12combustion

# 12combustion - Combustion combustion of dodecane a parafin...

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Combustion define some units combustion of dodecane a parafin of type C n H 2n kN := 10 3 N kPa := 10 3 Pa (represents diesel fuel) in stochiometric proportions: 6 3 MPa := 10 Pa kJ := 10 J yH 2 O + heat 3 C 12 H 26 + xO 2 = + zC O 2 kmol := overkill in this case but general method represented by solution of simultaneous equations from elements involved initial values for given x := 1 y := 1 z := 1 Given construct element: C z = 1 2 O x2 = y + 2 z 26 = 2 y H x y := Find x y , z) x 18.5 ( , y = 13 z z 12 so combustion equation is 37 C 12 H 26 + 2 = O + O 2 + heat C 12 H 26 + 2 O 2 = O + 12 C O 2 2 13 H 2 + heat O2 comes with nitrogen: 21% by volume and 23.3% by weight in air (79% N2 by volume and 76.7% N2 by weight - ~ 1% Ar lumped with N2) so . .. need 79/21 atoms (volume) N2 for each O2 79 18.5 = 69.6 21 and combustion is . .. this is on a stochiometric basis (mole basis- i.e. 1 mole of C 12 H 26 + 18.5 O 2 + = O + O 2 + C12H26 combines with 18.5 69.6 N 2 13 H 2 12 C 69.6 N 2 + heat moles of O2 etc.) or volume basis to convert to weight use molecular kg kg kg weights mw_O2 := 32 mw_C12_H26 := (144 + 26) mw_N2 := 28 kmol kmol kmol kg mw_H2_O := (2 + 16 ) kmol mw_C_O2 := (12 + 32) kg kmol 1kmolC 12 H 26 + O 2 + N 2 = H 2 O + O 2 + 18.5 kmol 69.6 kmol 13 kmol 12kmol C 69.6kmol N 2 + LHV 1kmol mw_C12_H26 ... = 13 mw_H2_O kmol H 2 O ... + LHV this is divided by 170 - the 170 C 12 H 26 170 molecular weight of C12H26 to 12kmol mw_C_O2 18.5 kmol.mw_O2 express on a per 1 kg fuel basis + O 2 ... + C O 2 ... 170 170 69.6kmol mw_N2 + 69.6 kmol mw_N2 N 2 + N 2 ) mw_C12_H26 = 170kg 170 170 (1kmol 10/11/2004 1

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for symbolic calculation result is . .. combustion of C12H26 by weight . .. 1kg C 12 3.48 kg 11.46 kg 1.38 kg = 11.46 kg weight of air : weight of fuel = air-fuel ratio air_fuel_ratio := 3.48 + 11.46 air_fuel_ratio H 26 O 2 N 2 H 2 O 3.11 k C O 2 N 2 + heat + + + + = 14.94 In order to insure complete combustion,air is usually supplied in excess, see example below. Products would include air i.e. O2 and N2 to analyze combustion process use first law . .. steady state, steady flow process . .. m_dot = flow_rate (5.46) 2 h e 2 gz e V i V e d d + d dt W c_v + (5.47) Q c_v m_dot i h i m_dot e + + + + 2 i n n = 2 t n n work, KE and PE = 0 . .
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12combustion - Combustion combustion of dodecane a parafin...

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