fluid_mechanics

6.5 Some Basic, Plane Potential Flows 293
E XAMPLE 6.6
Potential Flow—Free Vortex
GIVEN A liquid drains from a large tank through a small
opening as illustrated in Fig. E6.6a. A vortex forms whose velocity
distribution away from the tank opening can be approximated
as that of a free vortex having a velocity potential
f
2p u
FIND Determine an expression relating the surface shape to
the strength of the vortex as specified by the circulation .
SOLUTION
Since the free vortex represents an irrotational flow field, the
Bernoulli equation
p 1
g V 2 1
2g z 1 p 2
g V 2 2
2g z 2
can be written between any two points. If the points are selected
at the free surface, p 1 p 2 0, so that
V 2 1
2g z s V 2 2
2g
where the free surface elevation, z s , is measured relative to a datum
passing through point 112 as shown in Fig. E6.6b.
The velocity is given by the equation
v u 1 0f
r 0u
2pr
We note that far from the origin at point 112, V 1 v u 0 so that
Eq. 1 becomes
z s
8p 2 r 2 g
which is the desired equation for the surface profile.
2
(1)
(Ans)
F I G U R E E6.6a
p = p atm
F I G U R E E6.6b
x
z
(2)
z s
y
COMMENT The negative sign indicates that the surface falls
as the origin is approached as shown in Fig. E6.6. This solution is
not valid very near the origin since the predicted velocity becomes
excessively large as the origin is approached.
r
(1)
A doublet is formed
by an appropriate
source–sink pair.
6.5.4 Doublet
The final, basic potential flow to be considered is one that is formed by combining a source and
sink in a special way. Consider the equal strength, source–sink pair of Fig. 6.22. The combined
stream function for the pair is
c m 2p 1u 1 u 2 2
y
P
r 2
r
r 1
Source
a
θ 2
θ 1
θ
a
Sink
x
F I G U R E 6.22 The combination of
a source and sink of equal strength located along
the x axis.