� �
�
�
�
�
�
�
�
�
�
�
Performing
∂θ
∂θ ∂η
C 2
"
=
= θ √
∂y
∂η ∂y
x
∂ 2 θ
∂ ∂θ
∂
C 2
∂
C 2
∂η
C 2
"
"
2 ""
=
=
√ θ =
√ θ ·
=
θ
∂y 2
∂y ∂y
∂y
x
∂η
x
∂y
x
∂θ
∂θ ∂η
−η
"
=
·
= θ ·
∂x
∂η ∂x
2x

Inserting this into the energy equation we obtain

1

""
+
"
θ
Pr f θ = 0
(7.23)
2

Boundary conditions:

θ (0) = 0
(7.24)
θ (∞) = 1
Equation 7.23 can be solved as
θ ""
Pr
dη = −
f dη
θ "
2
η
Pr
"
ln θ = −
f dη + C
2
0
Pr
"
−
�
0
η
f dη
θ = e
2
· C 1
η
Pr f dη
θ = C 1 e
−
�
0
η
2
dη + C 2
0
at η = 0, θ(0) = 0, C 2 = 0

at η = 0, θ(∞) = 1,

1

C 1 = � η
� η
− 0 Pr/2( f dη) dη
0 e
Therefore,
� η
� η − 0 Pr/2( f dη) dη
θ =
0 e
� η
(7.25)
� ∞ − 0 Pr/2( f dη) dη
e
0 � η
� η Pr
"
θ = θ (0) exp −
f dη dη
(7.26)
0
0 2
147
External Forced Convection