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Untitled - Aerobib - Universidad Politécnica de Madrid

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6.9. SOLUTION OF THE FLAME EQUATIONS 157<br />

1) We neglect 1 − θ respect to (1 − θ 0 ) (1 − ε), in agreement with the procedure<br />

used in obtaining (6.85).<br />

2) We substitute (1 − Y ) by its approximation as a function of (1 − θ), given by<br />

Eq. (6.86).<br />

3) The remaining terms of (6.67) are substituted by the corresponding values at the<br />

hot boundary.<br />

Once these approximation are introduced into (6.67) one obtains the following<br />

approximation for this equations, which is valid at the neighborhood of the hot<br />

boundary<br />

1 − θ ≪ (1 − ε)(1 − θ 0 ) :<br />

dε<br />

dθ ≃<br />

Λ (<br />

) n<br />

L (1 − θ) n<br />

1 − θ 0 (1 − θ 0 )(1 + a) 1 − ε . (6.91)<br />

After integration the following approximation 1 − θ vs 1 − ε, is obtained<br />

( n + 1<br />

1 − θ ≃<br />

2<br />

) 1<br />

1 − θ 0<br />

Λ<br />

n + 1 ( (1 − θ 0 )(1 + a)<br />

L<br />

which when taken into (6.75) and integrated, gives for J<br />

J = n + 1 ( n + 1<br />

n + 3 2<br />

) 1<br />

1 − θ 0<br />

Λ<br />

n + 1 ( (1 − θ 0 )(1 + a)<br />

L<br />

) n<br />

n + 1 (1 − ε)<br />

2<br />

n + 1 , (6.92)<br />

) n<br />

n + 1 . (6.93)<br />

By taking this value of J and the one for I given by Eq. (6.74) into Eq. (6.73), the<br />

following equation results for the calculation of Λ<br />

1 − θ 0<br />

2<br />

− n + 1 ( n + 1<br />

n + 3 (1 − θ 0)<br />

2<br />

) 1<br />

n + 1 ( 1 + a<br />

L<br />

) n<br />

n + 1 Λ<br />

− 1<br />

n + 1 = ΛI, (6.94)<br />

which may be readily solved, either with a graphical method or by iteration, using for<br />

this last case as a first approximation the one given by (6.90) and as the correcting<br />

term for successive iterations the right hand si<strong>de</strong> of Eq. (6.94).<br />

The result reached with (6.94) are represented in Figs. 6.5 and 6.6 where the<br />

corresponding curves are named Kármán 2.<br />

The approximation for J may still be improved by writing it in the form<br />

J =<br />

∫ 1<br />

θ<br />

(1 − θ) dε dθ, (6.95)<br />

dθ<br />

and using for dε an approximation of (6.67) <strong>de</strong>rived as follows<br />

dθ<br />

(1 − ε) dε<br />

dθ ≃<br />

Λ (<br />

L<br />

1 − θ 0 (1 − θ 0 )(1 + a)<br />

) n<br />

λ<br />

λ f<br />

θ δ−n e −θ a<br />

1 − θ<br />

θ (1 − θ) n , (6.96)

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