College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 390
Fluid Mechanics
Spring 2008 Number: 11971 Instructor: Larry Caretto
April 15 Homework Solutions
8.5
Carbon dioxide at 20oC and a pressure of 550 kPa (abs) flows in a pipe at a rate of 0.04 N/s.
Determine the maximum diameter allowed if the flow is to be turbulent.
To ensure turbulent flow, the Reynolds number must be greater than 4000. This gives:
Re 
VD
 4000

We are given the weight flow rate which is the mass flow rate times g. We can use this to get a
relationship between the velocity and the area as follows:

W  m g  VAg  V D 2 g  V 
4
W


4
D2 g
Substituting this expression for velocity into the Reynolds number (and the condition for turbulent
flow) gives.
W

VD
Re 




4
D
2
D g


4 W
 4000 
 Dg
D
1 4 W
W

4000  g 1000g
The dynamic viscosity of carbon dioxide at 20oC and atmospheric pressure is found in Table 1.8
as  = 1.47x10-5 Ns/m2. Since the dyanmic viscosity does not have a significant dependence on
pressure, except for very large pressures, we can use this value at the given pressure of 550
kPa(abs). Substituting this value and the other given data into the inequality for diameter gives
W
D

1000g
8.6
0.04 N
100 cm
s
= 8.83 cm
5
1.47 x10 N  s 9.80655 m m
1000
m2
s2
The pressure distribution measured along a straight horizontal portion of a 50-mmdiameter pipe attached to a tank is shown in the table below. Approximately how long is
the entrance length? In the fully developed portion of the flow, what is the wall shear
stress?
The data in the table below have been used to compute a finite-difference approximation to the
pressure gradient. That is we compute p/x as an approximation to dp/dx according to the
following equation.
Jacaranda (Engineering) 3333
E-mail: lcaretto@csun.edu
Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062
April 15 homework solutions
ME 390, L. S. Caretto, Spring 2008
dp
dx

xi 1 / 2
Page 2
pi 1  pi
xi  t  xi
Locating the value of the estimated gradient at the mid point of the measured pressure data gives
a so-called cnetral difference approximation to the deivative which is more accurate than placing
the gradient at either of the end points. The gradient data are also shown in table below along
with a plot of the data. The plot also shows a straight line fitted through the last few data poitns
and extrpolated back to x = 0..
x (m)
0.00
0.25
0.50
0.75
1.00
1.25
1.50
1.75
2.00
2.25
2.50
2.75
3.00
3.25
3.50
3.75
4.00
4.25
4.50
4.75
5.00
p (mm
H2O)
520
p/x
mm
H2O/m
-186
427
-152
351
-126
288
-104
236
-96
188
-86
145
-72
109
-72
73
-74
36
-72
0
We see that the magnitude of the pressure gradient decreases with distance and eventually
reaches a nearly constant value after x = 3.25 m. In the plotted data the straight line fit appears
to match the data to a point slightly below x = 3.0 m. Thus a reasonable estimate of the entry
length, to within the accuracy of the data is that the entry length is 3 meters .
The relationship between the wall shear stress and the pressure gradient in fully-developed flow
can be used to compute the wall shear stress.
p 
4 w 
p D
 w 
D
 4
The average pressure gradient in the region where the gradient is nearly constant is 72.5 mm
H2O/m. To get this into conventional units we have to multiply by the specific weight of water.
Thus, 1 mm H2O = (0.001 m)(9800 N/m 3) = 9.80 N/m2. With this conversion factor we can find
the wall shear stress as follows
w 
9.80 N
50 mm
p D 72.25 mm H 2 O
m

 8.88 N/m2
2
 4
m
m  mm H 2 O  4 1000 mm
April 15 homework solutions
8.15
ME 390, L. S. Caretto, Spring 2008
Page 3
A fluid of density  = 1000 kg/m3 and viscosity  = 0.30 Ns/m2 flows steadily down a
vertical 0.10-m-diameter pipe and exits as a free jet from the lower end. Determine the
maximum pressure allowed in the pipe at a location 10 m above the pipe exit if the flow is
to be laminar.
The maximum pressure allowed in the pipe will be the pressure that produces the largest flow
rate that remains laminar. Since laminar flow requires the Reynolds number to be less than
2100, we have to satisfy the following inequaltiy.
0.30 N  s 1 kg  m
VD

m2
N  s2
Re 
 2100  V  2100
 2100
1000 kg  m

D

m3
0.10 m 

6.30 m
s
So we have the result that V < 6.30 m/s; we need to connect this result to the presure. The
relationship for pressure drop in laminar flow in a pipe that makes an angle  with the horizontal is
given by equation 8.11 on page 412 of the text, where p is defined as pin – pout.

p   sin  D 2
V
32

 pin  pout   sin  D 2
32
Setting the inequality that V < 6.30 m/s and solving for the inlet pressure gives.
V
 pin  pout   sin  D 2
32

6.30 m

s
pin 
6.30 m 32
 pout   sin 
s
D2
Since the exit is a free jet, pout = 0. The coordinate system for  takes = 0 when the flow is in a
horizontal direction to the right. A pipe that has a vertical upflow would have  = 90o; a pipe, like
the one here, that has a vertical downflow has  = –90o. and sin() = –1. We are given the
density of the fluid as 1000 kg/m 3, so its specific weight,  = g = (1000 kg/m3)(9.80655 m/s2)
(1 Ns2/kgm) = 9807 N/m3.
pin 
6.30 m 32
6.30 m 3210 m 0.30 N  s
9807 N
10 m 1
 pout   sin  
0
2
2
2
s
s 0.10 m
D
m
m3
pin < –3.759x104 Pa = –37.6 kPa
April 15 homework solutions
8.20
ME 390, L. S. Caretto, Spring 2008
Page 4
Oil flows through the horizontal pipe
shown in Figure P8.20 (copied from
the text at the right) under laminar
conditions. All sections of the pipe
are the same except one. Which
section of the pipe (A, B, C, D, or E)
is slightly smaller diameter than the
others? Explain.
For laminar flow in a horizontal pipe the
pressure drop is related to the diameter
by the following equation.
p 
128Q
D 4

p 128Q


D 4
This equation tells us that the pressure gradient, p/l, varies as D-4 if all other terms in the
equation are constant. If we assume that the temperature in the pipe is constant, the viscosity
and density in the pipe will be constant, and the constant density means that the volume flow
rate, Q, will be constant, even if the diameter changes, according to the continuity equation
We can compute the gradient in head measured by the piezometer tubes as a measure of the
pressure gradient. This is shown in the table below.
Section
Distance, x (ft)
Head, h (in)
h/x
A
15
60
B
25
56
-0.4
C
50
46
-0.4
D
66
39
-0.438
E
96
26
-0.433
We see that the gradient is the same (-0.4 in/ft) for the differences between section A and B and
section B and C; the first change occurs in section D, where the magnitude of the gradient
increases. This indicates that section D is the one with the smaller diameter. To confirm this we
can compute the gradient in section E to see if it is the same as the previous gradients, -0.4 in/ft.
To compute the gradient in section E, we have to start at the measurement point in section C,
where the total distance, x = 50 ft. The distance to the end of section C (which the same as the
start of section D) is 10 ft. (We know this because we are told that each section is 20 ft long.)
Thus with the gradient of -0.4 in/ft, the head at the start of section D would be 46 in – (0.4in/ft)(10
ft) = 42 in. We can then compute the gradient in D because we know that the head is 39 in over
a 6 ft length of section D. This gives the gradient in section D as (39 in – 42 in)/(6 ft) = -0.5 in/ft.
This gradient would reduce the head over the remaining 14 ft section D to 39 in – (0.5 in/ft)(14 ft)
= 32 in. With this head at the end of section D, which is the same as the start of section E, we
can compute the gradient in section E from the measured head of 26 in, 6 ft into section E as (26
in – 32 in)/(6 ft) = -0.4 in/ft, the same as at the start of the pipe. This confirms our preliminary
conclusion that section D has the smaller diameter .
April 15 homework solutions
ME 390, L. S. Caretto, Spring 2008
Oil of SG = 0.78 and a kinematic viscosity,  = 2.2x10-4
m2/s flows through the vertical pipe shown in Figure
P8.22 (copied at the right) at a rate of 4x10-4 m3/s.
Determine the manometer reading, h.
8.22
Page 5
1
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 downstream point (2) and the
upstream point (1).
h1
V22
p1 V12
p1 V12
 V2
z2 

 z1 

 hL  z1 

f
 2g
 2g
 2g
D 2g
p2
2
h2
We have to compute the Reynolds number as the first
step in determining the friction factotr. For this pipe D =
20 mm = 0.02 m and the area = D2/4 = (0.02 m)2 =
0.001257 m2. The velocity is found from the volume flow
rate by the usual equation and then used to compute the Reynolds number.
4 x10 4 m3
Q
1.273 m
s
V 

2
A 0.01257 m
s
VD VD
Re 




1.237 m
0.02 m
s
 115.7
2.2 x10  4 m 2
s
Since Re < 2,100, the flow is laminar and we find the friction factor as 64/Re. With this friction
factor, we can find the pressure difference from the energy equation after cancelling the equal
velocity terms.
2
 1.237 m 


2
p2  p1
64  V
64 4 m 
s
  5.14 m
 z1  z2 
4m
9
.
80665
m

Re D 2 g
115.7 0.02 m
s2
The manometer reading can be analyzed by starting with the lower level on the right of the
manometer. At this point, the pressures on both sides of the mamometer are the same and can
be equated to pressures in the pipe by the manometer formula.
pright  p1   oil h1  h   pleft  p2   oil h2   manoh 
p2  p1
 oil
 h1  h2  h 
 mano
h
 oil
From the diagram we see that the sum h1 + h2 is the length of 4 m between the pressure taps.
Furthermore, the ratio of the specific weights of the oil and manometer fluid can also be written as
the ratio of the specific gravity of each fluid. Thus we can solve for the unknown height, h, as
follows.
April 15 homework solutions
h
8.29
ME 390, L. S. Caretto, Spring 2008
h1  h2 
p2  p1
 oil
  mano 

 1
  oil


Page 6
4 m   5.14 m 
= 18.5 m
1.3
1
.87
Air at standard conditions flows through an 8-in-diameter, 14.6-ft-long, straight duct with
the velocity versus pressure drop data indicated in the table below. Determine the
average friction factor over this range of data.
The usual equation for pressure drop in terms of friction factor can be solved for the friction factor
as follows.
p  f
 V 2
D 2

f 
2 Dp
V 2
The data shown in the table below have the velocity in ft/min and pressure drop in inches of
water. To get the pressure drop in lbf/ft2 we have to multiply the pressure drop in inches of water
by the specific weight of water. Doing this and inserting the necessary conversion factors gives
the following calculation formula for the friction factor. (The density of air at standard conditions,
0.00238 slug/ft3, is taken from Table 1.7.)
2 Dp
f 

V 2
28 in 
62.4 lb f
ft 3
h in H 2O 
ft 2 slug  ft
144 in 2 1 lb f  s 2
14.6 ft  0.002383 slug  V ft min 
ft

 min 60 s 
2

718315h in H 2O 
 V ft 


 min 
2
Applying this computational equation to all the data in the table below gives a set of data for
friction factors from which we can compute the average value.
faverage = 0.0162
V ft/min
3950
3730
3610
3430
3280
3000
2700
p (in H2O)
0.35
0.32
0.30
0.27
0.24
0.20
0.16
Average f
f
0.0161
0.0165
0.0165
0.0165
0.0160
0.0160
0.0158
0.0162