Solutions Manual for
Introduction to Flight 9th Edition
By John Anderson, Mary
Bowden (All Chapters 1-10,
100% Original Verified, A+
Grade)
There are No Questions for
Solutions in Chapter 1.
This is The Only Original and
Complete Solutions Manual for
9th Edition, All Other Files in the
Market are Fake/Old/Wrong
Edition.
,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
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)
= 1.41 kg/m 2
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
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
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 = 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
Totalinternal energy = (energy per atom)(number of atoms)
= (1.035 ´ 10- 20
)(1.505 ´ 1026 ) = 1.558 ´ 106 J
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 2116 slug
= = = 0.00237 3
RT (1716)(460 + 59) ft
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 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 3
RT (1716)(460 - 10) ft
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 LLC.
, slug
= 0.00274 - 0.00237 = 0.00037
ft 3
0.00037
% change = = ´ (100) = 15.6% increase
0.00237
2.5 If 1500 lbm of air is pumped into a previously empty 900 ft 3 storage tank and the air
temperature in the tank is uniformly 70°F, what is the air pressure in the tank in
atmospheres?
First, calculate the density from the known mass and volume, = = 1.67 lb m /ft 3
In consistent units, = 1.67/32.2 = 0.052slug/ft 3. Also, T = 70 F = 70 + 460 = 530 R.
Hence,
p = RT = (0.52)(1716)(530)
p = 47, 290 lb/ft 2
or p = 47, = 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.
p = RT
np = np + nR + nT
Differentiating with respect to time,
1 dp 1 d 1 dT
= +
p dt dt T dt
dp p d p dT
or, = +
dt dt T dt
dp d dT
or, = RT + R (1)
dt dt dt
At the instant there is 1000 lbm of air in the tank, the density is
= = 1.11lb m /ft 3
= 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 ft 3, we also have
d 0.5 lb m /sec
= = 0.000556 lb m /(ft 3 )(sec)
dt 900 ft 3
Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw Hill LLC.
, d 0.000556
= = 1.73 10 −5 slug/(ft 3 )(sec)
dt 32.2
Thus, from equation (1) above,
d
= (1716)(510)(1.73 10−5 ) + (0.0345)(1716)(0.0167)
dt
16.1
= 15.1 + 0.99 = 16.1 lb/(ft 2 )(sec) =
2116
= 0.0076 atm/sec
2.7 Assume that, at a point on the wing of the Concorde supersonic transport, the air
temperature is −10°C and the pressure is 1.7 x 104 N/m2. Calculate the density at this point.
In consistent units,
T = −10 + 273 = 263 K
Thus,
= p/RT = (1.7 104 )/(287)(263)
= 0.225 kg/m3
2.8 At a point in the test section of a supersonic wind tunnel, the air pressure and temperature
are 0.5 x 105 N/m2 and 240 K, respectively. Calculate the specific volume.
= p/RT = 0.5 105 /(287)(240) = 0.726 kg/m3
v = 1/ = 1/0.726 = 1.38 m3 /kg
2.9 Consider a flat surface in an aerodynamic flow (say a flat sidewall of a wind tunnel). The
dimensions of this surface are 3 ft in the flow direction (the x direction) and 1 ft
perpendicular to the flow direction (the y direction). Assume that the pressure distribution
(in pounds per square foot) is given by p = 2116 − 10 x and is independent of y . Assume also
that the shear stress distribution (in pounds per square foot) is given by τw = 90/( x + 9)1/2
and is independent of y as shown in figure below. In these expressions, x is in feet, and x = 0
at the front of the surface. Calculate the magnitude and direction of the net aerodynamic
force on the surface.
3 3
Fp = Force due to pressure =
ò0 p dx =
ò0 (2116 - 10 x) dx
= [2116 x - 5x 2 ] 30 = 6303 lb perpendicular to wall.
Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw Hill LLC.
Introduction to Flight 9th Edition
By John Anderson, Mary
Bowden (All Chapters 1-10,
100% Original Verified, A+
Grade)
There are No Questions for
Solutions in Chapter 1.
This is The Only Original and
Complete Solutions Manual for
9th Edition, All Other Files in the
Market are Fake/Old/Wrong
Edition.
,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
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)
= 1.41 kg/m 2
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
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
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 = 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
Totalinternal energy = (energy per atom)(number of atoms)
= (1.035 ´ 10- 20
)(1.505 ´ 1026 ) = 1.558 ´ 106 J
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 2116 slug
= = = 0.00237 3
RT (1716)(460 + 59) ft
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 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 3
RT (1716)(460 - 10) ft
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 LLC.
, slug
= 0.00274 - 0.00237 = 0.00037
ft 3
0.00037
% change = = ´ (100) = 15.6% increase
0.00237
2.5 If 1500 lbm of air is pumped into a previously empty 900 ft 3 storage tank and the air
temperature in the tank is uniformly 70°F, what is the air pressure in the tank in
atmospheres?
First, calculate the density from the known mass and volume, = = 1.67 lb m /ft 3
In consistent units, = 1.67/32.2 = 0.052slug/ft 3. Also, T = 70 F = 70 + 460 = 530 R.
Hence,
p = RT = (0.52)(1716)(530)
p = 47, 290 lb/ft 2
or p = 47, = 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.
p = RT
np = np + nR + nT
Differentiating with respect to time,
1 dp 1 d 1 dT
= +
p dt dt T dt
dp p d p dT
or, = +
dt dt T dt
dp d dT
or, = RT + R (1)
dt dt dt
At the instant there is 1000 lbm of air in the tank, the density is
= = 1.11lb m /ft 3
= 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 ft 3, we also have
d 0.5 lb m /sec
= = 0.000556 lb m /(ft 3 )(sec)
dt 900 ft 3
Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw Hill LLC.
, d 0.000556
= = 1.73 10 −5 slug/(ft 3 )(sec)
dt 32.2
Thus, from equation (1) above,
d
= (1716)(510)(1.73 10−5 ) + (0.0345)(1716)(0.0167)
dt
16.1
= 15.1 + 0.99 = 16.1 lb/(ft 2 )(sec) =
2116
= 0.0076 atm/sec
2.7 Assume that, at a point on the wing of the Concorde supersonic transport, the air
temperature is −10°C and the pressure is 1.7 x 104 N/m2. Calculate the density at this point.
In consistent units,
T = −10 + 273 = 263 K
Thus,
= p/RT = (1.7 104 )/(287)(263)
= 0.225 kg/m3
2.8 At a point in the test section of a supersonic wind tunnel, the air pressure and temperature
are 0.5 x 105 N/m2 and 240 K, respectively. Calculate the specific volume.
= p/RT = 0.5 105 /(287)(240) = 0.726 kg/m3
v = 1/ = 1/0.726 = 1.38 m3 /kg
2.9 Consider a flat surface in an aerodynamic flow (say a flat sidewall of a wind tunnel). The
dimensions of this surface are 3 ft in the flow direction (the x direction) and 1 ft
perpendicular to the flow direction (the y direction). Assume that the pressure distribution
(in pounds per square foot) is given by p = 2116 − 10 x and is independent of y . Assume also
that the shear stress distribution (in pounds per square foot) is given by τw = 90/( x + 9)1/2
and is independent of y as shown in figure below. In these expressions, x is in feet, and x = 0
at the front of the surface. Calculate the magnitude and direction of the net aerodynamic
force on the surface.
3 3
Fp = Force due to pressure =
ò0 p dx =
ò0 (2116 - 10 x) dx
= [2116 x - 5x 2 ] 30 = 6303 lb perpendicular to wall.
Copyright 2022 © McGraw Hill LLC. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw Hill LLC.
