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New Developments in Combustion Technology

New Developments in Combustion Technology 5
New Developments in Combustion Technology Geo. A. Richards, Ph.D. National Energy Technology Laboratory U. S. Department of Energy 2014 PrincetonCEFRC Summer School On Combustion Course Length: 6 hrs June 2324, 2014 Energy Everyday, Everywhere Healthcare, education, infrastructure, water, transportation, communication, agriculture, recreation……… ‹› Energy Everyday, Everywhere…. ….Except…… From the International Energy Agency Website: “….Based on this updated analysis, we estimate that in 2009 the number of people without access to electricity was 1.3 billion or almost 20 of the world’s population…..” http://www.worldenergyoutlook.org/resources/energ ydevelopment/accesstoelectricity/ ‹› Electricity Demand 2011 Electricity Demand 2035 4,084 BkWh / Year 4,979 BkWh / Year 68 Fossil Energy 68 Fossil Energy Coal Coal + 22 34 Oil 42 Gas 1 Gas United States Oil 25 34 1 Nuclear Nuclear Renewables 19 16 16 Renewables 13 2,171 mmt CO 2,247 mmt CO 2 2 22,113 BkWhr / Year 39,854 BkWh / Year 68 Fossil Energy 65 Fossil Energy Gas + 80 Gas Coal Coal 22 23 41 40 World Nuclear Nuclear 12 10 Oil Oil 5 2 Renewables Renewables 25 20 ‹› 12,954 mmt CO 19,122 mmt CO 2 2 Sources: U.S. data from EIA, Annual Energy Outlook 2014 ‘er; World data from IEA, World Energy Outlook 2013, slide courtesy Peter Balash, NETL Why not my way Coal Wind Hydro Gas Nuclear Solar “It’s all regional. It’s all local. And we just have to descend to that level to judge it.” – Vaclav Smil, discussing preferable energy resources, Wall Street Journal, Wed April 9, 2014, pp. R1, Business and Environment special section. ‹› Why not my way One size does not fit all ‹› Energy and Carbon Dioxide • Carbon dioxide capture and storage – costly, but not explicitly required. • Carbon dioxide utilization in enhanced oil recovery (EOR) is needed, now. • Carbon dioxide costs from natural source anthropogenic sources. Delivered CO2 prices today 1040/tonne Domestic EOR CO Use 2 CO supply for North America EOR 2 million tonne/yr Can we develop efficient affordable methods to supply CO 2 A typical 550 MW coal plant emits 3.5 million tonne/ year Graphics and information NETL. Reference: DiPietro, J. P., Next generation Enhanced Oil Recovery, Presented at CO2;http://www.netl.doe.gov/energyanalyses/refshelf/PubDetails.a the Carbon Dioxide Utilization Congress, San Diego, California, Feb. 19, 2014. spxAction=ViewPubId=348 Available at http://netl.doe.gov/research/energyanalysis/publications/detailspub=68d576c3a2ac4e778963 5ada3b04984f ‹› Greenhouse Gas New Source Performance Standard (Proposed standard for new sources; comments accepted to May 9, 2014. Differs from proposed emission standards for existing sources, released June 2, 2014, http://www2.epa.gov/carbonpollutionstandards) 2,500 Existing Subcritical PC 2,000 New 1,500 SC PC NSPS Limit 1,000 New Uncontrolled New 500 NGCC Supercritical PC New 90 CCS NGCC 90 CCS 0 ‹› “Cost and Performance Baseline for Fossil Energy Plants, Volume 1: Bituminous Coal and Natural Gas to Electricity”, Revision 2a, September 2013, Lb CO /MWh 2Cost of CO Capture, /ton CO 2 2 140 123/ton CO 2 120 105/ton CO 2 32 100 27 80 21 First of a kind 18 60 Next of a kind (high range) Next of a kind (low range) 40 70 60 20 0 Subcritical PC New Supercritical, Retrofit, 90 CCS 1,100 Lb CO2/MWh “Cost and Performance Baseline for Fossil Energy Plants, Volume 1: Bituminous Coal and Natural Gas to Electricity”, Revision 2a, September 2013 /ton CO 2 Carbon dioxide capture options • Existing options: costly (1) Add a flue gas CO 2 • Next generation options: scrubber (e.g. photo below). depend on thermal (2) Convert hydrocarbons to science research: hydrogen and CO /w 2 – Supercritical carbon dioxide separation/capture. power cycles: 300 bar combustion (3) Separate oxygen from air – Pressurized oxyfuel. and use oxyfuel – Chemical looping combustion. combustion. – Pressuregain combustion for efficiency. – “Direct Power Extraction” via MHD 240 MWe slipstream at NRG Energy’s W.A. Parish power plant – Note the size of CCS process area. Photo courtesy Mike Knaggs, NETL ‹› The role of capture AND generator efficiency • A simple Define: heat/energy a = (kg CO produced) / (kg fuel burned ) 2 balance defines w = (separation work, Joules ) / (kg CO ) CO2 2 CO 2 the overall efficiency h with ov a carbon separation unit. h g Generator Efficiency • Reducing the Q = m DH f W 1 W Fuel Heat o Net penalty from Gross Input Generator Output Generator Carbon carbon capture Work Separation comes from Unit BOTH: – Decreasing w CO2 – Increasing h g Approx Ranges: (30 – 60) (610) ‹› This presentation Updated, expanded from 2012 CEFRC lecture: – Inherent carbon capture: chemical looping combustion (Day 1) – Stepchange in generator efficiency: pressure gain combustion (Day 2) – Frontier approach (): making oxyfuel an efficiency advantage (Day 2) Sampling Diagnostics RDC Flow Pgain rig NETL ‹› Disclaimer This report was prepared as an account of work sponsored by an agency of the United States Government. Neither the United States Government nor any agency thereof, nor any of their employees, makes any warranty, express or implied, or assumes any legal liability or responsibility for the accuracy, completeness, or usefulness of any information, apparatus, product, or process disclosed, or represents that its use would not infringe privately owned rights. Reference herein to any specific commercial product, process, or service by trade name, trademark, manufacturer, or otherwise does not necessarily constitute or imply its endorsement, recommendation, or favoring by the United States Government or any agency thereof. The views and opinions of authors expressed herein do not necessarily state or reflect those of the United States Government or any agency thereof. ‹› Chemical Looping Combustion ‹› Oxyfuel background Oxy fuel achieves carbon capture very easily: AirCombustion: CH + 5/4(O + 3.8N )  CO + 1/2H O +4.7 N 2 2 2 2 2 Costly to extract the CO from the N with amines 2 2 OxyCombustion: CH + 5/4(O )  CO + 1/2H O 2 2 2 Easy to extract the CO from the H O via condensation 2 2 “Usual” oxyfuel approach: oxygen diluted with CO or H O added to an existing boiler cycle. 2 2 – Dilution used to keep the temperatures same as Meridosia Illinois – Future Gen 2.0 planned site existing cycle. – Efficiency of the plant is penalized by the energy needed to make oxygen. Significant oxyfuel demonstration projects are occurring around the world. – See for example: http://www.newcastle.edu.au/project/oxyfuel workinggroup/demonstrations.html – More than 14 demos 10MWth listed Courtesy University of Utah – oxyfuel burner tests ‹› Making oxygen for oxyfuel • Oxygen can be supplied today by commercial Air Separation Units (ASU) based on established cryogenic separation. • The energy needed to separate oxygen from air is significant (see below). • In conventional oxycombustion, we dilute the purified oxygen to maintain the same boiler flame temperature as in aircombustion. 1 mole of air 0.21 moles oxygen Dilute again Air p = 1 atm with CO or steam O2 2 0.21 moles oxygen p O2 Separation = 0.21 atm Unit 0.79 moles nitrogen p (ASU) N2 0.79 moles nitrogen = 1 atm p = 0.79 atm N2 C + O  CO 2 2 DH DG = 394 kJ/gmol (C or O 2) Reversible separation work: 6 kJ/gmol O produced In efficient powerplants we convert 2 less than ½ of DH to work. Thus200kJ/gmol O2 work produced Current actual process: 18kJ/gmol O produced 2 Roughly 1/10 of that is needed for ASU. e.g, the change in gibbs energy for ideal mixing (Sandler, Chemical Engineering Thermodynamics (1989) pp. 313. See Trainier et al., “Air Separation Unit…..” Clearwater Coal Conference, 2010. ‹› Chemical Looping • Shares advantages of oxyfuel – Product is just CO and H O 2 2 • No separate oxygen production is needed • Schemes for H production, carbon capture… 2 Avoiding confusion N + O 2 2 with nomenclature: (vitiated air) Today’s Refer only to the CO + H O 2 2 discussion: CO + H O FUEL REACTOR 2 2 Focus on AIR REACTOR Seal air reactor Ash Here’s why it can otherwise be confusing: the air reactor “burns” or oxidizes the metal. Recycle Fuel CO + H O 2 2 The fuel reactor “reduces” the metal oxide but Seal oxidizes the fuel. Air Thus, you could call the fuel reactor an oxidizer for the fuel OR a reducer for the metal oxide. Carbon + metal oxide = CO + metal 2 Metal + air (oxygen) = metal oxide ‹› Not quite new • Chemical looping has been around – but for different CO 2 reasons and applications. – 1954 patent to manufacture CO 2 • Similar process: ironsteam HX route to hydrogen (circa 1920) M Reduce iron with fuel and oxidize it with steam: MO 2 Air Reactor M + (O + 3.8N ) F O + 4 CO  3Fe +4 CO 2 2 3 4 2 MO + 3.8 N 2 2 3 Fe + 4 H O  Fe O + 4H 2 3 4 2 Fuel Reactor MO + C  2 CO + M 2 • And, before that….respiration. “Production of Pure Carbon Dioxide” US Patent 2,665,972 (1954) Notice the heat exchangers (HX) in BOTH fuel and air reactors. Hemoglobin “loops” Should have made it a boiler to carry oxygen from lungs for hydrocarbon Hurst, S. (1939). “Production of Hydrogen by the IronSteam Method”, oxidation in cells. Journal of the American Oil Chemist’s Society, 16 (2), pp. 2936. ‹› Basic Thermodynamics In CL combustion, the overall reaction (1) of fuel with oxygen is split into two steps (23) , which add to the overall. Consider an example of carbon and a metal/metal oxide (M/MO): 1) C + O  CO DH Overall fuel oxidation exothermic DH = DH + DH 2 2 1 1 2 3 2) C+ MO  CO + M DH Metal oxide reduction Fuel reactor 2 2 2 fuel oxidation – can be endothermic OR exothermic Air Reactor 3) M + O  MO DH Metal oxidation exothermic 2 2 3 Nomenclature used in this talk: Exothermic carriers Endothermic carriers Neutral carriers DH 0, DH 0 DH 0 2 2 2 Fuel reactor Fuel reactor Fuel reactor does not consume or releases heat consumes heat release heat ‹› Chemical Looping Heat Release Air reactor: always exothermic; it releases heat. N + O 2 2 (vitiated air) Exothermic carriers: some heat will also come out in the fuel reactor CO + H O 2 2 CO + H O 2 2 Endothermic carriers: fuel reactor needs heat to reduce the metal. Seal • If you don’t put heat into the fuel reactor, the temperature will drop as reaction (2) proceeds, and the reactions will stop. Ash • You add the heat by carrying it with the oxygen Recycle carrier. Thus, the temperature of the carrier drops Fuel CO + H O 2 2 some DT from inlet to exit of the fuel reactor. Seal • The heat flow rate is then (mass flow) x (C DT ) and p (3) (2) must balance the heat used by reaction (2) • Note that steam pipes won’t work to transfer heat into Air the fuel reactor (800C input, typically exceeds steam piping temperature limits). 1) C + O  CO DH Overall fuel oxidation exothermic 2 2 1 (2) C+ MO  CO + M DH Metal oxide reduction fuel oxidation – 2 2 2 can be endothermic OR exothermic (3) M + O  MO DH Metal oxidation exothermic 2 2 3 ‹› A useful concept: Chemical looping creates a “metafuel” for the air reactor • A useful way to think about chemical looping combustion : – The fuel reactor takes the hydrocarbon fuel (coal, oil, biomass, natural gas) and uses that to make a different fuel (Cu, Fe, FeO, Fe O , CaS…etc; the reduced oxygen carrier). 3 4 – For convenience, call the reduce oxygen carrier a “metafuel” “metafuel” Burns w/o any CO 2. Recyclable Coal Iron oxide This is not a new phrase; a websearch reveals that the phrase was used to describe (inedible) tablets of fuel sold for camp stoves dating back to the 1920s: “…..‘Meta Fuel’ is now extensively used to replace methylated spirit for such purposes as are fulfilled by small spirit lamps and stoves. It acts as an efficient substitute in such circumstances, and has the advantage of being a solid substance, and thus easily portable and specially convenient. It is sold in small lamps and stoves, and refills are dispensed in the form of white tablets or cakes. Judging from my experience in connection with the first case detailed below, it appears that many who sell this material are ignorant of its composition and nature. Its poisonous properties on ingestion can hardly be too widely known……” R. Miller (1928). Archives of Diseases in Childhood. 3(18): 292–295. ‹› Potential oxygen carriers Many oxygen carriers have been studied to date: Iron: Fe O Hematite = Iron (III), Fe O Magnetite = Iron (II,III), FeO= Iron(II), Wusite, Fe 2 3 3 4 Copper: CuO Copper oxide, Cu O Cupric Oxide, Cu 2 What large Nickel: NiO, Ni difference in Manganese: MnO , MnO, Mn O , Mn O , Mn 2 2 3 3 4 system Cobalt: Co O , CoO, Co 3 4 configuration SulfatesSulfides: CaSO CaS, MnSO MnS, FeSFeSO 4 4 4 must exist for And others: Sb, Pb, Cd… copper versus Thermodynamics for iron and copper: iron carriers Methane overall ½ CH + O  ½ CO + H O DH = 402kJ Exothermic overall reaction 4 2 2 2 1000K Copper carrier 8 CuO + CH  4Cu O +CO + 2H O DH = 283kJ Exothermic metal reduction 4 2 2 2 1000C 4 CuO + CH  4Cu +CO + 2H O DH = 211kJ Exothermic metal reduction 4 2 2 1000C 2 Cu + O  2CuO DH = 274kJ Exothermic metal oxidation 2 1000K Iron carrier 12Fe O + CH 8Fe O + CO + 2H O DH = +154kJ Endothermic metal reduction 2 3 4 3 4 2 2 1000C 4Fe O + CH 8FeO + CO + 2H O DH = +303kJ Endothermic metal reduction 2 3 4 2 2 1000C 4/3Fe O + CH 8/3Fe + CO + 2H O DH = +154kJ Endothermic metal reduction 2 3 4 2 2 1000C Hint: where does the heat 4/3Fe + O  2/3 Fe O DH = 539kJ Exothermic metal oxidation 2 2 3 1000C go, above Fan, L. S., (2010). Chemical Looping Systems for Fossil Energy Conversions , John Wiley and Sons Publishers, see pp. 61 ff ‹› Thermodynamic limits on conversion How much oxygen, CO, H will exist in the products of CL combustion 2 CO Enhanced Oil Recovery specification establishes potential requirements 2 Notice there is no “excess air” to N + O 2 2 consume unused fuel. (vitiated air) CO + H O 2 2 Seal Ash Recycle Fuel CO + H O 2 2 Seal Air QUALITY GUIDELINES FOR ENERGY SYSTEM STUDIES CO2 Impurity Design Parameters, DOE/NETL341/011212, Jan 2012. http://www.netl.doe.gov/energyanalyses/pubs/QGESSSec3.pdf ‹› Understanding equilibrium limits on conversion Notice that at any temperature MeO if P P defined by (i), the 2 O2 O2 O (g) 2 metal oxide (MO )is reduced to 2 CH CO, the Metal (M). 4, Me CO , H , 2 2 Quiz for grad students: H O Your chemical looping 2 combustor is making 30 ppm CO. The metal/oxide reaction (i) M + O  MO ; DG = DG ˚ + RTln(1/P ) 2 2 T(i) T(i) O2 Your professor wants you At equilibrium, DG = 0, denote P ; DG ˚/(2.3RT)= log(P ) T(i) O2 T(i) O2 to add more metal oxide to improve CO burnout. The gasphase reactions 2 Will it work (ii) 2CH + O  2CO + 4 H ; DG ˚/(2.3RT)= 2log(P / P P )+ log(P ) 4 2 2 T(ii) CH4 CO H2 O2 (iii) 2CO+O 2CO ; DG ˚/(2.3RT)= 2log(P / P )+ log(P ) 2 2 T(iii) CO CO2 O2 A) Yes because…. (iv) 2H +O 2H O ; DG ˚/(2.3RT)= 2log(P / P )+ log(P ) 2 2 2 T(iv) H2 H2O O2 B) No because…. C) Maybe because…. If you have the values for DG ˚’s you can solve immediately for P , T O2 D) I just want to (P / P ) and (P / P ). You can get absolute concentrations of CO and H H2 H2O CO CO2 2 graduate. by noting the fuel is mostly converted to CO and H O. 2 2 ‹› Fe O reduction to Fe O with H /CO The metal/oxide reaction 2 3 3 4 2 Fe O = hematite; higher oxidation, 2 3 can reduce to magnetite. (i) 6Fe O O + 4Fe O 2 3 3 4 2 Fe O = magnetite; lower 3 4 oxidation, can oxidize to hematite, i.e., reverse of (i), or reduce further to FeO (Wustite), not discussed here. If you fix temperature, (i) will proceed forward or backward depending on the oxygen pressure established with gas phase fuel reactions, below: The gasphase reactions (ii) 2CO+O 2CO 2 2 (iii) 2H +O 2H O 2 2 2 Figure shows the reduction of Fe OFe O with H or CO. Even at equilibrium, there are ppm levels of residual H CO 2 3 3 4 2 2 making a slightly reducing environment. Note the combination of (ii) and (iii) is water gas shift CO + H O  CO +H . 2 2 2 Residual CO increases with temperature because the CO Fe2O3 reduction reaction is slightly exothermic (H Fe2O3 reduction 2 here is endothermic). Results are based on simulations using HSC Chemistry 7.1. Courtesy Mike Gallagher, NETL. ‹› Fe O reduction to Fe O with H /CO The metal/oxide reaction 2 3 3 4 2 Fe O = hematite; higher oxidation, 2 3 can reduce to magnetite. Notice what happens if you add (i) + (ii) : (i) 6Fe O O + 4Fe O 2 3 3 4 2 2CO + 6Fe O  4Fe O + 2 CO Fe O = magnetite; lower 2 3 3 4 2 3 4 oxidation, can oxidize to hematite, i.e., reverse of (i), or reduce further to FeO (Wustite), not discussed here. And similar for (i) + (iii): If you fix temperature, (i) will proceed forward or backward depending on the 2H + 6Fe O  4Fe O + 2 H O oxygen pressure established with gas 2 2 3 3 4 2 phase fuel reactions, below: The gasphase reactions The oxygen “disappears”. Why bother referencing oxygen (ii) 2CO+O 2CO 2 2 As will be seen, it is easier to understand… (iii) 2H +O 2H O 2 2 2 ‹› Ellingham diagrams1 These diagrams help quickly assess what species will oxidize or reduce. Write any reaction with gaseous RT ln (p ) oxygen as the reaction with one mole of O , e.g.: o2 2 8 p = 10 atm o2 4Fe O + O  6Fe O (rxn 1) 3 4 2 2 3 12 10 atm Then, at equilibrium, D G = 0 implies: rxn1 o o o 6G 4G – G –RT ln (p ) = 0 Fe2O3 Fe3O4 O2 o2 16 Temperature (K) 10 atm ZERO o STD. GIBBS FREE ENERGY CHANGE FOR (rxn1), DG rxn1 LOOK UP IN TABLES, DEPENDS ONLY ON TEMPERATURE Then, rearranging the above equation Fe O o O (g) DG = RT ln (p ) (eqn 1) 2 3 rxn1 o2 2 The equilibrium expressed by (eqn 1) is shown in the CH CO, 4, graph as the intersection of the lines for different values Fe O 3 4 CO , H , 2 2 of oxygen partial pressure. H O If the oxygen pressure is higher than equilibrium, (rxn 1) 2 will go forward. If it is less, (rxn 1) goes backward. What determines p The gas reactions Note: You want it to go backward in the fuel reactor O2 ‹› o DG or RT ln (p ) kJ/gmol O rxn o2 2 Ellingham diagrams2 These diagrams help quickly assess what species will oxidize or reduce. Write any reaction with gaseous RT ln (p ) oxygen as the reaction with one mole of O , e.g.: o2 2 8 p = 10 atm o2 4Fe O + O  6Fe O (rxn 1) a 3 4 2 2 3 12 10 atm Then, at equilibrium, D G = 0 implies: rxn1 b What if you want to consider other o o o 6G 4G – G –RT ln (p ) = 0 metal/ Fe2O3 oxide Fe3re O4actions O2 o2 16 Temperature (K) 10 atm ZERO (e.g. Me + O2  MeO) o STD. GIBBS FREE ENERGY CHANGE FOR (rxn1), DG rxn1 LOOK UP IN TABLES, DEPENDS ONLY ON TEMPERATURE o Plot the DG versus T for the metal of Then, rearranging the above equation interest; examples shown above (a) Fe O o O (g) DG = RT ln (p ) (eqn 1) 2 3 and below (b) the Fe O / Fe O line. rxn1 o2 3 4 2 3 2 The equilibrium expressed by (eqn 1) is shown in the CH CO, 4, graph as the intersection of the lines for different values Fe O 3 4 CO , H , 2 2 of oxygen partial pressure. H O If the oxygen pressure is higher than equilibrium, (rxn 1) 2 will go forward. If it is less, (rxn 1) goes backward. What determines p The gas reactions O2 ‹› o DG or RT ln (p ) kJ/gmol O rxn o2 2 Ellingham diagrams 3 4Fe O + O  6Fe O (rxn 1) 3 4 2 2 3 RT ln (p ) o2 8 p = 10 atm o2 Follow the exact same procedure to write the Ellingham diagram for CO or H 2 12 10 atm 2CO + O  2CO (rxn 2) 2 2 2H + O  2H O (rxn 3) 2 2 2 Temperature (K) 16 10 atm 500 1000 1500 Consider (rxn 2); you can treat (rxn 3) in exactly the same manner (not covered here). Then, at equilibrium, D G = 0 implies: rxn2 Higher CO o o o 2 2 2G 2G G RT ln (p p / p ) = 0 CO2 CO O2 o2 co co2 o DG – 2RT ln(p /p ) = RT ln (p ) rxn2 co co2 o2 RT ln (p ) o2 p = Notice you need to know the ratio of CO to CO partial o2 2 8 10 atm pressure to plot the left side versus temperature. 12 10 atm Assume some values for the ratio, and plot the left side Lower 2 5 7 (10 , 310 and 10 ). The intersection with the O2 16 oxygen lines represents an equilibrium condition. 10 atm The plots match your chemical intuition – higher CO levels have lower O at equilibrium 2 ‹› o DG or RT ln (p ) kJ/gmol O rxn o2 2 o DG or RT ln (p ) kJ/gmol O rxn o2 2 Ellingham diagrams 4 The last step is to add the equilibrium for the metal and oxide reaction to the CO reaction, i.e. combine RT ln (p ) o2 8 p = 10 atm o2 rxn1 and rxn2 plots as shown in the bottom graph 12 10 atm a 4Fe O + O  6Fe O (rxn 1) 3 4 2 2 3 b 2CO + O  2CO (rxn 2) 2 2 Temperature (K) 16 10 atm 500 1000 1500 The intersection of lines at (1000K, 225kj/gmol), circled, represents equilibrium. • What is the oxygen partial pressure • The ratio of CO to CO2 Higher CO/CO2 The graph provides insight into how changing parameters affects the equilibrium of (rxn 1): • What happens if you raise the temperature 12 10 atm 12 10 atm • Add more CO Lower • What happens if you consider a Me/MeO rxn that O2 P “sits” above or below rxn 1 on the plot 16 10 atm You can use published Ellingham diagrams with CO/CO , 2 H /H O, O nomographs and various A +O  AO 2 2 2 2 2 ‹› o DG or RT ln (p ) kJ/gmol O o rxn o2 2 DG or RT ln (p ) kJ/gmol O rxn o2 2 Quiz • Your classmate wants to make an oxygen carrier from aluminum because it is very energetic when it burns • How would you argue from experience if this is a good idea • How would you use an Ellingham diagram to figure out if that is a good idea 14 10 ‹› Solid Carbon Formation • What happens if any solid carbon is left on the oxygen carrier when it leaves the fuel reactor (Red arrow, below) • Carbon formation via equilibrium (chart, right) and also hydrocarbon cracking. • Notice that solid carbon on a metal oxide may not be a problem Boudard Reaction Equilibrium C (s) +CO (g) CO(g), 1atm, only CO/CO gases 2 2 N + O 2 2 1.E+00 (vitiated air) Carbon 1.E01 Forms CO + H O 2 2 1.E02 Seal Carbon 1.E03 Gasified Ash to CO 1.E04 Recycle Fuel CO + H O 2 2 Seal 1.E05 0 200 400 600 800 1000 1200 Temperature (C) Air th Gaskell, D. R. (2008) Introduction to the Thermodynamics of Materials, 5 ed, Taylor and Francis, pp. 365366 ‹› Volume Fraction CO OXYGEN CARRIER CAPACITY AND CIRCULATION RATES Fully oxidized Fully reduced Partially oxidized active species active species active species (e.g., CuO, FeO, etc.) (e.g., Cu, Fe, etc.) Define conversion X for a “supported” Inert support, metal oxide mass: m inrt carrier active mass: m active mass: m active mass: m red ox mm red X = 0 X = X=1 m m ox red ‹› Nomenclature active DX Inert X X 1 2 mass: m support, 1.00 mass: m m m m inrt red ox 0.80 C A 0.60 0.40 0.20 B Conversion 0.00 0 0.2 0.4 0.6 0.8 1 1.00 A= Active mass, oxidized 0.98 B = Inert mass 0.96 Oxygen C C = Working oxygen capacity 0.94 transport 0.92 R = C/A capability o 0.90 0 0.2 0.4 0.6 0.8 1 c = A/(A+B) o c R = C/(A+B) c = o o o ‹› Mass of carrier (1 kg oxidized state) Values for Oxygen Transport Capability (R ) for o Some Metal/Oxide Pairs Fe2O3 / Fe3O4 0.034 Mn2O3 / MnO 0.100 Mn2O3 / Mn3O4 0.034 Cu2O/ Cu 0.110 Inexpensive CuAl2O4 / CuAlO2 0.044 CuO / Cu 0.200 Fe2O3Al2O3 / FeAl2O4 0.045 CoO / Co 0.210 Good capacity Co3O4 /CoO 0.067 NiO / Ni 0.210 Mn3O4 / MnO 0.070 Co3O4 /Co 0.270 CuAl2O4 /CuAl2O3 0.089 ZnSO4 / ZnS 0.396 NiAl2O4 / Ni Al2O3 0.091 CuS04 / CuS 0.401 CuO / Cu2O 0.100 MnSO4 / MnS 0.424 Fe2O3 / FeO 0.100 FeSO4 / FeS 0.425 CaSO4 / CaS 0.470 Very inexpensive. Throwaway option Table 1. Values of R for some potential oxygen carrier reactions, arranged small to large. o ‹› Establishing Carrier Requirements from FCC Experience Some FCC units operate with 41,000 kg/min solid circulation rate. Petrochemical Proven Process Technology Fluid Catalytic Cracking (FCC) Cyclone Vessel Flue Gas Stripper Stripping Stream Catalyst Regenerator Stripper Standpipe Riser Reactor Regenerator Air Standpipe Air Heater Lower Feed Injection Dispersion Steam ‹› Properties of the Oxygen Carrier • Assuming a solids circulation rate operating fluid catalytic crackers (41,000 kg/min) • Calculate the thermal output possible for a chemical looping system for different carriers/conversions • This only accounts for supplying oxygen, not thermal balance (next slide). 2000 6 40 Cu, 60Al2O3: CuO→Cu Pure Cu: CuO→Cu 1800 5 1600 Pure Cu: CuO→Cu 1400 4 Ilmenite: Fe2O3→FeO 1200 3 Ilmenite: Fe2O3→Fe3O4 1000 Pure Fe: Fe2O3→FeO 2 Pure Fe: Fe2O3→FeO 800 Cu, 60Al: CuO→Cu 600 1 Pure Fe: Fe2O3→Fe3O4 Ilmenite: Fe2O3→FeO 400 Pure Fe: Fe2O3→Fe3O4 0 500 1000 1500 2000 2500 200 Ilmenite: Fe2O3→Fe3O4 Thermal Output MW 0 0.5 0.7 0.9 Conversion ‹› Thermal Output MW Acceptable Operating Range Determining the solids circulation rate • The oxygen carrier circulation rate is determined by: 1) The oxygen needed for the fuel flow rate 2) Endothermic carriers: the heat needed to drive the fuel reactor • How are requirements (1) and (2) compatible – It turns out the required circulation rate is sometimes entirely dominated by the need to supply heat to the fuel reactor. – In that case, a higher capacity oxygen carrier is not needed. ‹› Circulation rate analysis Vitiated Air Out • Solids circulation shown A by the double curved arrows, A and B. B • There is only ONE N + O 2 2 circulation rate – why (vitiated air) show as two arrows CO + H O 2 2 – Emphasize that you CO + H O 2 2 must satisfy two CO 2 Seal conditions: H O 2 A. Supply enough oxygen to react with all the fuel. Ash B. Endothermic carriers: Heat Recycle supply enough heat to Out Fuel CO + H O keep the carrier “hot”. 2 2 (steam) Seal • Note that the mass flows Fuel flow that satisfy A has a Air for 1MWth definite minimum for a (assume CH ) 4 fixed fuel flow rate, but no maximum. Which is bigger A or B Air in ‹› Circulation rate analysis, continued A – the mass flow needed to supply enough oxygen to react with the fuel. Vitiated Air Out B the mass flow needed to carry enough A heat to support the endothermic reaction B via a 50°C temperature drop. A B B/A Fuel Air Reactor Reactor CO 2 H O Min Circ Circ rate 2 Carrier Pair R o Rate O to supply Ratio: 2 supply heat Heat/O2 DT = 50°C Heat (kg/sec) (kg/sec) Rate Out Fe2O3 / Fe3O4 0.034 2.35 3.36 1.4 (steam) Fuel flow Fe2O3 / FeO 0.1 0.80 6.75 8.4 for 1MWth Fe2O3/Fe 0.3 0.27 6.10 22.9 CuO / Cu 0.2 0.40 n/a n/a Air in CaSO4 / CaS 0.47 0.17 3.07 18.1 Analysis is for methane fuel; could be repeated for coal, etc. Note that the exothermic carrier CuO has a much lower circulation rate than all the others. ‹› KINETIC RATES AND REACTOR SIZE ‹› KINETIC RATES AND REACTOR SIZE A “bubbling” fluid bed (BFB) is one possible reactor configuration (left). BFB is approximately like a stirred reactor (right). The usual combustor design concepts apply: 1) heat release rate balances the incoming rate of cold reactants or it will blow out 2) for a given throughput, the reactor volume is inversely proportional to the reaction rate. Reactor Temperature Carrier T State X 1 Fluid Bed Inputs Outputs Carrier m X oc,i i m X oc,o Reactor State X m 2 air,i m air,o Volume Air T i T V R ‹› Experimental measurement of kinetics Thermo gravimetric Analysis (TGA) thermocouple For gaseous reactions, sample weight describes the conversion: Cycling the fuel and oxidizer over the sample pan will give many “cycles” to analyze. Solid fuels (coal and biomass): must reload the pan every cycle. ‹› Example of kinetics measurement (CH ) 4 TGA Data for Cubased carrier 4CuO CH  4Cu  CO  2H O 4 2 2 1 44 1000 42 0.8 800 100 CH 4 40 0.6 600 38 o T ( C) 750 36 0.4 400 800 34 850 o 100 CH , 800 C 4 200 0.2 900 mm  r 32 d =150250 mm X = 1 p r mm  ox r 30 0 0 0 500 1000 1500 2000 0 0.4 0.8 1.2 1.6 2 Time (min) Time (min) Typical mass and temperature measurement for CuO/bentonite Effect of reaction temperature on CuO/bentonite particle and 100 CH for reduction and air for oxidation 4 particle and CH reaction 4 reactions 1 1 dX Results fit to JonsonMehlAvarmi (JMA) rate equation m n = ky n(1 X )ln(1 X ) CH 4 dt Monazam et al. (2012) “Kinetics of the Reduction of CuO/Bentonite by Methane (CH4) During Chemical Looping Combustion”, to appear Energy and Fuels. ‹› Mass (mg) o Temperature ( C) Conversion (X)A visual representation of reaction rates • An informative/interesting way to see the metalmetal oxide cycle. • Combustion Quiz – What type of flame does a propane torch use (e.g., diffusion, premixed, partially premixed) – What is the partial pressure of oxygen inside the flame Movie of copper reduction and oxidation ‹› Kinetics in a stirred reactor bed (1/2) • With kinetic rates, write energy and mass balances Air Reactor (Cu2OCuO) 2000 with an “efficiency” of 10 sec 1800 800C Input Temp conversion h (defined I 1600 below). 1400 1200 • Does the bed have “lightoff” 1000 behavior 800 600 T at steady out 400 state 200 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Efficiency (NI) Green line depicts efficiency as a function of output temp (from mass balance using kinetic rates), dashed line shows same using energy balance equations. Point of intersection is the desired steadystate solution. ‹› Reactor Internal Temperature (C)Kinetics in a stirred reactor bed (2/2) Air Reactor (Cu2OCuO) • With kinetic rates, write 2000 10 sec energy and mass balances 1800 800C Input Temp with an “efficiency” of 1600 Different inlet 1400 conversion h (defined I temperature, mass 1200 below). flow, etc. 1000 • Does the bed have “lightoff” 800 behavior 600 T at steady out 400 state 200 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 Efficiency (NI) Green line depicts efficiency as a function of output temp (from mass balance using kinetic rates), dashed line shows same using energy balance equations. Point of intersection is the desired steadystate solution. Detailed approach: kinetics inserted into CFD models (described later). ‹› Reactor Internal Temperature (C)How quickly do the reactions need to occur Answer: it depends on several parameters Simple estimate: m = mass flow rate The extent of conversion is X. “active” species Assume constant rate (dX/dt) Residence time is t. M = “active” mass , r V a t (dX/dt) = DX (2) t = M/m, res. time Assume we want DX 1 and combine (1) and (2): (3) DH = heat release/mass Thermal output r DH (dX/dT) a = Volume m DH Thermal output Using bubbling fluid bed boilers as an example, you = can “check” this with reasonable parameters: V Volume 3 r = (2) r ; r 500kg/m 1 a bed bed t 60 sec, Basu pp. 123 or, multiplying by r /r : DH = 23MJ/kg, coal HHV a a 3 Leads to volume heat release 3.8MJ/m ; “reasonable” see Basu pp. 236 r m DH a = r DH / t (1) Note: (1) you can adjust r a a (2) t is tens of seconds, or more r V a In fluid bed boilers, the solids flow will include significant amounts of inert material like ash , limestone, sand . st 1 Steam, 41 edition, Babcock and Wilcox, see Figures 4, 12 in chapter 17 for bed density; (2 )discussed later. ‹› Solid fuel combustion ‹› o o T Te em mp pe er ra at tu ur re e ( ( C) C) o Reaction temperature ( C) Can We Use Coal Directly TGA Profile of Coal in N TGA Profile of Coal +CuO in N 2 2 Nitrogen Nitrogen Oxy Oxygen gen 210 210 1000 1000 40 1000 N O 2 2 35 200 200 800 800 800 30 44.88 25 190 190 600 600 600 20 400 180 180 15 400 400 combustion 55.12 10 200 combustion Volatiles 170 170 200 200 5 mositure out 0 0 160 160 0 50 100 150 200 250 0 0 20 20 40 40 60 60 80 80 100 100 120 120 140 140 160 160 180 180 o o Reaction time (min) R Re ea ac ct tio ion n T Te em mp pe era rat tu ure re ( ( C) C) Reaction time (min) Rates were higher than expected. Why ‹› weight (mg) W We eiig gh ht t ( (m mg g) )Reaction Pathways for Solid Fuel CLC • Coal CLC with metal oxides via gaseous intermediates: In N : 2 Coal Coal pyrolysis CO/H + CuO Cu +CO /H O 2 2 2 CO + C 2CO 2 In CO : 2 C+CO 2CO 2 CO+CuO Cu+CO 2 • CLOU mechanism (Chemical Looping Oxygen Uncoupled): CuO Cu/Cu O +O 2 2 Discussed Coal + O CO 2 2 next • Solidsolid interaction: MeO+C MeO +CO 2 ‹› Possible Reasons for Rapid, LowTemperature reaction of solid carbon with CuO o At 500 C 2CuO = Cu O + 1/2O 2 2 2 9 10 = P 10 = P o 9 O2 O2 At 500 C, P is 1.110 O2 Reaction 9 not possible Will removal of oxygen (1.110 ) at low O No 2 continuously by carbon, facilitate the Reaction CuO decomposition Carbon CuO CuO Carbon • Reacted C with various oxygen partial Significant pressures Reaction vitiated air o o – With air: modest reaction at 500600 C. at 500 C o – No reaction at 500600 C with oxygen at low O partial pressure (2, vitiated air) 2 9 – Confirms that P is 1.110 not sufficient to O2 o react with C at 500 C to facilitate the forward reaction at sufficient rates ‹› o Reaction temperature ( C) Combustion Rates of Coal (100 micron) with Various Particle Sizes of CuO in TGA 0.10 1000 5 micron o 713 C 0.08 800 63177 micron o 780 C 0.06 600 354595 micron o 874 C 0.04 400 0.02 200 0.00 0 0 20 40 60 80 100 Reaction time (min) Higher combustion temperature with increasing particle size ‹› 1 Reaction rate (min )Effect of Dilution by Quartz Powder on the TGA Combustion Performance of Carbon and CuO • Mixed CuO, Quartz and C powders • CuO/C ratio was kept constant • Reaction T increased with increased dilution These data and others (flow tests + DFT calculations) suggest a solidphase reaction between carbon and the oxygen carrier (Fe, too). Reduces coking automatically Siriwardane, R. Tian, H., .Miller, D., Richards, G., Simonyi, T., Poston, J. (2010). Evaluation of reaction mechanism of coalmetal oxide interactions in chemicallooping combustion, Combustion and Flame, Combustion and Flame, Volume 157, Issue 11, November 2010, Pages 21982208 ‹› Reactor and System Design ‹› An introduction to fluidization • Fluidization is used Main topics widely in the • Physical description of chemical industry fluidization. – Fluid Catalytic Cracking (Hydrocarbons) • Key velocity parameters: minimum fluidization, – Catalytic reactions. bubbling, and terminal. – Drying and calcining. – Many reactions. • Fluidization regimes, application, and • Coal, biomass, and components. waste combustion • Effect of particle or gasification. morphology • Chemical looping. • Reactor configurations. ‹› References Basu, P.(2006), Combustion and Gasification in Fluidized Beds , CRC Press, Taylor and Francis Group, Boca Raton, FL. Excellent treatment of practical issues in design of fluid bed combustors and gasifiers. nd Kunii D., Levenspiel, O. (1991). Fluidization Engineering, 2 ed., ButterworthHeinemann, Newton MA. A classic text on fluidization – get a copy if you work on multiphase flows. ‹› Physical Description of Fluidization • A granular material (b) (a) does not typically flow like a fluid; it can form a “pile”; (a). mesh • But, if you supply No gas gas fluidizing gas, granular material can (d) (c) behave as a liquid: – Horizontal surface (b) – Flow from holes (c) – Equalizes levels (d) – Floats light objects (next No gas gas gas slide) ‹› Physical description of fluidized beds (cont.) – Lighter objects float. – The bed volume is larger in a Floats fluid state: Sinks void fraction e = V /(V + V ) gas gas solid – The gas flow rate is typically described by the superficial gas velocity U: (Gas volume flow rate) (Crosssection area, no solids present) DP – The bed pressure drop L Larger b balances the overhead weight Volume DP A = A L (1e ) (r r ) mesh xc xc b mf solid gas The mf subscript: voidage at minimum fluidization, explained next No gas gas Typical e values( ), U values m/s: Packed Bed (0.40.5) 13, Bubbling Bed (0.50.85)0.52.5, ‹› Circulating Bed (0.850.99) 46, Transport Reactor(0.980.998)1530, Basu pp. 22 Velocity parameters • Minimum Fluidization Velocity U : mf – The superficial velocity that “just” fluidizes; the point where the bed weight balances the pressure drop. – Typically you measure U to get an mf accurate value for a new granular material. gas • The minimum bubbling velocity U : mb – The velocity where bubbles first appear. – Can be equal or greater than U mf. gas • The particle terminal velocity U : t – Note the velocity is NOT a uniform profile. – What happens if U U t Shown smaller Fluidization regimes: smooth, bubbling, turbulent, to with more height fast, pneumatic transport gas above the bed ‹› 3D multiphase Application to Air Exhaust CFD modeling Colors are Separation chemical looping Crossover void fraction: cyclone Blue = e 0.999 • Air reactor (right), fuel dilute Red = e 0.6, reactor (left). dense • Solids are transported from Loop Seal the air reactor where UU . t Riser • Solids are fluidized in the CO fuel reactor where U UU . 2 mf t H O 2 Fuel Exhaust Solids Reactor Questions to discuss: transport Where is the pressure the greatest and why Air Reactor LValve What controls the solids circulation rate Solids Flow Fluidizing Gas + Fuel Fluidizing U UU What does the loop seal do, and mf t Air UU t how does it work ‹› Loop seal and Lvalve components Loop seal – Isolates process gas LValve controls the solid flow above (A) from process gas below (B). delivered to the right. Solid flow (A) Fluidizing gas Fluidizing gas (B) ‹› The particle properties: effect on fluidization Fluidization regimes depend on the particle morphology, size, density. A Geldart Classification A, B, C, D A = Aeratable. Can achieve smooth 3 fluidization, low density (1.4 g/cm ). B B= Bubbles form at U . Behaves like sand, mf particle sizes 40500 microns, density 1.4 – 3 . 4gm/cm C = Cohesive: hard to fluidize, very fine (40 C microns). Behaves like flour, starch. D = Dense, large particles (+500 microns). Can be difficult to fluidize, will channel and D spout. Drying coffee beans, grain, etc. Photo courtesy Mixtures of particle sizes properties are complicated Jonathan Tucker NETL Can segregate, defluidize, and may not react as expected. ‹› Types of gassolid reactors Circulating fluid bed combustor • Fixed bed: Not fluidized, UU mf Ex: Stoker. • Bubbling fluid bed: operates U U , the particles stay in the tr bed. Ex: BFB boiler. • Circulating fluid bed: operates UU , the particles are carried tr out of the bed, and are recycled. Ex: CFB Boiler, FCC for hydrocarbon cracking. • Moving bed: Not neccesarily fluidized, but the solid moves countercurrent to the process gas. Ex: Lurgi gasifier. • Entrained: no “bed”, dilute phase. Ex: Pulverized coal boiler ‹› Discussion • The preceding fluidization introduction has covered just hydrodynamics. Also need to consider reactions, conversion, heat transfer. • If you want to design a solid fuel chemical looping boiler, what combination of gassolid reactors is the best • What are the tradeoffs for small or large oxygen carrier particles • Some of these issues are best addressed with validated CFD simulations and system models. ‹› Modeling of Fluidized Beds (Courtesy: F. Shaffer, NETL) Movie DNS LBM DEM MPPIC MultiFluids FilteredEqs ROM www.mfix.netl.doe.gov ‹› Comparison of CFD and Cold Flow Rig EXPERIMENT SIMULATION Oxygen Lighter ash carrier carried out with fluidizing steam or CO 2 Movie Air reactor Oxygen carrier Solid fuel into this bed ‹› Design consideration – the air reactor (fuel reactor discussed later) ‹› Where do you take the heat out Endothermic carriers: you don’t need need to remove heat from the fuel side. Vitiated Air Out Exothermic carriers, you may need to remove heat from the fuel side to keep from overheating the carrier. Potentially use the recycle fluidizing gas as a heat exchange media. Fuel Air Reactor Reactor CO 2 All carriers: must manage the air reactor H O 2 exotherm. Endo or Exo Heat thermic Out (steam) Fuel flow for 1MWth Air in ‹› (Cont) where do you take the heat out – endothermic carrier Must allow equal temperature rise in the air reactor DT as ar temperature drop in the fuel reactor DT fr Vitiated Air Out A simple enthalpy balance will show that the air needed to transport the carrier has a minor Heat Out effect on the carrier temperature Fuel Air (steam) Reactor Reactor even w/o exothermic reactions. CO 2 H O 2 DT = 50°C fr DT = + 50°C You will need to take a lot of heat ar out as the reactions occur. How Fuel flow for 1MWth Air in ‹› Conventional Circulating Fluid Bed Combustion (CFB) • Fuel is added to the “riser” and reacts with air. • Unburned fuel may circulate around the “loop” several times. • The circulating “bed” of material is mostly inert, and provides a large thermal mass. • Heat can be removed in several places. • Note the similarity to chemical looping combustion. “…..the mass flow rate of recycled solids is many times the mass flow rate of incoming air, fuel, and limestone….the bed solid temperature remains relatively uniform” Steam, Edition 41, pp. 179, The Babcock and Wilcox Company. ‹› Circulating Fluid Bed (CFB) Combustion versus Chemical Looping Air Reactor • The CFB case uses significant inert flow that moderates the temperature rise. • The CLC case Meta appears more like Solid Fuel pulverized coal – Fuel Particles Particles potential for significant Inert bed temperature rise Particles • Compare reaction (ash, sand) enthalpies (i.e., HHV) Air Air for metafuel versus carbon (next slide). CLC CFB Air reactor Combustion ‹› Circulating Fluid Bed (CFB) Combustion versus Chemical Looping Reaction DH per kg 'fuel', "meta" fuel or carbon fuel Reaction MJ/kg Meta Solid 1 Fe O + 1/4O 3/2 Fe O 0.51 3 4 2 2 3 Fuel Fuel Iron carriers 2 FeO+1/4O 1/2Fe O 1.96 2 2 3 Particles Particles 3 Fe+3/4O 1/2Fe O 7.39 2 2 3 Inert bed 4 Cu+1/4O2 1/2Cu2O 1.34 Copper carriers Particles 5 Cu+1/2O CuO 2.46 2 (ash, sand) Calcium sulphide carrier 6 CaS + 2O CaSO 13.24 2 4 Air Air Carbon fuel 7 CLC C+O CO 32.76 CFB 2 2 Air reactor Combustion 35 Reactions 1 – 7 compare DH for metafuel (CLC) and Carbon 30 carbon fuel (CFB). 25 Per mass basis because CFB circulation is 20 2 CaS expressed as mass flux (kg/s/m ). 15 Iron 10 Copper The metafuel particles produce 360 times less heat 5 per mass than carbon fuel. 0 1 2 3 4 5 6 7 But, there can be significantly more mass flux of Reaction meta fuel versus carbon fuel (how much more). ‹› Standard reaction enthalpy at 298K; data calculated from NIST web book and Lange’s Handbook of Chemistry DH, J/kg CFB Typical Flow Parameters Solid Mass Flux G s • The chart below provides typical parameters for CFB operation in boiler and Fluidized Catalytic Cracking applications. Solid • Using the superficial gas velocity for the boiler 6 Fuel m/s and multiplying by the air density (800°C, 1 2 Particles atm) provides a mass flux of air : G = 2 kg/m /s. a Inert bed • With the smallest circulation rate, from chart below, 2 Particles G = 10 kg/m /s. Solids to air mass ratio is 10/2 = 5. s (ash, sand) Thus, G /G 5 typical solid/air mass ratio s a Air Mass Flux G a Parameter Boiler FCC Reactor 2 External Circulation Rate (kg/m /s) 1050 5001000 Superficial Gas Velocity (m/s) 6 25 3 Suspension Density, upper region (kg/m ) 110 10100 Particle Size (microns) 200 70 ‹› Basu, P.(2006), Combustion and Gasification in Fluidized Beds , pp. 42, CRC Press, Taylor and Francis Group, Boca Raton, FL. This ratio can cover a wide range of values; this value uses the parameters indicated. The relation between fuel/air ratio and G /G s a If the solids flow is metafuel (or, pure solid fuel): Solid Mass Flux G s fuel/air mass ratio = G /G s a If the solids flow has a mass fraction of fuel Y : f Y = (fuel mass flux)/(solid mass flux) f Meta Solid Fuel fuel/air mass ratio = Y ·G /G f s a Fuel Particles Particles The chart below lists the stoichiometric fuelair Inert bed ratios for several metafuels and carbon fuel: Particles (ash, sand) "meta" fuel or Stoich. f/a Air Air Air Mass Flux G carbon fuel Reaction mass ratio a CLC 1 Fe O + ¼(O +3.76N ) 3/2 Fe O + 0.94N 6.73 CFB 3 4 2 2 2 3 2 Air reactor Combustion Iron carriers 2 FeO+¼(O +3.76N ) 1/2Fe O + + 0.94N 2.09 2 2 2 3 2 3 Fe+3/4(O +3.76N ) 1/2Fe O +2.28N 0.542 2 2 2 3 2 Most of the f/a ratios are less than G /G = 5 s a (except reaction 1). 4 Cu+1/4(O +3.76N ) 1/2Cu O + 0.94N 1.85 2 2 2 2 Copper carriers Thus, you need to operate very dilute OR don’t 5 Cu+1/2(O +3.76N ) CuO + 1.88N 0.926 2 2 2 convert much of the material. Calcium sulphide Coal CFBs accomplish dilute fuel operation with 6 CaS + 2(O +3.76N ) CaSO + 7.52N 0.266 2 2 4 2 carrier the addition of lots of inert materials ( 2 of the riser mass is solid fuel) Carbon fuel 7 C+(O +3.76N ) CO +3.76N 0.0874 2 2 2 2 ‹› For carbon fuel example, f/a= Y ·G /G is 0.0874 = Y ·5. Thus, for stoichiometric conditions, Y = .0175. Actual plants operate with excess air, lowering this value further. f s a f fPractical issues CLC Air reactor Outlet The air reactor residence time is not necessarily Fe O or FeO 3 4 long enough to allow full oxidation. + Fe O 2 3 • CFB riser residence time 5 seconds, max. • Full oxidation takes 10 seconds for some carriers. Options: A. Supply approximately 10/5 = 2x more carrier than stoichiometric; convert part of the carrier (figure A, left) and use external heat exchanger. B. Add a bubbling bed at the bottom with inbed heat removal (figure B, below). C. Arrange two interconnected loops (figure C, below): Fe O or FeO 3 4 D. Your idea How about Moving Bed Air, Inlet T External HX exit External HX Air reactor Secondary Fuel reactor T bed carrier circulation carrier circulation Transport Air Steam Generation C B A Exchange between reactors Air ‹› Advantages of case C (last slide) Follow the mole flow rate of reduced species (subscript r) at stations 1 – 5 shown in the diagram. Some definitions: F = single pass fractional conversion; N =(1F)N r3 r2. R = recycle rate; R = N /N (see note, below) . Air reactor Fuel reactor r5 r4 carrier circulation carrier circulation F = 0 implies no oxidation occurs in the reactor. F = 1 implies no reduced material emerges at station 3 – full oxidation. R = 2 (for example) means the molar flow in recycle is twice the outflow. 1 (1F) Recycle Mole balances imply: N = N r1 r4 R (1RF) 0.8 R=0 0.6 R=1 N r4 4 0.4 R=2 3 R=4 0.2 N r4 5 R=8 0 R F 0 0.2 0.4 0.6 0.8 1 N r1 F Single pass fractional conversion 2 N = moles/sec reduced state input r1 1 N = moles/sec reduced state output r4 N r1 Main point: Modest singlepass fractional conversion can be offset by recirculation; e.g.: Air reactor process flow Equivalent topology nomenclature F = 0.2, R = 8 implies N /N = 0.3 r4 r1 Note that since the flow splits at 3 to 45, the recycle/output mole flow ratio is the same for reduced or total ( Only 30 is not reduced, 70 is oxidized) molar flow (oxidized + reduced + inert, tot subscript); N /N = (Y N )/(Y N ). The mole fractions Y r5 r4 r5 tot5 r4 tot4 r5 = Y because they are split from the same stream at station 3. Thus, N /N = N /N = R. Disadvantage: large R promotes attrition. r4 r5 r4 tot5 tot4 ‹› N / N r4 r1 Redued state flow out/in Recent Research Results ‹› Chemical Looping – work in progress at NETL – Experiments: Designed, built, and producing validation data. • Labscale to fullyintegrated 50kWth loop. – Simulations: Developed models, simulation tools . • ZeroD to fullloop 3D. – Developed novel performance/lowcost carriers. • ”Promoted” ironore, and “thermally neutral” CuFe. – Technoeconomic results: results for NG steam. • Confirms lowcost guides research (shown later). Separation cyclone Crossover Loop Seal Riser Fuel Reactor Air Reactor LValve Solids Flow NETL O Carriers 2 Cold flow validation rig, 50kWth Chemical Maximal O capacity. 2 Looping Reactor, Single fluid bed, attrition Models: zeroD, 3D loop, fuel reactor “Thermally neutral”. test, microwave solids flow sensor. Std. materials manufacturing Key Findings Chemical Looping – A few more details • Thermally neutral carrier development: (2) – Allows independent control of circulation rate. – Large O capacity means low circulation rate 2 Cyclone – reduced attrition C1200 Test Section Upper • Hydrodynamic predictions versus C1250 Riser R1150 measurements: – Cold flow predictions significantly affected by Loop Seal R1300 chosen model parameters. – Emphasizes need for model validation and Lower Riser diagnostics in hot flow. R1100 – Microwave sensor being developed for hot Fuel Reactor flow diagnostic. R1400 • Lessons learned and progress made • Multiple solids flow issues resolved – see (1) and (2) recent test had smooth LValve Air Housing solids circulation. Reactor R1450 R1000 • Batch mode oxidation/reduction on baseline (raw) hematite; follows (1) expected model results. Air Preheater and Tee Air Preheater and Tee H1800 H1850 H1800 H1850 • Bubbles matter See next slide Industrial Carbon Management Initiative ‹› Fuel Conversion and Bubbles Cyclone C1200 Test Section Upper C1250 Riser R1150 Loop Seal Movie (cold flow) aa cut R1300 Lower Riser R1100 Fuel a Reactor a’ R1400 a a’ LValve Air Housing Reactor R1450 R1000 Air Preheater and Tee Air Preheater and Tee H1800 H1850 H1800 H1850 Porous plate different injectors Video and CFD courtesy Doug Straub, A. Konan, NETL ‹› Fuel conversion: CFB versus chemical looping fuel reactor Circulating fluid bed combustor Chemical looping: air reactor and fuel reactor N + O 2 2 (vitiated air) CO + H O 2 2 Seal Ash Recycle Fuel CO + H O 2 2 Seal Air • Gasphase combustion for CO H2 • Gasphase combustion for CO requires oxygen carrier (gaseous 8 H2 has 3 oxygen. oxygen P 10 atm). O2 • Solid char is recycled. • Solid char cannot go to air reactor. ‹› Techno Economic Studies : Parameter Potential Impacts Expected Relative to a CLC Base case (assumes changes in parameters are within operating range) Vessel Vessel Circ. Auxiliary CO Cost of 2 Boiler Eff. Equip. Cost Height Diameter Rate Power Capture Steam Large Small Small Small Carrier Reactivity (literature) + = + = + = + = Carrier Loss (0 ) and Medium This is why we want to focus on this issue going forward Price (0/lb) + = + Carrier Size (0.28mm) and Small Small Small Small 3 Density (203 lb/ft ) + = + = + = + = Carrier Conversion (from Medium Large Small Small Small reducer 47; from oxidizer + = + + = + = + + = + + = + 95) Small Small Small Small Small Reactor Temperature (1700 F) + = + = + + = + = + = Reactor Velocities Large Large Small Small Small (reducer outlet 33.6 fps; + = + + = + = + + = + + = + oxidizer outlet 29.4 fps) Medium Small Small Large Small Small Natural Gas Conversion (97.5) + = + + = + + = + + = + + = + + = + Small Small Small Small Small Small Oxidizer XS O (3.8mol 2 in offgas) + = + = + + = + = + + = + + = + Industrial Carbon Management Initiative Chemical Looping – Discussion/ Thinking Question • Relative to the Define: earlier discussion a = (kg CO produced) / (kg fuel burned ) 2 of the figure at w = (separation work, Joules ) / (kg CO ) CO2 2 CO 2 the left: – Does chemical looping “fit” this description h g Generator – What are the Efficiency Q = m DH f W potential benefits 1 W Fuel Heat o Net Gross and drawbacks Input Generator Output Generator Carbon of chemical Work Separation looping w/r to Unit efficiency • What applications of chemical looping are most attractive Approx Ranges: (30 – 60) (610) ‹› Chemical Looping Summary • Not completely new, but new interest because of CO2 capture. • Various metal/metal oxide pairs are candidate oxygen carriers. • For a given heat output, reactor circulation rates depend on the oxygen capacity and the thermal balance. • Practical experience from CFB and FCC applications suggest the range of application. • In progress: N + O 2 2 (vitiated air) – kinetic improvements – reactor design and optimization CO + H O 2 2 – model validation Seal – attrition Ash Recycle Fuel CO + H O 2 2 Seal Air ‹›
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