FLIGHT MECHANICS
Exercise Problems
CHAPTER 4
Problem 4.1
• Consider the incompressible flow of water
through a divergent duct. The inlet velocity
and area are 5 ft/s and 10 ft2, respectively.
If the exit area is 4 times the inlet area,
calculate the water flow velocity at the exit.
Solution 4.1
m  1 A1V1   2 A2V2
A1
1
V2  V1
 5  1.25 ft / s
A2
4
Problem 4.2
• 4.2 In the above problem calculate the
pressure difference between the exit and
the inlet. The density of water is 62.4
Ibm/ft3.
Solution 4.2
p2
v2
p1
v1
 dp    VdV  0
 V12  V22 

p2  p1   
2 

62.4

 1.94 slug / ft 3
32.2
 52  1.252 
  22.7lb / ft 2
p2  p1  1.94
2


Problem 4.5
Consider the flow of air through a
convergent-divergent duct, such as the
venturi described in Prob. 4.4. The inlet,
throat, and exit areas are 3, 1.5, and 2 m2
respectively. The inlet and exit pressures
are 1.02 x 105 and 1.00 x 105 N/m2,
respectively. Calculate the flow velocity at
the throat. Assume incompressible flow
with standard sea-level density.
Solution 4.5
V32
V12
p1  
 p3  
2
2
A
V3  V1 1
A3
V1 
2( p1 p3 )
 A  2 
  1   1
 A3 

A1
3
V2  V1 
A2
1.5
Note that only a
pressure change of
0.02 atm produce this
high speed
2(1.02  1.00)105
 102.22m / s
2
 3 

1.225   1
 2 

Problem 4.6
An airplane is flying at a velocity of 130
mi/h at a standard altitude of 5000 ft. At a
point on the wing, the pressure is 1750.0
Ib/ft2. Calculate the velocity at that point
assuming incompressible flow.
Solution 4.6
88
V1  130mph  130
 197.7 ft / s
60
p1  V12  p2  V22
V 
2
2
2 p1  p2 

V2  216.8 ft / s
21760.9  1750 
4
V 
 190.7
0.0020482
2
1
Problem 4.7
Imagine that you have designed a low-speed
airplane with a maximum velocity at sea level of
90 m/s. For your airspeed instrument, you plan
to use a venturi tube with a 1.3 : 1 area ratio.
Inside the cockpit is an airspeed indicator—a
dial that is connected to a pressure gauge
sensing the venturi tube pressure difference p1p2 and properly calibrated in terms of velocity.
What is the maximum pressure difference you
would expect the gauge to experience?
Solution 4.7
V12
V22
p1  
 p2  
2
2
A1
V2  V1
A2
Maximum when maximum
velocity 90 m/s and sea level
density; however better
design for over speed during
diving
2


V  A1 
   1
p1  p2  
2  A2 


90 2
2
1.3  1  3423N / m 2
p1  p2  1.225
2
2
1


Problem 4.9
Derive an expression for the exit velocity of
a supersonic nozzle in terms of the
pressure ratio between the reservoir and
exit po/pe and the reservoir temperature To.
Solution 4.9
1
1
c pTo  V02  c pTe  Ve2
2
2
1
ho  he  Ve2
2
Ve  2c p (T0  Te )
Te  pe 
  
T0  p0 
 1


 pe 

Ve  2c pT0 1   
  p0 

 1





Note that the velocity
increases as To goes up
or pressure ratio goes
down; used for rocket
engine performance
analysis
Problem 4.11
The mass flow of air through a supersonic
nozzle is 1.5 Ibm/s. The exit velocity is
1500 ft/s, and the reservoir temperature
and pressure are 1000°R and 7 atm,
respectively. Calculate the area of the
nozzle exit. For air, Cp = 6000 ft •
lb/(slug)(°R).
Solution 4.11
1
c pT0  c pTe  Ve2
Energy eq.
2
Ve2
1500 2
Te  T0 
 1000 
 812.5 R
2c p
26000 
0 
7 2116  0.0086
p0

RT0 1716 1000 
 Te 

 To 
 e   0 
1
 1
 812.5 
 0.0086

1000


1
1.4 1
 0.0051
1.5
Continuity eq.
  e AeVe
32.2
m
1.5
Ae 

 0.0061 ft 2
 eVe 32.2 0.00511500
m 
No shock wave,
isentropic
relationship
Problem 4.12
A supersonic transport is flying at a
velocity of 1500 mi/h at a standard altitude
of 50,000 ft. The temperature at a point in
the flow over the wing is 793.32°R.
Calculate the flow velocity at that point.
Solution 4.12
1 2
1 2
c pT1  V1  c pT2  V2
2
2
V22  2c p T1  T2   V12
 88 
V1  1500mi / h  1500  ft / s  2200 ft / s
 60 
V22  26000389.99  7993.32  2200 2
V2  6.3 ft / s
Very low value, almost a stagnant point
Problem 4.14
Calculate the Mach number at the exit of
the nozzle in Prob. 4.11.
Solution 4.14
Ve  1500 ft / s
Te  812.5 R
ae  RTe 
1.41716812.5  1397 ft / s
Ve 1500
Me 

 1.07
ae 1397
Problem 4.17
Calculate the flight Mach number for the
supersonic transport in Prob. 4.12.
Solution 4.17
V  2200 ft / s
a  RT 
1.41716389.99  967.94 ft / s
V
2200
M 
 2.27
a 967.94
Problem 4.18
Consider a low-speed subsonic wind
tunnel with a nozzle contraction ratio of 1 :
20. One side of a mercury manometer is
connected to the settling chamber, and the
other side to the test section. The
pressure and temperature in the test
section are 1 atm and 300 K, respectively.
What is the height difference between the
two columns of mercury when the test
section velocity is 80 m/s?
Solution 4.18
p
1.01*105


 1.173kg / m 3
RT 287 300 
2

 A2  
V
1    
p1  p2  
2   A1  


p1  p2  h  1.33 *105 h
2
2
Manometer reading
2
2
2 


 A2 
V
1.173 80
 1  
1     
h 
1    
5
 2   A1   1.33 *10 2   20  


 0.028m  2.8cm
2
2
Problem 4.19
We wish to operate a low-speed subsonic wind tunnel so
that the flow in the test section has a velocity of 200 mi/h
at standard sea-level conditions. Consider two different
types of wind tunnels: (a) a nozzle and a constant-area
test section, where the flow at the exit of the test section
simply dumps out to the surrounding atmosphere, that
is, there is no diffuser, and (b) a conventional
arrangement of nozzle, test section, and diffuser, where
the flow at the exit of the diffuser dumps out to the
surrounding atmosphere. For both wind tunnels (a) and
(b) calculate the pressure differences across the entire
wind tunnel required to operate them so as to have the
given flow conditions in the test section.
For tunnel (a) the cross-sectional area of
the entrance is 20 ft2, and the crosssectional area of the test section is 4 ft2.
For tunnel (b) a diffuser is added to (a)
with a diffuser area of 18 ft2. After
completing your calculations, examine
and compare your answers for tunnels (a)
and (b). Which requires the smaller
overall pressure difference? What does
this say about the value of a diffuser
on a subsonic wind tunnel?
Solution 4.19 (a)
V12
V22
p1  
 p2  
2
2
A2
V1  V2
A1
2

 A2  
V
1    
p1  p2  
2   A1  


2
2
293.32
p1  p2  0.002377
2
  4 2 
2
1      98.15lb / ft
  20  
Solution 4.19 (b)
V32
V12
p1  
 p3  
2
2
A
A
V1  V2 2 , V3  V2 2
A1
A3
2
2





V
A2
A2 
     
p1  p3  
2  A3   A1  


2
2
293.32
p1  p2  0.002377
2
 4  2  4  2 
2
       0.959lb / ft
 18   20  
Economical to use diffuser (running compressor
or vacuum pump)
Problem 4.20
A Pitot tube is mounted in the test section
of a low-speed subsonic wind tunnel. The
flow in the test section has a velocity,
static pressure, and temperature of 150
mi/h, 1 atm, and 70°F, respectively.
Calculate the pressure measured by the
Pitot tube.
Solution 4.20
p
2116


 0.00233slug / ft 3
RT 171670  460 
p0  p 
V 2
2
2
0.00233 
88 
p0  2116 
150 * 
2
60 

0.00233
2202  2172lb / ft 2
p0  2116 
2
Problem 4.22
The altimeter on a low-speed airplane
reads 2 km. The airspeed indicator reads
50 m/s. If the outside air temperature is
280 K, what is the true velocity of the
airplane?
Solution 4.22
p
7.95 *10 4


 0.989kg / m 3
RT 287 280 
Vtrue

Veq
0

1.225
Vtrue  50
 56m / s
0.989
Problem 4.23
A Pitot tube is mounted in the test section
of a high-speed subsonic wind tunnel.
The pressure and temperature of the
airflow are 1 atm and 270 K, respectively.
If the flow velocity is 250 m/s, what is the
pressure measured by the Pitot tube?
Solution 4.23
a  RT  1.4 * 287 * 270  329m / s
V 250
M 
 0.76
a 329

p0  (  1) M 2   1

 1 
p 
2

1 .4
1.4 1
 (1.4  1)0.76 

 1 
 1.47
2


p0  1.47 p  1.47 * 1.01*105  1.48 *105
2


Problem 4.24
A high-speed subsonic Boeing 777
airliner is flying at a pressure altitude of
12 km. A Pitot tube on the vertical tail
measures a pressure of 2.96 x 104 N/m2.
At what Mach number is the airplane
flying?
Solution 4.24
p  1.94 *10 4
 1





2  p0
2
   1
M1 

  1  p1 


1.4 1


4
1
.
4
2  2.96 *10 





1

1.4  1  1.94 *10 4 


note; at altitude 12 km, p  1.94 *10 4 N/m 2
M 1  0.801
Problem 4.25
A high-speed subsonic airplane is flying at
Mach 0.65. A Pitot tube on the wing tip
measures a pressure of 2339 Ib/ft2. What
is the altitude reading on the altimeter?
Solution 4.25

p0  (  1) M 2   1

 1 
p 
2

1 .4
1.4 1
 (1.4  1)0.652 

 1 
 1.328
2


p0
2339
p

 1761
1.328 1.328
Appendix B, pressure altitude reads 5000 ft
Problem 4.27
An airplane is flying at a pressure altitude
of 10 km with a velocity of 596 m/s.
The outside air temperature is 220 K.
What is the pressure measured by a Pitot
tube mounted on the nose of the airplane?
Solution 4.27
a1  RT  1.4 * 287 * 220  297m / s
M1 
V1 596

 2.0
a1 297
Use Rayleigh Pitot tube formula

p02  (  1) 2 M 12   1 1    2M 12

 
2
p1  4M 1  2(  1) 
 1


(1.4  1) 2 2 2

 
2
 4 *1.4 * 2  2(1.4  1) 
as p1  2.65 *10 4
1.4
1.4 1
1  1.4  2 *1.4 * 2 2
 5.64
1.4  1
p02  5.64 * 2.65 *10 4  1.49 *105 N / m 2
Problem 4.28
The dynamic pressure is defined as q =
0.5V2. For high-speed flows, where
Mach number is used frequently, it is
convenient to express q in terms of
pressure p and Mach number M rather
than  and V. Derive an equation for q =
q(p,M).
Solution 4.28
as
so
1
1 p
p    2
2
2
q  V 
V   V
2
2 p
2  p 

dp
d
c

p
2
 1
a 

 c 
d
d

 
p    2 p  V 2  p 2
q   V   2   M
2  p 
2 a  2
Problem 4.29
After completing its mission in orbit around
the earth, the Space Shuttle enters the
earth's atmosphere at very high Mach
number and, under the influence of
aerodynamic drag, slows as it penetrates
more deeply into the atmosphere. (These
matters are discussed in Chap. 8.) During
its atmospheric entry, assume that the
shuttle is flying at Mach number M
corresponding to the altitudes h:
h,
km
M
60
50
40
30
20
17
9.5
5.5
3
1
Calculate the corresponding values of the
freestream dynamic pressure at each one
of these flight path points. Suggestion:
Use the result from Prob. 4.28. Examine
and comment on the variation of q∞ as the
shuttle enters the atmosphere.
Solution 4.29
h, km
60
50
40
30
20
p∞
25.6
87.9
299.8
1.19*103
5.53*103
M
17
9.5
5.5
3
1
q∞
5.2*103
5.6*103
6.3*103
7.5*103
3.9*103
q
p
2
M
2
Problem 4.30
Consider a Mach 2 airstream at standard
sea-level conditions. Calculate the total
pressure of this flow. Compare this result
with (a) the stagnation pressure that would
exist at the nose of a blunt body in the flow
and (b) the erroneous result given by
Bernoulli's equation, which of course does
not apply here.
Solution 4.30
Total pressure when the flow is isentropically stopped (true for
supersonic and subsonic)

p0  (  1) M   1  (1.4  1)2
  1 
 1 
p 
2
2


p0  7.824 p  7.8242116  16560
2
2



1.4
1.4 1
 7.824
But there must be a shockwave at the nose (at the stagnation point)

  1 1    2M 12
p02  (  1) M

 
2
p1  4M 1  2(  1) 
 1
2
2
1
1.4
1.4 1


(1.4  1) 2
1  1.4  2 *1.4 * 2 2

 
 5.64
2
1.4  1
 4 *1.4 * 2  2(1.4  1) 
p02  5.64 * 2.116  1.193 *10 4 lb / ft 2
2
2
If Bernoulli’s equation is used accidentally
p0  p 
V 2
 p 
p M  2
2
2
1.4
p0  2.116 
* 2.116 * 2 2  0.804 *10 4 lb / ft 2
2
51% error
Problem 4.31
Consider the flow of air through a
supersonic nozzle. The reservoir pressure
and temperature are 5 atm and 500 K,
respectively. If the Mach number at the
nozzle exit is 3, calculate the exit
pressure, temperature, and density.
Solution 4.31
 (  1) M e
pe  p0 1 
2

2





 1
1
 (1.4  1)32 

 5 * 1.01*10 1 
2



5

 (  1) M e 
  500 * 0.357  178.6 K
Te  T0 1 

2


p0
1.37 *10 4
0 

 0.267 kg / m 3
RT0 287 178.6 
2
1.4
1.4 1
 1.37 *10 4
Problem 4.32
• Consider a supersonic nozzle across
which the pressure ratio is pe/po = 0.2.
Calculate the ratio of exit area to throat
area.
Solution 4.32
pe  (  1) M e2 

 1 
p0 
2


 1
1







pe
2
 0.286
2




Me 

1

5
0
.
2
 1  2.92
 

(  1)   p0 



M e  1.71

Ae

At
1
1.712
1
2
Me

 2    1 2 
   1 1  2 M e  



 2  1. 4  1
2 
1

1
.
71

1.4  1 
2



 1
 1
1.4 1
1.4 1

 1.35
Problem 4.33
• Consider the expansion of air through a
convergent-divergent supersonic nozzle. The
Mach number varies from essentially zero in the
reservoir to Mach 2.0 at the exit. Plot on graph
paper the variation of the ratio of dynamic
pressure to total pressure as a function of Mach
number; that is, plot q/ po versus M from M = 0
to M = 2.0.
Solution 4.33
1
p  V 2  p 2
2
q  V   2   M
2
2 a  2
q M  p  M    1 2 

 

M 
1 
p
2  p 
2 
2

3.5
q
 0.7 M 2 1  0.2M 2
p
2
2


 1

The graph shows that the local dynamic
pressure has a peak value at M=1.4
Problem 4.35
In Prob. 4.34, assume the flow is
completely turbulent. Calculate the
boundary layer thickness at the trailing
edge and the total skin friction drag.
Compare these turbulent results with the
above laminar results.
Solution 4.35
0.37 L
0.37 * 3

0.033m  3.3cm
0.2
7 0 .2
Re L
4.10 *10
 turb 3.3

 13.75
 lar 0.24

Cf 


0.0074
0.0074

0.2
Re L
4.10 *107


0 .2
 0.0022
D f  q SC f  2.45 *10 4 * 3 *17.5 * 0.0022  2830 N
top and bottom
D f  2 * 2830 N  5660 N
10.5 times larger than laminar flow
assumption
Problem 4.36
• If the critical Reynolds number for
transition is 106, calculate the skin friction
drag for the wing in Prob. 4.34.
Laminar Flow A
Xcr
Turbulent Flow B
Solution 4.36
 V xcr
Re cr 

Re cr  106 *1.7894 *10 5
xcr 

 7.3 *10  2 m
 V
1.225 * 200
D f turb  q SC f 
0.074
0.074
q
S

q S
0.2 
6 0.2 
Re cr
10
 
1
1
2
q   V  1.225 * 200 2  2.45 *10 4 N / m 2
2
2
S  7.3 *10  2 m *17.5m
D f turb  146 N
Drag of one side
Calculate drag
force if the
laminar flow
portion A were
turbulent flow
D f total turbulent  2830 N
D f B  D f total  D f A
 2830  146  2684 N
turb
Df A
laminar

1328
 q SC f 
q S
0.5 
Re cr
135
10 
6 0.2
2.45 *10 7.3 *10
4
2

*17.5  42 N
D f  42 N  2684 N  5452 N
On the wing, it is mostly turbulent flow
Problem 4.37
Let us reflect back to the fundamental equations
of fluid motion discussed in the early sections of
this chapter. Sometimes these equations were
expressed in terms of differential equations, but
for the most pan we obtained algebraic relations
by integrating the differential equations.
However, it is useful to think of the differential
forms as relations that govern the change in
flowfield variables in an infinitesimally small
region around a point in the flow.
(a) Consider a point in an inviscid flow, where
the local density is 1.1 kg/m3. As a fluid
element sweeps through this point, it is
experiencing a spatial change in velocity of
two percent per millimeter. Calculate the
corresponding spatial change in pressure per
millimeter at this point if the velocity at the
point is 100 m/sec. (b) Repeat the calculation
for the case when the velocity at the point is
1000 m/sec. What can you conclude by
comparing your results for the low-speed flow
in part (a) with the results for the high-speed
flow part (b).
Solution 4.37
dp   VdV
 dV 


dp
dV
V 
  V
  V 2 
ds
ds
ds
 dV 


V

  0.02 / mm
ds
dp
 1.1 100 2 0.02   220 N / m 2 .mm
ds
dp
 1.1 1000 2 0.02   22000 N / m 2 .mm
ds




It requires a much larger pressure gradient in a
high-speed flow
Problem 4.38
The type of calculation in Problem 4.3 is a
classic one for low-speed, incompressible flow,
i.e., given the freestream pressure and velocity,
and the velocity at some other point in the flow,
calculate the pressure at that point. In a highspeed compressible flow, Mach number is more
fundamental than velocity. Consider an airplane
flying at Mach 0.7 at a standard altitude of 3 km.
At a point on the wing, the airflow Mach
number is 1.1. Calculate the pressure at this
point. Assume an isentropic flow.
Solution 4.38

p0  (  1) M
 1 
p 
2
2

p0  (  1) M
 1 
p 
2
2
  1  (1.4  1)0.7
  1 
2



  1  (1.4  1)1.1
  1 
2


2
2






1.4
1.4 1
1. 4
1.4 1
 1.387
 2.135
 p0 


p 
1.387

p
p 
p  0.65 * 7.0121*10 4  4.555 *10 4
2.135
 p0 
 
Pressure at 3 km altitude
 p
Problem 4.39
• Consider an airplane flying at a standard
altitude of 25,000 ft at a velocity of 800
ft/sec. To experience the same dynamic
pressure at sea level, how fast must the
airplane be flying?
Solution 4.39
Ve

V

0
1.0663 *10 3
Ve  800
 535.8 ft / s
3
2.3769 *10
Problem 4.40
In Section 4.9, we defined hypersonic flow
as that flow where the Mach number is five
or greater. Wind tunnels with a test section
Mach number of five or greater are called
hypersonic wind tunnels. From Eq. (4.88),
the exit-to-throat area ratio for supersonic
exit Mach numbers increases as the exit
Mach number increases. For hypersonic
Mach numbers, the exit-to-throat ratio
becomes extremely large, so hypersonic
wind tunnels are designed with long, highexpansion ratio nozzles.
In this and the following problems, let us
examine some special characteristics of
hypersonic wind tunnels. Assume we wish
to design a Mach 10 hypersonic wind
tunnel using air as the test medium. We
want the static pressure and temperature
in the test stream to be that for a
standard altitude of 55 km. Calculate: (a)
the exit-to-throat area ratio, (b) the required
reservoir pressure (in atm), and (c) the
required reservoir temperature. Examine
these results. What do they tell you about
the special (and sometimes severe)
operating requirements for a hypersonic
wind tunnel.
Solution 4.40
Ae

At
1
10 2
1
2
Me
 2    1 2 
   1 1  2 M e  



 2  1.4  1 2 
1.4  1 1  2 10 



1.4 1
1.4 1
 1
 1

 535.9

3.5
 (1.4  1)10 
po  (  1) M 
  1 
  4.224 *10 4
 1 
pe 
2
2



po  4.224 *10 4 48.373  2.053 *106  20.3atm \
2
e

 1
2

 (  1) M e2 
 (  1)10 2 
To  Te 1 
  275.781 
  5791K
2
2




The surface of the sun is about 6000k; sacrifice
accuracy because of temperature
Problem 4.41
• Calculate the exit velocity of the
hypersonic tunnel in Problem 4.40.
Solution 4.41
1.4287 275.78  332.9m / s
Ve  M e ae  10332.9   3329m / s
ae  RTe 
Problem 4.42
Let us double the exit Mach number of the
tunnel in Problem 4.40 simply by adding a
longer nozzle section with the requisite
expansion ratio. Keep the reservoir
properties the same as those in Problem
4.40. Then we have a Mach 20 wind
tunnel, with test section pressure and
temperature considerably lower than in
Problem 4.40, i.e., the test section flow no
longer corresponds to conditions at a
standard altitude of 55 km. Be that as it
may, we have at least doubled the Mach
number of the tunnel.
• Calculate: (a) the exit-to-throat area ratio
of the Mach 20 nozzle, (b) the exit
velocity. Compare these values with
those for the Mach 10 tunnel in Problems
4.40 and 4.41. What can you say about
the differences? In particular, note the
exit velocities for the Mach 10 and Mach
20 tunnels. You will see that they are not
much different. What is then giving the
big increase in exit Mach number?
Solution 4.42
Ae

At
1
20 2
1
2
Me
 2    1 2 
   1 1  2 M e  



 2  1 .4  1 2  
1.4  1 1  2 20 



1.4 1
1.4 1
 1
 1

28.7 times
increase of exit
area
 15377
1
1
 (  1) M 
 (  1)20 
5791

Te  T0 1 

  71.5 K
1 
2
2




2
e
2
Ve  M e ae  M e RTe  20 1.4 287 71.5  3390m / s
Not much increase in velocity