COMBUSTION  PROBLEM  NO  29-

Ocatne  is burnt with  10% excess  air. Calculate-

A - air/fuel  ratio by weight

b- air /fuel ratio by volume

c- weight  of dry exhaust gas formed per unit of fuel.

d- mole  of oxygen  in the exhaust  gas per unit weight of fuel.

e-  mole of water  vapour  in  exhaust  gas per unit weight  of fuel.

 F - volume  of exhaust  gas at 1 atmosphere   and 260 oc  per unit  weight of fuel.

The  specific  gravity  of octane  may be  taken as 0.7.

SOLUTION -

 Basis  -1 kg   mole  of octane  burnt-

Reaction  -  C8H18 +25/2  O2  =  8CO2 +9H2O

PART  a- Theoretical   oxygen  demand  =  12.5  kg mole

Oxygen  supplied  by  10% excess  air  = 12.5*110/100  = 13.75  kg- moles =13.75 *32   =440  kg

Nitrogen  supplied  by air = 13.75*79/21  =  51.73  kg - mole   =51.73 *28 = 1448.4 kg

Amount   of air supplied      =13.75 +51.73 = 65.48  kg - mole =65.48 *29 kg

 Molecular  wt  of air  =29 kg/kg mole

 Molecular  wt  of nitrogen =  28 kg/kg-mole

Molecular  wt  of oxygen = 32  kg/kg-mole

 Molecular wt of   octane ( fuel) = C8H18  = 8*12 +1*18  =96 +18 = 114 kg/kg mole

Weight  of air/weight  of fuel ( octane  )  = 1888.4/114  =  16.56

Part   - b  -  specific   gravity  of octane =0.7

 Density   of octane          =0.7  g /cc = 0.7*1000  =700 kg /m³

Volume   of fuel  =  114/700  =0.163 m³

 Assuming ideal gas   law  , volume  of air  at N.T.P

    =  65.48 *22.4 = 1466.75 NM³

Volume of air/volume  of fuel  = 1466.75/0.163 = 8998.5

Part   -c  

 Excess  O₂  =  supplied   O₂  --  used  O2  =  13.75 -12.50 =1.25 kg mole

  Dry    flue  gas analysis-

 Constituents                   amount   kg mole                           molecular wt                              amout   ,kg  

 CO₂                                       8.0                                                           44                      44*8 =352

 O₂                                          1.25                                                          32                      1.25*32 =40

 N₂                                          51.73                                                       28                       28*51.73=1448.4

 TOTAL                                    60.98                                                                                         1840.40

 WEIGHT  OF DRY  EXHAUST  GAS/WEIGHT  OF FUEL    =1840.4/114 = 16.4

Part  - d  -

 Mole  of O₂  in the  exhaust   gas/weight  of fuel  =  1.25/114  = 0.011

Part    - e -

  Mole   of water  vapour  in the exhaust  gas/ weight  of fuel  =  9.0/114   = 0.079

Part - f -

 Moles   of  exhaust  gas ( wet)  = 60.98 +9.0  = 69.98

  By ideal   gas law-

 Volume  at  260^oc  and 1 atmosphere

 V = n.RT/P    = 69.98 *0.082 *(260 +273)/1.0 = 3060.8  M₃

VOLUME   OF EXHAUST GAS ( WET)/WEIGHT  OF FUEL  =  3060.8/114 =  26.85