Untitled - Aerobib - Universidad Politécnica de Madrid

6.15. FLAME PROPAGATION IN HYDROGEN-BROMINE MIXTURES 195
This expression permits the integral of the left-hand side of Eq. (6.273) to be written
as follows
∫ ε1f
(
1 − θ+q4 (ε 4f − ε 4 ) ) dε 1
0
√
∫
Λ 1 (
)
f(θ)
=
1 − θ + q 4 (ε 4f − ε 4 ) √
2q 1 ∫ dθ.
θ 1
0
θ f(θ′ ) dθ ′
(6.280)
By using these expressions in Eq. (6.273), the following equation for Λ is obtained
ε 2 ∫ 1
1f
q 1
2 − Λ f(θ) dθ
θ
√
0
∫
Λ 1 (
M
)
4
f(θ)
−
1 − θ + q 4 (X 4f − X 4 ) √
2q 1 θ 0
M m ∫ dθ = 0.
1
θ f(θ′ ) dθ ′
Here Eq. (6.251) was used in order to eliminate ε 4 . 14
notation will be used
and
I =
∫ 1
θ 0
f(θ) dθ =
∫ 1
(6.281)
For simplicity the following
X 3 X 3/2
2 e −θ 1 − θ
a
θ
dθ, (6.282)
θ 0
θ(X 2 + 0.119X 1 )
∫ 1 (
M
)
4
f(θ)
J = (1 − θ) + q 4 (X 4f + X 4 ) √
θ 0
M m ∫ dθ. (6.283)
1
θ f(θ′ ) dθ ′
If these expressions are used in Eq. (6.281) and the resulting equation is solved, the
following is obtained for √ Λ
√
Λ =
√q 1 ε 2 1f
2I
+ 1 ( ) 2 J
+ 1
8q 1 I 2 √ J
2q 1 I . (6.284)
This value, when substituted in Eq. (6.272), gives the corresponding flame velocity.
Computation of X 2 and X 4
Equation (6.284) shows that when computing Λ it is only necessary to know the values
for I and J. In order to compute I, one must know f(θ) whereas for the calculation
of J one must know f(θ) and X 4 as functions of θ. Finally, Eq. (6.278) shows that
to obtain f(θ), one must know X 1 , X 2 , and X 3 as function of θ. Consequently, the
problem reduces now to the computation of X 1 , X 2 , X 3 , and X 4 as functions of θ.
14 In Eq. (6.281) when changing from ε 4 to X 4 , it has also been assumed that M m is constant through
the flame. Actually its variation is very small.