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 ft

2

, respectively. If the exit area is

4 times the inlet area, calculate the
water flow velocity at the exit.

Solution 4.1

s

ft

A

A

V

V

V

A

V

A

m

/

25

.

1

4

1

5

2

1

1

2

2

2

2

1

1

1

















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/ft

3

.

Solution 4.2

2

2

2

1

2

3

2

2

2

1

1

2

/

7

.

22

2

25

.

1

5

94

.

1

/

94

.

1

2

.

32

4

.

62

2

0

2

1

2

1

ft

lb

p

p

ft

slug

V

V

p

p

VdV

dp

v

v

p

p

























































Problem 4.3

• Consider an airplane flying with a

velocity of 60 m/s at a standard
altitude of 3 km. At a point on the
wing, the airflow velocity is 70 m/s.
Calculate the pressure at this point.
Assume incompressible flow.

Solution 4.3

H.W.

Problem 4.4

• An instrument used to measure the airspeed

on many early low-speed airplanes, principally

during 1919 to 1930, was the venturi tube.

This simple device is a convergent - divergent

duct (The front section's cross-sectional area

A decreases in the flow direction, and the back

section's cross-sectional area increases in the

flow direction. Somewhere in between the

inlet and exit of the duct, there is a minimum

area, called the throat.) Let A

1

and A

2

denote

the inlet and throat areas, respectively. Let p

1

and p

2

be the pressures at the inlet and

throat, respectively.

The venturi tube is mounted at a specific

location on the airplane (generally on the

wing or near the front of the fuselage),

where the inlet velocity V, is essentially the

same as the freestream velocity that is,

the velocity of the airplane through the air.

With a knowledge of the area ratio A

2

/A

1

(a

fixed design feature) and a measurement
of the pressure difference p

1

- p

2

the

airplane's velocity can be determined. For
example, assume A

2

/A

1

=1/4 and p

1

- p

2

=

80 Ib/ft

2

. If the airplane is flying at standard

sea level, what is its velocity?

Solution 4.4

H.W.

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 m

2

respectively. The inlet and exit pressures
are 1.02 x 10

5

and 1.00 x 10

5

N/m

2

,

respectively. Calculate the flow velocity
at the throat. Assume incompressible
flow with standard sea-level density.

Solution 4.5





s

m

V

A

A

V

A

A

p

p

V

A

A

V

V

V

p

V

p

/

22

.

102

1

2

3

225

.

1

10

)

00

.

1

02

.

1

(

2

5

.

1

3

1

)

(

2

2

2

2

5

1

2

1

2

2

3

1

3

1

1

3

1

1

3

2

3

3

2

1

1





















































































Note that only a
pressure change of
0.02 atm produce
this high speed

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/ft

2

. Calculate the velocity at

that point assuming incompressible flow.

Solution 4.6









s

ft

V

V

p

p

V

V

p

V

p

s

ft

mph

V

/

8

.

216

7

.

190

0020482

.

0

1750

9

.

1760

2

2

/

7

.

197

60

88

130

130

2

4

2

1

2

1

2

2

2

2

2

2

1

1

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 p

1

- p

2

and properly

calibrated in terms of velocity. What is the
maximum pressure difference you would
expect the gauge to experience?

Solution 4.7

 





2

2

2

2

1

2

2

1

2

1

2

1

2

1

1

2

2

2

2

2

1

1

/

3423

1

3

.

1

2

90

225

.

1

1

2

2

2

m

N

p

p

A

A

V

p

p

A

A

V

V

V

p

V

p























































Maximum when
maximum velocity 90 m/s
and sea level density;
however better design for
over speed during diving

Problem 4.8

A supersonic nozzle is also a

convergent-divergent duct, which is fed

by a large reservoir at the inlet to the

nozzle. In the reservoir of the nozzle, the

pressure and temperature are 10 atm

and 300 K, respectively. At the nozzle

exit, the pressure is 1 atm. Calculate the

temperature and density of the flow at

the exit. Assume the flow is isentropic

and, of course, compressible.

Solution 4.8

H.W.

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 p

o

/p

e

and the

reservoir temperature To.

Solution 4.9













































































1

0

e

0

1

0

e

0

e

0

2

2

2

0

1

2

)

(

2

2

1

2

1

2

1

p

p

T

c

V

p

p

T

T

T

T

c

V

V

h

h

V

T

c

V

T

c

p

e

e

p

e

e

e

o

e

e

p

o

p

Note that the velocity
increases as T

o

goes

up or pressure ratio
goes down; used for
rocket engine
performance analysis

Problem 4.10

Consider an airplane flying at a standard

altitude of 5 km with a velocity of 270

m/s. At a point on the wing of the

airplane, the velocity is 330 m/s.

Calculate the pressure at this point.

Solution 4.10

H.W.

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





 

















2

1

4

.

1

1

1

1

o

e

0

0

0

0

2

2

0

2

0

0061

.

0

1500

0051

.

0

2

.

32

5

.

1

2

.

32

5

.

1

0051

.

0

1000

5

.

812

0086

.

0

0086

.

0

1000

1716

2116

7

5

.

812

6000

2

1500

1000

2

2

1

ft

V

m

A

V

A

m

T

T

RT

p

R

c

V

T

T

V

T

c

T

c

e

e

e

e

e

e

e

p

e

e

e

e

p

p

















































































Energy eq.

Continuity
eq.

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











s

ft

V

V

s

ft

s

ft

h

mi

V

V

T

T

c

V

V

T

c

V

T

c

p

p

p

/

3

.

6

2200

32

.

7993

99

.

389

6000

2

/

2200

/

60

88

1500

/

1500

2

2

1

2

1

2

2

2

2

1

2

1

2

1

2

2

2

2

2

2

1

1







































Very low value, almost a stagnant point

Problem 4.13

For the airplane in Prob. 4.12, the total cross-
sectional area of the inlet to the jet engines is
20 ft

2

. Assume that the flow properties of the

air entering the inlet are those of the
freestream ahead of the airplane. Fuel is
injected inside the engine at a rate of 0.05 Ib
of fuel for every pound of air flowing through
the engine (i.e., the fuel-air ratio by mass is
0.05). Calculate the mass flow (in slugs/per
second) that comes out the exit of the
engine.

Solution 4.13

H.W.

Problem 4.14

Calculate the Mach number at the
exit of the nozzle in Prob. 4.11.

Solution 4.14

 





07

.

1

1397

1500

/

1397

5

.

812

1716

4

.

1

5

.

812

/

1500

e

e

















a

V

M

s

ft

RT

a

R

T

s

ft

V

e

e

e

e

e



Problem 4.15

A Boeing 747 is cruising at a velocity
of 250 m/s at a standard altitude of
13 km. What is its Mach number?

Solution 4.15

H.W.

Problem 4.16

A high-speed missile is traveling at

Mach 3 at standard sea level. What
is its velocity in miles per hour?

Solution 4.16

H.W.

Problem 4.17

Calculate the flight Mach number for the
supersonic transport in Prob. 4.12.

Solution 4.17

 





27

.

2

94

.

967

2200

/

94

.

967

99

.

389

1716

4

.

1

/

2200















a

V

M

s

ft

RT

a

s

ft

V



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



 

cm

m

A

A

V

h

h

h

p

p

A

A

V

p

p

m

kg

RT

p

8

.

2

028

.

0

20

1

1

2

80

10

*

33

.

1

173

.

1

1

2

10

*

33

.

1

1

2

/

173

.

1

300

287

10

*

01

.

1

2

2

5

2

1

2

2

2

5

2

1

2

1

2

2

2

2

1

3

5































































































































Manometer reading

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 ft

2

, and the cross-

sectional area of the test section is 4

ft

2

. For tunnel (b) a diffuser is added to

(a) with a diffuser area of 18 ft

2

. 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)

2

2

2

2

1

2

1

2

2

2

2

1

1

2

2

1

2

2

2

2

1

1

/

15

.

98

20

4

1

2

3

.

293

002377

.

0

1

2

2

2

ft

lb

p

p

A

A

V

p

p

A

A

V

V

V

p

V

p



















































































Solution 4.19 (b)

2

2

2

2

2

1

2

1

2

2

3

2

2

2

3

1

3

2

2

3

1

2

2

1

2

3

3

2

1

1

/

959

.

0

20

4

18

4

2

3

.

293

002377

.

0

2

,

2

2

ft

lb

p

p

A

A

A

A

V

p

p

A

A

V

V

A

A

V

V

V

p

V

p













































































































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







 

2

2

0

2

0

2

0

3

/

2172

220

2

00233

.

0

2116

60

88

*

150

2

00233

.

0

2116

2

/

00233

.

0

460

70

1716

2116

ft

lb

p

p

V

p

p

ft

slug

RT

p







































Problem 4.21

The altimeter on a low-speed Piper
Aztec reads 8000 ft. A Pitot tube
mounted on the wing tip
measures a pressure of 1650 Ib/ft

2

. If

the outside air temperature is 500°R,
what is the true velocity of the
airplane? What is the equivalent
airspeed?

Solution 4.21

H.W.

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?



 

s

m

V

V

V

m

kg

RT

p

true

eq

true

/

56

989

.

0

225

.

1

50

/

989

.

0

280

287

10

*

95

.

7

0

3

4



















Solution 4.22

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





5

5

0

1

4

.

1

4

.

1

2

1

2

0

10

*

48

.

1

10

*

01

.

1

*

47

.

1

47

.

1

47

.

1

2

76

.

0

)

1

4

.

1

(

1

2

)

1

(

1

76

.

0

329

250

/

329

270

*

287

*

4

.

1





























































p

p

M

p

p

a

V

M

s

m

RT

a









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 10

4

N/m

2

. At what Mach

number is the airplane flying?

Solution 4.24

801

.

0

N/m

10

*

94

.

1

p

km,

12

altitude

at

note;

1

10

*

94

.

1

10

*

96

.

2

1

4

.

1

2

1

1

2

10

*

94

.

1

1

2

4

4

.

1

1

4

.

1

4

4

1

1

0

2

1

4























































































M

p

p

M

p







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/ft

2

. What is the altitude reading on the

altimeter?

Solution 4.25

1761

328

.

1

2339

328

.

1

328

.

1

2

65

.

0

)

1

4

.

1

(

1

2

)

1

(

1

0

1

4

.

1

4

.

1

2

1

2

0

















































p

p

M

p

p







Appendix B, pressure altitude reads 5000 ft

Problem 4.26

A high-performance F-16 fighter is
flying at Mach 0.96 at sea level. What is
the air temperature at the
stagnation point at the leading edge of
the wing?

Solution 4.26

H.W.

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

2

5

4

02

4

1

2

1

4

.

1

4

.

1

2

2

2

2

1

1

2

1

2

1

2

1

02

1

1

1

1

/

10

*

49

.

1

10

*

65

.

2

*

64

.

5

10

*

65

.

2

64

.

5

1

4

.

1

2

*

4

.

1

*

2

4

.

1

1

)

1

4

.

1

(

2

2

*

4

.

1

*

4

2

)

1

4

.

1

(

1

2

1

)

1

(

2

4

)

1

(

0

.

2

297

596

/

297

220

*

287

*

4

.

1

m

N

p

p

as

M

M

M

p

p

a

V

M

s

m

RT

a































































































Use Rayleigh Pitot tube formula

Problem 4.28

The dynamic pressure is defined as

q = 0.5V

2

. 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

 

2

2

2

2

1

2

2

2

2

2

2

2

2

2

1

2

1

M

p

a

V

p

V

p

p

q

p

c

d

c

d

d

dp

a

V

p

p

V

p

p

V

q





































































































so

as

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:

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.

h,

km

60

50

40

30

20

M

17

9.5

5.5

3

1

Solution 4.29

2

2







M

p

q



h, km

60

50

40

30

20

p

∞

25.6

87.9

299.8

1.19*10

3

5.53*1

0

3

M

17

9.5

5.5

3

1

q

∞

5.2*10

3

5.6*10

3

6.3*10

3

7.5*10

3

3.9*10

3

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







16560

2116

824

.

7

824

.

7

824

.

7

2

2

)

1

4

.

1

(

1

2

)

1

(

1

0

1

4

.

1

4

.

1

2

1

2

0





















































p

p

M

p

p







Total pressure when the flow is isentropically stopped (true
for supersonic and subsonic)

2

4

02

2

1

4

.

1

4

.

1

2

2

2

2

1

1

2

1

2

1

2

1

02

/

10

*

193

.

1

116

.

2

*

64

.

5

64

.

5

1

4

.

1

2

*

4

.

1

*

2

4

.

1

1

)

1

4

.

1

(

2

2

*

4

.

1

*

4

2

)

1

4

.

1

(

1

2

1

)

1

(

2

4

)

1

(

ft

lb

p

M

M

M

p

p















































































But there must be a shockwave at the nose (at the
stagnation point)

2

4

2

0

2

2

0

/

10

*

804

.

0

2

*

116

.

2

*

2

4

.

1

116

.

2

2

2

ft

lb

p

M

p

p

V

p

p



























If Bernoulli’s equation is used accidentally

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











3

4

0

0

0

1

2

0

4

1

4

.

1

4

.

1

2

5

1

2

0

/

267

.

0

6

.

178

287

10

*

37

.

1

6

.

178

357

.

0

*

500

2

)

1

(

1

10

*

37

.

1

2

3

)

1

4

.

1

(

1

10

*

01

.

1

*

5

2

)

1

(

1

m

kg

RT

p

K

M

T

T

M

p

p

e

e

e

e























































































Problem 4.32

• Consider a supersonic nozzle across

which the pressure ratio is p

e

/p

o

=

0.2. Calculate the ratio of exit area
to throat area.

Solution 4.32

 





35

.

1

71

.

1

2

1

4

.

1

1

1

4

.

1

2

71

.

1

1

2

1

1

1

2

1

71

.

1

92

.

2

1

2

.

0

5

1

)

1

(

2

2

)

1

(

1

1

4

.

1

1

4

.

1

2

2

1

1

2

2

286

.

0

1

0

2

1

2

0







































































































































































e

e

t

e

e

e

e

e

e

M

M

A

A

M

p

p

M

M

p

p

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/ p

o

versus M from M = 0 to M = 2.0.

Solution 4.33





5

.

3

2

2

1

2

2

2

2

2

2

2

2

.

0

1

7

.

0

2

1

1

2

2

2

2

2

1



































































M

M

p

q

M

M

p

p

M

p

q

M

p

a

V

p

V

q

















The graph shows that the local
dynamic pressure has a peak value
at M=1.4

Problem 4.34

The wing of the Fairchild Republic A-10A

twin-jet close-support airplane is

approximately rectangular with a wingspan

(the length perpendicular to the flow

direction) of 17.5 m and a chord (the length

parallel to the flow direction) of 3 m. The

airplane is flying at standard sea level with

a velocity of 200 m/s. If the flow is

considered to be completely laminar,

calculate the boundary layer thickness at

the trailing edge and the total skin friction

drag. Assume the wing is approximated by

a flat plate. Assume incompressible flow.

Solution 4.34

H.W.

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









N

N

D

bottom

and

top

N

SC

q

D

C

cm

m

L

f

f

f

L

f

lar

turb

L

5660

2830

*

2

2830

0022

.

0

*

5

.

17

*

3

*

10

*

45

.

2

0022

.

0

10

*

10

.

4

0074

.

0

Re

0074

.

0

75

.

13

24

.

0

3

.

3

3

.

3

033

.

0

10

*

10

.

4

3

*

37

.

0

Re

37

.

0

4

2

.

0

7

2

.

0

2

.

0

7

2

.

0



































10.5 times larger than laminar
flow assumption

Problem 4.36

• If the critical Reynolds number for

transition is 10

6

, calculate the skin

friction drag for the wing in Prob. 4.34.

Laminar Flow
A
Turbulent Flow B

X

cr

Solution 4.36

 

N

D

m

m

S

m

N

V

q

S

q

S

q

SC

q

D

m

V

x

x

V

turb

f

cr

f

turb

f

cr

cr

cr

cr

146

5

.

17

*

10

*

3

.

7

/

10

*

45

.

2

200

*

225

.

1

2

1

2

1

10

074

.

0

Re

074

.

0

10

*

3

.

7

200

*

225

.

1

10

*

7894

.

1

*

10

Re

Re

2

2

4

2

2

2

.

0

6

2

.

0

2

5

6

































































Drag of one side

Calculate
drag force if
the laminar
flow portion
A were
turbulent
flow

  





N

N

N

D

N

S

q

SC

q

D

N

D

D

D

N

D

f

cr

f

A

f

A

f

total

f

B

f

turbulent

total

f

turb

5452

2684

42

42

5

.

17

*

10

*

3

.

7

10

*

45

.

2

10

135

Re

1328

2684

146

2830

2830

2

4

2

.

0

6

5

.

0

laminar

































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/m

3

. 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

 

 





 









mm

m

N

ds

dp

mm

m

N

ds

dp

mm

ds

V

dV

ds

V

dV

V

ds

dV

V

ds

dp

VdV

dp

.

/

22000

02

.

0

1000

1

.

1

.

/

220

02

.

0

100

1

.

1

/

02

.

0

2

2

2

2

2





























































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 high-speed 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

4

4

0

0

1

4

.

1

4

.

1

2

1

2

0

1

4

.

1

4

.

1

2

1

2

0

10

*

555

.

4

10

*

0121

.

7

*

65

.

0

135

.

2

387

.

1

135

.

2

2

1

.

1

)

1

4

.

1

(

1

2

)

1

(

1

387

.

1

2

7

.

0

)

1

4

.

1

(

1

2

)

1

(

1

























































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

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

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

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





















p

p

p

p

p

p

p

M

p

p

M

p

p













Pressure at 3 km
altitude

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

s

ft

V

V

V

e

e

/

8

.

535

10

*

3769

.

2

10

*

0663

.

1

800

3

3

0















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,

high-expansion 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









K

M

T

T

atm

p

M

p

p

M

M

A

A

e

e

o

o

e

e

o

e

e

t

e

5791

2

10

)

1

(

1

78

.

275

2

)

1

(

1

\

3

.

20

10

*

053

.

2

373

.

48

10

*

224

.

4

10

*

224

.

4

2

10

)

1

4

.

1

(

1

2

)

1

(

1

9

.

535

10

2

1

4

.

1

1

1

4

.

1

2

10

1

2

1

1

1

2

1

2

2

6

4

4

5

.

3

2

1

2

1

4

.

1

1

4

.

1

2

2

1

1

2

2

































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

















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

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



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



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

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

















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

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





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

































































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

 





 



s

m

a

M

V

s

m

RT

a

e

e

e

e

e

/

3329

9

.

332

10

/

9

.

332

78

.

275

287

4

.

1















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

 





s

m

RT

M

a

M

V

K

M

T

T

M

M

A

A

e

e

e

e

e

e

e

e

e

t

e

/

3390

5

.

71

287

4

.

1

20

5

.

71

2

20

)

1

(

1

5791

2

)

1

(

1

15377

20

2

1

4

.

1

1

1

4

.

1

2

20

1

2

1

1

1

2

1

1

2

1

2

0

1

4

.

1

1

4

.

1

2

2

1

1

2

2











































































































































Not much increase in velocity

28.7 times
increase of exit
area