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Chapter 7
•
Flow Past Immersed Bodies
7.1
For flow at 20 m/s past a thin flat plate, estimate the distances
x
from the leading
edge at which the boundary layer thickness will be either 1 mm or 10 cm, for (a) air; and
(b) water at 20
°
C and 1 atm.
Solution:
(a) For air, take
ρ
=
1.2 kg/m
3
and
µ
=
1.8E
−
5 kg/m
⋅
s. Guess laminar flow:
22
1/2
5.0
(0.001) (1.2)(20)
,:
(
—
1
)
25
25(1.8
5)
laminar
x
U
or
x
Ans. air
mm
xE
Re
δδ
==
=
=
−
0.0533 m
1.2(20)(0.0533)/1.8
5
71,000
,
x
Check
Re
E
OK laminar flow
=−
=
(a) For the thicker boundary layer, guess turbulent flow:
1/7
0.16
(
a
—
1
0
)
(/
)
turb
solve for
Ans.
cm
x
Ux
δ
ρµ
=
x6
.
0
6m
=
8.1 6,
,
x
Check
Re
E
OK turbulent flow
=
(b) For water, take
=
998 kg/m
3
and
=
0.001 kg/m
⋅
s. Both cases are probably turbulent:
=
1 mm:
x
turb
=
0.0442 m
, Re
x
=
882,000 (barely turbulent)
Ans.
(
water—1 mm
)
=
10 cm:
x
turb
=
9.5 m
, Re
x
=
1.9E8 (OK, turbulent)
Ans.
(
water—10 cm
)
7.2
Air, equivalent to a Standard Altitude of 4000 m, flows at 450 mi/h past a wing
which has a thickness of 18 cm, a chord length of 1.5 m, and a wingspan of 12 m. What is
the appropriate value of the Reynolds number for correlating the lift and drag of this
wing? Explain your selection.
Solution:
Convert 450 mi/h
=
201 m/s, at 4000 m,
=
0.819 kg/m
⋅
s, T
=
262 K,
=
1.66E
−
5 kg/m
⋅
s. The appropriate length is the
chord
, C
=
1.5 m, and the best parameter to
correlate with lift and drag is
Re
C
=
(0.819)(201)(1.5)/1.66E
−
5
=
1.5E7
Ans.
7.3
Equation (7.1
b
) assumes that the boundary layer on the plate is turbulent from the
leading edge onward. Devise a scheme for determining the boundarylayer thickness
more accurately when the flow is laminar up to a point Re
x
,crit
and turbulent thereafter.
Apply this scheme to computation of the boundarylayer thickness at
x
=
1.5 m in 40 m/s
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Solutions Manual
•
Fluid Mechanics, Fifth Edition
flow of air at 20
°
C and 1 atm past a flat plate. Compare your result with Eq. (7.1
b
).
Assume Re
x
,crit
≈
1.2E6.
Fig. P7.3
Solution:
Given the transition point x
crit
, Re
crit
, calculate the laminar boundary layer thick
ness
δ
c
at that point, as shown above,
c
/x
c
≈
5.0/Re
crit
1/2
. Then find the “apparent” distance
upstream, L
c
, which gives the same
turbulent
boundary layer thickness,
c
1/7
cc
L
/L
0.16/Re
.
≈
Then begin x
effective
at this “apparent origin” and calculate the remainder of the turbulent
boundary layer as
/x
eff
≈
0.16/Re
eff
1/7
. Illustrate with a numerical example as requested.
For air at 20
°
C, take
ρ
=
1.2 kg/m
3
and
µ
=
1.8E
−
5 kg/m
⋅
s.
c
crit
c
c
1/2
1/6
7/6
c
c
1.2(40)x
5.0(0.45)
Re
1.2E6
if x
0.45 m,
then
0.00205 m
1.8E 5
(1.2E6)
U
0.00205
1.2(40)
Compute
L
0.0731 m
0.16
0.16
1.8E 5
==
=
=
≈
−
±²
³
´
≈
µ¶
µ
¶
·¸
−
¹º
»
¼
Finally, at x
=
1.5 m, compute the effective distance and the effective Reynolds number:
eff
c
c
eff
eff
1.5 m
eff
1.2(40)(1.123)
x
x
L
x
1.5
0.0731 0.45
1.123 m,
Re
2.995E6
1.8E 5
0.16x
0.16(1.123)
Re
(2.995E6)
Ans.
=+ − = +
−
=
=
≈
−
≈=
≈

0.0213 m
Compare with a straight allturbulentflow calculation from Eq. (7.1
b
):
x1
.
5
m
1.2(40)(1.5)
0.16(1.5)
Re
4.0E6, whence
(25% higher)
.
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 Spring '09
 Lobban

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