?
ED
OV
PR
_AP
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, TABLE OF CONTENTS
Solutions Manual: Introduction to Flight, 9th Edition
ST
Authors: John Anderson, Mary Bowden
1. The First Aeronautical Engineers
UV
2. Fundamental Thoughts
3. The Standard Atmosphere
4. Basic Aerodynamics
5. Airfoils, Wings, and Other Aerodynamics Shapes
IA
6. Elements of Airplane Performance
7. Principles of Stability and Control
8. Space Flight (Astronautics)
_
9. Propulsion
10. Hypersonic Vehicles
AP
PR
OV
ED
?
, Chapter 2 – Introduction to Flight, 9th ed., Solutions
ST
2.1 Consider the low-speed flight of the Space Shuttle as it is nearing a landing. If the air
pressure and temperature at the nose of the shuttle are 1.2 atm and 300 K, respectively,
what are the density and specific volume?
UV
= p/RT = (1.2)(1.01105 )/(287)(300)
= 1.41 kg/m2
v = 1/ = 1/1.41= 0.71 m3/kg
2.2 Consider 1 kg of helium at 500 K. Assuming that the total internal energy of helium is due to
I
the mean kinetic energy of each atom summed over all the atoms, calculate the internal
energy of this gas. Note: The molecular weight of helium is 4. Recall from chemistry that the
A_
molecular weight is the mass per mole of gas; that is, 1 mol of helium contains 4 kg of mass.
Also, 1 mol of any gas contains 6.02 x 1023 molecules or atoms (Avogadro’s number).
3 3
Mean kinetic energy of each atom = (1.38 10−23 ) (500) = 1.035 10−20J
kT=
2 2
One kg-mole, which has a mass of 4 kg, has 6.02 × 1026 atoms. Hence 1 kg has
AP
1
(6.02 1026 ) = 1.505 1026 atoms
4
Total internal energy = (energy per atom)(number of atoms)
= (1.035´ 10- 20)(1.505´ 1026) = 1.558 ´ 106 J
PR
2.3 Calculate the weight of air (in pounds) contained within a room 20 ft long, 15 ft wide, and
8 ft high. Assume standard atmospheric pressure and temperature of 2116 lb/ft2 and 59°F,
respectively.
p slug
= = 2116 = 0.00237
OV
RT (1716)(460 + 59) ft3
Volume of the room = (20)(15)(8) = 2400 ft3
Total mass in the room = (2400)(0.00237) = 5.688slug
Weight = (5.688)(32.2) = 183lb
2.4 Comparing with the case of Prob. 2.3, calculate the percentage change in the total weight of
ED
air in the room when the air temperature is reduced to −10°F (a very cold winter day),
assuming that the pressure remains the same at 2116 lb/ft 2.
p 2116 slug
= = = 0.00274
RT (1716)(460 - 10) ft3
?
Since the volume of the room is the same, we can simply compare densities between the two
problems.
Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw Hill
, slug
= 0.00274 - 0.00237 = 0.00037
ft3
0.00037
% change = = ´ (100) = 15.6% increase
ST
0.00237
2.5 If 1500 lbm of air is pumped into a previously empty 900 ft3 storage tank and the air
temperature in the tank is uniformly 70°F, what is the air pressure in the tank in
atmospheres?
UV
First, calculate the density from the known mass and volume, = 1500/ 900 = 1.67 lbm /ft3
In consistent units, = 1.67/32.2 = 0.052slug/ft3. Also, T = 70 F = 70 + 460 = 530 R.
Hence,
p = RT = (0.52)(1716)(530)
I
p = 47, 290 lb/ft2
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or p = 47, = 22.3 atm
2.6 In Prob. 2.5, assume that the rate at which air is being pumped into the tank is 0.5 lbm/s.
Consider the instant in time at which there is 1000 lbm of air in the tank. Assume that the
air temperature is uniformly 50°F at this instant and is increasing at the rate of 1°F/min.
Calculate the rate of change of pressure at this instant.
AP
p = RT
Differentiating with respect to time,
1 dp 1 d 1 dT
= +
PR
p dt dt T dt
or, dp p d p dT
= +
dt dt T dt
dp d + R dT
or, = RT (1)
dt dt dt
OV
At the instant there is 1000 lbm of air in the tank, the density is
= = 1.11lb m /ft3
= 1.11/32.2 = 0.0345slug/ft3
Also, in consistent units, is given that
T = 50 + 460 = 510 R
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and that
dT
= 1F/min = 1R/min = 0.016R/sec
dt
From the given pumping rate, and the fact that the volume of the tank is 900 ft3, we also have
?
d 0.5 lbm /sec
= = 0.000556 lb /(ft3 )(sec)
3 m
dt 900 ft
Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw Hill
ED
OV
PR
_AP
IA
UV
ST
, TABLE OF CONTENTS
Solutions Manual: Introduction to Flight, 9th Edition
ST
Authors: John Anderson, Mary Bowden
1. The First Aeronautical Engineers
UV
2. Fundamental Thoughts
3. The Standard Atmosphere
4. Basic Aerodynamics
5. Airfoils, Wings, and Other Aerodynamics Shapes
IA
6. Elements of Airplane Performance
7. Principles of Stability and Control
8. Space Flight (Astronautics)
_
9. Propulsion
10. Hypersonic Vehicles
AP
PR
OV
ED
?
, Chapter 2 – Introduction to Flight, 9th ed., Solutions
ST
2.1 Consider the low-speed flight of the Space Shuttle as it is nearing a landing. If the air
pressure and temperature at the nose of the shuttle are 1.2 atm and 300 K, respectively,
what are the density and specific volume?
UV
= p/RT = (1.2)(1.01105 )/(287)(300)
= 1.41 kg/m2
v = 1/ = 1/1.41= 0.71 m3/kg
2.2 Consider 1 kg of helium at 500 K. Assuming that the total internal energy of helium is due to
I
the mean kinetic energy of each atom summed over all the atoms, calculate the internal
energy of this gas. Note: The molecular weight of helium is 4. Recall from chemistry that the
A_
molecular weight is the mass per mole of gas; that is, 1 mol of helium contains 4 kg of mass.
Also, 1 mol of any gas contains 6.02 x 1023 molecules or atoms (Avogadro’s number).
3 3
Mean kinetic energy of each atom = (1.38 10−23 ) (500) = 1.035 10−20J
kT=
2 2
One kg-mole, which has a mass of 4 kg, has 6.02 × 1026 atoms. Hence 1 kg has
AP
1
(6.02 1026 ) = 1.505 1026 atoms
4
Total internal energy = (energy per atom)(number of atoms)
= (1.035´ 10- 20)(1.505´ 1026) = 1.558 ´ 106 J
PR
2.3 Calculate the weight of air (in pounds) contained within a room 20 ft long, 15 ft wide, and
8 ft high. Assume standard atmospheric pressure and temperature of 2116 lb/ft2 and 59°F,
respectively.
p slug
= = 2116 = 0.00237
OV
RT (1716)(460 + 59) ft3
Volume of the room = (20)(15)(8) = 2400 ft3
Total mass in the room = (2400)(0.00237) = 5.688slug
Weight = (5.688)(32.2) = 183lb
2.4 Comparing with the case of Prob. 2.3, calculate the percentage change in the total weight of
ED
air in the room when the air temperature is reduced to −10°F (a very cold winter day),
assuming that the pressure remains the same at 2116 lb/ft 2.
p 2116 slug
= = = 0.00274
RT (1716)(460 - 10) ft3
?
Since the volume of the room is the same, we can simply compare densities between the two
problems.
Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw Hill
, slug
= 0.00274 - 0.00237 = 0.00037
ft3
0.00037
% change = = ´ (100) = 15.6% increase
ST
0.00237
2.5 If 1500 lbm of air is pumped into a previously empty 900 ft3 storage tank and the air
temperature in the tank is uniformly 70°F, what is the air pressure in the tank in
atmospheres?
UV
First, calculate the density from the known mass and volume, = 1500/ 900 = 1.67 lbm /ft3
In consistent units, = 1.67/32.2 = 0.052slug/ft3. Also, T = 70 F = 70 + 460 = 530 R.
Hence,
p = RT = (0.52)(1716)(530)
I
p = 47, 290 lb/ft2
A_
or p = 47, = 22.3 atm
2.6 In Prob. 2.5, assume that the rate at which air is being pumped into the tank is 0.5 lbm/s.
Consider the instant in time at which there is 1000 lbm of air in the tank. Assume that the
air temperature is uniformly 50°F at this instant and is increasing at the rate of 1°F/min.
Calculate the rate of change of pressure at this instant.
AP
p = RT
Differentiating with respect to time,
1 dp 1 d 1 dT
= +
PR
p dt dt T dt
or, dp p d p dT
= +
dt dt T dt
dp d + R dT
or, = RT (1)
dt dt dt
OV
At the instant there is 1000 lbm of air in the tank, the density is
= = 1.11lb m /ft3
= 1.11/32.2 = 0.0345slug/ft3
Also, in consistent units, is given that
T = 50 + 460 = 510 R
ED
and that
dT
= 1F/min = 1R/min = 0.016R/sec
dt
From the given pumping rate, and the fact that the volume of the tank is 900 ft3, we also have
?
d 0.5 lbm /sec
= = 0.000556 lb /(ft3 )(sec)
3 m
dt 900 ft
Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw Hill
