TF06_source_03.xmcd
TF06
- Source - Exercice 3Source de chaleur
0 e x
h
2h
1λ
T
1T
2q 34.9 10× 4 W m3
×
:= λ 17.5 W
m K×
×
:= h1 34.5 W
m2×K
×
:= h2 11.6 W
m2×K
× :=
e:=10×cm T1:= 200×°C T2:= 30°C
∇2T q
æç
λè ö÷ ø
-
= q débit de chaleur par unité de volume
d2T dx2
q
æç
λè ö÷ ø
-
= dT
dx q
æç
λè ö÷ ø
- ×x+C1
= T x( ) q
2×λ
æç è ö÷ ø
- ×x2+C1 x× +C2
=
Conditions aux limites
· x=0 h1 T1 T×
(
- ( )0)
= -λ×T'( )0 h1 T1 C2×(
-)
= -λ×C1 C2 T1 λ×C1 + h1=
· x=e -λ×T' e( ) = h2 T e×
(
( )-T2)
q e× -λ×C1 h2 q 2×λæç è ö÷ ø
- ×e2+C1 e× +C2-T2
éê ë ùú û
×
=
C1
T2 T1- q e× 1 h2
e 2×λ
æ
+ç è
ö ÷
×
ø
+
e λ + h1 λ
+h2
:= C2 T1
λ×C1 + h1
:= C1 1388.7 °C
×m
= C2= 904.4×°C
q 2×λ
æç è ö÷ ø
- -9971.429 °C m2
×
= T x( ) q
2×λ
æç è ö÷ ø
- ×x2+C1 x× +C2 :=
T0:= T(0m) T0= 904.4×°C T x( ) = -9971.4´x2+1388.7´x+904.4 Te:=T e( ) Te= 943.6×°C
T' x( ) q
æç
λè ö÷ ø
- ×x+C1
:= On a un maximum pour : xm λ
q×C1
:= T xm
( )
=952.8×°Cxm= 7.0×cm
0 1 2 3 4 5 6 7 8 9 10
900 910 920 930 940 950 960
Profil de température
distance de [1] vers [2] (cm)
température (°C)
Tm
x 0 1 2 3 4 5 6 7 8 9 10
×cm
= T x( )
904 917 928 937 944 949 952 953 952 949 944
×°C
=
MH 1/1 01/04/2011