fluid_mechanics

398 Chapter 8 ■ Viscous Flow in Pipes
E XAMPLE 8.3
Laminar Pipe Flow Properties
GIVEN The flowrate, Q, of corn syrup through the horizontal
pipe shown in Fig. E8.3a is to be monitored by measuring the pressure
difference between sections 112 and 122. It is proposed that
Q K ¢p, where the calibration constant, K, is a function of temperature,
T, because of the variation of the syrup’s viscosity and
density with temperature. These variations are given in Table E8.3.
FIND (a) Plot K1T 2 versus T for 60 °F T 160 °F. (b) Determine
the wall shear stress and the pressure drop,
¢p p for Q 0.5 ft 3 1 p 2 , s and T 100 °F. (c) For the conditions
of part 1b2, determine the net pressure force, 1pD 2 42 ¢p,
and the net shear force, pD/t w , on the fluid within the pipe between
the sections 112 and 122.
SOLUTION
(a)
or
If the flow is laminar it follows from Eq. 8.9 that
where the units on Q, ¢p, and m are ft 3 s, lbft 2 , and lb # sft 2 , respectively.
Thus
(1)
(Ans)
where the units of K are ft 5 lb # s. By using values of the viscosity
from Table E8.3, the calibration curve shown in Fig. E8.3b is obtained.
This result is valid only if the flow is laminar.
COMMENT As shown in Section 8.5, for turbulent flow the
flowrate is not linearly related to the pressure drop so it would not
be possible to have Q K ¢p. Note also that the value of K is independent
of the syrup density 1r
was not used in the calculations2
since laminar pipe flow is governed by pressure and viscous effects;
inertia is not important.
(b) For T 100 °F, the viscosity is m 3.8 10 3 lb # sft 2
so that with a flowrate of Q 0.5 ft 3 s the pressure drop 1according
to Eq. 8.92 is
12813.8 103 lb # sft 2 216 ft210.5 ft 3 s2
p1 3 12 ft2 4
119 lbft 2
provided the flow is laminar. For this case
so that
¢p 128m/Q
pD 4
Q pD4 ¢p
128m/ p1 3
12 ft2 4 ¢p
128m16 ft2
Q K ¢p
K
V Q A 0.5 ft3 s
10.2 fts
p
4 1 3
12 ft2 2
Re rVD
m 12.05 slugs ft 3 2110.2 fts21 3
12 ft2
13.8 10 3 lb # sft 2 2
1380 6 2100
1.60 105
¢p
m
1.60 105
m
(Ans)
Q
is
K, ft 5 /(lb • s)
10 0
10 –1
10 –2
10 –3
(1)
F I G U R E E8.3
TABLE E8.3
T ( F) R (slugs ft 3 ) M ( lb sft 2 )
60 2.07 4.0 10 2
80 2.06 1.9 10 2
100 2.05 3.8 10 3
120 2.04 4.4 10 4
140 2.03 9.2 10 5
160 2.02 2.3 10 5
Hence, the flow is laminar. From Eq. 8.5 the wall shear stress
t w ¢pD
4/ 1119 lb ft 2 21 3 12 ft2
1.24 lbft 2
416 ft2
(Ans)
(c) For the conditions of part 1b2, the net pressure force, F p , on
the fluid within the pipe between sections 112 and 122 is
F p p 4 D2 ¢p p 4 a 3
12 ftb 2
1119 lbft 2 2 5.84 lb
(Ans)
Similarly, the net viscous force, F v , on that portion of the fluid is
F v 2p a D 2 b /t w
6 ft
(a)
10 –4 60 100 140
T, °F
(b)
(2)
3-in.
diameter
180
3
2p c
21122 ft d16 ft211.24 lb ft 2 2 5.84 lb
(Ans)
COMMENT Note that the values of these two forces are the
same. The net force is zero; there is no acceleration.