�
�
�
�
�
�
Therefore,
1
y +
1
5
T
+
− T 5
+
= 5 ln
+
− 1 − 5 ln
+ − 1
Pr
5
Pr 5
(1/Pr) + (y + /5) − 1
= 5 ln
(1/Pr)
Pr y +
= 5 ln 1 +
− Pr
(10.47)
5
For the region 30 ≤ y + with u + = 2.5 ln y + + 5.0, (du + /dy + ) = (2.5/y + ) and
ε H
ε m
1 − (y + /R + )
=
=
− 1
ν
ν
(2.5/y + )
− 1
{[
}
]
Therefore,
T
+
− T
+
= 2.5 ln y
+
− 2.5 ln 30
(10.48)
30
The next question is how to determine heat transfer coefficient from the law
of wall for temperature profile. This is shown in the following.
One assumes the simple velocity and temperature profiles as
( ) 1/7 (
) 1/7
u
y
r
∼ =
= 1 −
U max
R
R
( ) 1/7 (
) 1/7
T − T w
y
r
∼ =
= 1 −
T c − T w
R
R
Therefore,
∫ u2πr dr
u b or V =
= 0.82U max
∫ 2πr dr
∫ R
0 (T w − T c )(1 − (r/R)) 1/7 U max (1 − (r/R)) 1/7 · r dr
T w − T b =
∫ R
0 U max (1 − (r/R)) 1/7 · r dr
15
∼ =
(T w − T c )
18

= 0.833(T w − T c )

212
Analytical Heat Transfer