�
�
�
�
�
Follow a similar procedure as in the previous case; one can obtain three-region
temperature profile.
""
where c = (−q /ρC p ).
w
Therefore,
""
∂T
q
− w
(α + ε H )
=
∂y
ρC p
""
∂T
− q /ρC p
w
=
∂y
α + ε H
y
""
∗
1
u
q
− w
ρC p
T − T w =
dy u ∗
υ((1/Pr) + (ε H /υ))
0
y
""
1
((1/Pr) + (ε H /υ))
dy
+
q
T w − T =
w
ρC p u ∗
(10.51)
0
where ε H � ε m .
And from Equation 10.20,
ε m
1 − (y + /δ + )
=
− 1
υ
(du + /dy + )
Follow the same procedure as outlined before:
+
+
+
0 < y < 5, u = y
T
+
= Pr y
+
(10.52)
+
5 < y
+
≤ 30, u = 5 ln y
+
− 3.05
Pr y +
T
+
− T
+
= 5 ln 1 +
− Pr
(10.53)
5
5
+
+
30 ≤ y , u = 2.5 ln y
+
+ 5.0
T
+
− T
+
= 2.5 ln y
+
− 2.5 ln 30
(10.54)
30
214
Analytical Heat Transfer
�
�
�
�
Follow a similar procedure as in the previous case; one can obtain three-region
temperature profile.
""
where c = (−q /ρC p ).
w
Therefore,
""
∂T
q
− w
(α + ε H )
=
∂y
ρC p
""
∂T
− q /ρC p
w
=
∂y
α + ε H
y
""
∗
1
u
q
− w
ρC p
T − T w =
dy u ∗
υ((1/Pr) + (ε H /υ))
0
y
""
1
((1/Pr) + (ε H /υ))
dy
+
q
T w − T =
w
ρC p u ∗
(10.51)
0
where ε H � ε m .
And from Equation 10.20,
ε m
1 − (y + /δ + )
=
− 1
υ
(du + /dy + )
Follow the same procedure as outlined before:
+
+
+
0 < y < 5, u = y
T
+
= Pr y
+
(10.52)
+
5 < y
+
≤ 30, u = 5 ln y
+
− 3.05
Pr y +
T
+
− T
+
= 5 ln 1 +
− Pr
(10.53)
5
5
+
+
30 ≤ y , u = 2.5 ln y
+
+ 5.0
T
+
− T
+
= 2.5 ln y
+
− 2.5 ln 30
(10.54)
30
214
Analytical Heat Transfer