DM4 Mécanique – Diffusion thermique

R0= 1,5·1011 m
vTG MSR0
30 km ·s−1
RT= 6,4·106mO
vbvr
vr
vbvT−→ vb=−→ vr+−→ vT
mb
~vr=vr~ex
b

Em
v r mb
MTG
A dmin =OA
~vA
−→
OA
vrb dmin vA=||~vA||
dmin G MTvrb
b
bmax RTG MTvr
vi
vi=qv2
r+v2
l
vlG MTRT
vl
vi
rb= 80 m
ρb= 2,5·103kg ·m−3
vi= 20 km ·s−1
4,18 ·109J
103106
rb= 80 m ρb= 2,5e3kg ·m−3
(~ex, ~ey, ~ez)
(x, y, z)
~g =−g~ezg= 9,0 m ·s−2
T0
ρ(z)P(z)
z ρ0P0z= 0
P(z)ρ(z)T0R
Mair
ρ(z)
ρ(z) = ρ0exp −z
Ha
HaMair T0R g

ρ0
HaP0= 1,0·105Pa T0= 290 K
vi= 20 km ·s−1
Ft=1
2Cρπr2
bv2C= 2
ρ v
ρ0
θ ~ez
y= 0
a=dv/dt C ρbrbρ(z)
z v(t)
dv/dz a =dv/dt
vz=dz/dt vz=−vcos θ dv/dz C ρb
rbρ(z)v θ
limz→+∞v(z) = vi
z v =f(z)
θ= 45
z= 0
1·104K
D
hv'8 MJ ·kg−1

Ecoh
D
Y Y
G0δ=3D
16
Eg
Egρc
D g
D
Y= 3.107ρc= 2,7·103kg ·m−3
D
[2 km; 8 km]
MT= 6,0·1024 kg
MS= 2,0·1030 kg
G= 6,67 ·10−11 N·m2·kg−2
R= 8,31 J ·K−1·mol−1
Mair = 29 g ·mol−1
e S λV
Ox
x= 0−T(x=
0−) = T0x=e−T(x=e−) = Te
Tint

T0=Tint Te=Text P
P=Gc(Tint −Text) = Tint −Text
Rc
GcRc
P Tint −Text Gc
S e λV
T0Tint
hint(Tint −T0)
x=e
hint(Te−Text)
Gconv,int Gconv,ext Gc
Tint = 293 K Text = 308 K
ϕ
T σ
Pr,int
x= 0
Pr,int =σS(T4
int −T4 0)
Pr,int
Tint Text 
Tint −T0
Tint 
(T0
Tint
)4= (1 −Tint −T0
Tint
)4
Pr,int =Gr,int(Tint −T0)
Gr,int σ Tint S
x=e
Pr,ext
Pr,ext Text Te
Gr,ext