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

sftA

AVV

VAVAm

/25.14

15

2

112

222111

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

222

12

3

22

21

12

/7.222

25.1594.1

/94.12.32

4.62

2

02

1

2

1

ftlbpp

ftslug

VVpp

VdVdpv

v

p

p

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

smV

A

AV

AA

ppV

A

AVV

Vp

Vp

/22.102

123

225.1

10)00.102.1(2

5.1

3

1

)(2

22

2

5

12

12

2

3

1

311

3

113

23

3

21

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/ft2. Calculate the velocity at that point assuming incompressible flow.

Solution 4.6

sftV

Vpp

V

VpVp

sftmphV

/8.216

7.1900020482.0

17509.176022

/7.19760

88130130

2

421

2122

222

211

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 p1-

p2 and properly calibrated in terms of velocity. What is the maximum pressure difference you would expect the gauge to experience?

Solution 4.7

222

21

2

2

12

121

2

112

22

2

21

1

/342313.12

90225.1

12

22

mNpp

A

AVpp

A

AVV

Vp

Vp

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

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

0

e0

1

0

e

0

e

0

2

220

12

)(2

2

12

1

2

1

p

pTcV

p

p

T

T

TTcV

Vhh

VTcVTc

pe

epe

eeo

eepop

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

2

14.1

11

1

o

e0

0

00

22

0

20

0061.015000051.02.32

5.12.32

5.1

0051.01000

5.8120086.0

0086.010001716

21167

5.81260002

15001000

2

2

1

ftV

mA

VAm

T

T

RT

p

Rc

VTT

VTcTc

eee

eee

e

p

ee

eepp

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

sftV

V

sftsfthmiV

VTTcV

VTcVTc

p

pp

/3.6

220032.799399.38960002

/2200/60

881500/1500

2

2

1

2

1

2

222

1

2121

22

222

211

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

07.11397

1500

/13975.81217164.1

5.812

/1500

e

e

a

VM

sftRTa

RT

sftV

e

ee

e

e

Problem 4.17

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

Solution 4.17

27.294.967

2200

/94.96799.38917164.1

/2200

a

VM

sftRTa

sftV

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

cmm

A

AVh

hhpp

A

AVpp

mkgRT

p

8.2028.0

20

11

2

80

10*33.1

173.11

2

10*33.1

12

/173.1300287

10*01.1

22

5

2

1

22

2

521

2

1

22

221

35

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 ft2, and the cross-sectional 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)

222

21

2

1

22

221

1

221

22

2

21

1

/15.9820

41

2

3.293002377.0

12

22

ftlbpp

A

AVpp

A

AVV

Vp

Vp

Solution 4.19 (b)

2222

21

2

1

2

2

3

22

231

3

223

1

221

23

3

21

1

/959.020

4

18

4

2

3.293002377.0

2

,

22

ftlbpp

A

A

A

AVpp

A

AVV

A

AVV

Vp

Vp

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

220

2

0

2

0

3

/21722202

00233.02116

60

88*150

2

00233.02116

2

/00233.0460701716

2116

ftlbp

p

Vpp

ftslugRT

p

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?

smV

V

V

mkgRT

p

true

eq

true

/56989.0

225.150

/989.0280287

10*95.7

0

34

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

550

14.1

4.12

120

10*48.110*01.1*47.147.1

47.12

76.0)14.1(1

2

)1(1

76.0329

250

/329270*287*4.1

pp

M

p

p

a

VM

smRTa

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

801.0

N/m 10*94.1p km, 12 altitudeat note;

110*94.1

10*96.2

14.1

2

11

2

10*94.1

1

24

4.1

14.1

4

4

1

1

021

4

M

p

pM

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/ft2. What is the altitude reading on the altimeter?

Solution 4.25

1761328.1

2339

328.1

328.12

65.0)14.1(1

2

)1(1

0

14.1

4.12

120

pp

M

p

p

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

25402

41

214.1

4.1

2

22

21

1

21

21

2

1

02

1

11

1

/10*49.110*65.2*64.5

10*65.2

64.514.1

2*4.1*24.11

)14.1(22*4.1*4

2)14.1(

1

21

)1(24

)1(

0.2297

596

/297220*287*4.1

mNp

pas

M

M

M

p

p

a

VM

smRTa

Use Rayleigh Pitot tube formula

Problem 4.28

The dynamic pressure is defined as q = 0.5rV2. 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 r and V. Derive an equation for q = q(p,M).

Solution 4.28

22

22

12

222

222

22

1

2

1

Mp

a

VpV

p

pq

pc

d

cd

d

dpa

Vp

pV

p

pVq

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

pq

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

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

165602116824.7824.7

824.72

2)14.1(1

2

)1(1

0

14.1

4.1212

0

pp

M

p

p

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

2402

214.1

4.1

2

22

21

1

21

21

2

1

02

/10*193.1116.2*64.5

64.514.1

2*4.1*24.11

)14.1(22*4.1*4

2)14.1(

1

21

)1(24

)1(

ftlbp

M

M

M

p

p

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

2420

22

0

/10*804.02*116.2*2

4.1116.2

22

ftlbp

Mpp

Vpp

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

00

12

0

414.1

4.12

512

0

/267.06.178287

10*37.1

6.178357.0*5002

)1(1

10*37.12

3)14.1(110*01.1*5

2

)1(1

mkgRT

p

KM

TT

Mpp

ee

ee

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

35.171.12

14.11

14.1

2

71.1

1

2

11

1

21

71.1

92.212.051)1(

2

2

)1(1

14.1

14.1

22

1

1

2

2

286.0

1

0

2

12

0

e

et

e

e

ee

ee

MMA

A

M

p

pM

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/ po versus M from M = 0 to M = 2.0.

Solution 4.33

5.322

12

22

22

22

2.017.0

2

11

22

222

1

MMp

q

MM

p

pM

p

q

Mp

a

VpVq

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

NND

bottomand top

NSCqD

C

cmmL

f

ff

L

f

lar

turb

L

56602830*2

28300022.0*5.17*3*10*45.2

0022.010*10.4

0074.0

Re

0074.0

75.1324.0

3.3

3.3033.010*10.4

3*37.0

Re

37.0

4

2.072.0

2.072.0

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

Turbulent Flow B

Xcr

Solution 4.36

ND

mmS

mNVq

SqSqSCqD

mV

x

xV

turbf

cr

fturbf

crcr

crcr

146

5.17*10*3.7

/10*45.2200*225.12

1

2

1

10

074.0

Re

074.0

10*3.7200*225.1

10*7894.1*10Re

Re

2

2422

2.062.0

256

Drag of one side

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

NNND

N

SqSCqD

NDDD

ND

f

cr

fAf

AftotalfBf

turbulenttotalf

turb

5452268442

425.17*10*3.710*45.210

135

Re

1328

26841462830

2830

242.06

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/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

mmmNds

dp

mmmNds

dp

mmdsVdV

dsVdV

Vds

dVV

ds

dp

VdVdp

./2200002.010001.1

./22002.01001.1

/02.0

22

22

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

44

0

0

14.1

4.1212

0

14.1

4.1212

0

10*555.410*0121.7*65.0135.2

387.1

135.22

1.1)14.1(1

2

)1(1

387.12

7.0)14.1(1

2

)1(1

pp

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

sftV

V

V

e

e

/8.53510*3769.2

10*0663.1800

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

KM

TT

atmp

M

p

p

MMA

A

eeo

o

e

e

o

e

et

e

57912

10)1(178.275

2

)1(1

\3.2010*053.2373.4810*224.4

10*224.42

10)14.1(1

2

)1(1

9.535102

14.11

14.1

2

10

1

2

11

1

21

22

64

4

5.3212

14.1

14.1

22

1

1

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

smaMV

smRTa

eee

ee

/33299.33210

/9.33278.2752874.1

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

smRTMaMV

KM

TT

MMA

A

eeeee

ee

e

et

e

/33905.712874.120

5.712

20)1(15791

2

)1(1

15377202

14.11

14.1

2

20

1

2

11

1

21

1212

0

14.1

14.1

22

1

1

2

2

Not much increase in velocity

28.7 times increase of exit area