SOLUTIONS MANUAL TO ACCOMPANY
INTRODUCTION TO FLIGHT
7th Edition
By
John D. Anderson, Jr.
Chapter 2
2.1
ρ = p/RT = (1.2)(1.01× 105 )/(287)(300)
ρ = 1.41 kg/m 2
v = 1/ρ = 1/1.41 = 0.71 m3 /kg
2.2
3
3
k T = (1.38 × 10−23 ) (500) = 1.035 × 10−20 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
Mean kinetic energy of each atom =
Totalinternal energy = (energy per atom)(number of atoms)
= (1.035 ´10-20 )(1.505 ´1026 ) = 1.558 ´ 106 J
2.3
ρ=
2116
slug
p
=
= 0.00237 3
ft
RT (1716)(460 + 59)
Volume of the room = (20)(15)(8) = 2400 ft 3
Total mass in the room = (2400)(0.00237) = 5.688slug
Weight = (5.688)(32.2) = 183lb
2.4
ρ=
p
2116
slug
=
= 0.00274 3
RT
(1716)(460 - 10)
ft
Since the volume of the room is the same, we can simply compare densities between the two
problems.
Δρ = 0.00274 - 0.00237 = 0.00037
% change =
2.5
Δρ
ρ
=
slug
ft 3
0.00037
´ (100) = 15.6% increase
0.00237
First, calculate the density from the known mass and volume, ρ = 1500 / 900 = 1.67 lb m /ft 3
In consistent units, ρ = 1.67/32.2 = 0.052 slug/ft 3. Also, T = 70 F = 70 + 460 = 530 R.
Hence,
p = ρ RT = (0.52)(1716)(530)
p = 47, 290 lb/ft 2
or
p = 47, 290 / 2116 = 22.3 atm
2
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2.6
p = ρ RT
np = np + nR + nT
Differentiating with respect to time,
1 dp 1 d ρ 1 dT
=
+
p dt ρ dt T dt
or,
dp p d ρ p dT
=
+
dt ρ dt T dt
or,
dp
dρ
dT
= RT
+ ρR
dt
dt
dt
(1)
At the instant there is 1000 lbm of air in the tank, the density is
ρ = 1000 / 900 = 1.11lb m /ft 3
ρ = 1.11/32.2 = 0.0345slug/ft 3
Also, in consistent units, is given that
T = 50 + 460 = 510 R
and that
dT
= 1F/min = 1R/min = 0.016 R/sec
dt
From the given pumping rate, and the fact that the volume of the tank is 900 ft3, we also have
d ρ 0.5 lb m /sec
=
= 0.000556 lb m /(ft 3 )(sec)
dt
900 ft 3
d ρ 0.000556
=
= 1.73 × 10−5 slug/(ft 3 )(sec)
dt
32.2
Thus, from equation (1) above,
dρ
= (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,
ρ = p/RT = (1.7 × 104 )/(287)(263)
ρ = 0.225 kg/m3
2.8
ρ = p/RT = 0.5 ×105 /(287)(240) = 0.726 kg/m3
v = 1/ρ = 1/0.726 = 1.38 m3 /kg
3
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2.9
Fp = Force due to pressure =
ò
3
p dx =
0
3
ò
(2116 - 10 x) dx
0
= [2116 x - 5 x 2 ] 30 = 6303 lb perpendicular to wall.
Fτ = Force due to shear stress =
ò
3
τ dx =
0
ò
3
0
90
(x +
1
9) 2
dx
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
æ 83.5 ÷ö
= 0.76º
çè 6303 ÷÷ø
θ = Arc Tan çç
3
V∞ sin θ
2
Minimum velocity occurs when sin θ = 0, i.e., when θ = 0° and 180°.
2.10 V =
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,
Vmax =
3
(85)(1) = 127.5 mph at θ = 90°,
2
i.e., the entire rim of the sphere in a plane perpendicular to the freestream direction.
4
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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, 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,
p4 = ρ4 RT4 = (11.3)(287)(4000) = 1.3 ´ 107 N/m 2
or,
p4 =
1.3 ´ 107
1.01 ´ 105
= 129 atm
2.13 The area of the piston face, where the diameter is 9 cm = 0.09 m, is
A=
(a)
π (0.09) 2
4
= 6.36 ´ 10-3 m 2
The pressure of the gas mixture at the beginning of combustion is
p3 = ρ3 RT3 = 11.3(287)(625) = 2.02 ´ 106 N/m 2
The force on the piston is
F3 = p3 A = (2.02 ´ 106 )(6.36 ´10-3 ) = 1.28 ´ 104 N
Since 4.45 N = l lbf,
F3 =
(b)
1.28 ´ 104
= 2876 lb
4.45
p4 = ρ4 RT4 = (11.3)(287)(4000) = 1.3 ´ 107 N/m 2
The force on the piston is
F4 = p4 A = (1/3 ´107 ) (6.36 ´10-3 ) = 8.27 ´ 104 N
F4 =
8.27 ´ 104
= 18,579 lb
4.45
5
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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/m 2
(a)
ρ3 =
p3
4 ´ 106
= 15.49 kg/m3
=
RT3
(287)(900)
(b)
ρ4 =
p4
4 ´ 106
= 9.29 kg/m3
=
RT4
(287)(1500)
2.15 1 mile = 5280 ft, and 1 hour = 3600 sec.
So:
æ miles ÷öæ 5280 ft ÷öæ 1 hour ÷ö
çç 60
÷çç
÷çç
÷ = 88 ft/sec.
èç hour ÷øèç mile ÷øèç 3600 sec ÷ø
A very useful conversion to remember is that
60 mph = 88 ft/sec
also,
1 ft = 0.3048 m
æ
ö
æ
÷öç 0.3048 m ÷÷ = 26.82 m
ç 88 ft ÷ç
÷÷
÷èç 1 ft
çèç sec ø÷ç
sec
ø
Thus
2.16
88
ft
m
= 26.82
sec
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
Ff = p f A = (1.0715 ×105 )(2) = 2.143 ×105 N
On the back face
Fb = pb A = (1.01 × 105 )(2) = 2.02 × 105 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,
F=
0.123 ×105
= 2765 lb
4.448
This force acts in the same direction as the flow (i.e., it is aerodynamic drag.)
6
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displayed, reproduced, or distributed in any form or by any means, without the prior written permission of the publisher, or used
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student using this manual is using it without permission.
2.18 Wing loading =
W 10,100
=
= 43.35 lb/ft 2
233
s
In SI units:
W 
lb   4.448 N   1 ft 
= 43.35 2  


s 
ft   1 lb   0.3048 m 
2
W
N
= 2075.5
s
m2
In terms of kilogram force,
kg f
N   1kf 
W 
=  2075.5 2  
= 211.8 2

m   9.8 N 
m
s 
miles   5280 ft   0.3048 m 

5 m = 703.3 km
2.19 V =  437
 
 
 = 7.033 × 10

hr
mile
1 ft
hr
hr
 0.3048 m 
= 7620 m = 7.62 km
Altitude = (25, 000 ft) 
 1 ft 
ft   0.3048 m 

3 m = 7.925 km
2.20 V =  26, 000
 
 = 7.925 × 10

sec
1 ft
sec
sec
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
7
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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
ρ2
p
-( g /RT )(h 2- h 1)
= 2 = e 0
ρ1
p1
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
(6000)
ρ2
p
= 2 = e (287)(216.66)
= 0.3884
ρ1
p1
Hence:
p2 = (0.3884)(1.9399 ´ 104 ) = 7.53 ´ 103 N/m 2
ρ2 = (0.3884)(3.1194 ´ 10-1) = 0.121 kg/m3
and, of course,
T2 = 216.66 K
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
ρ =
p
2.65 ´ 104
=
= 0.419 kg/m3
RT
(287)(220)
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,
T =
p
535.89
=
= 378 R
ρR
(8.2704 ´ 10-4 )(1716)
8
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3.5
h - hG
h
= 0.02 = 1 -
hG
h
From Eq. (3.6), the above equation becomes
æ r + hG
1 - çç
çè
r
÷÷ö = 1 - 1 - hG = 0.02
÷ø
r
hG = 0.02 r = 0.02(6.357 ´ 106 )
hG = 1.27 ´ 105 m = 127 km
3.6
T = 15 − 0.0065h = 15 − 0.0065(5000) = −17.5°C = 255.5°K
a =
dT
= -0.0065
dh
From Eq. (3.12)
æ T ö÷-g0 /aR
æ 255.5 ÷ö-(2.8) /(-0.0065)(287)
p
= ççç ÷÷
= çç
= 0.533
çè 288 ÷÷ø
çè T1 ø÷
p1
p = 0.533 p1 = 0.533 (1.01 ´ 105 ) = 5.38 ´ 104 N/m 2
3.7
n
p
g
(h - h1)
=p1
RT
h - h1 = -
1
p
1
RT n
=(4157)(150) n 0.5
g
p1
24.9
Letting h1 = 0 (the surface)
h = 17,358 m = 17.358 km
9
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student using this manual is using it without permission.
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:
g
æ T ö- aR
p
= ççç ÷÷÷
çè T1 ÷ø
p1
(1)
T = T1 + a (h – h1)
(2)
and
Differentiating Eq. (1) with respect to time:
g
æ
g
ö
æ 1 ö- aR æ g ö çççè - aR -1÷÷÷ø dT
1 dp
÷÷ T
çç = ççç ÷÷÷
çè T1 ÷ø
èç AR ÷ø
p1 dt
dt
(3)
Differentiating Eq. (2) with respect to time:
dT
dh
= a
dt
dt
(4)
Substitute Eq. (4) into (3)
g
æ g
ö÷
æ g ö -ççç +1÷÷ dh
dp
= - p1(T1) aR çç ÷÷÷ T è aR ø
çè R ø
dt
dt
(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
a =
T2 - T1
483.04 - 518.69
°R
=
= -0.00357
h2 - h1
10, 000 - 0
ft
Also from Appendix B, p1= 2116.2 lb/ft2 at sea level, and T = 429.64 °R at 25,000 ft.
Thus,
g
32.2
=
= -5.256
aR
(-0.00357)(1716)
Hence, from Eq. (5)
æ 32.2 ö÷
dp
(429.64)4.256 (500)
= -(2116.2)(518.69)-5.256 çç
çè 1716 ø÷÷
dt
dp
lb
= -17.17 2
dt
ft sec
10
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3.9
From the hydrostatic equation, Eq. (3.2) or (3.3),
or
dp = -ρ g0 dh
dp
dh
= -ρ g0
dt
dt
The upward speed of the elevator is dh/dt, which is
dp
dp/dt
=
dt
-ρ g0
At sea level, ρ = 1.225 kg/m3. Also, a one-percent change in pressure per minute starting from
sea level is
dp
= -(1.01 ´ 105 )(0.01) = -1.01 ´ 103 N/m 2 per minute
dt
Hence,
dh
-1.01 ´ 103
=
= 84.1meter per minute
dt
(1.225)(9.8)
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
æ 535.89 - 530 ö÷
33,500 + 500 çç
= 33737 ft
çè 535.89 - 523.47 ø÷÷
The density at the altitude at which the airplane is flying is
ρ =
ρ
RT
=
530
= 7.919 ´ 10-4 slug/ft 3
(1716)(390)
From Appendix B:
At 33,000 ft: ρ = 7.9656 ´ 10-4 slug/ft 3
At 33,500 ft: ρ = 7.8165 ´ 10-4 slug/ft 3
Hence, the density altitude is
æ 7.9656 - 7.919 ö÷
33, 000 + 500 çç
= 33,156 ft
çè 7.9656 - 7.8165 ø÷÷
11
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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
ρ
òp
dp = ρ g
a
or,
h
ò0 dh
p − pa = ρ g h
or,
p = ρ g h + pa
Eq. (2) gives the water pressure exerted on the wall at an arbitrary depth h.
(2)
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
F =
ò0
F
dF =
ò0
d
p  dh
(3)
where p is given by Eq. (2). Inserting Eq. (2) into (3),
F =
or,
ò0
d
(ρ g h + pa )  dh
æ d 2 ö÷
ç
F = ρ g  çç
÷÷ + pa  d
çè 2 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),
(10) 2
+ (2116)(30)(10)
2
F = 93, 600 + 634,800 = 728, 400 lb
F = (62.4)(30)
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
F =
728, 400
= 364.2 tons
2000
In the case of a freestanding wall, the net force, that due only to the weight of the water, is
F =
93, 600
= 46.8 tons
2000
12
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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.
3.12 For the exponential atmosphere model,
ρ
= e-g0 h/(RT )
ρ0
ρ
= e-(9.8)(45,000) /(287)(240) = e-6.402
ρ0
Hence,
ρ = ρ0e-6.402 = (1225)(1.6575 ´ 10-3 ) = 2.03 ´ 10-3 kg/m3
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
from App. A, T = 268.02 K, p = 6.9235 × 104 N/m2, and
using linear interpolation:
= 0.90926 kg/m3. At 3.1 km,
= 0.89994 kg/m3. At h = 3.035 km,
 0.035 
T = 268.67 − (268.67 − 268.02) 
= 268.44 K
 0.1 
 0.035 
p = 7.0121 × 10 4 − (7.0121 × 10 4 − 6.9235 × 10 4 ) 
 0.1 
= 6.98099 × 10 4
N
m2
kg
 0.035 
= 0.906
 0.1 
m3
ρ = 0.09026 − (0.90926 − 0.89994) 
13
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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),
T = T1 + a (h − h1 ) = Ts + a(h − 0) = 288.16 − (0.0065)(3035) = 268.43 K
From Eq. (3.12),
p T
=
p1  T1 
− g0 / aR
− g0
−9.8
=
= 5.25328
aR ( −0.0065)(28.7)
So
p T
=
ps  Ts 
5.25328
 268.43 
=
 288.16 
5.25328
= 0.688946
p = 0.688946(1.01325 × 105 ) = 6.9807 × 10 4
N
m2
By comparison the approximate result from the solution of Prob. 3.13 differs from the exact
result above by
6.980990 − 06.9807
= 4.15 × 10−5 = 0.00415%
6.9807
A very small amount.
From Eq. (3.13),
ρ T
=
ρ1  T1 
−[( g0 /aR )+1]
g0
+ 1 = −5.25328 + 1 = −4.25328
aR
ρ T
=
ρs  Ts 
4.2538
 268.43 
=
 288.16 
4.24328
= 0.73958
ρ = 0.73598 ρs = 0.73958 (1.2250) = 0.90599
14
kg
m3
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be
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3.15 We want to calculate hG when
hG − h
h
= 0.01 = 1 −
hG
hG
From Eq. (3.6)
h=
rhG
, where r = 6.356766 × 106 m
r + hG
Thus,
0.01 = 1 −
r
r + hG
r
= 1 − 0.01 = 0.99
r + hG
0.99 (r + hG ) = r
 1 − 0.99 
hG = 
r = 0.0101 (6.356766 × 105 )
 0.99 
hG = 64.21 × 103 m = 64.21 km
3.16 From Eq. (3.1),
 r 
g = g0 
 r + hG 
2
where r = 6.356766 × 106 m, g0 = 9.8 m/sec2, and
 0.3048 m 
hG = (70,000 ft) 
= 21,336 m
 1 ft 
Thus,


6.356766 × 106
m
g = 9.8 
 = 9.767
6
sec2
 6.356766 × 10 + 21,336 
 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.
15
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Chapter 4
4.1
AV
1 1 = A2V2
Let points 1 and 2 denote the inlet and exit conditions, respectively. Then,
æA ö
æ1ö
V2 = V1 ççç 1 ÷÷÷ = (5) çç ÷÷ = 1.25ft/sec
ç
è 4 ø÷
èç A2 ø÷
4.2
From Bernoulli’s equation,
V2
V2
p1 + ρ 1 = p2 + ρ 2
2
2
p2 - p1 =
ρ
2
(V12 - V22 )
In consistent units,
ρ =
62.4
= 1.94 slug/ft 3
32.2
Hence,
p2 - p1 =
1.94
[(5) 2 - (1.25)2 ]
2
p2 - p1 = 0.97(23.4) = 22.7 lb/ft 2
4.3
From Appendix A; at 3000m altitude,
p1 = 7.01 ´ 104 N/m 2
ρ = 0.909 kg/m3
From Bernoulli’s equation,
p2 = p1 +
ρ
2
(V12 - V22 )
p2 = 7.01 ´ 104 +
0.909
[602 - 702 ]
2
p2 = 7.01 ´ 104 - 0.059 ´ 104 = 6.95 ´ 104 N/m 2
16
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student using this manual is using it without permission.
4.4
ρ
ρ
V12 = p2 + V22
2
2
Also from the incompressible continuity equation
p1 +
From Bernoulli’s equation,
V2 = V1( A1/A2 )
ρ
p1 +
Combining,
2
V12 = p2 +
ρ
2
( A1/A2 )2
2( p1 - p2 )
é
ρ ê ( A1 / A2 )2 - 1ùú
ë
û
V1 =
At standard sea level, p = 0.002377 slug/ft3. Hence,
2(80)
V1 =
(.002377)[(4)2 - 1]
= 67 ft/sec
æ 60 ö
Note that also V1 = 67 çç ÷÷÷ = 46 mi/h. (This is approximately the landing speed of World
çè 80 ø
War I vintage aircraft).
4.5
p1 +
1
1
ρ V 2 = p3 + ρ V 2
2
2
V12 =
2( p3 - p1)
ρ
+ V32
A1 V1 = A3 V3 , or V3 =
(1)
A1
V1
A3
(2)
Substitute (2) into (1)
V12 =
2( p3 - p1)
ρ
2( p3 - p1)
é
æ A ö÷2 ùú
ê
ρ ê 1 - ççç 1 ÷÷ ú
ê
èç A3 ÷ø úú
êë
û
or,
V1 =
Also,
A1 V1 = A2 V2
or,
æ A ö2
+ ççç 1 ÷÷÷ V12
çè A3 ÷ø
(3)
æA ö
V2 = ççç 1 ÷÷÷ V1
çè A2 ÷ø
(4)
Substitute (3) into (4)
V1 =
A1
A2
2( p3 - p1)
é
æ A ö÷2 ùú
ê
ρ ê 1 - ççç 1 ÷÷ ú
çè A3 ÷ø ú
ê
êë
úû
3 2(1.00 - 1.02) ´ 105
V2 = 102.22 m/sec
é
1.5
æ 3 ö÷2 ùú
ê
(1.225) ê 1 - çç ÷÷ ú
çè 2 ø ú
êë
û
Note: It takes a pressure difference of only 0.02 atm to produce such a high velocity.
V2 =
17
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student using this manual is using it without permission.
4.6
æ 88 ö
V1 = 130 mph = 130 çç ÷÷÷ = 190.7 ft/sec
çè 60 ø
p1 +
1
1
ρ V12 = p2 + ρ V22
2
2
2
V22 = ( p1 - p2 ) + V12
ρ
2(1760.9 - 1750.0)
V22 =
+ (190.7)2
0.0020482
V2 = 216.8ft/sec
4.7
From Bernoulli’s equation,
ρ
(V22 - V12 )
2
And from the incompressible continuity equation,
p1 - p2 =
V2 = V1 ( A1/A2 )
ρ
V12 [( A1/A2 )2 - 1]
2
Hence, the maximum pressure difference will occur when simultaneously:
1. V1 is maximum
p1 - p2 =
Combining:
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,
1.225
(90)2[(1.3)2 - 1] = 3423 N/m 2
2
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.
p1 - p2 =
4.8
The isentropic relations are
γ
æ ρ ÷öγ
æ T öγ -1
pc
= ççç e ÷÷ = ççç c ÷÷÷
çè ρ0 ÷ø
p0
èç T0 ÷ø
γ
Hence,
14-1
1.4
æ ρ öγ -1
æ 1 ö
= (300) çç ÷÷
Te = T0 ççç e ÷÷÷
çè 10 ÷ø
çè ρ0 ÷ø
= 155 K
From the equation of state:
ρ0 =
p0
(10)(1.01 ´ 105 )
=
= 11.73kg/m3
(287)(300)
RT0
1
Thus,
1
æ p öγ
æ 1 ö1.4
ρe = ρ0 ççç e ÷÷÷ = 11.73çç ÷÷÷ = 2.26 kg/m3
çè 10 ø
èç p0 ÷ø
As a check on the results, apply the equation of state at the exit.
pe = ρe RTe ?
1.01 ´ 105 = (2.26)(287)(155)
1.01 ´ 105 = 1.01 ´ 105
It checks!
18
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student using this manual is using it without permission.
4.9
Since the velocity is essentially zero in the reservoir, the energy equation written between the
reservoir and the exit is
V2
h0 = he + e or, Ve2 = 2(h0 - he )
2
However, h = cpT. Thus Eq. (1) becomes
(1)
Ve2 = 2c p (T0 - Te )
æ
Ve2 = 2c p T0 ççç1 èç
Te ÷ö
÷÷
T0 ÷ø
(2)
However, the flow is isentropic, hence
γ -1
æp ö γ
Tc
= ççç c ÷÷÷
çè p0 ÷ø
T0
(3)
Substitute (3) into (1).
Ve =
γ -1 ù
é
ê
æ
ö
γ úú
p
÷
2 c p T0 êê 1 - ççç e ÷÷
ú
çè p0 ÷ø
ê
ú
êë
úû
(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,
V2
V2
c p T1 + 1 = c p T2 - 2
2
2
At a standard altitude of 5 km, from Appendix A,
(1)
p1 = 5.4 ´ 104 N/m 2
T1 = 255.7 K
Also, for air, cp = 1005 joule/(kg)(K). Hence, from Eq. (1) above,
V 2 - V22
T2 = T1 + 1
2 cp
T2 = 255.7 +
(270)2 - (330)2
2(1005)
T2 = 255.7 - 17.9 = 237.8 K
Since the flow is also isentropic,
γ -1
æT ö γ
p2
= ççç 2 ÷÷÷
çè T1 ø÷
p1
γ -1
Thus,
æT ö γ
p2 = p1 ççç 2 ÷÷÷
çè T ÷ø
1
1.4
æ 237.8 ö÷1.4-1
= 5.4 ´ 104 çç
çè 255.7 ø÷÷
p2 = 4.19 ´ 104 N/m 2
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.
19
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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.
4.11 From the energy equation
or,
V2
c pT0 = c pTe + e
2
Ve2
Te = T0 2 cp
Te = 1000 -
15002
= 812.5 R
2(6000)
In the reservoir, the density is
p0
(7)(2116)
=
= 0.0086 slug/ft 3
(1716)(1000)
RT0
ρ0 =
From the isentropic relation,
1
æ T öγ -1
ρc
= ççç e ÷÷÷
çè T0 ø÷
ρ0
1
æ 812.5 ö÷1.4-1
ρe = 0.0086 çç
= 0.0051 slug/ft 3
çè 1000 ÷÷ø
From the continuity equation,
m = ρe AeVe
Thus,
Ae =
m
ρeVe
In consistent units,
m =
1.5
= 0.047 slug/sec.
32.2
Hence,
Ae =
m
0.047
=
= 0.061ft 2
(0.0051)(1500)
ρcVe
æ 88 ö
4.12 V1 = 1500 mph = 1500 çç ÷÷÷ = 2200 ft/sec
çè 60 ø
V2
V2
C pT1 + 1 C pT2 2
2
2
V22 = 2C p (T1 - T2 ) + V12
V22 = 2(6000)(389.99 - 793.32) + (2200) 2
V2 = 6.3ft/sec
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.
20
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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.
4.13 At the inlet, the mass flow of air is
m air = ρ AV = (3.6391 ´ 10-4 )(20)(2200) = 16.0/slug/sec
m fuel = (0.05)(16.01) = 0.8slug/sec
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,
γ RTe =
ae =
(1.4)(1716)(812.5)
= 1397 ft/sec
V
1500
Mc = e =
= 1.07
ae
1397
Thus,
Note that the nozzle of Problem 4.11 is just barely supersonic.
4.15 From Appendix A,
T∞ = 216.66 K
Hence,
a∞ =
=
Thus,
γRT
(1.4)(287)(216.66) = 295m/sec
V
250
M∞ = ∞ =
= 0.847
a∞
295
4.16 At standard sea level, T∞ = 518.69 R
a∞ =
γ RT =
(1.4)(1716)(518.69) = 1116 ft/sec
V∞ = M ∞ a∞ = (3)(1116) = 3348ft/sec
Since 60 mi/hr - 88 ft/sec., then
V∞ = 3348(60/88) = 2283mi/h
4.17 V = 2200 ft/sec
a =
γRT =
M =
V
2200
=
= 2.27
a
967.94
(1.4)(1716)(389.99) = 967.94 ft/sec
21
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student using this manual is using it without permission.
4.18 The test section density is
ρ =
p
1.01 ´ 105
=
= 1.173kg/m3
(287)(300)
RT
Since the flow is low speed, consider it to be incompressible, i.e., with the above density
throughout.
ρ
p1 - p2 =
2
V22[1 - ( A2 /A1) 2 ]
(1)
In terms of the manometer reading,
p1 - p2 = ωΔh
(2)
where ω = 1.33 ´ 105 N/m3 for mercury.
Thus, combining Eqs. (1) and (2),
Δh =
ρ 2
V2 [1 - ( A2 /A1) 2 ]
2ω
1.173
=
5
(80)2[1 - (1/20)2 ]
(2)(1.33 ´ 10 )
Δh = 0.028m = 2.8 cm
æ 88 ö
4.19 V2 = 200 mph = 300 çç ÷÷÷ = 293.3 ft/sec
çè 60 ø
(a)
p1 +
1
1
ρ V12 = p2 + ρ V22
2
2
A1 V1 = A2V2
V1 =
A2
V2
A1
2
1 æç A2 ÷ö 2
1
2
÷
p1 + ρ çç
÷÷ V2 = p2 + ρ V2
ç
2 è A1 ø
2
é
æ A ÷ö2 ùú 2
1 ê
p1 - p2 = ρ ê 1 - ççç 2 ÷÷ ú V2
çè A1 ÷ø ú
2 ê
ëê
ûú
p1 - p2 =
0.002377
2
2ù
é
ê 1 - æç 4 ÷ö ú (293.3)2
çç ÷÷ ú
ê
è 20 ø ú
ëê
û
p1 - p2 = 98.15 lb/ft 2
22
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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.
(b)
p1 +
1
1
ρ V12 = p3 + ρ V33
2
2
AV
1 1 = A2V2 : V1 =
A2
V2
A1
A2V2 = A3V3 : V3 =
A2
V2
A3
p1 - p3 =
2
é
æ A ö2 ùú
1 ê çæ A2 ÷ö
ρ ê çç ÷÷ - ççç 2 ÷÷÷ ú V22
çè A1 ÷ø ú
2 ê çè A3 ÷ø
úû
ëê
p1 - p3 =
2
é
æ 4 ö÷2 ùú
0.002377 ê æç 4 ÷ö
çç ÷ ú (293.3)2
÷
ê çç ÷
ç 20 ø÷ ú
è
ø
è
2
18
ëê
û
p1 - p3 = 0.959 lb/ft 2
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
ρ =
p
2116
=
= 0.00233slug/ft 3
(1716)(70 + 460)
RT
The flow velocity is low enough so that incompressible flow can be assumed. Hence, from
Bernoulli’s equation,
1
ρV 2
2
1
p0 = 2116 + (0.00233)[150(88 / 60)]2
2
(Remember that 88 ft/sec = 60 mi/h.)
p0 = p +
1
(0.00233)(220)2
2
p0 = 2172 lb/ft 2
p0 = 2116 +
4.21 The altimeter measures pressure altitude. Thus, from Appendix B, p = 1572 lb/ft2. The air
density is then
ρ =
p
1572
=
= 0.00183slug/ft 3
(1716)(500)
RT
Hence, from Bernoulli’s equation,
Vtrue =
2( p0 - p)
ρ
=
2(1650 - 1572)
0.00183
Vtrue = 292 ft/sec
The equivalent airspeed is
Ve =
2( p0 - p)
ρs
=
2(1650 - 1572)
0.002377
Ve = 256 ft/sec
23
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student using this manual is using it without permission.
4.22 The altimeter measures pressure altitude. Thus, from Appendix A,
p = 7.95 × l04 N/m2.
Hence,
ρ =
p
7.95 ´ 104
=
= 0.989 kg/m3
(287)(280)
RT
The relation between Vtrue and Ve is
ρs / ρ
Vtrue /Ve =
Hence,
Vtrue = 50 (1.225) / 0.989 = 56 m/sec
4.23 In the test section,
a =
γ RT =
(1.4)(287)(270) = 329 m/sec
M = V/a = 250/329 = 0.760
γ
æ
p0
γ - 1 2 ÷öγ -1
= çç1 +
= [1 + 0.2(0.760)2 ]3.5 = 1.47
M ÷÷
çè
ø
p
2
Hence, p0 = 1.47 p = 1.47(1.01 ´ 105 ) = 1.48 ´ 105 N/m 2
4.24
p = 1.94 ´ 104 N/m 2 from Appendix A.
γ -1
é
ù
0.286
é
ù
êæ p ö γ
ú
ú
2
2 ê çæ 2.96 ´ 104 ÷÷ö
2
÷
0
ê
ú
ç
÷
ê çç
- 1ú =
- 1ú
M1 =
çç
÷÷
÷
ê
4
ú
1.4 - 1 ê çè 1.94 ´ 10 ÷ø
γ - 1 ê è p ø÷
ú
êë
úû
ëê
ûú
M12 = 0.642
M1 = 0.801
γ
4.25
æ
p0
γ - 1 2 ö÷γ -1
= çç1 +
M ÷÷
ç
è
ø
p
2
p0
= [1 + 0.2(0.65)2 ]3.5 = 1.328
p
p =
p0
2339
=
= 1761 lb/ft 2
1.328
1.328
From Appendix B, this pressure corresponds to a pressure altitude, hence altimeter reading
of 5000 ft.
24
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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.
4.26 At standard sea level,
T = 518.69 R
T0
γ -1 2
=1+
M = 1 + 0.2(0.96) 2 = 1.184
T
2
T0 = 1.184T = 1.184 (518.69)
T0 = 614.3R = 154.3F
4.27
γ RT1 =
a1 =
(1.4)(287)(220) = 297 m/sec
M1 = V1 / a1 = 596 /197 = 2.0
The flow is supersonic. Hence, the Rayleigh Pitot tube formula must be used.
p02
p1
γ
é (γ + 1)2 M 2 ù γ -1 é 1 - γ + 2γ M 2 ù
1
1 ú
ú
ê
= êê
ú
ê
ú
2
+
γ
1
êë 4γ M1 - 2(γ - 1) úû
êë
úû
p1
é
ù 3.5 é 1 - 1.4 + 2(1.4)(2)2 ù
(2.4)2 (2) 2
ê
ú ê
ú
= ê
ú ê
ú
2
2.4
êë 4(1.4)(2) - 2(0.4) úû êë
úû
p02
= 5.64
p02
p1
p1 = 2.65 ´ 104 N/m 2 from Appendix A.
Hence,
p02 = 5.64(2.65 ´ 104 ) = 1.49 ´ 105 N/m 2
4.28
q =
1
1 æ γ p ö÷
γ æ p ö÷ 2
γ V2
÷÷ ρ V 2 = p çç
÷÷V = p
ρ V 2 = çç
2
2 çè γ p ø÷
2 èç γ p ø÷
2 a2
Hence:
q =
4.29
q∞ =
γ
2
γ
2
pM 2
p∞ M ∞ 2 = 0.7 p∞ M ∞ 2
(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
2
p∞(N/m )
25.6
M
17
2
q∞(N/m )
5.2 × 10
50
40
87.9
299.8
9.5
3
5.6 × 10
30
1.19 × 10
5.5
3
6.3 × 10
20
3
3
7.5 × 10
3
5.53 × 103
1
3
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.
25
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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.
γ /(γ -1)
æ
p0
γ - 1 2 ÷ö
= çç1 +
= [1 + 0.2(2)2 ]1.4/0.4 = 7.824
M ∞ ÷÷
çè
ø
p∞
2
Hence:
p0 = 7.824 p∞ = 7.824(2116) = 1.656 ´ 104
lb
ft 2
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 nonisentropic. 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).
p02
p1
γ
é (γ + 1) 2 M 2
ù γ -1 é 1 - γ + 2γ M 2 ù
∞
1 ú =
ú
ê
= êê
ú
ê
ú
2
γ
1
+
êë 4γ M 2 - 2(γ - 1) úû
êë
úû
é
ù1.4/0.4 é 1 - 1.4 + 2(1.4)(2) 2 ù
(2.4)2 (2) 2
ê
ú
ê
ú = 5.639
ê
ú
ê
ú
2
2.4
4(1.4)(2)
2(0.4)
ëê
ûú
ëê
ûú
Hence,
p02 = 5.639 p∞ = 5.639(2116) = 1.193 ´ 104
lb
ft 2
If Bernoulli’s equation is used, the following wrong result for total pressure is obtained.
1
γ
ρ V∞ 2 = p∞ + p∞ M ∞ 2
2
2
2
4 lb
p0 = 2116 + 0.7(2116)(2) = 0.804 ´ 10 2
ft
p0 = p∞ + q∞ = p∞ +
Compared to the correct result of 1.656 ´ 104
4.31
lb
ft 2
, this leads to an error 51%.
æ
pe
γ - 1 2 ö÷ -γ
M e ÷÷
= çç1 +
èç
øγ - 1
p0
2
pe = 5(1.01 ´ 105 )[1 + 0.2(3) 2 ]-3.5
pe = 1.37 ´ 104 N/m 2
-1
æ
Te
γ - 1 2 ö÷
M e ÷÷ = [1 + 0.2(3)2 ]-1
= çç1 +
çè
ø
T0
2
Te = (500)(0.357) = 178.6 K
ρe =
pe
1.37 ´ 104
=
= 0.267 kg/m3
RTc
(287)(178.6)
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-γ
4.32
æ
pe
γ - 1 2 ö÷γ -1
M e ÷÷
= çç1 +
ç
è
ø
p0
2
Hence,
1-γ
é
ù
êæ p ö γ
ú
2
÷
ê çç e ÷
M e2 =
- 1úú
÷÷
ê çç
γ - 1 ê è p0 ø
ú
êë
úû
M e2 = 5[(0.2)-0.286 - 1] = 2.92
M e = 1.71
æ A ö÷2
ççç e ÷÷ =
èç A ø÷
t
(γ +1)/(γ -1)
1 éê 2 æç
γ - 1 ö÷ 2 ùú
÷
1
M
+
ç
ú
γ ø÷÷
M e 2 ëêê γ + 1 èç
ûú
æ A ö2
1
[(0.833)(1 + 0.2 (1.71)2 )]6
ççç e ÷÷÷ =
2
çè At ÷ø
(1.71)
Ae
= 1.35
At
4.33
-γ /(γ -1)
é
q
γ p 2
γ
γ -1 2ù
M = M 2 ê1 +
M ú
=
= 0.7M 2 (1 + 0.2M 2 )-3.5
êë
úû
p0
2 p0
2
2
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.
27
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4.34 First, calculate the value of the Reynolds number.
Re L =
ρ∞V∞ L
(1, 225)(200)(3)
=
= 4.10 ´ 107
μ∞
(1.7894 ´ 10-5 )
The dynamic pressure is
q∞ =
1
1
ρ∞V∞ 2 = (1.225)(200)2 = 2.45 ´ 104 N/m 2
2
2
δL =
5.2L
=
Re L
Cf =
1.328
=
Re L
Hence,
5.2(3)
= 0.0024 m = 0.24 cm
4.1 ´ 107
and
1.328
4.1 ´ 107
= 0.00021
The skin friction drag on one side of the plate is:
D f = q∞ Sc f = (2.45 ´ 104 )(3)(17.5)(0.00021)
D f = 270 N
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 δ =
0.37 L
(Re L )
0.2
=
0.37(3)
(4.1 ´ 107 )0.2
= 0.033 m = 3.3 cm
From Problem 4.24, we find
δ turbulent /δ laminar =
3.3
= 13.75
0.24
The turbulent boundary layer is more than an order of magnitude thicker than the laminar
boundary layer.
Cf =
0.074
(Re L )
0.2
=
0.074
(4.1 ´ 107 )0.2
= 0.0022
The skin friction drag on one side is then
D f = q∞ Sc f = (2/45 ´ 104 )(3)(17.5)(0.0022)
D f = 2830 N
The total, accounting for both top and bottom is
Total Df = 5660 N
From Problem 4.24, we find
( D f turbulent ) /( D flaminar ) =
5660
= 10.5
540
The turbulent skin friction drag is an order of magnitude larger than the laminar value.
28
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4.36
Rex
cr
=
ρ∞V∞ xcr
μ∞
æ μ ö
(106 )(1.789 ´ 10-5 )
xcr = Re xcr ççç ∞ ÷÷÷ =
çè ρ∞V∞ ø÷
(1.225)(200)
xcr = 7.3 ´ 10-2 m
The turbulent drag that would exist over the first 7.3 × 10−2 m of chord length from the leading
edge (area A) is
D fA =
D fA =
0.074
(Recτ )0.2
0.074
(106 )0.2
q∞ S A (on one side)
(2.45 ´ 104 )(7.3 ´ 10-2 )(17.5)
D f A = 146 N
(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:
D f B = 2830 - 146 - 2684 N (turbulent)
The laminar drag on area A is
D fA =
DfA
D fA
1.328
q∞ S
(Recr )0.5
1.328
(2.45 ´ 104 )(7.3 ´ 10-2 )(17.5)
=
(106 )0.5
= 42 N (laminar)
Hence, the skin friction drag on one side, assuming area A to be laminar and area B to be
turbulent is
D f = D f A (laminar) + D f B (turbulent)
D f = 42 + 2684 = 2726 N
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.
29
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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)
dp = -ρVdV
Letting s denote distance along the streamline through the point, Eq. (4.8) can be written as
dp
dV
= -ρ V
ds
ds
dp
(dV/V )
= -ρ V 2
ds
ds
or,
(a)
(dV/V )
= 0.02 per millimeter
ds
Hence,
dp
N
= -(1.1)(100)2 (0.02) = 220 2 per millimeter
ds
m
(b)
dp
N
= -(1.1)(1000) 2 (0.02) = 22, 000 2 per millimeter
ds
m
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.
γ
æ
öγ -1
p0
γ -1
= ççç1 +
= [1 + 0.2 (0.7)2 ]3.5 = 1.387
M ∞ 2 ÷÷÷
p∞
γ
è
ø÷
From Eq. (4.74) applied at the point on the wing,
γ
æ
p0
γ - 1 2 ÷öγ -1
= çç1 +
= [1 + 0.2 (1.1)2 ]3.5 = 2.135
M ÷÷
çè
ø
p
2
Hence,
é æ p ö æ p öù
æ 1.387 ö÷
p == 0.65 p∞
p = êê ççç 0 ÷÷÷ / ççç 0 ÷÷÷ úú p∞ = çç
çè 2.135 ÷÷ø ∞
÷
÷
ç
ëê è p∞ ø è p ø úû
At a standard altitude of 3 km, from Appendix A, p∞ = 7.0121 ´ 104 N/m 2 .
Hence,
p = (0.65)(7.0121 ´ 104 ) = 4.555 ´ 104 N/m 2
4.39 This problem is simply asking what is the equivalent airspeed, as discussed in Section 4.12.
Hence,
1/2
1/2
-3 ö
æ
÷÷ö = (800) çç 1.0663 ´ 10 ÷÷ = 535.8ft.sec
ç
-3 ÷÷
÷÷
èç 2.3769 ´ 10 ø÷
sø
æ ρ
Ve = V ççç
èç ρ
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4.40 (a) From Eq. (4.88)
æ A ÷ö2
çç e ÷ =
çèç A ÷÷ø
t
=
γ +1
1 éê 2 æç
γ - 1 2 ö÷ ùú γ -1
+
1
M e ÷÷
ç
ø úû
2
M e 2 êë γ + 1 èç
6
1 ì
ï 2 é
ï
2 ùü
ï
ï
+
= 2.87 ´ 105
1
0.2
(10)
êë
úû ý
2í
ï
ï
2.4
(10) î
ï
ï
þ
Hence,
Ae
=
At
(b)
2.87 ´ 105 = 535.9
Form Eq. (4.87)
γ
æ
p0
γ - 1 2 ÷öγ -1
= çç1 +
= [1 + 0.2 (10)2 ]3.5 = 4.244 ´ 104
M e ÷÷
çè
ø
pe
2
At a standard altitude of 55 km, p = 48.373 N/m2. Hence,
p0 = (4.244 ´ 104 )(48.373) = 2.053 ´ 106 N/m 2 = 20.3atm
(c)
From Eq. (4.85)
T0
γ -1 2
=1+
M e = 1 + 0.2 (10)2 = 21
Te
2
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
ae =
γ RTe =
(1.4)(287)(275.78) = 332.9 m/sec.
Hence,
Ve = M eae = 10(332.9) = 3329 m/sec
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4.42 (a) From Eq. 4.88, for Me = 20
æ A ÷ö2
çç e ÷ =
çèç A ÷÷ø
t
=
γ +1
1 éê 2 æç
γ - 1 2 ö÷ ùú γ -1
+
1
M e ÷÷
ç
ø úû
2
M e 2 êë γ + 1 èç
6
ì
ï 2
ï
2 ü
8
ï
ï
+
[1
0.2
(20)
]
ý = 2.365 ´ 10
2í
ï
ï
2.4
(20) î
ï
ï
þ
1
Hence,
Ae
= 15,377
At
(b)
From Eq. (4.85)
-1
æ
T0
γ - 1 2 ÷ö
= çç1 +
M e ÷÷ = [1 + (0.2)(20) 2 ]-1 = 0.01235
çè
ø
Te
2
Hence,
Te = (5791)(0.01235) = 71.5 K
ae =
γ RTe =
(1.4)(287)(71.5) = 169.5 m/sec
Ve = M eae = 20(169.5) = 3390 m/sec
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.
32
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4.43 We will use the sketch shown in Figure 4.47, identifying the separate plan form areas A and B.
(a)
q∞ = 1 2 ρ∞V∞ 2 = 1 2 (1.23)(20) 2 = 246 N/m 2
ρ V x
Re xcr = ∞ ∞ cr = 5 ´ 105
μ∞
Hence:
xcr =
μ∞ (5 ´ 105 )
(1.789 ´ 10-5 )(5 ´ 105 )
=
= 0.3636 m
(1.23)(20)
ρ∞V∞
Assuming turbulent flow over areas A + B,
C f A+ B =
0.074
Re L 0.2
Where
Re L =
ρ∞V∞ L
(1.23)(20)(4)
=
= 5.5 ´ 106
-5
μ∞
(1.788 ´ 10 )
0.074
C f A+ B =
D f A+ B
= 0.0032
(5.5 ´ 106 )0.2
= q∞ SC f = (246)(4)(4)(0.00332) = 13.07 N
The turbulent drag on area A is obtained from
C fA =
0.074
(Re xcr )
0.2
=
0.074
(5 ´ 105 )0.2
= 0.00536
D f A = q∞ S C f A = (246)(4)(0.3636)(0.00536) = 1.92 N
The turbulent drag on area B is
D f B = D f A + B - D f A = 13.07 - 1.92 = 11.15 N
The laminar drag on area A is obtained from
C fA =
1.328
(Re xcr )
0.5
=
1.328
(5 ´ 105 )0.5
= 0.00188
(Laminar) D f A = q∞ S C f A = (246)(4)(0.3636)(0.00188) = 0.67 N
The total friction drag is
D f = D f A + D f B = 0.67 + 11.15 = 11.82 N
33
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(b)
V∞ = 40 m/sec
q∞ =
1
Re L =
2 ρ∞ V∞
2
ρ∞ V ∞ L
μ∞
(Turbulent)
2
= 984 N/m 2
(1.23)(40)(4)
= 11 ´ 106
1
=
=
2 (1.23)(40)
(1.789 ´ 10-5 )
C f A+ B =
0.074
Re L
0.2
=
0.074
= 0.00289
(11 ´ 106 )0.2
D f A + B = q∞ S C f A + B = (984)(4)(4)(0.00289) = 45.5 N
(Turbulent)
C fA =
0.074
(Re xcr )
0.2
=
0.074
(5 ´ 105 )0.2
= 0.00536
Note that, since Re xcr remains the same, but V∞ is doubled, xcr is half that from
part (a)
0.3636
= 0.1818 m
2
xcr =
D f A = q∞ S C f A = (984)(4)(0.1818)(0.00536) = 3.84 N
Hence,
D f A = D f A + B - D f A = 45.5 - 3.84 = 41.66 N
(Laminar) C f A =
1.328
(Re xcr )
0.5
=
1.328
(5 ´ 105 )0.5
= 0.00188
D f A = q∞ S C f A = (984)(4)(0.1818)(0.00188) = 1.345 N
Thus,
D f = D f A + D f B = 1.345 + 41.66 = 43 N
(c)
n
D f ∝ V¥
Using the results from parts (a) and (b)
D fa
D fb
æ V ö÷n
= ççç a ÷÷
çè V ø÷
b
æ 20 ön
11.82
= çç ÷÷÷
çè 40 ø
43
0.275 = (0.5)n
Taking the log of both sides
−0.56 = n (−0.30)
0.56
n=
= 1.87
0.30
Note: Skin friction drag does not follow the velocity squared law; rather skin friction varies with
velocity at a power slightly less than 2.
34
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4.44 (a) For turblent flow, C f ∝
1
Re
0.2
. Since, Re =
ρ∞V∞ L
1
, then C f ∝ 0.2
μ
V¥
æ 1 ÷ö
ç
D f = q∞ S C f ∝ V∞ 2 çç 0.2 ÷÷÷
çè V∞ ÷ø
or,
Df ∝ V∞1.8
(b)
For laminar flow, C f ∝
1
Re
0.5
∝
1
V∞ 0.5
æ
ö
2 çç 1 ÷÷
D f ∝ V¥
ç 0.2 ÷÷
çè V¥ ÷ø
or,
Df ∝ V∞1.5
4.45 (a) M∞ = 1. Since a∞ = 340.3 m/sec at sea level,
V∞ = M∞ a∞ = (1)(340.3)=340.3 m/sec
q¥ =
1
Re L =
C finc =
2
2 ρ V∞
=
1
2 (1.23)(340.3)
2
= 7.12 ´ 104 N/m 2
ρ∞V∞ L
(1.23)(340.3)(4)
=
= 9.36 ´ 107
-5
μ∞
(1.789 ´ 10 )
0.074
Re L
0.2
=
0.074
(9.36 ´ 107 )0.2
= 0.00188
From Figure 4.44, for the turbulent case, approximate reading of the graph shows,
for M∞ = 1,
Cf
C finc
= 0.91
Thus,
Cf = 0.91(0.00188) = 0.00171
Df = q∞ S Cf (7.12 × 104)(4)(4)(0.00171)
D f = 1948 N
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(b)
M∞ = 3
V∞ = M ∞ a∞ = 3(340.3) = 1021m/sec
q∞ =
1
2 ρ∞V∞
2
=
1
2 (1.23)(1021)
2
= 6.41 ´ 105 N/m 2
ρ∞V∞ L
(1.23)(1021)(4)
=
= 2.81 ´ 108
μ∞
(1.789 ´ 10-5 )
Re L =
C finc =
0.074
(2.81 ´ 108 )0.2
= 0.00151
From Figure 4.44, for the turbulent case, approximate reading of the graph shows, for
M∞ = 3,
Cf
C finc
= 0.56
C f = 0.56(0.00151) = 0.000846
D f = q∞ S C f = (6.41 ´ 105 )(4)(4)(0.000846)
D f = 8677 N
(c)
D f1
D f2
æ V ön
= ççç 1 ÷÷÷
çè V2 ø÷
æ 340.3 ö÷n
1948
= çç
çè 1020.9 ø÷÷
8677
0 .2245 = (0.333)n
−0.6488 = n (−0.4772)
-0.6488
n=
= 1.36
-0.4772
1.36
This implies D f ∝ 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 V∞n where n is an exponent less than 2.
4.46 (a) The waves travel at the local speed of sound.
a =
γ RT =
(1.4)(287)(288) = 340.2 m/sec.
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.
36
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(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 rightrunning 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.
37
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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
ρ1 = 1.1117 kg/m3
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)
1
1
æ 7.9501 ´ 104 ÷ö1.4
æ p öγ
÷÷
ρ2 = ρ1 ççç 2 ÷÷÷ = 1.1117 ççç
çè 8.9876 ´ 104 ÷ø÷
èç p1 ø÷
ρ2 = 1.0184 kg/m3
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
A1V1 = A2V2
V1 =
A2
0.5
V =
(120) = 15 mph
A1 2
4
From the answer to Problem 2.15, 60 mph = 26.82 m/sec.
 26.82 
= (15) = 6.705 m/sec
V1 = 
 60 
4.50
p1 + 1 2 ρ V12 = p2 + 1 2 ρ V22
(
p2 = p1 + 1 2 ρ V12 − V22
)
p1 = 1 atm = 1.01 × 105 N/m 2
ρ = 1.23 kg/m3
V1 = 6.705 m/sec
 26.82 
= 53.64 m/sec
V2 = (120) 
 60 
p2 = 1.01 × 105 + 1 2 (1.23)[(6.705)2 − (53.64)2 ]
p2 = 0.99258 × 105 N/m 2
38
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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
A = (2) [hD + 2 x tan θ ]
Thus,
dA
= 4 tan θ = 4 tan15° = 1.0718 m
dx
4.52 The flow velocity at the entrance to the diffuser is the same as in the test section
V2 = 120 mph = 53.64 m/sec
At the exit of the diffuser, since
V2 A2 = V3 A3
A 
 0.5 
(53.64) = 7.66 m/sec
V3 =  2  V2 = 
 3.5 
 A3 
4.53 AV = const
dv
dA
+V
=0
dx
dx
dv
V  dA 
=−  
dx
A  dx 
A
From the solution of Problem 4.51,
dA
= 1.0718 m
dx
Thus,
dV
V
= −1.0718  
 A
dx
(a)
At the entrance, V2 = 53.64 m/sec (from Prob. 4.52)
A2 = (0.5)(2) = 1 m 2
dV
 53.64 
= −1.0718 
= −57.49 sec −1
 1 
dx
(b)
At the exit, V3 = 7.66 m/sec (from Prob. 4.52)
V 
dV
 7.66 
= −1.0718  3  = −1.0718 
= −1.173 sec −1
 7 
dx
 A3 
39
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4.54 From the Euler equation, Eq. (4.8) in the text,
dp = − ρVdV
dp
dV
= − ρV
dx
dx
(a)
At the inlet, from the solutions of Problems 4.52 and 4.53,
V2 = 53.64 m/sec and
dV
= −57.49 sec −1
dx
Thus,
dp
= −(1.23)(53.64)( −57.49) = 3.793 × 103 N/m3
dx
(b)
At the exit,
V3 = 7.66 m/sec and
dV
= −1.173 sec −1
dx
Thus,
dp
= −(1.23)(7.66)( −1.173) = 11.05 N/m3
dx
4.55
tan 15° =
(3.5 − 0.5) / 2 1.5
=
x
x
x=
1.5
1.5
=
= 5.6 m
tan15° 0.2679
4.56 From the solution of Problem 4.54,
 dp 
3
3
  = 3.793 × 10 N/m
dx 2
 dp 
3
  = 11.05 N/m
dx 3
3.793 × 103 + 11.05
N
 dp 
=
= 1902
 
dx ave
2
m2
xs =
183
183
=
= 0.096 m
1902
 dp 
 
dx ave
4.57 From the solution of Problem 4.56, (xs)laminar = 0.096 m.
−1
( xs )turbulent
=
( xs )laminar
 dp 
506  
 dx  ave
−1
 dp 
183  
 dx  ave
= 2.765
( xs )turbulent = (2.765)(0.096) = 0.265 m
40
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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.
a∞ = γ RT∞ = (1.4)(1716)(491.95 = 1087 ft/sec
 88 
V∞ = 229 mph   = 335.9 ft/sec
 60 
M∞ =
V∞ 335.9
=
= 0.309
a∞ 1087
p0 = ρ∞ + 1 2 ρ∞V∞2 = 1602.3 + (0.5)(1.8975 × 10−3 )(335.9)2
= 1709 lb/ft 2
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.
a∞ = γ RT∞ = (1.4)(1716)(429.64) = 1016 ft/sec
 88 
V∞ = 610   = 894.67 ft/sec
 60 
M∞ =
V∞ 894.67
=
= 0.88
a∞
1016
The flow is subsonic, but compressible. From Eq. (4.74),
γ
p0  γ − 1 2  γ −1
M∞ 
= 1 +

p∞ 
2
p0
= [1 + 0.2(0.88)2 ]3.5 = (1.15488)3.5 = 2.303
p∞
ρ0 = 2.303 p∞ = 2.303 (786.33) = 1811
41
lb
ft 2
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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.
a∞ = γ RT∞ = (1.4)(1716)(394.08) = 973 ft/sec
 88 
V∞ = 1328   = 1947.7 ft/sec
 60 
V
1947.7
M∞ = ∞ =
= 2.00
a∞
973
The flow is supersonic. From Eq. (4.79),
γ
 (γ + 1)2 M ∞2  γ −1 1 − γ + 2γ M ∞2
=

p∞  4γ M ∞2 − 2(γ − 1) 
γ +1
p02
1.4

 0.4 1 − 1.4 + 2(1.4)(2) 2 
(2.4) 2 (2)2
=


 = 5.64
2
2.4
 4(1.4) (2) − 2(0.4) 


p02 = 5.64(499.34) = 2817 lb/ft 2
42
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Chapter 5
5.1
Assume the moment is governed by
M = f (V∞ , ρ∞ , S , μ∞ , a∞ )
More specifically:
M = Z (V∞ a , ρ∞b , S d , a∞e , μ∞ f )
Equating the dimensions of mass, m, length, ℓ, and time t, and considering Z dimensionless,
æ  ö÷a æç m ö÷b 2
çç ÷ ç ÷ 
=
çè t ø÷ çè 3 ø÷
t2
m2
d æ  öe æ
( )
çç
èç t
÷÷ çç m ö÷÷
ø÷ çè t ÷ø
f
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,
and,
b=1−f
a=2−e−f
and,
2 = 2 − e − f − 3 + 3f + 2d + e − f
or
0 = −3 + f + 2d
d =
Hence,
3- f
2
M = Z V∞ 2-e- f ρ∞1- f S (3- f ) / 2 a∞ e μ∞ f
2
= Z V∞ ρ∞ SS
ö÷ f
ö÷e æç
μ
∞
÷
÷÷ çç
1/2 ÷÷÷
÷
ç
ø
V
ρ
S
è
ø
∞
∞ ∞
æ
çç
èç V
1/2 ç a∞
Note that S l/2 is a characteristic length; denote it by the chord, c.
æ a öc æ μ
öf
2
∞ ÷÷
M = ρ∞ V¥
S c Z ççç ∞ ÷÷÷ ççç
÷
èç V∞ ø÷ èç V∞ ρ∞c ø÷
However, a∞ /V∞ = 1/M ∞
and
μ∞
1
=
Re
V∞ ρ∞c
Let
æ 1 ö÷e æ 1 ö f
÷÷ = cm
÷÷ çç
Z ççç
ç
÷
2
èç M ∞ ø è Re ÷ø
where cm is the moment coefficient. Then, as was to be derived, we have
M =
or,
1
ρ∞ V∞ 2c cm
2
M = q∞ S c cm
43
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5.2
From Appendix D, at 5° angle of attack,
c = 0.67
cmc/4 = -0.025
(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 cmc/4 is different than that for cℓ—be careful
in reading the data.)
With regard to cd, first check the Reynolds number,
Re =
ρ∞V∞c
(0.002377)(100)(3)
=
μ∞
(3.7373 ´ 10-7 )
Re = 1.9 ´ 106
In the airfoil data, the closest Re is 3 × 106. Use cd for this value.
cd = 0.007 (for c = 0.67)
The dynamic pressure is
q∞ =
1
1
ρ∞V∞ 2 = (0.002377)(100) 2 = 11.9lb/ft 2
2
2
The area per unit span is S = 1(c) = (1)(3) = 3ft 2
Hence, per unit span,
L = q∞ S c = (11.9)(3)(0.67) = 23.9 lb
D = q∞ S cd = (11.9)(3)(0.007) = 0.25lb
M c/4 = q∞ S c cmc/4 = (11.9)(3)(3)(-0.025) = -2.68 ft.lb
5.3
ρ∞ =
p∞
(1.01 ´ 105 )
=
RT∞
(287)(303) = 1.61kg/m3
From Appendix D,
c = 0.98
cmc/4 = -0.012
Checking the Reynolds number, using the viscosity coefficient from the curve given in
Chapter 4,
μ∞ = 1.82 ´ 10-5 kg/m sec
at T = 303 K,
ρ V c
(11.57)(42)(0.3)
Re = ∞ ∞ =
= 8 ´ 105
5
μ∞
1.82 ´ 10
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
1
1
q∞ = ρ∞V∞ 2 = (1.161)(42)2 = 1024 N/m 2
2
2
The area per unit span is S = (1)(0.3) = 0.3m 2 . Hence,
L = q∞ S c = (1024)(0.3)(0.98) = 301N
D = q∞ S cd = (1024)(0.3)(0.01) = 3.07 N
M c/4 = q∞ S c cm = (1024)(0.3)(0.3)(-0.012) = -1.1Nm
44
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From the previous problem, q∞ = 1020 N/m2
5.4
L = q∞ S c
Hence,
L
q∞ S
c =
The wing area S = (2)(0.3) = 0.6 m 2
Hence,
c =
200
= 0.33
(1024)(0.6)
From Appendix D, the angle of attack which corresponds to this lift coefficient is
α = 2
From Appendix D, at α = 4,
5.5
c = 0.4
Also,
æ 88 ö
V∞ = 120 çç ÷÷÷ = 176ft/sec
çè 60 ø
q∞ =
1
1
ρ∞ V∞ 2 = (0.002377)(176)2 = 36.8lb/ft 2
2
2
L = q∞ S c
L
29.5
S =
=
= 2 ft 2
q∞c
(36.8)(0.4)
L = q∞ S c
5.6
D = q∞ S cd
Hence,
L
q Sc
c
= ∞  = 
D
q∞ S cd
cd
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 α = 00 ,
c = 0.25
cd = 0.006
Hence
L
c
0.25
=  =
= 4.17
D
cd
0.006
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,
æ L ö÷
çç ÷
» 112
çè D ø÷
max
45
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5.7
At sea level
ρ∞ = 1.225 kg/m3
ρ∞ = 1.01 ´ 105 N/m 2
Hence,
q∞ =
1
1
ρ∞V∞ 2 = (1.225)(50) 2 = 1531 N/m 2
2
2
From the definition of pressure coefficient,
Cp =
5.8
p - p∞
(0.95 - 1.01) ´ 105
=
= -3.91
q∞
1531
The speed is low enough that incompressible flow can be assumed. From Bernoulli’s equation,
1
1
ρ V∞ 2 = p∞ + ρ∞V∞ 2 = p∞ + q∞
2
2
1
1
q∞ - ρV 2
ρV 2
p - p∞
2
2
Cp =
=
= 11
q∞
q∞
ρ∞ V∞ 2
2
p+
Since ρ = ρ∞(constant density)
æ V ö÷2
æ 62 ö2
÷÷ = 1 - çç ÷÷ = 1 - 1.27 = -0.27
C p = ççç
çè 55 ø÷
çè V∞ ø÷
5.9
The flow is low speed, hence assumed to be incompressible. From Problem 5.8,
æ V ö÷2
æ 195 ÷ö2
÷÷ = 1 - çç
C p = 1 - ççç
= -0.485
çè 160 ø÷÷
èç V∞ ø÷
5.10 The speed of sound is
a∞ =
γ RT∞ =
M∞ =
V∞
700
=
= 0.63
a∞
1107
(1.4)(1716)(510) = 1107 ft/sec
Hence,
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
C p0 = -0.485
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
Cp =
C p0
1 - M ∞2
=
-0.485
1 - (0.63) 2
46
= -0.625
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5.11 The formula derived in Problem 5.8, namely
æ V ö÷2
÷ ,
C p = 1 - ççç
çè V∞ ø÷÷
utilized Bernoulli’s equation in the derivation. Hence, it is not valid for compressible flow.
In the present problem, check the Mach number.
a∞ =
M∞ =
γ RT∞ =
(1.4)(1716)(505) = 1101 ft/sec
780
= 0.708.
1101
The flow is clearly compressible! To obtain the pressure coefficient, first calculate ρ∞ from the
equation of state.
p∞
2116
=
= 0.00244 slug/ft 3
RT∞
(1716)(505)
ρ∞ =
To find the pressure at the point on the wing where V = 850 ft/sec, first find the temperature
from the energy equation
c pT +
V2
V 2
= c pT∞ + ∞
V
2
V 2 -V2
T = T∞ + ∞
2c p
The specific heat at constant pressure for air is
cp =
γR
=
γ -1
(1.4)(1716)
ft lb
= 6006
(1.4 - 1)
slug R
Hence,
T = 505 +
7802 - 8502
= 505 - 9.5 = 495.5 R
2(6006)
Assuming isentropic flow
γ
æ T ö÷γ -1
p
÷
= ççç
çè T∞ ÷÷ø
p
æ 495.5 ÷ö3.5
p = (2116) çç
= 1980 lb/ft 2
çè 505 ÷÷ø
From the definition of Cp
p - p∞
p - p∞
1980 - 2116
=
=
1
1
2
q∞
ρ∞ V∞
(0.00244)(780)2
2
2
C p = -0.183
Cp =
47
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5.12 A velocity of 100 ft/sec is low speed. Hence, the desired pressure coefficient is a low speed
value, C p0 .
Cp =
C p0
1 - M ∞2
From problem 5.11,
C p = -0.183 and M ∞ = 0.708. Thus, 0.183 =
C p0
1 - (0.708)2
C p0 = (-0.183)(0.706) = -0.129
5.13 Recall that the airfoil data in Appendix D is for low speeds. Hence, at α = 4,
c 0 = 0.58.
Thus, from the Prandtl-Glauert rule,
c =
c 0
1 - M∞
2
=
0.58
1 - (0.8)2
= 0.97
5.14 The lift coefficient measured is the high speed value, c . Its low speed counterpart is c 0 ,
where
c =
c 0
1 - M ∞2
Hence,
c 0 = (0.85) 1 - (0.7)2 = 0.607
For this value, the low speed data in Appendix D yield
α = 2
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5.15 First, obtain a curve of Cp,cr versus M∞, from
2
C p,cr =
γ M ∞2
γ /(γ -1)
éæ
ù
ê ç 2 + (γ - 1)M ∞2 ö÷
ú
÷÷
ê çç
ú
1
ê çè
ú
γ +1
÷÷ø
êë
úû
Some values are tabulated below for γ = 1.4 .
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
C p0 = -0.90. From the Prandtl-Glauert rule,
Cp =
Cp =
C p0
1 - M ∞2
-0.90
1 - M ∞2
Some tabulated values are:
M∞
Cp
0.4
−0.98
0.5
−1.04
0.6
−1.125
0.7
−1.26
0.8
−1.5
0.9
−2.06
A plot of the two curves is given on the next page.
From the intersection point,
M cr = 0.62
49
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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:
Cp =
M∞
Cp
C p0
1 - M∞
0.4
−0.71
2
=
0.5
−0.75
-0.65
1 - M ∞2
0.6
−0.81
0.7
−0.91
0.8
−1.08
0.9
−1.49
The results are plotted below.
From the point of intersection,
M cr = 0.68
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°
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5.18
æ 1
èç M
μ = Sin-1 çç
d = h/Tanμ =
ö÷
-1 æ 1 ö
÷ = Sin ççç ÷÷÷ = 23.6
ø÷
è 25 ø
10km
= 22.9 km
0.436
5.19 At 36,000 ft, from Appendix B,
T∞ = 390.5R
ρ∞ = 7.1 ´ 10-4 slug/ft 3
Hence,
a∞ =
γ RT∞ =
(1.4)(1716)(390.5) = 969 ft/sec
V∞ = a∞ M ∞ = (969)(2.2) = 2132 ft/sec
1
1
ρ∞V∞ 2 = (7.1 ´ 10-4 )(2132) 2 = 1614 lb/ft 2
2
2
In level flight, the airplane’s lift must balance its weight, hence
L = W = 16,000 lb.
From the definition of lift coefficient,
q∞ =
CL = L/q∞ S = 16, 000 /(1614)(210) = 0.047
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
CL =
4α
M ∞2 - 1
Solve for α,
1
1
CL M ∞2 - 1 = (0.047) (2.2)2 - 1
4
4
α = 0.023 radians (or 1.2 degrees)
α =
The wave drag coefficient is approximated by
CDw =
Hence,
4α
M ∞2 - 1
=
4(0.023)2
(2.2)2 - 1
= 0.00108
Dw = q∞ S CDw = (1614)(210)(0.00108)
Dw = 366 lb
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5.20 (a) At 50,000 ft, ρ ∞ = 3.6391 × 10−4 slug/ft3 and T∞ = 390°R. Hence,
γ RT∞ =
a∞ =
and
(1.4)(1716)(390) = 968 ft/sec
V∞ = a∞ M ∞ = (968)(2.2) = 2130 ft/sec
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
dμ
(2.12 - 1.54) ´ 10-5
kg
=
= 5.8 ´ 10-8
dT
(350 - 250)
(m)(sec)(K)
Extrapolating from the sea level value of μ = 1.7894 ´ 10-5 kg/(m)(sec), we have
at T∞ = 216.7 K.
μ∞ = 1.7894 ´ 10-5 - (5.8 ´ 10-8 )(288 - 216.7)
μ∞ = 1.37 ´ 10-5 kg/(m)(sec)
Converting to English engineering units, using the information in Chapter 4, we have
μ∞ =
ö
1.37 ´ 10-5 æç
slug
-7 slug ÷÷
3.7373
10
´
= 2.86 ´ 10-7
ç
÷
-5 ççè
÷
ft sec ø
ft sec
1.7894 ´ 10
Finally, we can calculate the Reynolds number for the flat plate:
Re L =
ρ∞V∞ L
3.6391 ´ 10-4 (2130)(202)
=
= 5.47 ´ 108
μ∞
2.86 ´ 10-7
Thus, from Eq. (4.100) reduced by 20 percent
C f = (0.8)
0.07
(Re L )
0.2
= (0.8)
0.074
(5.74 ´ 108 )0.2
= 0.00106
The wave drag coefficient is estimated from
cd ,w =
where α =
Thus,
4α 2
M ∞2 - 1
2
= 0.035 rad.
57.3
cd ,w =
4(0.035) 2
(2.2)2 - 1
= 0.0025
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
cd ,w =
4(0.0873)2
(2.2)2 - 1
= 0.01556
Total drag coefficient = 0.01556 + 2(0.00106) = 0.0177
52
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(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.
æ
km ö÷æç 1h ö÷æç 1000m ö÷
÷÷ç
5.21 V∞ = 251 km/h = çç 251
֍
÷ = 69.7 m/sec
çè
h ÷øççè 3600sec ø÷èç 1km ø÷
ρ∞ = 1.225 kg/m3
1
1
ρ∞ V∞ 2 = (1.225)(69.7) 2 = 2976 N/m 2
2
2
L
9800
CL =
=
= 0.203
q∞ S
(2976)(16.2)
q∞ =
CDi =
CL 2
(0.203)2
=
= 0.002894
π e AR
π (0.62)(7.31)
Di = q∞ S CDi = (2976)(16.2)(0.002894) = 139.5 N
5.22 V∞ = 85.5 km/h = 23.75 m/sec
1
1
ρ∞ V∞ 2 = (1.225)(23.75)2 = 345 N/m 2
2
2
L
9800
CL =
=
= 1.75
q∞ S
(345)(16.2)
q∞ =
CDi =
CL 2
(1.75)2
=
= 0.215
π eAR
π (0.62)(7.31)
Di = q∞ S CDi = (345)(16.2)(0.215) = 1202 N
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,
C = 1.05 at α = 8
C = 0 at α L = 0 = -1.5°
Hence,
a0 =
1.05 - 0
= 0.11 per degree
8 - (-1.5)
The lift slope for the finite wing is
a =
a0
0.11
=
= 0.076 degree
57.3 a0
57.3(0.11)
1+
1+
π (9)(5)
π ei AR
At α = 6°, CL = a(α - α L = 0 ) = (0.076)[6 - (-1.5)] = 0.57
The total drag coefficient is
CL 2
(0.57)2
= (0.004) +
π eAR
π (0.9)(5)
= 0.004 + 0.023 = 0.027
CD = cd +
CD
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5.24
q∞ =
1
1
ρ∞ V∞ 2 = (0.002377)(100)2 = 11.9 lb/ft 2
2
2
at α = 10°, L = 17.9 lb. Hence
L
17.9
=
= 1.0
q∞ S
(11.9)(1.5)
CL =
at α = −2°, L = 0 lb. Hence αL = 0 = −2°
dCL
1.0 - 0
=
= 0.083 per degree
dα
[10 - (-2)]
a =
This is the finite wing lift slope.
a =
a0
57.3 a0
1+
π e AR
Solve for aο.
a0 =
a
0.083
=
57.3 a
57.3(0.083)
11+
π e AR
π (0.96)(6)
a0 = 0.11 per degree
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5.25 At α = −1°, the lift is zero. Hence, the total drag is simply the profile drag.
CL 2
= cd + 0 = cd
π eAR
1
1
= ρ∞ V∞2 = (0.002377)(130)2 = 20.1 lb/ft 2
2
2
CD = cd +
q∞
Thus, at α = α L = 0 = -1°
cd =
D
0.181
=
= 0.006
q∞ S
(20.1)(1.5)
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°,
CL =
L
5
=
= 0.166
q∞ S
(20.1)(1.5)
CD =
D
0.23
=
= 0.00763
q∞ S
(20.1)(1.5)
However:
CD = cd +
CL 2
π eAR
0.00763 = 0.006 +
(0.166)2
0.00146
= 0.006 +
π e(6)
e
e = 0.90
To obtain the lift slope of the airfoil (infinite wing), first calculate the finite wing lift slope.
(0.166 - 0)
= 0.055 per degree
[2 - (-2)]
a
0.055
a0 =
=
57.3a
57.3(0055)
11π eAR
π (0.9)(6)
a =
a0 = 0.068 per degree
5.26 Vstall =
2W
=
ρ∞ S CLmax
2(7780)
(1.225)(16.6)(2.1)
Vstall = 19.1 m/sec = 68.7 km/h
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5.27 (a)
α =
5
= 0.087 radians
57.3
c = 2πα = 2π (0.087) = 0.548
(b)
Using the Prandtl-Glauert rule,
c =
(c)
c 0
1 - M ∞2
=
0.548
1 - (0.7)2
= 0.767
From Eq. (5.50)
c =
4α
2
M∞ - 1
=
4(0.087)
(2)2 - 1
= 0.2
5.28 For V∞ = 21.8 ft/sec at sea level
1
1
q∞ = ρ∞ V∞ 2 = (0.002377)(21.8)2 = 0.565 lb/ft 2
2
2
1 ounce = 1/16 lb = 0.0625 lb.
L
0.0625
CL =
=
= 0.11
q∞ S
(0.565)(1)
For a flat plate airfoil
c = 2πα = 2 ⋅ π (3 / 57.3) = 0.329
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:
CL = 0.8 cos 6° - 0.06 sin 6° = 0.7956 - 0.00627 = 0.789
CD = 0.8 sin α + 0.06 cos α = 0.0836 + 0.0597 = 0.1433
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.
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5.30 First solve for the angle of attack and the profile drag coefficient, which stay the same in this
problem
cL = a α =
or,
a0 α
1 + 57.3 a0 /(π e1 AR)
CL
[1 + 57.3 a0 / π e1 AR]
a0
0.35
=
{1 + 57.3(0.11)/[π (0.9)(7)]} = 4.2°
0.11
α =
The profile drag can be obtained as follows
CL
0.35
CD =
=
= 0.012
(CL / CD )
29
or,
CD = cd +
CL2
π e AR
cd = CD -
CL2
(0.35)2
= 0.012 = 0.0062
π e AR
π (.9)(7)
Increasing the aspect ratio at the same angle of attack increases CL and reduces CD.
For AR = 10, we have
a0 α
CL = a α =
1 + 57.3 a0 /(π e1 AR)
=
(0.11)(4.2)
= 0.3778
1 + 57.3(0.11)/[π (0.9)(10)]
CD = cd +
CL2
(0.3778)2
= 0.062 = 0.0062 + 0.005048 = 0.112
π e AR
π (.9)(10)
Hence, the new value of L/D is
CL
0.3778
=
= 33.7
CD
0.0112
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5.31 For incompressible flow, from Bernoulli’s equation, the stagnation pressure is
1
p0 = p∞ +
2
2 ρ∞ V∞
Let S be the surface area of each face of the plate. The aerodynamic force exerted on the front
face is
Ffront = p0 S
and the aerodynamic force on the back face is
Fback = p∞ S
The net force, which is the drag, is
D = Ffront - Fback = ( p0 - p∞ )S
From Bernoulli’s equation,
p0 - p∞ =
1
2
2 ρ ∞ V∞
Thus,
D =
CD
1
2 ρ ∞ V∞
2
S
1
ρ∞V∞ 2 S
2
=
=
1
1
ρ∞V∞ 2 S
ρ∞V∞ 2 S
2
2
D
Hence,
CD = 1
5.32 The drag of the airplane is
D = q∞ S CD
(1)
The drag of a flat plate at 90° to the flow, with an area f and drag coefficient of 1, is
D = q∞ f CDpblc = q∞ f (1)
(2)
Equating Eqs. (1) and (2)
q∞ S CD = q∞ f
or,
f = CD S
5.33
f = CD S = (0.0163)(233) = 3.8ft 2
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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)
Cd , f =
0.074
Re0.2
L
=
0.074
(8.9 ´ 106 )0.2
= 0.023
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,
cd , f = 2(0.003) = 0.006
Since from Section 5.12
cd = cd , f + cd , p
we have: cd , p = cd - cd , f = 0.006 - 0.006 = 0. 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°,
cd = 0.0075
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
cd , f = 0.006
Thus,
cd , p = cd - cd , f = 0.0075 - 0.006 = 0.0015
For this case, the percentage of drag due to pressure drag is
cd , p
cd
=
0.0015
= 0.2, or 20%
0.0075
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.
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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
0.074
0.074
0.074
(cd , f ) xcr =
=
=
= 0.00536
0.2
0.2
13.8
(Re x,cr )
(500, 000)
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
æ 500, 000 ÷ö
0.00536 çç
÷ = 3.01 ´ 10-4
çè 8.9 ´ 106 ÷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:
(cd , f ) turb = 0.003 - 3.01 ´ 10-4 = 0.0027
The laminar skin friction drag coefficient for the flow from the leading edge to the transition
point is, from Eq. (4.98)
C f , laminar
1.328
=
Re x , cr
1.328
1.328
=
= 1.878 ´ 10-3
707.1
500, 000
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
æ 500, 000 ö÷
(cd , f )lam = 1.878 ´ 10-3 çç
÷ = 1.055 ´ 10-4
çè 8.9 ´ 106 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
cd , f = 1.055 ´ 10-4 + 0.0027 = 0.0028
Including both the upper and lower surfaces, then we have
cd , f = 2(0.0028) = 0.0056
cd = cd , f + cd , p
Thus,
cd , p = cd - cd , f = 0.006 - 0.0056 = 0.004
The percentage of the total drag due to pressure drag due to flow separation is then
cd , p
cd
=
0.0004
= (100) = 6.7%
0.006
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 xcr / L = 500, 000 / 8.9 ´ 106 = 0.0056.
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.
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5.37 (a)
From Eq. (4.98),
Cf =
1.328
Re
1.328
=
= 4.43 × 10−4
9 × 106
This is for one side of the flat plate. The total skin friction drag coefficient including both sides
of the plate is
C f = 2(4.43 × 10−4 ) = 0.000885
(b) From Eq. (4.101)
Cf =
0.074
(Re)0.2
=
0.074
(9 × 106 )0.2
=
0.074
= 0.003
24.59
For both sides of the plate,
C f = 2(0.003) = 0.006
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,
C f = 0.006 (from Problem 5.37).
Transition takes place at location
xcr 650,000
=
= 0.072
c
9 × 106
The turbulent drag coefficient on the area upstream of transition would be


0.074
 0.074 
Cf = 
 2 = 
 2 = 0.0102
0.2
14.54 
(650,
000)


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 
Cf = 
 2 =  806  2 = 0.0033
 650, 000 
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%
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5.39 Assuming all turbulent flow,
Cf =
0.074
0.074
=
= 0.003749
0.2
(Re)
(3 × 106 )0.2
Taking into account both sides of the plate
C f = (0.003749) 2 = 0.0075
Transition takes place at location
xcr 650,000
=
= 0.2167
c
3 × 106
The turbulent drag coefficient on the area upstream of transition would be


0.074
Cf = 
 2 = 0.0102
0.2
 (650, 000) 
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 
Cf = 
 2 = 0.0033
 650, 000 
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.
 0.006 
Friction drag percentage = 
100 = 88%
 0.0068 
Pressure drag percentage = 11.8%
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Chapter 6
6.1
(a)
V¥ = 350 km/hr = 97.2 m/sec
1
1
2
ρ¥V¥
= (1.225)(97.2)2 = 5787 N/m 2
2
2
38220
W
CL =
=
= 0.242
(5787)(27.3)
q¥S
q¥ =
(0.242)2
CL 2
= 0.03 +
π eAR
π (0.9)(75)
CD = 0.03 + 0.0028 = 0.0328
CD = C D 0 +
CL / CD = 0.242 / 0.0328 = 7.38
TR =
(b)
W
38, 220
=
= 5179 N
CL /C D
7.38
q¥ =
CL =
1
1
2
ρ¥V¥
= (0.777)(97.2) 2 = 3670 N/m2
2
2
W
38, 220
=
= 0.38
q¥ S
(3670)(27.3)
CL 2
(0.38)2
= 0.03 +
π eAR
π (0.9)(7.5)
CD = 0.03 + 0.0068 = 0.0368
Cd = CD0 +
CL / CD = 0.38 / 0.0368 = 10.3
TR =
6.2
V¥ = 200
W
38,220
=
= 3711N
CL /CD
10.3
88
= 293.3ft/sec
60
1
1
2
ρ¥V¥
= (0.002377)(293.3)2 = 102.2 lb/ft 2
2
2
5000
L
W
CL =
=
=
= 0.245
(102.2)(200)
q¥ S
q¥ S
q¥ =
CDi =
(0.245)
CL 2
=
= 0.0024
π e AR
π (0.93)(8.5)
Since the airplane is flying at the condition of maximum L/D, hence minimum thrust
required, C Di = C De . Thus,
CD = CD0 + CDi = 2CDi = 2(0.0024) = 0.1048
D = q¥ SCD = (102.2)(200)(0.0048) = 98.1lb.
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6.3
(a) Choose a velocity, say V∞ =100 m/sec
q¥
1
1
2
ρ¥V¥
= (1.225)(100) 2 = 6125 N/m 2
2
2
103,047
W
CL =
=
= 0.358
(6125)(47)
q¥S
(0.358)2
CL 2
= 0.032 +
π eAR
π (0.87)(6.5)
= 0.032 + 0.007 = 0.0392
CD = CD0 +
CD
TR =
103, 047
W
=
= 11287 N
9.13
CL / CD
PR = TRV¥ = (11, 287)(100) = 1.129 ´ 106 watts
PR = 1129 kw
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)
PA = TAV¥ = (2)(40298)V¥ = 80596V¥
The power required and power available curves are plotted below.
From the intersection of the PA and PR curves, we find,
Vmax = 295 m/sec at sea level
64
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(c)
At 5 km standard altitude, ρ = 0.7364 kg/m3
Hence,
( ρ 0 /ρ )1/2 = (1.225/0.7364)1/2 = 1.29
Valt = ( ρ0 /ρ )1/2V0 = 1.29 V0
PRalt = ( ρ0 /ρ )1/2 PR 0 = 1.29 PR 0
From the results from part (a) above,
PR 0
V0
(m/sec)
(kw)
100
130
160
190
220
250
(d)
PRak
Valt
(m/sec)
(kw)
129
168
206
245
284
323
1456
2815
5028
8281
12,771
18,770
1129
2182
3898
6419
9900
14,550
æ ρ ö
TAall =TA 0 ççç ÷÷÷ = 0.601TA 0
çè ρ0 ÷ø
PAalt = TAalt V¥ = (0.601)(80596)V¥ = 48438V¥
Hence,
The power required and power available curves are plotted below.
From the intersection of the PR and PA curves, we find Vmax = 290 m/sec at 5 km
Comment: The mach numbers corresponding to the maximum velocities in parts (b) and (d) are
as follows:
At sea level
a¥ =
γ RT¥ =
M¥ =
V¥
295
=
= 0.868
a¥
340
(1.4)(287)(288.16) = 340 m/sec
At 5 km altitude
a¥ =
M¥ =
γ RT¥ =
(1.4)(287)(255.69) = 321m/sec
V¥
290
=
= 0.90
a¥
321
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.
65
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6.4
(a) Choose a velocity, say V¥ = 100 ft/sec
1
1
ρ¥V¥2 = (0.002377)(100)2 = 11.89 lb/ft 2
2
2
W
3000
CL =
=
= 1.39
q¥ S
(11.89)(181)
q¥ =
CD = CD0 +
CL 2
(1.39)2
= 0.027 +
π e AR
π (0.91)(6.2)
CD = 0.027 + 0.109 = 0.136
TR =
W
3000
3000
=
=
= 293.5 lb
CL /CD
(1.39) /(0.136)
10.22
PR = TR ⋅ V¥ = (293.5)(100) = 29350 ft lb/sec
In terms of horsepower,
PR =
29350
= 53.4 hp
550
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,
Vmax = 295ft/sec = 201mph at sea level
66
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(c)
At a standard altitude of 12,000 ft,
ρ = 1.648 ´ 10-3 slug/ft 3
Hence,
( ρ 0 /ρ )1/2 = (0.002377 / 0.001648)1/ 2 = 1.2
Valt = ( ρ0 /ρ )1/2V0 = 1.2V0
PRak = ( ρ0 /ρ )1/2 PR0 = 1.2 PR0
Using the results from part (a) above,
PR0
V0
(ft/sec)
70
100
150
200
250
300
(hp)
64.9
53.4
64.3
106
182
300
PRah
Valt
(ft/sec)
(hp)
84
120
180
240
300
360
77.9
64.1
76.9
127
218
360
(d) Assuming that the power output of the engine is proportional to p∞,
æ 0.001648 ÷ö
PAalt = (ρ /ρ0 ) PA 0 = çç
P = 0.693 PA 0
çè 0.002377 ÷÷ø A 0
At 12,000 ft,
PA = 0.693(286) = 198 hp
The power required and power available curves are plotted below.
From the intersection of the PA and PR curves,
Vmax = 290ft/sec =198 mph at12,000 ft.
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
excess power
9 ´ 106 watts
=
= 87.3m/sec
W
1.0307 ´ 105 N
and at 5 km altitude,
R/C =
R/C =
5 ´ 106
1.03047 ´ 105
= 48.5 m/sec
67
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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,
R/C =
excess power
(232)(550)
=
= 42.5ft/sec
W
3000
and at 12,000 ft altitude,
(134)(550)
= 24.6 ft/sec
3000
R/C =
6.7
Assuming (R/C)max varies linearly with altitude,
(R/C)max = ah + b
From the two (R/C)max values from Problem 6.5,
87.3 = a(o) + b
48.5 = a(5000) + b
Hence,
b = 87.3
a = -0.00776
(R/C)max = -0.00776h + 87.3
To obtain the absolute ceiling, set (R/C)max = 0, and solve for h.
h=
87.3
= 11, 250 m
0.00776
Hence,
absolute ceiling ≈ 11.3 km
6.8
(R/C)max = ah + b
From the results of Problem 6.6,
42.5 = a(0) + b
24.6 = a(12, 000) + b
Hence,
b = 42.5
a = -0.00149
(R/C)max = -0.00149h + 42.5
To obtain the absolute ceiling, set (R/C)max = 0, and solve for h.
42.5
= 28,523ft
0.00149
absolute ceiling ≈ 28,500 ft
h=
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.
68
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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.
6.9
R = h (L/D) max = 5000(7.7) = 38,500 ft = 729 miles
6.10 First, we have to calculate the CL corresponding to maximum L/D.
æC
çç L
ççè C
D
ö÷
÷÷
=
÷ø
max
C D,0
C D,Oπ e AR
2C D,0
1æ C
= ççç L
4 çè CD
ö÷
÷÷
÷ø
=
π e AR
4C D,0
-2
π e AT =
(0.25) π (0.7)(4.11)
(7.7)2
max
= 0.038
At maximum L/D, C D,0 = CD,i = 0.038
Hence: CD = 2(0.038) =0.076
æC
CL = CD ççç L
çè CD
ö÷
÷÷ = (0.076)(7.7) = 0.585
ø÷
Also:
æ L ö÷-1
= Arc Tan (0.13)
çè D ø÷÷
max
θ min = Arc Tan çç
θmin = 7.4
V¥ =
2 cos θ çæ W
ç
ρ C èç S
¥ L
ö
æ 1400 ö÷
2 cos(7.4)
çç
÷÷÷ =
÷ = 97.2 ft/sec
ø
(0.002175)(0.585) èç 231 ø÷
6.11 From Eq. (6.85),
(
π eARCD0
æLö
=
ççç ÷÷÷
è D ømax
2CD0
t/2
)
Putting in the numbers:
1/2
æ L ö÷
[ π (0.9)(6.72)(0.025) ]
ç
=
ççè D ø÷÷
0.050
max
69
= 13.78
PROPRIETARY MATERIAL. © 2012 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be
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6.12 Aviation gasoline weighs 5.64 Ib per gallon,
Hence,
WF = (44)(5.64) = 248lb.
Thus, the empty weight is
W1 = 3400 - 248 = 3152lb.
The specific fuel consumption, in consistent units, is
c = 0.42/(550)(3600) = 2.12 ´ 10-7 ft-1
The maximum L/D can be found from Eq. (6.85).
(π e AR CDc )1/2
æ L ö÷
[ π (0.91)(6.2)(0.027) ]1/2
çç ÷
=
=
= 12.8
çè D ø÷
2 CD
2(0.027)
max
c
Thus, the maximum range is:
æ η öæ C ö
R = çç ÷÷÷ççç L ÷÷÷
çè c ÷øèç C ø÷
D
max
æW ö
n ççç 0 ÷÷÷
èç W1 ø÷
æ
ö
æ 3400 ö÷
0.83
R = çç
÷÷÷ (12.8) n çç
÷
7
çè 2.12 ´ 10 ø
èç 3152 ø÷
R = 3.8 ´ 106 ft
In terms of miles,
R =
3.8 ´ 106
= 719 miles
5280
To calculate endurance, we must first obtain the value of (CL3/2 /CD ) max From Eq. (6.87),
æ C 3/2 ÷ö
(3 CD 0 π e AR)3/4
çç L ÷
=
ç C ÷÷
4 CDc
çè D ÷ø
max
Putting in the numbers:
æ C 3/2 ö÷
[ 3(0.027) π (0.91)(6.2) ]1/2
çç L ÷
=
= 11.09
çç C ÷÷
4(0.027)
è D ø÷max
Hence, the endurance is
æ η ö C3 / 2
E = çç ÷÷÷ L (2 ρ¥ S )1/2 (W1-1/ 2 - W0-1/ 2 )
çè c ø C
D
æ
ö÷
0.83
1/ 2
E = çç
÷ (11.09) [ 2(0.002377)(181) ] (3152-1/ 2 - 3400-1/ 2 )
7
çè 2.12 ´ 10 ø÷
E = 2.67 ´ 104 sec = 7.4 hr
70
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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,
W f = (1900)(29.67) = 56,370 N
Wt = W0 - W f = 136,960 - 56,370 = 80,590 N
In consistent units,
ct = 1.0
N
= 2.777 ´ 10-4 sec-1
(N)(hr)
Also, at a standard altitude of 8 km.
ρ¥ = 0.526 kg/m3
Since maximum range for a jet air craft depends upon maximum CL1/2 /C D , we must use
Eq. (6.86).
æ C 1/2 ö÷
çç L ÷
÷÷
çç
è CD ø÷
max
æ1
ö1/4
çç CD π e AR ÷÷
÷ø
çè 3 c
=
4
CDc
3
Putting in the numbers,
é1
ù1/4
ê (0.032)(0.87)(6.5) ú
æ C 1/2 ÷ö
ê3
ûú
ççç L ÷÷
= ë
= 15.46
÷
4
çè CD ø÷
(0.032)
max
3
For a jet airplane, the range is
R = 2
æ1
çç
ρ¥ S çèç ct
2
ö÷æç C 1/2 ö÷ 1/2
÷ç L ÷÷ (W0 - W11/2 )
֍
ø÷çè CD ø÷÷
é
ù1/2 æ
2
1
÷÷ö (15.46) é (136,960)1/2 - (80,590)1/2 ù
ú çç
R = 2ê
êë
úû
ê (0.526)(47) ú çè 2.777 ´ 10-4 ÷ø
ë
û
R = 2.73 ´ 106 m = 2730 km
The endurance depends on CLCD. From Eq. (6.85),
(π e AR CDc )1/2
æ L ö÷
[ π (0.87)(6.5)(0.032) ]1/2
çç ÷
=
=
= 11.78
çè D ø÷
2 CDc
2 (0.032)
max
The endurance for a jet aircraft is
E =
E =
1 æç C L ÷ö çæ W0 ÷ö
÷÷ n ç
÷÷
ç
ct ççè CD ÷ø ççè W1 ÷ø
1
-4
æ 136,960 ÷ö
(11.78) n çç
÷
çè 80,590 ÷÷ø
2.777 ´ 10
E = 22, 496 sec = 6.25 hr
71
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6.14
CL3/ 2
=
CD
CL3/ 2
CD0 +
CL2
π e AR
-2
éæ
ù
æ
d (CL3 / 2 / CD )
CL2 ö÷÷æç 3 12 ö÷
CL ö÷ ú æç
CL2 ö÷÷
çç
3/ 2 ç
ê
= ê ç CD0
÷ç C ÷ - C L çç 2
÷÷ ú çç CD + π e AR ÷÷ = 0
π e AR ø÷÷çè 2 L ø÷
d CL
èç π e AR ÷ø ú èç 0
êë çè
ø÷
û
3 12
1 CL5 / 2
CL CD0 = 0
2
2 π e AR
3 CD0 -
CL2
=0
π e AR
CD0 =
Hence,
1
CD
3 i
This is Eq. (6.80)
To obtain Eq. (6.81)
C1/2
L =
CD
C1/2
L
CD0 +
CL2
π e AR
-2
éæ
ùæ
2
d (C1/
CL2 ö÷÷ 1 - 12
CL2 ö÷÷
- 12 æç 2 CL ö÷ ú ç
L /C D ) = ê çç C
÷÷ ú çç CD +
÷ C - C L çç
÷ =0
ê çç D0
π e AR ÷÷ø 2 L
π e AR ÷÷ø
d CL
èç π e AR ÷ø ú èç 0
êë è
û
1
3 CL5/2
-1
= 0
CD0 CL 2 2
2 π e AR
CD0 - 3
CL3/2
=0
π e AR
CD0 = 3 C Di
6.15 From Eq. (6.81), CD0 = 3 C Di
CD0 = 3
CL2
π e AR
1
æ1
ö2
CL = çç π e AR C D0 ÷÷÷
çè 3
ø
Thus,
1
æ 12 ö÷
çç CL ÷
÷÷
çç
çç CD ÷÷÷
è
ø
max
æ1
ö4
çç π e AR CD ÷÷
÷ø
0
çè 3
This is Eq. (6.86)
=
4
CD0
3
From Eq. (6.80), CD0 =
1
CD
3 i
CD0 =
CL2
3π e AR
1
Thus, CL = (3π e AR CD0 ) 2
æ 32 ö÷
çC ÷
ççç L ÷÷
çç C D ÷÷÷
è
ø
max
æ
ç 3π e AR C D0
= çç
çç 4 CD
è
0
3
ö4
÷÷÷ (This is Eq. (6.87)
÷÷ø
72
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6.16
AR = b 2 /S , hence b =
S AR =
(47)(6.5) = 17.48 m
h = 5 ft = (5 / 3.28) m = 1.524 m
h/b = 1.54 /17.48 = 0.08719
φ =
(16 h/b)2
1 + (16 h / b)
2
VL0 = 1.2Vstall = 1.2
=
1.946
= 0.66
2.946
2W
ρ¥S CL,max
= 1.2
2(103, 047)
= 80.3 m/sec
(1.255)(47)(0.8)
Hence, 0.7 0.7VL0 = 56.2 m/sec. This is the velocity at which the average force is evaluated.
1
1
2
ρ¥V¥
= (1.225)(56.2)2 = 1935 N/m 2
2
2
L = q¥ S C L = (1935) (47)(0.8) = 72, 760 N
q¥ =
æ
CL2 ÷÷ö
ç
D = q¥ S C D = q¥ S çç C D0 + φ
÷
çè
π e AR ÷÷ø
D = (1935)(47)[(0.032) + (0.66)(0.8)2 /π (0.87)(6.5)]
D = 5072 N
From Eq. (6.90)
s L0 =
=
1.44 W 2
g ρ¥CL,max { T - [ D + μr (W - L) ] ave }
1.44 (103047)2
(9.8)(1.225)(47)(0.8) { 80595 - [ 5072 + (0.2)(103047 - 72760) ]}
sL0 = 452 m
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6.17
b=
S AR =
(181)(6.2) = 33.5 ft.
h/b = 4 / 33.5 = 0.1194
φ =
(16h/b)2
1 + (16h/b)
VL0 = 1.2
2
=
3.65
= 0.785
4.65
2W
ρ¥ S CL,max
= 1.2
2(3000)
= 135 ft/sec
(0.002377)(181)(1.1)
0.7 = VL0 = 0.7(135) = 94.5 ft/sec.
q¥ =
1
1
2
ρ¥V¥
= (0.002377)(94.5) 2 = 10.6 1b/ft 2
2
2
L = q¥S CL = (10.6)(181)(1.1) = 2110 1b
æ
C L2 ö÷÷
ç
D = q¥ S CD = q¥ S çç CD,0 + φ
÷
çè
π e AR ÷ø÷
é
(0.785)(1.1)2 ùú
D = (10.6)(181) êê (0.027 +
= 154.6 lb
π (0.91)(6.2) úúû
êë
T = (550 HPA )/V¥ = (550)(285) / 94.5 = 1659 lb
S L0 =
1.44 W 2
g ρ¥ S C L,max {T - [ D + μr (W - L)]ave}
1.44 (3000)2
(32.2)(0.002377)(181)(1.1){1659 - [154.6 + (0.2)(3000 - 2110]}
S L0 = 572 ft.
=
6.18 VT = 1.3
2W
ρ¥ S CL,max
= 1.3
2(103, 047)
= 46.39 m/sec
(1.23)(47)(2.8)
0.7 VT = 32.47 m/sec.
q¥
1
2
ρ¥V¥
= (0.5)(1.23)(32.47)2 = 648.4 N/m 2
2
Since the lift is zero after touchdown, C D = C D,0⋅
D = q¥ S C D,0 = (648.4)(47)(0.032) = 975.2 N
SL =
1.69 W 2
g ρ¥CL,max [ [ D + μr (W - L) ]0.7V
T
SL =
1.69 (103047)2
= 268 m
(9.8)(1.23)(47)(2.8)[975.2 + (0.4)(103, 047)]
74
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6.19 VT = 1.3
2W
ρ¥S CL,max
= 1.3
2(3000)
= 114.4 ft/sec
(0.002377)(181)(1.8)
0.7 VT = 80.08 ft/sec.
q¥ =
1
1
2
ρ¥ V¥
= 0.5(0.002377)(80.08) 2 = 7.62 1b/ft 2
2
2
D = q¥ S C D,0 = (7.62)(181)(0.027) = 37.2 1b
SL =
1.69 W 2
g ρ¥S CL,max [ [ D + μr (W - L) ]0.7V
T
SL =
2
1.69 (3000)
= 493 ft
(32.2)(0.002377)(181)(1.8)[37.2 + 0.4 (3000)]
æ 88 ö
6.20 V¥ = 250 mph = 250 çç ÷÷÷ ft/sec = 366.6 ft/sec
çè 60 ø
= 366.6 (0.3048) m/sec = 111.7 m/sec.
q¥ =
1
2
ρ¥V¥
= (0.5)(1.23)(111.7)2 = 7673 N/m 2
2
L = q¥ S CL,max = (7673)(47)(1.2) = 4.358 ´ 105 N
n =
R =
w=
L
4.328 ´ 105
=
= 4.2
W
103047
2
V¥
g n2 - 1
=
(111.7)2
9.8 (4.2)2 - 1
= 312 m
V¥
111.7
=
= 0.358 rad/sec
R
312
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T = D = q¥S CD
6.22 From Eq. (6.13)
CL2
π e AR
From Eq. (6.1 c)
C D = C D, 0 +
Combining (1) and (2)
é
CL2 ùú
T = q¥ S êê CD, 0 +
π e AR úúû
êë
L = W = q¥ S CL
From Eq. (6.14)
CL =
W
q¥ S
Substitute (4) into (3)
é
ù
W2
W2
ú = q SC
T = q¥S êê CD,0 + 2
+
¥
D, 0
ú
q¥ Sπ e AR
q¥ Sπ e AR ûú
ëê
Multiply by q∞
W2
Sπ e AR
2
q¥T = q¥
SCD, 0 +
or
2
q¥
SCD, 0 - q¥T +
W2
=0
Sπ e AR
From the quadratic formula:
T  T2 q¥ =
q¥ =
2
V¥
=
V¥
4 SCD, 0W 2
Sπ e AR
2 SCD, 0
T æç W
ç
W èç S
æT
çç
èç W
2
4 C D, 0
÷÷ö  W æçç T ÷÷ö ÷ø
π e AR
S èç W ÷ø
1
2
= ρ¥V¥
2 C D, 0
2
÷÷öæçç W
÷øèç S
é
ê æç T
ê çç
ê èW
= ê
ê
ê
ê
êë
2
4 C D, 0
÷÷ö + W ççæ T ÷÷ö ÷ø
π e AR
S çè W ÷ø
ρ¥CD, 0
÷÷öæçç W
÷øèç S
2
4 C D, 0
÷÷ö + W ççæ T ÷÷ö ÷ø
S çè W ÷ø
π e AR
ρ¥CD, 0
ù1/ 2
ú
ú
ú
ú
ú
ú
ú
úû
For Vmax : T = (TA )max
(V¥ )max
é
æW
ê æç TA ö÷
çç
ê çç ÷÷
ê è W ømax èç S
= ê
ê
ê
ê
êëê
ù1/ 2
4 C D, 0 ú
ö÷ W æ TA ÷ö 2
ç
ú
÷÷ø + S èçç W ÷÷ø
π e AR ú
max
ú
ú
ρ¥CD, 0
ú
ú
úûú
76
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beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. A
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6.23 From Figure 6.2, ( L/D) max = 18.5, and CD, 0 = 0.015. From Eq. (6.85)
ö
(CD, 0π e AR)1/2
÷÷÷
=
÷
2 CD,0
D ømax
æC
çç L
ççè C
4 CD, 0 (CL /CD )2 max
e=
or,
e=
π AR
4 (0.015)(18.5)2
= 0.83
π (7.0)
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: D = q¥ SCD
Power Available: PA = η Ps
Also, PA = TAV¥
Hence TA =
η Ps
V¥
(a) At an altitude of 30,000 ft, ρ∞ = 0.00089068 slug/ft3 and T∞ = 411.86°R. The speed of sound is
a¥ =
γ RT¥ =
(1.4)(1718)(411.86) = 994.7 ft/sec
Hence, at Mach one, the flight velocity is V∞ = 994.7 ft/sec
q¥ =
1
1
2
ρ¥V¥
= (0.00089068)(994.7)2 = 440.6 1b/ft 2
2
2
The drag coefficient at Mach one, as given in the problem statement is
CD, 0 (at M ¥ = 1) = 10[CD, 0 (at low speed)] = 10(0.0211) = 0.211
Hence, drag at Mach one is
D = q¥S CD = (440.6)(334)(0.211) = 31, 0511b
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
HP =
ρ¥ (at 30,000 ft)
0.00089068
(1500) =
(1500) = 969 HP
0.0013781
ρ¥ (at 17,500 ft)
From Eq. (3), above
TA =
η Ps
V¥
=
(0.3)(969)(550)
= 1611b
994.7
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.
77
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(b) At an altitude of 20,000 ft,
ρ¥ = 0.0012673 slug/ft 3
T¥ = 447.43R
a¥ = V¥ =
q¥ =
(1.4)(1716)(447.43) = 1036.8 ft/sec
1
1
2
ρ¥ V¥
= (0.0013672)(1036.8) 2 = 6811b/ft 3
2
2
D = q¥ S CD = (681)(334)(0.211) = 47,993 lb
The thrust available is obtained as follows
HP =
ρ¥ (at 20,000 ft)
0.0012673
(1500) =
(1500) = 1379 HP
0.0013781
ρ¥ (at 17,500 ft)
Hence,
TA =
η Ps
V¥
=
(0.3)(1379)(550)
= 219 lb
1036.8
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)
PA =
1
W2
3
S C D, 0 +
ρ¥V¥
1
2
ρ V¥ Sπ e AR
2
b2
(14.45) 2
=
= 19.3
S
11.45
W = 1020 kg f = (1020)(9.8) = 9996 N
AR =
PA = η (bhp) = (0.9)(85) = 76.5 hp
Using consistent units, noting that 1 hp = 746 Watts
PA = (76.5)(746) = 5.71 ´ 10 4Watts
1
1
ρ¥ S CD, 0 = (1.23)(11.45)(0.03) = 0.2113
2
2
1
1
ρ¥ Sπ e AR = (1.23)(11.45) π (0.7)(19.3) = 298.9
2
2
Inserting these numbers into Eq. (1)
3
5.71 ´ 104 = 0.2113 V¥
+
or,
3
5.71 ´ 104 = 0.2113V¥
+
Solving Eq. (2) for V∞
(9996)2
298.9 V¥
3.343 ´ 105
V¥
V¥ = Vmax = 62, 6 m/sec.
78
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6.26 From Eq. (6.67)
R =
η CL
c CD
n
W0
W1
ö÷
(CD, 0 π e AR)1/2
[ (0.03) π (0.7)(19.3) ]1/ 2
÷÷
18.8
=
=
÷
2 CD,0
2(0.03)
D ømax
æC
çç L
ççè C
Using consistent units,
c = 0.2
é kg f ù çæ 9.8N
ú çç
= 0.2 êê
úç
(hp)(hr)
ëê (hp)(hr) ûú èç 1 kg f
kg f
ö÷ é 1 hp ùæ 1 hr ö
÷÷
÷÷ ê
úç
÷ ê 746 Nm/s ú çççè 3600 sec ÷÷ø
÷øë
û
7
= 7.3 ´ 10m-1
W0 = 1020 kg f = 1020 (9.8) = 9996 N
W1 = W0 - Wfuel = 9996 - (295)(9.8) = 7105 N
From Eq. (6.67)
R =
η CL
W
n 0 =
c CD W1
æ
ö÷
æ 9996 ö÷
0.9
÷ (18.8) n çç
÷
ççç
7
÷
èç 7105 ø÷
è 7.3 ´ 10 ø
R = 3.44 ´ 106m = 3440 km
æ C 3/2 ÷ö
ç
6.27 çç L ÷÷÷
çè CD ÷ø
=
(3 CD, 0 π e AR)3/4
4 CD, 0
max
=
[ 3(0.03) π (0.7)(19.3) ]3/4
4(0.03)
= 22, 77
From Eq. (6.68)
æ η öæç C 3/2 ÷ö
E = çç ÷÷÷çç L ÷÷÷ (2ρ¥ S )1/2 (W1-1/2 - W0-1/2 )
èç c øçè CD ÷ø
Using results from Problem 6.26,
æ
ö÷
0.9
E = çç
÷ 22.77[2(1.23) (11.45)]1/2 [(7105)-1/2 - (9996)-1/2 ]
çè 7.3 ´ 10-7 ø÷
E = 14.9 ´ 107[0.01186 - 0.01000]
E = 2.77 ´ 105 sec
or,
E =
2.77 ´ 105
= 77 hours
3600
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student using this manual is using it without permission.
6.28
Di = Q¥ S CD, i = q¥ S
CL 2
π e AR
In steady level flight, L = W, and we have
æ L ÷ö2
æ W ÷ö2
÷÷ = çç
CL2 = ççç
÷÷
çè q¥ S ÷ø
èçç q¥ S ø÷
Substitute (2) into (1).
æ W ÷ö2 1
÷
Di = Q¥ S ççç
çè Q¥ S ÷÷ø π e AR
But AR =
b2
, hence Eq. (3) becomes
S
æ W ÷ö2 1 æ S ö
çç ÷÷
÷÷
Di = q¥S ççç
÷
èç d¥ S ÷ø πe èç b2 ø
Di =
6.29 (a)
q¥ =
1
1 æç W
ç
π eq¥ çè b
2
2 ρ¥V¥
=
q¥ = 198.26 1b/ft
1
ö÷2
÷÷ QED
ø
-3 éê æç 88 ö÷
2ù
÷÷ (300) ú
2 (2.0482 ´10 ) ê èçç
ú
ø
ë 60
û
2
b = 37 ft
W = 10,100 1b
Di =
1 çæ W
ç
π eq¥ çè b
2
æ
ö2
1
÷÷ö =
çç 10,100 ÷÷
ø÷
π (0.8)(198.26) èç 37 ø÷
Di = 149.5 lb
(b)
CDi =
CL2
π e AR
CL =
W
10,100
=
= 0.2186
q¥ S
(198.26)(233)
AR =
b2
(37)2
=
= 5.87
S
233
CL2
(0.2186) 2
=
= 0.003245
π eAR
π (0.8)(5.86)
Di = q¥ S CDi = (168.26)(233)(0.003245)
Di = 149.9 lb
CDi =
The results from parts (a) and (b) agree to within 3-place round off error on the author’s
hand calculator.
80
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6.30 (a) From (6.45), copied below, for rate-of-climb at the climb angle θ,
T = D + W sin θ
we have
T
D
=
+ sin θ
W
W
Assume L = W. Thus,
T
D
=
+ sin θ
W
L
or,
T
1
=
+ sin θ
W
L/D
(b) At a standard altitude of 31,000 ft, T∞= 408.3 R.
a∞ =
γRT∞ =
(1.4)(1716)(408.3) = 990 ft/sec
V∞ = M ∞ a∞ = 0.85(990) = 841.5 ft/sec.
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,
sin θ =
R /C
5
=
+ 5.94 ´ 10-3
841.5
V∞
Thus,
θ = 0.34 - very small.
Hence,
T
1
1
=
+ sin θ =
+ 5.94 ´ 10-3
W
L /D
18
or
T
= 0.0615
W
Thus
T = 0.0615 W = 0.0615(550, 000) = 33,825 lb.
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 31, 000 ft, ρ∞ = 8.5841 ´ 10-4 slug/ft 3. Assuming the engine thrust is proportional to
density, as discussed in Section 6.7, we have for the thrust at sea level,
æ 0.002377 ÷ö
T = çç
÷ 33825 = 93, 666 lb.
çè 8.5841 ´ 10-4 ÷ø
This is comparable to the 84,000 lb sea level static thrust of the particular Trent engine
mentioned in the problem.
81
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student using this manual is using it without permission.
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.
CD = 0.016(2.3) = 0.368 at Mach 1.
Let us assume standard sea-level conditions, where ρ∞ = 1.23 kg/m3 and a∞ = 340.3 m/sec.
At march 1, V∞ = a∞ = 340.3 m/sec. Thus,
q¥ = 1 2 p¥V¥ = 1 2 (1.23)(340.3)2 = 7.12 ´ 104 N/m 2
D = q¥ S CD = (7.12 ´ 104 )(27.84)(0.0368) = 7.3 ´ 104 N
The weight is given as 8,273 kgf; in terms of newtons
W = (8273)(9.8) = 8.11 ´ 104 N
Hence,
D + W = (7.3 + 8.11) ´ 104 = 1.541 ´ 105 N
Compare this with the thrust at sea level,
T = 131.6 k N = 1.316 ´ 105 N
For this case, T < D + W . The airplane can not accelerate through Mach one going straight
up at sea level.
Since D ∝ ρ∞ and T ∝ ρ∞ , 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
At 2000 m, T¥ = 275.16 K and ρ¥ = 1.0066 kg/m3
a¥ =
γRT¥ =
(1.4)(287)(275.16) = 332.5 m/sec.
V
100
M¥ = ¥ =
= 0.3
a¥
332.5
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
q¥ = 1 2 ρ¥ V¥
= 1 2 (1.0066)(100)2 = 5033 N/m 2
D = q¥ S C D = (5033)(27.87)(0.016) = 2243 N
The acceleration is obtained from Newton’s 2nd Law, F = ma. The net force acting on the
airplane in the upward direction is
T - W - D = 131.6 ´ 103 - (8273)(9.8) - 2.243 ´ 103 = 131.6 ´ 103
- 81.08 ´ 103 - 2.243 ´ 103 = 48.28 ´ 103 N
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,
a=
F
48.28 ´ 103
=
= 5.84 m/sec2
m
8273
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6.33 From the definition of TSFC, we have
W f = (TSFC) T t = (TSFC) D t =
w
(TSFC) t
L/ D
(1)
Also, the new thrust specific fuel consumption results in a new fuel weight
W f , new =
=
Wnew
L
D
(TSFC)(1 - ε f ) t
W (1 + ε w )
L D
(TSFC) (1 - ε f ) t
(2)
Substituting (1) into (2), we have
W f ,new = W f (1 + ε w )(1 - ε f )
(3)
Increase in engine weight = Wnew - W = W (1 + ε w ) - W = ε wW
Decrease in fuel weight = W f - W f , new = W f - W f (1 + ε w )(1- ε f ) (using Eq.(3))
At the break-even point, we have
Wnew - W = W f - W f , new
ε wW = W f - W f (1 + ε w )(1 - ε f )
εw
W
= 1 - (1 + ε w )(1 - ε f )
Wf
εw
W
= -ε w + ε f + ε wε f
Wf
Since ε w and ε f are very small fractions, we can ignore the product ε wε f . Thus,
æ
ç
ççè
ε f = ε w çç1 +
W
Wf
ö÷
÷÷
÷
ø÷
and from Eq. (1), we have
εf
é
ù
L
ê
ú
ú
D
= ε w êê 1 +
ú
(TSFC) t ú
ê
êë
úû
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6.34 Weight at beginning of flight = 1.35 ´ 106 1b.
Weight at end of flight = 1.35 ´ 106 - 0.5 ´ 106 = 0.85 ´ 106 lb.
1.35 ´ 106 + 0.85 ´ 106
= 1.1 ´ 106 lb.
2
From the result of Problem 6.33, we have
Average weight during flight =
εw =
εf
1+
W
Wf
where ε f = 0.01, W = 1.1 ´ 106 1b, and W f = 0.5 ´ 106 lb.
0.01
εw =
1+
1.1 ´ 10
6
=
0.01
= 0.00313
3.2
0.5 ´ 106
Wnew = W (1+ε w ) =1.1 ´ 106 (0.00313) = 3438 1b.
The weight increase allowed for each engine is therefore
3438
= 859 1b.
4
6.35

C2 
TA = D = q∞ S C D = q∞ S  CD,0 + L 
π eAR 


C2 
PA = TAV∞ = q∞ S V∞  CD,0 + L 
π eAR 

CL =
W
q∞ S


W 2V∞
W2
PA = q∞ S V∞  C D,0 +
 = q∞ S V∞ CD,0 + q S π eAR
2
q∞ S π eAR 

∞
CD,0
PA
(W/S )
= 1 2 ρ∞V∞3
+
W
(W/S ) 1 2 p∞V∞π eAR
CD,0
PA
2(W/S )
= 1 2 ρ∞V∞3
+
W
(W/S ) p∞V∞π eAR
(1)
PA = η P where P is the shaft power from the engine.
At maximum velocity, P = Pmax. Equation (1) then becomes
η Pmax
W
=
CD ,0
1
2 (W/S)
3
+
ρ∞ Vmax
2
(W/S) ρ∞Vmax π eAR
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6.36 For the CP-1,
Pmax = 230 hp = (230)(550) = 1.265 × 105 ft lb/sec
Pmax / W = 42.88 ft/sec
η = 0.8
W = 2950 lb
CD ,0 = 0.025
W/S = 2950 /174 = 16.954 lb/ft 2
e = 0.8
AR = b2 /S = (35.8) 2 /174 = 7.366
ρ∞ = 0.002377 slug/ft 3 at sea level
From the equation obtained as the result of Problem 6.35,
η Pmax
W
=
CD,0
1
2 (W/S )
3
+
ρ∞ Vmax
2
(W/S ) ρ∞Vmax π eAR
3
0.8(42.88) = 1 2 (0.002377) Vmax
3 +
34.304 = 1.75254 × 10−6 Vmax
(0.025)
2(16.954)
+
16.954 (0.002377)Vmax π (0.3)(7.366)
770.55
Vmax
(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.
34.304 = 1.75254 × 10 −6 (260)3 +
770.55
265
= 32.614 + 2.9077 = 35.52
Difference = 1.216
Let us try a slightly smaller value, say Vmax = 260 ft/sec.
34.304 = 1.75254 × 10 −6 (260)3 +
770.55
260
= 32.80 + 2.96 = 33.76
Difference = −0.544
Try Vmax = 262 ft/sec.
34.304 = 1.75254 × 10 −6 (262)3 +
770.55
262
= 31.519 + 2.941 = 34.46
Difference = 0.156
Try Vmax = 261 ft/sec.
34.304 = 1.75254 × 10−6 (261)3 +
770.55
261
= 31.159 + 2.952 = 34.111
Difference = −0.192
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Try Vmax = 261.6 ft/sec.
34.304 = 1.75254 × 10 −6 (26.61)3 +
770.55
261.6
= 31.3748 + 2.9455 = 34.32
Close enough! The analytical value for Vmax is
Vmax = 261.6 ft/sec
Compared with the numerical value of 265 ft/sec, a 1.3% difference.
6.37 For the CJ-1
TA = (3650) 2 = 7300 lb
W = 19,815 lb
TA
7300
=
= 0.368
W 19,815
S = 318 ft 2
W 19,815
=
= 62.3 lb/ft 2
S
318
CD ,0 = 0.02
e = 0.81
AR = (53.3) 2 /318 = 8.934
From Eq. (6.44)
Vmax
2 4C

  TA   W  +  W   TA  − D ,0
 W   S   S   W 
π eAR
=
ρ∞ CD,0



1/ 2







1/ 2


4(0.02)
2
 (0.368)(62.3) + (62.3) (0.368) −

(0.81)(8.934)
π

=


(0.002377)(0.02)




= 978.8 ft/sec
This is to be compared with the numerical value of 975 ft/sec, a 0.3% difference.
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6.38 From Eq. 6.53:
 η P
− 0.8776
( R/C )max = 
 W  max
W
S
1
/2
ρ∞ cD,0 ( L / D )3max
 0.001648 
= 159.5 hp
For the CP-1, at 12,000 ft, hp = 230 
 0.002377 
 η P  (0.8)(1595) (550)
= 23.79

=
W 
2950
W 2950
=
= 16.954
s
174
ρ∞ CD,0 = (0.001648)(0.025) = 4.12 × 10−5
( L/D) max =
(CD,0π eAR)1/ 2
2 CD,0
=
[(0.025) π (0.8)(7.366)]1/2
= 13.606
2(0.025)
16.954
1
5
−
6.12 × 10
(13.606)3 / 2
1
= 23.79 − 0.8776(641.49)
50.188
= 23.79 = 11.217 = 12.573 ft/sec
(R/C )max = (12.573)(60) = 754.4 ft/min
( R/C )max = 23.79 − 0.8776
The numerical result from Section 6.10 is 755 ft/min. The difference is 0.08%.
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6.39 From Eq. (6.52)
1/ 2
( R/C )max
 (W/S ) z 
=

 3 ρ∞ CD ,0 
3/ 2 

2
3
T
1 − −

 
2
2
W max  6 2(T/W )max ( L / D)max Z 
For the CJ-1
(53.3) 2
= 8.934
318
W/S = 19,815 / 318 = 62.31
AR =
CD,0 = 0.02
(3650)(2)  0.0011043 
T
=
 

 = 0.171
19.815  0.002378 
W max
(CD ,0π eAR)1/ 2 [(0.02)π (0.81)(8.934)]1/ 2 0.673
 L
=
=
=
= 16.858
 
2 CD ,0
2(0.02)
0.04
D max
Z = 1+ 1+
= 1+ 1+
3
( L/D)2max (T/W ) 2max
3
(16.858)2 (0.1711) 2
= 1 + 1 + 0.3606 = 2.166
1/ 2
( R/C )max
 (62.31)(2.166) 
(0.1711)3/ 2 ×
=

 3(0.0011043)(0.02) 
 2.166

3
−
1 −

2
2
6
2(0.1711) (16.858) 2.166 

= (2.0369 × 106 )1/ 2 (0.07077)(0.5558)
= 56.14 ft/sec = (56.14)(60) ft/min = 3368 ft/min
This is compared to the numerical answer of 3369 ft/min given in Section 6.10.
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
C2 
6.40 T = D = 1 2 ρ∞V∞2 S  C D,0 + L 
π eAR 

(1)
 88 
Vmax = 229 mph = 229   = 335.87 ft/sec.
 60 
ft lb
 0.0018975 
(0.9) = 9.4795 × 105
Pmax = Pη = (1200)(550) 
 0.002378 
sec
Tmax =
Pmax 9.4795 × 105
=
= 2822 lb
335.87
Vmax
S = 987 ft 2
AR = 9.14
e = 0.8
1 2 ρ∞V∞2 S = 1 2 (0.0018975)(335.87)2 (987) = 1.05636 × 105 lb
CL =
CL2
π eAR
=
W
25, 000
=
= 0.2366
5
1
1.05636
×
10
2
ρ V S
2 ∞ ∞
(0.23666) 2
= 0.0024382
π (0.8)(9.14)
From EQ. (1),
2822 = 1.05636 × 105 (CD,0 = 0.0024382)
CD ,0 + 0.0024382 =
2822
= 0.0267
1.05636 × 105
CD ,0 = 0.0243
Note: This answer is consistent with the value of CD,0 for the DC-3 given in Figure 6.77.
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Chapter 7
7.1
CM cg = CM ac + CL (h - hac )
CM ex = CM cg - CL (h - hac ) = 0.005 - 0.05(0.03) = -0.01
7.2
1
1
2
ρ¥V¥
= (1.225)(100)2 = 6125 N/m 2
2
2
At zero lift, the moment coefficient about c.g. is
q¥ =
CM cg =
M cg
q¥ S c
=
-12.4
= -0.003
(6125)(1.5)(0.45)
However, at zero lift, this is also the value of the moment coefficient about the a.c.
CM ac = -0.003
At the other angle of attack,
CL =
CM cg =
and
L
3675
=
= 0.4
q¥ S
(6150)(1.5)
M cg
q¥ S c
=
20.67
= 0.005
(6125)(1.5)(0.45)
Thus, from Eq. (7.9) in the text,
CM cg = CM ac + CL (h - hac )
Thus,
h - hac =
CM cg - CM ac
CL
=
0.005 - (-0.003)
0.4
h - hac = 0.02
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,
q¥ = 6125 N/m 2
CM ac = -0.003
h - hac = 0.02
In the present problem, the c.g. has been shifted 0.2c rearward. Hence,
h - hac = 0.02 + 0.2 = 0.22
Also, CL =
L
q¥ S
=
4000
= 0.435
(6125)(1.5)
Thus, CM cg = CM ac + CL (h - hac )
CM cg = -0.003 + 0.435(0.22) = 0.0927
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7.4
From Problem 7.2, we know
q¥ = 6125 N/m 2
CM ac = -0.003
h - hac = 0.02
From information provided in the present problem, at α a = 5,
CL =
L
4134
=
= 0.45
q¥ S
(6125)(1.5)
Hence,
a =
Also, VH =
dCL
0.45
=
= 0.09 per degree
dα
5
 t St
(1.0)(0.4)
=
= 0.593
cS
(0.45)(1.5)
From Eq. (7.26) in the text,
é
a æ
¶ε ÷ö ùú
CM cg = CM ac + aα a ê ( h - hac ) - VH t çç1 ÷ + VH at (it + ε o )
ç
ê
¶α ÷ø úû
aè
ë
é
ù
æ 0.12 ÷ö
(1 - 0.42) ú + (0.593)(0.12)(2.0)
CM cg = -0.003 + (0.09)(5) ê (0.02) - (0.593) çç
çè 0.09 ÷÷ø
ê
ú
ë
û
CM cg = -0.052
Hence, the moment is
M cg = q¥ S c CM cg = (6125)(1.5)(0.45)(-0.052)
M cg = -215 Nm
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7.5
CM cg
é
a æ
¶ε ÷ö ùú
= a ê (h - hac ) - VH t çç1 ÷
ê
¶α a
¶α ÷ø úû
a èç
ë
where, from Problems 7.4 and 7.2,
a = 0.09 per degree
h - hac = 0.02
CM ac = -0.003
VH = 0.593
at = 0.12
¶ε
= 0.42
¶α
it = 2.08
CM cg
é
ù
æ 0.12 ÷ö
ú = -0.039
= (0.09) ê (0.02) - (0.593) çç
(1
0.42)
÷
ê
ú
èç 0.09 ÷ø
¶α a
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 CM 0 .
Thus,
CM 0 = CM ac + VH at (it + ε 0 ) = -0.003 + (0.593)(0.12)(2.0 + 0) = 0.139
The trim angle of attack can be found from
CM cg
αe = 0
¶α
α e = -CM 0 /(¶CM cg / ¶α ) = -0.139 /(-0.039) = 3.56
CM cg = CM 0 +
7.6
This is a reasonable angle of attack, falling within the normal flight range. Hence, the airplane
model is also balanced.
a æ
¶ε ÷ö
hn = hacwb + VH t çç1 ÷
ç
¶ε ÷ø
aè
From Problem 7.2, hn - hacwb = 0.02
Hence,
hacwb = h - 0.02 = 0.26 - 0.02 = 0.24
æ 0.12 ÷ö
hn = 0.24 + 0.593çç
(1 - 0.42)
çè 0.09 ÷÷ø
hn = 0.70
By definition,
static margin = hn - h = 0.70 - 0.26 = 0.44
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7.7
δ trim =
CM 0 + (¶CM cg / ¶α a )α n
VH (¶CL1 / ¶δ e )
CM 0 = 0.139 (from Problem 7.5)
¶CM cg / ¶α a = -0.039 (from Problem 7.5)
α n = 8°
VH = 0.593 (from Problem 7.4)
¶CLt / ¶δ e = 0.04 (given)
Thus, δ trim =
7.8
0.139 + (-0.039)(8)
= -7.29°
(0.593)(0.04)
F = 1-
1 æç ¶CL1
ç
at ççè ¶δ e
F = 1-
1
(-0.007)
(0.04)
= 0.806
0.12
(-0.012)
¢ = CM
CM
0
ac
ab
÷÷ö ( ¶Che / ¶α1 )
÷÷
ø ( ¶Che / ¶δ e )
+ F VH at (it + ε 0 )
¢ = -0.003 + (0.806)(0.593)(0.12)(2 + 0) = 0.112
CM
0
This is to be compared with CM 0 = 0.139 from Problem 7.5 for stick-fixed stability.
a æ
¶ε ÷ö
hn¢ = hacwb + F VH 1 çç1 ÷
a çè
¶α ø÷
(0.12)
hn¢ = 0.24 + (0.806)(0.593)
(1 - 0.42) = 0.609
(0.09)
This is to be compared with hn = 0.70 from Problem 7.6 for stick-fixed stability.
hn¢ - h = 0.609 - 0.26 = 0.349
Note that the static margin for stick-free is 79% of that for stick-fixed.
¢
¶ CM
cg
¶α a
= -a(hn¢ - h) = -(0.09)(0.349) = -0.031
This is to be compared with a slope of −0.039 obtained from Problem 7.5 for the stick-fixed case.
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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.
M cg, t =  t Lt
(1)
Thus, Eq. (7.17) becomes
CM , cg, t = VH CL,t
(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
αt = α wb + it
(3)
Therefore, Eq. (7.22) is replaced by
CM , cg, t = VH atα wb + VH at it
(4)
In turn, Eq. (7.24) for the total pitching moment becomes
CM , cg = CM ,acwb + CLwb (h - hacwb ) + VH CL, t
(5)
Eq. (7.25) becomes
CM , cg = CM , acwb + awbα wb[(h - hacwb ) + VH
at
] + VH at it
awb
(6)
Differentiating Eq. (6) with respect to angle of attack, the form that replaces Eq. (7.28) is
¶CM , cg
¶α a
For static stability,
é
a ù
= a ê h - hacwb + VH t ú
êë
a úû
¶ CM , cg
¶αa
(7)
must be negative.
From Eq. (7), therefore,
(h - hac wb ) + VH
or,
(h - hac wb ) < VH
at
<0
a
at
a
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.
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Chapter 8
8.1
hG = 400 miles = 0.644 ´ 106
rb = re + hG = 6.4 ´ 106 + 0.644 ´ 106 = 7.044 ´ 106 m
k 2 = 3 / 986 ´ 1014 m3/sec2 from the text
h = r 2θ = rV θ = rbV cos βb = (7.044 ´ 106 )(13 ´ 103 ) cos10°
h = 9.018 ´ 1010 m 2 /sec
h2 = 8.133 ´ 1021 m 4 /sec2
p = h 2 / k 2 = 8.133 ´ 1021 / 3.986 ´ 107 m
This is the numerator of the orbit equation. We now proceed to find the eccentricity. The kinetic
energy per unit mass is:
T / m = V 2 /2 = (13 ´ 103 )2 /2 = 8.45 ´ 107 m 2 /sec2
The potential energy per unit mass is:
φ / m = k 2 / rb = 3.98 ´ 1014 / 7.044 ´ 106 = 5.659 ´ 107 m 2 /sec
Hence,
H / m = ( T - φ ) / m = (8.45 - 5.659) ´ 107 = 2.791 ´ 107 m 2 /sec2
Thus, the eccentricity is:
e = 1+
2h2 æç H ö÷
2(8.133 ´ 1021)(2.791 ´ 107 )
=
+
= 1.96
1
÷
ç
÷
k 4 çè m ø
(3.986 ´ 1014 )2
Obviously, the trajectory is a hyperbola because e > 1, and also because T > | φ |. The orbit
equation is
r =
p
1 + e cos(θ - C )
r =
2.04 ´ 107
1 + 1.96 cos(θ - C )
The phase angle, C, is calculated as follows. Substitute the burnout location
(rb = 7.044 ´ 106 m and θ = 0°) into the above equation.
2.04 ´ 107
1 + 1.96 ´ cos(-C )
cos (-C ) = 0.967
95
7.044 ´ 106 =
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Thus C = -14.76°.
Hence, the complete equation of the trajectory is
r =
2.04 ´ 107
1 + 1.96 cos (θ + 14.76)
where θ is in degrees and r in meters.
8.2
Escape velocity = V =
For Venus:V =
(2)(3.24 ´ 1014 ) / 6.16 ´ 106 = 1.03 ´ 104 m/sec = 10.3 km/sec
For Earth:V =
(2)(3.96 ´ 1013 ) / 6.39 ´ 106 = 1.11 ´ 104 m/sec = 11.3 km/sec
For Mars:V =
(2)(4.27 ´ 1013 ) /106 = 5.02 ´ 103 m/sec = 5.02 km/sec
For Jupiter:V =
8.3
2k 2 / r
(2)(1.27 ´ 1017 ) / 7.14 ´ 107 = 5.96 ´ 104 m/sec = 56.6 km/sec
G = 6.67 ´ 10-11 m3/(kg)(sec)2
M = 7.35 ´ 1022 kg
GM = k 2 = (6.67 ´ 10-11)(7.35 ´ 1022 ) = 4.9 ´ 1012 m3/sec2
Orbital velocity is V =
Vorbital =
k 2 /r
4.9 ´ 1012 /1.74 ´ 106 = 1678 m/sec = 1.678 km/sec
Escape velocity is larger by a factor of (2)1/2.
Vescape =
8.4
2(1.678) = 2.37 km/sec
From Kepler’s 3rd law,
æ τ ö÷2
æ ö3
çç 1 ÷ = çç a1 ÷÷
÷
ççè τ 2 ø÷÷
èçç a2 ø÷
a23 = (a1)3 (τ 2 / τ1)2
a2 = (a1)3 (τ 2 / τ1)2/3
a2 = (1.495 ´ 1011)(29.7 /1.0)2/3 = 1.43 ´ 1012 m
Please note: The “distant planet” is in reality Saturn.
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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
æ 4π 2 ÷ö
æ 2ö
÷÷ a3 = çç 4π ÷÷÷ r 3
ç 2 ÷
çè k 2 ÷÷ø
èç k ø÷
τ 2 = ççç
ç
(Remember, the orbit is a circle.)
Hence,
æ k 2 ÷ö1/3
ç
r = çç 2 ÷÷÷ τ 2/3
çè 4π ÷ø
æ 3.956 ´ 1014 ö÷1/3
ç
÷÷ (8.64 ´ 104 ) 2/3 = 4.21 ´ 107 m
r = çç
÷
çè
4π 2
ø÷
The radius of the Earth is 6.4 × 106 m. Hence, the altitude above the surface of the Earth is
hG = 4.21 ´ 107 - 6.4 ´ 106 = 3.57 ´ 107 m = 35, 700 km
V =
Circular velocity is
8.6
k2
=
r
3.956 ´ 1014
4.21 ´ 107
= 3065 m/sec
æ4
ö
æ4 ö
m = ρ v = ρ çç π r 3 ÷÷ = (6963) çç π ÷÷ (0.8)3 = 1.4933 ´ 104 kg
÷ø
çè 3
çè 3 ÷ø
S = π r 2 = π (0.8) 2 = 2.01 m 2
m
1.4933 ´ 104
=
= 7429 km/m3
(1)(2.01)
CD S
Z = g 0 /RT = 9.8 /(287)(29/88) = 0.000118 m-1
From Eq. (8.97), the density at the altitude for maximum deceleration is
ρ =
m
kg
Z sin θ = (7429)(0.000118) sin 30° = 0.4383 3
CD S
m
From Eq. (8.73)
h =-
æ 0.4383 ö÷
1
1
n(ρ /ρ0 )= n çç
÷ = 8710 m
0.000118 çè 1.225 ø÷
Z
From Eq. (8.100)
dV
V 2 Z sinθ
(8000)2 (0.000118)(sin 30°)
= E
=
= 694.6 m/sec 2
2e
2e
dt max
In terms of gs:
dV
694.6
=
= 70.88 gs
9.8
dt max
From Eq. (8.87)
V/VE =
é
ù
(1.225)
ú
-êê
ú
´
°
2(7429)
(0.000118)
sin
30
ê
ë
ûú
e
= 0.247
V = 0.247 VE = 0.247(8000) = 1978 m/sec
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8.7
3
Aerodynamic heating varies as V¥
. Hence:
æ 36, 000 ö÷3
q2
÷ = 2.37
= çç
çè 27, 000 ø÷÷
q1
q2 = (100)(2.37) = 237 Btu/(ft 2 )(sec)
8.8
Assume the mean velocity of the earth around the sun is essentially its circular orbital velocity
given by
Vearth =
k2
= 29.77 ´ 103 m/sec
r
where k2 = GM, and M is the mass of the sun. From Eq. (1)
k 2 = V 2r = (29.77 ´ 103 )2 (147 ´ 109 ) = 1.30 ´ 1020 m3/sec2
The asteroid is moving at 0.9 times the escape velocity from the sun.
Vasteroid = 0.9
2k 2
r
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,
Vasteroid = 0.9
2(1.3 ´ 1020 )
(147 ´ 109 )
= 37.850 ´ 108 m/sec
The relative head-on collision velocity, which is the velocity at which the asteroid would enter
the earth's atmosphere, is
Ventry = Vearth + Vasteroid = 29.77 ´ 103 + 37.85 ´ 103 = 67.62 ´ 103 m/sec
= 67.62 km/sec
8.9
Earth’s radius = R = 6.4 ´ 106 m
Following the nomenclature in Figure 8.18,
rmin = R + (altiude above Earth’s surface)
rmin = 6.4 ´ 106 + 4.17 ´ 105 = 6.817 ´ 106 m
From Eq. (8.62)
rmin =
h2 / k 2
1+ e
(1)
h2 / k 2
1- e
(2)
and from Eq. (8.61)
rmax =
Combining (1) and (2)
rmax =
1.00132
1+ e
=
= 1.0026
1- e
0.99868
rmax = rmin (1.0026) = (6.817 ´ 106 )(1.0026) = 6.8347 ´ 106 m
Altitude at apogee = 6.8347 ´ 106 - 6.4 ´ 106 = 0.4347 ´ 106 m = 434.7 km
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8.10 From Eq. (8.71),
τ2 =
4π 2 3
a
k2
k 2 = 3.956 ´ 1014 m3/sec2
From the solution of Problem 8.9,
rmax = 6.8347 ´ 106 m
rmin = 6.817 ´ 106 m
r
+ rmin
(6.8347 + 6.817) ´ 106
a = max
=
= 6.82585 ´ 106 m
2
2
Inserting these numbers into Eq. (8.71):
τ2 =
4π2 3
4π 2 (6.82585 ´ 106 )3
a
=
3.956 ´ 1014
k2
τ 2 = 3.1738 ´ 107
τ = 5634 sec
τ =
or,
5634
= 1.56 hr
3600
8.11 The angular momentum per unit mass is h = r 2θ . To calculate h, proceed as follows.
From Eq. (8.62),
rmin =
h2 / k 2
1+ e
or,
h 2 = rmin (1 + e)k 2
From the solution of Problem 8.9, we have rmin = 6.817 ´ 106 m. Thus,
h 2 = (6.817 ´ 106 )(1.00132)(3.956 ´ 1014 ) = 2.7 ´ 1021
h = 5.1965 ´ 1010
or,
Because h = r 2θ , at perigee
θ =
5.1965 ´ 1010
=
= 1.1182 ´ 10-3 rad/sec
2
6 2
(rmin )
(6.817 ´ 10 )
h
Vperigee = rminθ = (6.817 ´ 106 )(1.1182 ´ 10-3 ) = 7623 m/sec= 7.623 km/sec
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8.12 (a)
Et =
V2
k2
r
2
Evaluate this equation at the perigee, where from Example 8.3, rp = 7.169 ´ 106 m ,
and V p = 9.026 km/sec
Et =
V p2
2
-
k2
(9.026 ´ 103 )2
3.986 ´ 1014
=
=
rp
2
7.169 ´ 106
4.0734 ´ 10-7 - 5.56 ´ 107 = -1.486 ´ 10-7
joules
kg
(b) From Example 8.3, a = 1.341 ´ 107 m. Thus
Et = -
k2
3.986 ´ 1014
joules
== -1.486 ´ 10-7
7
2a
kg
2(1.341 ´ 10 )
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, Vθ = 3432 m/sec.
ævö
ΔV = 2Vθ sin çç ÷÷÷ = 2(3432) sin 10° = 1191.9 m/sec
çè 2 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)
ævö
v
sin çç ÷÷÷ »
çè 2 ø
2
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8.14 From Example 8.7, we already have
V1 = 6794 m/sec
and
r 2 = r1 = 1.0506 ´ 107 m
a2 =
rp.2
1- e
=
107
= 5 ´ 107 m
1 - 0.8
Thus,
2k 2
k2
2(3.986 ´ 10)14
3.986 ´ 1014
- 2 =
r2
a
1.0506 ´ 107
5 ´ 107
V2 = 8241 m/sec
e1 sin θ A
(0.4654) sin90°
=
= 0.4654
Tan β1 =
1 + e1 cos θ A
1 + 0.4654 cos90°
V2 =
β1 = 24.957°
For orbit 2, the true anomaly of point A is found from
cos θ A =
cos θ A =
rp, z (1 + e2 )
e2 r2
-
1
e2
1 ´ 107 (1 + 0.8)
7
(0.8)(1.0506 ´ 10 )
-
1
= 2.1416 - 1.25 = 0.8916
0.8
θ A = 26.93°
Thus,
Tan β 2 =
e2 sin θ A
(0.8)(0.4529)
0.3623
=
=
= 0.2115
1 + e2 cos θ 2
1 + (0.8)(0.8916)
1.7133
β 2 = 11.94°
α = β1 - β 2 = 24.957 - 11.94 = 13.017°
From Eq. (8.89),
(ΔV ) 2 = V12 + V2 2 - 2V1V2 cosα
= (6794)2 + (8241) 2 - 2(6794)(8241) cos (13.017°)
= 14.379 ´ 107 - 10.91 ´ 107 = 3.469 ´ 107
Thus,
ΔV = 5890 m/sec
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8.15 From Example 8.8,
V1 = 7771 m/sec
and
r1 = 6.6 ´ 106 m
r2 = 6.4 ´ 106 + 5 ´ 105 = 6.9 ´ 106 m
r + r2
6.6 ´ 106 + 6.9 ´ 106
13.5 ´ 106
a = 1
=
=
= 6.75 ´ 106 m
2
2
2
From Eq. (8.78),
Vpt =
2k 2
k2
=
r1
a
2(3.986 ´ 10)14
6.6 ´ 10
6
-
(3.986 ´ 1014 )
6.75 ´ 106
= 1.2079 ´ 108 - 0.5905 ´ 108 = 7857.5 m/sec
At point 1, the required impulse to enter the Hohmann transfer orbit is
ΔV1 = Vpt - V1 = 7857.5 - 7771 = 86.5 m/sec.
At point 2 on orbit 2, the required spacecraft velocity is
V2 =
k2
=
r1
3.986 ´ 1014
6.9 ´ 106
= 7600 m/sec
At point 2 on the Hohmann transfer orbit,
Vat =
2k 2
k2
=
r1
a
2(3.986 ´ 1014 )
6.9 ´ 106
-
3.986 ´ 1014
6.75 ´ 106
= 1.1554 ´ 108 - 0.5905 ´ 108 = 7516 m/sec
At point 2,
ΔV2 = V2 - Vat = 7600 - 7516 = 84 m/sec
The total impulse required is
ΔV = ΔV1 + ΔV2 = 86.5 + 84 = 170.5m/sec
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8.16 From Problem 8.2, for Mars,
k 2 = 4.27 ´ 1013 m3/sec2
k2
=
r1
V1 =
a =
4.27 ´ 1013
8 ´ 106
= 2310 m/sec
r1 + r2
8 ´ 106 + 15 ´ 106
=
= 11.5 ´ 106 m
2
2
2k 2
k2
=
r1
a
Vpt =
2(4.27 ´ 1013
8 ´ 106
-
4.27 ´ 1013
11.5 ´ 106
= 1.0675 ´ 107 - 0.3713 ´ 107 = 2638.6 m/sec
At point 1,
ΔV1 = Vpt - V1 = 2638.6 - 2310 = 328.6 m/sec
At point 2 on orbit 2,
k2
V2 =
r2
4.27 ´ 1013
=
15 ´ 106
= 1687 m/sec
At point 2 on the Hohmann transfer orbit,
Vat =
=
2k 2
r2
-
k2
=
a
2(4.27 ´ 1013 )
15 ´ 106
-
4.27 ´ 1013
11.5 ´ 106
5.693 ´ 106 - 3.713 ´ 106 = 1407 m/sec
ΔV2 = V2 - Vat = 1687 - 1407 = 280 m/sec.
The total required impulse is
ΔV = ΔV1 + ΔV2 = 328.6 + 280 = 608.6 m/sec
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8.17 Relative to the center of mercury, at periapsis,
rmin = 2, 440 + 200 = 2640 km
rmax = 2, 440 + 15,193 = 17, 633 km
The semi-major axis of the elliptical orbit is
r
+r
2640 + 17, 633
a = min max =
= 10,137 km
2
2
= 1.0137 × 107 m

m3 
23
k 2 = GM =  6.67 × 10−11
 (3.3 × 10 kg)
2

kg sec 
k 2 = 2.20 × 1013
m3
sec2
From Eq. 8.71 in the text:
τ2 =
4 π 2 a3
k2
=
4 π 2 (1.0137 × 107 )3
2.20 × 1013
= 1.87 × 109 sec2
Hence,
τ = 43243 sec
Since 3600 sec = 1 hr, we have
τ=
43243
= 12.01 hr
3600
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,
V=
2k 2 k 2
−
r
a
For r = rmin, the velocity at periapsis is
V=
2(2.20 × 1013 )
2.64 × 106
−
2.20 × 1013
1.0137 × 107
= 16.67 × 106 − 2.17 × 106
= 3809 m/sec = 3.809 km/sec
For r = rmax, the velocity at apoapsis is
V=
2(2.20 × 1013 )
1.7633 × 10
7
− 2.17 × 106 = 2.495 × 106 − 2.17 × 106
= 570 m/sec = 0.57 km/sec
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8.19 From the solution of Problem 8.18 and 8.19, at periapsis V = 3809 m/sec and rmin = 2.64 × 106m.
Thus,
θ =
V
3809
=
= 1.44 × 10−3 rad/sec
rmin 2.64 × 106
The angular momentum per unit mass is h = r2 θ. Thus,
h = (rmin )2 θ = (2.64 × 106 )2 (1.44 × 10−3 )
h = 1.00 × 1010 m 2 / sec
The same value is obtained at the apoapsis, where
V
570
= 3.2326 × 10 −5 rad/sec
1.7633 × 107
h = (rmax )2 θ = (1.7633 × 107 )2 (3.2326 × 10 −5 )
θ =
rmax
=
h = 1.00 × 1010 m 2 / sec.
This, of course, is simply verification that the angular momentum is constant for the orbit.
8.20 Eq. (8.63) is
a=
h 2 /k 2
1 − e2
or
e = 1−
h 2 /k 2
a
From Problems 8.17 and 8.19,
h2
k2
=
(1.00 × 1010 )2
2.20 × 1013
e = 1−
= 0.4545 × 107
0.4545 × 107
1.0137 × 107
= 1 − 0.448 = 0.743
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.
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Chapter 9
9.1
Following the nomenclature in the text;
æ V ÷öγ
p3
= ççç 2 ÷÷ = 6.751.4 = 13.0
çè V3 ÷ø
p2
p3 = (13.0)(1.0) = 13.0 atm
æ V ÷öγ -1
T3
= ççç 2 ÷÷
= (6.75)0.4 = 2.15
÷
ç
T2
è V3 ø
T3 = (2.15)(285) = 613K
The heat released per kg of fuel-air mixture is established in the text as 2.43 ´ 106 joule/kg.
Hence,
T4 =
q
+ T3 where cv = 720 joule/kg°K
cv
2.43 ´ 106
+ 613 = 3988°K
720
p4
T
= 4
p3
T3
T4 =
æT ö
æ 3988 ö÷
Thus, p4 = p3 ççç 4 ÷÷÷ = (13.0) çç
÷ = 84.6 atm
ç
çè T3 ÷ø
è 613 ø÷
æ V öγ
æ 1 ö÷1.4
p5
= ççç 4 ÷÷÷ = çç
= 0.069
çè 6.75 ø÷÷
çè V5 ÷ø
p4
p5 = (84.6)(0.069) = 5.84 atm
Wcompression =
p2V2 - p3V3
1- γ
The volumes V2 and V3 can be obtained as follows
( x + 9.84) / x = 6.75, Thus, x = 1.71 cm
Thus,
V2 =
π b2
4
(9.84 + 1.71) where b = bore = 11.1 cm
V2 = 1118 cm3 = 1.118 ´ 10-3 m3
V3 = V2 / 6.75 = 1.118 ´ 10-3 / 6.75 = 1.66 ´ 10-4 m3
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Thus,
p2V2 - p3V3
1-γ
é (1.0)(1.118 ´ 10-3 ) - (13.0)(1.66 ´ 10-4 ) ù 1.01 ´ 105
ê
úû
= ë
= 263 joule
-0.4
p V - p4V4
= 5 5
1-γ
é (5.84)(1.118 ´ 10-3 ) - (84.6)(1.66 ´ 10-4 ) ù 1.01 ´ 105
ê
úû
= ë
-0.4
= 1897 joule
Wcompression =
Wpower
Wpower
W = Wpower - Wcomperssion = 1897 - 263 = 1634 joule
1
ηηmech (RPM) N W
120
(0.85)(0.83)(2800)(4)(1634)
PA =
= 1.076 ´ 105 watts
120
PA =
PA =
or
9.2
PA =
1.076 ´ 105 watts
= 144 hp
746 watts/hp
1
ηηmech (RPM) Dpe
120
where D =
π b2
4
s N=
π (11.1)2 (984)(4)
4
D = 3809 cm = 3.809 ´ 10-3 m3
3
pe =
pe =
120 PA
ηη mech (RPM) D
120(1.076 ´ 105 )
(0.85)(0.83)(2800)(3.809 ´ 10-3 )
pe = 1.72 ´ 106 N/m 2 = 17 atm
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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,
ρe =
pe
1.01 ´ 105
=
= 0.469 kg/m3
RTe
287 (750)
m = ρ eAeVe = (0.469)(0.45)(400) = 84 kg/sec
At the inlet, assume standard sea level conditions.
m = ρ¥ AV
i i = 84 kg/sec
Hence,
Vj =
m
84
=
= 152 m/sec
(1.225)(0.45)
ρ¥ Ai
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 Vi . To obtain the thrust,
T = m (Ve - V j ) + ( pe - p¥ ) Ae = 84(400 - 152) + (0) Ae = 20,832 N
Since 1 lb = 4.448 N, we also have
T =
9.4
20,832
= 4684 lb
4.448
At a standard altitude of 40,000 ft,
p¥ = 393.12 lb/ft 2
ρ¥ = 5.8727 ´ 10-4 slug/ft 3
The free stream velocity is
æ 88 ö
V¥ = 530 çç ÷÷÷ = 777 ft/sec
çè 60 ø
Hence,
m = ρ¥V¥ A j = (5.87 ´ 10-4 )(777)(13) = 5.93 slug/sec
T = m (Ve - V¥ ) + ( pe - p¥ ) Ae
T = (5.93)(1500 - 777) + (450 - 393)(10)
T = 4287 + 570 = 4587 lb
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9.5
Assume the Mach number at the end of the diffuser (hence at the entrance to the compressor) is
close to zero, hence p2 can be assumed to be total pressure.
M1 =
V1
800
=
= 0.716
1117
a1
γ /γ -1
æ
p2
γ - 1 2 ÷ö
M1 ÷÷
= çç1 +
= [1 + (0.2)(0.616) 2 ]3.5 = 1.41
çè
ø
p1
2
Hence,
p3
p p
= 3 2 = (12.5)(1.41) = 17.6
p1
p2 p1
As demonstrated on the pressure-volume diagram for an ideal turbojet, the compression process
is isentropic. Hence,
æT ö
p3
= ççç 3 ÷÷÷
çè T1 ø÷
p1
γ
γ -1
æp ö
t3 = t1 ççç 3 ÷÷÷
çè p1 ø÷
γ
γ -1
= 519(17.6)0.286 = 1178°
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
(1.4 ´ 107 )(32.3) = 4.51 s 108 ft lb/slug . Hence, the heat released per slug of fuel-air
mixture is
q =
(4.51 ´ 108 )(0.05)
= 2.15 ´ 107 ft lb/slug
1.05
Because the heat is added at constant pressure,
q = c p (T4 - T3 )
q
cp
T4 = T3 +
For air, c p =
γR
γ -1
=
T4 = 1178 +
1.4(1716)
ft lb
= 6006
0.4
slug°R
2.15 ´ 107
= 1178 + 3580 = 4758° R
6006
Again, from the p-v diagram for an ideal turbojet,
p6
p
1
= 1 =
= 0.0568
17.6
p4
p3
γ
Also,
æ T öγ -1
p6
= ççç 6 ÷÷÷
çè T4 ø÷
p4
γ
æ p öγ -1
T6 = T4 ççç 6 ÷÷÷
= 4758 (0.0568)0.286 = 2095° R
çè p4 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.
109
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9.6
T = m (Ve - V¥ ) + ( pe - p¥ ) Ae
1000 = m (2000 - 950) + 0
m = 0.952 slug/sec
However, m = ρ ¥V¥ Ai
m
0.952
Thus, Ai =
=
= 0.42 ft 2
(0.002377)(950)
ρ¥V¥
9.7
At 50 km, p¥ = 87.9 N/m 2
 e + ( pe - p¥ ) Ae = (25)(4000) + (2 ´ 104 - 87.9)(2)
T = mV
= 100, 000 + 39,824 = 139,824 N
Since 1 lb = 4.48 N
139824
= 31, 435 lb
4.448
T =
9.8
(a) At a standard altitude of 25km,
p¥ = pe = 2527 N/m 2
I sp
I sp
0.18 ù
é
æ 2527 ÷ö1.18 ú
2(1.18)(8314)(3756) êê
ú
ç
÷
ê 1 - ççè
ú
6 ÷ø
(0.18)(20)
´
3.03
10
ê
ú
êë
úû
= 375sec
1
=
g0
(b) From the isentropic relations,
γ
æ T öγ -1
pe
= ççç e ÷÷÷
çè T0 ø÷
p0
γ -1
æp ö γ
Te = T0 ççç e ÷÷÷
çè p0 ø÷
cp =
æ 2527 ÷ö0.1525
= 3756 çç
= 1274°K
÷
çè 3.03 ´ 106 ÷ø
(1.18)(8314)
γR
=
= 2725joule/kg°K
γ -1
(0.18)(20)
From the energy equation
Ve =
(c)
pe =
2c p (T0 - Te ) =
2(2725)(3756 - 1274) = 3678 m/sec
pe
R
8314
joule
where R =
=
= 415.7
20
kgK
RTe
M
2527
= 0.00477 kg/m3
(415.7)(1274)
m = pe AeVe =(0.00477)(15)(3678)=263.5 kg/sec
pe =
110
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(d) From the definition of specific impulse,
T
where w is the weight flow
w
 0 = (263.5)(9.8) = 2582 N/sec
w = mg
Hence,
T = w Isp = (2582)(375) = 968, 250N
I sp =
968250
= 217682 lb
4.448
or, T =
(e)
m =
263.5 =
(γ +1) /(γ -1)
p0 A* γ æç 2 ö÷
÷÷
çç
To R è γ + 1 ø÷
12.11
1.18 çæ 2 ÷ö
÷
ç
415.7 çè 2.18 ø÷
(30(1.01 ´ 105 )) A*
(3756)1/ 2
A* = 0.169 m 2
9.9
m = 87.6 / 32.2 = 2.72 slug/sec
(γ +1) /(γ -1)
p0 A* γ æç 2 ö÷
ç
÷÷
T0 R èç γ + 1 ø÷
m =
p0(0.5)
6000
2.72 =
10.52
1.21 æç 2 ö÷
÷÷
çç
2400 è 2.21 ø
p0 = 31, 736 lb/ft 2
In terms of atmospheres,
p0 =
9.10 Vb = g0 I sp n
9.11
31, 736
= 15 atm
2116
Mi
= (9.8)(240) n 5.5 = 4009.6 m/sec
Mf
M f = Mi - M p
Mi
Mi
=
=
Mf
Mi - M p
1-
1
Mp
Mi
Mi
V /g I
= e b 0 sp = e(11,200) /(9.8)(360) = 23.9
Mf
where escape velocity is 11.2 km/sec = 11,200 m/sec.
23.9 =
1Mp
Mi
1
Mp
Mi
= 0.958
111
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9.12 From Eq. (9.43)
r = ap0n
log r = log a + n log p0
At p0 = 500 lb/in 2: log (0.04) = log a + n log (500)
log a = -1.3979 - 2.69897n
or:
(1)
At p0 = 1000 lb/in 2: log (0.058) = log a + n log (1000)
log a = -1.23657 - 3n
or:
(2)
Solving Eqs. (1) and (1) for a and n, we have
0 = -0.16133 + 0.03103n
0.16133
n=
= 0.5359
0.30103
log a = -1.23657 - 3(0.5359) = -2.84427
a = 0.0014313
Hence, the equation for the linear burning rate is
r = 0.0014313 p00.5359
For p0 = 1500 lb/in 2:
r = 0.0014313(1500)0.5359 = 0.0721 in/sec.
In 5 seconds, the total distance receded by the burning surface is 0.0721 (5) = 0.36 in.
9.13
éM + M + M + M + M ù
p1
s1
p2
s2
Lú
Vb1 = g0 Isp n êê
ú
M s1 + M p2 + M s2 + M L
êë
úû
7200 + 800 + 5400 + 600 + 60
= (9.8)(275) n
800 + 5400 + 600 + 60
= 1934 m/sec
éM + M + M ù
p2
s2
Lú
Vb2 - Vb1 = g0 Isp n êê
ú
+
M
M
êë
úû
s2
L
5400 + 600 + 60
= (9.8)(275) n
600 + 60
= 5975 m/sec
Hence:
Vb2 = 5975 + 1934 = 7909 m/sec
9.14 The velocity of the jet relative to you is (Ve - 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
1
(Ve - V¥ ) 2.
2
Hence, the kinetic energy deposited per unit time is
1
m (Ve - V¥ )2.
2
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9.15
Power available = T V∞
(1)
From Eq. (9.24), ignoring the pressure thrust term and assuming that m is essentially
m aira
T = m (Ve - V∞ )
(2)
The total power generated by the propulsive device is the useful power plus the wasted power.
Hence,
Total power generated = T V∞ +
1
m (Ve - V∞ )2
2
(3)
Thus,
ηp =
useful power available
T V∞
=
1
2
total power generated
T V∞ + m ( Ve - V∞ )
2
(4)
Substituting Eq. (2) into (4),
ηp =
m (Ve - V∞ ) V∞
1
2
m (Ve - V∞ ) V∞ + m ( Ve - V∞ )
2
(5)
Dividing numerator and denominator by m (Ve - V¥ )V¥ , Eq. (5) becomes
ηp =
1
1
=
æ
1
Ve ö÷
1ç
1 + (Ve - V¥ ) / V¥
÷
1
+
ç
2
V¥ ø÷÷
2 ççè
or,
ηp =
2
1 + Ve / V¥
9.16 From Example 9.3,
V¥ = 500 miles/hour = 733 ft/sec
And
Ve = 1600 ft/sec
ηp =
2
=
1+Ve /V¥
2
= 0.63
æ 1600 ö÷
1 + çç
÷
çè 733 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.
113
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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
γ −1
0.4
T3  p3  γ
=
= (11.7) 1.4 = (11.7)0.286 = 2.02
T2  p2 
T3 = 2.02 T2 = 2.02 (585) = 1182° R
Since the compressor does work, per unit mass of gas, wc, the energy equation (Eq. 4.42) is
modified as
V2
V2
C p T2 + 2 + Wc = C p T3 + 3
2
2
For V2 = V3, this becomes
Wc = cp (T3 – T2)
cp =
γ R (1.4)(1716)
ft lb
=
= 6006
γ −1
0.4
slug°R
Wc = 6006 (1182 − 585) = 3.58 × 106
ft lb
slug
lb 200
slug
=
= 6.21
.
sec 322
sec
The total work done per second is
The mass flow is 220
slug  

6 ft lb 
7 ft lb
 6.201
  3.58 × 10
 = 2.22 × 10
sec  
slug 
sec
Since one HP = 550
ft lb
, the horsepower provided by the compressor is
sec
HP =
2.22 × 107
= 40,364
550
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
q = c p (T4 − T3 ) = 6006(2110 − 1182)
q = 5.57 × 106 ft lb/slug
or,
ft lb   1 slug 
ft lb

q =  5.57 × 106
= 1.73 × 105




slug   32.2 lb m 
lb m
Since the heat released per unit mass of fuel is given as 1.4 × 107
ft lb ,
lbm
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
(0.0124)(200) = 2.48 lb m / sec
114
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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,
C p T4 +
V2
V42
− WT = CPT5 + 5
2
2
Since V5 = V4,
c p T5 = c p T4 − WT
Since there are no mechanical losses, the work provided by the turbine is the same as the work done
ft lb
. Thus
by the compressor, wT = wc. From the solution of Problem 9.18, wc = 3.58 × 106
slug
c p T5 = c p T4 − 3.58 × 106
or,
T5 = T4 −
3.58 × 106
3.58 × 106
= 2110 −
6006
cp
T5 = 1514° R
9.21 First, calculate the pressure at the entrance to the nozzle, p5. From Problem 9.18,
p3
= 11.7
p2
p3 = 11.7 p2 = 11.7 (2116) = 24, 757
lb
ft 2
The pressure is constant through the combustor, p4 = p3 = 24,757
lb
. The flow through the
ft 2
turbine is isentropic. From the solution of Problem 9.20, T5 = 1514oR. Hence
y
1.4
p5  T5  y −1  1514  0.4
=
=
= (0.71753)3.5 = 0.3129
 2110 
p4  T4 
p5 = (0.3129) p4 = (0.3129)(24757) = 7746
lb
ft 2
The expansion through the nozzle is isentropic.
T6  p6 
=
T5  p5 
y −1
y
0.4
2116  1.4
= 
= (0.273)0.286 = 0.690
 7746 
T6 = 0.690 T5 = (0.690)(1514) = 1045° R
c p T5 +
V52
V2
= c p T6 + 6
2
2
V62 = 2 c p (T5 − T6 ) + V52 = 2(6006)(1514 − 1045) + (1500)2
= 5.63 × 106 + 2.25 × 106 = 7.88 × 106
V6 = (7.88 × 106 )1/ 2 = 2807 ft/sec
115
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9.22 From Problem 9.21, T6 = 1045oR and V6 = 2807 ft/sec
a6 = γ RT6 = (1.4)(1716)(1045) = 1584 ft/sec
M6 =
2807
= 1.77
1584
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
T = m (Ve − V∞ ) + ( pe − p∞ ) Ae .
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.
a6 = γ RT∞ = (1.4)(1716)(390.53) = 968.6 ft/sec
At M ∞ = 2,
V∞ = 2(968.6) = 1937 ft/sec
The diameter of the exit is 28 inches = 2.33 ft. Hence
Ae =
πd2
4
=
π (2.33)2
4
m = 200 lb/sec =
= 4.26 ft 2
200
= 6.21 slug/sec.
32.2
Thus, from Eq. (9.25),
T = (6.21)(2807 − 1937) + (2116 − 476)(4.26)
= 5403 + 6986 = 12,389 lb .
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
T
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,
Specific thrust =
Specific thrust =
12,389
= 61.9 sec
200
116
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Chapter 10
10.1 Since δ ∝
M2
, we have
Re
æ 20 ÷ö2
= 0.3(100) = 30 inches = 2.5ft
çè 2 ÷÷ø
δ M = 20 = (δ M = 2 ) çç
This dramatically demonstrates that boundary layers at hypersonic Mach numbers can be very
thick.
10.2 From Chapter 4 we have
T0
γ -1 2
1.4 - 1
M¥ = 1 +
(20)2 = 81.
=1+
2
2
T¥
At 59 km, T¥ = 258.1 K
Thus: T0 = (81)(258.1) = 20,906 K
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,
φ = 57.3(s/R) = 57.3(6.12) = 28.65°
θ = 90° − φ = 61.5°
(a)
é
ù γ / γ -1 é 1 - γ + 2γ M 2
(γ + 1)2 M 2¥
¥
ê
ú
ê
= ê
ú
ê
2
p¥
γ +1
êë 4 γ M ¥ - 2(γ - 1) úû
êë
p0 2
ù
ú
ú
úû
é
ù1.4 / 0.4 é 1 - 1.4 + 2(1.4)(18) 2 ù
(2.4) 2 (18)2
ú
ê
ú = 417.6
= êê
ú
ê
ú
2
2.4
ëê 4(1.4)(18) - 2(0.4) ûú
ëê
ûú
C p, max =
æ p0
ö
2
çç 2 - 1÷÷ =
(417.6 - 1) = 1.836
÷
ç
2 ç p
÷
γ M¥ è ¥
ø (1.4)(18) 2
2
(Note: This value of C p, max is only slightly smaller than the value calculated in Example
10.1, which was 1.838 for M ¥ = 25 . 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:
C p = C p, maxsin 2θ = (1.836 ) sin 2 ( 61.5° ) = 1.418
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10.4 From Eq. (10.11),
CL = 2sin 2 α cos α
dCL
= (2sin 2 α )(- sin α ) + 4 cos 2 α sin α = 0
dα
sin 2α = 2cos 2α = 2(1 - sin 2α )
α = 54.7°
CL, max = 2 sin 2 (54.7) cos (54.7)=0.77
10.5 (a)
CL = 2 sin 2α cos α
CD = 2 sin 3 α + CD, 0
For small α , these become
CL = 2α 2
(1)
CD = 2α 3 + CD,0
(2)
CL
2α 2
=
CD
2 α 3 + C D, 0
(3)
æC ö
d ççç L ÷÷÷
(2 α 3 + C D, 0 ) 4α - 2α 2 (6α 2 )
çè CD ø÷
=
=0
dα
(2α 3 + CD, 0 )
8α 4 + 4α CD, 0 - 12α 4 = 0
4α 3 = 4CD, 0
α = (CD, 0 )1/3
(4)
Substituting Eq. (4) into Eq. (3):
ö÷
2(CD, 0 )2/3
2/3
0.67
÷÷
=
=
=
1/3
÷
C
C
2
+
(C D, 0 )
(CD, 0 )1.3
D ømax
D, 0
D, 0
æC
çç L
ççè C
Hence:
æ L ÷ö
çç ÷
= 0.67 /(CD, 0 )1/3 and it occurs at
çè D ÷ø
max
α = (CD, 0 )1/ 3.
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(b)
Repeating Eq.(2):
CD = 2α 3 + CD, 0
(2)
From the results of part (a), at (L/D)max we have α = (C, D, 0 )1/3 . Substituting this into
Eq.(2),
C D = 2 C D , 0 +C D , 0 = 3 C D , 0
Since CD =CD,0 +CD, 0
where CD,W is the wave drag, we have
CD = CD,W + CD, 0 = 3CD,0
or,
CD,W = 2 CD, 0
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.
119
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.