fluid_mechanics

414 Chapter 8 ■ Viscous Flow in Pipes
The following equation from Colebrook is valid for the entire nonlaminar range of the Moody
chart
The turbulent portion
of the Moody
chart is represented
by the Colebrook
formula.
1
1f 2.0 log ae D
3.7 2.51
Re1f b
(8.35a)
In fact, the Moody chart is a graphical representation of this equation, which is an empirical
fit of the pipe flow pressure drop data. Equation 8.35 is called the Colebrook formula. A difficulty
with its use is that it is implicit in the dependence of f. That is, for given conditions
1Re and eD2, it is not possible to solve for f without some sort of iterative scheme. With the
use of modern computers and calculators, such calculations are not difficult. A word of caution
is in order concerning the use of the Moody chart or the equivalent Colebrook formula.
Because of various inherent inaccuracies involved 1uncertainty in the relative roughness, uncertainty
in the experimental data used to produce the Moody chart, etc.2, the use of several
place accuracy in pipe flow problems is usually not justified. As a rule of thumb, a 10% accuracy
is the best expected. It is possible to obtain an equation that adequately approximates
the ColebrookMoody chart relationship but does not require an iterative scheme. For example,
an alternate form (Ref. 34), which is easier to use, is given by
where one can solve for f explicitly.
1.11
1
2f 1.8 log cae D
3.7 b 6.9
Re d
(8.35b)
E XAMPLE 8.5
Comparison of Laminar or Turbulent Pressure Drop
GIVEN Air under standard conditions flows through a 4.0-mmdiameter
drawn tubing with an average velocity of V 50 ms.
For such conditions the flow would normally be turbulent. However,
if precautions are taken to eliminate disturbances to the flow
1the entrance to the tube is very smooth, the air is dust free, the tube
does not vibrate, etc.2, it may be possible to maintain laminar flow.
FIND (a) Determine the pressure drop in a 0.1-m section of
the tube if the flow is laminar.
(b) Repeat the calculations if the flow is turbulent.
SOLUTION
Under standard temperature and pressure conditions the density
and viscosity are r 1.23 kgm 3 and m 1.79 10 5
N # sm 2 . Thus, the Reynolds number is
which would normally indicate turbulent flow.
(a) If the flow were laminar, then f 64Re 6413,700
0.00467 and the pressure drop in a 0.1-m-long horizontal section
of the pipe would be
or
Re rVD
m
¢p f / D
11.23 kg m 3 2150 ms210.004 m2
1.79 10 5 N # sm 2 13,700
1
2 rV 2
10.1 m2 1
10.004672
10.004 m2 2 11.23 kg m 3 2150 ms2 2
¢p 0.179 kPa
(Ans)
COMMENT Note that the same result is obtained from Eq. 8.8:
(b) If the flow were turbulent, then f f1Re, eD2, where
from Table 8.1, e 0.0015 mm so that eD 0.0015 mm
4.0 mm 0.000375. From the Moody chart with Re 1.37
10 4 and eD 0.000375 we obtain f 0.028. Thus, the pressure
drop in this case would be approximately
or
¢p f / D
¢p 32m/
D 2 V
3211.79 105 N # sm 2 210.1 m2150 ms2
10.004 m2 2
179 Nm 2
1
2 rV 2 10.0282
10.1 m2
10.004 m2
¢p 1.076 kPa
1
2 11.23 kg m 3 2150 ms2 2
(Ans)