Use only the quantities
, E, and A to nondimensionalize y, x, and t, and rewrite the differential
equation in dimensionless form. Do any parameters remain? Could they be removed by further
manipulation of the variables?
Solution 5.47
The appropriate dimensionless variables are
Problem 5.48
A smooth steel (SG = 7.86) sphere is immersed in a stream of ethanol at 20C moving at 1.5 m/s.
Estimate its drag in N from Fig. 5.3a. What stream velocity would quadruple its drag? Take
D = 2.5 cm.
Solution 5.48
For ethanol at 20C, take
789 kg/m3 and
0.0012 kg/ms. Then
Problem 5.49
The sphere in Prob. 5.48 is dropped in gasoline at 20C. Ignoring its acceleration phase, what will
be its terminal (constant) fall velocity, from Fig. 5.3a?
Solution 5.49
For gasoline at 20C, take
680 kg/m3 and
2.92E−4 kg/ms. For steel take
7800 kg/m3. Then, in “terminal” velocity, the net weight equals the drag force:
Problem 5.50
The parachute in the chapter-opener photo is, of course, meant to decelerate the payload on
Mars. The wind tunnel test gave a drag coefficient of about 1.1, based upon the projected area of
the parachute. Suppose it was falling on earth and, at an altitude of 1000 m, showed a steady
descent rate of about 18 mi/h. Estimate the weight of the payload.
Solution 5.50
Let’s convert everything to metric. The diameter is 55 ft = 16.8 m. Standard air density at
Problem 5.51
A ship is towing a sonar array which approximates a submerged cylinder 1 ft in diameter and 30 ft
long with its axis normal to the direction of tow. If the tow speed is 12 kn (1 kn = 1.69 ft/s), estimate
the horsepower required to tow this cylinder. What will be the frequency of vortices shed from the
cylinder? Use Figs. 5.2 and 5.3.
Solution 5.51
For seawater at 20C, take
1.99 slug/ft3 and
2.23E−5 slug/ft·s. Convert V = 12 knots
20.3 ft/s. Then the Reynolds number and drag of the towed cylinder is
Problem 5.52
When fluid in a long pipe starts up from rest at a uniform acceleration a, the initial flow is
laminar. The flow undergoes transition to turbulence at a time t* which depends, to first
approximation, only upon a,
, and
Experiments by P. J. Lefebvre, on water at 20C starting
from rest with 1-g acceleration in a 3-cm-diameter pipe, showed transition at t* = 1.02 s. Use
this data to estimate (a) the transition time, and (b) the transition Reynolds number ReD for water
flow accelerating at 35 m/s2 in a 5-cm-diameter pipe.
Solution 5.52
For water at 20C, take
= 998 kg/m3 and m = 0.001 kg/m-s. There are four variables. Write out
their dimensions:
Problem 5.53
Vortex shedding can be used to design a vortex flowmeter (Fig. 6.34). A blunt rod stretched across
the pipe sheds vortices whose frequency is read by the sensor downstream. Suppose the pipe
diameter is 5 cm and the rod is a cylinder of diameter 8 mm. If the sensor reads 5400 counts per
minute, estimate the volume flow rate of water in m3/h. How might the meter react to other liquids?
Solution 5.53
5400 counts/min = 90 Hz = f.
Problem 5.54
A fishnet is made of 1-mm–diameter strings knotted into 2 2 cm squares. Estimate the horsepower
required to tow 300 ft2 of this netting at 3 kn in seawater at 20C. The net plane is normal to the
flow direction.
Solution 5.54
5400 counts/min = 90 Hz = f.
Problem 5.55
The radio antenna on a car begins to vibrate wildly at 8 Hz when the car is driven at 45 mi/h over a
rutted road that approximates a sine wave of amplitude 2 cm and wavelength
= 2.5 m. The
antenna diameter is 4 mm. Is the vibration due to the road or to vortex shedding?
Solution 5.55
For seawater at 20C, take
1025 kg/m3 and
0.00107 kg/m·s.
Problem 5.56
Flow past a long cylinder of square cross-section results in more drag than the comparable round
cylinder. Here are data taken in a water tunnel for a square cylinder of side length b = 2 cm:
V, m/s:
1.0
2.0
3.0
4.0
Drag, N/(m of depth):
21
85
191
335
(a) Use these data to predict the drag force per unit depth of wind blowing at 6 m/s, in air at 20C,
over a tall square chimney of side length b = 55 cm. (b) Is there any uncertainty in your estimate?
Solution 5.56
Convert U = 45 mi/h = 20.1 m/s. Assume sea level air,
= 1.2 kg/m3,
Problem 5.57
The simply supported 1040 carbon-steel rod of Fig. P5.57 is subjected to a crossflow stream of air at
20C and 1 atm. For what stream velocity U will the rod center deflection be approximately 1 cm?
Solution 5.57
For air at 20C, take
1.2 kg/m3 and
1.8E−5 kg/m·s. For carbon steel take Young’s
modulus E 29E6 psi 2.0E11 Pa.
Problem 5.58
For the steel rod of Prob. 5.57, at what airstream velocity U will the rod begin to vibrate laterally in
resonance in its first mode (a half sine wave)? Hint: Consult a vibration text [34, 35] under “lateral
beam vibration.”
Solution 5.58
From a vibrations book, the first mode frequency for a simply-supported slender beam is
given by
Problem 5.59
A long, slender, 3-cm-diameter smooth flagpole bends alarmingly in 20 mi/h sea-level winds,
causing patriotic citizens to gasp. An engineer claims that the pole will bend less if its surface is
deliberately roughened. Is she correct, at least qualitatively?
Solution 5.59
For sea-level air, take
= 1.2255 kg/m3 and
= 1.78E-5 kg/m-s. Convert 20 mi/h = 8.94 m/s.
Problem 5.60*
The thrust F of a free propeller, either aircraft or marine, depends upon density
, the rotation
rate n in r/s, the diameter D, and the forward velocity V. Viscous effects are slight and neglected
here. Tests of a 25-cm-diameter model aircraft propeller, in a sea-level wind tunnel, yield the
following thrust data at a velocity of 20 m/s:
Rotation rate, r/min
4800
6000
8000
Measured thrust, N
6.1
19
47
(a) Use this data to make a crude but effective dimensionless plot. (b) Use the dimensionless data to
predict the thrust, in newtons, of a similar 1.6-m-diameter prototype propeller when rotating at
3800 r/min and flying at 225 mi/h at 4000 m standard altitude.
Solution 5.60
The given function is F = fcn(
, n, D, V), and we note that j = 3. Hence we expect 2 pi
groups. The writer chose (
, n, D) as repeating variables and found this:
Problem 5.61
If viscosity is neglected, typical pump flow results from Example 5.3 are shown in Fig. P5.61 for a
model pump tested in water. The pressure rise decreases and the power required increases with the
dimensionless flow coefficient. Curve-fit expressions are given for the data. Suppose a similar pump
of 12-cm diameter is built to move gasoline at 20C and a flow rate of 25 m3/h. If the pump rotation
speed is 30 r/s, find (a) the pressure rise and (b) the power required.
Solution 5.61
For gasoline at 20C, take
680 kg/m3 and
2.92E−4 kg/ms.
Convert Q = 25 m3/hr = 0.00694 m3/s. Then we can evaluate the “flow coefficient”:
Problem 5.62
For the system of Prob. P5.22, assume that a small model wind turbine of diameter 90 cm,
rotating at 1200 r/min, delivers 280 watts when subjected to a wind of 12 m/s. The data is to be
used for a prototype of diameter 50 m and winds of 8 m/s. For dynamic similarity, estimate
(a) the rotation rate, and (b) the power delivered by the prototype. Assume sea level air density.
Solution 5.62
If you worked Prob. P5.22, you would arrive at two Pi groups, like this:
Problem 5.63*
The Keystone Pipeline in the Chapter 6 opener photo has D = 36 in. and an oil flow rate
Q = 590,000 barrels per day (1 barrel = 42 U.S. gallons). Its pressure drop per unit length, Δp/L,
depends on the fluid density ρ , viscosity μ , diameter D , and flow rate Q . A water-flow model
test, at 20ºC, uses a 5-cm-diameter pipe and yields Δp/L = 4000 Pa/m. For dynamic similarity,
estimate Δp/L of the pipeline. For the oil take ρ = 860 kg/m3 and μ = 0.005 kg/m∙s.
Solution 5.63
First write out the dimension s of the variables:
2 2 3 3
{ / { / { / } { } { / }
/
}}
M L T M L L T L M LT
p L Q D
Problem 5.64
The natural frequency
of vibration of a mass M attached to a rod, as in Fig. P5.64, depends
only upon M and the stiffness EI and length L of the rod. Tests with a 2-kg mass attached to a
1040 carbon-steel rod of diameter 12 mm and length 40 cm reveal a natural frequency of 0.9 Hz.
Use these data to predict the natural frequency of a 1-kg mass attached to a 2024 aluminum-alloy
rod of the same size.
Solution 5.64
For steel, E 29E6 psi 2.03E11 Pa. If
= f(M, EI, L), then n = 4 and j = 3 (MLT), hence
we get only 1 pi group, which we can evaluate from the steel data:
Problem 5.65
In turbulent flow near a flat wall, the local velocity u varies only with distance y from the wall, wall
shear stress
w, and fluid properties
and
. The following data were taken in the University of
Rhode Island wind tunnel for airflow,
= 0.0023 slug/ft3,
= 3.81E−7 slug/(ft·s), and
w = 0.029 lbf/ft2:
y, in
0.021
0.035
0.055
0.080
0.12
0.16
u, ft/s
50.6
54.2
57.6
59.7
63.5
65.9
(a) Plot these data in the form of dimensionless u versus dimensionless y, and suggest a suitable
power-law curve fit. (b) Suppose that the tunnel speed is increased until u = 90 ft/s at y = 0.11 in.
Estimate the new wall shear stress, in lbf/ft2.
Solution 5.65
Given that u = fcn(y,
w,
,
), then n = 5 and j = 3 (MLT), so we expect
n − j = 5 − 3 = 2 pi groups, and they are traditionally chosen as follows (Chap. 6, Section
6.5):
Problem 5.66
A torpedo 8 m below the surface in 20C seawater cavitates at a speed of 21 m/s when atmospheric
pressure is 101 kPa. If Reynolds number and Froude number effects are negligible, at what speed
will it cavitate when running at a depth of 20 m? At what depth should it be to avoid cavitation at
30 m/s?
Solution 5.66
For seawater at 20C, take
= 1025 kg/m3 and pv = 2337 Pa. With Reynolds and Froude
numbers neglected, the cavitation numbers must simply be the same:
Problem 5.67
A student needs to measure the drag on a prototype of characteristic length dp moving at velocity
Up in air at standard atmospheric conditions. He constructs a model of characteristic dimension dm,
such that the ratio dp/dm is some factor f. He then measures the drag on the model at dynamically
similar conditions. The student claims that the drag force on the prototype will be identical to that
measured on the model. Is this claim correct? Explain.
Solution 5.67
Assuming no compressibility effects, dynamic similarity requires that
Problem 5.68
For the rotating-cylinder function of Prob. P5.20, if L >> D, the problem can be reduced to only
two groups, F/(
U2LD) versus (D/U). Here are experimental data for a cylinder 30 cm in
diameter and 2 m long, rotating in sea-level air, with U = 25 m/s.
, rev/min
0
3000
6000
9000
12000
15000
F, N
0
850
2260
2900
3120
3300
(a) Reduce this data to the two dimensionless groups and make a plot. (b) Use this plot to
predict the lift of a cylinder with D = 5 cm, L = 80 cm, rotating at 3800 rev/min in water at
U = 4 m/s.
Solution 5.68
(a) In converting the data, the writer suggests using in rad/s, not rev/min. For sea-level air,
= 1.2255 kg/m3. Take, for example, the first data point, = 3000 rpm x (2/60) = 314 rad/s,
and F = 850 N.