432 Chapter 8 Viscous Flow in Pipes

The turbulent

portion of the

Moody chart is

represented by the

Colebrook formula.

EXAMPLE 8.5

or

Re=

chart

for ove

Ap=

The following equation from Colebrook is valid for the entire nonlaminar range of the Moody

GIVEN Air under standard conditions flows through a

4.0-mm-diameter drawn tubing with an average velocity of

V = 50 m/s. For such conditions the flow would normally be

turbulent. However, if precautions are taken to eliminate distur-

bances to the flow (the entrance to the tube is very smooth, the

air is dust free, the tube does not vibrate, etc.), it may be possible

to maintain laminar flow.

SOLUTION

Under standard temperature and pressure conditions the density

and viscosity are p = 1.23 kg/m¹ and 1.79 × 103

N s/m². Thus, the Reynolds number is

pVD (1.23 kg/m³) (50 m/s) (0.004 m)

1.79 x 10 N s/m²

which would normally indicate turbulent flow.

"

.

(a) If the flow were laminar, then f= 64/Re 64/13,700=

0.00467, and the pressure drop in a 0.1-m-long horizontal section

of the pipe would be

= (0.00467) -

√

(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 (Re and e/D), 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 (uncertainty in the relative roughness, uncertainty in the experimental data used to produce

the Moody chart, etc.), the use of several place accuracy (i.e., more than about two significant

digits) 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 Colebrook/Moody

chart relationship but does not require an iterative scheme. For example, the Haaland equation (Ref. 34).

which is easier to use, is given by

1

VS

where one can solve for f explicitly.

(0.1 m) 1

(1.23 kg/m³) (50 m/s)²

(0.004 m) 2

Ap= 0.179 kPa

13,700

= -2.0 log

Comparison of Laminar or Turbulent Pressure Drop

(Ans)

-1.8 log

(e/D+

3.7

1.11

[(12) ***

+

3.7.

2.51

Re √f

or

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.

Ap=!

6.9

Re

COMMENT Note that the same result is obtained from

Eq. 8.8:

Ap

32μl

D²

32(1.79 x 10N s/m²) (0.1 m) (50 m/s)

(0.004 m)²

V

D2P

= 179 N/m²

(b) If the flow were turbulent, then f= (Re, e/D), where

from Table 8.1, e= 0,0015 mm so that e/D = 0.0015 mm/

4.0 mm 0.000375. From the Moody chart with Re 1.37 x

10 and e/D 0.000375 we obtain f= 0.028. Thus, the pres-

sure drop in this case would be approximately

(8.35b)

= (0.028) (0.004 m) 2

(0.1 m) 1

(0.028)

Ap 1.076 kPa

(1.23 kg/m²) (50 m/s)²

(Ans)