College of Engineering and Computer Science
Mechanical Engineering Department
Mechanical Engineering 390
Fluid Mechanics
Spring 2008 Number: 11971 Instructor: Larry Caretto
February 26 Homework Solutions
3.23
A differential pressure gage attached to a Pitot-static tube (See Video 3.4) is calibrated to
give speed rather than the difference between stagnation and static pressures. The
calibration is done so that the speed indicated on the gage is the actual fluid speed if the
fluid flowing past the Pitot-static tube is air at standard sea level conditions. Assume the
same device is used in water and the gage indicates a speed of 200 knots. Determine the
water speed.
A Pitot-static tube measures the difference between the static pressure, ps, where the fluid has a
velocity, V, and the stagnation pressure, p0 = ps + V2/2, which occurs when the original velocity,
V, is reduced to zero.. (See the discussion of the Pitot-static tube in the text and lecture
presentations for more information.) We can obtain the velocity, V, from the difference in static
and stagnation pressures by solving the equation p0 = ps + V2/2 for V.
In the problem given here, the gage has been calibrated to translate the pressure difference
directly into speed when the fluid is air. When we have a speed reading of 200 knots in any fluid
the Pitot-static tube thinks that it is in air so the pressure difference measured by the Pitot-static
tube can be found from the following formula:
p0  p s 
 airVair2
2

 air 200 knots2
2
When we put the Pitot-static tube in another fluid and measure this velocity we are actually
measuring the pressure diference that the Pitot-static tube would show in air at that velocity. The
other fluid, water in this problem, with a different density, will give the same Pitot-static tube
pressure difference if it has the following velocity relationship.
p0  p s 
2
 water Vwater
 200 knots2
 air
2
2
So the velocity of water is given by the following formula where we use the tabulated values for
the standard densities of air and water.
2
Vwater

 air 200 knots2
 water
 Vwater  200 knots
 air
0.00238 slugs ft 3
 200 knots
 water
1.94 slugs ft 3
Vwater = 7.01 knots
3.24
When an airplane is flying 200 mph at 5000-ft altitude in a standard atmosphere, the air
velocity at a certain point on the wing is
273 mph relative to the airplane. What
suction pressure is developed on the wing
at that point? What is the pressure at the
leading edge (a stagnation point) of the
wing?
Jacaranda (Engineering) 3333
E-mail: lcaretto@csun.edu
Mail Code
8348
Phone: 818.677.6448
Fax: 818.677.7062
February 26 homework solutions
ME 390, L. S. Caretto, Spring 2008
Page 2
The diagram above (taken from the solutions manual for the text) shows the flow along two
streamlines, both of which start far from the airplane where the velocity, relative to the airplane is
200 mph and the gage pressure can be taken as zero. At an elevation of 5000 ft the density in
the standard atmosphere (see Table C-1 in Appendix C on page 764) is 0.002048 slugs/ft3. The
Bernoulli equation can be written in the following form
V2
gz  
 C  Constant along a streamline
 2
p
Although there may be slight elevation differences in the wing, they will have a negligible impact
on the force-momentum balance because of the low specific weight of air. Consequently we can
write that P/ + V2/2 = C along a streamline. Before starting the calculations we can convert the
velocities given in the problem from mph to ft/s using the conversion factor that 30 mph = 44 ft/s.
This gives the velocity far from the airplane as 293 ft/s and the velocity over the airplane wing as
400 ft/s.
We first consider the streamline that starts far from the airplane and passes over the wing at a
point where the velocity is 273 mph = 400 ft/s. The Bernoulli equation with no effect of elevation
gives us
p1


V12 p3 V32


2

2

p3  p1  
V12  V32
2
Substituting data for the velocities and the definition of the reference gage pressure as p1 = 0
gives
p3  p1  
2
2
V12  V32
1 0.002048 slugs  293 ft   400 ft  
 0


 
 
2
2
ft 3
 s   s  
p3 = –70.6 lbf/ft2
Applying the Bernoulli equation along a stagnation streamline to point 2, where V 2= 0 gives
p1 V12 p 2 V22




2

2

p 2  p1  
V12  0
1 0.002048 slugs  293 ft 
 0


2
2
 s 
ft 3
2
p2 = 88.0 lbf/ft2
Both p3 and p2 are gage pressures relative to a pressure far from the airplane at the same
elevation as the airplane.
3.25
Water flows steadily downward through the pipe shown in the
figure at the right. Viscous effects are negligible, and the
pressure gage indicates the pressure is zero at point (1).
Determine the pressure and flow rate at point (2).
To solve this problem we have to apply the Bernoulli equation (and the
continuity equation) two times. First we apply it to points one and the
exit, which we will label as point 3. Since the exit is a free jet we know
that it has zero gage pressure. We also know the gage pressure at
point 1 is zero. Thus, we can use these pressure data to find the flow
rate in the pipe. Once, we know the flow rate, we can apply the
Bernoulli equation and the continuity equation between points (1) and
(3)
February 26 homework solutions
ME 390, L. S. Caretto, Spring 2008
Page 3
(2) to get the desired pressure at point (2). Applying the Bernoulli equation between point (1) and
the exit (3) gives.
gz1 
p
V2
p1 V12

 gz 3  3  3

2

2
The two velocities are related by the continuity equation: V1A1 = V3A3. This equation can be used
to eliminate V1. Doing this and substituting z2 – z3= 3 ft as shown in the diagram, p3 = 0 because
it is open to the atmosphere, and p1 = 0 as given in the problem statement gives the following
result.
g  z1  z 3  
p1  p3 V12  V32
0  0 V32 A32 / A12  V32

 g 3 ft  

0

2

2
We can solve the resulting equation for V3, using the fact that the area ratio is the diameter ratio
squared in the intermediate step below.
V3 
2 g 3 ft 
1
A32
/
A12

2 g 3 ft 
1
D34
/ D14
2

32.174 ft
s2
3 ft 
 0.1 ft 

1  
 0.12 ft 
4

19.3 ft
s
We can now compute the flow rate as Q = V3A3 From the continuity equation we know that his
flow rate will be the same throughout the pipe and gives the desired result for Q 2.
0.152 ft 3
19.3 ft 
2
0.1 ft  
Q  Q1  Q2  Q3  V3 A3 
s
s 4
To get the pressure at point (2) we can apply the Bernoulli equation between points (1) and (2).
Since the flow area is the same at these points we must have V1 = V2; we know that p1 = 0 and,
from the diagram, we see that z1 – z2 = –2 ft. This gives the following result for the Bernoulli
equation.
z1  z 2  
p1  p 2 V12  V22
0  p2

 2 ft  1
00 
g
2g

p 2   2 ft 
Using the specific weight of water as g = 62.4 lbf/ft3 gives the desired pressure: p2 = [(62.4
lbf)/ft3]·(-2 ft), so p2=–125 lbf/ft2 .
3.32
Water flows through the pipe contraction shown in
Figure P3.32 at the right. For the given 0.2-m
difference in manometer level, determine the flow
rate as a function of the diameter of the small pipe,
D.
Apply the Bernoulli equation for incompressible,
inviscid flows, shown below, between two points (1)
and (2) along a streamline in the center of the pipe.
z 2  z1  
(1)


p 2  p1 V22  V12

0
g
2g
(2)
February 26 homework solutions
ME 390, L. S. Caretto, Spring 2008
Page 4
The diagram shows that the measurement tube at point 2 is facing the flow; this means that it will
measure the stagnation pressure. The Piezometer tube at point 1 is measuring the static
pressure at that point. The measured difference h (= 0.2 m) times the specific weight of the fluid
is the difference between the stagnation pressure at point (2) (p2 + V22/2) and the static
pressure, p1 at point (1). This gives the following interpretation of the height difference.
h  gh  p2 
V22
 p1 
2
p1  p2 
V22
 gh
2
We can substitute this expression for p1 into our Bernoulli equation along with the fact that at the
center of the pipe z2 = z1. This gives the Bernoulli equation as
z 2  z1  

p 2  p1 V22  V12

g
2g



V 2
p 2   p 2  2  gh 

 V 2 V 2
2

 2
1
 0
0
g
2g


We see that the terms in p2 and V22 cancel leaving the following result.
h
V12
0
2g

V1  2 gh
The flow rate is simply V1A1 = V1D12/4.
Q  A1V1  A1
0.01556 m 3
9.81 m
 2

2
2 gh  D1 2 gh  0.1 m  20.2 m  2 
s
4
4
s
From the equation for Q, we see that the flow rate does not depend on the smaller
diameter D.
3.33
The speed of an airplane through the air is obtained by use of a Pitot-static tube that
measures the difference between the stagnation and static pressures. (See Video 3.4.)
Rather than indicating this pressure difference (psi or N/m 2) directly, the indicator is
calibrated in speed (mph or knots). This calibration is done using the density of standard
sea level air. Thus the air speed displayed (termed the indicated air speed) is the actual air
speed only at standard sea level conditions. If the aircraft is flying at an altitude of 20,000
ft and the indicated air speed is 220 knots, what is the actual air speed?
A Pitot-static tube measures the difference between the static pressure, ps, where the fluid has a
velocity, V, and the stagnation pressure, p0 = ps + V2/2, which occurs when the original velocity,
V, is reduced to zero.. (See the discussion of the Pitot-static tube in the text and lecture
presentations for more information.) We can obtain the velocity, V, from the difference in static
and stagnation pressures by solving the equation p0 = ps + V2/2 for V.
In the problem given here, the Pitot-static has been calibrated to translate the pressure difference
directly into speed when the fluid is air at standard sea-level conditions. When we have a speed
reading of 220 knots at an altitude of 20,000 ft., the pressure difference in the Pitot-static tube is
pressure difference it would see if it were travelling at that speed in sea level air. That pressure
difference is given by the following equation:
p0  p s 
2
 sea level Vindicated
2

 sea level 220 knots2
2
February 26 homework solutions
ME 390, L. S. Caretto, Spring 2008
Page 5
At 20,000 ft, that same pressure difference is generated at a different density and speed. The
pressure difference can then be written for both the actual and indicated speeds as follows.
p0  p s 
2
 20,000 ft Vactual
2

2
 sea level Vindicated
2
 sea level 220 knots2

2
So the actual velocity at velocity at 20,000 ft is given by the following formula.
 sea level
Vactual  Vindicated
 20,000
ft
We can find the air density at sea level and an elevation of 20,000 ft for the standard atmosphere
from Table C-1 in Appendix C on page 764: sea level = 0.00238 slugs/ft3 and 20,000 ftl = 0.001267
slugs/ft3. So, the actual air speed for an indicated air speed of 220 knots is found as follows.
Vactual  Vindicated
 sea level
 20,000
 220 knots
ft
0.00238 slugs ft 3
0.001267 slugs ft 3
Vactual = 302 knots
3.39
Water is siphoned from the tank shown in Figure
P3.39. The water barometer indicates a reading
of 30.2 ft. Determine the maximum value of h
allowed without cavitation occurring. Note that
the pressure of the vapor in the closed end of
the barometer equals the vapor pressure.
(2)
In order to avoid cavitation the pressure at the top of
the siphon tube must be greater than the vapor
pressure. We have to get a relationship for the
(1)
pressure at this point.
For this problem we will work with absolute
pressures because the vapor pressure is an
absolute pressure. The pressure, p1, is the
(absolute) atmospheric pressure, patm. This is also
the pressure at point (3). From the barometer
equation we have the following value for the
atmospheric pressure.
(3)
patm   water hbaro  pv
hbaro  30.2 m
We can use the Bernoulli equation for incompressible, inviscid flows, applied between the water
level in the tank, point (1), and the exit, point (3), to obtain an equation for the velocity in the
siphon.


p1  p3 V12  V32
g  z1  z 3  

0

2
We are not given the vapor pressure or any temperature data from which we can determine the
vapor pressure.
February 26 homework solutions
ME 390, L. S. Caretto, Spring 2008
Page 6
In applying the Bernoulli equation between the liquid level at point (1) and the exit at point (3), we
have p1 = p3 = patm, and z1 – z3 = h. Because the tank is large we assume that V12 << V22 and can
be neglected. This gives the follow results for the Bernoulli equation.
g z1  z 3  




p1  p3 V12  V32
p p
0  V32

 gh  atm atm 
0

2

2
This gives the simple relationship between the velocity V 3 and the height h.
V3  2 gh
The velocity at point (2), the top of the siphon, is related to V3 by the continuity equation. In the
last step below we use the fact that the area ratio for a circular pipe is the diameter ratio squared.
V2 A2  V3 A3  A3 2 gh
 V2 
A3 2 gh D32 2 gh

A2
D22
We can now apply the Bernoulli equation between the liquid surface in the tank, point (1) and the
top of the siphon, point (3) to determine the pressure there. Here we choose to write the equation
using the “head” terms that have dimensions of length.
z1  z 2  


p1  p 2 V12  V22

0

2g
We can solve this equation for p2 using (i) the previous assumption that V1 is negligible, (ii) the
equation derived above for V2 in terms of the height, h, and (iii) the height difference, z2 – z1 = 6 ft
from the diagram.
p 2   z1  z 2   p1  
V
2
1

2 ghD34 
 V22
 
  z1  z 2   p atm 
0
2g
2 g 
D24 
Simplifying this equation and using the result of the barometer equation that
patm  hbaro  pv
we can obtain the following inequality expressing the fact that the pressure at point (2) must be
greater than the vapor pressure to avoid cavitation.
p 2    z1  z 2   hbaro  p v 
hD34
D24
 pv
The pv terms cancel and we have the following inequalities for the height, h. The second is
obtained by dividing the first by –, which changed the direction of the inequality

D34
D24
h    z 2  z1   hbaro
D34

D24
h  hbaro   z 2  z1 
We can now solve for the height, h and enter the data to obtain the maximum value for h.
h
D24
D34
hbaro
4
 3 in 
 30.2 ft  6 ft   3.14 ft
 z 2  z1   
 5 in 
So there restriction to avoid cavitation is that h < 3.14 ft .