10-1 Air at 23 ºC, 100 kPa with a free-stream velocity of 100 km/hr flows along a flat plate.

e. How long does the

plate have to be to obtain a boundary layer thickness of 8 mm?

Approach:
For the distance x from the leading edge to where the
boundary layer thickness equals 8 mm, if we assume a
turbulent flow, we can use Eq. 10-8 to calculate the
distance. We must check the turbulent flow
assumption.

Assumptions:
1. The boundary layer is turbulent from the leading
edge.

Solution:
Assuming a turbulent boundary layer from the leading edge, the boundary layer thickness can be calculated with
Eq. 10-8:
δ 0.37 0.37
= =
x Re1x 5 ( ρ V∞ x µ )1 5
Solving for the distance x:
0.25
⎡ δ 5ρ V ⎤
x=⎢ ∞
⎥
⎢⎣ ( 0.37 ) µ ⎥⎦
5

From Appendix A-7 the air properties at 23ºC are: ρ = 1.177 kg/m3 and µ = 1.817 × 10−5 Ns m 2 .
⎡ ( 0.008m )5 (1.177 kg m3 ) (100000 m hr )(1hr 3600s ) ⎤
0.25

x=⎢ ⎥ =0.304m Answer

⎢
⎣ ( 0.37 )
5
(1.817×10 -5
Ns m 2
)( kgm Ns 2
) ⎥
⎦
Now we check the Reynolds number:
ρ V x (1.177 kg m ) (100,000 m hr )(1hr 3600s )( 0.304m )
3

Rex = = = 547, 000

µ 1.817×10-5 Ns m 2
This is turbulent flow as assumed (if turbulence begins at 500,000).

10- 1
10-2 A large cruise ship has a length, L = 250 m, beam (width), W = 65 m, and draft (depth), D = 20 m. It cruises
at 25 km/hr. Assume the flow over the hull can be approximated as that over a flat plate. Estimate:
a. the total skin friction drag (in N) and the power (in kW) to propel the ship on a voyage in the
Caribbean where the water temperature is 28 ºC
b. the total skin friction drag (in N) and the power (in kW) to propel the ship on a voyage to Alaska
where the water temperature is 4 ºC

Approach:
Assuming we can treat the sides and bottom as flow
over a flat plate, and ignoring the form drag at the
front, the skin friction can be calculated with the basic
drag equation and the appropriate drag coefficient.
Power is force times velocity.

Assumptions:
1. The sides and bottom can be treated as flat plates.
2. Seawater properties can be approximated with pure
water properties.

Solution:
a) The friction drag force is defined by Eq. 10-6:
FD = CD ρ V∞2 A 2
where A is the planform area: A = LW + 2 LD = ( 250m ) ⎡⎣ 65m+2 ( 20m ) ⎤⎦ =26,250m 2
We need the Reynolds number to evaluate the drag coefficient. We assume we can treat this area as flow
over a flat plate. Assuming seawater properties can be approximated by those of pure water from Appendix A-6
at 28 ºC, ρ = 996.1 kg/m3 and µ = 8.16×10-4 Ns/m2.
The length Reynolds number is:
ρ V∞ L ( 996.1kg m ) ( 9.62 m s )( 250m )
3

ReL = = = 2.936×109
µ 8.16×10-4 Ns m 2
This is turbulent flow, so ignoring the laminar contribution and using Eq. 10-8 (even though the Reynolds number
is larger than the applicable range):
0.455 0.455
CD = = = 0.00138
⎡⎣ log ( ReL ) ⎤⎦ ⎡ log ( 2.936 × 109 ) ⎤
2.58 2.58

⎣ ⎦
FD = ( 0.00138 ) ( 996.1kg m3 ) ( 9.62 m s ) ( 26250m 2 )( Ns 2 kgm ) 2 =1.670×106 N
2

Power is obtained from:

W = FD V∞ = (1.670×106 N ) ( 9.62 m s )(1kW 1000 J s ) =16,060 kW Answer
b) The only difference in part (b) is the temperature at which properties are evaluated. At 4 ºC, ρ = 1000 kg/m3
and µ = 15.5×10-4 Ns/m2.
ρ V∞ L (1000 kg m ) ( 9.62 m s )( 250m )
3

ReL = = = 1.552×109
µ 15.5×10-4 Ns m 2
0.455 0.455
CD = = = 0.00149
⎡⎣ log ( ReL ) ⎤⎦ ⎡ log (1.552 × 109 ) ⎤
2.58 2.58

⎣ ⎦
FD = ( 0.00149 ) (1000 kg m3 ) ( 9.62 m s ) ( 26250m 2 )( Ns 2 kgm ) 2 =1.809×106 N
2

W = FD V∞ = (1.809×106 N ) ( 9.62 m s )(1kW 1000 J s ) =17,410kW Answer

Comments:
Just by changing the water temperature, the power increase was 8.4%. This is almost entirely due to the change in
the water viscosity.

10- 2
10-3 A 5-mm × 1.5-m × 4-m plastic panel (SG = 1.75) is lowered from a ship to a construction site on a lake floor
at a rate of 1.5 m/s. Determine the tension in the cable lowering the panel:
a. assuming the panel descends vertically with its wide end down (in N)
b. assuming the panel descends vertically with its narrow end down (in N)

Approach:
This problem requires a force balance on the plastic
panel. The buoyancy, drag, tension, and weight forces
must be evaluated. We will assume that we can treat
the flow as if it is over a flat plate.

Assumptions:
1. The plastic panel is treated as a flat plate.
2. The properties can be approximated with pure
water properties.

Solution:
a) A force balance on the plate is:
∑ F = 0 = T + FD + Fbuoy − W → T = W − FD − Fbuoy
For the weight:
W = ρ plateVg = ρ water SG plateVg = (1000 kg m3 ) (1.75 )( 0.005m )( 4m )(1.5m ) ( 9.81m s 2 ) =515N
Buoyancy force is:
Fbuoy = ρ waterVg = (1000 kg m3 ) ( 0.005m )( 4m )(1.5m ) ( 9.81m s 2 ) =294N
The drag force is defined by Eq. 10-6. Accounting for drag on both sides:
FD = 2 CD ρ V 2 A 2
where A is the planform area: A = LH = (1.5m )( 4m ) =6.0 m 2
We need the Reynolds number to evaluate the drag coefficient. We assume we can treat this area as flow
over a flat plate. Assuming properties can be approximated by those of pure water from Appendix A-6 at 10 ºC:
ρ = 999.6 kg/m3 and µ = 12.9×10-4 Ns/m2. The length Reynolds number is:
ρ V L ( 999.6 kg m ) (1.5 m s )(1.5m )
3

ReL = = = 1.744×106
µ 12.9×10-4 Ns m 2
This is turbulent flow, but taking into account the laminar contribution at the leading edge:
0.074 1740 0.074 1740
CD = − = − = 0.00318
Re1L 5 ReL (1.744 × 106 )1 5 1.744 × 106

FD = 2 ( 0.00318) ( 999.6 kg m3 ) (1.5 m s ) ( 6m 2 )( Ns 2 kgm ) 2 =42.9 N

2

Therefore, T = 515N-42.9N-294 N=178 N Answer

b) For the narrow end down, the only change is in the drag force:
ρ V L ( 999.6 kg m ) (1.5 m s )( 4m )
3

ReL = = = 4.649×106
µ 12.9×10-4 Ns m 2
0.074 1740 0.074 1740
CD = − = − = 0.00306
ReL15
ReL ( 4.649×10 )
6 15
4.649×106

FD = 2 ( 0.00306 ) ( 99.6 kg m3 ) (1.5 m s ) ( 6m 2 )( Ns 2 kgm ) 2 =41.3N

2

T = 515N-41.3N-294.3N=179.7N Answer

Comments:
Drag plays a minor role in this problem compared to weight and buoyancy.

10- 3
10-4 Your car has broken down on an Interstate highway, and you are on the medium strip between the lanes. By
the time you decide you need to cross to the other side of the empty road, a stream of cars moving bumper to
bumper at 120 km/hr passes only 1 m away from you. For an air temperature of 25 ºC, determine the
velocity (in km/hr) of the wind that will hit you 10 seconds after the first car has passed.

Approach:
Assuming the bumper-to-bumper traffic acts as a flat
plate moving through air at 120 km/hr, a boundary
layer forms. We can calculate the distance the first
car travels in 10 s. We can check the Reynolds
number to determine if the flow is laminar or
turbulent, and using the appropriate equation, we can
calculate the boundary layer thickness. Finally, using
an expression for the velocity profile, we can then
determine the velocity.

Assumptions:
1. The traffic acts as a flat plate moving through still
air.
2. Pressure is one atmosphere.

Solution:
The distance, x, the first car travels is:
x = V t = (120,000 m hr )(1hr 3600s )(10s ) =333m
For air from Appendix A-7 at 25 ºC , µ = 1.832 × 10−5 Ns m 2 and ρ = 1.181kg m3
ρ VD (1.181kg m ) (120, 000 m hr )(1hr 3600s )( 333m )
3

Re = = =715,500
µ 1.832 × 10−5 Ns m 2
This is turbulent flow. Assuming turbulence begins at the start of the first car, by 10 s later the boundary layer
thickness is:
δ 0.37 ( 333m )( 0.37 )
= → δ= =2.09m
x Re1x 5 ( 715,500 )
15

Assuming a 1/7th power law velocity profile:

17 17
Vx ⎛ y ⎞ ⎛ km ⎞⎛ 1m ⎞ km
=⎜ ⎟ → Vx = ⎜ 120 ⎟⎜ ⎟ =108 Answer
V∞ ⎝ δ ⎠ ⎝ hr ⎠⎝ 2.09m ⎠ hr
Comments:
The main assumption (traffic acts as flat plate) is somewhat weak, because the cars act as blunt bodies and the
flow around them are different than over a flat plate. Nevertheless, we needed an answer (however approximate),
so we needed to make assumptions that would let us determine an answer.

10- 4
10-5 A flat-bottomed river barge 60-m long and 12-m wide is towed through still water (at 25 ºC) at 10 km/hr.
Determine:
a. the force required to overcome the drag (in N)
b. the power required by the towboat (in kW)
c. the boundary layer thickness at the end of the barge (in mm)

Approach:
Assuming the bottom can be treated as flow over a flat
plate, and the skin friction can be calculated with the
basic drag equation and the appropriate drag
coefficient. Power is force times velocity. Depending
on whether the flow is laminar or turbulent at the end
of the plate, the appropriate boundary layer thickness
equation can be used.

Assumptions:
1. The plate can be treated as flow over a flat plate.

Solution:
a) The drag force is defined by Eq. 10-6:
FD = CD ρ V∞2 A 2
where A is the planform area: A = LW = (12m )( 60m ) =720m 2
We need the Reynolds number to evaluate the drag coefficient. We assume we can treat this area as flow
over a flat plate. For water from Appendix A-6 at 12 ºC, ρ = 1000 kg/m3 and µ = 1.225×10-3 Ns/m2.
The length Reynolds number at 10 km/hr = 2.78 m/s is:
ρ V∞ L (1000 kg m ) ( 2.78 m s )( 60m )
3

ReL = = = 1.36×108
µ 1.225×10-3 Ns m 2
This is turbulent flow, so ignoring the laminar contribution and using Eq. 10-8:
0.455 0.455
CD = = = 0.00204
⎡⎣ log ( ReL ) ⎤⎦ ⎡ log (1.36 × 108 ) ⎤
2.58 2.58

⎣ ⎦
FD = ( 0.00204 ) (1000 kg m3 ) ( 2.78 m s ) ( 720m 2 )( Ns 2 kgm ) 2 =5674N
2
Answer

b) Power is obtained from:

W = FD V∞ = ( 5674N )( 2.78 m s )(1kW 1000 J s ) =15.8kW Answer

c) For the turbulent boundary layer thickness at the end of the plate:
δ 0.037 ( 60m )( 0.37 )
= → δ= =0.50m Answer
(1.38×108 )
2 15
L ReL

10- 5
10-6 Outboard racing boats are designed for part of the hull to rise completely out of the water when a high speed
is reached. Then the boat “planes” on the remainder of the hull. At 50 mi/hr on 60 ºF water the area of the
hull planing is 6 ft long and 5 ft wide. Determine the power required to overcome the friction drag (in hp).

Approach:
Assume the boat bottom can be treated as flow over a
flat plate, and the skin friction can be calculated with
the basic drag equation and the appropriate drag
coefficient. Power is force times velocity.

Assumptions:
1. The plate can be treated as flow over a flat plate.

Solution:
The drag force is defined by Eq. 10-6:
FD = CD ρ V∞2 A 2
where A is the planform area: A = LW = ( 6ft )( 5ft ) =30ft 2
We need the Reynolds number to evaluate the drag coefficient. We assume we can treat this area as flow
over a flat plate. Note that V = 50 mi/hr = 73.3 ft/s. For water from Appendix B-6 at 60 ºF, ρ = 62.3 lbm/ft3 and
µ = 76×10-5 lbm/fts.
ρ V∞ L ( 62.3lbm ft ) ( 73.3ft s )( 6ft )
3

ReL = = = 3.61×107
µ 76×10-5 lbm ft s
This is turbulent flow, so ignoring the laminar contribution and using Eq. 10-8:
0.455 0.455
CD = = = 0.00247
⎡⎣ log ( ReL ) ⎤⎦ ⎡ log ( 3.61× 107 ) ⎤
2.58 2.58

⎣ ⎦
FD = ( 0.00247 ) ( 62.3lbm ft 3 ) ( 73.3ft s ) ( 30ft 2 )( lbf s 2 32.2 ft lbm ) 2 =384 lbf
2

Power is obtained from:

W = FD V∞ = ( 385lbf )( 73.3ft s )(1 hp s 550 ft lbf ) =51.2 hp Answer

10- 6
10-7 A ceiling fan has five thin blades each 55-cm long and 15-cm wide. Assume the blades can be approximated
as flat plates. Air temperature is 27 ºC. For rotational speeds of 50 RPM, 100 RPM, and 150 RPM,
determine the power (in W) needed to overcome the drag force. (Hint: Because velocity varies with distance
from the center of rotation, you must integrate the drag coefficient.)

Approach:
Power is force time velocity, or torque times angular
velocity. Velocity varies from zero at the center of
rotation to a maximum at the blade tip. Drag force
depends on velocity and drag coefficient, both of
which vary along the length of the blade. Hence, we
must integrate to obtain the total power.

Assumptions:
1. Flow over the blade behaves as flow over a flat
plate.
2. Air is at one atmosphere.

Solution:
Using the differential element shown on 1 blade:
δ W = 2 VdFD
where the factor of 2 is used to take into account both sides of the blade. The velocity is V = rω . The
differential drag force is: dFD = CD ρ V 2Wdr 2
Because the drag coefficient is a function of velocity and, hence, distance along the blade, we must determine if
the flow is laminar or turbulent. The maximum Reynolds number will occur at the blade tip at the maximum
rotational speed. The air properties from Appendix A-7 at 300 K are ρ = 1.173 kg/m3 and µ = 1.846×10-5 Ns/m2.
ρ VW ρ rωW (1.173kg m ) (150 rev min )(1min 60s )( 2πrad rev )( 0.55m )( 0.15m )
3

Remax = = = = 8.24×104
µ µ 1.846×10-5 Ns m 2
Because this is less than 500,000, we assume the whole blade is in laminar flow, and we will use the laminar flow
drag coefficient for flow over a flat plate.
1.328 1.328 1.328
CD = = =
( ρ rωW µ ) ( ρωW µ ) r 0.5
0.5 0.5 0.5
ReW
⎡ 1.328 ⎤⎛ 1 ⎞
δ W = 2 ( rω ) ⎢ ⎥ ⎜ ⎟ ρ ( rω ) Wdr
2
Combining all the expressions:
⎢⎣ ( ρωW µ ) r ⎥⎦ ⎝ 2 ⎠
0.5 0.5

L 1.328W ρω 2 1.328W ρω 3 7 2
3
Simpifying and integrating: W = ∫ δ W = ∫ r 5 2 dr = L
( ρωW µ ) 7 ( ρωW µ )0.5
0 0.5

Substituting in values:
⎛ 2 ⎞ 1.328 ( 0.15m ) (1.173kg m ) (ω rad s )
3 3
7 2 ⎛ Ns ⎞ ⎛ 1J ⎞ ⎛ 1W ⎞
2

0.5 (

W =⎜ ⎟ 0.55m ) ⎜ ⎟⎜ ⎟⎜ ⎟ = 0.0000844ω
52

⎝ 7 ⎠ ⎡ (1.173kg m ) (ω rad s )( 0.15m ) ⎤

3
⎝ kgm ⎠ ⎝ Nm ⎠ ⎝ 1J s ⎠
⎢ ⎥
⎢⎣ (1.846×10 Ns m )( kgm Ns ) ⎥⎦
-5 2 2

This is for one blade. Finally,

At ( 50 rev min )(1min 60s )( 2πrad rev ) =5.24 rad s
W = 5 ( 0.0000844 )( 5.24 ) = 0.0265W
52
Answer
At (100 rev min )(1min 60s )( 2πrad rev ) =10.47 rad s
W = 5 ( 0.0000844 )(10.47 ) = 0.15W
52
Answer
At (150 rev min )(1min 60s )( 2πrad rev ) =15.71rad s
W = 5 ( 0.0000844 )(15.71) = 0.413W
52
Answer

10- 7
10-8 The high-speed trains in France and Japan are streamlined to reduce drag forces. Consider a 120-m long
train, whose outer surface can be approximated by a flat plate with a width of 10 m. At 101 kPa, 20 ºC,
determine the drag (in N) due to skin friction only and the power required (in kW) to overcome this drag at:
a. 100 km/hr, 200 km/hr, and 300 km/hr.
b. The front of the train can be approximated by a hemisphere facing forward with a circumference of
15 m. Estimate the drag force caused by the front of the train and the power required to overcome it.

Approach:
This is a straightforward application of the drag force
equation. We assume we can treat the plate as flow
over a flat plate, and the skin friction can be calculated
with the basic drag equation and the appropriate drag
coefficient. Power is force times velocity.

Assumptions:
1. The plate can be treated as flow over a flat plate.

Solution:
a) The drag force is defined by Eq. 10-6:
FD = CD ρ V 2 A 2
where A is the planform area: A = LW = (120m )(10m ) =1200m 2
We need the Reynolds number to evaluate the drag coefficient. We assume we can treat this area as flow
over a flat plate. For air from Appendix A-7 at 20 ºC, ρ = 1.206 kg/m3 and µ = 1.80×10-5 Ns/m2.
The length Reynolds number at 100 km/hr = 27.8 m/s is:
ρ V L (1.206 kg m ) ( 27.8 m s )(120m )
3

ReL = = = 2.24×108
µ 1.80×10-5 Ns m 2
This is turbulent flow, so ignoring the laminar contribution and using Eq. 10-10:
0.455 0.455
CD = = = 0.00191
⎡⎣ log ( ReL ) ⎤⎦ ⎡ log ( 2.24 × 108 ) ⎤
2.58 2.58

⎣ ⎦
FD = ( 0.00191) (1.206 kg m3 ) ( 27.8 m s ) (1200m 2 )( Ns 2 kgm ) 2 =1070 N
2

Power is obtained from:

W = FD V = (1070N )( 27.8 m s )(1 J 1 Nm ) =29,750W Answer
b) At velocity of 200 km/hr = 55.6 m/s:
ReL = 4.48×108 CD = 0.00174
F = 3890N
D W = 216,300W Answer
At velocity of 300 km/hr = 83.4 m/s:
ReL = 6.72×108 CD = 0.00165
F = 8300N
D W = 692, 200W Answer
c) The front of the train can be approximated as a hemisphere with a circumference of 15m. Hence, the diameter
is: π D = 15m → D = 4.77m
From Table 10-2, CD = 0.4 , so at 100 km/hr = 27.8 m/s:
FD = ( 0.4 ) (1.206 kg m3 ) ( 27.8 m s ) (π 4 )( 4.77m ) ( Ns 2 kgm ) 2 =3330 N
2 2

W = FD V = ( 3330N )( 27.8 m s )(1J 1 Nm ) =92,600 W Answer

At 200 km/hr F = 13,320 N and W = 740, 600 W
D

At 300 km/hr FD = 29,970 N and W = 2,500, 000 W Answer

Comments:
The blunt nose of the train increases the power requirement by a dramatic amount, even though its area is
relatively small compared to the flat areas of the train. This illustrates well the effect of different shapes on drag.

10- 8
10-9 In large electric power plants cool water flows through condensers downstream of the steam turbines which
drive the electric generators. This water is recirculated, so it is often cooled in cooling towers, such as shown
on the figure below. The design specification is that the tower must withstand a 100 mi/hr wind at 70 ºF.
Approximating the drag coefficient from information given in one of the tables, determine:
a. the drag force on the tower (in lbf)
b. the moment that must be resisted by the foundation of the tower (in ft-lbf).

Approach:
Drag force is calculated with the basic drag. We will
need to approximate the drag coefficient because we
do not have one specifically for this shape. The
moment is force times distance; we will assume the
drag force acts through the center of the cooling
tower.

Assumptions:
1. The flow over the cooling tower can be
approximated as flow perpendicular to a cylinder
with a small L/D ratio.
2. The flow is uniform over the tower.
3. The air is at one atmosphere.

Solution:
a) The drag force is defined by Eq. 10-6:
FD = CD ρ V∞2 A 2
We assume the velocity is uniform over the whole tower.
We assume this shape can be approximated as flow over a cylinder with a small L/D ratio. Therefore, the area is:
A = π L2 4 and Davg = ( Dmax + Dmin ) 2 = ( 230 + 200 ) 2 = 215ft .
From Table 10-2, L Davg = 300 215 = 1.395 , so we let CD ≈ 0.64 . The velocity is 100 mi/hr = 146.7 ft/s.
For air at 70 ºF, ρ = 0.0754 lbm/ft3
FD = ( 0.64 ) ( 0.0754 lbm ft 3 ) (146.7 ft s ) (π 4 )( 300ft ) (1lbf s 2 32.2 ft lbm ) 2
2 2

= 2.28 × 106 lbf Answer

b) The moment around the base of the tower is:

∑ M = Fd = FD ( L 2 + 10ft )
= ( 2.28 × 106 lbf ) ( 300ft 2+10ft ) = 3.65 × 108 ft-lbf Answer

Comments:
Several approximations were made. These were necessary since we needed an answer and we did not have a drag
coefficient for this specific shape.

10- 9
10-10 A child releases a helium filled balloon that is spherical in shape into 80 ºF, 14.7 psia air. If the balloon
weighs 0.01 lbf and has a diameter of 1 ft, determine its terminal velocity.

Approach:
Terminal velocity occurs when there is a balance of
forces among weight, buoyancy, and drag. Drag force
depends on velocity and is calculated with the basic
drag force equation.

Assumptions:
1. The balloon is spherical and smooth.
2. Helium and air are ideal gases at the same
conditions.

Solution:
A force balance on the balloon is:
∑ F = 0 = Fbuoy − FD − WHe − Wballoon
The weight of the helium in the sphere depends on the helium density. Assuming helium is an ideal gas:
⎛ lbf ⎞ ⎛ lbm ⎞ ⎛ in.2 ⎞
⎜14.7 2 ⎟ ⎜ 4.003 ⎟ ⎜144 2 ⎟
PM He ⎝ in. ⎠ ⎝ lbmol ⎠ ⎝ ft ⎠ lbm
ρ He = = =0.0102 3
RT ⎛ ft lbf ⎞ ft
⎜1545 ⎟ ( 80+460 ) R
⎝ lbmolR ⎠
WHe = ρ HeVg = ( 32.2 ft s 2 )( 0.0102 lbm ft 3 ) π (1ft ) ( lbf s 2 32.2ft lbm ) 6 =0.00532lbf
3

The buoyancy force is the weight of the displaced air. Air density is:
ρ a = ( 0.0102 lbm ft 3 ) ( 28.97 4.003) =0.0738lbm ft 3
Fbuoy = ρ aVg = ( 32.2 ft s 2 )( 0.0738lbm ft 3 ) π (1ft ) ( lbf s 2 32.2ft lbm ) 6 =0.0387lbf
3

The drag force is defined by Eq. 10-6:

FD = CD ρ V 2 A 2
where A = π D 2 4 . The drag coefficient depends on the Reynolds number, which requires the velocity. For air
from Appendix B-7 at 80 ºF , µ = 1.25 × 10−5 lbf fts
ρ VD ( 0.0738lbm ft ) ( V ft s )(1ft )
3

Re = = =5900 ( V ft s )
µ 1.25 × 10 −5 lbf fts
Substituting the known quantities into the force balance and solving for velocity:
FD = CD ρ V 2 A 2 = Fbuoy − WHe − Wballoon
CD ( 0.0738lbm ft 3 ) V 2 ⎡π (1ft ) 4 ⎤ 2 = 0.0387 − 0.00532 − 0.01 = 0.0234lbf
2
⎣ ⎦
0.5

V =⎢
( ) ⎥ = ⎛ 26.0 ⎞0.5 ft
⎡ 2 ( 0.0234lbf ) 32.2ft lbm lbfs 2 ⎤
⎜ ⎟
⎣⎢ D ( )
⎢ C 0.0738lbm ft 3 ⎡π (1ft )2 4 ⎤ ⎥
⎣ ⎦ ⎦⎥
⎝ CD ⎠ s

From Figure 10-10, guess CD ≈ 0.4 . Therefore,

0.5
⎛ 26.0 ⎞ ft ft
V =⎜ ⎟ = 8.06 Re = 5900 ( 8.06 ) = 47, 600
⎝ 0.4 ⎠ s s
At this Reynolds number, again from Figure 10-10, CD ≈ 0.5
0.5
⎛ 26.0 ⎞ ft ft
V =⎜ ⎟ = 7.21 Re = 5900 ( 7.21) = 42, 600
⎝ 0.5 ⎠ s s
At this Reynolds number, again from Figure 10-10, CD ≈ 0.5 which is the same as previously calculated, so the
velocity has been determined. V = 7.21ft s Answer

10- 10
10-11 You hike to the top of a mountain and climb the fire tower. The wind is blowing at 80 km/hr. The air
temperature is 17 ºC and the pressure is 94 kPa. Estimate the wind force (in N) that would act on you.

Approach:
The basic drag force equation is used. We need to
estimate the frontal area of a person.

Assumptions:
1. An average person is 1.8-m tall and 40-cm wide.
2. Air is an ideal gas.

Solution:
The drag force is defined by Eq. 10-6:
FD = CD ρ V 2 A 2
From Table 10-2, for a standing person, CD ≈ 1.2 . For the frontal area, assume an average person is 1.8-m tall
and 40-cm wide.
A = (1.8m )( 0.40m ) =0.72m 2

For the air density, use the ideal gas equation:

⎛ kN ⎞ ⎛ kg ⎞
⎜ 94 2 ⎟ ⎜ 28.97 ⎟
PM kg
= ⎝
m ⎠⎝ kmol ⎠
ρ= =1.129 3
RT ⎛ kJ ⎞ m
⎜ 8.314 ⎟ (17+273) K
⎝ kmolK ⎠
FD = (1.2 ) (1.129 kg m 3 ) ( 22.2 m s ) ( 0.72m 2 )( Ns 2 kgm ) 2 =241N
2
Answer

10- 11
10-12 In the western United States, empty boxcars sometimes are blown over by strong crosswinds. Shown in the
figure are the dimensions of one type of a 20,000-kg boxcar. Determine the minimum wind velocity (in
m/s and in mi/hr) normal to the side of the boxcar needed to blow it over. Evaluate the air at 22 ºC, 101
kPa.

Approach:
Assume the drag force acts through the center of the
side of the boxcar, and the weight acts through the
center of mass. The boxcar would just begin to tip
when the total moment around a wheel is zero.

Assumptions:
1. Drag force acts through the center of a side of the
boxcar.

Solution:
As shown in the schematic, taking a moment balance around the left hand wheel in the end view:
∑ M = 0 = FD ( S + H 2 ) − W ( D 2 )
The drag force is defined by Eq. 10-6:
FD = CD ρ V 2 A 2
From Figure 10-5 for a rectangle with an aspect ratio of 12.5/3.2 = 3.9, CD ≈ 1.2 .
For the air density, at one atmosphere at 22 ºC, ρ = 1.193kg m3
FD = (1.2 ) (1.193kg m3 ) ( V m s ) (12.5m )( 3.2m ) 2 = ( 28.63kg m )( V m s )
2 2

Substituting into the moment equation and solving for velocity:

W ( D 2)
0.5
⎡ ⎤
V =⎢ ⎥
⎣⎢ ( 28.63kg m )( S + H 2 ) ⎥⎦
⎡ ( 20000kg ) ( 9.81m s 2 ) (1.44m 2 ) ⎤
0.5
m mi
=⎢ ⎥ =44.4 =99.3 Answer
⎢⎣ ( 28.63kg m )( 0.9m+ 3.2m 2 ) ⎥⎦ s hr

10- 12
10-13 In a bicycle race, a bicyclist coasts down a hill with a 7 % grade to save energy. The mass of the bicycle and
rider is 85 kg, the projected area is 0.22 m2, and the drag coefficient is 0.9. Air temperature is 17 ºC.
Neglecting rolling friction and bearing friction, determine:
a. the maximum velocity if the air is still (in m/s)
b. the maximum speed if there is a head wind of 5 m/s (in m/s)
c. the maximum speed if there is a tail wind of 5 m/s (in m/s)

Approach:
For a steady speed, there is a force balance between
the drag force and the weight in the direction of travel.

Assumptions:
1. Rolling resistance is ignored.
2. Air is at one atmosphere.

Solution:
As shown in the schematic, taking a force balance on the bike/rider combination:
∑ F = 0 = W sin θ − FD
The drag force is defined by Eq. 10-6:
FD = CD ρ V 2 A 2
The weight is W = mg .
Combining equations and solving for velocity:
0.5
⎡ 2mg sin θ ⎤
V =⎢ ⎥
⎣ CD ρ A ⎦
The angle for a 7% grade is:
θ = tan −1 ( 0.07 ) = 4.00o
For the air density, at one atmosphere at 17 ºC, ρ = 1.214 kg m3
⎡ 2 ( 85kg ) ( 9.81m s 2 ) sin ( 4.0o ) ⎤
0.5
m
V =⎢ ⎥ =22.0
⎢⎣ ( 0.9 ) (1.214 kg m )( 0.22m ) ⎥⎦
3 2
s
The velocity calculated is the relative velocity between the object and the air. Therefore, the rider’s velocity
relative to the ground is:
In still air: 22.0 m/s
With a 5 m/s headwind: 22.0 - 5= 17.0 m/s Answers
With a 5 m/s tailwind: 22.0 +5 = 27.0 m/s

10- 13
10-14 A 2.5-cm sphere with a specific gravity of 0.25 is released into a fluid with a specific gravity of 0.71. The
sphere rises at a terminal velocity of 0.5 cm/s. Determine the dynamic viscosity of the fluid (in N·s/m2).

Approach:
Terminal velocity occurs when there is a balance of
forces among weight, buoyancy, and drag. Drag force
depends on velocity and is calculated with the basic
drag force equation.

Assumptions:
1. The sphere rises at a steady velocity.
2. The Reynolds number is < ~ 1.

Solution:
A force balance on the sphere is:
∑ F = 0 = Fbuoy − FD − W → FD = Fbuoy − FD
The buoyancy force is the weight of the displaced liquid:
Fbuoy = ρ LVg = g ρ w SGπ D 3 6
= ( 9.81m s 2 )(1000 kg m3 ) ( 0.71) π ( 0.025m ) ( Ns 2 kgm ) 6 =0.057N
3

The weight of the sphere is:

W = ρ sphereVg = g ρ w SGπ D 3 6
= ( 9.81m s 2 )(1000 kg m3 ) ( 0.25 ) π ( 0.025m ) ( Ns 2 kgm ) 6 =0.020N
3

The drag force is defined by Eq. 10-6:

FD = CD ρ V 2 A 2
where A = π D 2 4 . From Table 10-2, we assume the Reynolds number is small, so that
CD = 24 Re = 24 ( ρ VD µ )
Substituting this into the drag equation:
FD
FD = ⎡⎣ 24 ( ρ VD µ ) ⎤⎦ ρ V 2 A 2 = 3π VD µ → µ=
3π VD
Combining the equations for force balance, drag force, weight, and buoyancy, and simplifying, we obtain
FD = 0.057 − 0.020 = 0.037N
0.037N Ns
µ= =31.3 2 Answer
3π ( 0.005 m s )( 0.025m ) m
Now checking the Reynolds number:
( 0.71) (1000 kg m3 ) ( 0.005 m s )( 0.025m )
Re = =0.0028
31.3 Ns m 2
This Reynolds number is well within the valid range of the assumed drag coefficient equation, so the calculated
viscosity is correct.

10- 14
10-15 Many sports cars are convertibles. The air flow over such a car is significantly different depending on
whether the convertible top is up or down. The engine of the 1000-kg car delivers 135 kW to the wheels, the
car frontal area is 1.9 m2, and rolling resistance is 2.5% of the car weight. The drag coefficient when the top
is down is 0.43 and 0.31 when it is up. For 20 ºC air at one atmosphere, determine:
a. the maximum speed with the top up (in m/s)
b. the maximum speed with the top down (in m/s)

Approach:
Power is force times velocity. The total force consists
of rolling resistance ( FRR ) and drag force ( FD ) . The
basic equation can be used to calculate the drag force.

Assumptions:
1. Frontal area does not change with top up or down.

Solution:
Power is force times velocity;
W = Ftot V
where Ftot = FD + FRR .
The drag force is defined by Eq. 10-6:
FD = CD ρ V 2 A 2
For the air density, at one atmosphere at 20 ºC, ρ = 1.201kg m3 . Because we have two different drag
coefficients to evaluate, we will leave the drag equation in terms of the drag coefficient:
FD = CD (1.201kg m3 ) ( V m s ) (1.9m 2 )( Ns 2 kgm ) 2 =1.141CD ( V m s )
2 2

The drag force is in terms of N.

The rolling resistance is:
FRR = 0.025W = 0.025mg = ( 0.025 )(1000kg ) ( 9.81m s 2 )( Ns 2 kgm ) =245N
Combining the equations for drag force, rolling resistance, and power, and simplifying, we obtain the
following equation:
135000W = ( Ws Nm ) ⎡ 245N + 1.141CD ( V m s ) ⎤ ( V m s )
2
⎣ ⎦
Solving this equation for the two conditions:
Top up: V = 69.3m s =155 mi hr Answer
Top down: V = 62.5 m s =139 mi hr Answer

10- 15
10-16 Wind speed is measured with an anemometer. A home-made anemometer can be constructed from a thin
plate hinged on one end; when the plate is hung from the hinge, wind impinging on the plate will cause the
plate to rotate around the hinge. The angular deflection is a measure of the wind speed. For a brass plate 20-
mm wide and 50-mm long, derive a relationship between wind speed and angular deflection, θ. Assume the
drag force on the plate depends only on the velocity component normal to the surface for angles less that
about 40º and the air temperature is 25 °C. Determine:
a. the relationship between wind speed and angular deflection
b. the thickness of brass needed for θ = 30º at a wind speed of 60 km/hr (in mm).

Approach:
In a steady wind, the angle θ is set by a moment
balance between the wind force and the weight of the
plate. The appropriate component of each force must
be used.

Assumptions:
1. Air is at one atmosphere.

Solution:
a) A moment balance around the hinge, using the normal component of the drag force and the weight is:
∑ M = 0 = FD , N L 2 − WN L 2
The friction drag force is defined by Eq. 10-6, and using the normal component of velocity:
FD , N = CD ρ a ( V cos θ ) A 2
2

where A is the area: A = LD

The normal component of the weight is:
WN = W sin θ = mg sin θ = ρ B LDtg sin θ
Combining the expressions and solving for velocity:
12
⎛ 2 ρ B tg sin θ ⎞
V =⎜ ⎟ Answer
⎝ CD ρ a cos θ ⎠
2

b) Solving the above equation for thickness, t:

C ρ cos 2θ V 2
t= D a
2 ρ B g sin θ
The drag coefficient is obtained from Figure 10-5. With an aspect ratio of 0.05 0.02 = 2.5, → CD ≈ 1.15
For air at 25 ºC, ρa = 1.184 kg/m3. For brass, ρB = 8530 kg/m3.
(1.15 ) ⎛ 1.184 ⎞ cos 2 ( 30o ) ⎛ 60000m ⎞
2

t= ⎜ ⎟ ⎜ ⎟ = 0.00421m=4.21mm Answer
2 ⎝ 8530 ⎠ ( 9.81m s 2 ) sin ( 30o ) ⎝ 3600s ⎠
Comments:
If the angle becomes too large then this analysis will become invalid since the flow over the plate will become too
different than flow perpendicular to a plate, and the drag coefficient will be affected.

10- 16
10-17 A 70-kg bicycle racer in the Tour de France can maintain about 40 km/hr on a calm day over level ground.
The bike has a mass of about 10 kg, and has a rolling resistance of 1% of the weight of bicycle and rider.
The drag coefficient of the bike and rider is 1.1, and their frontal area is 0.24 m2. The air temperature is 25
°C. Determine:
a. the power output by the rider (in kW) on level ground
b. the velocity the rider could attain going up a hill that has a slope of 6º (in km/hr).

Approach:
Power is force times velocity. The forces in the x-
direction are drag, rolling resistance, and a component
of the weight. Each of these forces can be evaluated
with the given information.

Assumptions:
1. Air properties are at 25 ºC.

Solution:
a) Power is: W = Ftot V
The total force is: Ftot = FD + FRR + W sin θ
Drag force is:
FD = CD ρ V 2 A 2
For air at 25 ºC, ρa = 1.184 kg/m3. Also, V = ( 40000 mi hr )(1hr 3600s ) =11.1m s
FD = (1.1) (1.184 kg m 3 ) (11.1m s ) ( 0.24m 2 )( Ns 2 kgm ) 2 =19.3N
2

The force due to the rolling resistance and weight on horizontal ground ( θ = 0o )
FRR + W sin θ = FRR = 0.01mg = ( 0.01)( 80kg )( 9.81m s ) ( Ns 2 kgm ) =7.85N
Combining these expressions and solving for the power:
W = (19.3N+7.85N )(11.1m s ) =301W Answer
b) When traveling uphill ( θ = 6 ), the weight term is no longer zero
o

W = Ftot V = ( CD ρ V∞2 A 2 + 0.01mg + W sin θ ) V

Substituting in known values and using consistent units:
301W = ⎡⎣(1.1) (1.184 kg m 3 ) V 2 ( 0.24m 2 ) 2 + 7.85N+ ( 80kg ) ( 9.81m s 2 ) sin6o ⎤⎦ V
301 = 0.1564 V 3 + 7.86 V + 82.0 V
Solving for velocity
V = 3.29 m s = 11.8 km hr Answer

Comments:
Using the same power going up a hill as can be produced on a horizontal road results in a significantly lower
velocity, as anyone who has ridden a bicycle knows.

10- 17
10-18 Assume the bicycle rider in Problem P 10-17 adds a fairing to streamline his bike and body. The drag
coefficient is reduced to 0.24 but the frontal area is increased to 0.29 m2. From the power the rider can
produce, estimate the new speed (in km/hr) the rider can maintain on level ground.

Approach:
We use the equation developed in part (b) of Problem
P10-17 to obtain the new speed.

Assumptions:
1. Air properties are at 25 ºC, 1 atm.

Solution:
We developed the equation:
W = Ftot V = ( CD ρ V 2 A 2 + 0.01mg + W sin θ ) V
On horizontal ground, θ = 0o . Using the variable values obtained in Problem P10-17, with the new drag
coefficient and frontal area:
301W = ⎡⎣( 0.24 ) (1.184 kg m 3 ) V 2 ( 0.29m 2 ) 2 + 7.85N+ ( 80kg ) ( 9.81m s 2 ) sin0o ⎤⎦ V
301 = 0.0412 V 3 + 7.86 V
V = 16.2 m s = 58.2 km hr Answer

Comments:
This velocity is a significant increase from the 40 km/hr obtained without the fairing. The effect of streamlining is
well illustrated by this comparison.

10- 18
10-19 A parachutist controls her free-fall speed by falling spread-eagle (CD ≈ 1.2) to slow down or head down (CD
≈ 0.4) to speed up. The frontal areas in the two positions are about 0.70 m2 and 0.25 m2, respectively. For
a 55-kg skydiver at 3000 m (assume the density and temperature are approximately constant at this
elevation), determine:
a. the terminal speed in each position (in km/hr)
b. the time (in s) and distance (in m) to reach 95% of the terminal speed.

Approach:
Terminal velocity is achieved when the drag force
equals the weight of the parachutist, so a simple force
balance is used. When terminal velocity has not been
achieved, an additional term for the acceleration must
be included in the force balance.

Assumptions:
1. Air properties are evaluated at 3000 m and
assumed constant.
2. Drag coefficients are constant.

Solution:
a) A force balance on the parachutist in free fall is:
∑ F = 0 = FD − W → FD = W
Drag force is: FD = CD ρ V 2 A 2
12 12
⎛ 2W ⎞ ⎛ 2mg ⎞
Combining equation and solving for velocity: V =⎜ ⎟ =⎜ ⎟
⎝ CD ρ A ⎠ ⎝ CD ρ A ⎠
From Table 10-5 for air at 3000 m, ρa = 0.909 kg/m3. Using the drag coefficients given in the problem statement:
12
⎡ 2 ( 55kg )( 9.81m s ) ⎤ m
Spread eagle: V =⎢ ⎥ =37.6 Answer
⎢⎣ (1.2 ) ( 0.909 kg m3 )( 0.70m 2 ) ⎥⎦ s
12
⎡ 2 ( 55kg )( 9.81m s ) ⎤ m
Head down: V =⎢ ⎥ =109 Answer
⎢⎣ ( 0.4 ) ( 0.909 kg m )( 0.25m ) ⎥⎦
3 2
s
b) The time required to obtain 95% of the terminal velocity is determined by another force balance, but this time
another term involving the acceleration is included:
d V CD ρ V 2 A
∑ Fy = ma y = FD − W → FD = W → −m
dt
=
2
− mg

Rearranging this equation and separating the variables:

dV C ρ V 2A C ρ A ⎛ 2 2mg ⎞ V dV t C ρA

dt
=− D
2m
+g = D
2m ⎝
⎜ -V + ⎟
CD ρ A ⎠
→ ∫ 0 ⎛ 2 2mg ⎞
=∫ D
0 2m
dt
⎜ -V + ⎟
⎝ CD ρ A ⎠
2mg
Let a 2 = . Note that a 2 > V 2 and that a is the terminal velocity. Performing the integration:
CD ρ A
1 ⎛ V ⎞ C ρA
tanh ⎜ ⎟ = D t
a ⎝ a ⎠ 2m
Rearranging this equation:
12 12
⎛V ⎞ ⎛ C ρ A ⎞ ⎛ 2mg ⎞ ⎛ CD ρ A ⎞ ⎛ CD ρ Ag ⎞
tanh ⎜ ⎟ = a ⎜ D ⎟t = ⎜ ⎟ ⎜ ⎟t = ⎜ ⎟ t
⎝ a ⎠ ⎝ 2m ⎠ ⎝ CD ρ A ⎠ ⎝ 2m ⎠ ⎝ 2m ⎠
12
⎛ 2m ⎞ −1 ⎛ V ⎞
t =⎜ ⎟ tanh ⎜ ⎟
⎝ CD ρ Ag ⎠ ⎝ a ⎠
To reach 95% of terminal velocity:

10- 19
 Spread eagle:
12
⎡ 2 ( 55kg ) ⎤
t=⎢ ⎥ tanh -1 ( 0.95 ) = ( 3.832s ) tanh -1 ( 0.95 ) =7.02s Answer
⎢⎣ (1.2 ) ( 0.909 kg m3 )( 0.70m 2 ) ( 9.81m s ) ⎥⎦
Head down:
12
⎡ 2 ( 55kg ) ⎤
t=⎢ ⎥ tanh -1 ( 0.95 ) = (11.11s ) tanh -1 ( 0.95 ) =20.3s Answer
⎢⎣ ( 0.4 ) ( 0.909 kg m )( 0.25m ) ( 9.81m s ) ⎥⎦
3 2

For the distance traveled, we solve the equation above for velocity and integrate with respect to time:
⎡⎛ C ρ Ag ⎞1 2 ⎤
V = a tanh ⎢⎜ D ⎟ t⎥
⎣⎢⎝ 2m ⎠ ⎦⎥
Because V = dy dt
y t t ⎡⎛ C ρ Ag ⎞1 2 ⎤ a ⎧⎪ ⎡⎛ CD ρ Ag ⎞1 2 ⎤ ⎪⎫
y = ∫ dy = ∫ Vdt = ∫ a tanh ⎢⎜ D ⎟ t ⎥ = 12
log ⎨ cosh ⎢ ⎜ ⎟ t⎥⎬
0 0 0
⎢⎣⎝ 2m ⎠ ⎥⎦ ⎛ CD ρ Ag ⎞ ⎪⎩ ⎢⎣⎝ 2m ⎠ ⎥⎦ ⎭⎪
⎜ 2m ⎟
⎝ ⎠
37.6 m s ⎧ ⎡⎛ 1 ⎞ ⎤⎫
Spread eagle: y= log ⎨cosh ⎢⎜ ⎟ ( 7.02s ) ⎥ ⎬ = 72.8m Answer
⎛ 1 ⎞ ⎩ ⎣⎝ 3.832s ⎠ ⎦⎭
⎜ ⎟
⎝ 3.832s ⎠
109 m s ⎧ ⎡⎛ 1 ⎞ ⎤⎫
Head down: y= log ⎨cosh ⎢⎜ ⎟ ( 20.3s )⎥ ⎬ = 610m Answer
⎛ 1 ⎞ ⎩ ⎣⎝ 11.11s ⎠ ⎦⎭
⎜ ⎟
⎝ 11.11s ⎠
Comments:
The constant drag coefficient assumption is reasonable, since for blunt bodies, after a certain Reynolds number the
drag coefficient does not vary much.

10- 20
10-20 In the United States, the Bonneville Salt Flats in Utah are used by individuals trying to set land speed
records in various class vehicles. One challenger has developed a 1750 lbf car that has a 675-hp engine, a
streamlined body with a drag coefficient CD ≈ 0.29, a frontal area of 13.5 ft2, and rolling resistance of only
3% of the body weight. The car’s transmission has an efficiency of 88% (that is, 88% of the engine power
is transferred to the tires). On a day when the air temperature is 95 °F, determine the maximum speed of
the car (in mi/hr).

Approach:
Power is force times velocity. For this car, force
consists of drag force and rolling resistance, both of
which can be evaluated from the given information.

Assumptions:
1. The air is at one atmosphere.
2. The drag coefficient is constant.

Solution:
Power is obtained from:
W = Ftot V = ( FD + FRR ) V
The rolling resistance is:
FRR = ( 0.03)(1750lbf ) =52.5lbf
The drag force is defined by Eq. 10-6:
FD = CD ρ V 2 A 2
For air at 95 ºF, ρa = 0.0715 lbm/ft3
FD = ( 0.29 ) ( 0.0715lbm ft 3 ) ( V ft s ) (13.5ft 2 )( lbf s 2 32.2ft lbm ) 2 =0.00435V 2 lbf
2

Combining these equations and using the given information:

⎛ 1hp s ⎞
( 0.88 )( 675hp ) = ⎣⎡0.00435 V 2 + 52.5⎦⎤ lbf ( V ft s ) ⎜ ⎟
⎝ 550ft lbf ⎠
Solving for velocity: 326, 700 = 0.00435 V 3 + 52.5 V
V = 428ft s =292 mi hr Answer

Comments:
Note that if we ignore the rolling resistance, the top speed would be 437 ft/s.

10- 21
10-21 A BMW 520 has a drag coefficient of 0.31 and a frontal area of 22.5 ft2. It weighs 3,500 lbf. If rolling
resistance is 1.5% of the weight, determine:
a. the speed at which drag resistance becomes larger than the rolling resistance (in mi/hr)
b. the power (in kW and hp) required to cruise at 45 mi/hr and 75 mi/hr.

Approach:
Power is force times velocity. For this car, force
consists of drag force and rolling resistance, both of
which can be evaluated from the given information.

Assumptions:
1. The air is at 1 atm, 77 °F.

Solution:
a) The rolling resistance is:
FRR = 0.015W = ( 0.015 )( 3500lbf ) =52.5lbf
The drag force is defined by Eq. 10-6:
FD = CD ρ V 2 A 2
For air at 77 ºF, ρa = 0.074 lbm/ft3. Equating the drag and rolling resistance, and solving for velocity:
⎡ 2 ( 52.5lbf ) ( 32.2 ft lbf lbm s 2 ) ⎤
12
12
⎛ 2 FRR ⎞ ft mi
V =⎜ ⎟ =⎢ ⎥ =80.9 =55.2 Answer
⎝ CD ρ A ⎠ ⎢⎣ ( 0.31) ( 0.074 lbm ft 3 )( 22.5ft 2 ) ⎥⎦ s hr

b) Power is obtained from:

W = Ftot V = ( FD + FRR ) V
FD = ( 0.31) ( 0.074 lbm ft 3 ) ( V ft s ) ( 22.5ft 2 )( lbf s 2 32.2ft lbm ) 2 =0.00801V 2 lbf
2

Combining these equations and using the given information:

⎛ 1hp s ⎞
W = ⎡⎣0.00801 V 2 + 52.5⎤⎦ lbf ( V ft s ) ⎜ ⎟
⎝ 550ft lbf ⎠
At 45 mi/hr = 66 ft/s W = 10.5hp=7.822kW Answer
At 75 mi/hr = 110 ft/s W = 29.9hp=22.3kW Answer

Comments:
Note that the power required for the cruising speed is much small that the typical engine installed in cars. More
power is required to accelerate a car than is required to maintain a constant speed. Power requirements increase as
the time to reach a given speed decreases.

10- 22
10-22 Some military jets deploy parachutes when they land to reduce the distance required to stop. Suppose a
14,500-kg jet uses two 6-m diameter parachutes and lands at 300 km/hr in 20 °C air. Determine:
a. the total force the cables connecting the parachutes to the plane must withstand (in N)
b. the time (in s) and distance (m) required to decelerate the plane to 150 km/hr (without using brakes
and ignoring drag from the plane).

Approach:
For part (a), the drag force can be determined from the
basic drag equation. For part (b), all forces, including
deceleration forces must be included in a force
balance.

Assumptions:
1. Air is at one atmosphere.
2. There is no interaction between the two parachutes.

Solution:
a) Drag force is defined by Eq. 10-6:
FD = CD ρ V 2 A 2
where A is the projected area of the parachutes: A = π D 2 4
The drag coefficient is obtained from Table 10-2, CD ≈ 1.3 . For air at 20 ºC, ρ = 1.205 kg/m3.
FD = (1.3) (1.205 kg m 3 ) ( 0.5 )( 83.3m s ) π ( 6m ) 4 =153,800N
2 2

For two parachutes, Ftot = 2 FD = 307, 600N Answer

b) Using Newton’s second law of motion in the x-direction:

dV
∑ Fx = −2 FD = max = m dt
Drag force is as given above. Substituting that expression into the force balance, separating variables, and
integrating from the initial velocity to a second velocity:
t C ρA C ρA
V
V dV 1 1 1
∫ Vi V 2 ∫0 m
= − → − = − =− D
D
dt t
V Vi Vi V m
Solving for t:
m ⎛ 1 1 ⎞ (14500kg ) ⎛ 1 1 ⎞
t= ⎜⎜ − ⎟= 2 ⎜
- ⎟
CD ρ A ⎝ V f Vi ⎠ (1.3) (1.205 kg m ) ( π 4 )( 6m ) ⎝ 41.66 m s 83.33m s ⎠
⎟ 3

t = 3.93s Answer

To find the distance, we revisit the original force balance:

−2 FD = max → − CD ρ V 2 A = max
Note that ax = d V dt . Using the chain rule on the derivative:
d V dx dV
ax = =V
dx dt dx
Substituting this expression into the force balance, separating variables, and integrating:
dV xC ρA Vf d V
−CD ρ V 2 A = m V → − ∫ D dx = ∫
dx 0 m Vi V
C ρA ⎛ Vf ⎞ m ⎛ Vf ⎞
− D x = ln ⎜ ⎟ → x=− ln ⎜ ⎟
m ⎝ Vi ⎠ CD ρ A ⎝ Vi ⎠
14500kg ⎛ 150 ⎞
x=− ln ⎜ ⎟ =227m Answer
(1.3) (1.205 kg ) ( π 4 )( 6m ) 2
m 3
⎝ 300 ⎠

10- 23
10-23 In some automobiles, gas mileage (km/L) is calculated and displayed on the instrument panel. One day on
a long drive, a bored engineering student realizes that his gas mileage is 20% lower traveling into a head
wind than when there was no head wind. The road is level, the temperature is 7 ºC, and his speed is 120
km/hr. The driver (a car enthusiast) knows that the drag coefficient of his car is 0.35, frontal area is 2.1 m2,
mass is 950 kg, and rolling resistance is 3% of the body weight. To pass the time, he uses this information
to calculate the head wind velocity. What is it (in km/hr)?

Approach:
We assume gas mileage is inversely proportional to
power, which is calculated with force times velocity.
The total force consists of rolling resistance ( FRR )
and drag force ( FD ) . The velocity in the drag force is
the relative velocity between the car velocity and the
air velocity. The basic equation can be used to
calculate the drag force.

Assumptions:
1. Mileage is inversely proportional to power.
2. Air is at one atmosphere.
3. The quantity, C, is constant.

Solution:
Power is force times velocity;
W = Ftot V
where Ftot = FD + FRR . Using a constant of proportionality, C, to convert power to gas mileage, GM:
GM = C W = C ( F V )tot

The drag force is defined by Eq. 10-6:

FD = CD ρ V 2 A 2
For the air density, at one atmosphere at 7 ºC, ρ = 1.257 kg m3 . Therefore, the drag force is:
FD = ( 0.35 ) (1.257 kg m3 ) ( V m s ) ( 2.1m 2 )( Ns 2 kgm ) 2 =0.462 ( V m s )
2 2

The drag force is in terms of N.

The rolling resistance is:
FRR = 0.03W = 0.03mg = ( 0.03 )( 950kg ) ( 9.81m s 2 )( Ns 2 kgm ) =280N
Combining the equations for gas mileage, drag force, rolling resistance, and power, and simplifying, we obtain the
following equations:
C
In still air: GM 1 = (1)
⎡ 280N + 0.462 ( V1 m s )2 ⎤ ( V1 m s )
⎣ ⎦
C
With a head wind: GM 2 = (2)
⎡ 280N + 0.462 (V2 m s )2 ⎤ ( V2 m s )
⎣ ⎦
GM 2
We also know that = 0.80 (3)
GM 1
And V2 = V1 + Vheadwind = 33.3 + Vheadwind (4)
We have four equations and four unknowns. Solving them simultaneously:
V2 = 36.67 m s
The head wind is: Vheadwind = 36.67 − 33.3 = 3.33m s =12 km hr Answer

10- 24
10-24 If you have ever been hit by a hailstone, you know it can hurt because of its high speed. Consider a 4-cm
hailstone falling in 17 ºC, 96 kPa air. Assume the hailstone has a specific gravity of 0.84. Determine its
terminal velocity (in m/s and mi/hr):
a. for a smooth hailstone
b. for a hailstone with a surface roughness similar to that of a golf ball.

Approach:
Terminal velocity occurs when there is a balance of
forces among weight, buoyancy, and drag. The drag
force must be determined iteratively.

Assumptions:
1. Hailstone is a sphere.

Solution:
a) A force balance on the hailstone is:
∑ F = 0 = Fbuoy + FD − W → FD = W − Fbuoy
Assuming the hailstone is a sphere, its weight is:
W = ρ sphereVg = (1000 kg m3 ) ( 0.84 ) ( 9.81m s 2 ) π ( 0.04m ) ( Ns 2 kgm ) 6 =0.276N
3

The buoyancy force is the weight of the displaced air. For the air density, use the ideal gas equation:
⎛ kN ⎞ ⎛ kg ⎞
PM ⎜ 96 2 ⎟ ⎜ 28.97 ⎟ kg
ρ= = ⎝ m ⎠⎝ kmol ⎠
=1.153 3
RT ⎛ kJ ⎞ m
⎜ 8.314 ⎟ (17+273) K
⎝ kmolK ⎠
Fbuoy = ρVg = g ρπ D 3 6 = ( 9.81m s 2 )(1.153kg m3 ) π ( 0.04m ) ( Ns 2 kgm ) 6 =3.8 × 10-4 N
3

The drag force is defined by Eq. 10-6, FD = CD ρ V 2 A 2 , where A = π D 2 4 . The drag force depends
on the Reynolds number. From Appendix A-7, for air at 17 ºC, µ = 1.77 × 10−5 Ns m 2 .
Combining the equations for force balance, drag force, weight, and buoyancy, and simplifying, we obtain
⎡ 8 (W − Fbuoy ) ⎤ 8 ( 0.276-3.8×10-4 ) N
0.5 0.5
⎡ ⎤ 0.5
⎛ 381 ⎞ m
V =⎢ ⎥ =⎢ ⎥ =⎜ ⎟
⎢⎣ CD ρπ D ⎥⎦
2
⎢ CD π (1.153kg m3 ) ( 0.04m )2 ⎥ ⎝ CD ⎠ s
⎣ ⎦
Using Figure 10-10 for a smooth sphere, guess CD ≈ 0.5 . Calculating the velocity:
V = ( 381 0.5 )
0.5
= 27.6 m s
Now checking the Reynolds number:

Re =
(1.153kg m3 ) ( 27.6 m s )( 0.04m )
=71,900
1.77 × 10−5 Ns m 2
At this Reynolds number CD ≈ 0.51 , and recalculating the velocity
V = ( 381 0.51)
0.5
= 27.3m s Answer
This is close enough to the first calculation so that we do not need to iterate again.
b) Using Fig. 10-14 for the rough hailstone at the above Reynolds number, CD ≈ 0.23 ,
V = ( 381 0.23)
0.5
= 40.7 m s , the Reynolds number is 106,000.
V = ( 381 0.24 )
0.5
Reevaluating the drag coefficient, CD ≈ 0.24 , = 39.8 m s Answer
Comments:
The dramatic increase in the velocity is due to the roughness tripping the boundary layer into turbulence with its
attendant decrease in drag coefficient. Note that buoyancy could have been neglected with no loss in accuracy.

10- 25
10-25 A beginning bicyclist can produce 84 W for short periods of time. On a hot day (32 ºC) how fast can the
bicyclist travel if the projected area of the bike and cyclist is 0.5 m2, and the drag coefficient is 1.1?

Approach:
Power is force times velocity. Ignoring rolling
resistance, the only force is drag force, and the basic
equation can be used.

Assumptions:
1. Rolling resistance is ignored.
2. Air is at one atmosphere.

Solution:
Power is force times velocity;
W = FD V
The drag force is defined by Eq. 10-6:
FD = CD ρ V 2 A 2
Ignoring rolling resistance, combining equations, and solving for velocity:
13
⎡ W ⎤
V =⎢ ⎥
⎣ CD ρ A ⎦
For the air density, at one atmosphere at 32 ºC, ρ = 1.154 kg m3
⎡ 2 ( 84W ) ( kg m Ns 2 ) (1Nm Ws ) ⎤
13
m km
V =⎢ ⎥ =6.42 =23.1 Answer
⎢⎣ (1.1) (1.154 kg m )( 0.5m ) ⎥⎦
3 2
s hr

10- 26
10-26 A copper sphere 10-mm in diameter is dropped into a 1-m deep drum of asphalt. The asphalt has a density
of 1150 kg/m3 and a viscosity of 105 N·s/m2. Estimate the time (in hours) it takes for the sphere to reach
the bottom of the drum.

Approach:
Ignoring the time required for the sphere to accelerate
to its terminal velocity, the time required for the
particle to settle back to earth would be t = L V .
Terminal velocity occurs when there is a balance of
forces among weight, buoyancy, and drag.

Assumptions:
1. Ignore time required to accelerate to terminal
velocity.
2. The sphere is smooth.

Solution:
The time required for the sphere to fall 1m is calculated with t = L V , assuming we ignore the time required to
accelerate to terminal velocity, which we can determine from a force balance.
A force balance on the sphere is:
∑ F = 0 = Fbuoy + FD − W → FD = W − Fbuoy
With a density of copper of 8933 kg/m3, the weight of the sphere is:
W = ρ sphereVg = ( 9.81m s 2 )( 8933kg m3 ) π ( 0.01m ) ( Ns 2 kgm ) 6 =0.0459N
3

The buoyancy force is the weight of the displaced liquid.

Fbuoy = ρVg = g ρπ D 3 6
= ( 9.81m s 2 )(1150 kg m3 ) π ( 0.01m ) ( Ns 2 kgm ) 6 =5.91× 10-3 N
3

The drag force is defined by Eq. 10-6:

FD = CD ρ V 2 A 2
where A = π D 2 4 . From Table 10-2, because of the very high viscosity, we assume the Reynolds number is
small, so that CD = 24 Re = 24 ( ρ VD µ )
Substituting this into the drag equation:
FD
FD = ⎡⎣ 24 ( ρ VD µ ) ⎤⎦ ρ V 2 A 2 = 3π VD µ → V =
3πµ D
Combining the equations for force balance, drag force, weight, and buoyancy, and simplifying, we obtain

V =
( 0.0459-5.91×10-3 ) N =4.24×10-6 m
3π (105 Ns m 2 ) ( 0.01m ) s
With such a low velocity and high viscosity, there is no need to check the Reynolds number, since the Reynolds
number will be well within the valid range of the assumed drag coefficient equation. Now calculating the settling
time.
L 1m
t= = =2.358 × 105s=65.5 hr=2.73 days Answer
V 4.24 × 10-6 m s

10- 27
10-27 A meteorological balloon is to be filled with helium at 0 ºC, 100 kPa. The surrounding air is at the same
pressure and temperature. The instrument package the balloon must lift has a mass of 30 kg, and the
balloon material has a mass of 0.15 kg/m2. If an upward vertical velocity of 3 m/s is desired, what diameter
(in m) balloon is required?

Approach:
At terminal velocity there is a balance of forces
among weight, buoyancy, and drag. The drag and
buoyancy forces depend on the balloon diameter,
which is unknown. A force balance must be written,
and each force evaluated.

Assumptions:
1. The balloon is spherical and smooth.
2. Air and helium are ideal gases at the same
conditions.

Solution:
A force balance on the balloon is: ∑ F = 0 = Fbuoy − FD − WHe − Wballoon − Winstrum . Each term (except the weight of
the instruments) is written in terms of the unknown diameter.
The instrumentation package weight is:
Winstrum = mg = ( 30kg ) ( 9.81m s 2 )( Ns 2 kgm ) =294N
The balloon weight is:
Wballoon = mg = ( m A ) Ag = ( m A ) π D 2 g = ( 0.15 kg m 2 ) π ( Dm ) ( 9.81m s 2 )( Ns 2 kgm ) = 4.623D 2 N
2

The buoyancy force is the weight of the displaced air. Assuming air is an ideal gas:
PM (100 kN m ) ( 28.97 kg kmol )
2
kg
ρa = = =1.276 3
RT (8.314 kJ kmolK )( 0+273) K m
Fbuoy = ρ aVg = ( 9.81m s 2 )(1.276 kg m3 ) π ( Dft ) ( N s 2 kg m ) 6 =6.556D 3 N
3

The helium weight is:

PM ⎛ kg ⎞ ⎛ 4.003 ⎞ kg
ρ He = = ⎜ 1.276 3 ⎟ ⎜ ⎟ = 0.176 3
RT ⎝ m ⎠ ⎝ 28.97 ⎠ m
WHe = ρ HeVg = ( 9.81m s 2 )( 0.176 kg m3 ) π ( Dft ) ( N s 2 kg m ) 6 =0.906D 3 N
3

The drag force is defined by Eq. 10-6:

FD = CD ρ V 2 A 2
where A = π D 2 4 . The drag coefficient depends on the Reynolds number, which requires the velocity. For air
from Appendix A-7 at 0 ºC , µ = 1.653 × 10−5 Ns m 2
ρ VD (1.276 kg m ) ( 3m s )( Dm )
3

Re = = =231,600 ( Dm )
µ 1.653 × 10−5 Ns m 2
The drag force is:
FD = CD (1.276 kg m3 ) ( 3m s ) (π 4 )( Dm ) ( N s 2 kgm ) 2 =4.510CD D 2 N
2 2

Combining the force expressions: 0 = 6.556 D 3 − 4.510CD D 2 − 0.906 D 3 − 4.623D 2 − 294

The procedure to obtain a solution is: 1) guess a value of D, 2) calculate Re from the above equation, 3), at
this Re obtain CD from Figure 10-10, and 4) calculate D from the cubic equation and compare to the guessed
value. Continue the iteration until convergence is obtained. Performing the iteration:
CD ≈ 0.20 → D ≈ 4.09 m → Re ≈ 9.10 × 105 Answer

10- 28
10-28 In dry regions, wind storms can entrain much dust into the air. For a particle 0.05 mm in diameter with a
density of 1.8 g/cm3 raised to a height of 100 m in such a storm, estimate how long it will take the particle
to settle back to earth. Assume the air is at 27 ºC, 100 kPa and that the time required to reach the terminal
velocity is negligible.

Approach:
Ignoring the time required for the particle to accelerate
to its terminal velocity, the time required for the
particle to settle back to earth would be t = H V .
Terminal velocity occurs when there is a balance of
forces among weight, buoyancy, and drag.

Assumptions:
1. Ignore time required to accelerate to terminal
velocity.
2. The particle is a smooth sphere.
3. Reynolds number is < ~1.

Solution:
A force balance on the particle is:
∑ F = 0 = Fbuoy − FD − W → FD = Fbuoy − W
The weight of the sphere is:
W = ρ sphereVg = ( 9.81m s 2 )(1.8g cm3 ) π ( 5×10-5 m ) (100 cm 1m ) ( Ns 2 kgm ) (1kg 1000g ) 6 =1.156×10-9 N
3 3

The buoyancy force is the weight of the displaced air. For the air density, use the ideal gas equation:
⎛ kN ⎞ ⎛ kg ⎞
PM ⎜ 100 2 ⎟ ⎜ 28.97 ⎟ kg
ρ= = ⎝ m ⎠⎝ kmol ⎠
=1.162 3
RT ⎛ kJ ⎞ m
⎜ 8.314 ⎟ ( 27+273) K
⎝ kmolK ⎠
Fbuoy = ρVg = g ρπ D 3 6

= ( 9.81m s 2 )(1.162 kg m3 ) π ( 5 × 10-5 m ) ( Ns 2 kgm ) 6 =7.458 × 10-13 N

3

The drag force is defined by Eq. 10-6:

FD = CD ρ V 2 A 2
where A = π D 2 4 . From Table 10-2, we assume the Reynolds number is small (< ~1), so that
CD = 24 Re = 24 ( ρ VD µ )
Substituting this into the drag equation:
FD
FD = ⎡⎣ 24 ( ρ VD µ ) ⎤⎦ ρ V 2 A 2 = 3π VD µ → V =
3πµ D
From the appendix for air from Appendix A-7 at 27 ºC, µ = 1.846 × 10−5 Ns m 2
Combining the equations for force balance, drag force, weight, and buoyancy, and simplifying, we obtain

V =
(1.156×10-9 -7.458×10-13 ) N
=0.133
m
3π (1.846 × 10 Ns m )( 5×10 m )
−5 2 -5
s
Now checking the Reynolds number:

Re =
(1.162 kg m3 ) ( 0.133m s ) ( 5 × 10-5 m )
=0.42
1.846 × 10−5 Ns m 2
This Reynolds number is well within the valid range of the assumed drag coefficient equation, so we can now
calculate the settling time.
H 100m
t= = =752s=12.5min Answer
V 0.133m s

10- 29
10-29 A 5-mm iron sphere is dropped into a tank of 17 ºC unused engine oil. Determine the sphere’s terminal
velocity (in cm/s).

Approach:
Terminal velocity occurs when there is a balance of
forces among weight, buoyancy, and drag. The drag
force can be determined iteratively.

Assumptions:
1. Sphere is smooth.

Solution:
a) A force balance on the sphere is:
∑ F = 0 = Fbuoy + FD − W → FD = W − Fbuoy
3
With a density of iron of 7870 kg/m , the weight of the sphere is:
W = ρ sphereVg = ( 7870 kg m3 )( 9.81m s 2 ) π ( 0.005m ) ( Ns 2 kgm ) 6 =5.053 × 10-3 N
3

The buoyancy force is the weight of the displaced oil. From Appendix A-6 for oil at 20 ºC, ρ = 888.2 kg m3 ,
µ = 8450 × 10−4 Ns m 2
Fbuoy = ρVg = g ρπ D 3 6 = ( 9.81m s 2 )( 888.2 kg m3 ) π ( 0.005m ) ( Ns 2 kgm ) 6 =5.703 × 10-4 N
3

The drag force is defined by Eq. 10-6, FD = CD ρ V 2 A 2 , where A = π D 2 4 . The drag force depends
on the Reynolds number.
Combining the equations for force balance, drag force, weight, and buoyancy, and simplifying, we obtain
⎡ 8 (W − Fbuoy ) ⎤ ⎡ 8 ( 5.053×10-3 -5.703×10-4 ) N ⎤
0.5 0.5
0.5
⎛ 0.752 ⎞ m
V =⎢ ⎥ = ⎢ ⎥ = ⎜ ⎟
⎢⎣ CD ρπ D ⎥⎦
2
⎢ CD π ( 888.2 kg m3 ) ( 0.005m )3 ⎥ ⎝ CD ⎠ s
⎣ ⎦
The Reynolds number is:
(888.2 kg m3 ) ( V m s )( 0.005m ) =5.256 V m s
Re = ( )
8450 × 10−4 Ns m 2
To determine the terminal velocity, the procedure to use is: 1) guess CD using Figure 10-8, 2) calculate the
velocity, 3) calculate the Reynolds number, 4) evaluate CD using Figure 10-8 and compare to the initial guess, 50
repeat the proceeding steps until converged.
Guess CD ≈ 0.5 , calculate velocity V = ( 0.752 0.5 )
0.5
= 1.23m s , calculate Reynolds number
Re = 5.256 (1.23m s ) = 6.46 , reevaluate CD ≈ 6 ; recalculate velocity V = ( 0.752 7 )
0.5
= 0.354 m s , Reynolds
number Re = 1.86 . At this Reynolds number CD = 24 Re = 24 1.86 = 12.9 and V = ( 0.752 12.9 )
0.5
= 0.241m s .
Continuing several more steps in the iteration, we obtain
CD ≈ 27.6 , V = 0.165 m s , Re = 0.868 Answer

Comments:
Note that if we had initially assumed that the Reynolds number would be < ~1, then we could have used
CD = 24 Re to obtain an analytic solution. One the terminal velocity had been determined, we would have
needed to check the Reynolds number to confirm the assumption.

10- 30
10-30 The military sometimes needs to move large equipment into remote areas where there are no landing strips,
so the equipment is parachuted to the ground. To prevent damage, a bulldozer weighing 45 kN cannot
strike the ground at a velocity greater than 10 m/s. Determine how many 20-m diameter parachutes are
required when the air is at 17 ºC, 95 kPa?

Approach:
Terminal velocity occurs when there is a balance of
forces among weight, buoyancy, and drag. Buoyancy
is negligible compared to the drag and weight. We
need to determine the drag force on a parachute.

Assumptions:
1. Buoyancy is ignored.
2. Air is an ideal gas.

Solution:
A force balance on the bulldozer is:
∑ F = 0 = NFD − W → N = W FD
where N is the number of parachutes and FD is the drag force on one parachute.
The drag force is defined by Eq. 10-6:
FD = CD ρ V 2 A 2
where A = π D 2 4 . Using Table 10-2, the drag coefficient for a parachute is CD = 1.3
For the air density, use the ideal gas equation:
⎛ kN ⎞ ⎛ kg ⎞
95 28.97
PM ⎜⎝ m 2 ⎟⎠ ⎜⎝ ⎟
kmol ⎠ kg
ρ= = =1.1415 3
RT ⎛ kJ ⎞ m
⎜ 8.314 ⎟ ( 290K )
⎝ kmolK ⎠
The drag force is:
FD = (1.3) (1.1415 kg m3 ) (10 m s ) (π 4 )( 20m ) ( Ns 2 kgm ) 2 =23,300N
2 2

Solving for the number of parachutes:

45,000N
N= = 1.93
23,300N
Therefore, we use two parachutes and the terminal velocity would be slight under 10 m/s. Answer

10- 31
10-31 A helium-filled spherical balloon is released into air at 40 ºF, 14.0 psia. The combined weight of the
balloon and its payload is 300 lbf. If a vertical velocity of 10 ft/s is desired, what diameter balloon is
required (in ft)? Assume the helium is at the same temperature and pressure as the air. If this balloon is
tethered to the ground in a 10 mi/hr wind, what angle does the restraining cable make with the ground?

Approach:
At terminal velocity there is a balance of forces
among weight, buoyancy, and drag. The drag and
buoyancy forces depend on the balloon diameter,
which is unknown. A force balance must be written,
and each force evaluated. For part (b), there is a
horizontal velocity only. Again, a force balance must
be evaluated, which includes the tension in the
tethering cable.

Assumptions:
1. The balloon is spherical and smooth.
2. Air is an ideal gas.
3. The cable acts as a rigid link.

Solution:
a) A force balance on the balloon is: ∑F = 0 = F buoy − FD − W . Each term is written in terms of the
unknown diameter.
The buoyancy force is the weight of the displaced air. Assuming air is an ideal gas:
PM (14 lbf in. ) ( 28.97 lbm lbmol ) (144in ft )
2 2 2
lbm
ρa = = =0.0756 3
RT (1545ft lbf lbmolR )( 40+460 ) R ft
Fbuoy = ρ aVg = ( 32.2 ft s 2 )( 0.0756 lbm ft 3 ) π ( Dft ) ( lbf s 2 32.2ft lbm ) 6 =0.0396D 3 lbf
3

The drag force is defined by Eq. 10-6:

FD = CD ρ V 2 A 2
where A = π D 2 4 . The drag coefficient depends on the Reynolds number, which requires the diameter. For air
from Appendix B-7 at 40 ºF , µ = 1.179 × 10−5 lbm fts
ρ VD ( 0.0756 lbm ft ) (10 ft s )( Dft )
3

Re = = =64,100 ( Dm )
µ 1.179 × 10−5 lbm fts
FD = CD ( 0.0756 lbm ft 3 ) (10 ft s ) (π 4 )( Dft ) ( lbf s 2 32.2ft lbm ) 2 =0.0922CD D 2 lbf
2 2
The drag force is:
Combining the force expressions: 0 = 0.0396 D 3 − 0.0922CD D 2 − 300
The procedure to obtain a solution is: 1) guess a value of D, 2) calculate Re from the above equation, 3), at
this Re obtain CD from Figure 10-10, and 4) calculate D from the cubic equation and compare to the guessed
value. Continue the iteration until convergence is obtained. Performing the iteration:
CD ≈ 0.20 → D ≈ 19.8 ft → Re ≈ 1.27 × 106 Answer
b) For the stationary balloon tethered by a cable, two force balances are required, as was done in Example 10-4.
We need to calculate a new drag force for the 10 mi/hr = 14.7 ft/s wind.
Re = 1.27 × 106 (14.7 10 ) = 1.87 × 106 From the Figure 10-10, CD ≈ 0.25
FD = ( 0.25 ) ( 0.0756 lbm ft 3 ) (14.7 ft s ) (π 4 )(19.8ft ) ( lbf s 2 32.2ft lbm ) 2 =19.5lbf
2 2

From above, the buoyancy force is: Fbuoy = 0.0396 (19.8 ) =307 lbf
3

Using the expression given in Example 10-4 and adapting it to this problem:
⎛ F −W ⎞ −1 ⎛ 307 − 300 ⎞
θ = tan −1 ⎜ buoy ⎟ = tan ⎜ ⎟ = 19.7
o
Answer
⎝ F D ⎠ ⎝ 19.5 ⎠
Comment:
One assumption used in Example 10-4 was that the connecting cable acted as a rigid link. This is reasonable if the
angle is close to 90°. The present solution is not good because the angle is so small.
10- 32
10-32 A 40-mm ping pong ball weighing 0.025 N is released from the bottom of a 4-m deep swimming pool
whose temperature is 20 ºC. Ignoring the time to reach terminal velocity, how long does the ball take to
reach the pool surface (in s)?

Approach:
The time required for the ball to reach the surface of
the swimming pole is obtained by dividing the depth
by the velocity. Ignoring the time required to reach
terminal velocity, terminal velocity occurs when there
is a balance of forces among weight, buoyancy, and
drag. The drag force must be determined iteratively.

Assumptions:
1. Ignore the time required to accelerate to terminal
velocity.

Solution:
Ignoring the time required to accelerate the ping pong ball to its terminal velocity, the time required for the sphere
to reach the surface is t = H V . Hence, we need to determine the terminal velocity.
A force balance on the ping pong ball is:
∑ F = 0 = Fbuoy − FD − W → FD = Fbuoy − W
The buoyancy force is the weight of the displaced liquid. For water from Appendix A-6 at 20ºC, ρ = 998.2
kg/m and µ = 9.85 × 10−4 Ns m 2 .
3

Fbuoy = ρVg = g ρπ D 3 6 = ( 9.81m s 2 )( 998.2 kg m3 ) π ( 0.04m ) ( Ns 2 kgm ) 6 =0.328N

3

The drag force is defined by Eq. 10-6, FD = CD ρ V 2 A 2 , where A = π D 2 4 .

Combining the equations for force balance, drag force, weight, and buoyancy, and simplifying, we obtain
⎡ 8 ( Fbuoy − W ) ⎤
0.5 0.5
⎡ 8 ( 0.328-0.025 ) N ⎤ 0.5
⎛ 0.483 ⎞ m
V =⎢ ⎥ = ⎢ ⎥ =⎜ ⎟
⎢⎣ CD ρπ D ⎥⎦
2
⎢ CD π ( 998.2 kg m3 ) ( 0.04m )2 ⎥ ⎝ CD ⎠ s
⎣ ⎦
The drag coefficient depends on Reynolds number:

Re =
( 998.2 kg m3 ) ( V m s )( 0.04m ) =40540V
9.85 × 10 −4 Ns m 2
The procedure to use is:
Guess a drag coefficient CD
Calculate the velocity
Calculate Reynolds number
Evaluate the drag coefficient with Figure 10-10 and compare to the guessed value.
Continue until converged.
Guess CD ≈ 0.4
0.5
⎛ 0.483 ⎞ m
V =⎜ ⎟ = 1.10 Re = 40540 (1.1) = 44,560
⎝ 0.4 ⎠ s
From the figure, CD ≈ 0.5
0.5
⎛ 0.483 ⎞ m
V =⎜ ⎟ = 0.98 Re = 40540 ( 0.98 ) = 39,800
⎝ 0.5 ⎠ s
From the figure, CD ≈ 0.5 . Because of the difficulty in reading the figure more closely, another iteration is not
justified. Therefore, the time for the ball to reach the surface is:
H 4m
t= = =4.08s Answer
V 0.98 m s

10- 33
10-33 A 50 mi/hr, 60 °F wind blows perpendicular to an outdoor movie screen that is 70-ft wide and 35-ft tall, the
screen is supported on 10-ft tall pilings. Estimate:
a. the drag force on the screen (in lbf)
b. the moment at the base of the pilings (in ft-lbf)

Approach:
The basic drag force equation is used to calculate the
force. The drag coefficient must be evaluated. The
moment is calculated by assuming the force acts
through the center of the screen.

Assumptions:
1. The air is at one atmosphere.
2. Drag force on support poles is ignored.

Solution:
a) The drag force is defined by Eq. 10-6:
FD = CD ρ V 2 A 2
where A is the area: A = bH = ( 70ft )( 35ft ) =2450ft 2 .
Assuming the Reynolds number is greater than 1000, we use Figure 10-7 with b/H = 70/35 = 2. Therefore,
CD ≈ 1.2. From the appendix for air from Appendix B-7 at 60 °F, ρ = 0.077 lbm/ft3.
FD = (1.2 ) ( 0.077 lbm ft 3 ) ( 73.3ft s ) ( 2450ft 2 )( lbfs 2 32.2ft lbm ) 2 =18,900 lbf
2
Answer
b) Assuming the force acts through the center of area of the screen:
M = FD ( S + H 2 ) = (18,900lbf )(10+ 35 2 ) ft=519,000ft lbf Answer

10- 34
10-34 A hotdog company decides to create a giant helium-filled balloon of a hotdog to float in parades for
advertising purposes. It will float 75 ft above the street and will be controlled by people holding onto
tethering lines. The balloon is 50-ft long and 10-ft in diameter and can be approximated as a cylinder. Air
at 70 ºF, 14.7 psia is funneled down the street between the buildings at a velocity of 15 mi/hr. Determine
the drag force (in lbf).

Approach:
The drag force with a uniform approach velocity is
calculated with the basic equation.

Assumptions:
1. The balloon can be treated as a short cylinder.
2. Air is an ideal gas.

Solution:
The drag force is
FD = CD ρ V 2 A 2 .

For air at one atmosphere and 70 ºF, ρ = 0.0749 lbm/ft3.

From Table 10-2 for a short cylinder with its axis parallel to the flow and D/L = 50/10 = 5, CD ≈ 0.85

2 2 2
⎛ lbm ⎞⎛ mi ⎞ ⎛ 5280ft ⎞ ⎛ hr ⎞ π
FD = ( 0.85 ) ⎜ 0.0749 3 ⎟⎜ 25 ⎟ ⎜ ⎟ (10ft ) ( lbfs 32.2ft lbm ) 2
2 2
⎟ ⎜
⎝ ft ⎠⎝ hr ⎠ ⎝ mi ⎠ ⎝ 3600s ⎠ 4
=209 lbf Answer

10-35 A telephone wire 5-mm in diameter is suspended between telephone poles spaced 50 m apart. If the wind
velocity is 100 km/hr and the air is at 2 ºC, 1 atm, determine the horizontal force (in N) the wire exerts on
the poles.

Approach:
The basic drag force equation is used to calculate the
force. The drag coefficient must be evaluated.

Assumptions:
1. The cable acts as a long cylinder.
2. Flow is perpendicular to the cable.

Solution:
The drag force is defined by Eq. 10-6:
FD = CD ρ V 2 A 2
where A is the area: A = LD . From Appendix A-7, the air properties 2 ºC are: ρ = 1.280 kg/m3 and
µ = 1.67 × 10−5 Ns m 2 . To evaluate the drag coefficient, we need the Reynolds number:
ρ V∞ D (1.280 kg m ) (100,000 m hr )(1hr 3600 s )( 0.005m )
3

Re = = = 10, 650
µ 1.67×10-5 Ns m 2
From Figure 10-10, CD ≈ 1.4.
FD = (1.4 ) (1.280 kg m3 ) (100,000 m hr ) (1hr 3600s ) ( 50m )( 0.005m ) ( Ns 2 kgm ) 2 =173 N
2 2
Answer

10- 35
10-36 When parachuting, an Army Ranger and his gear may weigh as much as 250 lbf. To prevent injury, the
Ranger’s vertical landing speed must be less than 15 ft/s. If the parachute can be approximated as an open
hemisphere and the air is at 70 ºF, 14.7 psia, what diameter (in ft) parachute is required?

Approach:
Terminal velocity occurs when there is a balance
between the weight of the parachutist and drag.
Buoyancy is negligible compared to the drag and
weight. We need to determine the drag force on a
parachute.

Assumptions:
1. Buoyancy is ignored.
2. Air is an ideal gas.

Solution:
A force balance on the parachutist is:
∑ F = 0 = FD − W → FD = W
The drag force is defined by Eq. 10-6:
FD = CD ρ V 2 A 2
where A = π D 2 4 . Combining the three equations and solving for diameter:
0.5
⎛ 8W ⎞
D=⎜ 2 ⎟
⎝ CD ρπ V ⎠
Using Table 10-2, the drag coefficient for a parachute is CD = 1.4
For the air density, use the ideal gas equation:
⎛ lbf ⎞ ⎛ lbm ⎞ ⎛ 144in.2 ⎞
⎜ 14.7 2 ⎟ ⎜ 28.97 ⎟⎜ ⎟
PM ⎝ in. ⎠ ⎝ lbmol ⎠ ⎝ 1ft 2 ⎠ lbf
ρ= = =0.0749 3
RT ⎛ ft lbf ⎞ ft
⎜ 1545 ⎟ ( 530R )
⎝ lbmolR ⎠
⎡ 8 ( 250lbf ) ( 32.2 ft lbm lbf s 2 ) ⎤
0.5

D=⎢ ⎥ =29.5ft Answer

⎢ (1.4 ) ( 0.0749 lbm ft 3 ) π (15ft s )2 ⎥
⎣ ⎦

10- 36
10-37 An office building, approximately 90 m wide and 150 m tall, is to be built in a new development far from
any other building. Its drag coefficient is 1.4. Determine:
a. the drag force (in N) if the wind at 17 °C is uniform at 15 m/s
b. the drag force (in N) if the velocity profile can be approximated with the 1/7th power law
V x V ∞ = ( y δ ) with a boundary layer thickness of 100 m and a free-stream velocity of 15 m/s
17

(Hint: integrate to obtain the total drag force.)

Approach:
The drag force with a uniform velocity can be
calculated with the basic equation. If the velocity
profile is taken into account, then a differential
analysis must be performed and the resulting equation
integrated.

Assumptions:
1. The air is at one atmosphere.
2. The drag coefficient for a building is the same for
the whole structure and for a differential element.

Solution:
a) Drag force is:
FD = CD ρ V 2 A 2
For air at 17 ºC, ρa = 1.214 kg/m3.
FD = (1.4 ) (1.224 kg m3 ) (15 m s ) ( 90m )(150m ) ( Ns 2 kgm ) 2 =2.60 × 106 N
2

b) When the velocity profile is taken into account, begin with the differential force on a differential area, as shown
on the schematic:
dFD = ( CD ρ V 2 2 ) dA = ( CD ρ V 2 2 ) Wdy
Substituting in the expression for the velocity profile, we recognize that we must break this into two parts:
2
dFD = 0.5CD ρ ⎡ V∞ ( y δ ) ⎤ Wdy
17
0≤ y <δ
⎣ ⎦
dFD = 0.5CD ρ [ V∞ ] Wdy
2
δ ≤ y≤H
With the drag coefficient, density, and free-stream velocity all constant, both expressions can be integrated:
δ 2
FD = ∫ dFD = ∫ 0.5CD ρ ⎡ V∞ ( y δ ) ⎤ Wdy = 0.5 ( 7 9 ) CD ρ V∞2W δ
17
0≤ y <δ
0 ⎣ ⎦
FD = ( 0.5 )( 7 9 )(1.4 ) (1.214 kg m3 ) (15 m s ) ( 90m )(100m ) ( Ns 2 kgm ) =1.338 × 106 N
2

FD = ∫ dFD = ∫ 0.5CD ρ [ V∞ ] Wdy = 0.5CD ρ V∞2W ( H − δ )

H 2
δ ≤ y≤H
δ

FD = ( 0.5 )(1.4 ) (1.214 kg m3 ) (15 m s ) ( 90m )(150 − 100m ) ( Ns 2 kgm ) =0.860 × 106 N
2

The total drag force is:

Ftot = 1.338 × 106 N+0.860 × 106 N=2.198 × 106 N Answer

Comments:
Note that by taking into account the changing velocity over the height of the building, the total drag force is
reduced by 14.8%.

10- 37
10-38 The superintendent of a national cemetery wants to erect a larger than usual flagpole and flag. The flag
pole is 125 ft tall. The flag has a height H = 20 ft and a length L = 38 ft. Assume the flag pole must
withstand a wind of 60 mi/hr at 32 °F when the flag is flying. The drag coefficient of the flag can be
estimated by CD = 0.05L/H. If the pole has a diameter of 9 in., determine:
a. the total force exerted on the pole (in lbf)
b. the moment at the base of the pole (in ft-lbf).

Approach:
This is a composite body. We assume the combined
drag force is the sum of the individual drag forces.
Moment is force time distance, and we assume each
individual drag force acts through the center of its
object.

Assumptions:
1. The air is an ideal gas at one atmosphere.
2. The pole is a smooth cylinder.
3. The drag on each component is calculated as if it
stands alone.

Solution:
a) The drag force is defined by Eq. 10-6:
FD = CD ρ V 2 A 2
The drag coefficient depends on the Reynolds number. For air from Appendix B-7 at 40 ºF , ρ = 0.081lbf ft 3
µ = 1.179 × 10−5 lbf fts
For the flag, CD = 0.05 L H = ( 0.05 )( 38ft ) ( 20ft ) = 0.095
FD , flag = ( 0.095 ) ( 0.081lbm ft 3 ) ( 88ft s ) ( 20ft )( 38ft ) ( lbf s 2 32.2ft lbm ) 2 =703lbf
2

For the pole, we need the Reynolds number to evaluate the drag coefficient:
ρ VD ( 0.081lbm ft ) ( 88ft s )( 0.75ft )
3

Re = = =453, 000
µ 1.179 × 10 −5 lbf fts
Assuming the pole is a smooth cylinder, from Figure 10-10, CD ≈ 0.22
FD , pole = ( 0.22 ) ( 0.081lbm ft 3 ) ( 88ft s ) (125ft )( 0.75ft ) ( lbf s 2 32.2ft lbm ) 2 =201lbf
2

Therefore, the total force is: Ftot = 703 + 201 = 904lbf Answer

b) The bending moment at the base is:

M = FD , pole ( H pole 2 ) + FD , flag ( H pole − H flag 2 )
= ( 201lbf )(125ft ) 2+ ( 703lbf )(125ft- 20ft 2 )
= 93, 400ft lbf Answer

Comments:
Because the Reynolds number is at the “dip” in the drag coefficient-Reynolds number curve, a second calculation
should be performed at a lower velocity and a high velocity to see if the resulting force is much different. For
example, if the velocity were 50 mi/hr, then Re = 380,000 and CD ≈ 0.5, which would give a drag on the pole of
317 lbf, which is 50% greater than what was calculated previously. However, the flag drag force would decrease
to 488 lbf, and the total would be 805 lbf, which is 11% less than what was previously calculated.

10- 38
10-39 Antennas on old cars are vertical circular cylinders 0.25 in. in diameter and 4-ft long. Some people attach
objects to the top of their antenna so that their car is more easily found in crowded parking lots. If the car is
driven at 65 mi/hr, and the air is at 80 ºF, 14.7 lbf/in.2, determine:
a. the bending moment (in ft-lbf) at the base of the antenna without the object attached
b. the bending moment (in ft-lbf) at the base of the antenna if an object shaped like a sphere 3-in. in
diameter is attached to the top of the antenna.

Approach:
A moment is force times distance. Assume the drag
forces on the antenna and the sphere act through their
centers. For this composite body, assume the drag on
the individual components is calculated as if the other
part were not present.

Assumptions:
1. The drag force on each part is calculated as if the
other part were not present.

Solution:
a) For the antenna, the drag force is defined by Eq. 10-6:
FD = CD ρ V 2 A 2
where A is the area: A = Ld = ( 4ft )( 0.25in.)(1ft 12 in.) =0.0833ft 2 .
For the drag coefficient, we need the Reynolds number, so from Appendix B-7 for air at 80 ºF, ρ = 0.074
lbm/ft3 and µ = 1.25×10-5 lbm/fts.
ρ V d ( 0.074 lbm ft ) ( 95.3ft s )( 0.0208ft )
3

Re = = = 11, 740
µ 1.25×10-5 lbm fts
Using Figure 10-10, CD ≈ 1.2.
FD = (1.2 ) ( 0.074 lbm ft 3 ) ( 95.3ft s ) ( 0.0833ft 2 )( lbfs 2 32.2ft lbm ) 2 =1.04 lbf
2

Assuming the force acts through the center of the antenna

M = FD L 2 = (1.04lbf )( 4ft ) 2 =2.08ft lbf Answer
b) For the composite body, we need to calculate the force and moment created by the sphere. For the sphere:
ρ V D ( 0.074 lbm ft ) ( 95.3ft s )( 0.25ft )
3

Re = = = 141, 000
µ 1.25×10-5 lbm fts
From Figure 10-10 for a smooth sphere CD ≈ 0.5.
FD = ( 0.5 ) ( 0.074 lbm ft 3 ) ( 95.3ft s ) (π 4 )( 0.25ft ) ( lbfs 2 32.2ft lbm ) 2 =0.256 lbf
2 2

M = 2.08ft lbf + ( 0.256lbf )( 4+ 0.25 2 ) ft=3.14ft lbf Answer

10- 39
10-40 The external rearview mirrors (two each) on old cars were circular disks 10-cm in diameter. New cars use
streamlined rearview mirrors (two each) to reduce drag losses; these mirrors can be approximated as
hemispheres facing upstream. A car without mirrors has a drag coefficient of 0.36, a frontal area of 1.5 m2,
and rolling resistance can be ignored. For a car speed of 125 km/hr in air at 23 ºC, 100 kPa, what percent
increase in gas mileage could be obtained by replacing the old mirrors with two new ones of the same
diameter?

Approach:
The car plus mirror combination is a composite body.
For this composite body, the individual force/power
contributions from the car and the mirrors need to be
calculated. The total power with and without the
mirrors then can be compared. Power is force times
velocity

Assumptions:
1. The drag on a composite body can be calculated as
if the separate parts act independently.
2. Gas mileage is inversely proportional to power.

Solution:
The drag force is defined by Eq. 10-6:
FD = CD ρ V∞2 A 2
For composite bodies, we assume that the drag force contribution from each part is simply additive. Power is:
W = FD V∞ = CD ρ V∞3 A 2
For a composite body, we add the contributions from the parts. With velocity, air conditions, and engine
efficiency the same for the two mirror types on the car (two mirrors on the car)
Therefore, the ratio of power with new and old mirrors is:
Wnew ( CD ,car Acar + 2CD , mirror Amirror )new ρ V∞ 2 ( CD ,car Acar + 2CD , mirror Amirror )new
3

= =
Wold ( C A + 2C
D , car car A
D , mirror ) ρ V 3 2 ( C A + 2C
mirror old ∞ D , car car A
D , mirror )mirror old

From Table 10-2

For a disk, CD , mirror , old ≈ 1.1 and A = π D 2 4
For a hemisphere facing into the velocity, CD , mirror , new ≈ 0.4 and A = π D 2 4
Therefore,
Wnew ( 0.36 ) (1.5m ) +2 ( 0.4 )( π 4 )( 0.1m )
2 2

= = 0.980
Wold ( 0.36 ) (1.5m 2 ) +2 (1.1)( π 4 )( 0.1m )
2

Therefore, the new mirrors reduce the fuel consumption by about 2.0%. Answer

Comments:
By itself the changing of the mirror configuration does not affect the gas mileage dramatically. But many small
changes in the aerodynamics of a car plus decreasing the car weight can significantly increase the gas mileage.

10- 40
10-41 Taxicabs carry advertising signs on their roofs to generate extra income for the operator. If the sign is a
rectangular box 30-cm high, 1.2-m wide, and 1.2-m long, estimate the increased fuel cost caused by the
addition of the sign. Assume the taxi is driven 100,000 km annually at an average speed of 50 km/hr. Its
engine cycle thermal efficiency is 25%. A reasonable average air condition is 10 ºC, 100 kPa. The fuel
costs $0.40/L, its specific gravity is 0.82, and its energy content is 40,000 kJ/kg.

Approach:
Increased fuel cost is directly proportional to
increased power. The sign causes a drag force, and
power is force times velocity. The total time the taxi
is used is obtained from the average velocity and the
distance traveled. Combining time and power gives
total energy used, which is related to the gasoline
consumed.

Assumptions:
1. The sign can be treated as flow over a long
rectangular rod.
2. Air is at one atmosphere.

Solution:
The drag force on the sign is defined by Eq. 10-6:
FD = CD ρ V 2 A 2
From Table 10-1, we assume the sign acts as a long rectangular rod, with D/L – 1.2/0.3 = 4, so CD ≈ 1.3 .
For air at 10 ºC, ρ = 1.244 kg/m3.
FD = (1.3) (1.244 kg m3 ) (13.9 m s ) ( 0.3m )(1.2m ) ( Ns 2 kgm ) 2 =56.2 N
2

Power is obtained from:

W = FD V = ( 56.2 N )(13.9 m s )(1W 1J s ) =781W
The total energy used for the 100,000 km/yr requires the time:
distance (100,000 km yr )(1000 m km )
t= = =7.194×106 s yr
velocity 13.9 m s
Total energy used, including engine thermal efficiency, is:
Wt ( 789W ) ( 7.194×106 s yr ) (1kJ 1000J )
E= = = 2.247×107 kJ yr
ηcycle 0.25
The mass of fuel used is:
m = ( 2.247×107 kJ yr ) ( 40,000 kJ kg ) =562 kg yr
Volume of fuel used:
m ( 562 kg yr ) (1L 10 m )
-3 3
L
V= = =685
ρ ( )(
0.82 1000 kg m 3
) yr
Cost of fuel = ( $0.40 L )( 685 L yr ) = $274 yr Answer

Comments:
The drag coefficient is approximate, so this answer has a fair amount of uncertainty. The revenue generated by
carrying the sign would be balanced against the additional fuel cost.

10- 41
10-42 A thin flat plate 10-ft long and 2-ft wide is mounted horizontally on a 10-ft long, 3-in. diameter pole. Air
flows at 60 ºF, 14.7 psia along the 10-ft length of the plate. The velocity profile of the air flow varies from
0 at the base of the pole to 50 ft/s at 10 ft (along the top of the plate). Taking into account the variation in
velocity, determine the total drag force (in lbf) acting on the composite body.

Approach:
This is a drag force on a composite body. The
contribution from each part must be evaluated.
Because of the velocity variation on the pole, that
force will need to be obtained by integration.

Assumptions:
1. The contributions from each part of the composite
body are additive.
2. The pole drag coefficient is constant over its
length.

Solution:
The drag force is on the flat plate (both sides) is given by FD = 2CD ρ V 2 A 2 , where the plan area is A = LW .
The drag force depends on the Reynolds number.
From Appendix B-7 for air at 60 ºF, ρ = 0.077 lbm/ft3 and µ = 1.214×10-5 lbm/fts. The Reynolds number
is:

ReL =
( 0.077 lbm ft 3 ) ( 50 ft s )(10ft ) =3.171×106
1.214 × 10 −5 lbm fts
Assuming transition to turbulent occurs at a Reynolds number of about 500,000:
0.074 1740 0.074 1740
CD = − = − = 0.00316
Re1L 5 ReL ( 3.171× 106 )1 5 3.171× 106

FD , plate = 2 ( 0.00316 ) ( 0.077 lbm ft 3 ) ( 50 ft s ) (10ft )( 2ft ) ( lbf s 2 32.2ft lbm ) 2 =0.38lbf
2

For the drag on the pole, begin with a differential area:

dFD = 0.5CD ρ V 2 dA = 0.5CD ρ V 2 Ddy
The velocity varies linearly with distance from the ground. Using the given information V = V∞ ( y H ) . For
the drag coefficient, the pole Reynolds number varies from zero at the ground to a maximum at H = 10 ft.
Therefore,

Remax =
( 0.077 lbm ft 3 ) ( 50 ft s )( 0.25ft ) =79, 280
1.214 × 10 −5 lbm fts
From Figure 10-10, we assume a reasonable value of the drag coefficient over the complete pole CD ≈ 1.3 .
Therefore, integrating the expression for the pole drag force:
2
H ⎛ y⎞
FD , pole = ∫ dFD = ∫ 0.5CD ρ ⎜ V∞ ⎟ Ddy = CD ρ V∞2 HD 6
0
⎝ H⎠
= (1.3) ( 0.077 lbm ft 3 ) ( 50 ft s ) (10ft )( 0.25ft ) ( lbf s 2 32.2ft lbm ) 6 =3.24lbf
2

Finally, the total force is

FD = 3.24 + 0.38 = 3.62lbf Answer

10- 42
10-43 A large family is going on a vacation in their minivan that has a drag coefficient of 0.44 and a frontal area
of 3.5 m2. Because they need more room for their luggage, they will use a rectangular car top carrier that is
1.5-m wide, 30-cm high, and 2-m long. Estimate the increase in power required to drive at 100 km/hr with
the car top carrier compared to without it.

Approach:
For this composite body, the individual force/power
contributions from the car and the car top carrier need
to be calculated. The total power with and without the
carrier then can be compared. Power is force times
velocity

Assumptions:
1. The drag on a composite body can be calculated as
if the separate parts act independently.
2. The car top carrier is approximated as a flat
rectangular plate perpendicular to the velocity.

Solution:
The drag force is defined by Eq. 10-6:
FD = CD ρ V∞2 A 2
For composite bodies, we assume that the drag force contribution from each part is simply additive.
Power is:
W = FD V∞ = CD ρ V∞3 A 2

Therefore, the ratio of power with and without the carrier is:
Ww
=
( CD,car Acar + CD,carrier Acarrier ) ρ V∞3 2 = CD,car Acar + CD ,carrier Acarrier
W w/o C ρ V 3A 2
D , car ∞ car C A
D , car car

Assuming we can treat the car top carrier as a flat rectangular plate, from Fig. 10-7, CD ,carrier ≈ 1.2
Ww ( 0.44 )( 3.5 ) + (1.2 )( 0.30 )(1.5)
= = 1.35 or a 35% increase in power required. Answer
W w/o ( 0.44 )( 3.5 )

10- 43
10-44 A small aircraft has a wing area of 27 m2, a take-off mass of 2500 kg, a lift coefficient at take-off of 0.49,
and a drag coefficient at take-off of 0.0074. For standard atmospheric conditions, determine:
a. the take-off speed at sea level (in km/hr)
b. the power required at take-off (in kW)
c. the maximum mass (in kg) possible at take-off speed using the power from part (b) if the airport is at
2500 m.

Approach:
The take-off speed is obtained by knowing that lift
force equals weight at take-off. The basic equations
are used, and a standard atmosphere is assumed.

Assumptions:
1. The air conditions equal the standard atmosphere.

Solution:
a) At take-off speed, lift force equals weight: FL = W = mg
The lift force is calculated with FL = CL ρ V A 2 . Combining these two expressions and solving for velocity:
2

0.5
⎛ 2mg ⎞
V =⎜ ⎟
⎝ CL ρ A ⎠
From Table 10-5, ρ = 1.225 kg m3
⎡ 2 ( 2500kg ) ( 9.81m s 2 ) ⎤
0.5
m km
V =⎢ ⎥ =55.0 = 198 Answer
⎢⎣ ( 0.49 ) (1.225 kg m )( 27m ) ⎥⎦
3 2
s hr

b) Power is obtained from: W = FD V = CD ρ V 2 A 2 . Note that FD = CD ρ V 2 A 2 . Dividing the drag force by

the lift force: FD FL = CD CL . Therefore,
W = ( C C ) mg V
D L

= ( 0.0074 0.49 )( 2500kg ) ( 9.81m s 2 ) ( 55.0 m s ) (1Ns 2 kgm ) (1Ws Nm ) =20,370W Answer
c) At higher elevation, we have two equations:
FL = W → CL ρ V 2 A 2 = mg
F V = W → C ρ V 3 A 2 = W
D D
From the second equation, we can obtain the maximum velocity at the higher elevation using the power from part
(b). Solving for velocity:
13
⎛ 2W ⎞
V =⎜ ⎟
⎝ CD ρ A ⎠
From Table 10-5 by interpolation, ρ ≈ ( 0.782 ) (1.225 kg m3 ) = 0.958 kg m3

⎡ 2 ( 20370W )( Nm Ws ) ( kgm Ns 2 ) ⎤
13
m km
V =⎢ ⎥ =59.7 = 215
⎢⎣ ( 0.0074 ) ( 0.958 kg m )( 27m ) ⎥⎦
3 2
s hr
From the first equation:
CL ρ V 2 A ( 0.49 ) ( 0.958 kg m ) ( 59.7 m s ) ( 27m )
3 2
2

m= = =2300kg Answer
2g 2 ( 9.81m s 2 )

10- 44
10-45 For a small plane, the lift coefficient at the landing speed is 1.15, and the maximum lift coefficient (at the
stall speed) is 1.42. The landing speed of the airplane is 8 m/s faster than its stall speed. Determine both
the landing and stalling speeds (in m/s).

Approach:
At landing and stall, the lift force just equals the
weight of the plane. We can use the given
information and the basic definitions of lift force to
determine the two speeds.

Assumptions:
1. The air is an ideal gas.
2. The pole is a smooth cylinder.

Solution:
Let V = the speed at landing and V S = the speed at stall.
L These two speeds are related by
VL = VS + 8 m s
At landing and at stall, lift force equals weight:
Landing: FL = CL , L ρ VL2 A 2 = mg
Stall: FL = CL , S ρ VS2 A 2 = mg
Equating the two equations and simplifying:
CL , S VS2 = CL , L VL2
Substituting in the relationship between stall and landing speed:
CL , S VS2 = CL , L ( VS + 8 m s )
2

(1.42 ) VS2 = (1.15)( VS + 8 m s )

2

Solving for the velocities:

VS = 71.9 m s Answer
VL = 71.9 +8=79.8 m s Answer

10- 45
10-46 A 250-kg glider with a wing area of 22 m2 has a minimum glide angle of 1.7º. Its lift coefficient is 1.1. For a
still day at 15 ºC, 100 kPa, determine:
a. the total horizontal distance for the glider to descend from 1500 m to sea level (in km)
b. the time required (in min).

Approach:
Part (a) is a geometry problem; for a right triangle of
height H and angle θ, the use of the tangent will give us
the base length, L. For part (b), we must determine the
plane speed by recognizing that at steady flight, lift force
equals weight. Once we have the velocity, then time is
determined noting that velocity is distance divided by
time.

Assumptions:
1. The air is an ideal gas at one atmosphere.

Solution:
a) From geometry:
tanθ = H L → L = H tanθ = 1500m tan (1.7 o ) =50,540 m=50.5 km Answer

b) We can calculate the time required for the glider to reach the ground with:
t=D V
where D, the distance traveled is obtained from geometry:
D = H sin θ = 1500m sin (1.7o ) =50,563m
Velocity is determined from the fact that in steady flight, weight equals lift:
FL = W
CL ρ V 2 A 2 = mg
For air at 15 ºC, ρ = 1.222 kg m3 . Solving for velocity;
⎡ 2 ( 250kg ) ( 9.81m s 2 ) ⎤
0.5
0.5
⎡ 2mg ⎤ m
V =⎢ ⎥ =⎢ ⎥ =12.9
⎣ CL ρ A ⎦ ⎢⎣ (1.1) (1.222 kg m3 )( 22m 2 ) ⎥⎦ s
50,563m
t= =3907s=65.1min Answer
12.9 m s

10- 46
10-47 When a plane glides at its shallowest angle, lift, drag, and weight forces are all in equilibrium. Show that
the glide slope angle, θ, is given by θ = tan −1 ( CD CL ) .

Approach:
At equilibrium for unpowered flight, lift, drag, and
weight forces balance out. Force balances in the x and
y directions must be written.

Solution:
a) Force balances in the two coordinate directions are:
∑ Fx = FD − W sin θ → FD = W sin θ
∑F y = FL − W cos θ → FL = W cos θ
Dividing these two expressions:
FD W sin θ
= = tan θ
FL W cos θ
Drag force is:
FD = CD ρ V 2 A 2
Lift force is:
FL = CL ρ V 2 A 2
Combining these two expressions:
FD CD
=
FL CL
Substituting this into the ratio of the force balances:
FD CD
= = tan θ
FL CL
Therefore,
⎛C ⎞
θ = tan −1 ⎜ D ⎟ Answer
⎝ CL ⎠

10- 47
10-48 A hydrofoil is a watercraft that rides above the surface of the water on foils, which are essentially wings
attached to the bottom of struts connecting the foils to the hull of the boat. Suppose the area of the foils in
contact with the water on a 2000-kg hydrofoil is 1.1 m2. Their lift and drag coefficients are 1.72 and 0.45,
respectively. Determine:
a. the minimum speed required for the foils to support the hydrofoil (in km/hr)
b. the power required to propel the hydrofoil at the speed calculated in part (a) (in kW)
c. the top speed if the boat has a 175 kW engine (in km/hr). (Note that at higher speeds the hydrofoil
rises further out of the water, and the lifting area is decreased.)

Approach:
The minimum speed for the foils to support the
hydrofoil occurs when the lift force matches the
weight. We have sufficient information given to
evaluate the speed. Power is drag force times
velocity.

Assumptions:
1. Water density is 1000 kg/m3.

Solution:
a) At the minimum speed:
FL = W = mg
where lift force is: FL = CL ρ V 2 A 2
12
⎛ 2mg ⎞
Combining equations and solving for velocity: Vminimum = ⎜ ⎟
⎝ CL ρ A ⎠
We assume for water, ρ = 1000 kg/m3.
12
⎡ 2 ( 2000kg )( 9.81m s ) ⎤ m km
Vminimim =⎢ ⎥ =4.55 = 16.4 Answer
⎢⎣ (1.72 ) (1000 kg m 3 )(1.1m 2 ) ⎥⎦ s hr

b) The power required at the minimum speed is:

W = FD V = CD ρ V 3 A 2 = ( 0.45 ) (1000 kg m3 ) ( 4.55 m s ) (1.1m 2 ) 2 =23,310W
3
Answer

c) At the maximum speed, FL = W = mg . Comparing the lift and drag equations, we can see that:
CC C
FD = FL = C mg
CL CL
In addition, power is: Wmax = FD Vmax . Combining these equations:
W ⎛ C ⎞ W ⎛ 1.72 ⎞ (175000W )(1J 1Ws ) m km
Vmax = max = ⎜ L ⎟ max = ⎜ ⎟ =34.1 =123 Answer
FD C
⎝ D⎠ mg ⎝ 0.45 ⎠ ( 2000 kg m 3
)( 9.81m s 2
) s hr

10- 48
10-49 Consider a U-2 reconnaissance plane that loses power at 35,000 ft over hostile territory. If its lift and drag
characteristics are similar to those given in Figure 10-19, determine:
a. the optimum angle of attack for maximum glide distance
b. if it can make it to an airport 425 miles away. (Ignore initial velocity.)

Approach:
At equilibrium for unpowered flight, lift, drag, and
weight forces balance out. Force balances in the x and
y directions must be written.

Solution:
a) Force balances in the two coordinate directions are:
∑ Fx = FD − W sin θ → FD = W sin θ
∑F y = FL − W cos θ → FL = W cos θ
Dividing these two expressions:
FD W sin θ
= = tan θ
FL W cos θ
Drag force is: FD = CD ρ V 2 A 2
Lift force is: FL = CL ρ V 2 A 2
Combining these two expressions:
FD CD
=
FL CL
Substituting this into the ratio of the force balances:
FD CD
= = tan θ
FL CL
Therefore,
⎛C ⎞
θ = tan −1 ⎜ D ⎟ Answer
⎝ CL ⎠
b) From Figure 10-16, we can obtain CD and CL. We want the minimum angle θ to maximize the distance
traveled. Choosing values of drag and lift coefficients for different angles of attack:

Angle of attack CD CL CD/CL

(degrees)
0 0.0061 0.13 0.0469
2 0.0063 0.32 0.0197
4 0.0066 0.55 0.0120
6 0.0072 0.75 0.0096
8 0.0082 1.00 0.0082
10 0.010 1.2 0.0083
12 0.0125 1.4 0.0089

So using an angle of attack of about 8º, the maximum glide distance will be obtained. Answer
c) Using the given geometry
H H H 35,000ft ⎛ 1mi ⎞
tan θ = → L= = = ⎜ ⎟ =808mi
L tan θ ( CD CL ) 0.0082 ⎝ 5280ft ⎠
The plane can reach the airport. Answer

Comments:
This is an optimistic answer. In reality, the plane could not glide this far.

10- 49
 10-50 If you have ever flown a kite in a strong wind, you know that the pull on the string can be quite strong.
Consider a 1.2-m by 0.8-m kite that has a mass of 0.5 kg. Its lift coefficient can be approximated by
CL = 2π sin α where α is the angle of attack. In a 45 km/hr wind, the kite’s string has an angle of 50° to
the horizontal, and the kite has an angle of attack of 5°. Determine the force on the string (in N).

Approach:
The tension in the string can be obtained from a force
balance on the kite.

Assumptions:
1. Air is an ideal gas at one atmosphere.

Solution:
A force balance on the kite in the y-direction is:
FL − Wkite FL − mg
∑F y = 0 = FL − T sin θ − Wkite → T=
sin θ
=
sin θ
The lift force is defined by:
FL = CL ρ V 2 A 2
Assuming air is an ideal gas at 25 ºC, one atmosphere, ρ = 1.181 kg/m3.
For the drag coefficient:
CL = 2π sin ( 5o ) = 0.548
The lift force is:
FL = ( 0.548 ) (1.181kg m3 ) (12.5 m s ) (1.2m )( 0.8m ) ( Ns 2 kgm ) 2 =48.5N
2

The tension force is:

48.5N- ( 0.5kg ) ( 9.81m s 2 )( Ns 2 kgm )
T= =56.9N Answer
sin ( 50o )

10-51 A small experimental plane has a mass of 750 kg, a drag coefficient of 0.063, and a lift coefficient of 0.4. In
level flight, it is flown at 175 km/hr. For standard conditions (1 atm, 25 ºC), determine the effective lift area
of the plane (in m2).

Approach:
The basic equation for lift force is applied directly to
this problem. For steady flight, weight equals lift.
Sufficient information is given to solve for the plan
area.

Assumptions:
1. Air properties are at 25 ºC, 1 atm.

Solution:
The lift force is calculated with the following equation:
FL = CL ρ V 2 A 2
In steady flight, lift equals weight: FL = W = mg
Combining the equations and solving for effective area:
2mg
A=
CL ρ a V 2
For air at 25 ºC, ρ = 1.184 kg/m3.
2 ( 750kg ) ( 9.81m s 2 ) ( 3600s hr )
2

A= =13.1m 2 Answer
( 0.4 ) (1.184 kg ) (175000 m hr )
3 2
m

10- 50
10-52 Because of the decrease in density and temperature with increasing elevation in the atmosphere, lift and
drag forces change. Consider a plane flying at velocity V at sea level. For the same lift and drag
coefficients, determine:
a. the speed required at 10,000 m to generate the same lift force
b. the change in drag force.

Approach:
At steady flight, lift equals weight. We can compare
the velocity requirements at two different elevations
using the basic definition of lift force.

Assumptions:
1. The lift coefficient is constant.

Solution:
a) In steady flight, FL = W
The lift force is defined by:
FL = CL ρ V 2 A 2 → CL ρ V 2 A 2 = W
For two different elevations, we can find the relative velocities required by taking a ratio of the equation above
evaluated at the two conditions:
CL ,1 ρ1 V1 2 A1 2 W1
=
CL ,2 ρ 2 V22 A2 2 W2
Because the weight, area, and lift coefficient are the same at the two conditions:
0.5 0.5
V22 ρ V2 ⎛ ρ1 ⎞ ⎛ρ ρ ⎞
= 1 → =⎜ ⎟ =⎜ 1 0 ⎟
V1 2 ρ2 V1 ⎝ ρ 2 ⎠ ⎝ ρ2 ρ0 ⎠
Obtaining the values of density at different elevations from Table 10-5:
0.5
V2 ⎛ 1 ⎞
=⎜ ⎟ = 1.72 Answer
V1 ⎝ 0.3376 ⎠

b) Drag force is FD = CD ρ V 2 A 2 . Taking the ratio of the forces at the two conditions:
FD ,2 CD ,2 ρ 2 V22 A2 2 ρ 2 V22
= =
FD ,1 CD ,1 ρ1 V1 A1 2
2
ρ1 V1 2
Using the relation given above for the ratio of velocities
FD ,2 ⎛ ρ 2 ⎞⎛ ρ1 ⎞
= ⎜ ⎟⎜ ⎟ = 1 Answer
FD ,1 ⎝ ρ1 ⎠⎝ ρ 2 ⎠

10- 51
10-53 When planes take-off and land at airports at higher elevations, the lower density air (due to reduced
atmospheric pressure) must be taken into account because of its effect on lift and drag. If a plane requires
15 s to reach its take-off speed of 220 km/hr at sea level in a distance of 500m, estimate for an airport at
2000 m:
a. the take-off speed (in km/hr)
b. the take-off time (in s)
c. the additional runway length required (in m). Assume the same constant acceleration for both cases.

Approach:
The take-off speed is obtained by knowing that lift force
equals weight at take-off; the take-off speed at the second
elevation can be obtained by comparison. The additional
time and distance required at the second elevation can be
approximated by a force balance that now takes into account
the acceleration. (Use Newton’s second law of motion.)

Assumptions:
1. Lift coefficient is the same at both elevations and speeds.

Solution:
a) At take-off speed, lift force equals weight: FL = W = mg
The lift force is calculated with FL = CL ρ V A 2 . Combining these two expressions and solving for velocity:
2

0.5
⎛ 2mg ⎞
V =⎜ ⎟
⎝ CL ρ A ⎠
Assuming the lift coefficient is the same at both elevations and speeds, the ratio of the speeds is:
0.5 0.5
V2 ⎛ ρ1 ⎞ ⎛ρ ρ ⎞
=⎜ ⎟ =⎜ 1 0 ⎟
V1 ⎝ ρ 2 ⎠ ⎝ ρ 2 ρ0 ⎠
From Table 10-5, at 2000 m, ρ 2 ρ 0 = 0.8217 → V2 = ( 220 km hr )(1 0.8217 )
0.5
= 243km hr Answer
b) To find the additional time, we need to take into account the acceleration of the plane. Using Newton’s second
law of motion: ∑ Fx = − FD = −max = −m d V dt
Drag force is FD = CD ρ V 2 A 2 . Substituting this into the force balance, separating variables, and integrating
from zero (the initial velocity) to the take-off velocity:
t C ρA C ρA 2m ⎛ 1 1 ⎞
V
Vf d V 1 f 1 1
∫ Vi V ∫0 2m
= − = − = D → t=− ⎜⎜ − ⎟
D
dt → t
2
V Vi Vi V f 2m CD ρ A ⎝ V f Vi ⎟⎠
If we let the initial velocity go to zero, then its inverse goes to infinity. Therefore, assume the initial velocity is 1
km/hr, and the ratio of times is:
t2 (1 ρ 2 )(1 V2 − 1) ⎛ 1 ⎞ ⎛ 1 243 − 1 ⎞
= → t2 = ⎜ ⎟⎜ ⎟ (15s ) =18.3s Answer
t1 (1 ρ1 )(1 V1 − 1) ⎝ 0.8217 ⎠ ⎝ 1 220 − 1 ⎠
c) For the distance, we revisit the original force balance: − FD = −max → CD ρ V 2 A 2 = max
d V d V dx dV
Note that ax = d V dt , and using the chain rule on the derivative: ax =
= =V
dt dx dt dx
Substituting this expression into the force balance, separating variables, and integrating:
dV xC ρA Vf d V CD ρ A ⎛ Vf ⎞ 2m ⎛ Vf ⎞
CD ρ V 2 A 2 = m V → ∫ D dx = ∫ → x = ln ⎜ ⎟ → x= ln ⎜ ⎟
dx 0 2m Vi V 2m ⎝ Vi ⎠ CD ρ A ⎝ Vi ⎠
x2 (1 ρ 2 ) ln ( V2 Vi )
Therefore, = . Assuming the initial velocity is 1 km/hr:
x1 (1 ρ1 ) ln ( V1 Vi )
ln ( V2 ) ⎛ 1 ⎞ ln ( 243)
x2 = x1 ( ρ1 ρ 2 ) = ( 500m ) ⎜ ⎟ =620m Answer
ln ( V1 ) ⎝ 0.8217 ⎠ ln ( 220 )

10- 52
10-54 An airplane is to be designed that will fly at 650 km/hr at an altitude where the density is 0.655 kg/m3 and
the kinematic viscosity is 2 × 10-5 m2/s. A 1/15th scale model (that is, the model is geometrically similar to
the prototype but is 1/15th its size) is built for use in a wind tunnel whose velocity is 650 km/hr. The air in
the wind tunnel is at 55 ºC, and viscosity is independent of pressure. Determine:
a. the test section pressure (in kPa) so that the model data are useful in designing the prototype (Hint:
match Reynolds numbers)
b. what is the relation between the drag on the prototype and that on the model?

Approach:
To match the flow conditions, the Reynolds number of
the prototype and model should be equal. For the
relationship between the drag of the prototype and the
drag of the model, we assume because the Reynolds
numbers are the same, the drag coefficients are equal.
We then can use the drag coefficient to determine the
relationship.

Assumptions:
1. Flow similarity occurs when Reynolds numbers
match.
2. Drag coefficient depends only on Reynolds
number.
3. Air is an ideal gas.

Solution:
a) For flow similarity, we set the Reynolds numbers equal to each other:
Re p = Rem
ρ p V p L p ρ m Vm Lm
=
µp µm
The velocities and viscosities are the same, and Lm L p = 1 15 . Therefore,
ρ m = ρ p ( L p Lm ) = ( 0.655 kg m3 ) (15 ) =9.825 kg m3
For the pressure, we use the ideal gas law:

ρ RT
( 9.825 kg m3 ) ⎛⎜⎝ 8.314 kmolK
kJ ⎞
⎟ ( 55+273) K
⎠
P= = =925kPa Answer
M ⎛ kg ⎞
⎜ 28.97 ⎟
⎝ kmol ⎠
b) To find the relationship between the drag forces on the prototype and model, we take the ratio of the drag force
expressions:
FD , p CD , p ρ p V p2 Ap 2
=
FD , m CD , m ρ m Vm2 Am 2

Canceling like terms and noting that Am Ap = ( Lm L p ) = (1 15 )

2 2

2
FD , p ⎛ ρ p ⎞ ⎛ L p ⎞ ⎛ 0.655 ⎞
⎟ (15 ) = 15
2
=⎜ ⎟⎜ ⎟ =⎜ Answer
FD , m ⎝ ρ m ⎠ ⎝ Lm ⎠ ⎝ 9.825 ⎠

10- 53
10-55 A 1-m by 1.5-m plate moves through still water at 3 m/s and is at an angle of 12º to the velocity vector. For
this situation the drag coefficient is 0.17 and the lift coefficient is 0.72. Determine: a) the resultant force on
the plate (in N), b) the angle at which this force acts on the plate, c)) the power required to move the plate
(in kW), and d) the drag force (in N) and power (in kW) if the plate moves through 20 °C air instead of
water.

Approach:
The resultant forced is due to the vector sum of the lift
and drag forces. The angle if makes with the
horizontal can be determined by geometry.

Assumptions:
1. The water density is 1000 kg/m3.
2. Air is at one atmosphere.

Solution:
a) The magnitude of the resultant force is the vector sum of the lift and drag forces:
FR = ( FL2 + FD2 )
0.5

The lift force is FL = CL ρ V 2 A 2 . With a water density of 1000 kg/m3:

FL = ( 0.72 ) (1000 kg m3 ) ( 3m s ) (1.5m )(1m ) ( Ns 2 kgm ) 2 =4860N
2

The drag force is FD = CD ρ V 2 A 2 .

FD = ( 0.17 ) (1000 kg m3 ) ( 3m s ) (1.5m )(1m ) ( Ns 2 kgm ) 2 =1150N
2

The resultant force is

(
FR = [1150] + [ 4860]
2
)
2 0.5
= 4994N Answer
b) The angle the resultant force makes with the horizontal is obtained with:
F ⎛F ⎞ ⎛ 1148 ⎞
tan θ R = L → θ R = tan −1 ⎜ L ⎟ = tan −1 ⎜ ⎟ = 13.3
o
Answer
FR ⎝ FR ⎠ ⎝ 4860 ⎠
c) Power is force times velocity. The force is the drag force only (the force in the direction of motion):
W = FD V = (1150N )( 3m s )( J Nm )(1W 1J s ) =3450W Answer
d) The only difference in the two situations is the fluid density. For air at 20 ºC, ρ = 1.201kg m3
⎛ 1.201 ⎞
FD = (1150N ) ⎜ ⎟ =1.38N
⎝ 1000 ⎠
W = (1.38N )( 3m s )( J Nm )(1W 1J s ) =4.14W Answer

10- 54
10-56 An expression for the laminar velocity profile on a flat plate is:
V x = C1 sin ( C2 y ) + C3
where the argument of the sine function is in radians. Using the three common physical conditions that the
velocity profile should satisfy, determine:
a. the constants C1, C2, and C3
b. the non-dimensional velocity profile
c. the boundary thickness (δ x ) as a function of length
d. the skin friction coefficient, Cf, as a function of length.

Approach:
The velocity profile is given by Vx = C1 sin ( C2 y ) + C3 .
Using the approach given in the momentum integral
example problem, the constants can be evaluated using
the boundary conditions listed after Eq. 10-21. The
boundary layer thickness and the skin friction coefficient
can be evaluated with Eqs. 10-27 and 10-28,
respectively.

Assumptions:
1. The momentum integral approach is valid.
2. The assumed profile is reasonable.

Solution:
a) Using the three boundary conditions listed after Eq. 10-21:
1. At y = 0, Vx = 0 → 0 = C1 sin ( 0 ) + C3 → C3 = 0
2. At y = δ, Vx = V∞ → V∞ = C1 sin ( C2δ )
3. At y = δ, d ( Vx ) dy = 0 → d ( Vx ) dy = C1C2 cos ( C2δ ) = 0
From boundary condition 3, cos ( C2δ ) = 0 → C2δ = π 2 → C2 = π 2δ
Substituting this into boundary condition 2, V∞ = C1 sin ([π 2δ ] δ ) = C1 sin (π 2 ) = C1
⎛π y ⎞ Vx ⎛π ⎞
Therefore, Vx = V∞ sin ⎜ ⎟ → F (η ) = = sin ⎜ η ⎟ Answer
⎝2δ⎠ V∞ ⎝2 ⎠
δ 2β γ
b) The boundary layer thickness is defined by Eq. 10-27: = . From Eq. 10-24:
x Rex
⎛π ⎞⎡ ⎛ π ⎞⎤
1 1
γ = ∫ F (η ) ⎣⎡1 − F (η )⎦⎤ dη = ∫ sin ⎜ η ⎟ ⎢1 − sin ⎜ η ⎟ ⎥ dη
0
2 0 ⎝ 2 ⎠⎣ ⎝ ⎠⎦
1
2⎡ ⎛π ⎞ 1 π 1 ⎤ 2⎡ π ⎤
= − cos ⎜ η ⎟ − η + sin (πη ) ⎥ = ⎢0 + 1 − + 0 + 0 − 0 ⎥ = 0.137
π ⎣⎢ ⎝2 ⎠ 22 4 ⎦0 π ⎣ 4 ⎦
dF d ⎡ ⎛π ⎞ π ⎛π ⎞ π
Using Eq. 10-25: β = = sin ⎜ η ⎟ = cos ⎜ η ⎟ =
dη η =0
dη ⎢⎣ ⎝ 2 ⎠ η = 0 2 ⎝ 2 ⎠ η =0 2

δ 2 (π 2 ) 0.137 4.80
Using Eq. 10-27: = = Answer
x Rex Rex
c) The skin friction coefficient is defined by Eq. 10-28
2 βγ 2 (π 2 )( 0.137 ) 0.656
Cf = = = Answer
Rex Rex Rex
Comments:
Note that the boundary layer thickness constant of 4.80 is a good approximation to the Blasius solution of 5.0, as
is the skin friction coefficient of 0.656 compared to the Blasius solution of 0.664.

10- 55
10-57 The velocity distribution in a laminar boundary layer on a smooth flat plate is given by
V x V ∞ = 3 ( y δ ) − 2 ( y δ ) . Develop an expression for the drag coefficient.
2

Approach:
With the given laminar velocity profile
Vx V∞ = 3η − 2η 2 , where η = y δ , the drag
coefficient can be evaluated with Eq. 10-29.

Assumptions:
1. The momentum integral approach is valid.
2. The assumed velocity profile is reasonable.

Solution:
From the momentum integral solution for flow in a boundary layer, the drag coefficient can be evaluated with Eq.
10-29:
2 2 βγ
CD =
ReL
The two constants are evaluated with the appropriate equations. From Eq. 10-24:
1 1
γ = ∫ F (η ) ⎡⎣1 − F (η ) ⎤⎦ dη = ∫ ( 3η − 2η 2 ) ⎡⎣1 − ( 3η − 2η 2 ) ⎤⎦ dη
0 0
1
1
= ∫ ( 3η − 11η 2 + 12η 3 − 4η 4 ) dη =
0 30
dF d
Using Eq. 10-25: β = = ⎡3η − 2η 2 = ( 3 − 4η ) η = 0 = 3
dη η = 0 dη ⎣ η =0

Therefore,
2 2 ( 3)(1 30 ) 0.894
CD = =
ReL ReL

Comments:
This has poor agreement with the Blasius solution, whose constant is 1.328. We can determine why by
considering Assumption No. 2 (the velocity profile is reasonable). Three boundary conditions are given in the
text that a velocity profile should satisfy. We satisfy the first two boundary conditions. However, we do not
satisfy the third boundary condition. Hence, this velocity profile is not reasonable.

10- 56
10-58 Measurements from flow over a flat plate results in the turbulent velocity profile V x V ∞ = ( y δ )1 9 and skin
friction coefficient C f = 0.046 Re1x 5 . Develop a relationship that describes the growth of the boundary layer
thickness. (Note that Eq. 10-8 cannot be used.)

Approach:
We can follow the development used with the
momentum integral equation, but we need to bring in
the information about the skin friction coefficient.

Assumptions:
1. The momentum integral approach is valid.
2. The assumed velocity profile is reasonable.
3. The measured/correlated Cf is valid.

Solution:
Letting η = y δ , we are given
Vx V∞ = η1 9
15
and C f = 0.046 Re x
Note that the skin friction coefficient is defined as:
τw
Cf = (1)
0.5 ρ V∞2
From the momentum integral analysis, Eq. 10-23
τw d ⎡ ⎤
1
= ⎢ ∫ F (η ) ⎡⎣1 − F (η ) ⎤⎦ dη ⎥
δ (2)
ρ V∞ dx ⎣ 0
2
⎦
Rewriting equation 1 and incorporating the skin friction expression:
τw 1 1 ⎛ 0.046 ⎞ 0.023
= Cf = ⎜ 1 5 ⎟ =
ρ ∞
V 2
2 2 ⎝ Rex ⎠ Re1x 5
Using this expression in equation 2, and substituting in the velocity profile:
0.023 d ⎡ ⎤ 9 dδ
1
= ⎢δ ∫ η ⎡⎣1 − η ⎤⎦ dη ⎥ =
19 19
15
Rex dx ⎣ 0 ⎦ 110 dx
0.023 9 dδ
=
( ρ V∞ x µ )
15
110 dx
Separating variables and integrating:
x δ
0.023 9
( ρ V∞ µ ) ∫0 110 ∫0
15
x −1 5
= dδ

⎛ 5 ⎞ 0.023 9
⎜ ⎟ x4 5 = δ
⎝ 4 ⎠ ( ρ V∞ µ )
15
110
Simplifying:
δ 0.351 0.351
= = Answer
( ρ V∞ x µ )
15
x Re1x 5

Comments:
Note that this solution is close to the correlation obtained from a curve fit of experimental data
δ 0.37
=
x Re1x 5

10- 57
10-59 Assume a cubic velocity profile for flow over a flat plate of the form V x V ∞ = C0 + C1η + C2η 2 + C3η 3
where η = y δ . The profile should satisfy the three common physical conditions given in Section 10.7. A
fourth condition can be determined from the differential momentum equation and is: at y = 0,
d 2 V x dy 2 = 0 . Determine:
a. the constants C0, C1, C2, and C3
b. the boundary thickness (δ x ) as a function of length
c. the skin friction coefficient, Cf, as a function of length.

Approach:
The velocity profile on a flat plate is given by
F (η ) = Vx V∞ = C0 + C1η + C2η 2 + C3η 3 where
η = y δ . Using the approach given in the
momentum integral example problem, the constants
can be evaluated using the boundary conditions listed
after Eq. 10-21. The boundary layer thickness and the
skin friction coefficient can be evaluated with Eqs. 10-
27 and 10-28, respectively.

Assumptions:
1. The momentum integral approach is valid.
2. The assumed profile is reasonable.

Solution:
a) Using the three boundary conditions listed after Eq. 10-18 and the fourth one given in the problem statement:
1. At y = 0, Vx = 0 → y δ = η = 0, Vx V∞ = 0 → 0 = C0 + C1 ( 0 ) + C2 ( 0 ) + C3 ( 0 ) → C0 = 0
2. At y = δ, Vx = V∞ → y δ = η = 1, Vx V∞ = 1 → 1 = C1 (1) + C2 (1) + C3 (1)
3. At y = δ, d ( Vx ) dy = 0 → y δ = η = 1, d ( Vx V∞ ) dη = 0 → 0 = C1 (1) + 2C2 (1) + 3C3 (1)
y δ = η = 0, d 2 ( Vx V∞ ) dη = 0 → 0 = 2C2 + 6C3 ( 0 )
2
4. At y = 0, d 2 Vx dy 2 = 0 → → C2 = 0
Solving boundary conditions 2 and 3: , C1 = 3 2 and C3 = −1 2
3 1
F (η ) = Vx V∞ = η − η 3 Answer
2 2
δ 2β γ
b) The boundary layer thickness is defined by Eq. 10-27: =
x Rex
1 1
⎛3 1 ⎞⎡ ⎛3 1 ⎞⎤ 39
From Eq. 10-24: γ = ∫ F (η ) ⎣⎡1 − F (η ) ⎦⎤ dη = ∫ ⎜ η − η 3 ⎟ ⎢1 − ⎜ η − η 3 ⎟ ⎥ dη =
0 0⎝2 2 ⎠⎣ ⎝ 2 2 ⎠⎦ 280
dF d ⎡3 1 ⎛3 3 ⎞ 3
Using Eq. 10-25: β= = ⎢ η − η3 = ⎜ − η2 ⎟ =
dη η = 0 dη ⎣ 2 2 η =0 ⎝ 2 2 ⎠ η =0 2

δ 2 ( 3 2 ) ( 39 280 ) 4.64
Using Eq. 10-27: = = Answer
x Rex Rex
c) The skin friction coefficient is defined by Eq. 10-28

2 βγ 2 ( 3 2 )( 39 280 ) 0.646
Cf = = = Answer
Rex Rex Rex
Comments:
Note that the boundary layer thickness constant of 4.64 is a relatively good approximation to the Blasius value of
5.0, as is the skin friction coefficient constant of 0.646 compared to 0.664.

10- 58