�
�
�
�
�
�
� � �
df
u
"
f =
=
= velocity profile
(7.14)
dη
U ∞
d 2 f
d(u/U ∞ )
""
f =
=
= velocity gradient
(7.15)
dη 2
dη
Boundary conditions:
at
u
"
y = 0, u = v = 0, ⇒ η = 0, f =
= 0; v = 0, ⇒ f = 0 ⇒ f (0) = 0 (7.16)
U ∞
at y = ∞, u = U ∞ , ⇒ f " (∞) = 1
1
f
"""
""
= − ff
(7.17)
2
Equation 7.17 may be solved numerically by expressing f (η) in a power
series (1908 Blasius Series Expansion) with the above-mentioned BCs as
α 2 η 5
α 3 η 8
α 4 η 11
αη 2
1
11
375
f = f i =
−
+
−
+ · · · for small η
(7.18)
2!
2 5!
4 8!
8 11!
∞ �
�
�
�
�
� 2
η − β
dη dη . . . for large η
(7.19)
f = f o = η − β − γ
exp −
2
η
where α = 0.332, β = 1.73, and γ = 0.231. Then u and v can be determined.
Equation 7.17 can also be solved by numerical integration as
""
d f
1
2
= −
f dη
""
f
η
1
2
ln f
""
= −
f dη + C
0
� η
f
""
− 0 1/2(f dη)
= e
· C 1
η
f
"
= df
"
=
(
)
� η
e
− 0 1/2( f dη) dη · C 1 + C 2
0
� η
� ∞
"
−
"
where C 2 = 0 at η = 0, f = 0; C 1 = (1/
e 0 1/2( f dη) dη) at η = ∞, f = 1.
0
η η
f = c 1
e
− 1/2( f dη) dη dη + c 3
0 0
where C 3 = 0 at η = 0, f = 0.
144
Analytical Heat Transfer
�
�
�
�
�
� � �
df
u
"
f =
=
= velocity profile
(7.14)
dη
U ∞
d 2 f
d(u/U ∞ )
""
f =
=
= velocity gradient
(7.15)
dη 2
dη
Boundary conditions:
at
u
"
y = 0, u = v = 0, ⇒ η = 0, f =
= 0; v = 0, ⇒ f = 0 ⇒ f (0) = 0 (7.16)
U ∞
at y = ∞, u = U ∞ , ⇒ f " (∞) = 1
1
f
"""
""
= − ff
(7.17)
2
Equation 7.17 may be solved numerically by expressing f (η) in a power
series (1908 Blasius Series Expansion) with the above-mentioned BCs as
α 2 η 5
α 3 η 8
α 4 η 11
αη 2
1
11
375
f = f i =
−
+
−
+ · · · for small η
(7.18)
2!
2 5!
4 8!
8 11!
∞ �
�
�
�
�
� 2
η − β
dη dη . . . for large η
(7.19)
f = f o = η − β − γ
exp −
2
η
where α = 0.332, β = 1.73, and γ = 0.231. Then u and v can be determined.
Equation 7.17 can also be solved by numerical integration as
""
d f
1
2
= −
f dη
""
f
η
1
2
ln f
""
= −
f dη + C
0
� η
f
""
− 0 1/2(f dη)
= e
· C 1
η
f
"
= df
"
=
(
)
� η
e
− 0 1/2( f dη) dη · C 1 + C 2
0
� η
� ∞
"
−
"
where C 2 = 0 at η = 0, f = 0; C 1 = (1/
e 0 1/2( f dη) dη) at η = ∞, f = 1.
0
η η
f = c 1
e
− 1/2( f dη) dη dη + c 3
0 0
where C 3 = 0 at η = 0, f = 0.
144
Analytical Heat Transfer
