Di usion thermique Les points du cours à connaître

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~
jQ
δQ =h(Ts−Ta)dSdt
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U=RRR ud3τ
dt dU = +δW.+Pthdt δW
Pth ~ jth
~
jth =−κ.−−→ gradT T V dU =
RRR c∂T
∂t dtd3τx
x+dx
x x + dx
d S d S
1 2
2 2
1 2
d S
3
2
•
•
•
•
κ ` S
Rth =`
κ.S
T1T2
Rth T2−T1=Rth.Φth Φth
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τ
L
1
τ
∼D
L2
cVµ κ
∂T
∂t =Dth.∆T Dth =κ µ.cV
cVµ κ
κ.∆T+µ.pm=µ.cV.∂T
∂t
x
cVµ κ
∂T (x,t)
∂t =Dth.∂2T(x,t) ∂x2Dth =κ µ.cV
R µ c
κ t r
r∈[r;r+dr]
∆T(r) = 1
r2.sin θ∂ ∂r r2.sin θ. ∂T ∂r  =1 r2h2.r. ∂T ∂r +r2∂2T ∂r2i
O a
b a < b κ c µ
T1r=a T2r=b
T1> T2
T
M t
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T(r)M
Pta b
Rt
κ. ∂
∂r r2.∂T ∂r =µ.c.r2∂T ∂t
T(r) = a.b
r
T1−T2
b−a−a.T1−b.T2 b−a
Pt= 4π.κ a.b
b−a(T1−T2)
Rt=1
4π.κ
b−a
a.b
r1Rl
r2κ
r
` r ∈[r1;r2]φth
I
2.π.rκ ∂T
∂r +Rl.I2= 0
d= 10cm κ = 40mW.m−1.K−1
x= 0 θ1= 100◦C x =d
θa= 25◦C θ2= 30◦C
h
h=κ
d
θ1−θ2
θ2−θa= 5,6W.m−2.K−1
κ= 4,0.10−2W.m−1.K−11 2
d= 2,0cm 1
θa= 20◦C2θ2= 50◦C
1θ1= 30◦C
h
h= 4,0W.m−2.K−1
Tint Text < Tint
Smκmem
Sv∼Smκvκmevem
Rth
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