??
ED
OV
PR
AP
A?_
UVI
ST
, TABLE OF CONTENTS
Solutions Manual: Introduction to Flight, 9th Edition
Authors: John Anderson, Mary Bowden
ST
1. The First Aeronautical Engineers
U
2. Fundamental Thoughts
3. The Standard Atmosphere
VI
4. Basic Aerodynamics
5. Airfoils, Wings, and Other Aerodynamics Shapes
6. Elements of Airplane Performance
A
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
2.1 Consider the low-speed flight of the Space Shuttle as it is nearing a landing. If the air
ST
pressure and temperature at the nose of the shuttle are 1.2 atm and 300 K, respectively,
what are the density and specific volume?
= p/RT = (1.2)(1.01105 )/(287)(300)
U
= 1.41 kg/m2
v = 1/ = 1/1.41= 0.71 m3/kg
VI
2.2 Consider 1 kg of helium at 500 K. Assuming that the total internal energy of helium is due to
the mean kinetic energy of each atom summed over all the atoms, calculate the internal
A
energy of this gas. Note: The molecular weight of helium is 4. Recall from chemistry that the
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
1
AP
(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.
OV
p slug
= = 2116 = 0.00237
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
ED
2.4 Comparing with the case of Prob. 2.3, calculate the percentage change in the total weight of
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
0.00237
ST
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?
U
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.
VI
Hence,
p = RT = (0.52)(1716)(530)
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
OV
or, = RT (1)
dt dt dt
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
ED
Also, in consistent units, is given that
T = 50 + 460 = 510 R
and that
?
dT
= 1F/min = 1R/min = 0.016R/sec
dt
?
From the given pumping rate, and the fact that the volume of the tank is 900 ft3, we also have
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
A?_
UVI
ST
, TABLE OF CONTENTS
Solutions Manual: Introduction to Flight, 9th Edition
Authors: John Anderson, Mary Bowden
ST
1. The First Aeronautical Engineers
U
2. Fundamental Thoughts
3. The Standard Atmosphere
VI
4. Basic Aerodynamics
5. Airfoils, Wings, and Other Aerodynamics Shapes
6. Elements of Airplane Performance
A
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
2.1 Consider the low-speed flight of the Space Shuttle as it is nearing a landing. If the air
ST
pressure and temperature at the nose of the shuttle are 1.2 atm and 300 K, respectively,
what are the density and specific volume?
= p/RT = (1.2)(1.01105 )/(287)(300)
U
= 1.41 kg/m2
v = 1/ = 1/1.41= 0.71 m3/kg
VI
2.2 Consider 1 kg of helium at 500 K. Assuming that the total internal energy of helium is due to
the mean kinetic energy of each atom summed over all the atoms, calculate the internal
A
energy of this gas. Note: The molecular weight of helium is 4. Recall from chemistry that the
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
1
AP
(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.
OV
p slug
= = 2116 = 0.00237
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
ED
2.4 Comparing with the case of Prob. 2.3, calculate the percentage change in the total weight of
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
0.00237
ST
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?
U
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.
VI
Hence,
p = RT = (0.52)(1716)(530)
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
OV
or, = RT (1)
dt dt dt
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
ED
Also, in consistent units, is given that
T = 50 + 460 = 510 R
and that
?
dT
= 1F/min = 1R/min = 0.016R/sec
dt
?
From the given pumping rate, and the fact that the volume of the tank is 900 ft3, we also have
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
