ABSTRACT
Dividing Eq. (5) by the amount of air mass ma, then we have
α λ
= ⎡⎣ ⎤⎦1− 3 2 1 32 22 ( / )
( )− . ( ) , m
Q 0+ 5 T T−f s
(6)
where λ is the excess air coefficient defined as λ = (ma/mf)/(ma/mf)s, (ma/mf) is air-fuel ratio and the subscripts a, f, and s, respectively, denotes air, fuel and the stoichiometric condition.
