Page 457 - Bird R.B. Transport phenomena

§14.3 Heat Transfer Coefficients for Forced Convection in Tubes 437

Note also that Fig. 14.3-2 somewhat resembles the friction-factor plot in Fig. 6.2-2, al-
though the physical situation is quite different. In the highly turbulent range (Re >
10,000) the heat transfer ordinate agrees approximately with //2 for the long smooth
3
pipes under consideration. This was first pointed out by Colburn, who proposed the fol-
lowing empirical analogy for long, smooth tubes:
torn * \f > 10,0001 (14.3-18)
in which
Nu 2/3
ln (14.3-19)
RePr 1/3 (pv)C ' wC
p n
where S is the area of the tube cross section, w is the mass rate of flow through the tube,
and//2 is obtainable from Fig. 6.2-2 using Re = DW/SJJL = 4W/TTDIJL. Clearly the analogy
of Eq. 14.3-18 is not valid below Re = 10,000. For rough tubes with fully developed tur-
bulent flow the analogy breaks down completely, because / is affected more by rough-
ness than j is.
H
One additional remark about the use of Fig. 14.3-2 has to do with the application to
conduits of noncircular cross section. For highly turbulent flow, one may use the mean
hydraulic radius of Eq. 6.2-16. To apply that empiricism, D is replaced by AR h every-
where in the Reynolds and Nusselt numbers.

EXAMPLE 14.3-1 Air at 70°F and 1 atm is to be pumped through a straight 2-in. i.d. tube at a rate of 70 lb /hr.
w
A section of the tube is to be heated to an inside wall temperature of 250°F to raise the air tem-
Design of a Tubular perature to 230°F. What heated length is required?
Heater
SOLUTION
The arithmetic average bulk temperature is T ba = 150°F, and the film temperature is 7y =
|(150 + 250) = 200°F. At this temperature the properties of air are /x, = 0.052 lb /ft • hr, C p = 0.242
m
Btu/lb • F, fc = 0.0180 Btu/hr • ft • F, and Pr = С /л/к = 0.70. The viscosities of air at 150°F and
р
m
250°F are 0.049 and 0.055 lb /ft - hr, respectively, so that the viscosity ratio is \x l ^ =
w
b
0.049/0.055 = 0.89.
The Reynolds number, evaluated at the film temperature, 200°F, is then
Dw 4w 4(70) = 1.02 X 10 4 (14.3-20)
TT(2/12)(0.052)
From Fig. 14.3-1 we obtain
~ T )
M 2 / 3
4 f P r ^ =0.0039 (14.3-21)
When this is solved for L/D we get
1 (Tt,o - Гм) / и Л 0 1 4
1 К1 Ь2 L b \ ) p 2/3 Mb
D
4(0.0039) (T - ГД„ V^°/
o
1 (230 - 70) (0.70) (0.89Г 014
2/3
4(0.0039) 72.2
1 160
4(0.0039) 72.8 (0.788X1.02) = 113 (14.3-22)
Hence the required length is
L = 113D = (113X2/12) = 19 ft (14.3-23)

If Re had been much smaller, it would have been necessary to estimate L/D before reading
b
Fig. 14.3-2, thus initiating a trial-and-error process.