fluid_mechanics

3.3 F ma Normal to a Streamline 101
Weight and/or pressure
can produce
curved streamlines.
V3.6 Free vortex
The physical interpretation of Eq. 3.10 is that a change in the direction of flow of a fluid
particle 1i.e., a curved path, r 6 2 is accomplished by the appropriate combination of pressure
gradient and particle weight normal to the streamline. A larger speed or density or a smaller radius
of curvature of the motion requires a larger force unbalance to produce the motion. For example,
if gravity is neglected 1as is commonly done for gas flows2 or if the flow is in a horizontal 1dzdn 02
plane, Eq. 3.10 becomes
0p
2
0n rV r
(3.10b)
This indicates that the pressure increases with distance away from the center of curvature
10p0n
is negative since rV 2 r is positive—the positive n direction points toward the “inside”
of the curved streamline2. Thus, the pressure outside a tornado 1typical atmospheric pressure2
is larger than it is near the center of the tornado 1where an often dangerously low
partial vacuum may occur2. This pressure difference is needed to balance the centrifugal
acceleration associated with the curved streamlines of the fluid motion. (See Fig. E6.6a in
Section 6.5.3.)
E XAMPLE 3.3
Pressure Variation Normal to a Streamline
GIVEN Shown in Figs. E3.3a,b are two flow fields with circular
streamlines. The velocity distributions are
V1r2 1V 0 /r 0 2r
for case (a)
y
V = (V 0 /r 0 )r
y
V = (V 0 r 0 )/r
and
V1r2 1V 0 r 0 2
r
where V 0 is the velocity at r r 0 .
for case (b)
r = n
(a)
x
(b)
x
FIND Determine the pressure distributions, p p(r), for each,
given that p p 0 at r r 0 .
6
4
(a)
SOLUTION
We assume the flows are steady, inviscid, and incompressible
with streamlines in the horizontal plane (dz/dn 0). Because the
streamlines are circles, the coordinate n points in a direction opposite
that of the radial coordinate, ∂/∂n ∂/∂r, and the radius
of curvature is given by r r. Hence, Eq. 3.9 becomes
For case (a) this gives
whereas for case (b) it gives
0p
0r V 2
r
0p
0r 1V 0 /r 0 2 2 r
0p
0r 1V 0 r 0 2 2
r 3
For either case the pressure increases as r increases since ∂p/∂r 0.
Integration of these equations with respect to r, starting with a
known pressure p p 0 at r r 0 ,gives
p p 0 1V 2 02231r/r 0 2 2 14
(Ans)
2
p – p 0
0
ρV 2 0 /2
2
4
6
0 0.5 1 1.5 2 2.5
r/r 0
(c)
F I G U R E E3.3
for case (a) and
p p 0 1rV 2 02231 1r 0/r2 2 4
(Ans)
for case (b). These pressure distributions are shown in Fig. E3.3c.
COMMENT The pressure distributions needed to balance the
centrifugal accelerations in cases (a) and (b) are not the same because
the velocity distributions are different. In fact, for case (a) the
(b)