Untitled - Aerobib - Universidad Politécnica de Madrid

292 CHAPTER 12. DIFFUSION FLAME
Therefore m r i
is given by
∑
m r i = Mi
r ν ijr ′ j , (i = 1, 2, . . . , l r ). (12.14)
j
Likewise if m p i is the mass of species Ap i produced per unit surface and per
unit time one has
m p i = M p i
∑
j
ν ′′
ijr j , (i = l r + 1, l r + 2, . . . , l a ), (12.15)
where l a = l r + l p is the number of active species (reactants plus products) and l p is
the number of products.
Reaction rates r j are unknown “a priori”. Elimination of the same between
equations (12.14) and (12.15) gives l a −r relations between Yi
r and Y p
i which enables
to express all mass fractions of active species as functions of r selected from them.
Thus the variables of the problem reduce in l a − r.
So far no solutions have been obtained for this general system of equations.
In order to make computation available, additional assumptions must be introduced
to simplify considerably the problem. Such simplifications will be performed in the
following paragraph.
12.4 Simplified equations
In this approximate study the number of species is assumed to be three. Namely, fuel
A 1 , oxidizer A 3 and products A 2 . Products include not only species resulting from
reactions, but also diluents of fuel and oxidizer initially mixed with them. 6 Flame surface
divides space into two regions: an interior region at the fuel side and an exterior
one at the oxidizer side. Only A 1 and A 2 species exist within the interior region. In
the exterior one only A 2 and A 3 . Therefore, the composition of the mixture within the
interior region is determined by the value of Y 1 which must satisfy equation (12.2)
ρ(¯v · ∇)Y 1 + ∇ · (ρY 1¯v d1 ) = 0. (12.16)
Similarly in the exterior region one has for Y 3
ρ(¯v · ∇)Y 3 + ∇ · (ρY 3¯v d3 ) = 0. (12.17)
6 A larger number of species could easily be included by distinguishing, for example , between diluents
and products. Such is done, among others, by Zeldovich [17] and Fay [15]. However, this does not represent
any fundamental advantage and computations become more elaborate. A differentiation between products
and diluent can be done once the problem is solved.