25 September 2023

stoichiometric fuel air ratio

Rate this post

stoichiometric fuel air ratio

Stoichiometric fuel air ratio :

All the available oxygen is used to burn the fuel completely or atleast to the best possible value. For Unit mass of fuel how much air theoretically required to burn all the combustible content of fuel. This ratio is called the stoichiometric air-fuel ratio.

Calculation of stoichiometric air fuel ratio for propane gas[C3H8] :

Propane           +       Oxygen                                      Carbon di-oxide          +                 Water
C3H8 +                      5O2                           =               3CO2                            +                 4H2O
3×12+1×8                  5x16x2                                       3(12+2×16)                                    4(2×1+16)                                                                                                                                                                                                                            44                               160                                             132                                                  72

It means 44gm propane required 160gm of oxygen, which is equivalent to 695.65(160/0.23) gm of Air.

Therefore 1 gm of propane required 15.81(695.65/44)gm of Air. Hence ration Stoichiometric air fuel ratio will be 15.81:1.

Leave a Reply

Your email address will not be published. Required fields are marked *