Solution Manual for Introduction to Flight 7th Edition by Anderson

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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 0.71m /kg
ρ
ρ
ρ
×
=
= p RT =
v= = =
2.2 Mean kinetic energy of each atom 3 3 (1.38 10 23 ) (500) 1.035 10 20J
2 2
= k T = × − = × −
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
4
× = × atoms.
20 26 6
Totalinternal energy (energy per atom)(number of atoms)
(1.035 10- )(1.505 10 ) 1.558 10 J
=
= ´ ´ = ´
2.3 3
2116 0.00237 slug
(1716)(460 59) ft
ρ= = =
+
p
RT
Volumeof the room (20)(15)(8) 2400ft3
Total mass in the room (2400)(0.00237) 5.688slug
Weight (5.688)(32.2) 183lb
= =
= =
= =
2.4 3
2116 0.00274 slug
(1716)(460 10) ft
p
RT
ρ = = =
-
Since the volume of the room is the same, we can simply compare densities between the two
problems.
3
0.00274 0.00237 0.00037 slug
ft
Δρ = - =
%change 0.00037 (100) 15.6% increase
0.00237
ρ
ρ
Δ = = ´ =
2.5 First, calculate the density from the known mass and volume, 3
ρ = 1500/900 = 1.67lbm/ft
In consistent units, ρ = 1.67/32.2 = 0.052slug/ft3. Also, T = 70 F = 70 + 460 = 530 R.
Hence,
p = ρ RT = (0.52)(1716)(530)
p = 47,290lb/ft2
or p = 47,290 / 2116 = 22.3 atm
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.
3
2.6 p = ρ RT
np = np + nR + nT
Differentiating with respect to time,
1 1 ρ 1
ρ
dp = d + dT
p dt dt T dt
or,
T
ρ
ρ
dp = p d + p dT
dt dt dt
or, dp = RT dρ +ρ R dT
dt dt dt
(1)
At the instant there is 1000 lbm of air in the tank, the density is
3
m ρ = 1000 / 900 = 1.11lb /ft
ρ = 1.11/32.2 = 0.0345slug/ft3
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 ft3, we also have
m 3
3 m
0.5 lb /sec
0.000556 lb /(ft )(sec)
900 ft
dρ = =
dt
0.000556 1.73 10 5 slug/(ft3 )(sec)
32.2
dρ = = × −
dt
Thus, from equation (1) above,
5
2
(1716)(510)(1.73 10 ) (0.0345)(1716)(0.0167)
15.1 0.99 16.1 lb/(ft )(sec) 16.1
2116
0.0076 atm/sec
ρ = × − +
= + = =
=
d
dt
2.7 In consistent units,
T = −10 + 273 = 263 K
Thus,
4
3
/ (1.7 10 )/(287)(263)
0.225 kg/m
ρ
ρ
= = ×
=
p RT
2.8 5 3
3
/ 0.5 10 /(287)(240) 0.726 kg/m
1/ 1/0.726 1.38 m /kg
ρ
ρ
= = × =
= = =
p RT
v
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.
4
2.9
3 3
0 0
2 3
0
Forcedue to pressure (2116 10 )
[2116 5 ] 6303 lb perpendicular to wall.
Fp p dx x dx
x x
= = = -
= - =
ò ò
3 3
τ 1 0 0
2
1
2 30
Force due to shear stress= τ 90
( 9)
[180 ( 9) ] 623.5 540 83.5 lb tangential towall.
F dx dx
x
x
= =
+
= + = - =
ò ò
Magnitude of the resultant aerodynamic force =
(6303)2 (835)2 6303.6 lb
Arc Tan 83.5 0.76º
6303
θ
R = + =
æç ÷ö = çç ÷÷÷ = è ø
2.10 3 sin
2
V = V∞ θ
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,
max
3 (85)(1) 127.5 mph at 90 ,
2
V = = θ = °
i.e., the entire rim of the sphere in a plane perpendicular to the freestream direction.
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.
5
2.11 The mass of air displaced is
M = (2.2)(0.002377) = 5.23´10-3 slug
The weight of this air is
3
Wair (5.23 10 )(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
(0.168) 4 0.0233 lb.
Hc 28.8 W = =
Hence, the maximum weight that can be lifted by the balloon is
0.168 − 0.0233 = 0.145 lb.
2.12 Let p3, ρ
3, and T3 denote the conditions at the beginning of combustion, and p4, ρ
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 = ρ
3 = 11.3 kg/m3. Thus, from the equation of state,
7 2
4 4 4 p = ρ RT = (11.3)(287)(4000) = 1.3´10 N/m
or,
7
4 5
1.3 10 129 atm
1.01 10
p ´
= =
´
2.13 The area of the piston face, where the diameter is 9 cm = 0.09 m, is
2
(0.09) 6.36 10 3m2
4
A = π = ´ -
(a) The pressure of the gas mixture at the beginning of combustion is
6 2
3 3 3 p = ρ RT = 11.3(287)(625) = 2.02 ´10 N/m
The force on the piston is
6 3 4
F3 p3A (2.02 10 )(6.36 10 ) 1.28 10 N = = ´ ´ - = ´
Since 4.45 N = l lbf,
4
3
1.28 10 2876 lb
4.45
F ´
= =
(b) 7 2
4 4 4 p = ρ RT = (11.3)(287)(4000) = 1.3´10 N/m
The force on the piston is
7 3 4
F4 p4A = (1/3 10 ) (6.36 10 ) = 8.27 10 N = ´ ´ - ´
4
4
8.27 10 18,579 lb
4.45
F ´
= =
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.
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 2
p3 = p4 = 4 ´10 N/m
(a)
6
3 3
3
3
4 10 = 15.49 kg/m
(287)(900)
ρ p
RT
´
= =
(b)
6
4 3
4
4
4 10 = 9.29 kg/m
(287)(1500)
ρ p
RT
´
= =
2.15 1 mile = 5280 ft, and 1 hour = 3600 sec.
So:
60 miles 5280 ft 1 hour 88 ft/sec.
hour mile 3600 sec
æç ÷öçæ ÷öçæ ÷ö çç ÷÷÷çç ÷÷÷çç ÷÷÷ = è øè øè ø
A very useful conversion to remember is that
60 mph = 88 ft/sec
also, 1 ft = 0.3048 m
88 ft 0.3048 m 26.82 m
sec 1 ft sec
æ ÷öçæ ÷ö çç ÷ç ÷÷ = èç ø÷÷çèç ø÷
Thus 88 ft 26.82 m
sec sec
=
2.16 692 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
(1.0715 105 )(2) 2.143 105 N f f F = p A = × = ×
On the back face
(1.01 105 )(2) 2.02 105 N b b F = p A = × = ×
The net force on the plate is
(2.143 2.02) 105 0.123 105 N f b F = F − F = − × = ×
From Appendix C,
1 4.448 N. f lb =
So,
0.123 105 2765 lb
4.448
F = × =
This force acts in the same direction as the flow (i.e., it is aerodynamic drag.)
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.
7
2.18 10,100 2
Wing loading 43.35 lb/ft
233
W
s
= = =
In SI units:
2
2
2
lb 4.448 N 1 ft
43.35
ft 1 lb 0.3048 m
2075.5 N
m
W
s
W
s
      =        
=
In terms of kilogram force,
2 2
N 1 k kg
2075.5 211.8
m 9.8 N m
W f f
s
    =     =  
2.19 437 miles 5280 ft 0.3048 m 7.033 105 m 703.3 km
hr mile 1 ft hr hr
V =       = × =
0.3048 m
Altitude (25,000 ft) 7620 m 7.62 km
1 ft
=   = =  
2.20 26,000 ft 0.3048 m 7.925 103 m 7.925 km
sec 1 ft sec sec
V =     = × =
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 m
ft
=   =  
wing span = 40 ft, 11.726 inches = 40.98 ft
0.3048 m
40.98 ft 12.49 m
ft
=   =  