College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 390
Fluid Mechanics
Spring 2008 Number: 11971 Instructor: Larry Caretto
Solutions to Exercise Nine – Flow in Pipes
1
A 70-ft-long, 0.5-in-diameter hose with a roughness of  = 0.00009 ft is fastened to a water
faucet where the pressure is p1. Determine p1 if there is no nozzle attached and the
average velocity in the hose is 6 ft/s. Neglect minor losses and elevation changes.
We start by computing the Reynolds number for this flow to determine if the flow is laminar or
turbulent. We use the properties for water from Table 1.6:  = 1.21x10-5 ft2/s and  = 1.94
slugs/ft3.
6 ft
0.5 in  1 ft
s
12 in
VD VD
Re 


 2.07 x10 4
5


1.21x10 ft 2
2
Therefore the flow is turbulent and we can find the friction factor, f, from the Moody diagram on
page 434. The relative roughness of the pipe, /D = (0.00009 ft) / (0.5/12 ft) = 0.0216. This value
of /D and Re = 2.07x104 gives f = 0.03. Checking this with the Colebrook equation gives the
following value.
 D
2.51
 2.0 log 10 

 3.7 Re f
f

1


2.51
  2.0 log 10  0.0216 
  5.78 

4

3
.
7
2
.
07
x
10
0
.
052



f  0.0299
The pressure drop in the 70-ft length of hose can now be found.
2
2
1755 lb f
70 ft 12 in 1 1.94 slug  6 ft  1 lb f  s
 V 2
p  f
 0.0299



3
D 2
0.5 in ft 2
ft
ft 2
 s  slug  ft
We assume that the hose flow exits to the atmosphere where the (gage) pressure is zero so the
pressure drop computed above is the same as the pressure, p1, at the water faucet. Dividing the
result by 144 to get psi gives p1 = 12.2 psi .
2
Blood (assume  = 4.5x10-5 lbfs/ft2 and SG = 1.0) flows through an artery in the neck of a
giraffe from its heart to its head at a rate of 2.5x10-4 ft3/s. Assume that the length of the
artery is 10 ft and that its diameter is 0.20 in. If the pressure at the beginning of the artery
(outlet of the heart) is equivalent to 0.70 ft Hg, determine the pressure at the end of the
artery when the head is (a) 8 ft above the heart or (b) 6 ft below the heart. Assume steady
flow. How much of this pressure difference is due to elevation effects and how much is
due to friction?
We can write the head loss in the energy equation as p/ = p/g =[ f (l/D) V2/2 ] / (g) =
f(l/D)V2/2g so that the energy can be written to give the following relationship between the end
position in the giraffe’s head (2) and the inlet at the heart (1).
z2 
p2


V22
p V2
p V2
 V2
 z1  1  1  hL  z1  1  1  f
2g
 2g
 2g
D 2g
Jacaranda (Engineering) 3333
E-mail: lcaretto@csun.edu
Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062
Exercise nine solutions
ME 390, L. S. Caretto, Spring 2008
Page 2
Since the area is assumed constant and the blood is assumed to have constant density, the
velocity at all points in the artery is the same. Thus we can cancel the V12 and V22 terms and
write the following equation for the pressure difference between the head (p 2) and the heart (p1).
p2  p1

 z1  z 2  f
 V2
D 2g
To compute the friction factor we first compute the Reynolds number to determine if the flow is
laminar or turbulent. Note that a specific gravity of 1 gives a density  = 1.94 slug/ft3 and a
specific weight of 62.4 lbf/ft3 for the blood. The diameter of 0.20 in = 0.01667 ft. giving a cross
sectional area of D2/4 = (0.01667 ft)2/4 = 0.000218 ft2. We can compute the velocity from the
flow rate: V = Q/A = (2.5x10-4 ft3/s) / (0.000218 ft2) = 1.146 ft/s.
1.94 slug 1.146 ft
0.01667 ft 
s
VD
ft 3
Re 

 823

4.5 x10 5 lb f  s
ft 2
Thus, the flow is laminar so f = 64/Re = 64/823 = 0.0777. We can now compute the head loss
term.
2
 1.146 ft 


2
10 ft 
V
s
  0.9517 ft
f
 0.0777
32
.
174
ft
D 2g
0.01667 ft
2
2
s
The pressure at the start of the artery is given as equivalent to 0.70 ft of Hg. Multiplying this
height by the specific weight of mercury, Hg = 847 lbf/ft3, from Table 1.5 gives p1 = 593 lbf/ft2.
Substituting this value for p1 and the value of 0.9517 ft found for the head loss into the energy
equation gives.

 V 2  593 lb f 62.4 lb f

 
z1  z 2  0.9517 ft 
p2  p1    z1  z 2  f

D
2
g
ft 2
ft 3


When the head is 8 ft above the heart, z2 – z1 = 8 ft and the value of p2 is
p2 
593 lb f
ft
2

62.4 lb f
ft
3
z1  z 2  0.9517 ft  
593 lb f
ft
2

62.4 lb f
ft
3
 8
ft  0.9517 ft  =
34.5 lb f
ft  0.9517 ft  =
908 lb f
ft 2
When the head is 6 ft below the heart, z1 – z2 = 6 ft and the value of p2 is
p2 
593 lb f
ft 2

62.4 lb f
ft 3
z1  z 2  0.9517 ft  
593 lb f
ft 2

62.4 lb f
ft 3
6
ft 2
In both cases, the head loss (due to frictional effects) is (0.9517 ft) = (0.9517 ft)(62.4 lbf/ft3) =
59.4 lbf/ft2. When the head is above the heart the initial pressure of 593 lbf /ft2 is reduced by 59.4
lbf/ft2 by frictional effects and an additional (8 ft)(62.4lbf/ft2) = 499 lbf/ft2 by the elevation change.
When the head is 6 ft below the heart, the usual pressure change due to the elevation difference
would be an increase of (6 ft)(62.4 lbf/ft2) = 374 lbf/ft2 from 593 lbf/ft2 to 967 lbf/ft2. However the
friction loss decreases the actual pressure gain so that the pressure is only 908 lbf/ft2 in this case.
Exercise nine solutions
3
ME 390, L. S. Caretto, Spring 2008
Page 3
Water flows downward through a vertical 10-mm-diameter galvanized iron pipe with an
average velocity of 5.0 m/s and exits as a free jet. There is a small hole in the pipe 4 m
above the outlet. Will water leak out of the pipe through the hole or will air enter into the
pipe through the hole? Repeat the problem if the average velocity is 0.5 m/s.
To answer this problem we have to find the pressure in the pipe at the location of the hole. If this
pressure is greater than atmospheric pressure (zero gage pressure) then water will flow out the
hole; if it is less, air will flow into the pipe.
We can write the head loss, hL, in the energy equation as p/ = p/g =[ f (l/D) V2/2 ] / (g) =
f(l/D)V2/2g so that the energy can be written to give the following relationship between the initial
location at the hole (1) and the outlet flow at the bottom of the pipe(2).
z2 
p2 V22
p V2
p V2
 V2

 z1  1  1  hL  z1  1  1  hs  f
 2g
 2g
 2g
D 2g
This pipe has no work effect so hs = 0. Since the area is assumed constant for the pipe and the
water has constant density, the velocity at all points in the pipe is the same. Thus we can cancel
the V12 and V22 terms. Furthermore the outlet pressure is a free jet so, p2 = 0. This leaves the
following result for the pressure at the hole, p1.

 V 2 
p1   z 2  z1  f

D 2 g 

We start by computing the Reynolds number for this flow to determine if the flow is laminar or
turbulent. We use the properties for water from Table 1.6:  = 1.12x10-6 ft2/s.
Re 
VD VD




5m
10 mm 1 m
s
1000 mm
1.12 x10
s
6
m
2
 4.45 x10 4
We find the relative roughness for the galvanized iron pipe from Table 8.1:  = 0.15 mm so that
/D = (0.15 mm)/(10 mm) = 0.015. From the Moody diagram, for these values of /D and Re, we
read f = 0.045.
Checking this with the Colebrook equation gives the following value.
 D
1
2.51
 2.0 log 10 


f
 3.7 Re f


2.51
  2.0 log 10  0.015 
  5.78 
 3.7
4

4
.
45
x
10
0
.
045



f  0.0447
Using the friction factor and the elevation difference, z2 – z1 = –4m can now compute p1.
2

 V 2  9800 N 
4 m  5 m  1 s 2  1.86 x105 N

p1   z 2  z1  f

 4 m  0.045




D 2 g 
0.01m  s  2 9.81 m 
m3 
m2

Since p1 is greater than zero water will leak out of the hole .
We have the same calculations if V = 0.5 m/s.
Exercise nine solutions
Re 
ME 390, L. S. Caretto, Spring 2008
VD VD




0.5 m
10 mm 1 m
s
1000 mm
1.12 x10
s
6
m
2
Page 4
 4.45 x103
For this Reynolds number and e/D = 0.015 we find f = 0.052 from the Moody diagram. Checking
this with the Colebrook equation gives.
 D
1
2.51
 2.0 log 10 

 3.7 Re f
f



2.51
  2.0 log 10  0.015 
  5.78 

3

4.45 x10 0.052 
 3.7


 V 2  9800 N
p1   z 2  z1  f



D
2
g
m3


f  0.0524
2

4 m  0.5 m  1 s 2  3.66 x10 4 N
 4 m  0.052



0.01m  s  2 9.81 m 
m2

Since p1 is less than zero air will leak into the pipe .