Problem 7.76
Models are commonly used to study the dispersion of a gaseous pollutant from an exhaust
stack located near a building complex. Similarity requirements for the pollutant source
involve the following independent variables: the stack gas speed, V; the wind speed, U; the
density of the atmospheric air, ; the difference in densities between the air and the stack
gas, − s; the acceleration of gravity, g; the kinematic viscosity of the stack gas, vs; and the
stack diameter, D. (a) Based on these variables, determine a suitable set of similarity
requirements for modeling the pollutant source. (b) For this type of model, a typical length
scale might be 1:200. If the same fluids were used in model and prototype, would the
similarity requirements be satisfied? Explain and support your answer with the necessary
calculations.
Solution 7.76
(a) Since
ρρρ
−−− − − −
−
1 1 42 42 2 2 1 ,
ss
VLT ULT FLT LTT gLT v LT DL
(b) For =1
200
m
D
D and vsm = vs the second similarity requirement is
Problem 7.77
As winds blow past buildings, complex flow patterns can develop due to various factors
such as flow separation and interactions between adjacent buildings. Assume that the local
gage pressure, p, at a particular location on a building is a function of the air density,
ρ
; the
wind speed, V; some characteristic length, ; and all other pertinent lengths, i, needed to
characterize the geometry of the building or building complex. (a) Determine a suitable set
of dimensionless parameters that can be used to study the pressure distribution. (b) An
eight-story building that is 100 ft tall is to be modeled in a wind tunnel. If a length scale of
1:300 is to be used, how tall should the model building be? (c) How will a measured
pressure in the model be related to the corresponding prototype pressure? Assume the same
air density in model and prototype. Based on the assumed variables, does the model wind
speed have to be equal to the prototype wind speed? Explain.
Solution 7.77
(a)
()
ρ
=,,,
i
p
fV
ρ
−− −
=2421
ii
p
FL FL T V LT L L
(b) For geometric similarity
and it follows that all pertinent lengths are scaled with the length scale m
. Thus, with
=1
300
m
==
100 ft
m
odel height 0.333ft
300 .
(c) With geometric similarity satisfied it follows that
Problem 7.78
Assume that the pump performance characteristics shown in the figure below are for a
pump running at 1200 rpm. Assuming that the pump head, hp; the pump impeller diameter,
D; and the flowrate, Q, through the pump are the other relevant parameters, calculate the
pump performance characteristics at speeds of 1800 rpm and 2400 rpm. Neglect viscous
effects.
Solution 7.78
The dimensionless performance curve for a pump is
12 12
or
2
22 22
11 11
and
hN QN
hN QN
==
0 25 50 75 100 125 150
0
50
100
150
200
Volume flowrate, Q (gal/min)
Head (ft)
h
p
T = 55°F
Similarity, the curve at 2400 rpm is found from
Curves are shown in the following figure.
600
700
800
2
4
0
0
r
p
m
Problem 7.79
A stream of atmospheric air is used to keep a ping-pong ball aloft by blowing the air
upward over the ball. The ping-pong ball has a mass of 2.5 g and a diameter D1 = 3.8 cm,
and the air stream has an upward velocity of V1 = 0.942 m/s. This system is to be modeled
by pumping water upward with a velocity V2 over a solid ball of diameter D2 and density
ρ
=
2
3
2710 kg/m
b. In both cases, the net weight of the ball Wb is equal to the air drag,
ρ
=
2
C,
2
b
AV
W
D
where CD = 0.60,
ρ
is the fluid density, A the ball’s projected area, and V the velocity of the
fluid upstream from the ball. Determine all possible combinations of V2 and D2. [Hint: A
force balance involving the drag on the ball, the buoyant force on the ball, and the weight
of the ball is needed.]
Solution 7.79
Both the ping-pong ball-air system and the other ball-water system must have the same
Reynolds number.
or
()
()
()
ρ
ρρ
== =
−−
2
2
3
22
2
2223
kg
0.60 1000
C
33 s
m0.0268 .
mkg
44 m
9.81 2710 1000
sm
b
D
V
D
Solving the second equation for D2 and substituting this into the first equation gives
Both the ping-pong ball-air system and the other ball-water system must have the same
Reynolds number or
Using kinematic viscosity values from tables of properties of air and water,
The net weight of the ball is the weight minus the buoyant force.
Problem 7.80
A 1-scale
10 model of an airplane is tested in a wind tunnel at 70 °F and 14.40 psia. The
model test results are:
Velocity (mph) 0 50 100 150 200
Drag (lb) 0 5 21 46 85
Find the corresponding airplane velocities and drags if only fluid compressibility is
important and the airplane is flying in the U.S. Standard Atmosphere at 30,000 ft. Assume
that the air is an ideal gas.
Solution 7.80
If only compressibility is important, the Mach Numbers (or Cauchy Numbers) must be
matched:
R
is a constant and at the temperatures involved
k
is nearly constant so
Using Eqs. (1) and (2) together with the given data:
Problem 7.81
A company manufactures geometrically similar airplane propellers up to 4.0 m in diameter.
Wind tunnel tests are run on geometrically similar propellers up to 0.33 m in diameter. In
the test of a 0.33-m model of a 4.0-m propeller, the air velocity was 50 m/s. The model
rotational speed was 2000 rpm, the thrust 100 N, and the propeller input torque 20 N · m.
Calculate the corresponding prototype rotational speed, thrust, and input torque for an
airplane speed of 100 m/s. The Reynolds numbers and Mach numbers are of minor
importance.
Solution 7.81
First, we need to find appropriate dimensionless parameters for the propeller.
The primary parameters are speed ratio:
pp mm
p
pp mmm ppp mmm
From (A)
or
()
==0.165 2000 rpm 330 rpm
p
N
From (B), noting m = p gives
Problem 7.82
A breakwater is a wall built around a harbor so the incoming waves dissipate their energy
against it. The significant dimensionless parameters are the Froude number F and the
Reynolds number Re. A particular breakwater measuring 450 m long and 20 m deep is hit
by waves 5 m high and velocities up to 30 m/s. In a 1
100-scale model of the breakwater, the
wave height and velocity can be controlled. Can complete similarity be obtained using
water for the model test?
Solution 7.82
The Reynolds and Froude number requirements are
Obviously, these cannot both be true unless S = 1.
Problem 7.83
A viscous fluid is contained between wide, parallel plates spaced a distance h apart as
shown in the figure below. The upper plate is fixed, and the bottom plate oscillates
harmonically with a velocity amplitude U and frequency
ω
. The differential equation for
the velocity distribution between the plates is
2
2
uu
ty
ρµ
∂∂
=
∂∂
where u is the velocity, t is time, and
ρ
and μ are fluid density and viscosity, respectively.
Rewrite this equation in a suitable nondimensional form using h, U, and
ω
as reference
parameters.
Solution 7.83
Let *
y
yh
=, *u
uU
=, and t* =
ω
t
so that:
Thus, the original differential equation becomes
*2*
*2*2
uUu
Uthy
µ
ρ
ω
∂∂
=
∂∂
yhu
u
=
U
cos
t
ω
x
Fixed plate
Problem 7.84
The deflection of the cantilever beam of the figure below is governed by the differential
equation.
()
=−
2
2
dy
E
IPx
dx
where E is the modulus of elasticity and I is the moment of inertia of the beam cross
section. The boundary conditions are y = 0 at x = 0 and dy / dx = 0 at x = 0. (a) Rewrite the
equation and boundary conditions in dimensionless form using the beam length, , as the
reference length. (b) Based on the results of part (a), what are the similarity requirements
and the prediction equation for a model to predict deflections?
Solution 7.84
(a) Let *
y
y= and = so that:
or
()
2* 2 *
*2 1
dy P x
EI
dx
=−
x
P
y
ℓ
The prediction equation is
=
**
m
y
y
or
Problem 7.85
A liquid is contained in a pipe that is closed at one end as shown in the figure below.
Initially the liquid is at rest, but if the end is suddenly opened the liquid starts to move.
Assume the pressure p1 remains constant. The differential equation that describes the
resulting motion of the liquid is
ρµ
∂∂∂
=+ +
∂∂
∂
2
1
2
1
zzz
vp v v
trr
r
where vz is the velocity at any radial location, r, and t is time. Rewrite this equation in
dimensionless form using the liquid density,
ρ
; the viscosity, μ; and the pipe radius, R, as
reference parameters.
Solution 7.85
Let *r
rR
=, *t
t
τ
=, and
*
*z
z
v
v
V
= where is some combinations of the parameters
ρ
, μ, and R
having the dimensions of time, and V is some combination of the same parameters having
the dimensions of a velocity. Let
With these dimensionless variables:
The original differential equation can be expressed as
p1
r
z
End initially
closed
vzR
ℓ
Problem 7.86
A student drops two spherical balls of different diameters and different densities. She has a
stroboscopic photograph showing the positions of each ball as a function of time. However,
she wants to express the velocity of each as a function of time in dimensionless form.
Develop the dimensionless groups. The equation of motion for each ball is:
2
D
C,
2
dV
mg AV m dt
ρ
−=
where m is ball mass, g is acceleration of gravity, CD is a dimensionless and constant drag
coefficient,
ρ
is air mass density, A is ball cross-sectional area 2
(4)D
π
= with D ball
diameter, V is ball velocity, and t is time.
Solution 7.86
ρ
−=
2
2
D
CdV
mg AV m dt
First put the equation in nondimensional form by dividing by mg
46
where
ρ
S is the density of the sphere material. Substituting and simplifying on the left-hand
side of the equation gives
On the right-hand side of the equation, multiply and divide by and
g
gD D to get
Problem 7.87
The basic equation that describes the motion of the fluid above a large oscillating flat
plate is
2
2,
uu
v
t
y
∂∂
=
∂∂
where u is the fluid velocity component parallel to the plate, t is time, y is the spatial
coordinate perpendicular to the plate, and v is the fluid kinematic viscosity. The plate
oscillating velocity is given by U = U0 sin(
ω
t). Find appropriate dimensionless parameters
and the dimensionless differential equation.
Solution 7.87
∂∂
=
∂∂
2
2
uu
v
t
y
We will nondimensionalize the equation by introducing reference value for the variables.
The reference value for u is obviously U0 and the reference value for t is obviously 1
ω
. There
is no reference length per se; however,
Problem 7.88
The dimensionless parameters for a ball released and falling from rest in a fluid are
ρ
ρ
2
D,,,and,
b
gt Vt
C
DD
where
C
D is a drag coefficient (assumed to be constant at 0.4), g is the acceleration of
gravity, D is the ball diameter, t is the time after it is released, is the density of the fluid in
which it is dropped, and is the density of the ball. Ball 1, an aluminum ball b1 =
2710kg/m3 having a diameter D1 = 1.0 cm, is dropped in water ( = 1000 kg/m3). The ball
velocity V1 is 0.733 m/s at t1 = 0.10 s. Find the corresponding velocity V2 and time t2 for
ball 2 having D2 = 2.0 cm and 21
bb
ρρ
=. Next, use the computed value of V2 and the
equation of motion,
ρρρ
−=
2
C,
2
bb
dV
Vg AV V dt
D
where V is the volume of the ball and A is its cross-sectional area, to verify the value of t2.
Should the two values of t2 agree?
Solution 7.88
The dimensionless group
2
gt
D gives
The dimensionless group Vt
D gives
or
=
2
m
1.04 s
V
Now write Eq. (1) as
rearranging, and integrating gives
α
=
−
22
00
1
Vt
dV gdt
V
or
α
αα
+
=
−
11
ln .
21
V
tgV
For the D2 = 2.0 cm ball
For ==
2
m
1.04 s
V
V,