�
13
3
−3/4 +
s = r = C 1 x
14 Pr
13
at x = 0, C 1 = 0 , r 3 = 14Pr
(
) /
δ T
13
−1/3
r =
=
1 3 ∼ 0.975(Pr )
∼ Pr −1/3
=
δ
14Pr
4.64
δ T = δ Pr −1/3 = √
· Pr −1/3
Re x
−k (∂T /∂y ) y =0
3 k
h =
=
T w − T ∞
2 δ T
3
k
3
k
=
=
√
2 δPr −1/3
2 x(4.64/ Re x )Pr −1/3
For this typical case,
hx
Nu x = k
= 0.323 Re x
1/2 Pr 1/3
(7.38)
From Figure 7.7,
at x = x o , δ T = 0, r = 0
1
13
13 3/4
0 = C 1
+
, C 1 = −
x o
3/4
14Pr
14Pr
x o
Therefore,
( ) 3/4
� 1/3
δ T
x o
r =
= 0.975Pr −1/3 1 −
δ
x
156
Analytical Heat Transfer
U ∞ ,T ∞
T w
δ
δ T
0
Apply heat
x
Apply heat
x
x 0
δ
δ T
U ∞ ,T ∞
0
FIGURE 7.7
Integral approximation method.
13
3
−3/4 +
s = r = C 1 x
14 Pr
13
at x = 0, C 1 = 0 , r 3 = 14Pr
(
) /
δ T
13
−1/3
r =
=
1 3 ∼ 0.975(Pr )
∼ Pr −1/3
=
δ
14Pr
4.64
δ T = δ Pr −1/3 = √
· Pr −1/3
Re x
−k (∂T /∂y ) y =0
3 k
h =
=
T w − T ∞
2 δ T
3
k
3
k
=
=
√
2 δPr −1/3
2 x(4.64/ Re x )Pr −1/3
For this typical case,
hx
Nu x = k
= 0.323 Re x
1/2 Pr 1/3
(7.38)
From Figure 7.7,
at x = x o , δ T = 0, r = 0
1
13
13 3/4
0 = C 1
+
, C 1 = −
x o
3/4
14Pr
14Pr
x o
Therefore,
( ) 3/4
� 1/3
δ T
x o
r =
= 0.975Pr −1/3 1 −
δ
x
156
Analytical Heat Transfer
U ∞ ,T ∞
T w
δ
δ T
0
Apply heat
x
Apply heat
x
x 0
δ
δ T
U ∞ ,T ∞
0
FIGURE 7.7
Integral approximation method.