fluid_mechanics

444 Chapter 8 ■ Viscous Flow in Pipes
1.00
0.98
C v
0.96
Range of values
depending on specific
geometry
0.94
10 4 10 5 10 6 10 7 10 8
Re = ρVD/
μ
F I G U R E 8.45 Venturi
meter discharge coefficient (Ref. 23).
Thus, the flowrate through a Venturi meter is given by
The Venturi discharge
coefficient
is a function of the
specific geometry
of the meter.
Q C v Q ideal CA T B
21p 1 p 2 2
r11 b 4 2
(8.40)
where A T pd 2 4 is the throat area. The range of values of C v , the Venturi discharge coefficient,
is given in Fig. 8.45. The throat-to-pipe diameter ratio 1b dD2, the Reynolds number, and the
shape of the converging and diverging sections of the meter are among the parameters that affect
the value of C v .
Again, the precise values of C n , C o , and C v depend on the specific geometry of the devices
used. Considerable information concerning the design, use, and installation of standard flow meters
can be found in various books 1Refs. 23, 24, 25, 26, 312.
E XAMPLE 8.15
Nozzle Flow Meter
GIVEN Ethyl alcohol flows through a pipe of diameter FIND Determine the diameter, d, of the nozzle.
D 60 mm in a refinery. The pressure drop across the nozzle
meter used to measure the flowrate is to be ¢p 4.0 kPa when
the flowrate is Q 0.003 m 3 s.
SOLUTION
From Table 1.6 the properties of ethyl alcohol are r 789 kgm 3 As a first approximation we assume that the flow is ideal, or
and m 1.19 10 3 N # sm 2 . Thus,
C n 1.0, so that Eq. 1 becomes
From Eq. 8.39 the flowrate through the nozzle is
or
Re rVD
m
4rQ
pDm
41789 kgm 3 210.003 m 3 s2
p10.06 m211.19 10 3 N # sm 2 2 42,200
Q 0.003 m 3 s C n p 4 d 2 B
214 10 3 Nm 2 2
789 kgm 3 11 b 4 2
1.20 10 3 C nd 2
21 b 4
where d is in meters. Note that b dD d0.06. Equation 1
and Fig. 8.43 represent two equations for the two unknowns d and
C n that must be solved by trial and error.
(1)
d 11.20 10 3 21 b 4 2 1 2
In addition, for many cases 1 b 4 1, so that an approximate
value of d can be obtained from Eq. 2 as
d 11.20 10 3 2 1 2 0.0346 m
Hence, with an initial guess of d 0.0346 m or b dD
0.03460.06 0.577, we obtain from Fig. 8.43 1using Re
42,2002 a value of C n 0.972. Clearly this does not agree with
our initial assumption of C n 1.0. Thus, we do not have the solution
to Eq. 1 and Fig. 8.43. Next we assume b 0.577 and
C n 0.972 and solve for d from Eq. 1 to obtain
12
1.20 103
d a 21 0.577 4 b
0.972
or d 0.0341 m. With the new value of b 0.03410.060
0.568 and Re 42,200, we obtain 1from Fig. 8.432 C n 0.972 in
(2)