Modeling of Combustion Systems A Practical Approach 1 pot
Modeling of Combustion Systems A Practical Approach 1 pot
... 17 .19 5 1. 1 3484 13 01 Gas 5.0 15 .0 Ethane C 2 H 6 22323 15 .899 1. 3 3540 968 11 66 Gas 3.0 12 .5 Propane C 3 H 8 216 69 15 .246 1. 3 3573 8 71 Gas 2 .1 10 .1 n-Butane C 4 H 10 213 21 14.984 1. 2 3583 7 61 ... (Dry) Waste Gases Natural Gas LPG Cracked Gas Coking Gas Reforming Gas FCC Gas Refinery Gas Sample 1 Refinery Gas Sample 2 PSA Gas Flexicoking Gas Tulsa Alaska N...
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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 (x k ), 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...
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Modeling of Combustion Systems A Practical Approach 13 potx
... ,, ,11 1 1 αΔ=+ () − () QKZk Z Z kz H 11 1 1 == = Q 1 K k fHkz fCTT aw ca awpg ref,, , ,11 11 1ααΔ−=+ () − () TT f f Hkz C ref aw aw ca pg 1 1 1 1 1 −= + − , , , , α α Δ Q 21 fHQ fCTT aw ca awpg ref,, , ,12 112 1 1 αΔ−=+ () − () © ... ref w NO NO, ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = + + − =α α ββα 0 01 1 11 0 β δ ββ 1 10 11 =− δ α ββα β 0 01 1 1 1 = + + − w ref w re...
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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 3 2 .1. 2.4 Flat-Flame Premix Burners 11 4 2 .1. 2.5 Flashback 11 5 2 .1. 2.6 Use of Secondary Fuel and Air 11 5 2 .1. 2.7 Round Combination Bu...
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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 &q...
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Modeling of Combustion Systems A Practical Approach 3 docx
... ft-lb/min 1 W = 1 J/sec 550 ft-lb/sec 1 W/m-K = 0.5778 Btu/ft-hr-°F 6 41. 4 kcal/hr 745.7 W © 2006 by Taylor & Francis Group, LLC 576 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 K 2 –2 1 Heat flux Btu/ft 2 W/m 2 1...
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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 6.67 3.0 6 .11 23. 81 1 41. 47 10 9.42 10 13.2 11 22.6 0. 016 300 11 8.73 3.5 7 .13 22.79 14 7.56 11 5. 51 1009.6 11 25 .1 0. 016 3 31 102.74 4.0 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...
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Modeling of Combustion Systems A Practical Approach 5 docx
... 2.07 1. 97 1. 89 1. 74 1. 60 1. 43 95% 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. 28 90% 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 17 0.4 .15 54 .15 91 .16 28 .16 64 .17 00 .17 36 .17 72 .18 08 .18 44 .18 79 0.5 .19...
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Modeling of Combustion Systems A Practical Approach 6 pdf
... 10 00 9 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 4 8 . Regrouping in twos gives 1, 10, 01, 00, which yields 12 ,10 4 . TABLE F.4 Base Equivalents Base 10 16 8 4 2 000 0 0 11 1 1 1 222 2 10...
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Modeling of Combustion Systems A Practical Approach 8 doc
... reactant, the double-headed arrow (↔) means that both the forward and reverse reactions occur (typically at different rates), k is an index from 1 to n products, p k is the number of moles of ... K eq in terms of mole fraction rather than concentration for gases, then, using the ideal gas law, we obtain (H.3) rr pp 11 22 11 22 RR PP++↔++$$ rp jj j m kk k n R == ∑∑ ↔ 11 P K pp rr...
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