fluid_mechanics

8.5 Pipe Flow Examples 433
The value of f is dependent on Re, which is dependent on V,
an unknown. However, from Table B.3, n 1.79 10 4 ft 2 s and
we obtain
or
where again V is in fts.
Also, since eD 10.0005 ft21412 ft2 0.0015 1see Table
8.1 for the value of e2, we know which particular curve of the
Moody chart is pertinent to this flow. Thus, we have three relationships
1Eqs. 2, 3, and the eD 0.0015 curve of Fig. 8.202
from which we can solve for the three unknowns f, Re, and V.
This is done easily by an iterative scheme as follows.
It is usually simplest to assume a value of f, calculate V from Eq.
2, calculate Re from Eq. 3, and look up the appropriate value of f in
the Moody chart for this value of Re. If the assumed f and the new f
do not agree, the assumed answer is not correct—we do not have the
solution to the three equations. Although values of either f, V, or Re
could be assumed as starting values, it is usually simplest to assume
a value of f because the correct value often lies on the relatively flat
portion of the Moody chart for which f is quite insensitive to Re.
Thus, we assume f 0.022, approximately the large Re limit
for the given relative roughness. From Eq. 2 we obtain
and from Eq. 3
Re VD n 1 4
12 ft2 V
1.79 10 4 ft 2 s
Re 1860 V
12
945
V c
6.0 6010.0222 d 11.4 fts
Re 1860111.42 21,200
With this Re and eD, Fig. 8.20 gives f 0.029, which is not
equal to the assumed solution f 0.022 1although it is close!2.
We try again, this time with the newly obtained value of
f 0.029, which gives V 11.0 fts and Re 20,500. With
these values, Fig. 8.20 gives f 0.029, which agrees with the assumed
value. Thus, the solution is V 11.0 fts, or
Q AV p 4 1 4
12 ft2 2 111.0 fts2 0.960 ft 3 s
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(Ans)
COMMENTS Note that the need for the iteration scheme is
because one of the equations, f f1Re, eD2, is in graphical
form 1the Moody chart2. If the dependence of f on Re and eD is
known in equation form, this graphical dependency is eliminated,
and the solution technique may be easier. Such is the case
if the flow is laminar so that the friction factor is simply
f 64Re. For turbulent flow, we can use the Colebrook equation
rather than the Moody chart. Thus, we keep Eqs. 2 and 3
and use the Colebrook equation 1Eq. 8.35a) with eD 0.0015
to give
1
1f 2.0 log ae D
3.7 2.51
Re1f b
2.0 log a4.05 10 4 2.51
Re1f b
From Eq. 2 we have V 394516.0 60 f 24 12 , which can be
combined with Eq. 3 to give
Re 57,200
16.0 60 f
The combination of Eqs. 4 and 5 provides a single equation for
the determination of f
1
2.0 log a4.05 104
1f
4.39 10 5 60 6.0 b
B f
By using a root-finding technique on a computer or calculator,
the solution to this equation is determined to be f 0.029, in
agreement with the above solution which used the Moody
chart.
Note that unlike the Alaskan pipeline example 1Example
8.92 in which we assumed minor losses are negligible, minor
losses are of importance in this example because of the relatively
small length-to-diameter ratio: /D 2014122 60.
The ratio of minor to major losses in this case is K L1 f/D2
6.0 30.029 16024 3.45. The elbows and entrance produce
considerably more loss than the pipe itself.
(4)
(5)
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E XAMPLE 8.11
Type II, Determine Flowrate
GIVEN The turbine shown in Fig. E8.11 extracts 50 hp from
the water flowing through it. The 1-ft-diameter, 300-ft-long
pipe is assumed to have a friction factor of 0.02. Minor losses
are negligible.
FIND
Determine the flowrate through the pipe and turbine.
(1)
z 1 = 90 ft
300-ft-long,
1-ft-diameter pipe
Turbine
Free jet
f = 0.02
F I G U R E E8.11
(2)
z 2 = 0