fluid_mechanics

436 Chapter 8 ■ Viscous Flow in Pipes
SOLUTION
The energy equation 1Eq. 8.212 can be applied between two points
on the surfaces of the reservoirs 1p 1 p 2 V 1 V 2 z 2 02
as follows:
p 1
g V 2 1
2g z 1 p 2
g V 2 2
2g z 2 h L
or
where V QA 4QpD 2 4126 ft 3 s2pD 2 , or
is the velocity within the pipe. 1Note that the units on V and D are
fts and ft, respectively.2 The loss coefficients are obtained from
Table 8.2 and Figs. 8.22 and 8.25 as K Lent 0.5, K Lelbow 0.2,
and K Lexit 1. Thus, Eq. 1 can be written as
or, when combined with Eq. 2 to eliminate V,
To determine D we must know f, which is a function of Re and
eD, where
and
44 ft
V 2
e 1700 f 3410.22 0.5 14 f
2132.2 fts 2 2 D
Re VD n
z 1 V 2
2g af / D a K Lb
V 33.1
D 2
f 0.00152 D 5 0.00135 D
3133.12 D 2 4D 2.74 106
5
1.21 10 D
e
D 0.0005
D
where D is in feet. Again, we have four equations 1Eqs. 3, 4, 5, and
the Moody chart or the Colebrook equation2 for the four unknowns
D, f, Re, and eD.
Consider the solution by using the Moody chart. Although
it is often easiest to assume a value of f and make calculations
to determine if the assumed value is the correct one, with the
inclusion of minor losses this may not be the simplest method.
For example, if we assume f 0.02 and calculate D from
Eq. 3, we would have to solve a fifth-order equation. With
only major losses 1see Example 8.122, the term proportional to
D in Eq. 3 is absent, and it is easy to solve for D if f is given.
With both major and minor losses included, this solution
for D 1given f 2 would require a trial-and-error or iterative
technique.
Thus, for this type of problem it is perhaps easier to assume
a value of D, calculate the corresponding f from Eq. 3, and with
the values of Re and eD determined from Eqs. 4 and 5, look up
the value of f in the Moody chart 1or the Colebrook equation2.
The solution is obtained when the two values of f are in agree-
(1)
(2)
(3)
(4)
(5)
ment. A few rounds of calculation will reveal that the solution is
given by
D 1.63 ft
(Ans)
COMMENTS Alternatively, we can use the Colebrook equation
rather than the Moody chart to solve for D. This is easily
done by using the Colebrook equation (Eq. 8.35a) with f as a
function of D obtained from Eq. 3 and Re and eD as functions of
D from Eqs. 4 and 5. The resulting single equation for D can be
solved by using a root-finding technique on a computer or calculator
to obtain D 1 .63 ft. This agrees with the solution obtained
using the Moody chart.
By repeating the calculations for various pipe lengths, /,
the results shown in Fig. E8.13b are obtained. As the pipe
length increases it is necessary, because of the increased friction,
to increase the pipe diameter to maintain the same
flowrate.
It is interesting to attempt to solve this example if all losses are
neglected so that Eq. 1 becomes z 1 0. Clearly from Fig. E8.13a,
z 1 44 ft. Obviously something is wrong. A fluid cannot flow
from one elevation, beginning with zero pressure and velocity,
and end up at a lower elevation with zero pressure and velocity
unless energy is removed 1i.e., a head loss or a turbine2 somewhere
between the two locations. If the pipe is short 1negligible
friction2 and the minor losses are negligible, there is still the kinetic
energy of the fluid as it leaves the pipe and enters the reservoir.
After the fluid meanders around in the reservoir for some
time, this kinetic energy is lost and the fluid is stationary. No matter
how small the viscosity is, the exit loss cannot be neglected.
The same result can be seen if the energy equation is written from
the free surface of the upstream tank to the exit plane of the pipe,
at which point the kinetic energy is still available to the fluid. In
either case the energy equation becomes z 1 V 2 2g in agreement
with the inviscid results of Chapter 3 1the Bernoulli
equation2.
D, ft
1.8
1.6
1.4
1.2
1.0
0.8
0.6
0.4
0.2
0.0
0 500 1000
, ft
F I G U R E E8.13b
(1700 ft, 1.63 ft)
1500 2000