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Untitled - Aerobib - Universidad Politécnica de Madrid

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192 CHAPTER 6. LAMINAR FLAMES<br />

Selection is ma<strong>de</strong> of the two equations corresponding to X 1 and X 3 because that<br />

corresponding to X 2 presents some difficulties, as will be seen later on. Thus<br />

and<br />

dX 1<br />

dx<br />

dX 3<br />

dx<br />

= RT 4∑<br />

(<br />

)<br />

f<br />

p<br />

θ 1 ε j ε 1<br />

X 1 − X j , (6.261)<br />

D<br />

j=1 1j M j M 1<br />

= RT 4∑<br />

(<br />

)<br />

f<br />

p<br />

θ 1 ε j ε 3<br />

X 3 − X j . (6.262)<br />

D<br />

j=1 3j M j M 3<br />

Hence the system of the flame equations has been reduced to the five relations<br />

given in Eqs. (6.229), (6.246), (6.251), (6.254) and (6.255) and to the four<br />

differential equations (6.250), (6.256), (6.261) and (6.262) for the nine unknowns ε i<br />

(i = 1, 2, 3, 4), X i (i = 1, 2, 3, 4) and θ. If, furthermore, the differential equations<br />

(6.250), (6.261) and (6.262) are divi<strong>de</strong>d by Eq. (6.256), the coordinate x is eliminated<br />

and θ is introduced as in<strong>de</strong>pen<strong>de</strong>nt variable. The results are<br />

dε 1<br />

dθ = λ ( ) 3/2<br />

√<br />

p<br />

m 2 2M 1 θ −3/2 k1<br />

k 2<br />

c p RT f k 5<br />

(<br />

X 3 X 3/2<br />

2 X 2 + k )<br />

4<br />

X 1 − 1<br />

k 3<br />

·<br />

θ − 1 + q 1 (ε 1f − ε 1 ) − q 4 (ε 4f − ε 4 ) . (6.263)<br />

dX 1<br />

dθ<br />

dX 3<br />

dθ<br />

(<br />

)<br />

4∑ 1 ε j ε 1<br />

= λ<br />

X 1 − X j<br />

RT f<br />

mc p p θ j=1D 1j M j M 1<br />

θ − 1 + q 1 (ε 1f − ε 1 ) − q 4 (ε 4f − ε 4 ) , (6.264)<br />

(<br />

)<br />

4∑ 1 ε j ε 3<br />

= λ<br />

X 3 − X j<br />

RT f<br />

mc p p θ j=1D 3j M j M 3<br />

θ − 1 + q 1 (ε 1f − ε 1 ) − q 4 (ε 4f − ε 4 ) . (6.265)<br />

Solution of the reaction equations<br />

Von Kármán and Penner [40] give the following expressions for k 2<br />

√<br />

k1<br />

k 5<br />

and k 4<br />

k 3<br />

and<br />

where<br />

k 2<br />

√<br />

k1<br />

k 5<br />

= 0.8 × 10 14 e −θ 1/θ , (6.266)<br />

k 4<br />

= 1 = 0.119, (6.267)<br />

k 3 8.4<br />

θ a =<br />

40 200<br />

RT f<br />

(6.268)

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