fluid_mechanics

304 Chapter 6 ■ Differential Analysis of Fluid Flow
Γ
Γ = 0 < 1
4 π Ua
(a)
(b)
Stagnation
point
Γ
4 π Ua
(c)
= 1
Γ
4 Ua π
(d)
> 1
F I G U R E 6.29
The location of stagnation
points on a circular cylinder:
(a) without circulation; (b, c, d)
with circulation.
Potential flow past
a cylinder with circulation
gives zero
drag but non-zero
lift.
A variety of streamline patterns can be developed, depending on the vortex strength, . For
example, from Eq. 6.121 we can determine the location of stagnation points on the surface of the
cylinder. These points will occur at u u stag where v u 0 and therefore from Eq. 6.121
(6.122)
If 0, then u stag 0 or p—that is, the stagnation points occur at the front and rear of the cylinder
as are shown in Fig. 6.29a. However, for 1 4pUa 1, the stagnation points will occur at
some other location on the surface as illustrated in Figs. 6.29b,c. If the absolute value of the
parameter 4pUa exceeds 1, Eq. 6.122 cannot be satisfied, and the stagnation point is located
away from the cylinder as shown in Fig. 6.29d.
The force per unit length developed on the cylinder can again be obtained by integrating the
differential pressure forces around the circumference as in Eqs. 6.117 and 6.118. For the cylinder
with circulation, the surface pressure, p s , is obtained from the Bernoulli equation 1with the surface
velocity given by Eq. 6.1212
or
(6.123)
Equation 6.123 substituted into Eq. 6.117 for the drag, and integrated, again yields 1Problem 6.742
That is, even for the rotating cylinder no force in the direction of the uniform flow is developed.
However, use of Eq. 6.123 with the equation for the lift, 1Eq. 6.1182, yields 1Problem 6.742
(6.124)
Thus, for the cylinder with circulation, lift is developed equal to the product of the fluid density,
the upstream velocity, and the circulation. The negative sign means that if U is positive 1in the
positive x direction2 and is positive 1a free vortex with counterclockwise rotation2, the direction
of the F y is downward.
Of course, if the cylinder is rotated in the clockwise direction (0) the direction of F y would
be upward. This can be seen by studying the surface pressure distribution (Eq. 6.123), which is plotted
in Fig. 6.30 for two situations. One has 4pUa 0, which corresponds to no rotation of the cylinder.
The other has 4pUa 0.25, which corresponds to clockwise rotation of the cylinder. With no
sin u stag
p 0 1 2 rU 2 p s 1 r a2U sin u
2
2 2pa b
p s p 0 1 2 rU 2 a1 4 sin 2 u
F x 0
4pUa
F y rU
2 sin u
paU 2
4p 2 a 2 U 2b
F y