Solution 7.62
For sea-level air, take
= 1.225 kg/m3 and
= 1.78E−5 kg/ms. Convert
Problem 7.63
For those who think electric cars are sissy, Keio University in Japan has tested a 22-ft long
prototype whose eight electric motors generate a total of 590 horsepower. The “Kaz” cruises at
180 mi/h (see Popular Science, August 2001, p. 15). If the drag coefficient is 0.35 and the frontal
area is 26 ft2, what percent of this power is expended against sea-level air drag?
Solution 7.63
For air, take
= 0.00237 slug/ft3. Convert 180 mi/h to 264 ft/s. The drag is
Problem 7.64
A parachutist jumps from a plane, using an 8.5-m-diameter chute in the standard atmosphere.
The total mass of chutist and chute is 90 kg. Assuming a fully open chute in quasi-steady motion,
estimate the time to fall from 2000-m to 1000-m.
Solution 7.64
For the standard altitude (Table A-6), read
= 1.112 kg/m3 at 1000 m altitude and
Problem 7.65
As soldiers get bigger and packs get heavier, a parachutist and load can weigh as much as
400 lbf. The standard 28-ft parachute may descend too fast for safety. For heavier loads, the U.S.
Army Natick Center has developed a 28-ft, higher drag, less porous XT-11 parachute (see the
URL http://www.paraflite.com/). This parachute has a sea-level descent speed of 16 ft/s with a
400-lbf load. (a) What is the drag coefficient of the XT-11? (b) How fast would the standard
chute descend at sea-level with such a load?
Solution 7.65
For sea-level air, take
= 0.00237 slug/ft3. (a) Everything is known except CD:
Problem 7.66
A sphere of density
s and diameter D is dropped from rest in a fluid of density
and viscosity
.
Assuming a constant drag coefficient
o,
d
C
derive a differential equation for the fall velocity V(t)
and show that the solution is
o
1/ 2
4 ( 1) tanh
3d
gD S
V Ct
C

−
=


o
1/2
2
3 ( 1)
4
d
gC S
CSD
−

=

where S =
s/
is the specific gravity of the sphere material.
Solution 7.66*
Newton’s law for downward motion gives
Problem 7.67
The Toyota Prius has a drag coefficient of 0.25, a frontal area of 23.4 ft2, and an empty weight of
3042 lbf. Its rolling resistance coefficient is Crr = 0.03, that is, the rolling resistance is 3 percent
of the normal force on the tires. If rolling freely down a slope of 8º at an altitude of 500 m,
calculate its maximum velocity, in mi/h.
Solution 7.67
From Table A.6 at 500 m, ρ = 1.1677 kg/m3 ÷ 515.38 = 0.00227 slug/ft3. Maximum velocity is
at zero acceleration, where the weight along the slope balances drag and rolling resistance:
Problem 7.68
The Mars roving-laboratory parachute, in the Chap. 5 opener photo, is a 55-ft-diameter disk-gap-
band chute, with a measured drag coefficient of 1.12 [59]. Mars has very low density, about
2.9E-5 slug/ft3, and its gravity is only 38% of earth gravity. If the mass of payload and chute is
2400 kg, estimate the terminal fall velocity of the parachute.
Solution 7.68
Convert D = 55 ft = 16.8 m and r = 2.9E-5 slug/ft3 = 0.015 kg/m3. At terminal velocity, the
parachute weight is balanced by chute drag:
Problem 7.69
Two baseballs, of diameter 7.35 cm, are connected to a rod 7 mm in diameter and 56 cm
long, as in Fig. P7.69. What power, in W, is required to keep the system spinning at
400 r/min? Include the drag of the rod, and assume sea–level standard air.
Solution 7.69
For sea-level air, take
= 1.225 kg/m3 and
= 1.78E−5 kg/ms. Assume a laminar drag
coefficient CD 0.47 from Table 7.3.
Problem 7.70
The Army’s new ATPS personnel parachute is said to be able to bring a 400-lbf load, trooper
plus pack, to ground at 16 ft/s in “mile–high” Denver, Colorado. If we assume that Table 7.3 is
valid, what is the approximate diameter of this new parachute?
Solution 7.70
Assume that Denver is 5280 ft = 1609 m standard altitude. From Table A.6, interpolate
Problem 7.71
The 2013 Toyota Camry has an empty weight of 3190 lbf, a frontal area of 22.06 ft2, and a drag
coefficient of 0.28. Its rolling resistance is Crr ≈ 0.035. Estimate the maximum velocity, in mi/h,
this car can attain when rolling freely at sea-level down a 4º slope.
Solution 7.71
The weight component down the slope is balances by rolling resistance and air drag. At sea-
Problem 7.72
A settling tank for a municipal water supply is 2.5 m deep, and 20C water flows through
continuously at 35 cm/s. Estimate the minimum length of the tank which will ensure that all
sediment (SG = 2.55) will fall to the bottom for particle diameters greater than (a) 1 mm and
(b) 100
m.
Solution 7.72
For water at 20C, take
= 998 kg/m3 and
= 0.001 kg/ms. The particles travel with the stream
Problem 7.73
A balloon is 4 m in diameter and contains helium at 125 kPa and 15C. Balloon material and
payload weigh 200 N, not including the helium. Estimate (a) the terminal ascent velocity in sea-
level standard air; (b) the final standard altitude (neglecting winds) at which the balloon will
come to rest; and (c) the minimum diameter (<4 m) for which the balloon will just barely begin
to rise in sea-level standard air.
Solution 7.73
For sea-level air, take
= 1.225 kg/m3 and
= 1.78E−5 kg/ms. For helium R = 2077 J/kgK.
Sea-level air pressure is 101350 Pa. For upward motion V,
Problem 7.74
It is difficult to define the “frontal area” of a motorcycle due to its complex shape. One then
measures the drag-area, that is, CDA, in area units. Hoerner [12] reports the drag-area of a
typical motorcycle, including the (upright) driver, as about 5.5 ft2. Rolling friction is typically
about 0.7 lbf per mi/h of speed. If that is the case, estimate the maximum sea-level speed (in
mi/h) of the new Harley-Davidson V-Rod cycle, whose liquid-cooled engine produces 115 hp.
Solution 7.74
For sea-level air, take
= 0.00237 slug/ft3. Convert 0.7 lbf per mi/h rolling friction to 0.477 lbf
per ft/s of speed. Then the power relationship for the cycle is
Problem 7.75
The helium-filled balloon in Fig. P7.75 is tethered at 20C and 1 atm with a string of negligible
weight and drag. The diameter is 50 cm, and the balloon material weighs 0.2 N, not including the
helium. The helium pressure is 120 kPa. Estimate the tilt angle
if the airstream velocity U is
(a) 5 m/s or (b) 20 m/s.
Solution 7.75
For air at 20C and 1 atm, take
= 1.2 kg/m3 and
= 1.8E−5 kg/ms.
For helium, R = 2077 J/kgK. The helium density = (120000)/[2077(293)] 0.197 kg/m3.
Problem 7.76
The 2005 movie The World’s Fastest Indian tells the story of Burt Munro, a New Zealander
who, in 1937, set a motorcycle record of 201 mi/h on the Bonneville Salt Flats. Using the data of
Prob. P7.74, (a) estimate the horsepower needed to drive this fast. (b) What horsepower would
have gotten Burt up to 250 mi/h?
Problem 7.74
It is difficult to define the “frontal area” of a motorcycle due to its complex shape. One then
measures the drag-area, that is, CDA, in area units. Hoerner [12] reports the drag-area of a
typical motorcycle, including the (upright) driver, as about 5.5 ft2. Rolling friction is typically
about 0.7 lbf per mi/h of speed. If that is the case, estimate the maximum sea-level speed (in
mi/h) of the new Harley-Davidson V-Rod cycle, whose liquid-cooled engine produces 115 hp.
Solution 7.76
Prob. P7.74 suggests CDA = 5.5 ft2 and Frolling = 0.7 lbf per mi/h of speed. Convert 201 mi/hr to
Problem 7.77
To measure the drag of an upright person, without violating human subject protocols, a life-sized
mannequin is attached to the end of a 6-m rod and rotated at = 80 rev/min, as in Fig. P7.77.
The power required to maintain the rotation is 60 kW. By including rod drag power, which is
significant, estimate the drag-area CDA of the mannequin, in m2.
Solution 7.77
For air, take
= 1.2 kg/m3 and
= 1.8E−5 kg/ms. The mannequin velocity is
Vm = L = [(80 2
/60)rad/s](6m) (8.38 rad/s)(6 m) 50.3 m/s. The velocity at mid–
Problem 7.78
On April 24, 2007, a French bullet train set a new record, for rail-driven trains, of 357.2 mi/h,
beating the old record by 12 per cent. Using the data in Table 7.3, estimate the sea-level
horsepower required to drive this train at such a speed.
Solution 7.78
Take sea-level density as 1.2255 kg/m3. Convert 357.2 mi/h to 159.7 m/s. From Table 7.3, the
Problem 7.79
Assume that a radioactive dust particle approximates a sphere of density 2400 kg/m3. How long,
in days, will it take such a particle to settle to sea level from an altitude of 12 km if the particle
diameter is (a) 1 μm or (b) 20 μm?
Solution 7.79
For such small particles, tentatively assume that Stokes’ law prevails:
Problem 7.80
A heavy sphere attached to a string should hang at an angle
when immersed in a stream of
velocity U, as in Fig. P7.80. Derive an expression for
as a function of the sphere and flow
properties. What is
if the sphere is steel (SG = 7.86) of diameter 3 cm and the flow is sea-level
standard air at U = 40 m/s? Neglect the string drag.
Solution 7.80
Problem 7.81
A typical U.S. Army parachute has a projected diameter of 28 ft. For a payload mass of 80 kg,
(a) what terminal velocity will result at 1000-m standard altitude? For the same velocity and
payload, what size drag-producing “chute” is required if one uses a square flat plate held
(b) vertically; and (c) horizontally? (Neglect the fact that flat shapes are not dynamically stable
in free fall.)
Solution 7.81
For air at 1000 meters, from Table A-3,
1.112 kg/m3. Convert D = 28 ft = 8.53 m. Convert
W = mg = 80(9.81) = 785 N. From Table 7-3 for a parachute, read CD 1.2, Then, for part (a),
Problem 7.82
Skydivers, flying over sea-level ground, typically jump at about 8000 ft altitude and free-fall
spread-eagled until they open their chutes at about 2000 ft. They take about 10 s to reach
terminal velocity. Estimate how many seconds of free-fall they enjoy if (a) they fall spread-
eagled; or (b) they fall feet first? Assume a total skydiver weight of 220 lbf.
Solution 7.82
From Table 7.3 use CDA = 9 ft2 spread-eagled and 1.2 ft2 when feet first. From Table A.6,
= 0.001869 slug/ft3 at 8000 ft standard altitude and 0.002242 slug/ft3 at 2000 ft. (a) Compute
each terminal velocity spread-eagled (CDA = 9 ft2):
Problem 7.83
A blimp approximates a 4:1 spheroid that is 196 ft long. It is powered by two 150 hp ducted
fans. Estimate the maximum speed attainable, in mi/h, at an altitude of 8200 ft.
Solution 7.83
At 8200 ft standard altitude (2500 m), from Table A.6, ρ = 0.9570 kg/m3 = 0.00186 slug/ft3. The
Problem 7.84
A Ping-Pong ball weighs 2.6 g and has a diameter of 3.8 cm. It can be supported by an air jet
from a vacuum cleaner outlet, as in Fig. P7.84. For sea-level standard air, what jet velocity is
required?
Solution 7.84
For sea-level air, take
= 1.225 kg/m3 and
= 1.78E−5 kg/ms. The ball weight must balance its
drag:
Problem 7.85
In this era of expensive fossil fuels, many alternatives have been pursued. One idea from
SkySails, Inc., shown in Fig. P7.85, is the assisted propulsion of a ship by a large tethered kite.
The tow force of the kite assists the ship propeller and is said to reduce annual fuel consumption
by 10%-35%. For a typical example, let the ship be 120 m long, with a wetted area of 2800 m2.
The kite area is 330 m2 and has a force coefficient of 0.8. The kite cable makes an angle of 25
with the horizontal. Let Vwind = 30 mi/h. Neglect ship wave drag. Estimate the ship speed
(a) due to the kite only; and (b) if the propeller delivers 1,250 hp to the water. [Hint: The kite
sees the relative velocity of the wind.]
Solution 7.85
Assume sea level air density,
a = 1.2255 kg/m3. For seawater, take
= 1025 kg/m3 and
= 0.00107 kg/m-s. The wind velocity is 30 mi/h = 13.4 m/s. (a) For a wind Vair and a ship
speed V, the kite force equals the friction drag of the ship:
Problem 7.86
Hoerner [Ref. 12 of Chap. 7, p. 3–25] states that the drag coefficient of a flag of 2:1 aspect ratio is
0.11 based on planform area. The University of Rhode Island has an aluminum flagpole 25 m
high and 14 cm in diameter. It flies equal-sized national and state flags together. If the fracture
stress of aluminum is 210 MPa, what is the maximum flag size that can be used without breaking
the flagpole in hurricane (75 mi/h) winds? Neglect the drag of the flagpole.
Solution 7.86
URI is approximately sea-level,
= 1.225 kg/m3. Convert 75 mi/h = 33.5 m/s. We will use the most
elementary strength of materials formula, without even a stress-concentration factor, since this is
just a fluid mechanics book:
Problem 7.87
A tractor-trailer truck has a drag area CDA = 8 m2 bare and CDA = 6.7 m2 with an aerodynamic
deflector added (Fig. 7.18b). Its rolling resistance is 50 N for each mile per hour of speed.
Calculate the total horsepower required at sea level with and without the deflector if the truck
moves at (a) 55 mi/h; and (b) 75 mi/h.
Solution 7.87
For sea-level air, take
= 1.225 kg/m3 and
= 1.78E−5 kg/ms. Convert V = 55 mi/h = 24.6 m/s
and 75 mi/h = 33.5 m/s. Take each speed in turn:
Problem 7.88
A pickup truck has a clean drag-area CDA of 35 ft2. Estimate the horsepower required to drive the
truck at 55 mi/h (a) clean and (b) with the 3- by 6-ft sign in Fig. P7.88 installed if the rolling
resistance is 150 lbf at sea level.
Solution 7.88
For sea-level air, take
= 0.00238 slug/ft3 and
= 3.72E−7 slug/fts.
Convert V = 55 mi/h = 80.7 ft/s. Calculate the drag without the sign: