11-Practice Problems.pdf

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ssm_ch11.pdf
Chapter 11
Drag and Lift
Problem 11.1
Air with a speed of ! ! ß ows over a long bar that has a 15 ! wedge-shaped cross-
section. The pressure variation, as represented using the coe ! cient of pressure, is
shown in the following sketch. On the west face of the bar, the coe ! cient of pres-
sure is everywhere equal to +1. On the northeast face, the coe ! cient of pressure
varies linearly from -2 to 0, and on the south face the variation is linear from +1 to
0. Determine the coe ! cient of drag and the coe ! cient of lift.
Solution
As shown in Fig. 1, the pressure distributions cause resultant forces to act on each
face of the bar. Identify faces 1, 2, and 3 as the west, northeast, and south sides
of the bar, respectively.
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CHAPTER 11. DRAG AND LIFT
Fig. 1 Resultant force diagram
On the west face of the bar, the average coe ! cient of pressure is " "1 = 1#0 Thus
the force $ 1 is given by
$ 1 = " "1 % 1
&! !
2
= 1#0% 1
&! !
2
(1)
Similarly, the force on the northeast face is
$ 2 = " "2 % 2
&! !
2
&! !
2
= +1#0% 2
(2)
where " "2 is given as +1 because of the direction of the pressure (outward), as
shown in the sketch in the problem statement. Finally,
$ 3 = " "3 % 3
&! !
2
&! !
2
= 0#5% 3
(3)
By de Þ nition, lift ($ # ) and drag ($ $ ) force are perpendicular and parallel, respec-
tively, to the free stream.
Fig. 2 Lift and drag
Equating drag force (Fig. 2) with forces due to pressure (Fig. 1) gives
$ $ = $ 1 +$ 2 SIN15 !
(4)
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101
Substituting Eqs. (1) and (2) into Eq. (4) gives
$ $ = $ 1 +$ 2 SIN15 !
= (1#0% 1 +1#0% 2 SIN15 ! ) &! !
2
Letting % 1 '% 2 = SIN15 ! gives
$ $ = 2% 1
&! !
2
(5)
Since % 1 is the projected area, Eq. (5) becomes
" $ =
³
$ $
´
%& !
2
% 1
= 2
From Figs. 1 and 2
$ # = $ 2 COS(15 ! )+$ 3
(6)
Substitute Eqs. (2) and (3) into Eq. (6).
$ # = % 2
&! !
2
COS(15 ! )+0#5% 3
&! !
2
&! !
2
&! !
2
= % 3
+0#5% 3
&! !
2
= (1#0+0#5)% 3
(7)
or
$ #
% 3 %& 2
= 1#5
(8)
Use area % 3 to de Þ ne the coe ! cient of lift.
" # =
³
$ #
´
(9)
%& 2
% 3
Combine Eqs. (8) and (9)
" # = 1#54
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102
CHAPTER 11. DRAG AND LIFT
Problem 11.2
Air with a speed of 30 m/s and a density of 1.25 kg/m 3 ß ows normal to a rec-
tangular sign of dimension 5.5 m by 7.5 m. Find the force of the air on the sign.
Solution
The drag force is
$ $ = " $ % " &! !
2
³
´ ¡
30 2 m 2 ' s 2
¢
1#25 kg/m 3
¡
¢
= " $
5#5×7#5 m 2
2
= " $ (23#2 kN )
The coe ! cient of drag from Table 11.1 with (') = 7#5'5#5 ! 1#0 is 1.18. Thus
$ $ = 1#18(23#2 kN )
= 27#4 kN
Problem 11.3
A student is modeling the drag force on the Þ ns of a model rocket. The rocket has
three Þ ns, each fabricated from 16 in.-thick balsa-wood to a dimension of 2#5 × 1
in. The coe ! cient of drag for each Þ n is 1.4, and the Þ n is subjected to air with a
speed of 100 mph and a density of 0.00237 slug/ft 3 # Determine the total drag force
on the Þ ns. Since the Þ ns are not streamlined, assume that drag on a given Þ n is
based on the projected area (not the planform area).
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103
Solution
The drag force for three Þ ns is
$ $ = 3" $ % " &! !
2
The projected area (normal to ß uid velocity) of one Þ n is
Μ
1
16 ×1
Μ
1 ft 2
144 in. 2
% " =
in. 2
= 4#34×10 !4 ft 2
The velocity is
Μ
1#467 ft/s
1 mph
! ! = (100 mph )
= 147 ft/s
Thus the drag force is
$ $ = 3" $ % " &! !
2
³
´³
147 2 ft 2 /s 2
´
0#00237 slug/ft 3
¡
¢
4#34×10 !4 ft 2
= 3(1#4)
2
= 0#0467 lbf
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