Modeling of Combustion Systems A Practical Approach 1 pot

... 17 .19 5 1. 1 3484 13 01 Gas 5.0 15 .0Ethane C2H622323 15 .899 1. 3 3540 968 11 66 Gas 3.0 12 .5Propane C3H8 216 69 15 .246 1. 3 3573 8 71 Gas 2 .1 10 .1 n-Butane C4H 10 213 21 14.984 1. 2 3583 7 61 ... (Dry) Waste GasesNatural Gas LPGCrackedGasCokingGasReformingGasFCCGasReﬁnery GasSample 1 Reﬁnery GasSample 2PSAGasFlexicokingGasTulsa Alaska Netherlands Algeria Propane ButaneCH493.4% ... 11 0.49 12 4.92 13 7.47 14 8.33 18 3.80 204.87C4H6 1, 3-Butadiene 75.54 86. 51 96. 61 114 . 41 129.34 14 1.76 15 2.06 18 5.05 222.02C4H8 1- Butene 79.54 91. 87 10 3. 41 124.30 14 2.47 15 8.20 17 1.75 216 .22...

Modeling of Combustion Systems A Practical Approach 7 pot

... 614 Modeling of Combustion Systems: A Practical Approach Usually, one deﬁnes a reaction coordinate known as the conversion (xk),having the property that for species k the reaction starts at ... ideal gas law applies:(G .12 )where are the total moles of the reaction. This gives(G .13 )We may also write as a function of conversion:(G .14 )Typically, we use Equation G.8 or Equation G .10 ... $ 11 0 1 12022,,⎡⎣⎤⎦⎛⎝⎜⎞⎠⎟=⎡⎣⎤⎦⎛⎝⎜⎞⎠⎟=pxpxPPP$xrrxRRRRPR 1 20 10 1 22 10 10 =⎡⎣⎤⎦⎡⎣⎤⎦==⎡⎣⎤⎦⎡,,,,$⎣⎣⎤⎦=⎡⎣⎤⎦⎡⎣⎤⎦=rpxrpx 1 1 1 20 10 1 22PPPR,,$© 2006 by Taylor & Francis Group, LLCKinetics Primer 615 We may also substitute mole fractions for concentrations using(G .11 )For combustion in furnaces,...

Modeling of Combustion Systems A Practical Approach 13 potx

... ,, ,11 1 1 αΔ=+()−()QKZkZZkzH 11 1 1== =Q 1 KkfHkz fCTTaw ca awpg ref,, , ,11 11 1ααΔ−=+()−()TTffHkzCrefawawcapg 1 1 1 1 1 −=+−,,,,ααΔQ 21 fHQ fCTTaw ca awpg ref,, , ,12 112 1 1 αΔ−=+()−()© ... refwNONO,⎛⎝⎜⎞⎠⎟=++−=ααββα0 01 1 11 0βδββ 1 10 11 =−δαββα β0 01 1 1 1=++−w refw ref,,βδαδδδ0 1 0 01 1 1= −++⎛⎝⎜⎞⎠⎟w ref,βδα δδ 1 1 01 111 1=+ +⎛⎝⎜⎞⎠⎟w ref,© 2006 by Taylor & Francis ... Group, LLC464 Modeling of Combustion Systems: A Practical Approach where TAFT ,1 is the adiabatic ﬂame temperature at the ﬁrst zone. We shall useto denote that our speciﬁc heat of combustion refers...

Modeling of Combustion Systems A Practical Approach 14 potx

... Analysis 6 1. 3.2 Dimensional Analysis 8 1. 3.3 Raleigh’s Method 9 1. 3.3 .1 Cautions Regarding Dimensional Analysis 11 1. 3.4 Function Shape Analysis 15 1. 3.5 The Method of Partial Fractions 18 1. 3.5 .1 Limitations ... Burners 11 32 .1. 2.4 Flat-Flame Premix Burners 11 42 .1. 2.5 Flashback 11 52 .1. 2.6 Use of Secondary Fuel and Air 11 52 .1. 2.7 Round Combination Burners 11 62 .1. 2.8 Burner Orientations 11 92 .1. 2.9 Upﬁred ... Upﬁred 11 92 .1. 2 .10 Downﬁred 12 02 .1. 2 .11 Side-Fired 12 12 .1. 2 .12 Balcony Fired 12 12 .1. 2 .13 Combination Side and Floor Firing 12 12.2 Archetypical Process Units 12 32.2 .1 Boilers 12 32.2 .1. 1 Firetube...

Modeling of Combustion Systems A Practical Approach 2 pptx

... 11 17 37. 01. 6 011 25.630526 .11 "8/5 11 0357.04.8 014 19.630057 .11 "4/3 11 19 67.08. 011 603.730578 .11 "8/7 11 4587. 01. 311 996.730000. 21& quot; 21 918 .09. 711 584.830052. 21& quot;4 /1 21 158.07.2 219 62.930005. 21& quot;2 /1 ... 0. 010 31 1 7 /16 " 1. 4375 4. 516 0 1. 6230 0. 011 27 1 1/2" 1. 5000 4. 712 4 1. 76 71 0. 012 27 1 9 /16 " 1. 5625 4.9087 1. 917 5 0. 013 32 1 5/8" 1. 6250 5 .10 51 2.0739 0. 014 40 1 11/ 16" 1. 6875 ... 8.89 3.68 1. 047 0.8 81 12. 51 3.85 6.28 3 .14 1. 307… … 5S 0.083 4.334 14 .75 1. 152 1. 178 1. 135 3.92 6.40 2. 811 1. 249 1. 562… … 10 S 0 .12 0 4.260 14 .25 1. 6 51 1 .17 8 1. 115 5. 61 6 .17 3.96 1. 762 1. 5494...

Modeling of Combustion Systems A Practical Approach 3 docx

... ft-lb/min 1 W = 1 J/sec550 ft-lb/sec 1 W/m-K = 0.5778 Btu/ft-hr-°F6 41. 4 kcal/hr745.7 W© 2006 by Taylor & Francis Group, LLC576 Modeling of Combustion Systems: A Practical Approach TABLE C.2 ... 1 1 2 –2 1 Heat Btu/h W 1 2 –3 Heat capacity, molar Btu/lbmol °F J/mol K 1 1 2 –2 1 Heat capacity, speciﬁc Btu/lbm °F kJ/kg K2 –2 1 Heat ﬂux Btu/ft2W/m 2 1 2 Heat transfer coefﬁcient ... 574 Modeling of Combustion Systems: A Practical Approach TABLE C .1 Common ConversionsTEMPERATURE CONVERSIONS°C = 5/9 (°F – 32) °F = 9/5°C + 32K = °C + 273 .15 °R = °F + 459.67From: Baukal,...

Modeling of Combustion Systems A Practical Approach 4 doc

... 24.22 13 8.78 10 6.73 10 14.7 11 21. 5 0. 016 287 12 6.673.0 6 .11 23. 81 1 41. 47 10 9.42 10 13.2 11 22.6 0. 016 300 11 8.733.5 7 .13 22.79 14 7.56 11 5. 51 1009.6 11 25 .1 0. 016 3 31 102.744.0 8 .14 21. 78 15 2.96 12 0.92 ... 81. 23 10 29.5 11 10.7 0. 016 178 243.02 1. 6 3.26 26.66 11 7.98 85.95 10 26.8 11 12.7 0. 016 196 214 .33 1. 8 3.66 26.26 12 2.22 90 .18 10 24.3 11 14.5 0. 016 213 19 1.852.0 4.07 25.85 12 6.07 94.03 10 22 .1 111 6.2 ... 880.4 11 89.6 0. 017 85 3.8 813 11 6.0 10 1.3 338.73 309.9 879.9 11 89.8 0. 017 86 3.8495 11 7.0 10 2.3 339.37 310 .6 879.3 11 89.9 0. 017 87 3. 818 3 11 8.0 10 3.3 340. 01 311 .3 878.8 11 90 .1 0. 017 87 3.7875 11 9.0 10 4.3...

Modeling of Combustion Systems A Practical Approach 5 docx

... 2.07 1. 97 1. 89 1. 74 1. 60 1. 4395% 3.94 3.09 2.70 2.46 2. 31 2 .19 2 .10 2.03 1. 97 1. 93 1. 77 1. 68 1. 62 1. 57 1. 48 1. 39 1. 2890% 2.76 2.36 2 .14 2.00 1. 91 1.83 1. 78 1. 73 1. 69 1. 66 1. 56 1. 49 1. 45 1. 42 1. 35 ... .12 55 .12 93 .13 31 .13 68 .14 06 .14 43 .14 80 .15 170.4 .15 54 .15 91 .16 28 .16 64 .17 00 .17 36 .17 72 .18 08 .18 44 .18 790.5 .19 15 .19 50 .19 85 .2 019 .2054 .2088 . 212 3 . 215 7 . 219 0 .22240.6 .2258 .22 91 .2324 ... 5.34 5. 31 5.28 5.27 5.25 5.24 5.23 5.20 5 .18 5 .17 5 .17 5 .15 5 .14 5 .13 4 99% 21. 20 18 .00 16 .69 15 .98 15 .52 15 . 21 14.98 14 .80 14 .66 14 .55 14 .20 14 .02 13 . 91 13.84 13 .69 13 .58 13 .4695% 7. 71 6.94...

Modeling of Combustion Systems A Practical Approach 6 pdf

... 10 009 9 11 21 10 01 10 A 12 22 10 10 11 B 13 23 10 11 12 C 14 30 11 00 13 D 15 31 110 1 14 E 16 32 11 10 15 F 17 33 11 11 16 10 20 10 0 10 000© 2006 by Taylor & Francis Group, LLC 610 Modeling of ... to 14 48. Regrouping in twos gives 1, 10, 01, 00, which yields 12 ,10 4.TABLE F.4Base EquivalentsBase 10 16 8 4 2000 0 0 11 1 1 1222 2 10 333 3 11 44 410 10 055 511 10 166 612 11 077 713 11 18 8 10 ... representation. For example, 11 10 .11 2 = 1, 110 .11 02 = 16 .68, because 1 2 = 1 8, 11 02 = 68. Conversion to base 4 follows the same pattern: base 4 = 22,so grouping 11 10 .11 2 in groups of...

Modeling of Combustion Systems A Practical Approach 8 doc

... reactant, the double-headed arrow (↔) meansthat both the forward and reverse reactions occur (typically at different rates),k is an index from 1 to n products, pk is the number of moles of ... Keq in terms of mole fraction rather than concentrationfor gases, then, using the ideal gas law, we obtain(H.3)rr pp 11 22 11 22RR PP++↔++$$rpjjjmkkknR==∑∑↔ 11 PKpprrk=⎡⎣⎤⎦⎡⎣⎤⎦⎡⎣⎤⎦⎡⎣⎤⎦=⎡⎣⎤⎦PPRRP 12 12 12 12 $$ppknjrjmkj==∏∏⎡⎣⎤⎦ 1 1RKPRTyyyyss ... 617 Appendix HEquilibrium PrimerConsider a general reaction:or equivalently(H .1) Here, j is an index from 1 to m reactants, rj refers to the number of moles of the jth reactant, Rj...

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