Introduction to Flight 8th Edition Anderson Solutions Manual Download:http://testbanklive.com/download/introduction-to-flight-8thedition-anderson-solutions-manual/ SOLUTIONS MANUAL TO ACCOMPANY INTRODUCTION TO FLIGHT 8th Edition By John D. Anderson, Jr.

Chapter 2 2.1

= p/RT = (1.2)(1.01105 )/(287)(300) 1.41 kg/m2 v = 1/ = 1/1.41= 0.71 m3 /kg

2.2

3 k T = (1.38 1023 ) (500) = 1.035 1020 J 2 2 One kg-mole, which has a mass of 4 kg, has 6.02 × 1026 atoms. Hence 1 kg has 1 (6.02 1026 ) = 1.505 1026 atoms. 4 Total internal energy = (energy per atom)(number of atoms)

Mean kinetic energy of each atom

3

= (1.035 ´ 10- 20 )(1.505 ´ 10 26 ) = 1.558 ´ 106 J

p RT

2.3

2116 0.00237 (1716)(460 59)

slug ft 3

Volume of the room = (20)(15)(8) = 2400 ft 3 Total mass in the room = (2400)(0.00237) = 5.688 slug Weight = (5.688)(32.2) = 183 lb

2.4

=

p = RT

2116 slug = 0.00274 3 ft (1716)(460 - 10)

Since the volume of the room is the same, we can simply compare densities between the two problems. 0.00274 - 0.00237 = 0.00037 slug ft 3

=

% change =

2.5

=

0.00037 ´ (100) = 15.6% increase 0.00237

First, calculate the density from the known mass and volume, = 1500/ 900 = 1.67 lb m /ft 3 In consistent units, =

1.67/32.2 = 0.052 slug/ft 3. Also, T = 70 F = 70 + 460 = 530 R.

Hence, p = RT =

(0.52)(1716)(530)

p = 47, 290 lb/ft 2 or

p = 47, 290 / 2116 = 22.3 atm

2.6

p = RT np np nR nT Differentiating with respect to time, 1 dp 1 d 1 dT p dt dt T dt

or,

dp p d ď€

p dT

dt

T dt

dp

or,

dt

dt d RT ď€

R

dT

dt

(1) dt

At the instant there is 1000 lb m of air in the tank, the density is 1000 / 900 1.11lb m /ft

3

1.11/32.2 0.0345 slug/ft 3

Also, in consistent units, is given that T = 50 + 460 = 510 R and that dT 1F/min 1R/min 0.016R/sec dt From the given pumping rate, and the fact that the volume of the tank is 900 ft 3, we also have d 0.5 lb m /sec 0.000556 lb /(ftm3 )(sec) dt 900 ft 3 d 0.000556 1.73105 slug/(ft 3 )(sec) dt 32.2 Thus, from equation (1) above, d (1716)(510)(1.73 10 5 ) (0.0345)(1716)(0.0167) dt 16.1 15.1 0.99 16.1 lb/(ft 2 )(sec) 2116 0.0076 atm/sec 2.7

In consistent units, T 10 273 263 K Thus, p/RT (1.7 10 4 )/(287)(263) 0.225 kg/m3

2.8

p/RT 0.5 105 /(287)(240) 0.726 kg/m3 v 1/ 1/0.726 1.38 m3 /kg

2.9

Fp = Force due to pressure =

ò

3

p dx =

ò

0

= [2116x -

5x 2 ] 30

3

(2116 - 10x) dx

0

= 6303 lb perpendicular to wall. 3

Fτ = Force due to shear stress =

ò

τ dx =

0

ò

3

0

90 1

dx

(x + 9) 2

1

= [180 (x + 9) 2 ] 30= 623.5 - 540 = 83.5 lb tangential to wall.

Magnitude of the resultant aerodynamic force = R =

(6303)2 + (835)2 = 6303.6 lb

= Arc Tan ç

2.10 V =

æ 83.5 ö÷ = ÷ 6303 ø

0.76º

3 V sin 2

Minimum velocity occurs when sin = 0, i.e., when = 0° and 180°. Vmin = 0 at = 0° and 180°, i.e., at its most forward and rearward points. Maximum velocity occurs when sin = 1, i.e., when = 90°. Hence, Vmax =

3 (85)(1) = 127.5 mph at = 90, 2

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

2.11 The mass of air displaced is M = (2.2)(0.002377) = 5.23 ´ 10- 3 slug The weight of this air is Wair = (5.23´ 10- 3)(32.2) = 0.168 lb 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 WH c = (0.168)

4 = 0.0233 lb. 28.8

Hence, the maximum weight that can be lifted by the balloon is 0.168 0.0233 = 0.145 lb. 2.12 Let p3, 3, and T3 denote the conditions at the beginning of combustion, and p4, 4, and T4 denote conditions at the end of combustion. Since the volume is constant, and the mass of the gas is constant, then p4 = 3 = 11.3 kg/m3. Thus, from the equation of state, (11.3)(287)(4000) = 1.3 ´ 107 N/m2

p4 = 4 RT4 = or,

p4 =

1.3 ´ 107 = 129 atm 5 1.01 ´ 10

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

(0.09) 2 = 4

6.36 ´ 10- 3 m2

The pressure of the gas mixture at the beginning of combustion is p3 = 3RT3 = 11.3(287)(625) =

2.02 ´ 106 N/m2

The force on the piston is F3 = p3 A = (2.02 ´ 106 )(6.36 ´ 10- 3) = 1.28 ´ 10 4 N Since 4.45 N = l lbf, 1.28 ´ 104 F3 =

(b)

= 2876 lb

4.45

p4 = 4 RT4 =

(11.3)(287)(4000) = 1.3 ´ 107 N/m2

The force on the piston is F4 = p4 A = (1/3 ´ 10 7 ) (6.36 ´ 10 - 3) = 8.27 ´ 10 4 N 8.27 ´ 104 F4 =

4.45

= 18, 579 lb

2.14 Let p3 and T3 denote conditions at the inlet to the combustor, and T4 denote the temperature at the exit. Note: p3 = p4 = 4 ´ 106 N/m2

(a)

3

=

p3

=

RT3 (b)

4

=

p4

4 ´ 106

= 15.49 kg/m3

(287)(900)

=

RT4

4 ´ 106

= 9.29 kg/m3

(287)(1500)

2.15 1 mile = 5280 ft, and 1 hour = 3600 sec. So: æ miles÷ öæ=5280 ft÷ öæ= 1 hour ö÷ ÷ç ÷ç ÷ = 88 ft/sec. ç 60 ÷ ÷ ç ÷ hour ø mile øè 3600 sec ø A very useful conversion to remember is that 60 mph = 88 ft/sec also,

1 ft = 0.3048 m æ ö ç88 ft 0.3048 m æ ö÷ç çè sec ÷ øçè

Thus

88

ft

1 ft

= 26.82

sec 2.16

÷= ø÷

26.82

m

sec

m sec

692 miles 88 ft/sec 1015 ft/sec hour 60 mph 692

miles 26.82 m/sec 309.3 m/sec hour 60 mph

2.17 On the front face F f p Af (1.0715 105 )(2) 2.143105 N On the back face 5 5 Fb p A b (1.01 10 )(2) 2.02 10 N

The net force on the plate is F Ff Fb

(2.143 2.02) 105 0.123 105 N

From Appendix C, 1 lb f 4.448 N.

So,

0 .

F

123 105 4.448

2765 lb

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

W 10,100

2.18 Wing loading

s

43.35 lb/ft 2

233

In SI units: W s

lb 4.448 N 43.35 2 1 lb ft

2

1 ft 0.3048 m

W N 2075.5 s m2 In terms of kilogram force, W N 1kf 2075.5 2 9.8 N s m 2.19 V 437

miles 5280 ft 0.3048 m hr

mile

kg f 211.8

7.033 105

m

m2

703.3

1 ft

km

hr

hr

0.3048 m 7620 m 7.62 km 1 ft 0.3048 m m km 7.925 103 7.925

Altitude (25, 000 ft) 2.20 V 26, 000

ft

sec

1 ft

sec

sec

2.21 From Fig. 2.16, length of fuselage = 33 ft, 4.125 inches = 33.34 ft 33.34 ft

0.3048 m 10.16 m ft

wing span = 40 ft, 11.726 inches = 40.98 ft 40.98 ft 2.22 (a)

0.3048 m 12.49 m ft

From App. C 1 ft. 0.3048 m. Thus, 354,200 ft (354,000)(0.3048) 107,960 m 107.96 km

(b) From Example 2.6: 60 mph 26.82 m/sec Thus,

4520

miles hr

4520

miles mi

m 26.82 sec 2020.4 m/sec hr 60 hr

2.23 m

34, 000 lb 1055.9 slug 32.2 lb/slug

From Newton’s 2nd Law F ma

a

F 57, 000 53.98 ft/sec2 M 1055.9

2.24 # of g’s

53.98 1.68 32.2

2.25 From Appendix C, one pound of force equals 4.448 N. Thus, the thrust of the Rolls-Royce Trent engine in pounds is T

373.7 103 N 84, 015 lb 4.448 N/ lb

2.26 6.762 106 N

(a)

F (690, 000)(9.8)

(b)

F 6.762 106 /4.448 1.5 106 lb

Introduction to Flight 8th Edition Anderson Solutions Manual Download:http://testbanklive.com/download/introduction-to-flight-8th-edition-anderson-solutionsmanual/ introduction to flight 8th edition pdf introduction to flight 8th edition solutions introduction to flight 7th edition pdf free download introduction to flight international edition introduction to flight anderson 8th edition download introduction to flight 6th edition sharif ~iae download introduction flight introduction to flight 8th pdf

Introduction to flight 8th edition anderson solutions manual

Download: http://testbanklive.com/download/introduction-to-flight-8th-edition-anderson-solutions-manual/ Introduction to Flight 8th Edition...

Introduction to flight 8th edition anderson solutions manual

Published on Sep 4, 2017

Download: http://testbanklive.com/download/introduction-to-flight-8th-edition-anderson-solutions-manual/ Introduction to Flight 8th Edition...

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