fluid_mechanics

9.1 General External Flow Characteristics 465
E XAMPLE 9.1
Drag from Pressure and Shear Stress Distributions
GIVEN Air at standard conditions flows past a flat plate as
is indicated in Fig. E9.1. In case 1a2 the plate is parallel to the
upstream flow, and in case 1b2 it is perpendicular to the upstream
flow. The pressure and shear stress distributions on
the surface are as indicated 1obtained either by experiment or
theory2.
FIND
Determine the lift and drag on the plate.
U = 25 ft/s
p = 0 (gage)
y
p = p(x) = 0
(a)
F I G U R E E9.1
b = width = 10 ft
4 ft
= (x) = (1.24 × 10 –3 )/ x lb/ft 2
where x is in feet
τ w τ w
U
x
y
p = 0.744 1 – __ y
( 2
lb/ft 2
4 )
where y is in feet
U = 25 ft/s
p = 0
(b)
τ w
p = –0.893 lb/ft 2
τ wτw (y) =
– (–y)
x
SOLUTION
For either orientation of the plate, the lift and drag are obtained
from Eqs. 9.1 and 9.2. With the plate parallel to the upstream flow
we have u 90° on the top surface and u 270° on the bottom
surface so that the lift and drag are given by
and
where we have used the fact that because of symmetry the shear
stress distribution is the same on the top and the bottom surfaces,
as is the pressure also [whether we use gage 1 p 02 or absolute
1 p p atm 2 pressure]. There is no lift generated—the plate does
not know up from down. With the given shear stress distribution,
Eq. 1 gives
or
(1)
(Ans)
With the plate perpendicular to the upstream flow, we have
u 0° on the front and u 180° on the back. Thus, from Eqs.
9.1 and 9.2
and
l p dA
top
p dA 0
bottom
d t w dA
top
t w dA 2
bottom
t w dA
top
d 2
4 ft
x0
1.24 103
a lbft 2 b 110 ft2 dx
x 1 2
d 0.0992 lb
l t w dA
front
t w dA 0
back
d p dA
front
p dA
back
symmetrical about the center of the plate. With the given relatively
large pressure on the front of the plate 1the center of the
plate is a stagnation point2 and the negative pressure 1less than
the upstream pressure2 on the back of the plate, we obtain the
following drag
or
d
2 ft
y2
c 0.744 a1 y 2
4 b lb ft 2
10.8932 lbft 2 d 110 ft2 dy
d 55.6 lb
(Ans)
COMMENTS Clearly there are two mechanisms responsible
for the drag. On the ultimately streamlined body 1a zero thickness
flat plate parallel to the flow2 the drag is entirely due to the shear
stress at the surface and, in this example, is relatively small. For
the ultimately blunted body 1a flat plate normal to the upstream
flow2 the drag is entirely due to the pressure difference between
the front and back portions of the object and, in this example, is
relatively large.
If the flat plate were oriented at an arbitrary angle relative to
the upstream flow as indicated in Fig. E9.1c, there would be both
a lift and a drag, each of which would be dependent on both the
shear stress and the pressure. Both the pressure and shear stress
distributions would be different for the top and bottom surfaces.
High p
Low p
τ w
≠ 0
τ w
≠ 0
Again there is no lift because the pressure forces act parallel to the
upstream flow 1in the direction of d not l2 and the shear stress is
(c)
F I G U R E E9.1
(Continued)