Chapter 5
•
Dimensional Analysis
and Similarity
5.1
For axial flow through a circular tube, the Reynolds number for transition to
turbulence is approximately 2300 [see Eq. (6.2)], based upon the diameter and average
velocity. If
d
=
5 cm and the fluid is kerosene at 20
°
C, find the volume flow rate in m
3
/h
which causes transition.
Solution:
For kerosene at 20
°
C, take
ρ
=
804 kg/m
3
and
µ
=
0.00192 kg/m
⋅
s. The only
unknown in the transition Reynolds number is the fluid velocity:
tr
Vd
(804)V(0.05)
Re
2300
,
solve for V
0.11m/s
0.00192
≈==
=
3
2
m
Then
Q
VA
(0.11)
(0.05)
2.16E 4
3600
4s
Ans.
π
==
=
−
×
≈
3
m
0.78
hr
5.2
In flow past a thin flat body such as an airfoil, transition to turbulence occurs at about
Re
=
1E6, based on the distance
x
from the leading edge of the wing. If an airplane flies at
450 mi/h at 8km standard altitude and undergoes transition at the 12 percent chord position,
how long is its chord (wing length from leading to trailing edge)?
Solution:
From Table A6 at z
=
8000 m,
≈
0.525 kg/m
3
, T
≈
236
°
K, hence
≈
1.53E
−
5 kg/m
⋅
s. Convert U
=
450 mi/hr
≈
201 m/s. Then the transition Reynolds no. is
6
tr
x,tr
Ux
(0.525)(201)(0.12C)
Re
10
,
solve for C
1.53E 5
Ans.
=
≈
−
1.21 m
5.3
An airplane has a chord length
L
=
1.2 m and flies at a Mach number of 0.7 in the
standard atmosphere. If its Reynolds number, based on chord length, is 7E6, how high is it
flying?
Solution:
This is harder than Prob. 5.2 above, for we have to search in the U.S. Stan
dard Atmosphere (Table A6) to find the altitude with the right density and viscosity
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•
Dimensional Analysis and Similarity
309
and speed of sound. We can make a first guess of T
≈
230 K, a
≈
√
(kRT)
≈
304 m/s,
U
=
0.7a
≈
213 m/s, and
µ
≈
1.51E
−
5 kg/m
⋅
s. Then our first estimate for density is
3
C
UC
(213)(1.2)
Re
7E6
,
or
0.44 kg/m
and Z
9500 m (Table A6)
1.51E 5
ρρ
ρ
==
≈
≈
≈
−
Repeat and the process converges to
≈
0.41 kg/m
3
or
Z
≈
10100 m
Ans
.
5.4
When tested in water at 20
°
C flowing at 2 m/s, an 8cmdiameter sphere has a
measured drag of 5 N. What will be the velocity and drag force on a 1.5mdiameter
weather balloon moored in sealevel standard air under dynamically similar conditions?
Solution:
For water at 20
°
C take
≈
998 kg/m
3
and
≈
0.001 kg/m
⋅
s. For sealevel
standard air take
≈
1.2255 kg/m
3
and
≈
1.78E
−
5 kg/m
⋅
s. The balloon velocity follows
from
dynamic similarity
, which requires identical Reynolds numbers:
balloon
model
model
proto
1.2255V
(1.5)
VD
998(2.0)(0.08)
Re
1.6E5
Re
0.001
1.78E 5
=
=
=
−

ρ
or V
balloon
≈
1.55 m/s. Then the two spheres will have identical drag coefficients:
balloon
D,model
D,proto
22
2
2
2
2
F
F5
N
C
0.196
C
V D
998(2.0) (0.08)
1.2255(1.55) (1.5)
=
=
=
Solve for
.
Ans
F
1.3 N
balloon
≈
5.5
An automobile has a characteristic length and area of 8 ft and 60 ft
2
, respectively.
When tested in sealevel standard air, it has the following measured drag force versus speed:
V
, mi/h:
20
40
60
Drag, lbf:
31
115
249
The same car travels in Colorado at 65 mi/h at an altitude of 3500 m. Using dimensional
analysis, estimate (a) its drag force and (b) the horsepower required to overcome air drag.
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 Winter '08
 Sarkar
 Fluid Dynamics, Drag coefficient, Fifth Edition, pi groups

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