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SOLUTIONS MANUAL TO ACCOMPANY

INTRODUCTION TO FLIGHT 7th Edition

By

John D. Anderson, Jr.

PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A student using this manual is using it without permission.

2

Chapter 2

2.1 5

2

3

/ (1.2)(1.01 10 )/(287)(300)

1.41 kg/m

1/ 1/1.41 0.71 m /kg

ρρ

ρ

×

=

= p RT =

v = = =

2.2 Mean kinetic energy of each atom 23 203 3(1.38 10 ) (500) 1.035 10 J

2 2− −= × ×k T = =

One kg-mole, which has a mass of 4 kg, has 6.02 × 1026 atoms. Hence 1 kg has

26 261(6.02 10 ) 1.505 10

4=× × atoms.

20 26 6

Totalinternalenergy (energy per atom)(number of atoms)

(1.035 10 )(1.505 10 ) 1.558 10 J-

=

= ´ ´ = ´

2.3 3

2116 slug0.00237

(1716)(460 59) ftρ = = =

+p

RT

3Volumeof the room (20)(15)(8) 2400ft

Total mass in the room (2400)(0.00237) 5.688slug

Weight (5.688)(32.2) 183lb

= == == =

2.4 3

2116 slug0.00274

(1716)(460 10) ft

p

RTρ = = =

-

Since the volume of the room is the same, we can simply compare densities between the two problems.

3

slug0.00274 0.00237 0.00037

ftρΔ = - =

0.00037

%change (100) 15.6% increase0.00237

ρρ

Δ= = ´ =

2.5 First, calculate the density from the known mass and volume, 3m1500 900 1.67 lb ftρ / /= =

In consistent units, 31.67/32.2 0.052slug ft .ρ /= = Also, T = 70 F = 70 + 460 = 530 R.

Hence,

(0.52)(1716)(530)ρp RT= =

247,290lb/ftp =

or 47,290 / 2116 22.3 atmp = =


3

2.6 ρp RT=

np np nR nT= + +

Differentiating with respect to time,

1 1 1ρ

ρ= +dp d dT

p dt dt T dt

or, T

ρρ

= +dp p d p dT

dt dt dt

or, ρ ρ= +dp d dT

RT Rdt dt dt

(1)

At the instant there is 1000 lbm of air in the tank, the density is

3m1000 / 900 1.11lb /ftρ = =

31.11/32.2 0.0345slug/ftρ = =

Also, in consistent units, is given that

T = 50 + 460 = 510 R

and that

1 /min 1 /min 0.016 /sec= = =dTF R R

dt

From the given pumping rate, and the fact that the volume of the tank is 900 ft3, we also have

3mm3

0.5 lb /sec0.000556 lb /(ft )(sec)

900 ft

ρ = =d

dt

5 30.0005561.73 10 slug/(ft )(sec)

32.2

ρ −= = ×d

dt

Thus, from equation (1) above,

5

2

(1716)(510)(1.73 10 ) (0.0345)(1716)(0.0167)

16.115.1 0.99 16.1 lb/(ft )(sec)

21160.0076 atm/sec

ρ −= × +

= + = =

=

d

dt

2.7 In consistent units,

10 273 263 K= − + =T

Thus,

4

3

/ (1.7 10 )/(287)(263)

0.225 kg/m

ρρ

= = ×

=

p RT

2.8 5 3

3

/ 0.5 10 /(287)(240) 0.726 kg/m

1/ 1/0.726 1.38 m /kg

ρρ

= = × =

= = =

p RT

v


4

2.9

3 3

0 0

2 30

Forcedue to pressure (2116 10 )

[2116 5 ] 6303 lb perpendicular to wall.

pF p dx x dx

x x

= = = -

= - =

ò ò

3 3

τ 10 0

2

1320

90Force due to shear stress = τ

( 9)

[180 ( 9) ] 623.5 540 83.5 lb tangential to wall.

F dx dx

x

x

= =

+

= + = - =

ò ò

Magnitude of the resultant aerodynamic force =

2 2(6303) (835) 6303.6 lb

83.5Arc Tan 0.76º

6303θ

R = + =

æ ö÷ç= =÷ç ÷÷çè ø

2.10 3

sin2

θ∞V V=

Minimum velocity occurs when sin θ = 0, i.e., when θ = 0° and 180°.

Vmin = 0 at θ = 0° and 180°, i.e., at its most forward and rearward points.

Maximum velocity occurs when sin θ = 1, i.e., when θ = 90°. Hence,

max3

(85)(1) 127.5 mph at 90 ,2

θ °V == =

i.e., the entire rim of the sphere in a plane perpendicular to the freestream direction.


5

2.11 The mass of air displaced is

3(2.2)(0.002377) 5.23 10 slugM -= = ´

The weight of this air is

3air (5.23 10 )(32.2) 0.168 lbW -= ´ =

This is the lifting force on the balloon due to the outside air. However, the helium inside the balloon has weight, acting in the downward direction. The weight of the helium is less than that of air by the ratio of the molecular weights

4

(0.168) 0.0233 lb.28.8cHW = =

Hence, the maximum weight that can be lifted by the balloon is

0.168 − 0.0233 = 0.145 lb.

2.12 Let p3, ρ3, and T3 denote the conditions at the beginning of combustion, and p4, ρ4, and T4 denote conditions at the end of combustion. Since the volume is constant, and the mass of the gas is constant, then p4 = ρ3 = 11.3 kg/m3. Thus, from the equation of state,

7 24 4 4 (11.3)(287)(4000) 1.3 10 N/mρp RT= = = ´

or,

7

4 51.3 10

129 atm1.01 10

p´

= =´

2.13 The area of the piston face, where the diameter is 9 cm = 0.09 m, is

2

3 2(0.09)6.36 10 m

4

πA -= = ´

(a) The pressure of the gas mixture at the beginning of combustion is

6 23 3 3 11.3(287)(625) 2.02 10 N/mρp RT= = = ´

The force on the piston is

6 3 43 3 (2.02 10 )(6.36 10 ) 1.28 10 NF p A -= = ´ ´ = ´

Since 4.45 N = l lbf,

4

31.28 10

2876 lb4.45

F´

= =

(b) 7 24 4 4 (11.3)(287)(4000) 1.3 10 N/mρp RT= = = ´

The force on the piston is

7 3 44 4 = (1/3 10 ) (6.36 10 ) = 8.27 10 NF p A -= ´ ´ ´

4

48.27 10

18,579 lb4.45

F´

= =


6

2.14 Let p3 and T3 denote conditions at the inlet to the combustor, and T4 denote the temperature at

the exit. Note: 6 23 4 4 10 N/mp p= = ´

(a) 6

333

3

4 10= 15.49 kg/m

(287)(900)ρ p

RT

´= =

(b) 6

344

4

4 10= 9.29 kg/m

(287)(1500)ρ p

RT

´= =

2.15 1 mile = 5280 ft, and 1 hour = 3600 sec.

So:

miles 5280 ft 1 hour

60 88 ft/sec.hour mile 3600 sec

æ öæ öæ ö÷ ÷ ÷ç ç ç =÷ ÷ ÷ç ç ç÷ ÷ ÷÷ ÷ ÷ç ç çè øè øè ø

A very useful conversion to remember is that

60 mph 88 ft/sec =

also, 1 ft = 0.3048 m

ft 0.3048 m m

88 26.82sec 1 ft sec

æ öæ ö ÷ç÷ç ÷ =÷çç ÷÷ç÷ç ÷÷çè øè ø

Thus ft m

88 26.82 sec sec

=

2.16 miles 88 ft/sec

692 1015 ft/sechour 60 mph

=

miles 26.82 m/sec

692 309.3 m/sechour 60 mph

=

2.17 On the front face

5 5(1.0715 10 )(2) 2.143 10 Nf fF p A= = × = ×

On the back face

5 5(1.01 10 )(2) 2.02 10 Nb bF p A= = × = ×

The net force on the plate is

5 5(2.143 2.02) 10 0.123 10 Nf bF F F= − = − × = ×

From Appendix C,

1 4.448 N.flb =

So,

50.123 10

2765 lb4.448

F×= =

This force acts in the same direction as the flow (i.e., it is aerodynamic drag.)


7

2.18 210,100Wing loading 43.35 lb/ft

233

W

s= = =

In SI units:

2

2

2

lb 4.448 N 1 ft43.35

1 lb 0.3048 mft

N2075.5

m

W

s

W

s

=

=

In terms of kilogram force,

2 2

1 k kgN2075.5 211.8

9.8 Nm mf fW

s

= =

2.19 5miles 5280 ft 0.3048 m m km437 7.033 10 703.3

hr mile 1 ft hr hrV

= = × =

0.3048 m

Altitude (25,000 ft) 7620 m 7.62 km1 ft

= = =

2.20 3ft 0.3048 m m km26,000 7.925 10 7.925

sec 1 ft sec secV

= = × =

2.21 From Fig. 2.16,

length of fuselage = 33 ft, 4.125 inches = 33.34 ft

0.3048 m

33.34 ft 10.16 mft

= =

wing span = 40 ft, 11.726 inches = 40.98 ft

0.3048 m

40.98 ft 12.49 mft

= =


8

Chapter 3

3.1 An examination of the standard temperature distribution through the atmosphere given in Figure 3.3 of the text shows that both 12 km and 18 km are in the same constant temperature region. Hence, the equations that apply are Eqs. (3.9) and (3.10) in the text. Since we are in the same isothermal region with therefore the same base values of p and ρ, these equations can be written as

0 2 1( / )( )2 2

1 1

ρρ

g RT h hpe

p

- -= =

where points 1 and 2 are any two arbitrary points in the region. Hence, with g0 = 9.8 m/sec2 and R = 287 joule/kgK, and letting points 1 and 2 correspond to 12 km and 18 km altitudes, respectively,

9.8

(287)(216.66) (6000)2 2

1 10.3884

ρρ

pe

p= = =

Hence:

4 3 2

2

1 32

(0.3884)(1.9399 10 ) 7.53 10 N/m

(0.3884)(3.1194 10 ) 0.121 kg/mρ

p-

= ´ = ´

= ´ =

and, of course,

2 216.66 KT =

These answers check the results listed in Appendix A of the text within round-off error.

3.2 From Appendix A of the text, we see immediately that p = 2.65 × 104 N/m2 corresponds to 10,000 m, or 10 km, in the standard atmosphere. Hence,

pressure altitude = 10 km

The outside air density is

4

32.65 100.419 kg/m

(287)(220)ρ p

RT

´= = =

From Appendix A, this value of ρ corresponds to 9.88 km in the standard atmosphere.

Hence,

density altitude = 9.88 km

3.3 At 35,000 ft, from Appendix B, we find that p = 4.99 × 102 = 499 lb/ft2.

3.4 From Appendix B in the text,

33,500 ft corresponds to p = 535.89 lb/ft2

32,000 ft corresponds to ρ = 8.2704 × 10−4 slug/ft3

Hence,

4

535.89378

(8.2704 10 )(1716)ρp

T RR -= = =

´


9

3.5 0.02 1G Gh h h

h h

-= = -

From Eq. (3.6), the above equation becomes

1 1 1 0.02G Gr h h

r r

æ ö+ ÷ç- = - - =÷ç ÷÷çè ø

6

5

0.02 0.02(6.357 10 )

1.27 10 m 127 km

G

G

h r

h

= = ´

= ´ =

3.6 T = 15 − 0.0065h = 15 − 0.0065(5000) = −17.5°C = 255.5°K

0.0065

dTa

dh= = -

From Eq. (3.12)

0/ (2.8) /( 0.0065)(287)

1 1

255.50.533

288

g aRp T

p T

- - -æ ö æ ö÷ç ÷ç÷= = =ç ÷ç÷ ÷ç ÷ç÷÷ç è øè ø

5 4 210.533 0.533 (1.01 10 ) 5.38 10 N/mp p= = ´ = ´

3.7 11

( ) p gn h h

p RT= - -

11

1 1(4157)(150) 0.5

24.9 p

h h RT n ng p

- = - = -

Letting h1 = 0 (the surface)

h = 17,358 m = 17.358 km


10

3.8 A standard altitude of 25,000 ft falls within the first gradient region in the standard atmosphere. Hence, the variation of pressure and temperature are given by:

1 1

g

aRp T

p T

-æ ö÷ç ÷= ç ÷ç ÷÷çè ø (1)

and

T = T1 + a (h – h1) (2)

Differentiating Eq. (1) with respect to time:

1

1 1

1 1g g

aR aRdp g dTT

p dt T AR dt

æ ö- ÷ç- - ÷ç ÷÷ççè øæ ö æ ö÷ç ÷ç÷= -ç ÷ç÷ ÷ç ÷ç÷÷ç è øè ø (3)

Differentiating Eq. (2) with respect to time:

dT dh

adt dt

= (4)

Substitute Eq. (4) into (3)

1

1 1( )

ggaRaRdp g dh

p T Tdt R dt

æ ö÷ç- + ÷ç ÷÷ççè øæ ö÷ç= - ÷ç ÷÷çè ø (5)

In Eq. (5), dh/dt is the rate-of-climb, given by dh/dt = 500 ft/sec. Also, in the first gradient region, the lapse rate can be calculated from the tabulations in Appendix B. For example, take 0 ft and 10,000 ft, we find

2 1

2 1

483.04 518.690.00357

10,000 0 ft

°T T Ra

h h

- -= = = -

- -

Also from Appendix B, p1= 2116.2 lb/ft2 at sea level, and T = 429.64 °R at 25,000 ft.

Thus,

32.2

5.256( 0.00357)(1716)

g

aR= = -

-

Hence, from Eq. (5)

5.256 4.25632.2(2116.2)(518.69) (429.64) (500)

1716

dp

dt- æ ö÷ç= - ÷ç ÷÷çè ø

2lb

17.17ft sec

dp

dt= -


11

3.9 From the hydrostatic equation, Eq. (3.2) or (3.3),

0ρdp g dh= -

or 0ρdp dhg

dt dt= -

The upward speed of the elevator is dh/dt, which is

0

/

ρdp dp dt

dt g=

-

At sea level, ρ = 1.225 kg/m3. Also, a one-percent change in pressure per minute starting from sea level is

5 3 2(1.01 10 )(0.01) 1.01 10 N/m per minutedp

dt= - ´ = - ´

Hence,

31.01 10

84.1meter per minute(1.225)(9.8)

dh

dt

- ´= =

3.10 From Appendix B:

At 35,500 ft: p = 535.89 lb/ft2

At 34,000 ft: p = 523.47 lb/ft2

For a pressure of 530 lb/ft2, the pressure altitude is

535.89 530

33,500 500 33737ft535.89 523.47

æ ö- ÷ç+ =÷ç ÷÷çè ø-

The density at the altitude at which the airplane is flying is

4 35307.919 10 slug/ft

(1716)(390)

ρρRT

-= = = ´

From Appendix B:

4 3

4 3

At 33,000 ft: 7.9656 10 slug/ft

At 33,500 ft: 7.8165 10 slug/ft

ρ

ρ

-

-

= ´

= ´

Hence, the density altitude is

7.9656 7.919

33,000 500 33,156 ft7.9656 7.8165

æ ö- ÷ç+ =÷ç ÷÷çè ø-


12

3.11 Let ℓ be the length of one wall of the tank, ℓ = 30 ft. Let d be the depth of the pool, d = 10 ft. At the water surface, the pressure is atmospheric pressure, pa. The water pressure increases with increasing depth; the pressure as a function of distance below the surface, h, is given by the hydrostatic equation

dp = ρ g dh (1)

Note: The hydrostatic equation given by Eq. (3.2) in the text has a minus sign because hG is measured positive in the upward direction. In Eq. (1), h is measured positive in the downward direction, with h = 0 at the surface of the water. Hence, no minus sign appears in Eq. (1); as h increases (as we go deeper into the water), p increases. Eq. (1) is consistent with this fact. Integrating Eq. (1) from h = 0 where p = pa to some local depth h where the pressure is p, and noting that ρ is constant for water, we have

0

ρρ

a

h

pdp g dh=ò ò

or, p − pa = ρ g h

or, p = ρ g h + pa (2)

Eq. (2) gives the water pressure exerted on the wall at an arbitrary depth h.

Consider an elementary small sliver of wall surface of length ℓ and height dh.

The water force on this sliver of area is

dF = p ℓ dh

Total force, F, on the wall is

0 0

F d

F dF p dh= =ò ò (3)

where p is given by Eq. (2). Inserting Eq. (2) into (3),

0

( + ) ρ d

aF g h p dh= ò

or, 2

2ρ a

dF g p d

æ ö÷ç ÷ç= +÷ç ÷ç ÷÷çè ø (4)

In Eq. (4), the product ρ g is the specific weight (weight per unit volume) of water; ρ g = 62.4 lb/ft3. From Eq. (4),

2(10)

(62.4)(30) (2116)(30)(10)2

F = +

93,600 634,800 728,400lbF = + =

Note: This force is the combined effect of the force due to the weight of the water, 93,600 lb, and the force due to atmospheric pressure transmitted through the water, 634,800 Ib. In this example, the latter is the larger contribution to the force on the wall. If the wall were freestanding with atmospheric pressure exerted on the opposite side, then the net force exerted on the wall would be that due to the weight of the water only, i.e., 93,600 Ib. In tons, the force on the side of the wall in contact with the water is

728,400

364.2 tons2000

F = =

In the case of a freestanding wall, the net force, that due only to the weight of the water, is

93,600

46.8 tons2000

F = =


13

3.12 For the exponential atmosphere model,

0 /( )

0

(9.8)(45,000) /(287)(240) 6.402

0

ρρρρ

g h RTe

e e

-

- -

=

= =

Hence,

6.402 3 3 30 (1225)(1.6575 10 ) 2.03 10 kg/mρ ρ e- - -= = ´ = ´

From the standard atmosphere, at 45 km, p = 2.02 × 10−3 kg/m3. The exponential atmosphere model gives a remarkably accurate value for the density at 45 km when a value of 240 K is used for the temperature.

3.13 At 3 km, from App. A, T = 268.67 K, p = 7.0121 × 104 N/m2, and = 0.90926 kg/m3. At 3.1 km,

from App. A, T = 268.02 K, p = 6.9235 × 104 N/m2, and = 0.89994 kg/m3. At h = 3.035 km,

using linear interpolation:

ρ

= − − =

= × − × − ×

= ×

= − − =

4 4 4

42

3

0.035268.67 (268.67 268.02) 268.44 K

0.1

0.0357.0121 10 (7.0121 10 6.9235 10 )

0.1

N6.98099 10

m

0.035 kg0.09026 (0.90926 0.89994) 0.906

0.1 m

T

p


14

3.14 The altitude of 3.035 km is in the gradient region. From Example 3.1, a = –0.0065 K/m. From Eq. (3.14),

1 1( ) ( 0) 288.16 (0.0065)(3035) 268.43 KsT T a h h T a h= + − = + − = − =

From Eq. (3.12),

0 /

1 1

0 9.85.25328

( 0.0065)(28.7)

g aRp T

p T

g

aR

− =

− −= =−

So

5.25328 5.25328

5 42

268.430.688946

288.16

N0.688946(1.01325 10 ) 6.9807 10

m

s s

p T

p T

p

= = =

= × = ×

By comparison the approximate result from the solution of Prob. 3.13 differs from the exact result above by

56.980990 06.98074.15 10 0.00415%

6.9807−−

= × =

A very small amount.

From Eq. (3.13),

ρρ

ρρ

− + =

+ = − + = −

= = =

0[( / ) 1]

1 1

0

4.2538 4.24328

1 5.25328 1 4.25328

268.430.73958

288.16

g aR

s s

T

T

g

aR

T

T

3

kg0.73598 0.73958 (1.2250) 0.90599

msρ ρ= = =


15

3.15 We want to calculate hG when

0.01 1G

G G

h h h

h h

−= = −

From Eq. (3.6)

6, where 6.356766 10 mG

G

rhh r

r h= = ×

+

Thus,

0.01 1

1 0.01 0.99

G

G

r

r h

r

r h

= −+

= − =+

0.99( )Gr h r+ =

5

3

1 0.990.0101 (6.356766 10 )

0.99

64.21 10 m 64.21 km

G

G

h r

h

− = = ×

= × =

3.16 From Eq. (3.1),

2

0G

rg g

r h

= +

where r = 6.356766 × 106 m, g0 = 9.8 m/sec2, and

0.3048 m

(70,000 ft) 21,336 m1 ftGh

= =

Thus,

6

6 2

6.356766 10 m9.8 9.767

6.356766 10 21,336 secg

×= = × +

9.8 9.767

100 0.34%98

− =

Thus, at 70,000 ft, the acceleration of gravity has decreased only by 0.34% compared to its sea level value.


16

Chapter 4

4.1 1 1 2 2AV A V=

Let points 1 and 2 denote the inlet and exit conditions, respectively. Then,

12 1

2

1(5) 1.25ft/sec

4

AV V

A

æ ö æ ö÷ç ÷ç÷= = =ç ÷ç÷ ÷ç ÷ç÷÷ç è øè ø

4.2 From Bernoulli’s equation,

2 21 2

1 2

2 22 1 1 2

2 2

( )2

ρ ρ

ρ

V Vp p

p p V V

+ = +

- = -

In consistent units,

362.41.94slug/ft

32.2ρ = =

Hence,

2 2

2 1

22 1

1.94[(5) (1.25) ]

2

0.97(23.4) 22.7 lb/ft

p p

p p

- = -

- = =

4.3 From Appendix A; at 3000m altitude,

4 21 7.01 10 N/mp = ´

30.909kg/mρ =

From Bernoulli’s equation,

2 22 1 1 2

4 2 22

4 4 4 22

( )2

0.9097.01 10 [60 70 ]

2

7.01 10 0.059 10 6.95 10 N/m

ρp p V V

p

p

= + -

= ´ + -

= ´ - ´ = ´


17

4.4 From Bernoulli’s equation, 2 21 1 2 22 2

ρ ρp V p V+ = +

Also from the incompressible continuity equation

2 1 1 2( / )V V A A=

Combining, 2 21 1 2 1 2( / )

2 2

ρ ρp V p A A+ = +

1 21 2

1 2

2( )

( / ) 1ρp p

VA A

-=

é ù-ê úë û

At standard sea level, p = 0.002377 slug/ft3. Hence,

1 22(80)

67 ft/sec(.002377)[(4) 1]

V = =-

Note that also 160

67 46mi/h.80

Væ ö÷ç= =÷ç ÷÷çè ø

(This is approximately the landing speed of World

War I vintage aircraft).

4.5 2 21 3

1 1

2 2ρ ρp V p V+ = +

2 23 11 3

2( )

ρp p

V V-

= + (1)

11 1 3 3 3 1

3, or

AA V A V V V

A= = (2)

Substitute (2) into (1)

2

2 23 1 11 1

3

2( )

ρp p A

V VA

æ ö- ÷ç ÷= + ç ÷ç ÷÷çè ø

or, 3 11 2

1

3

2( )

1ρ

p pV

AA

-=

é ùæ öê ú÷ç ÷- çê ú÷ç ÷÷çê úè øê úë û

(3)

Also, 1 1 2 2A V A V=

or, 12 1

2

AV V

A

æ ö÷ç ÷= ç ÷ç ÷÷çè ø (4)


3 111 2

21

3

2( )

1ρ

p pAV

A AA

-=

é ùæ öê ú÷ç ÷- çê ú÷ç ÷÷çê úè øê úë û

5

2 2

3 2(1.00 1.02) 10

1.5 3(1.225) 1

2

V- ´

=é ùæ öê ú÷ç- ÷çê ú÷÷çè øê úë û

2 102.22m/secV =

Note: It takes a pressure difference of only 0.02 atm to produce such a high velocity.


18

4.6 188

130 mph 130 190.7 ft/sec60

Væ ö÷ç= = =÷ç ÷÷çè ø

2 21 1 2 2

1 1

2 2ρ ρp V p V+ = +

2 22 1 2 1

2( )

ρV p p V= - +

2 22

2(1760.9 1750.0)(190.7)

0.0020482V

-= +

2 216.8ft/secV =

4.7 From Bernoulli’s equation,

2 21 2 2 1( )

2

ρp p V V- = -

And from the incompressible continuity equation,

2 1 1 2( / )V V A A=

Combining: 2 21 2 1 1 2[( / ) 1]

2

ρp p V A A- = -

Hence, the maximum pressure difference will occur when simultaneously:

1. V1 is maximum

2. ρ is maximum i.e., sea level

The design maximum velocity is 90 m/sec, and ρ = 1.225 kg/m3 at sea level. Hence,

2 2 21 2

1.225(90) [(1.3) 1] 3423N/m

2p p- = - =

Please note: In reality the airplane will most likely exceed 90 m/sec in a dive, so the airspeed indicator should be designed for a maximum velocity somewhat above 90 m/sec.

4.8 The isentropic relations are

1

0 0 0

γγγρ

ρc e cp T

p T

-æ ö æ ö÷ ÷ç ç÷ ÷= =ç ç÷ ÷ç ç÷ ÷÷ ÷ç çè ø è ø

Hence,

14 11 1.4

00

1(300) 155K

10

γγρ

ρe

eT T

--æ ö æ ö÷ç ÷ç÷= = =ç ÷ç÷ ÷ç ÷ç÷÷ç è øè ø

From the equation of state:

5

300

0

(10)(1.01 10 )11.73kg/m

(287)(300)ρ p

RT

´= = =

Thus,

1 1

1.4 30

0

111.73 2.26kg/m

10

γρ ρ e

ep

p


As a check on the results, apply the equation of state at the exit.

?ρe ep RTe=

51.01 10 (2.26)(287)(155)´ =

5 51.01 10 1.01 10´ = ´ It checks!


19

4.9 Since the velocity is essentially zero in the reservoir, the energy equation written between the reservoir and the exit is

2

0 2e

eV

h h= + or, 202( )e eV h h= - (1)

However, h = cpT. Thus Eq. (1) becomes

202 ( )e p eV c T T= -

20

02 1 e

e pT

V c TT

æ ö÷ç ÷= -ç ÷ç ÷÷çè ø (2)

However, the flow is isentropic, hence

1

0 0

γγc cT p

T p

-æ ö÷ç ÷= ç ÷ç ÷÷çè ø

(3)

Substitute (3) into (1).

1

00

2 1

γγe

e pp

V c Tp

-é ùê úæ ö÷ê úç ÷= - ç ÷ê úç ÷÷çè øê úê úë û

(4)

This is the desired result. Note from Eq. (4) that Ve increases as T0 increases, and as pe/p0

decreases. Equation (4) is a useful formula for rocket engine performance analysis.

4.10 The flow velocity is certainly large enough that the flow must be treated as compressible. From the energy equation,

2 2

1 21 22 2p p

V Vc T c T+ = - (1)

At a standard altitude of 5 km, from Appendix A,

4 2

1

1

5.4 10 N/m

255.7 K

p

T

= ´

=

Also, for air, cp = 1005 joule/(kg)(K). Hence, from Eq. (1) above,

2 2

1 22 1 2 p

V VT T

c

-= +

2 2

2(270) (330)

255.72(1005)

T-

= +

2 255.7 17.9 237.8KT = - =

Since the flow is also isentropic,

1

2 2

1 1

γγp T

p T

-æ ö÷ç ÷= ç ÷ç ÷÷çè ø

Thus,

1 1.4

1.4 1422 1

1

237.85.4 10

255.7

γγT

p pT

-

-æ ö æ ö÷ç ÷ç÷= = ´ç ÷ç÷ ÷ç ÷ç÷÷ç è øè ø

4 22 4.19 10 N/mp = ´

Please note: This problem and Problem 4.3 ask the same question. However, the flow velocities in the present problem require a compressible analysis. Make certain to examine the solutions of both Problems 4.10 and 4.3 in order to contrast compressible versus incompressible analyses.


20

4.11 From the energy equation

2

0 2e

p p eV

c T c T= +

or, 2

0 2e

ep

VT T

c= -

21500

1000 812.52(6000)eT R= - =

In the reservoir, the density is

300

0

(7)(2116)0.0086 slug/ft

(1716)(1000)ρ p

RT= = =

From the isentropic relation,

1

1

0 0

γρρ

c eT

T

-æ ö÷ç ÷= ç ÷ç ÷÷çè ø

1

1.4 1 3812.50.0086 0.0051 slug/ft

1000ρe

-æ ö÷ç= =÷ç ÷÷çè ø

From the continuity equation,

ρ e e em A V=

Thus, ρ

e

eeV

mA =


1.5

0.047 slug/sec.32.2

m = =

Hence,

2

c

0.0470.061ft

(0.0051)(1500)ρ

ee

mA

V= = =

4.12 188

1500mph 1500 2200ft/sec60


2 21 2

1 22 2p pV V

C T C T+

2 2

2 1 2 12 ( )pV C T T V= - +

2 2

2 2(6000)(389.99 793.32) (2200)V = - +

2 6.3ft/secV =

Note: This is a very small velocity compared to the initial freestream velocity of 2200 ft/sec.

At the point in question, the velocity is very near zero, and hence the point is nearly a stagnation point.


21

4.13 At the inlet, the mass flow of air is

4air AV (3.6391 10 )(20)(2200) 16.0/slug/secρm -= = ´ =

fuel (0.05)(16.01) 0.8slug/secm = =

Total mass flow at exit 16.01 0.8 16.81slug/sec= + =

4.14 From Problem 4.11,

Ve = 1500 ft/sec

Te = 812.5 R

Hence,

(1.4)(1716)(812.5)

1397ft/sec

γe ea RT= =

=

Thus, 1500

1.071397

ec

e

VM

a= = =

Note that the nozzle of Problem 4.11 is just barely supersonic.

4.15 From Appendix A,

T∞ = 216.66 K Hence,

(1.4)(287)(216.66) 295m/sec

∞ γa RT=

= =

Thus, 250

0.847295

∞∞

∞

VM

a= = =

4.16 At standard sea level, 518.69∞T R=

(1.4)(1716)(518.69) 1116ft/seca RTγ∞ = = =

(3)(1116) 3348ft/secV M a∞ ∞ ∞= = = Since 60 mi/hr - 88 ft/sec., then

3348(60/88) 2283mi/hV∞ = =

4.17 V = 2200 ft/sec

(1.4)(1716)(389.99) 967.94ft/seca RTγ= = =

22002.27

967.94

VM

a= = =


22

4.18 The test section density is

5

31.01 101.173kg/m

(287)(300)ρ p

RT

´= = =

Since the flow is low speed, consider it to be incompressible, i.e., with the above density throughout.

2 21 2 2 2 1[1 ( / ) ]

2

ρp p V A A- = - (1)

In terms of the manometer reading,

1 2 ωΔp p h- = (2)

where 5 31.33 10 N/mω = ´ for mercury.

Thus, combining Eqs. (1) and (2),

2 22 2 1

2 25

[1 ( / ) ]2

1.173(80) [1 (1/20) ]

(2)(1.33 10 )

0.028m 2.8 cm

ρω

Δ

Δ

h V A A

h

= -

= -´

= =

4.19 288

200mph 300 293.3 ft/sec60


(a) 2 21 1 2 2

1 1

2 2ρ ρp V p V+ = +

1 1 2 2A V A V=

21 2

1

AV V

A=

2

2 221 2 2 2

1

1 1

2 2ρ ρA

p V p VA

æ ö÷ç ÷+ = +ç ÷ç ÷÷çè ø

2

221 2 2

1

11

2ρ A

p p VA

é ùæ öê ú÷ç ÷- = - çê ú÷ç ÷÷çê úè øê úë û

2

21 2

0.002377 41 (293.3)

2 20p p

é ùæ öê ú÷ç- = - ÷çê ú÷÷çè øê úë û

21 2 98.15 lb/ftp p- =


23

(b) 2 31 1 3 3

1 1

2 2ρ ρp V p V+ = +

21 1 2 2 1 2

1:

AAV A V V V

A= =

22 2 3 3 3 2

3:

AA V A V V V

A= =

2 2

22 21 3 2

3 1

1

2ρ A A

p p VA A

é ùæ ö æ öê ú÷ ÷ç ç÷ ÷- = -ç çê ú÷ ÷ç ç÷ ÷÷÷ ççê úè øè øê úë û

2 2

21 3

0.002377 4 4(293.3)

2 18 20p p

é ùæ ö æ öê ú÷ ÷ç ç- = -÷ ÷ç çê ú÷ ÷÷ ÷ç çè ø è øê úë û

21 3 0.959lb/ftp p- =

Note: By the addition of a diffuser, the required pressure difference was reduced by an order of magnitude. Since it costs money to produce a pressure difference (say by running compressors or vacuum pumps), then a diffuser, the purpose of which is to improve the aerodynamic efficiency, allows the wind tunnel to be operated more economically.

4.20 In the test section

321160.00233slug/ft

(1716)(70 460)ρ p

RT= = =

+

The flow velocity is low enough so that incompressible flow can be assumed. Hence, from Bernoulli’s equation,

20

1

2ρp p V= +

20

12116 (0.00233)[150(88 / 60)]

2p = +

(Remember that 88 ft/sec = 60 mi/h.)

20

12116 (0.00233)(220)

2p = +

20 2172lb/ftp =

4.21 The altimeter measures pressure altitude. Thus, from Appendix B, p = 1572 lb/ft2. The air density is then

315720.00183slug/ft

(1716)(500)ρ p

RT= = =

Hence, from Bernoulli’s equation,

0true

2( ) 2(1650 1572)

0.00183ρp p

V- -

= =

true 292 ft/secV =

The equivalent airspeed is

02( ) 2(1650 1572)

0.002377ρes

p pV

- -= =

256 ft/seceV =


24

4.22 The altimeter measures pressure altitude. Thus, from Appendix A,

p = 7.95 × l04 N/m2. Hence,

4

37.95 100.989 kg/m

(287)(280)ρ p

RT

´= = =

The relation between Vtrue and Ve is

true/ /ρ ρe sV V =

Hence,

true 50 (1.225) / 0.989 56m/secV = =

4.23 In the test section,

(1.4)(287)(270) 329m/secγa RT= = =

/ 250/329 0.760M V a= = =

12 2 3.50 1

1 [1 0.2(0.760) ] 1.472

γγγp

Mp

-æ ö- ÷ç= + = + =÷ç ÷÷çè ø

Hence, 5 5 20 1.47 1.47(1.01 10 ) 1.48 10 N/mp p= = ´ = ´

4.24 4 21.94 10 N/mp = ´ from Appendix A.

1 0.28642 01 4

2 2 2.96 101 1

1 1.4 1 1.94 10

γγ

γp

Mp

-é ù é ùê ú æ öæ ö ê ú´ ÷ç÷ê úç ÷çê ú÷= - = -ç ÷ç÷ê ú ÷ç ÷ ê úç ÷ ç- - ÷÷çè ø ´ê ú è øê úë ûê úë û

21 0.642M =

1 0.801M =

4.25 120 1

12

γγγp

Mp

-æ ö- ÷ç= + ÷ç ÷÷çè ø

2 3.50 [1 0.2(0.65) ] 1.328p

p= + =

20 23391761 lb/ft

1.328 1.328

pp = = =

From Appendix B, this pressure corresponds to a pressure altitude, hence altimeter reading of 5000 ft.


25

4.26 At standard sea level,

T = 518.69 R

2 20 11 1 0.2(0.96) 1.184

2

TM

T

γ -= + = + =

T0 = 1.184T = 1.184 (518.69)

T0 = 614.3R = 154.3F

4.27 1 1 (1.4)(287)(220) 297 m/seca RTγ= = =

1 1 1/ 596 /197 2.0M V a= = =

The flow is supersonic. Hence, the Rayleigh Pitot tube formula must be used.

22 2 210 1 1

21 1

( 1) 1 2

14 2( 1)

p M M

p M

γγγ γ γ

γγ γ

-é ù é ù+ - +ê ú ê ú= ê ú ê ú+- -ê ú ê úë û ë û

2

3.52 2 20

21

(2.4) (2) 1 1.4 2(1.4)(2)

2.44(1.4)(2) 2(0.4)

p

p

é ù é ù- +ê ú ê ú= ê ú ê ú-ê ú ê úë û ë û

20

15.64

p

p=

4 21 2.65 10 N/mp = ´ from Appendix A.

Hence,

2

4 5 20 5.64(2.65 10 ) 1.49 10 N/mp = ´ = ´

4.28 2

2 2 22

1 1

2 2 2 2

γ γ γρ ργ γ

p p Vq V V p V p

p p a

æ ö æ ö÷ ÷ç ç÷ ÷= = = =ç ç÷ ÷ç ç÷ ÷ç ç÷ ÷è ø è ø

Hence:

2

2

γq pM=

4.29 2 20.72

γ∞ ∞ ∞ ∞ ∞q p M p M= = (1)

Use Appendix A to obtain the values of p∞ corresponding to the given values of h. Then use Eq. (1) above to calculate q∞.

h(km) 60 50 40 30 20

p∞(N/m2) 25.6 87.9 299.8 1.19 × 103 5.53 × 103

M 17 9.5 5.5 3 1

q∞(N/m2) 5.2 × 103 5.6 × 103 6.3 × 103 7.5 × 103 3.9 × 103

Note that q∞ progressively increases as the shuttle penetrates deeper into the atmosphere, that it peaks at a slightly supersonic Mach number, and then decreases as the shuttle completes its entry.


26

4.30 Recall that total pressure is defined as that pressure that would exist if the flow were slowed isentropically to zero velocity. This is a definition; it applies to all flows — subsonic or supersonic. Hence, Eq. (4.74) applies, no matter whether the flow is subsonic or supersonic.

/( 1)

2 1.4/0.40 11 [1 0.2(2) ] 7.824

2

γ γγ 2∞

∞

pM

p

-æ ö- ÷ç= + = + =÷ç ÷÷çè ø

Hence:

40 2

lb7.824 7.824(2116) 1.656 10

ft∞p p= = = ´

Note that the above value is not the pressure at a stagnation point at the nose of a blunt body, because in slowing to zero velocity, the flow has to go through a shock wave, which is non-isentropic. The stagnation pressure at the nose of a body in a Mach 2 stream is the total pressure behind a normal shock wave, which is lower than the total pressure of the freestream, as calculated above. This stagnation pressure at the nose of a blunt body is given by Eq. (4.79).

22 2 210 1

21 2

1.4/0.42 2 2

2

( 1) 1 2

14 2( 1)

(2.4) (2) 1 1.4 2(1.4)(2)5.639

2.44(1.4)(2) 2(0.4)

γγγ γ γ

γγ γ∞p M M

p M

-é ù é ù+ - +ê ú ê ú= =ê ú ê ú+- -ê ú ê úë û ë û

é ù é ù- +ê ú ê ú =ê ú ê ú-ê ú ê úë û ë û

Hence,

2

40 2

lb5.639 5.639(2116) 1.193 10

ft∞p p= = = ´

If Bernoulli’s equation is used, the following wrong result for total pressure is obtained.

2 20

1

2 2

γρ∞ ∞ ∞ ∞ ∞ ∞ ∞p p q p V p p M= + = + = +

2 40 2

lb2116 0.7(2116)(2) 0.804 10

ftp = + = ´

Compared to the correct result of 42

lb1.656 10 ,

ft´ this leads to an error 51%.

4.31 2

0

11

2 1

γ γγ

ee

pM

p

æ ö- -÷ç= + ÷ç ÷÷çè ø -

5 2 3.5

4 2

5(1.01 10 )[1 0.2(3) ]

1.37 10 N/m

e

e

p

p

-= ´ +

= ´

1

2 2 1

0

11 [1 0.2(3) ]

2

γee

TM

T

--æ ö- ÷ç= + = +÷ç ÷÷çè ø

(500)(0.357) 178.6 KeT = =

4

31.37 100.267 kg/m

(287)(178.6)ρ e

ec

p

RT

´= = =


27

4.32 12

0

11

2

γγγe

ep

Mp

--æ ö- ÷ç= + ÷ç ÷÷çè ø

Hence,

1

2

0

21

1

γγ

γe

ep

Mp

-é ùê úæ ö÷ê úç ÷= -ç ÷ê úç ÷÷ç- è øê úê úë û

2 0.2865[(0.2) 1] 2.92eM -= - =

1.71eM =

( 1)/( 1)2

22

1 2 11

1

γ γγ

γ γe

t e

AM

A M

+ -é ùæ ö æ ö-÷ ÷ç çê ú÷ ÷= +ç ç÷ ÷ç ê úç÷ ÷ç÷ ÷ç + è øè ø ê úë û

2

2 62

1[(0.833)(1 0.2(1.71) )]

(1.71)e

t

A

A

æ ö÷ç ÷ = +ç ÷ç ÷÷çè ø

1.35e

t

A

A=

4.33 /( 1)

2 2 2 2 2 3.5

0 0

11 0.7 (1 0.2 )

2 2 2

γ γγ γ γq pM M M M M

p p

- --é ù-ê ú= = + = +

ê úë û

M M2 q/p0

0 0 0

0.2 0.04 0.027

0.4 0.16 0.100

0.6 0.36 0.198

0.8 0.64 0.294

1.0 1.0 0.370

1.2 1.44 0.416

1.4 1.96 0.431

1.6 2.56 0.422

1.8 3.24 0.395

2.0 4.00 0.358

Note that the dynamic pressure increases with Mach number for M < 1.4 but decreases with Mach number for M > 1.4. I.e., in an isentropic nozzle expansion, there is a peak local dynamic pressure which occurs at M = 1.4.


28

4.34 First, calculate the value of the Reynolds number.

75

(1, 225)(200)(3)Re 4.10 10

(1.7894 10 )

ρμ

∞ ∞

∞L

V L-= = = ´

´

The dynamic pressure is

2 2 4 21 1(1.225)(200) 2.45 10 N/m

2 2ρ∞ ∞ ∞q V= = = ´

Hence,

7

5.2 5.2(3)0.0024 m 0.24 cm

Re 4.1 10δL

L

L= = = =

´

and

7

1.328 1.3280.00021

Re 4.1 10f

L

C = = =´

The skin friction drag on one side of the plate is:

4(2.45 10 )(3)(17.5)(0.00021)∞f fD q Sc= = ´

270 NfD =

The total skin friction drag, accounting for both the top and the bottom of the plate is twice this value, namely

Total Df = 540 N

4.35 0.2 7 0.2

0.37 0.37(3)0.033 m 3.3 cm

(Re ) (4.1 10 )δ

L

L= = = =

´

From Problem 4.24, we find

turbulent laminar3.3

/ 13.750.24

δ δ = =

The turbulent boundary layer is more than an order of magnitude thicker than the laminar boundary layer.

0.2 7 0.2

0.074 0.0740.0022

(Re ) (4.1 10 )f

L

C = = =´

The skin friction drag on one side is then

4(2/45 10 )(3)(17.5)(0.0022)

2830 N

∞f f

f

D q Sc

D

= = ´

=

The total, accounting for both top and bottom is

Total Df = 5660 N

From Problem 4.24, we find

turbulent laminar

5660( ) /( ) 10.5

540f fD D = =

The turbulent skin friction drag is an order of magnitude larger than the laminar value.


29

4.36 cr

crρμ

∞ ∞

∞xe

V xR =

cr

6 5

cr(10 )(1.789 10 )

Re(1.225)(200)

μρ

∞

∞ ∞xx

V

-æ ö ´÷ç ÷= =ç ÷ç ÷÷çè ø

2cr 7.3 10 mx -= ´

The turbulent drag that would exist over the first 7.3 × 10−2 m of chord length from the leading edge (area A) is

0.2

cτ

4 26 0.2

0.074(on one side)

(Re )

0.074(2.45 10 )(7.3 10 )(17.5)

(10 )

∞A

A

f A

f

D q S

D -

=

= ´ ´

146 NAfD = (on one side)

From Problem 4.25, the turbulent drag on one side, assuming both areas A and B to be turbulent, is 2830N. Hence, the turbulent drag on area B alone is:

2830 146 2684 NBf

D = - - (turbulent)

The laminar drag on area A is

0.5

cr

1.328

(Re )∞AfD q S=

4 26 0.5

1.328(2.45 10 )(7.3 10 )(17.5)

(10 )AfD -= ´ ´

42 NAfD = (laminar)

Hence, the skin friction drag on one side, assuming area A to be laminar and area B to be turbulent is

(laminar) (turbulent)

42 2684 2726 NA Bf f f

f

D D D

D

= +

= + =

The total drag, accounting for both sides, is

Total Df = 5452 N

Note: By comparing the results of this problem with those of Problem 4.25, we see that the flow over the wing is mostly turbulent, which is usually the case for real airplanes in flight.


30

4.37 The relation between changes in pressure and velocity at a point in an inviscid flow is given by the Euler equation, Eq. (4.8)

ρdp VdV= -

Letting s denote distance along the streamline through the point, Eq. (4.8) can be written as

ρdp dVV

ds ds= -

or, 2 ( / )ρdp dV VV

ds ds= -

(a) ( / )

0.02 per millimeterdV V

ds=

Hence,

22

N(1.1)(100) (0.02) 220 per millimeter

m

dp

ds= - =

(b) 22

N(1.1)(1000) (0.02) 22,000 per millimeter

m

dp

ds= - =

Conclusion: At a point in a high-speed flow, it requires a much larger pressure gradient to achieve a given percentage change in velocity than for a low speed flow, everything else being equal.

4.38 We use the fact that total pressure is constant in an isentropic flow. From Eq. (4.74) applied in the freestream.

12 2 3.50 1

1 [1 0.2(0.7) ] 1.387

γγγ

γ ∞∞

pM

p

-æ ö- ÷ç ÷= + = + =ç ÷ç ÷ç ÷è ø

From Eq. (4.74) applied at the point on the wing,

12 2 3.50 1

1 [1 0.2(1.1) ] 2.1352

γγγp

Mp

-æ ö- ÷ç= + = + =÷ç ÷÷çè ø

Hence,

0 0 1.387/ p 0.65

2.135∞ ∞ ∞∞

p pp p p

p p

é ùæ ö æ ö æ ö÷ ÷ç çê ú ÷ç÷ ÷= = ==ç ÷ç ç÷ ÷ê ú ÷ç ç ÷ç÷ ÷ç ÷÷ç è øè øè øê úë û

At a standard altitude of 3 km, from Appendix A, 4 27.0121 10 N/m∞p = ´ .

Hence,

4 4 2(0.65)(7.0121 10 ) 4.555 10 N/mp = ´ = ´

4.39 This problem is simply asking what is the equivalent airspeed, as discussed in Section 4.12. Hence,

1/21/2 3

3

1.0663 10(800) 535.8ft.sec

2.3769 10

ρρe

sV V

-

-

æ öæ ö ´ ÷ç÷ç ÷ç÷= = =ç ÷÷ çç ÷÷÷ çç ÷÷çè ø ´è ø


31

4.40 (a) From Eq. (4.88)

1212

2

62 5

2

1 2 11

1 2

1 21 0.2 (10) 2.87 10

2.4(10)

γγγ

γe

et e

AM

A M

+-æ ö é ùæ ö-÷ç ÷çê ú÷ = +ç ÷ç÷ ÷ç ÷çê ú÷÷ç è ø+è ø ë û

ì üï ïé ùï ï= + = ´í ýê úï ïë ûï ïî þ

Hence,

52.87 10 535.9e

t

A

A= ´ =

(b) Form Eq. (4.87)

12 2 3.5 40 1

1 [1 0.2(10) ] 4.244 102

γγγ

ee

pM

p-æ ö- ÷ç= + = + = ´÷ç ÷÷çè ø

At a standard altitude of 55 km, p = 48.373 N/m2. Hence,

4 6 2

0 (4.244 10 )(48.373) 2.053 10 N/m 20.3atmp = ´ = ´ =

(c) From Eq. (4.85)

2 20 11 1 0.2 (10) 21

2

γe

e

TM

T

-= + = + =

At a standard altitude of 55 km, T = 275.78 K. Hence,

T0 = 275.78 (21) = 5791 K Examining the above results, we note that:

1. The required expansion ratio of 535.9 is huge, but is readily manufactured.

2. The required reservoir pressure of 20.3 atm is large, but can be handled by proper design of the reservoir chamber.

3. The required reservoir temperature of 5791 K is tremendously large, especially when you remember that the surface temperature of the sun is about 6000 K. For a continuous flow hypersonic tunnel, such high reservoir temperatures can not be handled. In practice, a reservoir temperature of about half this value or less is employed, with the sacrifice made that “true temperature” simulation in the test stream is not obtained.

4.41 The speed of sound in the test stream is

(1.4)(287)(275.78) 332.9m/sec.γe ea RT= = =

Hence,

10(332.9) 3329m/sece e eV M a= = =


32

4.42 (a) From Eq. 4.88, for Me = 20

1212

2

62 8

2

1 2 11

1 2

1 2[1 0.2 (20) ] 2.365 10

2.4(20)

γγγ

γe

et e

AM

A M

+-æ ö é ùæ ö-÷ç ÷çê ú÷ = +ç ÷ç÷ ÷ç ÷çê ú÷÷ç è ø+è ø ë û

ì üï ïï ï= + = ´í ýï ïï ïî þ

Hence,

15,377e

t

A

A=

(b) From Eq. (4.85)

1

2 2 10 11 [1 (0.2)(20) ] 0.01235

2

γe

e

TM

T

--æ ö- ÷ç= + = + =÷ç ÷÷çè ø

Hence,

(5791)(0.01235) 71.5K

(1.4)(287)(71.5) 169.5m/sec

20(169.5) 3390m/sec

γe

e e

e e e

T

a RT

V M a

= =

= = =

= = =

Comments:

1. To obtain Mach 20, i.e., to double the Mach number in this case, the expansion ratio must be increased by a factor of 15,377/535.9 = 28.7. High hypersonic Mach numbers demand wind tunnels with very large exit-to-throat ratios. In practice, this is usually obtained by designing the nozzle with a small throat area.

2. Of particular interest is that the exit velocity is increased by a very small amount, namely by only 61 m/sec, although the exit Mach number has been doubled. The higher Mach number of 20 is achieved not by a large increase in exit velocity but rather by a large decrease in the speed of sound at the exit. This is characteristic of most conventional hypersonic wind tunnels — the higher Mach numbers are not associated with corresponding increases in the test section flow velocities.


33

4.43 We will use the sketch shown in Figure 4.47, identifying the separate plan form areas A and B.

(a)

cr

2 2 21 12 2

5cr

(1.23)(20) 246 N/m

Re 5 10

ρρ

μ

∞ ∞ ∞

∞ ∞

∞x

q V

V x

= = =

= = ´

Hence:

5 5 5

cr(5 10 ) (1.789 10 )(5 10 )

0.3636 m(1.23)(20)

μρ

∞

∞ ∞x

V

-´ ´ ´= = =

Assuming turbulent flow over areas A + B,

0.2

0.074

ReA BfL

C+

=

Where

65

6 0.2

(1.23)(20)(4)Re 5.5 10

(1.788 10 )

0.0740.0032

(5.5 10 )

(246)(4)(4)(0.00332) 13.07 N

ρμ

∞ ∞

∞

∞

A B

A B

L

f

f f

V L

C

D q SC

+

+

-= = = ´´

= =´

= = =

The turbulent drag on area A is obtained from

cr

0.2 5 0.20.074 0.074

0.00536(Re ) (5 10 )

(246)(4)(0.3636)(0.00536) 1.92 N∞

A

A A

fx

f f

C

D q S C

= = =´

= = =

The turbulent drag on area B is

13.07 1.92 11.15 NB A B Af f fD D D

+= - = - =

The laminar drag on area A is obtained from

cr

0.5 5 0.51.328 1.328

0.00188(Re ) (5 10 )Af

x

C = = =´

(Laminar) (246)(4)(0.3636)(0.00188) 0.67 N∞A Af fD q S C= = =

The total friction drag is

0.67 11.15 11.82 NA Bf f fD D D= + = + =


34

(b) 40m/sec∞V =

2 2 21 12 2 (1.23)(40) 984 N/mρ∞ ∞ ∞q V= = =

65

(1.23)(40)(4)Re 11 10

(1.789 10 )

ρμ

∞∞

∞L

V L-= = = ´

´

(Turbulent) 0.2 6 0.2

0.074 0.0740.00289

Re (11 10 )A BfL

C+

= = =´

(984)(4)(4)(0.00289) 45.5 N∞A B A Bf fD q S C+ +

= = =

(Turbulent)

cr

0.2 5 0.20.074 0.074

0.00536(Re ) (5 10 )Af

x

C = = =´

Note that, since cr

Rex remains the same, but V∞ is doubled, xcr is half that from

part (a)

cr0.3636

0.1818m2

x = =

(984)(4)(0.1818)(0.00536) 3.84 N∞A Af fD q S C= = =

Hence,

45.5 3.84 41.66 N

A A B Af f fD D D+

= - = - =

(Laminar)

cr

0.5 5 0.51.328 1.328

0.00188(Re ) (5 10 )Af

x

C = = =´

(984)(4)(0.1818)(0.00188) 1.345 N∞A Af fD q S C= = =

Thus,

1.345 41.66 43NA Bf f fD D D= + = + =

(c) ∝ nfD V¥

Using the results from parts (a) and (b)

11.82 20

43 40

0.275 (0.5)

a

b

nf a

f b

n

n

D V

D V

æ ö÷ç ÷= ç ÷ç ÷÷çè ø

æ ö÷ç= ÷ç ÷÷çè ø

=

Taking the log of both sides

−0.56 = n (−0.30)

0.56

1.870.30

n = =

Note: Skin friction drag does not follow the velocity squared law; rather skin friction varies with velocity at a power slightly less than 2.


35

4.44 (a) For turblent flow, 0.2

1

Re∝fC . Since, Re ,

ρμ

∞ ∞V L= then

0.21∝fC

V¥

20.2

1∞ ∞

∞∝f fD q S C V

V

æ ö÷ç ÷ç= ÷ç ÷ç ÷ç ÷è ø

or,

1.8∞∝Df V

(b) For laminar flow, 0.5 0.5

1

Re ∞

1∝ ∝fCV

20.2

1∝fD VV

¥¥

æ ö÷ç ÷ç ÷ç ÷ç ÷ç ÷è ø

or,

1.5∞∝Df V

4.45 (a) M∞ = 1. Since a∞ = 340.3 m/sec at sea level,

V∞ = M∞ a∞ = (1)(340.3)=340.3 m/sec

2 4 21 12 2 (1.23)(340.3) 7.12 10 N/mρ 2

∞q V¥ = = = ´

75

(1.23)(340.3)(4)Re 9.36 10

(1.789 10 )

ρμ

∞ ∞

∞L

V L-

= = = ´´

inc 0.2 7 0.2

0.074 0.0740.00188

Re (9.36 10 )f

L

C = = =´

From Figure 4.44, for the turbulent case, approximate reading of the graph shows, for M∞ = 1,

inc

0.91f

f

C

C=

Thus,

Cf = 0.91(0.00188) = 0.00171 Df = q∞ S Cf (7.12 × 104)(4)(4)(0.00171)

1948 NfD =


36

(b) 3∞M =

3(340.3) 1021m/sec∞ ∞ ∞V M a= = =

2 2 5 21 12 2 (1.23)(1021) 6.41 10 N/mρ∞ ∞ ∞q V= = = ´

85

(1.23)(1021)(4)Re 2.81 10

(1.789 10 )

ρμ

∞ ∞

∞L

V L-

= = = ´´

inc 8 0.2

0.0740.00151

(2.81 10 )fC = =

´

From Figure 4.44, for the turbulent case, approximate reading of the graph shows, for M∞ = 3,

inc

0.56f

f

C

C=

0.56(0.00151) 0.000846fC = =

5(6.41 10 )(4)(4)(0.000846)∞f fD q S C= = ´

8677 NfD =

(c)

1

2

1

2

nf

f

D V

D V


1948 340.3

8677 1020.9

næ ö÷ç= ÷ç ÷÷çè ø

0 .2245 = (0.333)n −0.6488 = n (−0.4772)

0.6488

1.360.4772

n-

= =-

This implies 1.36.∝fD V¥ This is considerably different than the velocity squared law. In fact, comparing this result, where n = 1.36, with the incompressible result in Problem 4.43, where n = 1.87, indicates that the added effect of compressibility (higher Mach numbers) continues to drive down the exponent. The comparison is only qualitative, however, because the Reynolds numbers in Problem 4.45 are much larger than those in Problem 4.43.

In any event, the bottom line of Problems 4.43,4.44, and 4.45 is that flows with friction and/or compressibility do not follow the velocity squared law, but rather for such flows, the skin friction drag varies as nV∞ where n is an exponent less than 2.

4.46 (a) The waves travel at the local speed of sound.

(1.4)(287)(288) 340.2m/sec.a RTγ= = =

This speed is relative to the gas. Since the gas velocity is zero, then this is also the wave velocity relative to the pipe. One wave travels to the right along the positive x-axis with the velocity 340.2 m/sec, and the other wave travels to the left along the negative x-axis with a velocity of −340.2 m/sec.


37

(b) The x-location of the right-running wave at t = 0.2 sec is

x = at = (340.2)(0.2) = 68 m

The x-location of the left-running wave at t = 0.2 sec is

x = at = (−340.2)(0.2) = −68 m

4.47 The waves still travel at the local speed of round relative to the gas. Since the gas itself is moving relative to the pipe, then the wave speed relative to the pipe is the sum of the wave speed relative to the gas and the speed of the gas relative to the pipe. The velocity of right-running wave is therefore

Vwave = Vgas + wave speed

(a) When Vgas = 30 m/sec,

Vwave = 30 + 340.2 = 370.2 m/sec

The right-running wave travels to the right at 370.2 m/sec relative to the stationary pipe. After 0.2 sec, the x-location of this wave is

x = (370.2)(0.2) = 74 m

The left-running wave travels to the left at the velocity

Vwave = 30 − 340.2 = −310.2 m/sec

After 0.2 sec, the x-location of this left-running wave is

x = (−310.2)(0.2) = −62 m

(b) The flow velocity relative to the pipe is 400 m/sec, which is faster than the speed of sound. The weak pressure distributions, however, continue to propagate to the right and left at the speed of sound relative to the gas. Thus, the right-running wave has a velocity relative to the pipe of

Vwave = 400 + 340.2 = 740.2 m/sec

After 0.2 sec, the location of this wave is:

x = (740.2) 0.2 = 148.02 m

The left-running wave has a velocity relative to the pipe of

Vwave = 400 − 340.2 = 59.8 m/sec

After 0.2 sec, the location of this wave is

x = (59.8) (0.2) = 11.96 m

Note: The left-running wave, although it is traveling along to the left at the speed of sound relative to the gas, is moving to the right relative to the pipe because the flow velocity in the pipe is higher than the speed of sound.


38

4.48 The element of air initially has a pressure and density at the standard altitude of 1000 m of, from Appendix A,

p1 = 8.9876 × 104 N/m2

and

31 1.1117 kg/mρ =

At a standard altitude of 2000 m, p2 = 7.9501 × 104 N/m2. The density of the element of air, assuming it has experienced an isentropic process, is, from Eq. (4.37)

111.44

22 1 4

1

32

7.9501 101.1117

8.9876 10

1.0184kg/m

γρ ρ

ρ

p

p

æ öæ ö ´ ÷ç÷ç ÷ç÷= =ç ÷÷ çç ÷÷÷ çç ÷÷çè ø ´è ø

=

Comparing with the standard density at 2000 m of 1.0066 kg/m3, we see that the raised element of air has a higher density and is therefore heavier than its surrounding neighbors. This situation is a stable atmosphere because the isentropically perturbed element will tend to sink back down to its originally lower altitude.

4.49 1 1 2 2A V A V=

21 2

1

0.5(120) 15 mph

4

AV V

A= = =

From the answer to Problem 2.15, 60 mph = 26.82 m/sec.

126.82

(15) 6.705 m/sec60

V = = =

4.50

( )ρ ρ

ρ

+ = +

= + −

2 21 21 2

2 22 1 1 2

1 2 1 2

1 2

p V p V

p p V V

ρ

= = ×

==

= =

= × + −

= ×

5 21

3

1

2

5 2 22

5 22

1 atm 1.01 10 N/m

1.23 kg/m

6.705 m/sec

26.82(120) 53.64 m/sec

60

1.01 10 1 2(1.23)[(6.705) (53.64) ]

0.99258 10 N/m

p

V

V

p

p


39

4.51 Let x be the distance along the diffuser, and hD the height at the diffuser entrance. At any location x, the distance from the top to the bottom wall is hD + 2x tan θ where θ is the angle of the wall. Thus, the cross-sectional area at any location x is

(2) [ 2 tan ]DA h x θ= +

Thus,

4 tan 4 tan15 1.0718 mdA

dxθ= = ° =

4.52 The flow velocity at the entrance to the diffuser is the same as in the test section

2 120 mph 53.64 m/secV = =

At the exit of the diffuser, since

=

= = =

2 2 3 3

23 2

3

0.5(53.64) 7.66 m/sec

3.5

V A V A

AV V

A

4.53 AV = const

0dv dA

A Vdx dx

dv V dA

dx A dx

+ =

= −

From the solution of Problem 4.51,

1.0718 mdA

dx=

Thus,

1.0718dV V

dx A = −

(a) At the entrance, V2 = 53.64 m/sec (from Prob. 4.52)

22

1

(0.5)(2) 1 m

53.641.0718 57.49 sec

1

A

dV

dx−

= =

= − = −

(b) At the exit, V3 = 7.66 m/sec (from Prob. 4.52)

− = − = − = − 3 1

3

7.661.0718 1.0718 1.173 sec

7

VdV

dx A


40

4.54 From the Euler equation, Eq. (4.8) in the text,

dp VdV

dp dVV

dx dx

ρ

ρ

= −

= −

(a) At the inlet, from the solutions of Problems 4.52 and 4.53,

12 53.64 m/sec and 57.49 sec

dVV

dx−= = −

Thus,

3 3(1.23)(53.64)( 57.49) 3.793 10 N/mdp

dx= − − = ×

(b) At the exit,

13 7.66 m/sec and 1.173 sec

dVV

dx−= = −

Thus,

3(1.23)(7.66)( 1.173) 11.05 N/mdp

dx= − − =

4.55 −

° = =(3.5 0.5) / 2 1.5

tan 15x x

1.5 1.5

5.6 mtan15 0.2679

x = = =°

4.56 From the solution of Problem 4.54,

= ×

=

× + = =

= = =

3 3

2

3

3

3

2ave

ave

3.793 10 N/m

11.05 N/m

3.793 10 11.05 N1902

2 m

183 1830.096 m

1902s

dp

dx

dp

dx

dp

dx

xdp

dx

4.57 From the solution of Problem 4.56, (xs)laminar = 0.096 m.

1

turbulent ave1

laminar

ave

turbulent

506( )

2.765( )

183

( ) (2.765)(0.096) 0.265 m

s

s

s

dp

dxx

x dp

dx

x

−

−

= =

= =


41

4.58 At h = 7,500 ft, from App. B, T∞ = 491.95oR, p∞ = 1602.3 lb/ft2, and ∞ = 1.8975 × 10–3 slug/ft3.

2 3 20

(1.4)(1716)(491.95 1087 ft/sec

88229 mph 335.9 ft/sec

60

335.90.309

1087

1 2 1602.3 (0.5)(1.8975 10 )(335.9)

a RT

V

VM

a

p V

γ

ρ ρ

∞ ∞

∞

∞∞

∞−

∞ ∞ ∞

= = =

= =

= = =

= + = + ×

21709 lb/ft=

4.59 At h = 25,000 ft, from App. B, T∞ = 429.64oR, p∞ = 786.33 lb/ft2, and ∞ =1.0663 × 10–3 slug/ft3.

(1.4)(1716)(429.64) 1016 ft/sec

88610 894.67 ft/sec

60

894.670.88

1016

a RT

V

VM

a

γ∞ ∞

∞

∞∞

∞

= = =

= =

= = =

The flow is subsonic, but compressible. From Eq. (4.74),

10 2

0 2 3.5 3.5

0 2

11

2

[1 0.2(0.88) ] (1.15488) 2.303

lb2.303 2.303 (786.33) 1811

ft

pM

p

p

p

p

γγγ

ρ

−∞

∞

∞

∞

− = +

= + = =

= = =


42

4.60 At h = 35,000 ft, from App. B, T∞ = 394.08oR, p∞ = 499.34 lb/ft2, and ∞ = 7.382 × 10–4 slug/ft3.

(1.4)(1716)(394.08) 973 ft/sec

881328 1947.7 ft/sec

60

1947.72.00

973

a RT

V

VM

a

γ∞ ∞

∞

∞∞

∞

= = =

= =

= = =

The flow is supersonic. From Eq. (4.79),

2

2

2 2 210

2

1.42 2 20.4

2

20

( 1) 1 2

14 2( 1)

(2.4) (2) 1 1.4 2(1.4)(2)5.64

2.44(1.4) (2) 2(0.4)

5.64(499.34) 2817 lb/ft

p M M

p M

p

γγγ γ γ

γγ γ−∞ ∞

∞ ∞

+ − += +− −

− += = −

= =


43

Chapter 5

5.1 Assume the moment is governed by

( , , , , )ρ μ∞ ∞ ∞ ∞M f V S a=

More specifically:

( , , , , )a b d e fM Z V S aρ μ∞ ∞ ∞ ∞=

Equating the dimensions of mass, m, length, ℓ, and time t, and considering Z dimensionless,

( )2

22 3

b fa ed

tm m m

t tt

æ ö æ öæ ö æ ö÷ ÷÷ ÷ç çç ç= ÷ ÷÷ ÷ç çç ç÷ ÷÷ ÷÷ ÷÷ ÷ç çç çç çè ø è øè ø è ø

1 = b + f (For mass) 2 = a – 3b + 2d + e – f (For length) −2 = −a − e − f (for time)

Solving a, b, and d in terms of e and f,

b = 1 − f and, a = 2 − e − f

and, 2 = 2 − e − f − 3 + 3f + 2d + e − f

or 0 = −3 + f + 2d

3

2

fd

-=

Hence,

2 1 (3 )

2 1/21/2

/ 2ρ μ

μρρ

∞ ∞ ∞ ∞

∞ ∞∞ ∞

∞ ∞ ∞

e f f f e f

fe

M Z V S a

aZ V SS

V V S

- - - -=

æ öæ ö ÷ç÷ç ÷ç÷= ç ÷÷ çç ÷÷÷ç ç ÷ç ÷è ø è ø

Note that S l/2 is a characteristic length; denote it by the chord, c.

2 μρρ

∞ ∞∞

∞ ∞ ∞

c fa

M V S c ZV V c¥

æ ö æ ö÷ ÷ç ç÷ ÷= ç ç÷ ÷ç ç÷ ÷÷ ÷ç çè ø è ø

However, / 1/∞ ∞ ∞a V M=

and 1

Re

μρ∞

∞ ∞V c=

Let

1 1

Re 2∞

e fmc

ZM

æ ö æ ö÷ç ÷ç÷ =ç ÷ç÷ ÷ç ÷ç÷÷ç è øè ø

where cm is the moment coefficient. Then, as was to be derived, we have

21

2ρ∞ ∞ mM V c c=

or, ∞ mM q S c c=


44

5.2 From Appendix D, at 5° angle of attack,

0.67c =

/4

0.025cmc = -

(Note: Two sets of lift and moment coefficient data are given for the NACA 1412 airfoil—with and without flap deflection. Make certain to read the code properly, and use only the unflapped data, as given above. Also, note that the scale for

/4cmc is different than that for cℓ—be careful

in reading the data.)

With regard to cd, first check the Reynolds number,

7

(0.002377)(100)(3)Re

(3.7373 10 )

ρμ∞ ∞

∞

V c-

= =´

6Re 1.9 10= ´

In the airfoil data, the closest Re is 3 × 106. Use cd for this value.

0.007 (for 0.67)dc c= =

The dynamic pressure is

2 2 21 1(0.002377)(100) 11.9lb/ft

2 2ρ∞ ∞ ∞q V= = =

The area per unit span is 21( ) (1)(3) 3ftS c= = =

Hence, per unit span,

/4/4

(11.9)(3)(0.67) 23.9 lb

(11.9)(3)(0.007) 0.25lb

(11.9)(3)(3)( 0.025) 2.68 ft.lb

∞

∞

∞

c

d

c m

L q S c

D q S c

M q S c c

= = =

= = =

= = - = -

5.3 5

3(1.01 10 )

(287)(303) 1.61kg/mρ ∞

∞∞

p

RT

´= =

=

From Appendix D,

/4

0.98

0.012

cm

c

c

=

= -

Checking the Reynolds number, using the viscosity coefficient from the curve given in Chapter 4,

51.82 10μ∞-= ´ kg/m sec at T = 303 K,

55

(11.57)(42)(0.3)Re 8 10

1.82 10

ρμ∞ ∞

∞

V c-= = = ´

´

This Reynolds number is considerably less than the lowest value of 3 × 106 for which data is given for the NACA 23012 airfoil in Appendix D. Hence, we can use this data only to give an educated guess; use cd ≈ 0.01, which is about 10 percent higher than the value of 0.009 given for Re = 3 × 106 The dynamic pressure is

2 2 21 1(1.161)(42) 1024 N/m

2 2ρ∞ ∞ ∞q V= = =

The area per unit span is 2(1)(0.3) 0.3mS = = . Hence,

/4

(1024)(0.3)(0.98) 301N

(1024)(0.3)(0.01) 3.07 N

(1024)(0.3)(0.3)( 0.012) 1.1Nm

∞

∞

∞

d

c m

L q S c

D q S c

M q S c c

= = =

= = =

= = - = -


45

5.4 From the previous problem, q∞ = 1020 N/m2

∞ L q S c=

Hence,

∞

L

cq S

=

The wing area 2(2)(0.3) 0.6mS = =

Hence,

200

0.33(1024)(0.6)c = =

From Appendix D, the angle of attack which corresponds to this lift coefficient is

2α =

5.5 From Appendix D, at 4 ,α =

0.4c =

Also,

2 2 2

88120 176ft/sec

60

1 1(0.002377)(176) 36.8lb/ft

2 2ρ

∞

∞ ∞ ∞

V

q V


= = =

∞ L q S c=

229.52 ft

(36.8)(0.4)∞

LS

q c= = =

5.6 ∞ L q S c=

dD q S c∞=

Hence, ∞

∞

d d

L q S c c

D q S c c= =

We must tabulate the values of cℓ /cd for various angles of attack, and find where the maximum

occurs. For example, from Appendix D, at 00 ,α =

0.25

0.006

d

c

c

=

=

Hence 0.25

4.170.006

d

L c

D c= = =

A tabulation follows.

α 0° 1° 2° 3° 4° 5° 6° 7° 8° 9°

cℓ 0.25 0.35 0.45 0.55 0.65 0.75 0.85 0.95 1.05 1.15

cd 0.006 0.006 0.006 0.0065 0.0072 0.0075 0.008 0.0085 0.0095 0.0105

cℓ 41.7 58.3 75 84.6 90.3 100 106 112 111 110

cd

From the above tabulation, max

112L

D

æ ö÷ç »÷ç ÷÷çè ø


46

5.7 At sea level

3

5 2

1.225 kg/m

1.01 10 N/m

ρ

ρ∞

∞

=

= ´

Hence,

2 2 21 1(1.225)(50) 1531 N/m

2 2ρ∞ ∞ ∞q V= = =

From the definition of pressure coefficient,

5(0.95 1.01) 10

3.911531

∞

∞p

p pC

q

- - ´= = = -

5.8 The speed is low enough that incompressible flow can be assumed. From Bernoulli’s equation,

2 21 1

2 2ρ ρ∞ ∞ ∞ ∞ ∞ ∞p V p V p q+ = + = +

2 2

2

1 12 21

12

ρ ρ

ρ

∞∞

∞ ∞ ∞ ∞p

q V Vp pC

q q V

--= = = -

Since ρ = ρ∞(constant density)

2 262

1 1 1.27 0.2755p

VC

V∞

æ ö æ ö÷ç ÷ç÷= = - = - = -ç ÷ç÷ ÷ç ÷ç÷÷ç è øè ø

5.9 The flow is low speed, hence assumed to be incompressible. From Problem 5.8,

2 2

1951 1 0.485

160∞p

VC

V

æ ö æ ö÷ç ÷ç÷= - = - = -ç ÷ç÷ ÷ç ÷ç÷÷ç è øè ø

5.10 The speed of sound is

(1.4)(1716)(510) 1107 ft/secγ∞ ∞a RT= = =

Hence,

700

0.631107

∞∞

∞

VM

a= = =

In Problem 5.9, the pressure coefficient at the given point was calculated as −0.485. However, the conditions of Problem 5.9 were low speed, hence we identify

0

0.485pC = -

At the new, higher free stream velocity, the pressure coefficient must be corrected for compressibility. Using the Prandtl-Glauert Rule, the high speed pressure coefficient is

0

2 2

0.4850.625

1 1 (0.63)∞

pp

CC

M

-= = = -

- -


47

5.11 The formula derived in Problem 5.8, namely

2

1 ,∞

pV

CV

æ ö÷ç ÷= - ç ÷ç ÷÷çè ø

utilized Bernoulli’s equation in the derivation. Hence, it is not valid for compressible flow. In the present problem, check the Mach number.

(1.4)(1716)(505) 1101 ft/sec

7800.708.

1101

γ∞ ∞

∞

a RT

M

= = =

= =

The flow is clearly compressible! To obtain the pressure coefficient, first calculate ρ∞ from the

equation of state.

321160.00244 slug/ft

(1716)(505)ρ ∞

∞∞

p

RT= = =

To find the pressure at the point on the wing where V = 850 ft/sec, first find the temperature from the energy equation

2 2

2 2

2

2

∞∞

∞∞

p p

p

V Vc T c T

V

V VT T

c

+ = +

-= +

The specific heat at constant pressure for air is

(1.4)(1716) ft lb

60061 (1.4 1) slug

γγp

Rc

R= = =

- -

Hence,

2 2780 850

505 505 9.5 495.52(6006)

T R-

= + = - =

Assuming isentropic flow

1

3.52495.5

(2116) 1980 lb/ft505

γγ

∞

p T

p T

p

-æ ö÷ç ÷= ç ÷ç ÷÷çè ø


From the definition of Cp

2 2

1980 21161 1

(0.00244)(780)2 2

0.183

ρ∞ ∞

∞ ∞ ∞p

p

p p p pC

q V

C

- - -= = =

= -


48

5.12 A velocity of 100 ft/sec is low speed. Hence, the desired pressure coefficient is a low speed value,

0.pC

0

21 ∞

pp

CC

M=

-

From problem 5.11,

0.183pC = - and 0.708.∞M = Thus, 0

20.183

1 (0.708)

pC=

-

0

( 0.183)(0.706) 0.129pC = - = -

5.13 Recall that the airfoil data in Appendix D is for low speeds. Hence, at 4 ,α =

0

0.58.c =

Thus, from the Prandtl-Glauert rule,

0

2 2

0.580.97

1 1 (0.8)∞

cc

M= = =

- -

5.14 The lift coefficient measured is the high speed value, .c Its low speed counterpart is 0,c

where

0

1 2∞

cc

M=

-

Hence,

0

2(0.85) 1 (0.7) 0.607c = - =

For this value, the low speed data in Appendix D yield

2α =


49

5.15 First, obtain a curve of Cp,cr versus M∞, from

/( 1)

,cr2 2 ( 1)

11

γ γγγγ

2∞

2∞

pM

CM

-é ùæ öê ú+ - ÷ç ÷çê ú= -÷ç ÷ê úç + ÷÷çè øê úë û

Some values are tabulated below for 1.4γ = .

M∞` 0.4 0.5 0.6 0.7 0.8 0.9 1.0

Cp,cr −3.66 −2.13 −1.29 −0.779 −0.435 −0.188 0

Now, obtain the variation of the minimum pressure coefficient, Cp, with M∞, where

00.90.pC = - From the Prandtl-Glauert rule,

0

2

2

1

0.90

1

∞

∞

pp

p

CC

M

CM

=-

-=

-

Some tabulated values are:

M∞ 0.4 0.5 0.6 0.7 0.8 0.9 Cp −0.98 −1.04 −1.125 −1.26 −1.5 −2.06

A plot of the two curves is given on the next page.

From the intersection point,

cr 0.62M =


50

5.16 The curve of Cp, cr versus M∞ has already been obtained in the previous problem; it is a universal curve, and hence can be used for this and all other problems. We simply have to obtain the variation of Cp with M∞ from the Prandtl-Glauert rule, as follows:

0

2 2

0.65

1 1

pp

CC

M M∞ ∞

-= =

- -

M∞ 0.4 0.5 0.6 0.7 0.8 0.9 Cp −0.71 −0.75 −0.81 −0.91 −1.08 −1.49

The results are plotted below.

From the point of intersection,

cr 0.68M =

Please note that, comparing Problems 5.15 and 5.16, the critical Mach number for a given airfoil is somewhat dependent on angle of attack for the simple reason that the value of the minimum pressure coefficient is a function of angle of attack. When a critical Mach number is stated for a given airfoil in the literature, it is usually for a small (cruising) angle of attack.

5.17 Mach angle = μ = arc sin (1/M)

μ = arc sin (1/2) = 30°


51

5.18

1 11 1Sin Sin 23.6

25

10km/Tan = 22.9 km

0.436

μ

μ

M

d h

- -æ ö æ ö÷ ÷ç ç= = = ÷ ÷ç ç÷ ÷÷ ÷ç çè ø è ø

= =

5.19 At 36,000 ft, from Appendix B,

4 3

390.5

7.1 10 slug/ftρ∞

∞

T R-

=

= ´

Hence,

2 4 2 2

(1.4)(1716)(390.5) 969 ft/sec

(969)(2.2) 2132 ft/sec

1 1(7.1 10 )(2132) 1614 lb/ft

2 2

γ

ρ

∞ ∞

∞ ∞ ∞

∞ ∞ ∞

a RT

V a M

q V -

= = =

= = =

= = ´ =

In level flight, the airplane’s lift must balance its weight, hence

L = W = 16,000 lb.

From the definition of lift coefficient,

/ 16,000 /(1614)(210) 0.047∞LC L q S= = =

Assume that all the lift is derived from the wings (this is not really true because the fuselage and horizontal tail also contribute to the airplane lift.) Moreover, assume the wings can be approximated by a thin flat plate. Hence, the lift coefficient is given approximately by

4

1

α2∞

LCM

=-

Solve for α,

21 11 (0.047) (2.2) 1

4 4α 2

∞LC M= - = -

0.023α = radians (or 1.2 degrees)

The wave drag coefficient is approximated by

2

2

4 4(0.023)0.00108

1 (2.2) 1

α2∞

wDCM

= = =- -

Hence, (1614)(210)(0.00108)

366 lb

∞ ww D

w

D q S C

D

= =

=


52

5.20 (a) At 50,000 ft, ρ ∞ = 3.6391 × 10−4 slug/ft3 and T∞ = 390°R. Hence,

(1.4)(1716)(390) 968 ft/seca RTγ∞ ∞= = =

and (968)(2.2) 2130 ft/sec∞ ∞ ∞V a M= = =

The viscosity coefficient at T∞ = 390°R = 216.7 K can be estimated from an extrapolation of the straight line given in Fig. 4.30. The slope of this line is

5

8(2.12 1.54) 10 kg5.8 10

(350 250) (m)(sec)(K)

μd

dT

--- ´

= = ´-

Extrapolating from the sea level value of 51.7894 10μ -= ´ kg/(m)(sec), we have

at 216.7 K.∞T =

5 8

5

1.7894 10 (5.8 10 )(288 216.7)

1.37 10 kg/(m)(sec)

μ

μ∞

∞

- -

-

= ´ - ´ -

= ´

Converting to English engineering units, using the information in Chapter 4, we have

5

7 75

1.37 10 slug slug3.7373 10 2.86 10

ft sec ft sec1.7894 10μ∞

-- -

-

æ ö´ ÷ç ÷= ´ = ´ç ÷ç ÷÷çè ø´

Finally, we can calculate the Reynolds number for the flat plate:

4

87

3.6391 10 (2130)(202)Re 5.47 10

2.86 10

ρμ

∞ ∞

∞L

V L -

-´

= = = ´´

Thus, from Eq. (4.100) reduced by 20 percent

0.2 8 0.2

0.07 0.074(0.8) (0.8) 0.00106

(Re ) (5.74 10 )f

L

C = = =´

The wave drag coefficient is estimated from

2

, 2

4

1

α

∞d wc

M=

-

where 2

0.03557.3

α = = rad.

Thus, 2

, 2

4(0.035)0.0025

(2.2) 1d wc = =

-

Total drag coefficient = 0.0025 + (2) (0.00106) = 0.00462

Note: In the above, Cf is multiplied by two, because Eq. (4.100) applied to only one side of the flat plate. In flight, both the top and bottom of the plate will experience skin friction, hence that total skin friction coefficient is 2(0.00106) = 0.00212.

(b) If α is increased to 5 degrees, where α = 5/57.3 − 0.00873 rad, then

2

, 2

4(0.0873)0.01556

(2.2) 1d wc = =

-

Total drag coefficient = 0.01556 + 2(0.00106) = 0.0177


53

(c) In case (a) where the angle of attack is 2 degrees, the wave drag coefficient (0.0025) and the skin friction drag coefficient acting on both sides of the plate (2 × 0.00106 = 0.00212) are about the same. However, in case (b) where the angle of attack is higher, the wave drag coefficient (0.0177) is about eight times the total skin friction coefficient. This is because, as α increases, the strength of the leading edge shock increases rapidly. In this case, wave drag dominates the overall drag for the plate.

5.21 km 1h 1000m

251 km/h 251 69.7 m/sech 3600sec 1km∞V

æ öæ ö æ ö÷ç÷ ÷ç ç÷= = =÷ç ÷ç ç÷÷ ÷ç÷ ÷ç ç÷÷çè ø è øè ø

31.225 kg/mρ∞ =

2 2 21 1(1.225)(69.7) 2976 N/m

2 2ρ∞ ∞ ∞q V= = =

9800

0.203(2976)(16.2)∞

LL

Cq S

= = =

2 2(0.203)

0.002894AR (0.62)(7.31)π πi

LD

CC

e= = =

(2976)(16.2)(0.002894) 139.5 N∞ ii DD q S C= = =

5.22 85.5 km/h 23.75 m/sec∞V = =

2 2 21 1(1.225)(23.75) 345 N/m

2 2ρ∞ ∞ ∞q V= = =

9800

1.75(345)(16.2)∞

LL

Cq S

= = =

2 2(1.75)

0.215(0.62)(7.31)π πi

LD

CC

eAR= = =

(345)(16.2)(0.215) 1202 N∞ ii DD q S C= = =

Note: The induced drag at low speeds, such as near stalling velocity, is considerably larger than at high speeds, near maximum velocity. Compare the results of Problems 5.20 and 5.21.

5.23 First, obtain the infinite wing lift slope. From Appendix D for a NACA 65-210 airfoil,

0

1.05 at 8

0 at 1.5

αα °

L

C

C =

= =

= = -

Hence, 01.05 0

0.11 per degree8 ( 1.5)

a-

= =- -

The lift slope for the finite wing is

0

0

0.110.076 degree

57.3 57.3(0.11)11

(9)(5)AR ππ i

aa

a

e

= = =++

At α = 6°, 0( ) (0.076)[6 ( 1.5)] 0.57α αL LC a == - = - - =

The total drag coefficient is

2 2(0.57)(0.004)

AR (0.9)(5)

0.004 0.023 0.027

π πL

D d

D

CC c

e

C

= + = +

= + =


54

5.24 2 2 21 1(0.002377)(100) 11.9 lb/ft

2 2ρ∞ ∞ ∞q V= = =

at α = 10°, L = 17.9 lb. Hence

17.9

1.0(11.9)(1.5)∞

LL

Cq S

= = =

at α = −2°, L = 0 lb. Hence αL = 0 = −2°

1.0 0

0.083 per degree[10 ( 2)]α

LdCa

d

-= = =

- -

This is the finite wing lift slope.

0

057.31

ARπ

aa

a

e

=+

Solve for aο.

00.083

57.3 57.3(0.083)1 1

AR (0.96)(6)π π

aa

ae

= =- +

a0 = 0.11 per degree


55

5.25 At α = −1°, the lift is zero. Hence, the total drag is simply the profile drag.

2

0ARπL

D d d dC

C c c ce

= + = + =

2 21 1(0.002377)(130) 20.1 lb/ft

2 2ρ 2

∞ ∞ ∞q V= = =

Thus, at 0 1α α °L== = -

0.181

0.006(20.1)(1.5)∞

dD

cq S

= = =

At α = 2°, assume that cd has not materially changed, i.e., the “drag bucket” of the profile drag curve (see Appendix D) extends at least from −1° to 2°, where cd is essentially constant. Thus, at α = 2°,

50.166

(20.1)(1.5)

0.230.00763

(20.1)(1.5)

∞

∞

L

D

LC

q S

DC

q S

= = =

= = =

However:

2

ARπL

D dC

C ce

= +

2(0.166) 0.00146

0.00763 0.006 0.006(6)π e e

= + = +

e = 0.90

To obtain the lift slope of the airfoil (infinite wing), first calculate the finite wing lift slope.

0

(0.166 0)0.055 per degree

[2 ( 2)]

0.05557.3 57.3(0055)

1 1AR (0.9)(6)π π

a

aa

ae

-= =

- -

= =- -

a0 = 0.068 per degree

5.26 max

stall2 2(7780)

(1.225)(16.6)(2.1)ρ∞ L

WV

S C= =

stall 19.1 m/sec 68.7 km/hV = =


56

5.27 (a) 5

0.087 radians57.3

α = =

2 2 (0.087) 0.548πα πc = = =

(b) Using the Prandtl-Glauert rule,

0

2 2

0.5480.767

1 1 (0.7)∞

cc

M= = =

- -

(c) From Eq. (5.50)

2 2

4 4(0.087)0.2

1 (2) 1

α

∞c

M= = =

- -

5.28 For V∞ = 21.8 ft/sec at sea level

2 2 21 1(0.002377)(21.8) 0.565 lb/ft

2 2ρ∞ ∞ ∞q V= = =

1 ounce = 1/16 lb = 0.0625 lb.

0.0625

0.11(0.565)(1)∞L

LC

q S= = =

For a flat plate airfoil

2 2 (3 / 57.3) 0.329πα πc = = ⋅ =

The difference between the higher value predicted by thin airfoil theory and the lower value measured by Cayley is due to the low aspect ratio of Cayley’s test wing, and viscous effects at low Reynolds number.

5.29 From Eqs. (5.1) and (5.2), writtten in coefficient form

CL = CN cos α − CA sin α CD = CN sin α + CA cos α

Hence:

0.8 cos 6 0.06 sin 6 0.7956 0.00627 0.789

0.8 sin 0.06 cos 0.0836 0.0597 0.1433L

D

C

C α α° °= - = - =

= + = + = Note: At the relatively small angles of attack associated with normal airplane flight, CL and CN

are essentially the same value, as shown in this example.


57

5.30 First solve for the angle of attack and the profile drag coefficient, which stay the same in this problem

0

0 11 57.3 /( AR)

ααπL

ac a

a e= =

+

or, 0 10

[1 57.3 / AR]α πLCa e

a= +

0.35

{1 57.3(0.11)/[ (0.9)(7)]} 4.20.11

π °= + =

The profile drag can be obtained as follows

2

0.350.012

( / ) 29

ARπ

LD

L D

LD d

CC

C C

CC c

e

= = =

= +

or, 2 2(0.35)

0.012 0.0062AR (.9)(7)π πL

d DC

c Ce

= - = - =

Increasing the aspect ratio at the same angle of attack increases CL and reduces CD. For AR = 10, we have

0

0 1

2 2

1 57.3 /( AR)

(0.11)(4.2)0.3778

1 57.3(0.11)/[ (0.9)(10)]

(0.3778)0.062 0.0062 0.005048 0.112

AR (.9)(10)

ααπ

π

π π

L

LD d

aC a

a e

CC c

e

= =+

= =+

= + = - = + =

Hence, the new value of L/D is 0.3778

33.70.0112

L

D

C

C= =


58

5.31 For incompressible flow, from Bernoulli’s equation, the stagnation pressure is

10 2 ρ 2

∞ ∞ ∞p p V= +

Let S be the surface area of each face of the plate. The aerodynamic force exerted on the front face is

front 0F p S=

and the aerodynamic force on the back face is

back ∞F p S=

The net force, which is the drag, is

front back 0( )∞D F F p p S= - = -

From Bernoulli’s equation,

10 2 ρ 2

∞ ∞ ∞p p V- =

Thus,

212

2

2 2

12

1 12 2

ρ

ρ

ρ ρ

∞ ∞

∞ ∞

∞ ∞ ∞ ∞D

D V S

V SDC

V S V S

=

= =

Hence,

1DC =

5.32 The drag of the airplane is

∞ DD q S C= (1)

The drag of a flat plate at 90° to the flow, with an area f and drag coefficient of 1, is

pblc

(1)∞ ∞DD q f C q f= = (2)

Equating Eqs. (1) and (2)

∞ ∞Dq S C q f=

or,

Df C S=

5.33 2(0.0163)(233) 3.8ftDf C S= = =


59

5.34 From Appendix D, for the NACA 2412 airfoil at zero angle of attack, Cd = 0.006. This is the profile drag coefficient, the sum of skin friction drag and pressure drag due to flow separation. To estimate the skin friction drag coefficient, from Eq. (4.101)

, 0.2 6 0.20.074 0.074

0.023Re (8.9 10 )

d fL

C = = =´

This is the skin friction coefficient due to flows over only one side of the plate. Including both the top and bottom surfaces of the plate,

, 2(0.003) 0.006d fc = =

Since from Section 5.12

, ,= +d d f d pc c c

we have: , , 0.006 0.006 0.d p d d fc c c= - = - = For this case, the pressure drag due to flow

separation is negligible.

5.35 From Appendix D, for the NACA 2412 airfoil at an angle of attack of 6°,

0.0075dc =

Assuming, as in Problem 5.34, that the skin friction drag is given by the flat plate result at zero angle of attack, we have

, 0.006d fc =

Thus,

, , 0.0075 0.006 0.0015d p d d fc c c= - = - =

For this case, the percentage of drag due to pressure drag is

, 0.00150.2,or 20%

0.0075d p

d

c

c= =

Clearly, as the angle of attack of the airfoil increases, the region of separated flow grows larger, and the pressure drag due to flow separation increases. The large increase in cd at higher values of cl, hence a, seen in Appendix D is due almost entirely to an increase in pressure drag due to flow separation.


60

5.36 In Problem 5.34, the total turbulent skin friction drag coefficient along one surface of the airfoil (top or bottom) was calculated to be 0.003. The transition Reynolds number is 500,000. This implies that the turbulent skin friction coefficient based on the running length of surface from the leading edge to the transition point, xcr, is

cr, 0.2 0.2

,cr

0.074 0.074 0.074( ) 0.00536

13.8(Re ) (500,000)d f x

x

c = = = =

The contribution of this turbulent skin friction drag coefficient to the total turbulent skin friction drag coefficient for the complete surface calculated to be 0.003 in Problem 5.34 is

46

500,0000.00536 3.01 10

8.9 10-æ ö÷ç = ´÷ç ÷ç ÷çè ø´

Hence, the contribution to the overall skin friction drag coefficient for one surface of the airfoil due to the turbulent flow from the transition point to the trailing edge of the airfoil is:

4, turb( ) 0.003 3.01 10 0.0027d fc -= - ´ =

The laminar skin friction drag coefficient for the flow from the leading edge to the transition point is, from Eq. (4.98)

3, laminar

,cr

1.328 1.328 1.3281.878 10

707.1Re 500,000f

x

C -= = = ´

Since the laminar skin friction is acting only on that part of the surface from x = 0 to x = xcr, then its contribution to the total overall skin friction coefficient for the whole top surface is

3 4, lam 6

500,000( ) 1.878 10 1.055 10

8.9 10d fc - -æ ö÷ç= ´ = ´÷ç ÷ç ÷çè ø´

Therefore, the total skin friction coefficient for one surface, including both the laminar flow from x = 0 to xcr, and turbulent flow from xcr to the trailing edge, is

4, 1.055 10 0.0027 0.0028d fc -= ´ + =

Including both the upper and lower surfaces, then we have

,

, ,

2(0.0028) 0.0056d f

d d f d p

c

c c c

= =

= +

Thus, , , 0.006 0.0056 0.004d p d d fc c c= - = - =

The percentage of the total drag due to pressure drag due to flow separation is then

, 0.0004(100) 6.7%

0.006d p

d

c

c= = =

In contrast to the result obtained in the solution to Problem 5.34, the more realistic result obtained here shows that the pressure dag is not negligible, although it is small compared to the skin friction drag.

Note: This problem is similar to worked Example 4.28 in the text. In Example 4.28, the respective skin friction drag forces on various parts of the surface are calculated and are used to solve the problem. In our solution here for Problem 5.36, we have used only the various skin friction coefficients to solve the problem. We do not need to calculate forces. Indeed, in the statement of the problem there is no velocity, density, or surface area given, so we can not calculate forces; none is needed. However, we have been careful in the present calculation to account for the specific regions of the surface to which the calculated coefficients apply, and to multiply the coefficients obtained over the portion of surface between x = 0 and xcr from Eqs. (4.98) and

(4.101) by the ratio 6cr / 500,000 / 8.9 10 0.0056.x L = ´ =

It is instructive to work out Example 4.28 in the text using only the various skin friction coefficients as we have used here, and then to calculate total drag Df at the very end of the calculation. You get the same answer as was obtained in Example 4.28.


61

5.37 (a) From Eq. (4.98),

46

1.328 1.3284.43 10

Re 9 10fC −= = = ×

×

This is for one side of the flat plate. The total skin friction drag coefficient including both sides of the plate is

42(4.43 10 ) 0.000885fC −= × =

(b) From Eq. (4.101)

0.2 6 0.2

0.074 0.074 0.0740.003

24.59(Re) (9 10 )fC = = = =

×

For both sides of the plate,

2(0.003) 0.006fC = =

The section drag coefficient for the NACA 2415 airfoil from App. D is 0.0064.

If the boundary layer were laminar, then the fraction of drag due to friction is (0.000885/0.0064) = 0.138, or 13%. If the boundary layer were turbulent, then the fraction of drag due to friction is 0.006/0.0064 = 0.938, or 93.8%. These results highlight the large effect of a turbulent boundary on the drag of an airfoil as compared to a laminar boundary layer.

5.38 (See Section 4.19 for the solution technique.)

Assuming all turbulent flow,

0.006 (from Problem 5.37).fC =

Transition takes place at location

cr6

650,0000.072

9 10

x

c= =

×

The turbulent drag coefficient on the area upstream of transition would be

0.2

0.074 0.0742 2 0.0102

14.54(650,000)fC

= = =

Thus, the relative turbulent drag contribution due to the area downstream of the transition point is

0.006 (0.0102)(0.072) 0.006 0.000734 0.005266− = − =

The laminar drag coefficient on the area upstream of transition is

1.328 1.328

2 2 0.0033806650,000

fC = = =

The laminar drag contribution over this area is

(0.0033)(0.072) = 0.000238

The total skin friction drag coefficient is

laminar contribution + turbulent contribution =

0.000238 0.005266 0.0055+ =

The total section drag coefficient is 0.0064 from App. D. Thus,

0.0055

Friction drag percentage 100 86%0.0064

= =

Pressure drag percentage %= 14


62

5.39 Assuming all turbulent flow,

0.2 6 0.2

0.074 0.0740.003749

(Re) (3 10 )fC = = =

×

Taking into account both sides of the plate

(0.003749) 2 0.0075fC = =

Transition takes place at location

cr6

650,0000.2167

3 10

x

c= =

×

The turbulent drag coefficient on the area upstream of transition would be

0.2

0.0742 0.0102

(650,000)fC

= =

The turbulent drag contribution due to the area downstream of the transition point is

0.0075 – (0.0102)(0.2167) = 0.00529

The laminar drag coefficient on the area upstream of transition is

1.328

2 0.0033650,000

fC

= =

The laminar drag contribution over this area is

(0.0033)(0.2167) = 0.000715

The total skin friction drag coefficient is

0.000715 0.00529 0.0060+ =

The experimental data for the total section drag coefficient from App. D is 0.0068.

0.006Friction drag percentage 100 88%

0.0068

Pressure drag percentage 11.8%

= =

=


63

Chapter 6

6.1 (a) 350km/hr 97.2m/secV¥ = =

0

2 2 2

2 2

1 1(1.225)(97.2) 5787 N/m

2 238220

0.242(5787)(27.3)

(0.242)0.03

AR (0.9)(75)

0.03 0.0028 0.0328

ρ

π π

L

LD D

D

q V

WC

q S

CC C

e

C

¥ ¥ ¥

¥

= = =

= = =

= + = +

= + =

/ 0.242 / 0.0328 7.38L DC C = =

38,220

= 5179 N/ 7.38R

L D

WT

C C= =

(b) 2 2 21 1(0.777)(97.2) 3670 N/m

2 2ρq V¥ ¥ ¥= = =

0

2 2

38,220= 0.38

(3670)(27.3)

(0.38)0.03

AR (0.9)(7.5)

0.03 0.0068 0.0368

π π

L

Ld D

D

WC

q S

CC C

e

C

¥= =

= + = +

= + =

/ 0.38 / 0.0368 10.3L DC C = =

38,220

3711N/ 10.3R

L D

WT

C C= = =

6.2 88

200 293.3ft/sec60

V¥ = =

2 2 2

2

1 1(0.002377)(293.3) 102.2 lb/ft

2 25000

0.245(102.2)(200)

(0.245)0.0024

AR (0.93)(8.5)

ρ

π πi

L

LD

q V

L WC

q S q S

CC

e

¥ ¥ ¥

¥ ¥

= = =

= = = =

= = =

Since the airplane is flying at the condition of maximum L/D, hence minimum thrust required, .

i eD DC C= Thus,

02 2(0.0024) 0.1048

(102.2)(200)(0.0048) 98.1lb.i iD D D D

D

C C C C

D q SC¥

= + = = =

= = =


64

6.3 (a) Choose a velocity, say V∞ =100 m/sec

0

2 2 2

2 2

6

1 1(1.225)(100) 6125 N/m

2 2103,047

0.358(6125)(47)

(0.358)0.032

AR (0.87)(6.5)

0.032 0.007 0.0392

103,04711287 N

/ 9.13

(11,287)(100) 1.129 10 watts

1129kw

ρ

π π

L

LD D

D

RL D

R R

R

q V

WC

q S

CC C

e

C

WT

C C

P T V

P

¥ ¥ ¥

¥

¥

= =

= = =

= + = +

= + =

= = =

= = = ´

=

A tabulation for other velocities follows on the next page:

V (m/sec) CL CD CL/CD PR (kw)

100 0.358 0.0392 9.13 1129

130 0.212 0.0345 6.14 2182

160 0.140 0.0331 4.23 3898

190 0.099 0.0325 3.05 6419

220 0.074 0.0323 2.29 9900

250 0.057 0.0322 1.77 14,550

280 0.046 0.0321 1.43 20,180

310 0.037 0.0321 1.15 27,780

(b) (2)(40298) 80596A AP T V V V¥ ¥ ¥= = =

The power required and power available curves are plotted below.

From the intersection of the AP and RP curves, we find,

max= 295m/sec at sea levelV


65

(c) At 5 km standard altitude, 3=0.7364kg/mρ

Hence,

alt 0 0

1/2 1/20

1/2alt 0 0 0

1/20

( / ) (1.225/0.7364) 1.29

( / ) 1.29

( / ) 1.29

ρ ρ

ρ ρ

ρ ρR R R

V V V

P P P

= =

= =

= =

From the results from part (a) above,

0

(m/sec)

V 0

(kw)

RP

alt

(m/sec)

V

(kw)akRP

100 1129 129 1456 130 2182 168 2815 160 3898 206 5028 190 6419 245 8281 220 9900 284 12,771 250 14,550 323 18,770

(d) all 0 0

0= 0.601

ρρA A AT T T

æ ö÷ç ÷ =ç ÷ç ÷÷çè ø

Hence, alt alt

(0.601)(80596) 48438A AP T V V V¥ ¥ ¥= = =


From the intersection of the PR and PA curves, we find max 290m/secat 5kmV =

Comment: The mach numbers corresponding to the maximum velocities in parts (b) and (d) are as follows:

At sea level

(1.4)(287)(288.16) 340m/sec

2950.868

340

γa RT

VM

a

¥ ¥

¥¥

¥

= = =

= = =

At 5 km altitude

(1.4)(287)(255.69) 321m/sec

2900.90

321

γa RT

VM

a

¥ ¥

¥¥

¥

= = =

= = =

These mach numbers are slightly larger than what might be the actual drag-divergence Mach number for an airplane such as the A-10. Our calculations have not taken the large drag rise at drag-divergence into account. Hence, the maximum velocities calculated above are somewhat higher than reality.


66

6.4 (a) Choose a velocity, say 100ft/secV¥ =

0

2 2 2

2 2

1 1(0.002377)(100) 11.89 lb/ft

2 23000

1.39(11.89)(181)

(1.39)0.027

AR (0.91)(6.2)

0.027 0.109 0.136

3000 3000293.5lb

/ (1.39) /(0.136) 10.22

(293.5)(100) 29

ρ

π π

L

LD D

D

RL D

R R

q V

WC

q S

CC C

e

C

WT

C C

P T V

¥ ¥ ¥

¥

⋅ ¥

= = =

= = =

= + = +

= + =

= = = =

= = = 350ft lb/sec

In terms of horsepower,

29350

53.4hp550RP = =

A tabulation for other velocities follows:

V∞ (ft/sec) CL CD CL/CD PR (hp)

70 2.85 0.485 5.88 64.9

100 1.39 0.136 10.22 53.4

150 0.62 0.0487 12.37 64.3

200 0.349 0.0339 10.29 106

250 0.223 0.0298 7.48 182

300 0.155 0.0284 5.46 300

350 0.114 0.0277 4.12 463

(b) At sea level, maximum PA = 0.83 (345) = 286 hp. The power required and power available are plotted below.

From the intersection of the PA and PR curves,

max 295ft/sec 201mph at sea levelV = =


67

(c) At a standard altitude of 12,000 ft,

3 31.648 10 slug/ftρ -= ´

Hence,

ak 0 0

1/2 1/ 20

1/2alt 0 0 0

1/20

( / ) (0.002377 / 0.001648) 1.2

( / ) 1.2

( / ) 1.2

ρ ρ

ρ ρ

ρ ρR R R

V V V

P P P

= =

= =

= =

Using the results from part (a) above,

0

(ft/sec)

V 0

(hp)RP

alt

(ft/sec)

V

ah

(hp)RP

70 64.9 84 77.9 100 53.4 120 64.1 150 64.3 180 76.9 200 106 240 127 250 182 300 218 300 300 360 360

(d) Assuming that the power output of the engine is proportional to p∞,

alt 0 0 00

0.001648( / ) 0.693

0.002377ρ ρA A A AP P P P

æ ö÷ç= = =÷ç ÷÷çè ø

At 12,000 ft,

0.693(286) 198hpAP = =


From the intersection of the PA and PR curves,

max = 290ft/sec =198mph at12,000ft.V

6.5 From the PA and PB curves generated in Problem 6.3, we find approximately:

excess power = 9000 kw at sea level excess power = 5000 kw at 5 km

Hence, at sea level

6

5excess power 9 10 watts

R/C= 87.3m/sec1.0307 10 NW

´= =

´

and at 5 km altitude,

6

55 10

R/C = 48.5m/sec1.03047 10

´=

´


68

6.6 From the PA and PR curves generated in Problem 6.4, we find approximately:

excess power = 232 hp at sea level excess power = 134 hp at 12,000 ft

Hence, at sea level,

excess power (232)(550)

R/C= 42.5ft/sec3000W

= =

and at 12,000 ft altitude,

(134)(550)

R/C 24.6ft/sec3000

= =

6.7 Assuming (R/C)max varies linearly with altitude,

max(R/C) ah b= +

From the two (R/C)max values from Problem 6.5,

87.3 (o)

48.5 (5000)

a b

a b

= += +

Hence,

max

87.3

0.00776

(R/C) 0.00776 87.3

b

a

h

== -

= - +

To obtain the absolute ceiling, set (R/C)max = 0, and solve for h.

87.3

11,250m0.00776

h = =

Hence,

absolute ceiling ≈ 11.3 km

6.8 max(R/C) ah b= +

From the results of Problem 6.6,

42.5 (0)

24.6 (12,000)

a b

a b

= += +

Hence,

max

42.5

0.00149

(R/C) 0.00149 42.5

b

a

h

== -

= - +

To obtain the absolute ceiling, set (R/C)max = 0, and solve for h.

42.5

28,523ft0.00149

h = =

absolute ceiling ≈ 28,500 ft

COMMENT TO THE INSTRUCTOR In the previous problems dealing with a performance analysis of the twin-jet and single-engine piston airplanes, the answers will somewhat depend on the precision and number of calculations made by the student. For example, if the PR curve is constructed from 30 points instead of the six or eight points as above, the subsequent results for rate-of-climb and absolute ceiling will be more accurate than obtained above. Some leeway on the students' answers is therefore advised. In my own experience, I am glad when the students fall within the same ballpark.


69

6.9 max( / ) 5000(7.7) 38,500ft 729milesR h L D= = = =

6.10 First, we have to calculate the CL corresponding to maximum L/D.

2

max

,O

,0 ,0max

,0 2

AR AR

2 4

1 (0.25) (0.7)(4.11)e AT 0.038

4 (7.7)

π π

ππ

DL

D D D

LD

D

C eC e

C C C

CC

C

-

æ ö÷ç ÷ = =ç ÷ç ÷÷çè ø

æ ö÷ç ÷= = =ç ÷ç ÷÷çè ø

At maximum ,0 ,i/ , 0.038D DL D C C= =

Hence: CD = 2(0.038) =0.076

(0.076)(7.7) 0.585LL D

D

CC C

C

æ ö÷ç ÷= = =ç ÷ç ÷÷çè ø

Also:

1

minmax

min

ArcTan ArcTan (0.13)

7.4

2cos 2cos(7.4 ) 140097.2ft/sec

(0.002175)(0.585) 231

θ

θ

θρ L

L

D

WV

C S

-

¥¥


=

æ ö æ ö÷ ÷ç ç= = =÷ ÷ç ç÷ ÷÷ ÷ç çè ø è ø

6.11 From Eq. (6.85),

( )0

0

t/2

max

AR

2

π D

D

e CL

D C

æ ö÷ç =÷ç ÷÷çè ø

Putting in the numbers:

[ ]1/2

max

(0.9)(6.72)(0.025)13.78

0.050

πL

D

æ ö÷ç = =÷ç ÷÷çè ø


70

6.12 Aviation gasoline weighs 5.64 Ib per gallon,

Hence,

(44)(5.64) 248lb.FW = =

Thus, the empty weight is

1 3400 248 3152lb.W = - =

The specific fuel consumption, in consistent units, is

7 1= 0.42/(550)(3600) = 2.12 10 ftc - -´

The maximum L/D can be found from Eq. (6.85).

[ ]c

c

1/2 1/2

max

( AR ) (0.91)(6.2)(0.027)12.8

2 2(0.027)

π πD

D

e CL

D C

æ ö÷ç = = =÷ç ÷÷çè ø

Thus, the maximum range is:

0

1max

7

6

0.83 3400(12.8)

31522.12 10

3.8 10 ft

η

L

D

WCR n

c C W

R n

R

-

æ ö æ öæ ö ÷ ÷ç ç÷ç ÷ ÷= ç ç÷ç ÷ ÷÷ç ç÷ç ÷ ÷÷ ÷ç çè øè ø è ø

æ ö æ ö÷ ÷ç ç= ÷ ÷ç ç÷ ÷÷ç ç÷ç è øè ø´

= ´

In terms of miles,

63.8 10

719 miles5280

R´

= =

To calculate endurance, we must first obtain the value of 3/2max( / )L DC C From Eq. (6.87),

0

3/43/2

max

(3 AR)

4

π

c

DL

D D

C eC

C C

æ ö÷ç ÷ç =÷ç ÷ç ÷÷çè ø

Putting in the numbers:

[ ]1/23/2

max

3(0.027) (0.91)(6.2)11.09

4(0.027)

πL

D

C

C

æ ö÷ç ÷ç = =÷ç ÷ç ÷÷çè ø

Hence, the endurance is

[ ]

3/ 21/2 1/ 2 1/ 2

1 0

1/ 2 1/ 2 1/ 27

4

(2 ) ( )

0.83(11.09) 2(0.002377)(181) (3152 3400 )

2.12 10

2.67 10 sec 7.4hr

η ρL

D

CE S W W

c C

E

E

- -¥

- --

æ ö÷ç= -÷ç ÷÷çè ø

æ ö÷ç= -÷ç ÷ç ÷çè ø´

= ´ =


71

6.13 One gallon of kerosene weighs 6.67 Ib. Since 1 Ib = 4.448 N, then one gallon of kerosene also weighs 29.67 N. Thus,

0

(1900)(29.67) 56,370 N

136,960 56,370 80,590 N

f

t f

W

W W W

= =

= - = - =


4 1N= 1.0 2.777 10 sec

(N)(hr)tc - -= ´

Also, at a standard altitude of 8 km.

30.526 kg/mρ¥ =

Since maximum range for a jet air craft depends upon maximum 1/2/ ,L DC C we must use

Eq. (6.86).

c

1/4

1/2

max

1AR

34

3

π

c

DL

DD

C eC

C C

æ ö÷ç ÷çæ ö ÷÷çè ø÷ç ÷ç =÷ç ÷ç ÷÷çè ø

Putting in the numbers,

1/4

1/2

max

1(0.032)(0.87)(6.5)

315.46

4(0.032)

3

L

D

C

C

é ùê úæ ö ê ú÷ç ë û÷ç = =÷ç ÷ç ÷÷çè ø

For a jet airplane, the range is

1/21/2 1/2

0 1

1/21/2 1/2

4

6

2 12 ( )

2 12 (15.46) (136,960) (80,590)

(0.526)(47) 2.777 10

2.73 10 m 2730km

ρL

t D

CR W W

S c C

R

R

¥

-

æ öæ ö ÷ç÷ç ÷ç÷= -ç ÷÷çç ÷÷÷çç ÷÷çè øè ø

é ù æ ö é ù÷çê ú= -÷ç ê ú÷çê ú ÷ç ë ûè ø´ë û

= ´ =

The endurance depends on CLCD. From Eq. (6.85),

[ ]1/2 1/2

max

( AR ) (0.87)(6.5)(0.032)11.78

2 2 (0.032)

π πc

c

D

D

e CL

D C

æ ö÷ç = = =÷ç ÷÷çè ø

The endurance for a jet aircraft is

0

1

4

1=

1 136,960(11.78)

80,5902.777 10

22,496 sec 6.25 hr

L

t D

WCE n

c C W

E n

E

-

æ ö æ ö÷ ÷ç ç÷ ÷ç ç÷ ÷ç ç÷ ÷÷ ÷ç çè ø è ø

æ ö÷ç ÷= ç ÷ç ÷çè ø´= =


72

6.14

0

3/ 2 3/ 2

2

ARπ

L L

D LD

C C

C CC

e

=

+

12

0 0

23 / 2 2 23/ 2( / ) 3

2 0AR 2 AR ARπ π π

L D L L LD L DL

L

d C C C C CC C C C

d C e e e

-é ùæ ö æ öæ öæ ö÷ ÷ç çê ú÷ç÷÷ ÷çç ç÷= - + =÷ ç÷ ÷çê úç ÷ ç÷ ç÷ ÷÷ç ÷÷çç çè ø÷ ÷÷ ÷ç çè øê úè ø è øë û

12

0

5 / 23 10

2 2 ARπL

DLC

C Ce

- =

0

23 0

ARπL

DC

Ce

- =

Hence, 0 i

1

3D DC C= This is Eq. (6.80)

To obtain Eq. (6.81)

0

1/2 1/2

2

ARπ

L L

D LD

C C

C CC

e

=

+

1 12 2

0 0

21/ 2 2 2( / ) 1 20

AR 2 AR ARπ π πL D L L L

D DL LL

d C C C C CC C C C

d C e e e

-- -

é ùæ ö æ öæ ö÷ ÷ç çê ú÷ç÷ ÷ç ç÷= - + =ç÷ ÷ê úç ÷ çç÷ ÷÷÷çç ç÷ ÷÷ ÷ç çè øê úè ø è øë û

12

0

5/21 30

2 2 ARπL

D LC

C Ce

- - =

0

3/23 0

ARπL

DC

Ce

- =

0 i

3D DC C=

6.15 From Eq. (6.81), 0 i

3D DC C=

0

23

ARπL

DC

Ce

=

Thus,

12

0

1AR

3πL DC e C


14

12 0

0max

1AR

343

π DL

DD

e CC

C C

æ ö÷çæ ö ÷ç÷ ÷÷ç çè ø÷ç ÷ç =÷ç ÷ç ÷÷çç ÷è ø This is Eq. (6.86)

From Eq. (6.80), 0 i

1

3D DC C= 0

2

3 ARπL

DC

Ce

=

Thus, 12

0(3 AR )πL DC e C=

33 420

0max

3 AR

4

π DL

D D

e CC

C C

æ ö æ ö÷ç ÷ç÷ç ÷÷ çç = ÷÷ çç ÷÷ ç ÷ç ÷ ÷ççè ø÷çç ÷è ø (This is Eq. (6.87)


73

6.16 2AR / , hence AR (47)(6.5) 17.48 mb S b S= = = =

,max

2

2

0 stall

5 ft (5 / 3.28) m 1.524 m

/ 1.54 /17.48 0.08719

(16 / ) 1.9460.66

2.9461 (16 / )

2 2(103,047)1.2 1.2 1.2 80.3 m/sec

(1.255)(47)(0.8)

φ

ρLL

h

h b

h b

h b

WV V

S C¥

= = == =

= = =+

= = = =

Hence, 0.7 00.7 56.2 m/sec.LV = This is the velocity at which the average force is evaluated.

2 2 21 1(1.225)(56.2) 1935 N/m

2 2ρq V¥ ¥ ¥= = =

(1935) (47)(0.8) 72,760 NL

L q S C¥= = =

0

2

ARφ

πL

D DC

D q S C q S Ce¥ ¥

æ ö÷ç ÷ç= = + ÷ç ÷ç ÷÷çè ø

2(1935)(47)[(0.032) (0.66)(0.8) / (0.87)(6.5)]πD = + 5072 ND =

From Eq. (6.90)

[ ]{ }

[ ]{ }

2

0,max r

2

0

1.44 W

T D ( ) ave

1.44 (103047)

(9.8)(1.225)(47)(0.8) 80595 5072 (0.2)(103047 72760)

452 m

ρ μLL

L

sg C W L

s

¥=

- + -

=- + -

=


74

6.17 AR (181)(6.2) 33.5 ft.b S= = =

/ 4 / 33.5 0.1194h b = =

2

2

(16 / ) 3.650.785

4.651 (16 / )φ h b

h b= = =

+

,max

02 2(3000)

1.2 1.2 135 ft/sec(0.002377)(181)(1.1)ρL

L

WV

S C¥= = =

00.7 0.7(135) 94.5 ft/sec.LV= = =

2 2 21 1(0.002377)(94.5) 10.6 1b/ft

2 2ρq V¥ ¥ ¥= = =

(10.6)(181)(1.1) 2110 1bLL q S C¥= = =

2

,0 ARφ

πL

D DC

D q S C q S Ce¥ ¥

æ ö÷ç ÷ç= = + ÷ç ÷ç ÷÷çè ø

2(0.785)(1.1)

(10.6)(181) (0.027 154.6 lb(0.91)(6.2)π

Dé ùê ú= + =ê úê úë û

2

0,max r ave

2

0

(550 HP )/ (550)(285) / 94.5 1659 lb

1.44 W

{ [ ( )] }

1.44 (3000)

(32.2)(0.002377)(181)(1.1){1659 [154.6 (0.2)(3000 2110]}

572 ft.

ρ μ

A

LL

L

T V

Sg S C T D + W L

S

¥

¥

= = =

=- -

=- + -

=

6.18 L,max

2W 2(103,047)1.3 1.3 46.39 m/sec

(1.23)(47)(2.8)ρTVS C¥

= = =

0.7 32.47 m/sec.TV =

2 2 21(0.5)(1.23)(32.47) 648.4 N/m

2ρq V¥ ¥ ¥ = =

Since the lift is zero after touchdown, ,0D DC C ⋅=

,0 (648.4)(47)(0.032) 975.2 NDD q S C¥= = =

[ ]

2

,max 0.7

1.69 W

[ ( )T

LL r V

Sg C D W Lρ μ¥

=+ -

21.69 (103047)

268 m(9.8)(1.23)(47)(2.8)[975.2 (0.4)(103,047)]LS = =

+


75

6.19 ,max

2 2(3000)1.3 1.3 114.4 ft/sec

(0.002377)(181)(1.8)ρTL

WV

S C¥= = =

0.7 80.08 ft/sec.TV =

2 2 21 10.5(0.002377)(80.08) 7.621b/ft

2 2ρq V¥ ¥ ¥= = =

,0 (7.62)(181)(0.027) 37.2 1bDD q S C¥= = =

[ ]

2

,max r 0.7

1.69 W

[ ( )ρ μT

LL V

Sg S C D W L¥

=+ -

21.69 (3000)

493 ft(32.2)(0.002377)(181)(1.8)[37.2 0.4 (3000)]LS = =

+

6.20 88

250 mph 250 ft/sec 366.6 ft/sec60

V =¥æ ö÷ç= = ÷ç ÷÷çè ø

366.6 (0.3048) m/sec 111.7 m/sec.= =

2 2 21(0.5)(1.23)(111.7) 7673 N/m

2ρq V¥ ¥ ¥= = =

5,max (7673)(47)(1.2) 4.358 10 NLL q S C¥= = = ´

54.328 10

4.2103047

Ln

W

´= = =

2 2

2 2

(111.7)312 m

1 9.8 (4.2) 1

VR

g n

¥= = =- -

111.7

0.358 rad/sec312

Vw

R¥= = =


76

6.22 From Eq. (6.13) DT D q S C¥= =

From Eq. (6.1 c) 2

,0 ARπL

D DC

C Ce

= +

Combining (1) and (2) 2

,0+ARπL

DC

T = q S Ce¥

é ùê úê úê úë û

From Eq. (6.14) L

L

L W q S C

WC

q S

¥

¥

= =

=


2 2

,0 ,02 ARAR ππD DW W

T = q S C q S Cq S eq S e

¥ ¥¥¥

é ùê ú+ = +ê úê úë û

Multiply by q∞

2

2,0 ARπD

Wq T q SC

S e¥ ¥= +

or 2

2,0 0

ARπDW

q SC q T +S e¥ ¥- =

From the quadratic formula:

2,02

,0

4

AR

2

πD

D

SC WT T

S eq

SC¥

-=

2,0

2

,0

4

AR 1

2 2

πρ

D

D

CT W W T

W S S W eq V

C¥ ¥ ¥

æ ö æ ö÷ ÷ç ç -÷ ÷ç ç÷ ÷÷ ÷ç çè ø è ø= =

2,0

2

,0

4

ARπρ

D

D

CT W W T

W S S W eV

C¥¥

æ öæ ö æ ö÷ ÷ ÷ç ç ç+ -÷ ÷ ÷ç ç ç÷ ÷ ÷÷ ÷ ÷ç ç çè øè ø è ø=

1/ 22

,0

,0

4

ARπρ

D

D

CT W W TW S S W e

VC¥

¥

é ùæ öæ ö æ öê ú÷ ÷ ÷ç ç ç+ -ê ú÷ ÷ ÷ç ç ç÷ ÷ ÷÷ ÷ ÷ç ç çè øè ø è øê úê ú=ê úê úê úê úë û

max maxFor : ( )AV T = T

1/ 22

,0

max maxmax

,0

4

AR( )

πρ

DA A

D

CT W W T

W S S W eV

C¥¥

é ùæ ö æ ö æ öê ú÷ ÷ ÷ç ç ç+ -ê ú÷ ÷ ÷ç ç ç÷ ÷ ÷÷ ÷ ÷ç ç çè ø è ø è øê úê ú= ê úê úê úê úê úë û


77

6.23 From Figure max ,06.2, ( / ) 18.5, and 0.015.DL D C= = From Eq. (6.85)

1/2,0

,0max

( AR)

2

πDL

D D

C eC

C C

æ ö÷ç ÷ =ç ÷ç ÷÷çè ø

or, 2

,0 max4 ( / )

ARπD L DC C C

e =

24 (0.015)(18.5)

0.83(7.0)π

e = =

Please note: Consistent with the derivation of Eq. (6.85) where a parabolic drag polar is assumed with the zero-lift drag coefficient equal to the minimum drag coefficient, for the value of CD,0 in this problem we read the minimum drag coefficient from Figure 6.2.

6.24 Drag: DD q SC¥=

Power Available: ηA sP P=

Also, A AP T V¥=

Henceη s

AP

TV¥

=

(a) At an altitude of 30,000 ft, ρ∞ = 0.00089068 slug/ft3 and T∞ = 411.86°R. The speed of sound is

(1.4)(1718)(411.86) 994.7 ft/seca RTγ¥ ¥= = =

Hence, at Mach one, the flight velocity is V∞ = 994.7 ft/sec

2 2 21 1(0.00089068)(994.7) 440.6 1b/ft

2 2ρq V¥ ¥ ¥= = =

The drag coefficient at Mach one, as given in the problem statement is

,0 ,0(at 1) 10[ (at low speed)] 10(0.0211) 0.211D DC M C¥ = = = =

Hence, drag at Mach one is

(440.6)(334)(0.211) 31,0511bDD = q S C¥ = =

The thrust available is obtained as follows. The engine produces 1500 horsepower supercharged to 17,500 ft. Above that altitude, we assume that the power decreases directly as the air density. At 17,500 ft, ρ∞ = 0.0013781 slug/ft3. Hence

(at 30,000 ft) 0.00089068

HP (1500) (1500) 969 HP(at 17,500 ft) 0.0013781

ρρ

= ¥

¥= =

From Eq. (3), above

s (0.3)(969)(550)1611b

994.7

ηA

PT

V¥= = =

Consider the airplane in a vertical dive at Mach one. The maximum downward vertical force is W + TA = 12,441 + 161 = 12,602 lb. However, the drag is the retarding force acting vertically upward, and it is 31,051 lb. At Mach one, the drag far exceeds the maximum downward force. Hence, it is not possible for the airplane to achieve Mach one.


78

(b) At an altitude of 20,000 ft,

30.0012673 slug/ft

447.43

ρT R¥

¥

=

=

(1.4)(1716)(447.43) 1036.8 ft/seca V¥ ¥= = =

2 2 31 1(0.0013672)(1036.8) 6811b/ft

2 2ρq V¥ ¥ ¥= = =

(681)(334)(0.211) 47,993 lbDD q S C¥= = =

The thrust available is obtained as follows

(at 20,000 ft) 0.0012673

HP (1500) (1500) 1379 HP(at 17,500 ft) 0.0013781

ρρ

= ¥

¥= =

Hence,

s (0.3)(1379)(550)219 lb

1036.8

ηA

PT

V¥= = =

In a vertical, power-on dive at 20,000 ft, the maximum vertical force downward is W + TA = 12,441 + 219 = 12,660 lb. However, the drag is the retarding force acting vertically upward, and it is 47,993 lb. At Mach one, the drag far exceeds the maximum downward force. Hence it is not possible for the airplane to achieve Mach 1.

6.25 Taking the analytical approach given in Example 6.19, from Eq. (E6.19.1)

2

3,0

112 AR2

ρρ π

A DW

P V S CV S e

¥ ¥

¥

= +

2 2(14.45)

AR 19.311.45

b

S= = =

1020 (1020)(9.8) 9996 NfW kg= = =

( hp) (0.9)(85) 76.5 hpηAP b= = =

Using consistent units, noting that 1 hp = 746 Watts

4

,0

(76.5)(746) 5.71 10 Watts

1 1(1.23)(11.45)(0.03) 0.2113

2 21 1

AR (1.23)(11.45) (0.7)(19.3) 298.92 2

ρ

ρ π π

A

D

P

S C =

S e =

¥

¥

= = ´

=

=

Inserting these numbers into Eq. (1)

2

4 3 (9996)5.71 10 0.2113

298.9V

V¥¥

´ = +

or, 5

4 3 3.343 105.71 10 0.2113V

V¥¥

´´ = +

Solving Eq. (2) for V∞ max 62,6 m/sec.V V¥ = =


79

6.26 From Eq. (6.67)

0

1

η L

D

WCR n

c C W=

[ ]1/ 21/2

,0

,0max

( AR) (0.03) (0.7)(19.3)18.8

2 2(0.03)

π πDL

D D

C eC

C C

æ ö÷ç ÷ = =ç ÷ç ÷÷çè ø

Using consistent units,

1

7m

0

1 0 fuel

kg kg 9.8N 1 hp 1 hr0.2 0.2

(hp)(hr) (hp)(hr) 1 kg 746 Nm/s 3600 sec

7.3 10

1020 kg 1020(9.8) 9996 N

9996 (295)(9.8) 7105 N

f f

f

f

c

W

W W W

--

æ öé ù é ù æ ö÷ç ÷ç÷ê ú ê úç ÷= = ç÷ç ÷÷ çê ú ê ú ÷÷ç ç÷÷ç è øê ú è ø ë ûë û

= ´

= = =

= - = - =

From Eq. (6.67)

0

71

6

0.9 9996(18.8)

71057.3 10

3.44 10 m 3440 km

η L

D

WCR n n

c C W

R

-

æ ö æ ö÷ ÷ç ç= = ÷ ÷ç ç÷ ÷÷ç ç÷ç è øè ø´

= ´ =

6.27 3/43/2

,0

,0max

(3 AR)

4

πDL

D D

C eC

C C

æ ö÷ç ÷ç =÷ç ÷ç ÷÷çè ø

[ ]3/43(0.03) (0.7)(19.3)

22,774(0.03)

π= =

From Eq. (6.68)

3/2

1/2 1/2 1/21 0(2 ) ( )

η ρL

D

CE S W W

c C- -

¥

æ öæ ö ÷ç÷ ÷ç ç= -÷ ÷ç ç÷ ÷÷ç çè ø ÷÷çè ø

Using results from Problem 6.26,

1/2 1/2 1/27

7

5

0.922.77[2(1.23) (11.45)] [(7105) (9996) ]

7.3 10

14.9 10 [0.01186 0.01000]

2.77 10 sec

E =

E

E =

- --

æ ö÷ç -÷ç ÷ç ÷çè ø´

= ´ -

´

or, 52.77 10

77 hours3600

E´

= =


80

6.28 2

, ARπL

i D iC

D Q S C q Se¥ ¥= =

In steady level flight, L = W, and we have

2 2

2L

L WC

q S q S¥ ¥

æ ö æ ö÷ ÷ç ç÷ ÷= =ç ç÷ ÷ç ç÷ ÷÷ ÷ç çè ø è ø

Substitute (2) into (1).

21

π ARiW

D Q SQ S e¥¥


But AR 2

,b

S= hence Eq. (3) becomes

2

21

πiW S

D q Sd S e b

¥¥

æ ö æ ö÷ç ÷ç÷= ÷ç ç÷ ÷ç ç ÷ç÷÷ç è øè ø

2

1

πiW

D QEDeq b¥


6.29 (a) 2 3 21 12 2

2

88(2.0482 10 ) (300)

60

198.26 1b/ft

ρq V

q

-¥ ¥ ¥

¥

é ùæ ö÷çê ú= = ´ ÷ç ÷÷çê úè øë û

=

37 ft

10,1001b

b

W

==

2 2

1 1 10,100

(0.8)(198.26) 37π πiW

Deq b¥

æ ö æ ö÷ ÷ç ç= =÷ ÷ç ç÷ ÷÷ ÷ç çè ø è ø

149.5 lbiD =

(b) 2

ARπ=

i

LD

CC

e

2 2

2 2

10,1000.2186

(198.26)(233)

(37)AR 5.87

233

(0.2186)0.003245

AR (0.8)(5.86)

(168.26)(233)(0.003245)

149.9 lb

π πi

i

L

LD

i D

i

WC

q S

b

S

CC

e

D = q S C

D

¥

¥

= = =

= = =

= = =

=

=

The results from parts (a) and (b) agree to within 3-place round off error on the author’s hand calculator.


81

6.30 (a) From (6.45), copied below, for rate-of-climb at the climb angle θ,

sin θT D W= +

we have

sin θT D

W W= +

Assume L = W. Thus,

sin θT D

W L= +

or, 1

sin/

θT

W L D= +

(b) At a standard altitude of 31,000 ft, T∞= 408.3 R.

(1.4)(1716)(408.3) 990 ft/sec

0.85(990) 841.5 ft/sec.

a RT

V M a∞ ∞

∞ ∞ ∞

γ= = =

= = =

The specified rate-of-climb of 300 ft/min, in terms of ft/sec, is 300/60 = 5 ft/sec. From Eq. (6.48) and Fig. 6.28,

3R /C 5sin 5.94 10

841.5θ

∞V-= = + ´

Thus,

0.34 very small.θ = -

Hence,

31 1sin 5.94 10

/ 18θT

W L D-= + = + ´

or

0.0615T

W=

Thus

0.0615 0.0615(550,000) 33,825 lb.T W= = =

This is the required combined thrust from both engines. The thrust of each engine at this flight condition is 33,825/2, or 16,912 Ib. At 4 331,000 ft, 8.5841 10 slug/ft .ρ∞

-= ´ Assuming the engine thrust is proportional to density, as discussed in Section 6.7, we have for the thrust at sea level,

4

0.00237733825 93,666 lb.

8.5841 10T = -

æ ö÷ç =÷ç ÷ç ÷çè ø´

This is comparable to the 84,000 lb sea level static thrust of the particular Trent engine mentioned in the problem.


82

6.31 To answer this question, we need to compare the thrust with the weight plus drag at Mach 1; in a vertical climb at Mach 1, if T > W + D, then the airplane will accelerate through Mach 1 in a vertical climb. For an airplane in a vertical climb, lift = 0. Hence, the total drag coefficient is that for zero lift.

0.016(2.3) 0.368 at Mach 1.DC = =

Let us assume standard sea-level conditions, where 31.23 kg/mρ∞ = and 340.3 m/sec.∞a = At march 1, 340.3 m/sec.V a∞ ∞= = Thus,

2 4 21 1

2 2

4 4

(1.23)(340.3) 7.12 10 N/m

(7.12 10 )(27.84)(0.0368) 7.3 10 ND

q p V

D = q S C

¥ ¥ ¥

¥

= = = ´

= ´ = ´

The weight is given as 8,273 kgf; in terms of newtons

4(8273)(9.8) 8.11 10 NW = = ´

Hence, 4 5(7.3 8.11) 10 1.541 10 ND + W = + ´ = ´

Compare this with the thrust at sea level,

5131.6 k N 1.316 10 NT = = ´

For this case, .T D W< + The airplane can not accelerate through Mach one going straight up at sea level.

Since ρ∞∝D and ,ρ∞∝T both D and T will decrease by the same factor with an increase in altitude. Because the weight is fixed,

T D W< + at all altitudes.

Hence, at all altitudes, the airplane can not accelerate through Mach one going straight up.

6.32 3At 2000m, 275.16 K and 1.0066 kg/mρT¥ ¥= =

(1.4)(287)(275.16) 332.5 m/sec.

1000.3

332.5

γa RT

VM

a

¥ ¥

¥¥

¥

= = =

= = =

Hence, the airplane is flying at a low Mach number, far below the drag-divergence Mach number. Thus, the zero-lift drag coefficient is CD = 0.016.

2 2 21 1

2 2 (1.0066)(100) 5033 N/m

(5033)(27.87)(0.016) 2243 N

ρ

D

q V

D q S C

¥ ¥ ¥

¥

= = =

= = =

The acceleration is obtained from Newton’s 2nd Law, F = ma. The net force acting on the airplane in the upward direction is

3 3 3

3 3 3

131.6 10 (8273)(9.8) 2.243 10 131.6 10

81.08 10 2.243 10 48.28 10 N

T W D- - = ´ - - ´ = ´

- ´ - ´ = ´

From the discussion in Section 2.4, the weight in kgf is the same number as the mass in kg. Hence, the mass of the F-16 is 8273 kg. Thus,

3

248.28 105.84 m/sec

8273

Fa =

m

´= =


83

6.33 From the definition of TSFC, we have

(TSFC) (TSFC) (TSFC)/fw

W T t Dt tL D

= = = (1)

Also, the new thrust specific fuel consumption results in a new fuel weight

new, new (TSFC)(1 )

(1 )(TSFC) (1 )

ε

ε ε

f fLD

wf

L D

WW t

Wt

= -

+= - (2)

Substituting (1) into (2), we have

,new (1 )(1 )ε εf f w fW W= + - (3)

Increase in engine weight new= (1 )ε εw wW W = W W = W- + -

Decrease in fuel weight new= , (1 )(1 ) (using Eq.(3))ε εf f f f w fW W = W W- - + -

At the break-even point, we have

new , new

(1 )(1 )

1 (1 )(1 )

ε ε ε

ε ε ε

ε ε ε ε ε

f f

w f f w f

w w ff

w w f w ff

W W = W W

W W W

W

W

W

W

- -

= - + -

= - + -

= - + +

Since εw and ε f are very small fractions, we can ignore the product .ε εw f Thus,

1ε εf wf

W

W

æ ö÷ç ÷ç= + ÷ç ÷ç ÷÷çè ø

and from Eq. (1), we have

1(TSFC)

ε εf w

L

Dt

é ùê úê ú= +ê úê úê úë û


84

6.34 Weight at beginning of flight 61.35 10 1b.= ´

Weight at end of flight 6 6 61.35 10 0.5 10 0.85 10 lb.= ´ - ´ = ´

Average weight during flight6 6

61.35 10 0.85 101.1 10 lb.

2

´ + ´= = ´

From the result of Problem 6.33, we have

1

εε f

w

f

WW

=+

where 6 60.01, 1.1 10 1b, and 0.5 10 lb.f fW Wε = = ´ = ´

6

6

0.01 0.010.00313

3.21.1 101

0.5 10

εw = = =´

+´

6new (1+ ) =1.1 10 (0.00313) 34381b.εwW W= ´ =

The weight increase allowed for each engine is therefore

34388591b.

4=

6.35 2

,0

2

,0

LA D D

LA A D

CT D q S C q S C

eAR

CP T V q S V C

eAR

π

π

∞ ∞

∞ ∞ ∞

= = = +

= = +

22

,0 ,02

L

A D D

WC

q S

W VWP q S V C q S V C

q S eARq S eAR ππ ∞

∞

∞∞ ∞ ∞

∞∞

=

= + = +

,03 ( / )1 2

( / ) 1 2DA CP W S

VW W S p V eAR

ρπ∞ ∞

∞ ∞= +

,03 2( / )1 2

( / )DA CP W S

VW W S p V eAR

ρπ∞ ∞

∞ ∞= + (1)

AP Pη= where P is the shaft power from the engine.

At maximum velocity, P = Pmax. Equation (1) then becomes

,0max 3max

max

1 2 (W/S)

2 (W/S)DCP

VW V eAR

ηρ

ρ π∞∞

= +


85

6.36 For the CP-1,

5max

max

,0

2

2 2

3

230 hp (230)(550) 1.265 10 ft lb/sec

/ 42.88 ft/sec

0.8

2950 lb

0.025

/ 2950 /174 16.954 lb/ft

0.8

/ (35.8) /174 7.366

0.002377 slug/ft at sea level

D

P

P W

W

C

W S

e

AR b S

η

ρ∞

= = = ×=

==

=

= ==

= = =

=

From the equation obtained as the result of Problem 6.35,

,0max 3max

max

1 2( / )

2 ( / )DCP W S

VW W S V eAR

ηρ

ρ π∞∞

= +

3max

max

(0.025) 2(16.954)0.8(42.88) 1 2(0.002377)

16.954 (0.002377) (0.3)(7.366)V

V π= +

6 3max

max

770.5534.304 1.75254 10−= × +V

V (1)

Eq. (1) is an equation for Vmax that yields an analytic solution for Vmax. But it must be solved by trial and error. We will shortcut the trial and error process for this solutions manual by first trying Vmax = 265 ft/sec as obtained from the numerical solution.

6 3 770.5534.304 1.75254 10 (260)

26532.614 2.9077 35.52

Difference 1.216

−= × +

= + ==

Let us try a slightly smaller value, say Vmax = 260 ft/sec.

6 3 770.5534.304 1.75254 10 (260)

26032.80 2.96 33.76

Difference 0.544

−= × +

= + == −

Try Vmax = 262 ft/sec.

6 3 770.5534.304 1.75254 10 (262)

26231.519 2.941 34.46

Difference 0.156

−= × +

= + ==

Try Vmax = 261 ft/sec.

6 3 770.5534.304 1.75254 10 (261)

26131.159 2.952 34.111

Difference 0.192

−= × +

= + == −


86

Try Vmax = 261.6 ft/sec.

6 3 770.55

34.304 1.75254 10 (26.61)261.6

31.3748 2.9455 34.32

−= × +

= + =

Close enough! The analytical value for Vmax is

max 261.6 ft/sec=V

Compared with the numerical value of 265 ft/sec, a 1.3% difference.

6.37 For the CJ-1

2

2

2

,0

2

(3650) 7300 lb

19,815 lb

73000.368

19,815

318 ft

19,81562.3 lb/ft

3180.02

0.81

(53.3) /318 8.934

A

A

D

T

W

T

W

S

W

SC

e

AR

= ==

= =

=

= =

=

=

= =

From Eq. (6.44)

1/ 22

,0

max,0

1/ 22

4

4(0.02)(0.368)(62.3) (62.3) (0.368)

(0.81)(8.934)

(0.002377)(0.02)

978.8 ft/sec

DA A

D

CT TW W

W S S W eARV

C

πρ

π

∞

+ − =

+ −

=

=

This is to be compared with the numerical value of 975 ft/sec, a 0.3% difference.


87

6.38 From Eq. 6.53:

max 3/ 2,0max max

1( / ) 0.8776

( / )

WS

D

PR C

W c L D

ηρ∞

= −

For the CP-1, at 12,000 ft, hp = 230 0.001648

159.5 hp0.002377

=

(0.8)(1595) (550)23.79

2950

295016.954

174

P

W

W

s

η = =

= =

5,0

1/ 2 1/2,0

max,0

max 5 3/ 2

(0.001648)(0.025) 4.12 10

( ) [(0.025) (0.8)(7.366)]( / ) 13.606

2 2(0.025)

16.954 1( / ) 23.79 0.8776

6.12 10 (13.606)

123.79 0.8776(641.49)

50.18823.79 11.217 12.573

D

D

D

C

C eARL D

C

R C

ρ

π π

−∞

−

= = ×

= = =

= −×

= −

= = =

max

ft/sec

( / ) (12.573)(60) 754.4 ft/minR C = =

The numerical result from Section 6.10 is 755 ft/min. The difference is 0.08%.


88

6.39 From Eq. (6.52)

1/ 2 3 / 2

max 2 2,0 max max max

( / ) 2 3( / ) 1

3 6 2( / ) ( / )D

W S z TR C

C W T W L D Zρ∞

= − −

For the CJ-1

2(53.3)8.934

318/ 19,815 / 318 62.31

AR

W S

= =

= =

,0

max

1/ 2 1/ 2,0

,0max

0.02

(3650)(2) 0.00110430.171

19.815 0.002378

( ) [(0.02) (0.81)(8.934)] 0.67316.858

2 2(0.02) 0.04

D

D

D

C

T

W

C eARL

D C

π π

=

= =

= = = =

2 2max max

2 2

31 1

( / ) ( / )

31 1

(16.858) (0.1711)

1 1 0.3606 2.166

ZL D T W

= + +

= + +

= + + =

1/ 23/ 2

max

2 2

6 1/ 2

(62.31)(2.166)( / ) (0.1711)

3(0.0011043)(0.02)

2.166 31

6 2(0.1711) (16.858) 2.166

(2.0369 10 ) (0.07077)(0.5558)

56.14 ft/sec (56.14)(60) ft/min 3368 ft/min

R C

= ×

− −

= ×

= = =

This is compared to the numerical answer of 3369 ft/min given in Section 6.10.


89

6.40 2

21,02

LD

CT D V S C

eARρ

π∞ ∞

= = +

(1)

max

5max

5max

maxmax

2

2 2

88229 mph 229 335.87 ft/sec.

60

0.0018975 ft lb(1200)(550) (0.9) 9.4795 10

0.002378 sec

9.4795 102822 lb

335.87

987 ft

9.14

0.8

1 2 1 2(0.0018975)(335.87) (98

V

P P

PT

V

S

AR

e

V S

η

ρ∞ ∞

= = =

= = = ×

×= = =

===

= 5

52

2 2

7) 1.05636 10 lb

25,0000.2366

1 1.05636 102

(0.23666)0.0024382

(0.8)(9.14)

L

L

WC

V S

C

eAR

ρ

π π

∞ ∞

= ×

= = =×

= =

From EQ. (1),

5,0

,0 5

,0

2822 1.05636 10 ( 0.0024382)

28220.0024382 0.0267

1.05636 10

0.0243

D

D

D

C

C

C

= × =

+ = =×

=

Note: This answer is consistent with the value of CD,0 for the DC-3 given in Figure 6.77.


90

Chapter 7

7.1 cg ac ac( )M M LC C C h h= + -

ex cg ac( ) 0.005 0.05(0.03) 0.01M M LC C C h h= - - = - = -

7.2 2 2 21 1(1.225)(100) 6125 N/m

2 2ρq V¥ ¥ ¥= = =

At zero lift, the moment coefficient about c.g. is

cg

cg 12.40.003

(6125)(1.5)(0.45)MM

Cq S c¥

-= = = -

However, at zero lift, this is also the value of the moment coefficient about the a.c.

ac

0.003MC = -

At the other angle of attack,

3675

0.4(6150)(1.5)L

LC

q S¥= = =

and cg

cg 20.670.005

(6125)(1.5)(0.45)MM

Cq S c¥

= = =

Thus, from Eq. (7.9) in the text,

cg ac ac( )M M LC C C h h= + -

Thus,

cg ac

ac

ac

0.005 ( 0.003)

0.4

0.02

M M

L

C Ch h

C

h h

- - -- = =

- =

The aerodynamic center is two percent of the chord length ahead of the center of gravity.

7.3 From the results of Problem 7.2,

ac

2

ac

6125 N/m

0.003

0.02

M

q

C

h h

¥ =

= -

- =

In the present problem, the c.g. has been shifted 0.2c rearward. Hence,

ac 0.02 0.2 0.22h h- = + =

Also, 4000

0.435(6125)(1.5)L

LC

q S¥= = =

Thus, cg ac

cg

ac( )

0.003 0.435(0.22) 0.0927

M M L

M

C C C h h

C

= + -

= - + =


91

7.4 From Problem 7.2, we know

ac

2

ac

6125 N/m

0.003

0.02

M

q

C

h h

¥ =

= -

- =

From information provided in the present problem, at 5 ,αa =

4134

0.45(6125)(1.5)L

LC

q S¥= = =

Hence,

0.45

0.09 per degree5α

LdCa

d= = =

Also, (1.0)(0.4)

0.593(0.45)(1.5)

t tH

SV

c S= = =

From Eq. (7.26) in the text,

( )cg ac

cg

cg

ac t o1 ( )

0.120.003 (0.09)(5) (0.02) (0.593) (1 0.42) (0.593)(0.12)(2.0)

0.09

0.052

εα εα

tM M a H H t

M

M

aC C a h h V V a i

a

C

C

é ùæ ö¶ ÷çê ú= + - - - + +÷ç ÷÷çê úè ø¶ë ûé ùæ ö÷çê ú= - + - - +÷ç ÷÷çê úè øë û

= -

Hence, the moment is

cgcg

cg

(6125)(1.5)(0.45)( 0.052)

215 Nm

MM q S c C

M

¥= = -

= -


92

7.5 cgac( ) 1

εα αM t

Ha

C aa h h V

a

é ùæ ö¶ ÷çê ú= - - - ÷ç ÷÷çê úè ø¶ ¶ë û

where, from Problems 7.4 and 7.2,

ac

ac

t

0.09 per degree

0.02

0.003

0.593

0.12

0.42

2.08

εα

M

H

t

a

h h

C

V

a

i

=- =

= -

=

=

¶=

¶=

Thus, cg 0.12(0.09) (0.02) (0.593) (1 0.42) 0.039

0.09αM

a

C é ùæ ö÷çê ú= - - = -÷ç ÷÷çê úè ø¶ ë û

The slope of the moment coefficient curve is negative, hence the airplane model is statically stable. To examine whether or not the model is balanced, first calculate

0MC .

0 ac 0( ) 0.003 (0.593)(0.12)(2.0 0) 0.139εM M H t tC C V a i= + + = - + + =

The trim angle of attack can be found from

cg

cg 0

cg0

0

/( / ) 0.139 /( 0.039) 3.56

αα

α α

MM M e

e M

CC C

CM C

= + =¶

= - ¶ ¶ = - - =

This is a reasonable angle of attack, falling within the normal flight range. Hence, the airplane model is also balanced.

7.6 wbac 1

εε

tn H

ah h V

a

æ ö¶ ÷ç= + - ÷ç ÷÷çè ø¶

From Problem 7.2, wbac 0.02nh h- =

Hence,

wbac 0.02 0.26 0.02 0.24

0.120.24 0.593 (1 0.42)

0.09

0.70

n

n

h h

h

h

= - = - =

æ ö÷ç= + -÷ç ÷÷çè ø

=

By definition,

static margin = 0.70 0.26 0.44nh h- = - =


93

7.7 0 cg

1

0

cg

trim

( / )

( / )

0.139 (from Problem 7.5)

/ 0.039 (from Problem 7.5)

8

0.593 (from Problem 7.4)

/ 0.04 (given)

α αδ

δ

α

α

δ

°

t

M M a n

H L e

M

M a

n

H

L e

C C

V C

C

C

V

C

+ ¶ ¶=

¶ ¶

=

¶ ¶ = -

=

=

¶ ¶ =

Thus, trim0.139 ( 0.039)(8)

7.29(0.593)(0.04)

δ °+ -= = -

7.8 ( )( )

1

0 acab

0

he 1

he

0

/11

/

1 ( 0.007)1 (0.04) 0.806

0.12 ( 0.012)

( )

0.003 (0.806)(0.593)(0.12)(2 0) 0.112

αδ δ

ε

L

t e e

M M H t t

M

C CF

a C

F

C C F V a i

C

æ ö¶ ¶ ¶÷ç ÷ç= - ÷ç ÷÷çç ¶ ¶ ¶è ø

-= - =

-¢ = + +

¢ = - + + =

This is to be compared with 0

0.139MC = from Problem 7.5 for stick-fixed stability.

wb

1ac 1

(0.12)0.24 (0.806)(0.593) (1 0.42) 0.609

(0.09)

εαn H

n

ah h F V

a

h

æ ö¶ ÷ç¢ = + - ÷ç ÷÷çè ø¶

¢ = + - =

This is to be compared with 0.70nh = from Problem 7.6 for stick-fixed stability.

0.609 0.26 0.349nh h¢ - = - =

Note that the static margin for stick-free is 79% of that for stick-fixed.

cg ( ) (0.09)(0.349) 0.031αM

na

Ca h h

¢¶¢= - - = - = -

¶

This is to be compared with a slope of −0.039 obtained from Problem 7.5 for the stick-fixed case.


94

7.9

Examining the above figure for a canard, let us trace through the pertinent equations in the text,

modifying them appropriately for the canard configuration. Starting with Eq. (7.12) for the moment generated about the center of gravity due to the tail, the

minus sign is replaced by a positive, because the tail is now ahead of the center of gravity, creating a positive moment.

cg, t t tM L= (1)

Thus, Eq. (7.17) becomes

, cg, ,M t H L tC V C= (2)

Note that the canard tail sees no downwash, and the canard is canted upward relative to the wing-body zero lift link, so Eq. (7.18) for the angle of attack of the tail becomes

wbα αt ti= + (3)

Therefore, Eq. (7.22) is replaced by

, cg, wbαM t H t H t tC V a V a i= + (4)

In turn, Eq. (7.24) for the total pitching moment becomes

wb wb wb,cg ,ac ac ,( )M M L H L tC C C h h V C= + - + (5)

Eq. (7.25) becomes

wb wb

t,cg ,ac wb wb ac

wb[( ) ]αM M H H t t

aC C a h h V V a i

a= + - + + (6)

Differentiating Eq. (6) with respect to angle of attack, the form that replaces Eq. (7.28) is

wb

,cgacα

M tH

a

C aa h h V

a

¶ é ùê ú= - +ê ú¶ ë û

(7)

For static stability, ,cg

αM

a

C¶

¶ must be negative.

From Eq. (7), therefore,

wbac( ) 0t

Ha

h h Va

- + <

or, wbac( ) t

Ha

h h Va

- <

From inequality (8), for the canard configuration to be statically stable, the center of gravity must be sufficiently far forward of the wing-body aerodynamic center to satisfy the inequality (8). This can readily be done in the design of a canard airplane. Indeed, to help shift the wing-body aerodynamic center rearward, helping to satisfy (8), is the reason that canard designs tend to have the wings shifted farther back on the fuselage compared to conventional rear-tail configurations.


95

Chapter 8

8.1

6

6 6 6

2 14 3 2

2 6 3

10 2

2 21 4 2

2 2 21 7

400miles 0.644 10

6.4 10 0.644 10 7.044 10 m

3/ 986 10 m /sec from the text

cos (7.044 10 )(13 10 ) cos10°

9.018 10 m /sec

8.133 10 m /sec

/ 8.133 10 / 3.986 10 m

θ θ β

G

b e G

b b

h

r r h

k

h r rV r V

h

h

p h k

= = ´

= + = ´ + ´ = ´

= ´

= = = = ´ ´

= ´

= ´

= = ´ ´

This is the numerator of the orbit equation. We now proceed to find the eccentricity. The kinetic energy per unit mass is:

2 3 2 7 2 2/ /2 (13 10 ) /2 8.45 10 m /secT m V= = ´ = ´

The potential energy per unit mass is:

2 14 6 7 2/ / 3.98 10 / 7.044 10 5.659 10 m /secφ bm k r= = ´ ´ = ´

Hence, ( ) 7 7 2 2/ / (8.45 5.659) 10 2.791 10 m /secφH m T m= - = - ´ = ´

Thus, the eccentricity is:

2 21 7

4 14 22 2(8.133 10 )(2.791 10 )

1 1 1.96(3.986 10 )

h He

mk

æ ö ´ ´÷ç= + = + =÷ç ÷÷çè ø ´

Obviously, the trajectory is a hyperbola because e > 1, and also because T > | φ |. The orbit equation is

7

1 cos( )

2.04 10

1 1.96 cos( )

θ

θ

pr

e C

rC

=+ -

´=

+ -

The phase angle, C, is calculated as follows. Substitute the burnout location 6( 7.044 10 m and 0 )θ °br = ´ = into the above equation.

76 2.04 10

7.044 101 1.96 cos( )

cos ( ) 0.967

C

C

´´ =

+ ´ -- =


96

Thus 14.76°.C = -

Hence, the complete equation of the trajectory is

72.04 10

1 1.96 cos ( 14.76)θr

´=

+ +

where θ is in degrees and r in meters.

8.2 2Escape velocity 2 /V k r= =

14 6 4

13 6 4

13 6 3

17

For Venus: (2)(3.24 10 ) / 6.16 10 1.03 10 m/sec 10.3 km/sec

For Earth: (2)(3.96 10 ) / 6.39 10 1.11 10 m/sec 11.3 km/sec

For Mars: (2)(4.27 10 ) /10 5.02 10 m/sec 5.02 km/sec

For Jupiter: (2)(1.27 10

V

V

V

V

= ´ ´ = ´ =

= ´ ´ = ´ =

= ´ = ´ =

= ´ 7 4) / 7.14 10 5.96 10 m/sec 56.6 km/sec´ = ´ =

8.3 11 3 2

22

6.67 10 m /(kg)(sec)

7.35 10 kg

G

M

-= ´

= ´

2 11 22 12 3 2k (6.67 10 )(7.35 10 ) 4.9 10 m /secGM -= = ´ ´ = ´

Orbital velocity is 2/V k r=

12 6orbital 4.9 10 /1.74 10 1678 m/sec 1.678 km/secV = ´ ´ = =

Escape velocity is larger by a factor of (2)1/2.

escape 2(1.678) 2.37 km/secV = =

8.4 From Kepler’s 3rd law,

2 3

1 1

2 2

3 3 22 1 2 1

3 2/32 1 2 1

11 2/3 122

( ) ( / )

( ) ( / )

(1.495 10 )(29.7 /1.0) 1.43 10 m

ττ

τ τ

τ τ

a

a

a a

a a

a

æ ö æ ö÷ ÷ç ç÷ ÷=ç ç÷ ÷ç ç÷ ÷÷ ÷ç çè ø è ø

=

=

= ´ = ´

Please note: The “distant planet” is in reality Saturn.


97

8.5 In order to remain over the same point on the Earth’s equator at all times (assuming the Earth is a perfect sphere), the satellite must have a circular orbit with a period of 24 hours = 8.64.104 sec. As part of the derivation of Kepler’s third law in the test, it was shown that

2 2

2 3 32 2

4 4π πτ a rk k

æ ö æ ö÷ ÷ç ç÷ ÷ç ç= =÷ ÷ç ç÷ ÷ç ç÷ ÷÷ ÷ç çè ø è ø (Remember, the orbit is a circle.)

Hence,

1/322/3

2

1/3144 2/3 7

2

4

3.956 10(8.64 10 ) 4.21 10 m

4

τπ

π

kr

r

æ ö÷ç ÷ç= ÷ç ÷ç ÷÷çè ø

æ ö´ ÷ç ÷ç= ´ = ´÷ç ÷ç ÷÷çè ø

The radius of the Earth is 6.4 × 106 m. Hence, the altitude above the surface of the Earth is

7 6 74.21 10 6.4 10 3.57 10 m 35,700 kmGh = ´ - ´ = ´ =

Circular velocity is 2 14

73.956 10

3065 m/sec4.21 10

kV

r

´= = =

´

8.6 3 3 44 4(6963) (0.8) 1.4933 10 kg

3 3ρ ρ π πm v r

æ ö æ ö÷ ÷ç ç= = = = ´÷ ÷ç ç÷ ÷÷ ÷ç çè ø è ø

2 2 2(0.8) 2.01 mπ πS r= = =

4

31.4933 107429 km/m

(1)(2.01)D

m

C S

´= =

10g / 9.8 /(287)(29/88) 0.000118 mZ RT -= = =

From Eq. (8.97), the density at the altitude for maximum deceleration is

3

kgsin (7429)(0.000118) sin 30

mρ θ ° = 0.4383

D

mZ

C S= =

From Eq. (8.73)

01 1 0.4383

( / )= 8710 m0.000118 1.225

ρ ρ h n nZ

æ ö÷ç= - - =÷ç ÷÷çè ø

From Eq. (8.100)

2 2

2

max

sin (8000) (0.000118)(sin 30 )694.6 m/sec

2 2

θ °EdV V Z

dt e e= = =

In terms of gs:

max

694.670.88

9.8

dVgs

dt= =

From Eq. (8.87)

(1.225)2(7429) (0.000118) sin 30

/ 0.247

0.247 0.247(8000) 1978 m/sec

°E

E

V V e

V V

é ùê ú-ê úê úë û´= =

= = =


98

8.7 Aerodynamic heating varies as 3 .V¥ Hence:

32

1

22

36,0002.37

27,000

(100)(2.37) 237 Btu/(ft )(sec)

q

q

q

æ ö÷ç ÷= =ç ÷ç ÷çè ø

= =

8.8 Assume the mean velocity of the earth around the sun is essentially its circular orbital velocity given by

2

3earth 29.77 10 m/sec

kV

r= = ´

where k2 = GM, and M is the mass of the sun. From Eq. (1)

2 2 3 2 9 20 3 2(29.77 10 ) (147 10 ) 1.30 10 m /seck V r= = ´ ´ = ´

The asteroid is moving at 0.9 times the escape velocity from the sun.

2

asteroid2

0.9k

Vr

=

To intersect with the earth, the asteroid's value of r in Eq. (2) is the same as far for the earth in Eq. (1). Hence,

20

8asteroid 9

2(1.3 10 )0.9 37.850 10 m/sec

(147 10 )V

´= = ´

´

The relative head-on collision velocity, which is the velocity at which the asteroid would enter the earth's atmosphere, is

3 3 3

entry earth asteroid 29.77 10 37.85 10 67.62 10 m/sec

67.62 km/sec

V V V= + = ´ + ´ = ´

=

8.9 Earth’s radius 66.4 10 mR= = ´

Following the nomenclature in Figure 8.18,

rmin = R + (altiude above Earth’s surface)

6 5 6min 6.4 10 4.17 10 6.817 10 mr = ´ + ´ = ´

From Eq. (8.62)

2 2

min/

1

h kr

e=

+ (1)

and from Eq. (8.61)

2 2

max/

1

h kr

e=

- (2)

Combining (1) and (2)

max

6 6max min

1 .001321.0026

1 0.99868

(1.0026) (6.817 10 )(1.0026) 6.8347 10 m

er

e

r r

+ 1= = =

-

= = ´ = ´

Altitude at apogee 6 6 66.8347 10 6.4 10 0.4347 10 m 434.7 km= ´ - ´ = ´ =


99

8.10 From Eq. (8.71),

22 3

2

2 14 3 2

4

3.956 10 m /sec

πτ ak

k

=

= ´

From the solution of Problem 8.9,

6

max

6min

6.8347 10 m

6.817 10 m

r

r

= ´

= ´

6

6max min (6.8347 6.817) 106.82585 10 m

2 2

r ra

+ + ´= = = ´

Inserting these numbers into Eq. (8.71):

2 2 6 3

2 32 14

2 7

4 4 (6.82585 10 )

3.956 10

3.1738 10

5634 sec

π πτ

ττ

ak

´= =

´

= ´=

or, 5634

1.56 hr3600

τ = =

8.11 The angular momentum per unit mass is 2θh r= . To calculate h, proceed as follows. From Eq. (8.62),

2 2

min/

1

h kr

e=

+

or,

2 2min(1 )h r e k= +

From the solution of Problem 8.9, we have 6min 6.817 10 m.r = ´ Thus,

2 6 14 21(6.817 10 )(1.00132)(3.956 10 ) 2.7 10h = ´ ´ = ´

or, 105.1965 10h = ´

Because 2 ,θh r= at perigee

10

32 6 2

min

5.1965 101.1182 10 rad/sec

( ) (6.817 10 )θ h

r-´

= = = ´´

6 3perigee min (6.817 10 )(1.1182 10 ) 7623 m/sec= 7.623 km/secθV r -= = ´ ´ =


100

8.12 (a) 2 2

2tV k

Er

= -

Evaluate this equation at the perigee, where from Example 8.3, 67.169 10 mpr = ´ ,

and 9.026 km/secpV =

2 2 3 2 14

6

7 7 7

(9.026 10 ) 3.986 10

2 2 7.169 10

joules4.0734 10 5.56 10 1.486 10

kg

pt

p

V kE

r

- -

´ ´= - = - =

´

´ - ´ = - ´

(b) From Example 8.3, 71.341 10 m.a = ´ Thus

2 14

77

3.986 10 joules1.486 10

2 kg2(1.341 10 )t

kE

a-´

= - = - = - ´´

The results are the same, which is to be expected. Eqs. (8.74) and (8.77) are numerically the same.

8.13 From Example 8.6, 3432 m/sec.θV =

2 sin 2(3432) sin 10 m/sec2θΔ ° = 1191.9v

V Væ ö÷ç= =÷ç ÷÷çè ø

Comparing with the impulse calculated in Example 8.6 for a change in inclination angle of 10°, the present impulse is 1191.9/598.2 = 1.99 larger − almost twice as large for double the inclination angle change. We would expect this because, for relatively small angles (in radians)

sin2 2

v væ ö÷ç »÷ç ÷÷çè ø


101

8.14 From Example 8.7, we already have

1 6794 m/secV =

and

72 1 1.0506 10 mr r= = ´

7

.2 72

105 10 m

1 1 0.8pr

ae

= = = ´- -

Thus,

2 2 14 14

2 2 7 72

2

11

1

1

2 2(3.986 10) 3.986 10

1.0506 10 5 10

8241 m/sec

sin (0.4654) sin90Tan 0.4654

1 cos 1 0.4654 cos90

24.957

θβθ

β

°°

°

A

A

k kV

r a

V

e

e

´ ´= - = -

´ ´=

= = =+ +

=

For orbit 2, the true anomaly of point A is found from

, 2

2 2 2

7

7

(1 ) 1cos

1 10 (1 0.8) 1cos 2.1416 1.25 0.8916

0.8(0.8)(1.0506 10 )

26.93

θ

θ

θ °

p zA

A

A

r e

e r e

+= -

´ += - = - =

´=

Thus,

22

2 2

2

1 2

sin (0.8)(0.4529) 0.3623Tan 0.2115

1 cos 1 (0.8)(0.8916) 1.7133

11.94

24.957 11.94 13.017

θβθ

βα β β

°°

Ae

e= = = =

+ +

=

= - = - =

From Eq. (8.89),

2 2 21 2 1 2

2 2

7 7 7

( ) 2 cos

(6794) (8241) 2(6794)(8241) cos (13.017 )

14.379 10 10.91 10 3.469 10

Δ α

°

V V V V V= + -

= + -

= ´ - ´ = ´

Thus,

5890 m/secΔV =


102

8.15 From Example 8.8,

1 7771 m/secV =

and

6

1

6 5 62

6.6 10 m

6.4 10 5 10 6.9 10 m

r

r

= ´

= ´ + ´ = ´

6 6 6

61 2 6.6 10 6.9 10 13.5 106.75 10 m

2 2 2

r ra

+ ´ + ´ ´= = = = ´

From Eq. (8.78),

2 2 14 14

pt 6 61

8 8

2 2(3.986 10) (3.986 10 )

6.6 10 6.75 10

1.2079 10 0.5905 10 7857.5 m/sec

k kV

r a

´ ´= - = -

´ ´

= ´ - ´ =

At point 1, the required impulse to enter the Hohmann transfer orbit is

1 pt 1Δ 7857.5 7771 86.5 m/sec.V V V= - = - =

At point 2 on orbit 2, the required spacecraft velocity is

2 14

2 61

3.986 107600 m/sec

6.9 10

kV

r

´= = =

´

At point 2 on the Hohmann transfer orbit,

2 2 14 14

at 6 61

8 8

2 2(3.986 10 ) 3.986 10

6.9 10 6.75 10

1.1554 10 0.5905 10 7516 m/sec

k kV

r a

´ ´= - = -

´ ´

= ´ - ´ =

At point 2,

2 2 atΔ 7600 7516 84 m/secV V V= - = - =

The total impulse required is

1 2Δ Δ Δ 86.5 84 170.5m/secV V V= + = + =


103

8.16 From Problem 8.2, for Mars,

2 13 3 2

2 13

1 61

4.27 10 m /sec

4.27 102310 m/sec

8 10

k

kV

r

= ´

´= = =

´

6 6

61 2 8 10 15 1011.5 10 m

2 2

r ra

+ ´ + ´= = = ´

2 2 13 13

pt 6 61

2 2(4.27 10 4.27 10

8 10 11.5 10

k kV

r a

´ ´= - = -

´ ´

7 71.0675 10 0.3713 10 2638.6 m/sec= ´ - ´ =

At point 1,

1 pt 1ΔV 2638.6 2310 328.6 m/secV V= - = - =

At point 2 on orbit 2,

2 13

2 2 64.27 10

1687 m/sec15 10

kV

r

´= = =

´

At point 2 on the Hohmann transfer orbit,

2 2 13 13

at 2 6 6

6 6

2 2 at

2 2(4.27 10 ) 4.27 10

15 10 11.5 10

5.693 10 3.713 10 1407 m/sec

ΔV 1687 1407 280 m/sec.

k kV

ar

V V

´ ´= - = -

´ ´

= ´ - ´ == - = - =

The total required impulse is

1 2ΔV ΔV ΔV 328.6 280 608.6 m/sec= + = + =


104

8.17 Relative to the center of mercury, at periapsis,

min

max

2,440 200 2640 km

2,440 15,193 17,633 km

r

r

= + == + =

The semi-major axis of the elliptical orbit is

min max

7

32 11 23

2

32 13

2

2640 17,63310,137 km

2 2

= 1.0137 10 m

m6.67 10 (3.3 10 kg)

kg sec

m2.20 10

sec

r ra

k GM

k

−

+ += = =

×

= = × ×

= ×

From Eq. 8.71 in the text:

2 3 2 7 3

2 9 22 13

4 4 (1.0137 10 )1.87 10 sec

2.20 10

a

k

π πτ ×= = = ××

Hence,

τ = 43243 sec

Since 3600 sec = 1 hr, we have

43243

12.01 hr3600

τ = =

8.18 From the solution of Problem 8.17, the semi-major axis is a = 1.0137 × 107m, and k2 = 2.20 × 1013 m3/sec2. At periapsis, rmin = 2640 km = 2.64 × 106 m. From the vis-viva equation, Eq. (8.78) in the text,

2 22k k

Vr a

= −

For r = rmin, the velocity at periapsis is

13 136 6

6 7

2(2.20 10 ) 2.20 1016.67 10 2.17 10

2.64 10 1.0137 10

3809 m/sec 3.809 km/sec

V× ×

= − = × − ×× ×

= =

For r = rmax, the velocity at apoapsis is

136 6 6

7

2(2.20 10 )2.17 10 2.495 10 2.17 10

1.7633 10

570 m/sec 0.57 km/sec

V×= − × = × − ××

= =


105

8.19 From the solution of Problem 8.18 and 8.19, at periapsis V = 3809 m/sec and rmin = 2.64 × 106m. Thus,

36

min

38091.44 10 rad/sec

2.64 10

V

rθ −= = = ×

×

The angular momentum per unit mass is h = r2 .θ Thus,

2 6 2 3

min

10 2

( ) (2.64 10 ) (1.44 10 )

1.00 10 m / sec

h r

h

θ −= = × ×

= ×

The same value is obtained at the apoapsis, where

57

max

2 7 2 5max

10 2

5703.2326 10 rad/sec

1.7633 10

( ) (1.7633 10 ) (3.2326 10 )

1.00 10 m / sec.

V

r

h r

h

θ

θ

−

−

= = = ××

= = × ×

= ×

This, of course, is simply verification that the angular momentum is constant for the orbit.

8.20 Eq. (8.63) is

2 2

2

/

1

h ka

e=

−

or

2 2/

1h k

ea

= −

From Problems 8.17 and 8.19,

2 10 27

2 13

7

7

(1.00 10 )0.4545 10

2.20 10

0.4545 101 1 0.448 0.743

1.0137 10

h

k

e

×= = ×

×

×= − = − =

×

8.21 Notice that the equations for a spacecraft’s trajectory in space as derived in Chapter 8 depend only on such quantities as momentum or energy per unit mass. The actual mass of the spacecraft plays a role only in the rocket booster burnout conditions at the end of launch, as shown in Section 9.10. Once the initial burnout conditions are achieved, the subsequent trajectory is governed only by the force of gravity and Newton’s second law, for which energy and momentum per unit mass are relevant. So, if you know the complete details of the orbit of a spacecraft, it is still not possible to back out the total mass of the spacecraft.


106

Chapter 9

9.1 Following the nomenclature in the text;

1.43 2

2 3

3

10.43 2

2 3

3

6.75 13.0

(13.0)(1.0) 13.0 atm

(6.75) 2.15

(2.15)(285) 613 K

γ

γ

p V

p V

p

T V

T V

T

-

æ ö÷ç ÷= = =ç ÷ç ÷÷çè ø

= =

æ ö÷ç ÷= = =ç ÷ç ÷÷çè ø

= =

The heat released per kg of fuel-air mixture is established in the text as 62.43 10´ joule/kg. Hence,

4 3

6

4

4 4

3 3

where 720 joule/kg K

2.43 10613 3988 K

720

°

°

vv

qT T c

c

T

p T

p T

= + =

´= + =

=

Thus, 44 3

3

3988(13.0) 84.6 atm

613

Tp p

T


1.45 4

4 5

5

2 2 3 3compression

10.069

6.75

(84.6)(0.069) 5.84 atm

1

γ

γ

p V

p V

p

p V p VW


= =

-=

-

The volumes 2V and 3V can be obtained as follows

( 9.84) / 6.75,x x+ = Thus, 1.71 cmx =

Thus,

2

2

3 3 32

3 4 33 2

(9.84 1.71) where bore 11.1 cm4

1118 cm 1.118 10 m

/ 6.75 1.118 10 / 6.75 1.66 10 m

πbV b

V

V V

-

- -

= + = =

= = ´

= = ´ = ´


107

Thus,

2 2 3 3compression

3 4 5

5 5 4 4power

3 4 5

power

power comperssion

1

(1.0)(1.118 10 ) (13.0)(1.66 10 ) 1.01 10263 joule

0.4

1

(5.84)(1.118 10 ) (84.6)(1.66 10 ) 1.01 10

0.41897 joule

γ

γ

p V p VW

p V p VW

W

W W W

- -

- -

-=

-é ù´ - ´ ´ê úë û= =

--

=-

é ù´ - ´ ´ê úë û=-

=

= - =

mech

5

1897 263 1634 joule

1(RPM) N W

120(0.85)(0.83)(2800)(4)(1634)

1.076 10 watts120

ηηA

A

P

P

- =

=

= = ´

or 51.076 10 watts

144 hp746 watts/hpAP

´= =

9.2 mech1

(RPM)120

ηηAP Dpe=

where 2 2(11.1) (984)(4)

=4 4

π πbD s N=

3 3 33809 cm 3.809 10 mD -= = ´

mech

5

3

6 2

120

(RPM)

120(1.076 10 )

(0.85)(0.83)(2800)(3.809 10 )

1.72 10 N/m 17 atm

ηηA

e

e

e

Pp

D

p

p

-

=

´=

´

= ´ =


108

9.3 First, calculate the mass flow through the engine. For simplicity, we will ignore the mass of the added fuel, and assume the mass flow from inlet to exit is constant. Evaluating conditions at the exit,

531.01 10

0.469 kg/m287 (750)

(0.469)(0.45)(400) 84 kg/sec

ρ

ρ

ee

e

e e

p

RT

m eA V

´= = =

= = =

At the inlet, assume standard sea level conditions.

84 kg/secρ i im AV¥= =

Hence,

84

152 m/sec(1.225)(0.45)ρ

j

i

mV

A¥= = =

Note: Even though the engine is stationary, it is sucking air into the inlet at such a rate that the streamtube of air entering the engine is accelerated to a velocity of 152 m/sec at the inlet. The Mach number at the inlet is approximately 0.45. Therefore, by making the assumption of standard sea level density at the inlet, we are making about a 10 percent error in the calculation of iV . To obtain the thrust,

( ) ( ) 84(400 152) (0) 20,832 N e j e e eT m V V p p A A¥= - + - = - + =

Since 1 lb = 4.448 N, we also have

20,832

4684 lb4.448

T = =

9.4 At a standard altitude of 40,000 ft,

2

4 3

393.12 lb/ft

5.8727 10 slug/ftρ

p¥-

¥

=

= ´

The free stream velocity is

88

530 777 ft/sec60

V¥æ ö÷ç= =÷ç ÷÷çè ø

Hence,

4(5.87 10 )(777)(13) 5.93 slug/sec

( ) ( )

(5.93)(1500 777) (450 393)(10)

4287 570 4587 lb

ρ

j

e e e

m V A

T m V V p p A

T

T

-¥ ¥

¥ ¥

= = ´ =

= - + -

= - + -= + =


109

9.5 Assume the Mach number at the end of the diffuser (hence at the entrance to the compressor) is close to zero, hence 2p can be assumed to be total pressure.

11

1/ 1

2 2 3.521

1

8000.716

1117

11 [1 (0.2)(0.616) ] 1.41

2

γ γγ

VM

a

pM

p

-

= = =

æ ö- ÷ç= + = + =÷ç ÷÷çè ø

Hence, 3 3 2

1 2 1(12.5)(1.41) 17.6

p p p

p p p= = =

As demonstrated on the pressure-volume diagram for an ideal turbojet, the compression process is isentropic. Hence,

1

1

3 3

1 1

0.28633 1

1519(17.6) 1178

γγ

γγ

°

p T

p T

pt t

p

-

-


æ ö÷ç ÷= = =ç ÷ç ÷÷çè ø

Combustion occurs at constant pressure. For each slug of air entering the combustor, 0.05 slug of fuel is added. Each slug of fuel releases a chemical energy (heat) of

7 8(1.4 10 )(32.3) 4.51 s 10 ft lb/slug´ = . Hence, the heat released per slug of fuel-air mixture is

8

7(4.51 10 )(0.05)2.15 10 ft lb/slug

1.05q

´= = ´

Because the heat is added at constant pressure,

4 3

4 3

( )p

p

q c T T

qT T

c

= -

= +

For air, 1.4(1716) ft lb

60061 0.4 slug

γγ °p

Rc

R= = =

-

7

42.15 10

1178 1178 3580 47586006

°T R´

= + = + =

Again, from the p-v diagram for an ideal turbojet,

6 1

4 3

10.0568

17.6

p p

p p= = =

Also, 16 6

4 4

γγp T

p T

-æ ö÷ç ÷= ç ÷ç ÷÷çè ø

1 0.2866

6 44

4758 (0.0568) 2095

γγ

°pT T R

p

-æ ö÷ç ÷= = =ç ÷ç ÷÷çè ø

Note: The temperatures calculated in this problem exceed those allowable for structural integrity. However, in real life, the losses due to heat conduction will decrease the temperature. Also, the fuel-air ratio would be decreased in order to lower the temperatures in an actual application.


110

9.6 ( ) ( )

1000 (2000 950) 0

0.952 slug/sec

e e eT m V V p p A

m

m

¥ ¥= - + -

= - +=

However, ρ im V A¥ ¥=

Thus, 20.9520.42 ft

(0.002377)(950)ρ

im

AV¥ ¥

= = =

9.7 At 50 km, 287.9 N/mp¥ =

4( ) (25)(4000) (2 10 87.9)(2)

100,000 39,824 139,824 N

e e eT mV p p A¥= + - = + ´ -

= + =

Since 1 lb 4.48 N=

139824

31, 435 lb4.448

T = =

9.8 (a) At a standard altitude of 25km,

2

0.18

1.186

0

2527 N/m

1 2(1.18)(8314)(3756) 25271

(0.18)(20) 3.03 10

375sec

e

sp

sp

p p

Ig

I

¥ = =

é ùê úæ ö÷ê úç= - ÷ç ÷ê úç ÷çè ø´ê úê úë û

=

(b) From the isentropic relations,

1

0 0

10.1525

0 60

25273756 1274 K

3.03 10

(1.18)(8314)2725joule/kg K

1 (0.18)(20)

γγ

γγ

γγ

°

°

e e

ee

p

p T

p T

pT T

p

Rc

-

-


æ ö æ ö÷ç ÷ç÷= = =÷ç ç÷ ÷ç ç ÷ç÷÷ç è øè ø ´

= = =-

From the energy equation

02 ( ) 2(2725)(3756 1274) 3678 m/sece p eV c T T= - = - =

(c) 8314 joule

where 415.720 kg K

ee

e

p Rp R

RT M= = = =

32527

0.00477 kg/m(415.7)(1274)

=(0.00477)(15)(3678)=263.5kg/sec

e

e e e

p

m p A V

= =

=


111

(d) From the definition of specific impulse,

sp

TI

w= where w is the weight flow

0 (263.5)(9.8) 2582 N/sec w mg= = =

Hence,

sp (2582)(375) 968, 250N

968250or, 217682 lb

4.448

T w I

T

= = =

= =

(e) ( 1) /( 1)

0

o

* 2

1

γ γγγ

p Am

RT

+ -æ ö÷ç ÷= ç ÷ç ÷ç ÷+è ø

12.115

1/ 2

2

(30(1.01 10 )) * 1.18 2263.5

415.7 2.18(3756)

* 0.169 m

A

A

æ ö´ ÷ç= ÷ç ÷÷çè ø

=

9.9 87.6 / 32.2 2.72 slug/secm = =

( 1) /( 1)0

0

10.52(0.5)0

20

* 2

R 1

1.21 22.72

2400 2.216000

31,736 lb/ft

γ γγγ

p Am

T

p

p

+ -æ ö÷ç ÷= ç ÷ç ÷ç ÷+è ø


=

In terms of atmospheres,

031,736

15atm2116

p = =

9.10 0 sp (9.8)(240) 5.5 4009.6 m/sec ib

f

MV g I n n

M= = =

9.11 f i pM M M= -

1

1

i i

pf i p

i

M MMM M M

M

= =-

-

0 sp/ (11,200) /(9.8)(360) 23.9bV g Ii

f

Me e

M= = =

where escape velocity is 11.2 km/sec = 11,200 m/sec.

123.9

1

0.958

p

i

p

i

M

M

M

M

=-

=


112

9.12 From Eq. (9.43)

0

0log log log

nr ap

r a n p

=

= +

20At 500 lb/in : log (0.04) log log (500)p a n= = +

or: log 1.3979 2.69897a n= - - (1)

20At 1000 lb/in : log (0.058) log log (1000)p a n= = +

or: log 1.23657 3a n= - - (2)

Solving Eqs. (1) and (1) for a and n, we have

0 0.16133 0.03103n= - +

0.16133

0.53590.30103

n = =

log 1.23657 3(0.5359) 2.84427a = - - = - 0.0014313a = Hence, the equation for the linear burning rate is

0.5359

0

20

0.0014313

For 1500lb/in :

r p

p

=

=

0.53590.0014313(1500) 0.0721 in/sec.r = =

In 5 seconds, the total distance receded by the burning surface is 0.0721 (5) = 0.36 in.

9.13 1 1 2 21

1 2 2

0 sp

7200 800 5400 600 60(9.8)(275)

800 5400 600 601934 m/sec

p s p s Lb

s p s L

M M M M MV g I n

M M M M

n

é ù+ + + +ê ú= ê ú+ + +ê úë û+ + + +

=+ + +

=

2 12

2 20 sp

5400 600 60(9.8)(275)

600 605975 m/sec

p s Lb b

s L

M M MV V g I n

M M

n

é ù+ +ê ú- = ê ú+ê úë û+ +

=+

=

Hence: 25975 1934 7909 m/secbV = + =

9.14 The velocity of the jet relative to you is ( )eV V¥- . This is the velocity you feel of the "wind"

left behind in the air after the device passes through your space. The energy it has is kinetic space. The energy it has is kinetic energy, which per unit mass is

21( ) .

2 eV V¥-

Hence, the kinetic energy deposited per unit time is

21 ( ) .

2 em V V¥-


113

9.15 Power available ∞T V= (1)

From Eq. (9.24), ignoring the pressure thrust term and assuming that m is essentially aairm

( )∞ eT m V V= - (2)

The total power generated by the propulsive device is the useful power plus the wasted power.

Hence,

Total power generated 21( )

2∞ ∞ eT V m V V= + - (3)

Thus,

( )2

useful power available1total power generated2

η ∞

∞ ∞p

e

T V

T V m V V= =

+ - (4)

Substituting Eq. (2) into (4),

( )2

( )1

( )2

η ∞ ∞

∞ ∞ ∞

e

p

e e

m V V V

m V V V m V V

-=

- + - (5)

Dividing numerator and denominator by m ( ) ,eV V V¥ ¥- Eq. (5) becomes

1 1

1 11 ( ) / 12 2

η pee

VV V VV

¥ ¥¥

= =æ ö÷ç+ - ÷+ç ÷ç ÷÷çè ø

or,

2

1 /peV V

η¥

=+

9.16 From Example 9.3,

500 miles/hour 733 ft/secV¥ = =

And

1600 ft/sec

2 20.63

16001+ / 1733

η

e

pe

V

V V¥

=

= = =æ ö÷ç+ ÷ç ÷÷çè ø

9.17 Briefly:

(a) The flow velocity increase across the propeller is small, hence Ve is only slightly larger than V¥ , and η p is high.

(b) The exit velocity of the gas from a rocket engine is supersonic. Hence, Ve is very high compared to the velocity of the rocket, V¥ , and η p is low.

(c) The exit velocity from a gas turbine jet engine is usually a very high subsonic value, and sometimes a supersonic value just above one. Hence, it is less efficient than a propeller, but considerably more efficient than a rocket engine.


114

9.18 The temperature entering the compressor is T2 = 585oR. The temperature at the exit of the compressor, T3, is obtained from the isentropic relation

1 0.40.2863 3 1.4

2 2

3 2

(11.7) (11.7) 2.02

2.02 2.02 (585) 1182

T p

T p

T T R

γγ−

= = = =

= = = °

Since the compressor does work, per unit mass of gas, wc, the energy equation (Eq. 4.42) is modified as

22

322 32 2p c p

VVC T W C T+ + = +

For V2 = V3, this becomes

Wc = cp (T3 – T2)

6

(1.4)(1716) ft lb6006

1 0.4 slug°

ft lb6006(1182 585) 3.58 10

slug

p

c

Rc

R

W

γγ

= = =−

= − = ×

The mass flow is 220 lb 200 slug

6.21 .sec 322 sec

= =

The total work done per second is

6 7slug ft lb ft lb6.201 3.58 10 2.22 10

sec slug sec × = ×

Since one HP = 550 ft lb

,sec

the horsepower provided by the compressor is

72.22 10

HP 40,364550

×= =

9.19 From the solution of Problem 9.18, the temperature at the entrance to the combustor is the same as the temperature at the exit of the compressor, namely T3 = 1182oR. The pressure is essentially constant through the combustor, so the total heat added per unit mass to the air (neglecting the small mass of fuel added) is

4 3

6

( ) 6006(2110 1182)

5.57 10 ft lb/slug

pq c T T

q

= − = −

= ×

or,

6 5

m m

ft lb 1 slug ft lb5.57 10 1.73 10

slug 32.2 lb lbq

= × = ×

Since the heat released per unit mass of fuel is given as 1.4 × 107 m

ft lblb

, the amount of fuel

mixed with each lbm of air is 1.73 × 105/1.4 × 107 = 0.0124 lbm. The mass flow of air given in Problem 9.18 is 200 lbm/sec. So the fuel rate is

m(0.0124)(200) 2.48 lb / sec=


115

9.20 In the flow through the turbine, work wT is extracted from the gas and provided to driving the turbine, and hence to the compressor. From the energy equation,

22

544 52 2p T P

VVC T W C T+ − = +

Since V5 = V4,

5 4p p Tc T c T W= −

Since there are no mechanical losses, the work provided by the turbine is the same as the work done

by the compressor, wT = wc. From the solution of Problem 9.18, wc = 3.58 × 106 ft lb

.slug

Thus

65 4 3.58 10p pc T c T= − ×

or,

6 6

5 4

5

3.58 10 3.58 102110

6006

1514

pT T

c

T R

× ×= − = −

= °

9.21 First, calculate the pressure at the entrance to the nozzle, p5. From Problem 9.18,

3

2

3 2 2

11.7

lb11.7 11.7 (2116) 24,757

ft

p

p

p p

=

= = =

The pressure is constant through the combustor, p4 = p3 = 24,757 2

lb.

ft The flow through the

turbine is isentropic. From the solution of Problem 9.20, T5 = 1514oR. Hence

1.41 0.4 3.55 5

4 4

5 4 2

1514(0.71753) 0.3129

2110

lb(0.3129) (0.3129)(24757) 7746

ft

y

yp T

p T

p p

− = = = =

= = =

The expansion through the nozzle is isentropic.

1 0.4

1.4 0.2866 6

5 5

6 5

2116(0.273) 0.690

7746

0.690 (0.690)(1514) 1045

yyT p

T p

T T R

− = = = =

= = = °

2 25 6

5 6

2 2 26 5 6 5

6 6 6

6 1/ 26

2 2

2 ( ) 2(6006)(1514 1045) (1500)

5.63 10 2.25 10 7.88 10

(7.88 10 ) 2807 ft/sec

p p

p

V Vc T c T

V c T T V

V

+ = +

= − + = − +

= × + × = ×

= × =


116

9.22 From Problem 9.21, T6 = 1045oR and V6 = 2807 ft/sec

6 6

6

(1.4)(1716)(1045) 1584 ft/sec

28071.77

1584

a RT

M

γ= = =

= =

Comment: The flow at the exit of the engine is supersonic, and because of this the nozzle must be a convergent-divergent shape. Because the pressure ratio across the compressor given in Problem 9.18 is high, namely 11.7, this is a high performance turbojet engine suitable for use on supersonic airplanes, and a supersonic exit flow is a characteristic of such an engine. However, the result obtained here, namely M6 = 1.77 is a relatively high value, influenced by our assumption of no mechanical or aerodynamic losses that in real life detract from the engine performance.

9.23 From Eq. (9.25), the thrust from the engine is

( ) ( ) .e e eT m V V p p A∞ ∞= − + −

From Problem 9.18, m = 200 lb/sec. From the solution of Problem 9.21, Ve = 2807 ft/sec. Also, from Problem 9.21, pe = 2116 lb/ft2.

At 36,000 ft, from App. B, T∞ = 390.53oR and p∞ = 476 lb/ft2.

6 (1.4)(1716)(390.53) 968.6 ft/seca RTγ ∞= = =

At 2,M∞ =

V∞ = 2(968.6) = 1937 ft/sec

The diameter of the exit is 28 inches = 2.33 ft. Hence

2 22(2.33)

4.26 ft4 4

200200 lb/sec 6.21 slug/sec.

32.2

ed

A

m

π π= = =

= = =

Thus, from Eq. (9.25),

(6.21)(2807 1937) (2116 476)(4.26)

5403 6986 12,389 lb .

T = − + −

= + =

Comment: For the engine and the conditions treated in Problems 9.18 – 9.23, we find that the term (pe – p∞) Ae in the thrust equation is larger than the term (Ve – V∞). This is usually not the case forturbojet engines, but is the case here for the high compression ratio engine treated and the conditions chosen.

9.24 Specific thrustT

W=

From Problem 9.23, T = 12,389 lb and the weight flow through the engine is 200 lb/sec (the same as the mass flow when the mass is expressed in pounds mass). Thus,

12,389

Specific thrust 61.9 sec200

= =


117

Chapter 10

10.1 Since 2

Reδ ∝ M

, we have

2

20 220

( ) 0.3(100) 30 inches 2.5ft2

δ δM M= =æ ö÷ç= = = =÷ç ÷÷çè ø

This dramatically demonstrates that boundary layers at hypersonic Mach numbers can be very thick.

10.2 From Chapter 4 we have

2 20 1 1.4 11 1 (20) 81.

2 2

γTM

T ¥¥

- -= + = + =

At 59 km, 258.1 KT¥ =

Thus: 0 (81)(258.1) 20,906 KT = =

Considering that the surface temperature of the sun is about 6000K, the above result is an extremely high temperature. This illustrates that hypersonic flows can be very high temperature flows. At such temperatures, the air becomes highly chemically reacting, and in reality the ratio of specific heats is no longer constant; in turn, the above equation, which assumes constant γ, is no longer valid. Because the dissociation of the air requires energy (essentially "absorbs" energy), the gas temperature at the stagnation point will be much lower than calculated above; it will be approximately 6000K. This is still quite high, and is sufficient to cause massive dissociation of the air.

10.3 Following the nomenclature of Example 10.1,

57.3( / ) 57.3(6.12) 28.65

90 61.5

φθ φ

°° − °

s R= = == =

(a) 2

/ 12 2 20

2

( 1) 1 2

14 2( 1)

γ γγ γ γ

γγ γ

p M M

p M

-¥ ¥

¥ ¥

é ù é ù+ - +ê ú ê ú= ê ú ê ú+- -ê ú ê úë û ë û

1.4 / 0.42 2 2

2

(2.4) (18) 1 1.4 2(1.4)(18)= 417.6

2.44(1.4)(18) 2(0.4)

é ù é ù- +ê ú ê ú =ê ú ê ú-ê ú ê úë û ë û

20, max 2 2

2 2 1 = (417.6 1) 1.836

(1.4)(18)γpp

CpM ¥¥

æ ö÷ç ÷ç= - - =÷ç ÷÷ççè ø

(Note: This value of , maxpC is only slightly smaller than the value calculated in Example 10.1, which was 1.838 for 25M¥ = . This is an illustration of the hypersonic Mach number independence principle, which states that pressure coefficient is relatively independent of Mach number at hypersonic speeds.)

(b) From modified Newtonian:

( ) ( )2 2, maxsin 1.836 sin 61.5 1.418θ °p pC C= = =


118

10.4 From Eq. (10.11),

2

2 2

2 2 2

2, max

2sin cos

(2sin )( sin ) 4cos sin 0

sin 2cos 2(1 sin )

54.7

2 sin (54.7)cos (54.7)=0.77

α α

α α α ααα α αα °

L

L

L

C

dC

d

C

=

= - + =

= = -=

=

10.5 (a) 22 sin cos α αLC =

3D, 02sin CαDC = +

For smallα , these become

22αLC = (1)

3,02αD DC C= + (2)

2

3,0

2

2

αα

L

D D

C

C C=

+ (3)

3 2 2

,03

,0

(2 C ) 4 2 (6 )0

(2 )

α α α αα α

L

DD

D

Cd

C

d C

æ ö÷ç ÷ç ÷ç ÷÷ç + -è ø= =

+

4 4,08 4 12 0α α αDC+ - =

3,04 4α DC=

1/3,0( )α DC= (4)

Substituting Eq. (4) into Eq. (3):

2/3

,01/3 1.3

,0 ,0 ,0 ,0max

2( ) 2/3 0.67

2 ( ) ( )

DL

D D D D D

CC

C C C C C

æ ö÷ç ÷ = = =ç ÷ç ÷÷ç +è ø

Hence:

1/3,0

max0.67 /( ) and it occurs atD

LC

D

æ ö÷ç =÷ç ÷÷çè ø

1/ 3,0( ) .α DC=


119

(b) Repeating Eq.(2):

3,02αD DC C= + (2)

From the results of part (a), at (L/D)max we have 1/3,0( , )α DC= . Substituting this into

Eq.(2),

,0 ,0 ,02 + 3 D D D DC C C C= =

Since ,0 ,0 = +D D DC C C

where ,D WC is the wave drag, we have

, ,0 ,0 3D D W D DC C C C= + =

or, , ,02 D W DC C=

Wave drag = 2 (friction drag) when L/D is maximum. Or, another way of stating this is that friction drag is one-third the total drag.

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