OchaPrayataWistara
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Aug 19, 2024
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About This Presentation
CFB course
Size: 3.41 MB
Language: en
Added: Aug 19, 2024
Slides: 43 pages
Slide Content
BED MATERIAL in CFB BOILER “ 7. Bed Material dan Heat Transfer ”
HEAT ABSORBING SECTION IN CFB 19/08/2024 2
TYPICAL HEAT TRANSFER COEFFICIENTS 19/08/2024 3
Gas t o Partic l e Heat Tra n s f er • Mech a nism o f Heat Tr a nsfer I n a CFB b o i l er, f i ne s o l i d p art i c l e s agg l o m erate a n d form c l u s ters or s t a n d i n a co n t i n u u m of gener a l l y u p -f l ow i ng g a s c o n t a i n i ng sp a rs e l y d i sper s ed sol i ds. T he c o nt i n u um i s cal l ed the d i sper s ed p h a se, w hi l e the agg l o mera t es are c a l l ed the c l uster p h ase. The he a t t ra n s f er to furn a ce w a ll occurs t hr o ugh c o nducti o n from p a rt i c l e c luste r s, c o nvection from dispe r s e d ph a s e , a n d r a diation from b oth ph a s e . 7 1
Heat T r an s fer in CFB B o i l e r • E f f e c t o f S uspension Densi t y and parti c le size
Heat T r an s fer in CFB B o i l e r • E f f e c t o f Fluidization Velocity 7 3
Heat T r an s fer in CFB B o i l e r • E f f e c t o f Fluidization Velocity
Heat T r an s fer in CFB B o i l e r • E f f e c t o f Verti c al Length o f Heat Tr a nsfer S urface
Heat T r an s fer in CFB B o i l e r • E f f e c t o f B ed T e m p e rature
Heat T r an s fer in CFB B o i l e r • E f f e c t o f B ed T e m p e rature
Heat T r an s fer in CFB B o i l e r • E f f e c t o f particle o f Heat Tr a nsfer S urface
Heat T r an s fer in CFB B o i l e r • Heat F lux on 300 M W CFB B o iler ( Z . M an, et. al)
Heat T r an s fer in CFB B o i l e r • Heat transfer to t he w a lls o f c ommercia l -size 7 9
Heat T r an s fer in CFB B o i l e r • Circum f er e ntial Distri b uti o n of H eat Tr a nsfer Coe f f i ci e nt
CONTOH SOAL
CONTOH SOAL
Using porous bed material Mechanism of carbon deposit: Carbon Capture Porous Solid Porous particles capture V.M. at high temperature ( capacitance effect ); carbon deposit is formed within pores increase residence time carbon deposit V.M. pore 19/08/2024 17
FB Apparatus for Devolatilization Experiments Gas Analyzers Pellet Feed Data Processor Filter Cooling Electric Furnace Flame sensor Flow meter O 2 Gas Cylinder N 2 N 2 Gas Cylinder O 2 Flow meter Fuel: PE pellet 1cm diam. X 1cm length Batch feed 19/08/2024 19
Effect of various bed materials on the onset of devolatilization 19/08/2024 20
Effect of solid type on formation of carbon deposit over bed material at high temperature V.M. evolution Deposit carbon combustion Capacitant effect 19/08/2024 21
Effect of solid type on formation of carbon deposit over bed material at high temperature V.M. evolution Deposit carbon combustion QS 19/08/2024 22
Relationship between onset of devolatilization and V.M. capture 19/08/2024 23
Effect of gas velocity and solid bed type on heat transfer coefficient at high temperature 19/08/2024 24
Onset of flame detection vs. heat transfer coefficient Due to high volatile matter capture by the fine AB, the concentration of volatile matter in the freeboard could not be sufficient to maintain flame combustion 19/08/2024 25
Onset of flame detection vs. onset of CO 2 detection 19/08/2024 26
Onset of flame, CO 2 CO detection vs. Heat Transfer Coefficient 19/08/2024 27
Chemical Reactions of carbon (II.1) (II.2) of sulfur (II.3) of (II.4) of (II.5) of sulfur (II.6) BEBERAPA PERSAMA PENTING
Air Required coal (II.7) burned. (II.8) (II.9) (II.10) (II.11) BEBERAPA PERSAMAAN PENTING
Solid Waste Produced (II.12) Appendix II: Stoichiometric Calculations , (II.13) Where unburnt carbon is approximated as Gaseous Waste Products Carbon Dioxide Carbon dioxide produced from fixed carbon in coal = 3.66 C. (II.14) (II.15) BEBERAPA PERSAMA PENTING
Calculate the amount of CO 2 produced per kilogram of the coal burned. Assume that calcium to sulfur molar ratio of 2.5. The limestone contains 88% CaCO3 by weight and 105 MgCO3 by weight. Solution Using Eq.II.14 , calculate the kg per kg fuel burnt as: Produced from coal combustion = 3.66 x 0.728 = 2.66 Produced from calcination (Eq. II. 15) = 0.0858 kg/kg fuel Total carbon dioxide produced = 2.66 + 0.858= 2.745 kg per kg of coal burned C H O N S ASH MOISTURE ULTIMATE TEST 72.8 4.8 6.2 1.5 2.2 9.0 3.5 CONTOH SOAL
Water Vapor (II.16) Nitrogen (II.17) Oxygen (II.18) Sulfur dioxide (II.19) Fly ash (II.20) BEBERAPA PERSAMA PENTING
burned(II.21) Heating Value of Fuels (II.22) (II.23) BEBERAPA PERSAMA PENTING
Fuel° Volatile matter (%) Fixed carbon (%) H 2 O (%) ASH (%) S (%) HHV kJ/kg Flue gas enthalpy at 850 °C (kJ)/fuel fired(kJ) Wood waste 36.5 13.1 50.0 0.4 0.01 10,519 0.55 Anthracite 12.8 73.7 1.3 12.2 0.0 31,602 0.45 Lignite 34.5 27.2 25.4 11.7 1.2 17,445 0.5 No. 6 oil 0.4 0.1 0.5 45,522 0.43 Petroleum coke 8.9 89.5 1.4 0.2 9.4 35,066 0.39 Bituminous coal 29.4 56.4 1.1 13.0 2.3 30,971 0.43 Natural gas 61,057 0.41 Fuel° Volatile matter (%) Fixed carbon (%) H 2 O (%) ASH (%) S (%) HHV kJ/kg Flue gas enthalpy at 850 °C (kJ)/fuel fired(kJ) Wood waste 36.5 13.1 50.0 0.4 0.01 10,519 0.55 Anthracite 12.8 73.7 1.3 12.2 0.0 31,602 0.45 Lignite 34.5 27.2 25.4 11.7 1.2 17,445 0.5 No. 6 oil 0.4 0.1 0.5 45,522 0.43 Petroleum coke 8.9 89.5 1.4 0.2 9.4 35,066 0.39 Bituminous coal 29.4 56.4 1.1 13.0 2.3 30,971 0.43 Natural gas 61,057 0.41 BEBERAPA KARAKTERISTIK BAHAN BAKAR
Contoh soal: thermal design of a CFB boiler Find the principal dimension of the furnace of a CFB boiler with thermal capacity of 406 MW. The flue gas temperature at boiler exit is 143 °C. Assume solid residues to leave the boiler at average temperature of 220 °C, and surface losses to the ambient to be 0.5%. Sulfur capture target is 90% with calcium to sulfur molar ratio of 2.0. Solid residues contain about 1.5% carbon. Take result of stoichiometric calculation from Appendix II. Other data as follows; Steam condition: 130 bar, 588 °C; saturation temperature, 363 °C; Feed water temperature 20 °C. Coal : Same as of Appendix II. Solution Heat duties of different boiler element as found using a Steam are as follows : Economizer = 80.0 MW Evaporator = 168.0 MW Superheater = 113.0 MW Reheater = 54.8 MW Total thermal duty, = 406 MW
2. Combustion calculations ; The stoichiometric calculation for above coal and limestone is carried out in Appendix II. So, we take values directly from them as below : Excess air = 20% Theoretical air = 8.05 kg/kg Total dry air = 9.66 kg/kg fuel Assuming 0.013 we get total wet air = = 9.79 kg/g fuel From table 6.6, we take the primary air to be 78% of the theoretical air Primary air = 6.28 kg/kg fuel From appendix II Limestone required per coal feed = 0.335 kg/kg fuel Solid waste produced = 0.514 kg/kg fuel Flue gas weight = 10.7 kg/kg fuel
3. Heat balance on the basis of higher heating value of coal, 24,657 kJ/kg, is worked out below: We use 20 °C as the base temperature. Enthalpy of water at 20 °C = 4,186 20 = 83.7 kJ/kg Enthalpy of water vapor (steam) at 143 °C above this base =2762 – 83.7 = 2678 kJ/kg Loss due to moisture in sorbent Los due to moisture in fuel Calcination loss from CaCO 3 (Eq. 6.2) Calcination loss from MgCO 3 (Eq. 6.3) Sulfation credit (Eq. 6.5) Unburnt carbon (Eq. 6.6) Heat in dry flue gas (Eq. 6.7) Moisture in hydrogen
Moisture in air Ash loss taking sp , heat of ash = 1.0 Radiation and convention loss (from Table 6.2) = 0.5% FD fan credit (assumed ~ 1%) = (-1.0%) Total heat loss = sum of above losses = 12.3 % 4. Fuel heat input 463 MW 5. Mass balance : Coal feed required, Total air flow Coal ash
Fly ash (19% of total ) Sorbent feed Total solid waste produced Using the result of combustion calculation in Sect. 2, we get: Primary air weight Flue gas weight 6. Furnace cross section Permissible heat release chosen from Table 6.4 = 3.51 MW/ Bed upper cross-section Density of flue gas at 850 °C = 0.326 kg/ Fluidization velocity at 850 °C Check it on for fast fluidization (refer to Chap. 2)
Primary area required to maintain same velocity Shape of cross section : Refer to sect. 6.3.2, Width =1/2 Length Width This is too wide and greater than 7.5 m. So we limit it to 7.5 m. So the new value Width =7.5 Length 7. Furnace height; Required gas- residence time for sulfur capture is assumed as 3.5 s Assuming the core velocity to exceed average velocity by 50%, the maximum core velocity, U = 1.5 4.67 = 7.0 m/s So, the height required to 3.5 s residence time = 7 3.5 = 24.5 m
The adequacy of the height for sulfur capture may be further checked . However, it is most important to check if this height would allow absorption of the heat required to maintain the furnace temperature at 850 °C. the average heat transfer coefficient in a typical CFB boiler is in the range of 130-200W/m 2 K. So, for preliminary calculation instead of following the rigorous calculation procedure we choose a mean value of 175 W/m 2 K. Metal temperature of evaporator tube is generally 25 °C hotter than the saturation temperature of the water. W e find the wall temperature as follow; Tube wall temperature From step 1, we have: Evaporator duty =168 MW 1000 = 168,000 kW Total evaporator area required For 2” tube in 3” pitch So, projected area required = 2078/1.38 = 1505 m 2 Ceiling area = Furnace cross section =131.89 m 2 Wall area required = 1505 131.89 = 1373.8 Exit area of the furnace is taken as 30% of bed cross section. Wall area unavailable for heat transfer =0.3 131.89 = 39.5 m 2
Gross wall area to be provide = 1373 + 39.5 = 1413 m 2 Water wall height required The new furnace height, 28.15 m, is greater than 24.5 m calculated earlier for gas residence. So, we can make the furnace 28.15 m tall above the refractory zone. This will provide a gas-residence time greater than 3.5 s required. This make the design adequate from both sulfur capture and heat absorption consideration.
Chemical Reactions of carbon (II.1) (II.2) of sulfur (II.3) of (II.4) of (II.5) of sulfur (II.6)