fluid_mechanics

434 Chapter 8 ■ Viscous Flow in Pipes
SOLUTION
The energy equation 1Eq. 8.212 can be applied between the surface
of the lake [point 112] and the outlet of the pipe as
where h T is the turbine head, p 1 V 1 p 2 z 2 0, z 1 90 ft,
and V 2 V, the fluid velocity in the pipe. The head loss is given by
where V is in fts. Also, the turbine head is
Thus, Eq. 1 can be written as
or
h L f / D
p 1
g V 1 2
2g z 1 p 2
g V 2 2
2g z 2 h L h T
V 2 1300 ft2 V 2
0.02
2g 11 ft2 2132.2 fts 2 2 0.0932V 2 ft
h T p a
gQ p a
g1p42D 2 V
150 hp231550 ft # lbs2hp4
561
162.4 lbft 3 231p4211 ft2 2 V4 V ft
90 V 2
2132.22 0.0932V 2 561
V
0.109V 3 90V 561 0
where V is in fts. The velocity of the water in the pipe is found as
the solution of Eq. 2. Surprisingly, there are two real, positive
roots: V 6.58 fts or V 24.9 fts. The third root is negative
(1)
(2)
1V 31.4 fts2 and has no physical meaning for this flow.
Thus, the two acceptable flowrates are
or
Q p 4 D2 V p 4 11 ft22 16.58 fts2 5.17 ft 3 s
Q p 4 11 ft22 124.9 fts2 19.6 ft 3 s
(Ans)
(Ans)
COMMENTS Either of these two flowrates gives the same
power, p a gQh T . The reason for two possible solutions can be
seen from the following. With the low flowrate 1Q 5.17 ft 3 s2, we
obtain the head loss and turbine head as h L 4.04 ft and
h T 85.3 ft. Because of the relatively low velocity there is a relatively
small head loss and, therefore, a large head available for the
turbine. With the large flowrate 1Q 19.6 ft 3 s2, we find
h L 57.8 ft and h T 22.5 ft. The high-speed flow in the pipe produces
a relatively large loss due to friction, leaving a relatively small
head for the turbine. However, in either case the product of the turbine
head times the flowrate is the same. That is, the power extracted
1p a gQh T 2 is identical for each case. Although either flowrate
will allow the extraction of 50 hp from the water, the details of the
design of the turbine itself will depend strongly on which flowrate is
to be used. Such information can be found in Chapter 12 and various
references about turbomachines 1Refs. 14, 19, 202.
If the friction factor were not given, the solution to the problem
would be much more lengthy. A trial-and-error solution similar
to that in Example 8.10 would be required along with the solution
of a cubic equation.
In pipe flow problems for which the diameter is the unknown 1Type III2, an iterative or numerical
root-finding technique is required. This is, again, because the friction factor is a function of the
diameter—through both the Reynolds number and the relative roughness. Thus, neither Re rVDm
4rQpmD nor eD are known unless D is known. Examples 8.12 and 8.13 illustrate this.
E XAMPLE 8.12 Type III without Minor Losses, Determine Diameter
GIVEN Air at standard temperature and pressure flows FIND Determine the minimum pipe diameter.
through a horizontal, galvanized iron pipe 1e 0.0005 ft2 at a
rate of 2.0 ft 3 s. The pressure drop is to be no more than 0.50 psi
per 100 ft of pipe.
SOLUTION
We assume the flow to be incompressible with r
0.00238 slugsft 3 and m 3.74 10 7 lb # sft 2 . Note that if the
pipe were too long, the pressure drop from one end to the other,
p 1 p 2 , would not be small relative to the pressure at the beginning,
and compressible flow considerations would be required.
For example, a pipe length of 200 ft gives 1p 1 p 2 2p 1
310.50 psi21100 ft241200 ft214.7 psia 0.068 6.8%, which is
probably small enough to justify the incompressible assumption.
With z 1 z 2 and V 1 V 2 the energy equation 1Eq. 8.212
becomes
p 1 p 2 f / rV 2
D 2
where V QA 4Q1pD 2 2 412.0 ft 3 s2pD 2 , or
V 2.55
D 2
(1)