fluid_mechanics

8.5 Pipe Flow Examples 429
Pipe flow problems
can be categorized
by what parameters
are given and what
is to be calculated.
■ TABLE 8.4
Pipe Flow Types
Variable Type I Type II Type III
a. Fluid
Density Given Given Given
Viscosity Given Given Given
b. Pipe
Diameter Given Given Determine
Length Given Given Given
Roughness Given Given Given
c. Flow
Flowrate or Given Determine Given
Average Velocity
d. Pressure
Pressure Drop or Determine Given Given
Head Loss
In a Type III problem we specify the pressure drop and the flowrate and determine the diameter
of the pipe needed. For example, what diameter of pipe is needed between the water heater and
dishwasher if the pressure in the water heater is 60 psi 1determined by the city water system2 and the
flowrate is to be not less than 2.0 galmin 1determined by the manufacturer2?
Several examples of these types of problems follow.
E XAMPLE 8.8
Type I, Determine Pressure Drop
GIVEN Water at 60 ºF flows from the basement to the second
floor through the 0.75-in. (0.0625-ft)-diameter copper pipe
(a drawn tubing) at a rate of Q 12.0 galmin 0.0267 ft 3 s
and exits through a faucet of diameter 0.50 in. as shown in Fig.
E8.8a.
FIND Determine the pressure at point (1) if
(a) all losses are neglected,
(b) the only losses included are major losses, or
(c) all losses are included.
SOLUTION
Since the fluid velocity in the pipe is given by V 1 QA 1
Q1pD 2 42 10.0267 ft 3 s2 3p10.0625 ft2 2 44 8.70 fts, and the
fluid properties are and
10 5 lb # r 1.94 slugsft 3 m 2.34
sft 2 1see Table B.12, it follows that Re
slugsft 3 218.70 fts210.0625 ft212.34 10 5 lb # rVDm 11.94
sft 2 2 45,000.
Thus, the flow is turbulent. The governing equation for either case
1a2, 1b2, or 1c2 is the energy equation given by Eq. 8.21,
p 1
g a 1 V 2 1
2g z 1 p 2
g a 2 V 2 2
2g z 2 h L
where z lbft 3 1 0, z 2 20 ft, p 2 0 1free jet2, g rg 62.4 ,
and the outlet velocity is V 2 QA 2 10.0267 ft 3 s2 3p10.50
122 2 ft 2 44 19.6 fts. We assume that the kinetic energy coefficients
a 1 and a 2 are unity. This is reasonable because turbulent velocity
profiles are nearly uniform across the pipe. Thus,
Q =
12.0
gal/min
where the head loss is different for each of the three cases.
(a) If all losses are neglected 1h L 02, Eq. 1 gives
or
0.75-in.-diameter
copper pipe
(1)
15 ft
10 ft
(4)
(6)
5 ft 10 ft
(3)
(5)
F I G U R E E8.8a
p 1 gz 2 1 2r1V 2 2 V 2 12 gh L
p 1 162.4 lbft 3 2120 ft2
10 ft 10 ft
(7) (8)
1.94 slugs ft 3
ca19.6 ft 2
2
s b a8.70 ft 2
s b d
11248 2992 lbft 2 1547 lbft 2
g
Threaded
90° elbows
p 1 10.7 psi
Wide open
globe valve
K L = 2 based on
pipe
velocity
(2)
0.50-in.
diameter
(1)
(Ans)