where
τ w =
1 ρU
2
· C f
∞
2
1 ρU ∞ x μC p
Nu = C f
·
2
μ
k
Nu
1
1
= C f ⇒
C f = St
(10.23)
Re · Pr
2
2
1 C f = St · Pr
2/3
(10.24)
2
For 0.7 ≤ Pr ≤ 60 for air, water, and oil.
where
""
Nu
(hx/k)
q
h
w
St =
=
=
=
Re · Pr
(ρU ∞ x/μ) · (μC p /k)
ρC p U ∞
ρC p U ∞ (T w − T ∞ )
(10.25)
There are two kinds of problems:
1. For given C f to determine St or h;
2. For given h or St, to determine C f

For a turbulent pipe flow

0.046
C f =
(10.26)
Re 0.2
D
1 0.046
Nu
=
Pr
2/3
Re 0.2
2
Re · Pr
D
Nu D =
hD
D Pr
1/3 for cooling
(10.27)
= 0.023 Re
0.8
k
Nu D =
hD = 0.023 Re D
0.8 Pr
0.4 for heating
(10.28)
k
For a turbulent boundary-layer flow,

0.0592

C f =
(10.29)
Re 0.2
x
1 0.0592
Nu x
=
Pr
2/3
Re 0.2
2
Re x Pr
x
Nu x =
hx = 0.0296 Re
0.8 Pr
1/3
(10.30)
x
k
204
Analytical Heat Transfer