Page 199 - Bird R.B. Transport phenomena

§6.2 Friction Factors for Flow in Tubes 183

5
This equation is stated to be accurate within 1.5%. As can be seen in Fig. 6.2-2, the fric-
tional resistance to flow increases with the height, k, of the protuberances. Of course, к
has to enter into the correlation in a dimensionless fashion and hence appears via the
ratio k/D.
For turbulent flow in noncircular tubes it is common to use the following empiricism:
First we define a "mean hydraulic radius" R as follows:
h
R h = S/Z (6.2-16)
in which S is the cross section of the conduit and Z is the wetted perimeter. Then we can
use Eq. 6.1-4 and Fig. 6.2-2, with the diameter D of the circular pipe replaced by 4R . That
h
is, we calculate pressure differences by replacing Eq. 6.1-4 by

and getting / from Fig. 6.2-2 with a Reynolds number defined as

Re,, = ^ ^ (6.2-18)

This estimation method of Eqs. 6.2-16 to 18 should not be used for laminar flow.

EXAMPLE 6.2-1 What pressure gradient is required to cause diethylaniline, C H N(C H5)2, to flow in a
2
5
6
horizontal, smooth, circular tube of inside diameter D = 3 cm at a mass rate of 1028 g/s at
3
Pressure Drop Required 0°C? At this temperature the density of diethylaniline is p = 0.935 g/cm and its viscosity is
2
for a Given Flow Rate ^ = 1.95 p.
C
SOLUTION The Reynolds number for the flow is
V(v )p Dw 4w
Re = z
2
(TTD /4)/X
4(1028 g/s)
= 2.24 X 10 4 (6.2-19)
2
7r(3cm)(1.95xlO- g/cm-s)
From Fig. 6.2-2, we find that for this Reynolds number the friction factor / has a value of
0.0063 for smooth tubes. Hence the pressure gradient required to maintain the flow is (ac-
cording to Eq. 6.1-4)
Po ~~ PL

2
2
_ 32w f = (32)(1028) (0.0063))
5
2
5
TT D P " 7T (3.0) (0.935)
2
= 95(dyne/cm )/cm = 0.071 (mm Hg)/cm (6.2-20)
2
EXAMPLE 6.2-2 Determine the flow rate, in pounds per hour, of water at 68°F through a 1000-ft length of hori-
zontal 8-in. schedule 40 steel pipe (internal diameter 7.981 in.) under a pressure difference of
Flow Rate for a Given 3.00 psi. For such a pipe use Fig. 6.2-2 and assume that k/D = 2.3 X 10" .
4
Pressure Drop
SOLUTION
We want to use Eq. 6.1-4 and Fig. 6.2-2 to solve for (v ) when p - p is known. However, the
z 0 L
quantity (v ) appears explicitly on the left side of the equation and implicitly on the right side
z
in /, which depends on Re = D(v )p//л. Clearly a trial-and-error solution can be found.
z