Untitled - Aerobib - Universidad Politécnica de Madrid

6.6. EXAMPLE 141
generally impossible to obtain explicit solutions for the equations. Hence, it is necessary
to resort to cumbersome numerical integrations or to analytical methods which
will be discussed further on. However, in order to illustrate the nature of the solutions
and the way in which the eigenvalue appears determined, the present paragraph offers
an example in which through an adequate simplification of both transport coefficients
and chemical reaction velocity, it is possible to obtain an explicit solution of the problem
giving correct qualitative predictions. A possible objection to this solution could
be that the results obtained depend on the ignition temperature of the mixture, which,
as before said, does not exist. But this is due to the fact that the simplification assumed
for the reaction velocity is excessive and consequently the inconvenient will vanish for
more realistic cases, as will be proven in the following.
For this example the following assumptions will be adopted:
1) The coefficient of thermal conductivity λ and the product ρD are constant.
2) The reaction is of first-order and its velocity is independent from temperature, in
which case it has the form 1 w = ρ 0 k(1 − Y ). (6.23)
It is precisely this simplified form of the reaction velocity and, in special, the fact
that in it the activation energy is assumed to be zero, an indispensable condition
for the integrability of the system, which makes the solution depends on the
ignition temperature, as aforesaid.
When expression (6.23) is taken into (6.2), it gives
The diffusion equation subsists in the form (6.3)
m dε
dx = ρ 0k(1 − Y ). (6.24)
ρD dY
dx
= m(Y − ε). (6.25)
Finally, when the dimensionless temperature θ = T/T f is introduced, the energy
equation (6.5.a) may be written
where is θ 0 = T 0 /T f .
λ dθ
mc p dx = θ − 1 + (1 − θ 0)(1 − ε), (6.26)
1 See §8 of Chap. 1.