Etude de la combustion de gaz de synthèse issus d'un processus de ...

Chapter 2
Where r b is the radius of the inner boundary of the unburned gas of density ρ u .
Combining the above relations results in
−1
dp Ab
ρu
⎛dx
⎞
= ⎜ ⎟ S
dt V ρi
⎝dp
⎠
u
(2.56)
Isentropic compression of the unburned gases gives
ρ ⎛
u p ⎞
= ⎜ ⎟
ρi
⎝pi
⎠
−1 γ u
(2.57)
Or, alternatively,
tel-00623090, version 1 - 13 Sep 2011
( γu
−1)
γu
⎛ p ⎞
Tu
= Ti
⎜ ⎟ p
i
⎝
⎠
.
(2.58)
Where γ u = c pu /c vu , is the isentropic exponent of the unburned mixture. Surface area
and volume of the flame are related to its radius as A b = 4πr 2 and
Vb
4 3
= π r ,
3
respectively. Realizing that m u = ρ i V (1 − x), and p i = ρ i R u T i , the volume of the
unburned mixture can be written as
Inserting
Vb
V=V u +V b yields:
V
mRT
p
T
( 1 )
= u u u i u
u
X V
p
= p T
− (2.59)
i
4 3
= πR
, R is the effective radius of the vessel and the above relations into
3
pi
Tu
rb
R ⎡
⎤
= ⎢1−( 1−x)
⎥
⎣ pT
i ⎦
1/ 3
(2.60)
The area-to-volume ratio in Eq. (2.56) now becomes:
A 3 ⎡
b
p 1 ( 1
i
T ⎤
u
= ⎢ − −x)
⎥
V R ⎣ p Ti
⎦
2/3
(2.61)
Inserting Eqs. (2.57), (2.58) and (2.61) into Eq.(2.56) we finally obtain
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