.671 1.3x8.234

  11. Example 2-6 : Orsat Analysis CO2 : 12.3%  12.3 kgmoles 100 kgmoles 기준 O2 : 5.1%  5.1 kgmoles N2 = 100 – (12.3+5.1) = 82.6% C12.3Hy+(x+5.1)(O2+3.76N2) 12.3CO2+5.1O2+82.6N2+ H2O for N2 : 3.76x+(5.1)(3.76) = 82.6 x = 16.9 for O2 : (16.9 + 5.1) = 12.3 + 5.1 + y = 18.4 fuel(waste)의 화학식 : C12.3H18.4 Weight ratio of H to C = = 0.125 Weight % of H = = 11.1% Weight % of C = = 88.9%

  12. Air requirement; = (16.9 + 5.1)(32+3.78  28) = 3020.2 kg air/100kgmoles of fuel or = (18.2/100)kg air/kg of fuel Excess air:  100 = 30.2% Exhaust Gas: Total moles = 12.3 + 5.1 + 82.6 + = 109.2kgmoles/100kgmoles of fuel

  13. (기체 연료의 이론 공기량) 기체연료 1Sm3중에 H2, CO, CH4, C2H4, C3H8, C4H10,‥‥, CxHy, O2, CO2가 각각 %(volume비)씩 들어 있는 경우 이론공기량, Ao는 단위: Sm3/Sm3 (ex) 분자식 CmHn인 탄화수소가스 1Sm3의 연소시 필요한 이론 공기량은? Sm3/Sm3 Sm3/Sm3

  14. 여기서 C, H, O, S는 중량비(wt%)임. or, (액체 및 고체연료의 경우) 이론 공기량: Sm3/kg or Kg/kg 단위는 Sm3/kg 이다. or, 단위는 kg/kg 이다.

  15. (ex) 탄소 85%, 수소13%, 황2%의 조성을 가진 중유의 이론 공기량은? (ex) 탄소, 수소, 산소, 황의 중량%가 86.6%, 4%, 8%, 1.4%인 중유의 연소에 필요한 이론산소량(Sm3/kg)과 이론 공기량(Sm3/kg)을 계산하라. 이론 산소량 이론 공기량

  16. (ex) 메탄올(CH3OH) 1kg이 연소하는데 필요한 이론 공기량을 Sm3/kg과 kg/kg으로 구하라.

  17. *CH3OH(메탄올) 1kg연소시의 이론 공기량은? 이론 공기량

  18. (ex) 탄소 85%, 수소13%, 황2%의 조성을 가진 중유의 이론 공기량은? 이론 공기량 (Sm3/kg)

  19. *이론 산소량 or, 이론 공기량 (kg/kg)

  20. 연소가스의 조성비가 N2, O2, CO로 구성 여기서, N2=100-{(CO2)+(O2)+(CO)} Excess Air(과잉공기) m=공기비(과잉공기계수)=A/Ao=실제공기량/이론공기량=m•(Ao) 과잉공기비율=(A-Ao)/Ao=m-1 과잉공기량=(m-1)Ao (ex)탄소, 수소의 중량 조성이 각각 86%, 14%인 액체 연료를 매시 100kg 연소한 경우 배기가스의 분석치는 다음과 같다. CO2:12.5%, O2:3.5%, N2:84% 이 경우에 매시 필요한 공기량(Sm3/h)은? 이론 공기량, Ao 공기비(과잉공기계수), 실제 공기량=(1.19)(11.38Sm3/kg=13.54Sm3/kg 100kg 연소시 실제 공기량=13.54100=1354Sm3/h

  21. 100 Kg Kgmole C 86 Kg 86/12 = 7.167 H 14 Kg 14/1 = 14 → C7.167H14 * 100 Kg/h 의 연료성분 C = 86%, H = 14%

More Related