Page 200 - Bird R.B. Transport phenomena

184 Chapter 6 Interphase Transport in Isothermal Systems

However, if one has to make more than a few calculations of (v z), it is advantageous to de-
velop a systematic approach; we suggest two methods here. Because experimental data are
often presented in graphical form, it is important for engineering students to use their origi-
nality in devising special methods such as those described here.
Method A. Figure 6.2-2 may be used to construct a plot 6 of Re versus the group ReVf,
which does not contain (v z):
D
fro ~ PL _ P /fro ~
)D
The quantity ReVf can be computed for this problem, and a value of the Reynolds number
can be read from the Re versus ReVf plot. From Re the average velocity and flow rate can
then be calculated.
Method B. Figure 6.2-2 may also be used directly without any replotting, by devising a
scheme that is equivalent to the graphical solution of two simultaneous equations. The two
equations are
/ = /(Re, k/D) curve given in Fig. 6.2-2 (6.2-22)
2
(ReVf)
/ = — straight line of slope 2 on log-log plot (6.2-23)
-
The procedure is then to compute ReVf according to Eq. 6.2-21 and then to plot Eq. 6.2-23 on
the log-log plot of / versus Re in Fig. 6.2-2. The intersection point gives the Reynolds number
of the flow, from which (v z) can then be computed.
For the problem at hand, we have
2
2
2
2
PO~PL = (3.00 lb f/in. )(32.17 lb m/ft lb, s )(144 in. /ft )
4
= 1.39X 10 lb w/ft • s 2
D = (7.981 in.)(£ ft/in.) = 0.665 ft
L = 1000 ft
p = 62.3 lb,,,/ft 3
4
/x = (1.03 cp)(6.72 x l(T (lb HI/ft • s)/cp)
4
= 6.93 X 10" lb,, ?/ft • s
Then according to Eq. 6.2-21,
4
Dp (po - p L)D (0.665X62.3) /(1.39 X 10 )(0.665)
4
2Lp (6.93 X 10" ) V 2(1000X62.3)
= 1.63 X 10 4 (dimensionless) (6.2-24)
4
The line of Eq. 6.2-23 for this value of ReVf passes through / = 1.0 at Re = 1.63 X 10 and
5
through/ = 0.01 at Re = 1.63 X 10 . Extension of the straight line through these points to the
curve of Fig. 6.2-2 for k/D = 0.00023 gives the solution to the two simultaneous equations:
= 2.4X10*
(6.2-25)
Solving for w then gives
w = (TT/4)D/I Re
5
4
= (0.7854X0.665X6.93 X 10" )(3600)(2.4 X 10 )
5
k
= 3.12 X 10 lb m /hr = 39 g/s (6.2-26)
6
A related plot was proposed by T. von Karman, Nachr. Ges. Wiss. Gottingen, Fachgruppen, I, 5, 58-76
(1930).