Page 183 - Bird R.B. Transport phenomena

§5.5 Turbulent Flow in Ducts 167

If this is compared with experimental data on flow rate versus pressure drop, it is found that
good agreement can be obtained by changing 2.5 to 2.45 and 1.75 to 2.0. This "fudging" of the
constants would probably not be necessary if the integration over the cross section had been
done by using the local expression for the velocity in the various layers. On the other hand,
there is some virtue in having a simple logarithmic relation such as Eq. 5.5-1 to describe pres-
sure drop vs. flow rate.
In a similar fashion the power law profile can be integrated over the entire cross section to
give (see Ref. 4 of §5.3)

3
6
in which a = 3/(2 In Re). This relation is useful over the range 3.07 X 10 < Re < 3.23 X 10 .
EXAMPLE 5.5-2 Show how Eqs. 5.4-4 and 5 can be used to describe turbulent flow in a circular tube.

Application of SOLUTION
Frandtl's Mixing
Length Formula to Equation 5.2-12 gives for the steadily driven flow in a circular tube,
Turbulent Flow in a
Circular Tube (5.5-3)

in which T = rj? + т£. Over most of the tube the viscous contribution is quite small; here we
rz
neglect it entirely. Integration of Eq. 5.5-3 then gives
-ш _ (5.5-4)
T 2L
where r is the wall shear stress and у = R - r is the distance from the tube wall.
0
According to the mixing length theory in Eq. 5.4-4, with the empirical expression in Eq.
5.4-5, we have for dvjdr negative
d\
(5.5-5)
dr dr
Substitution of this into Eq. 5.5-4 gives a differential equation for the time-smoothed velocity.
If we follow Prandtl and extrapolate the inertial sublayer to the wall, then in Eq. 5.5-5 it is ap-
propriate to replace т£ by r 0. When this is done, Eq. 5.5-5 can be integrated to give
= — In у + constant (5.5-6)
v z
Thus a logarithmic profile is obtained and hence the results from Example 5.5-1 can be used;
that is, one can apply Eq. 5.5-6 as a very rough approximation over the entire cross section of
the tube.

{t)
EXAMPLE 5.5-3 Determine the ratio fi /fi at у = R/2 for water flowing at a steady rate in a long, smooth,
round tube under the following conditions:
Relative Magnitude of
Viscosity and Eddy R = tube radius = 3 in. = 7.62 cm
Viscosity r 0 = wall shear stress = 2.36 X 10" lty/in. 2 = 0.163 Pa
5
p = density = 62.4 lb /ft 3 = 1000 kg/m 3
w
2
5
2
7
v = kinematic viscosity = 1.1 X 10~ ft /s = 1.02 X 10~ m /s
SOLUTION The expression for the time-smoothed momentum flux is
(5.5-7)