fluid_mechanics

424 Chapter 8 ■ Viscous Flow in Pipes
E XAMPLE 8.6
Minor Losses
GIVEN The closed-circuit wind tunnel shown in Fig. E8.6a is a
smaller version of that depicted in Fig. E8.6b in which air at standard
conditions is to flow through the test section [between sections
(5) and (6)] with a velocity of 200 ft/s. The flow is driven by a fan
that essentially increases the static pressure by the amount p 1 p 9
that is needed to overcome the head losses experienced by the fluid
as it flows around the circuit.
FIND Estimate the value of p 1 p 9 and the horsepower supplied
to the fluid by the fan.
V 5 = 200 ft/s
(4)
(5) (6) (7)
(3)
Test section
Flow-straightening
screens
(8)
(2)
Q (1)
(9)
F I G U R E E8.6a
(Photograph courtesy of
DELTALAB.France.)
Fan
F I G U R E E8.6b
SOLUTION
The maximum velocity within the wind tunnel occurs in the
test section 1smallest area; see Table E8.6 on the next page2.
Thus, the maximum Mach number of the flow is Ma 5 V 5c 5 ,
where V 5 200 fts and from Eq. 1.20 the speed of sound is
c 11716 ft # lbslug # 31460 592 °R46 12 5 1kRT 5 2 12 51.4
°R2
1117 fts. Thus, Ma 5 2001117 0.179. As was indicated in
Chapter 3 and discussed fully in Chapter 11, most flows can be considered
as incompressible if the Mach number is less than about 0.3.
Hence, we can use the incompressible formulas for this problem.
The purpose of the fan in the wind tunnel is to provide the necessary
energy to overcome the net head loss experienced by the
air as it flows around the circuit. This can be found from the energy
equation between points 112 and 192 as
p 1
g V 2 1
2g z 1 p 9
g V 2 9
2g z 9 h L19
where h L19 is the total head loss from 112 to 192. With z 1 z 9 and
V 1 V 9 this gives
p 1
g p 9
g h L 19
Similarly, by writing the energy equation 1Eq. 5.842 across the fan,
from 192 to 112, we obtain
p 9
g V 2 9
2g z 9 h p p 1
g V 2 1
2g z 1
(1)
h p
where is the actual head rise supplied by the pump 1fan2 to the
air. Again since z 9 z 1 and V 9 V 1 this, when combined with
Eq. 1, becomes
h p 1 p 1 p 9 2
g
h L19
The actual power supplied to the air 1horsepower, p a 2 is obtained
from the fan head by
p a gQh p gA 5 V 5 h p gA 5 V 5 h L19 (2)
Thus, the power that the fan must supply to the air depends on
the head loss associated with the flow through the wind tunnel. To
obtain a reasonable, approximate answer we make the following
assumptions. We treat each of the four turning corners as a mitered
bend with guide vanes so that from Fig. 8.31 K Lcorner 0.2. Thus,
for each corner
h Lcorner K V 2
L
2g 0.2 V 2
2g
where, because the flow is assumed incompressible, V V 5 A 5A.
The values of A and the corresponding velocities throughout the
tunnel are given in Table E8.6.
We also treat the enlarging sections from the end of the test
section 162 to the beginning of the nozzle 142 as a conical diffuser
with a loss coefficient of K Ldif 0.6. This value is larger than that
of a well-designed diffuser 1see Fig. 8.29, for example2. Since the