�
�
�
� �
�
�
�
� �
�
�
�
�
From the energy integral, we obtain
�
δ � T
� ( )
∂T �
3 1
d
3 y
−k
= k (T w − T ∞ )
=
ρC p
∂y
2 δ T
dx
2 δ
y =0
0
( ) 3
� 3
1 y
3 y
1 y
−
U ∞ (T w − T ∞ ) 1 −
+
dy
2 δ
2 δ T
2 δ T
Then,
� � 3
� �
3 1
δ T 1
δ T
1
d δ T
k
= ρC p U ∞ 3δ
−
2 δ T
δ 10
δ
70 dx δ
� � 2
� � 4
δ T
1
δ T
1
dδ
+3
−
δ
20
δ
280 dx
Let

� � 3 � � 4

δ T
δ T
δ T
r =
< 1,
∼
∼ 0
δ
δ
δ
We obtain

dδ
α

2 2 dr
3
2r δ
+ r δ
= 10
dx
dx
U ∞
From above

280 υx
dδ
140 υ

2
δ =
, δ
=
13 U ∞
dx
13 U ∞
We obtain
2 280 υx dr
3 140 υ
α
2r
+ r
= 10
13 U ∞ dx
13 U ∞
U ∞
Then,
dr
13
2
3
4r x
+ r =
dx
14Pr

3

Let s = r
( )
1 d
1 ds
2 dr
3
r
=
r =
dx
3 dx
3 dx

4 ds
13

s + x
=
3 dx
14Pr
155
External Forced Convection