Graebel W.P. Advanced Fluid Mechanics
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54 Inviscid Irrotational Flows
Streamlines
Potential lines
Figure 2.2.3 Source/sink streamlines and iso-potential lines
Line doublet (dipole)
Consider a source and sink pair of equal strengths, and let them approach one another
along a connecting line in such a manner that their strengths increase inversely as the
distance between them increases. The pair then has the combined potential
source
r +a +
sink
r −a
2
a
and since the sink is negative with respect to the source, in the limit as a goes to zero
this becomes a derivative. Therefore, define the potential of a line doublet as
doublet
=B·
source of strength 2
=
B
x
x −x
0
+B
y
y −y
0
r −r
0
2
(2.2.15)
where B gives the strength and direction of the doublet. Often a doublet is denoted by
a half-filled circle, the filled part representing the “source end” and the unfilled part the
“sink end,” to show the source-sink nature and the directionality. From its relation to
the source, we expect that there is no net discharge or vorticity at x
0
y
0
. This can
be verified by taking appropriate integrations around that point. To find what lines of
constant are like, rearrange equation (2.2.15) as
x −x
0
2
+y −y
0
2
=
B
x
x −x
0
+B
y
y −y
0
or equivalently
x −x
0
−
B
x
x −x
0
2
+
y −y
0
−
B
y
y −y
0
2
=
B
x
x −x
0
2
+
B
y
y −y
0
2
For constant values of this is the equation of a circle whose radius is B/2 centered
at x
0
+B
x
/2 y
0
+B
y
/2. Therefore, it is on a line through x
0
y
0
in the direction
of B. Constant and lines are shown in Figure 2.2.4. The constant lines are nested
circles centered at B
x
/2 B
y
/2 and with radius B
x
/2
2
+B
y
/2
2
1/2
. Constant
2.2 Irrotational Flows and the Velocity Potential 55
B
Streamlines
Potential Lines
Figure 2.2.4 Doublet streamlines and iso-potential lines
lines are a similar family of circles, but rotated 90 degrees with respect to the circles.
They are centered at B
y
/2 −B
x
/2 and have radius B
x
/2
2
+B
y
/2
2
1/2
.
The velocity field associated with the doublet is
v
x
=
x
=
B
x
y −y
0
2
−x −x
0
2
−2B
y
x −x
0
y −y
0
x −x
0
2
+y −y
0
2
v
y
=
y
=
B
y
x −x
0
2
−y −y
0
2
−2B
x
x −x
0
y −y
0
x −x
0
2
+y −y
0
2
(2.2.16)
Along a line in the direction of B, we see from equation (2.2.16) that the velocity
dies out as the square of the distance. Since
source
satisfies Laplace’s equation, and
since
2
source
=0 = 0, therefore
0 = B ·
2
source
=B ·
2
source
=
2
B·
source
=
2
doublet
(2.2.17)
The velocity potential for a line doublet therefore satisfies Laplace’s equation.
From equation (2.2.6) the stream function for a line doublet is
=
B
y
x −x
0
−B
x
y −y
0
r −r
0
2
(2.2.18)
Since we have also labeled the source and doublet as monopole and dipole, you
might wonder whether taking even higher derivatives would be useful. The derivative
of the dipole is the quadrapole, which is mainly of interest in acoustic problems. The
dipole is usually sufficient for fluid mechanics use.
Line vortex
The vortex is a “reverse analog” of the source, in that it has concentrated vorticity rather
than concentrated discharge and also because the constant lines are radial lines while
the lines of constant are concentric circles. Its velocity potential is
vortex
=
2
tan
−1
y −y
0
x −x
0
(2.2.19)
56 Inviscid Irrotational Flows
Streamlines Potential Lines
Figure 2.2.5 Vortex streamlines and iso-potential lines
with velocity components
v
x
=
x
=
−y −y
0
2
r −r
0
2
v
y
=
y
=
x −x
0
2
r −r
0
2
(2.2.20)
Thus, the velocity decreases inversely with the distance from x
0
y
0
. Constant (radial
lines) and (concentric circles) lines are shown in Figure 2.2.5.
Since
2
x
2
=
y −y
0
x −x
0
x −x
0
2
+y −y
0
2
2
and
2
y
2
=
y −y
0
x −x
0
x −x
0
2
+y −y
0
2
2
=−
2
x
2
Laplace’s equation is seen to be satisfied. The vortex is counterclockwise if is positive,
clockwise if is negative.
Since
vortex
is multivalued (in traversing a path around the point r
0
vortex
changes
by ), we anticipate that there is circulation associated with vortex flows. Checking this
by calculating the circulation arounda2by2square centered at x
0
y
0
, we have
Circulation =
y
0
+1
y
0
−1
v
y
x
0
−1 ydy +
x
0
−1
x
0
+1
v
x
x y
0
+1dx +
y
0
−1
y
0
+1
v
y
x
0
−1 ydy
+
x
0
+1
x
0
−1
v
x
x y
0
−1dx
=
2
y
0
+1
y
0
−1
dy
1+y −y
0
2
+
x
0
−1
x
0
+1
−dx
1+x −x
0
2
+
y
0
−1
y
0
+1
−dy
1+y −y
0
2
+
x
0
+1
x
0
−1
dx
1+x −x
0
2
=
a result that depends only on the fact that the path encircles x
0
y
0
, not on any other
details of the path such as shape or size. Any path not enclosing the vortex has zero
2.2 Irrotational Flows and the Velocity Potential 57
TABLE 2.2.1 Velocity potentials and stream functions for irrotational flows
Flow Element Two-dimensional
Uniform stream xU
x
+yU
y
yU
x
−xU
y
Source or sink
m
2
ln
x −x
0
2
+y −y
0
2
m
2
tan
−1
y −y
0
x −x
0
Doublet
B
x
x −x
0
+B
y
y −y
0
r −r
0
2
B
y
x −x
0
−B
x
y −y
0
r −r
0
2
Line vortex
2
tan
−1
y −y
0
x −x
0
−
2
tan
−1
y −y
0
x −x
0
Flow Element Three-dimensional
(axisymmetric)
Uniform stream xU
x
+yU
y
+zU
z
05U
z
r
2
Source or sink
−m
4
r −r
0
−mz −z
0
4
r
2
+z −z
0
2
Doublet
B · r −r
0
r −r
0
3
−B
z
r
2
r
2
+z −z
0
2
Line vortex None None
circulation, as can be verified by Stokes’s theorem. Therefore, a vortex has concentrated
vorticity but no concentrated mass discharge.
From equation (2.2.6) the stream function for a vortex is
vortex
=−
2
ln
r −r
0
=−
2
ln
x −x
0
2
+y −y
0
2
(2.2.21)
The velocity potentials and stream functions for all of these flows plus their three-
dimensional counterparts are summarized in Table 2.2.1.
2.2.3 Hele-Shaw Flows
The four solutions we have just seen—uniform stream, source/sink, doublet, and
vortex—are the fundamental solutions for two-dimensional potential flow. Before con-
sidering combinations of these to obtain physically interesting flows, it is useful to see
how these basic flows can be produced in a laboratory.
One way of producing them is by means of a Hele-Shaw table, which consists of
a flat horizontal floor with a trough at one end for introducing a uniform stream and
another trough at the other end for removing it. Holes in the bottom of the floor are
connected to bottles that may be raised or lowered. These provide the sources and sinks,
with the elevation of the bottle controlling their strength. Doublets can be produced
by putting a source and sink pair almost together, having the source height above the
table being equal to the sink height below. Sometimes a top transparent plate is added
to ensure that the flow is of the same depth everywhere. The vorticity component
perpendicular to the floor will be essentially zero; components in the plane of the floor
will be nonzero.
58 Inviscid Irrotational Flows
Vortices are more difficult to produce in Hele-Shaw flows than are sources or sinks.
One way of producing them is to use a vertical circular rod driven by an electric motor.
Hele-Shaw flows are slow flows, and because the velocity will vary roughly parabol-
ically with the coordinate in the direction normal to the floor inherent to these flows,
there is a great deal of vorticity. However, this vorticity is largely parallel to the bed,
the vorticity component normal to the floor being virtually zero. Thus, Hele-Shaw flows
viewed perpendicular to the floor are good models of two-dimensional irrotational flows.
Streamlines can be traced by inserting dye into the flow. A permanent record of
these streamlines can be made by photography. Alternatively, one method that has been
used is to cast the bed of the table out of a hard plaster, with inlets for the source and
sink permanently cast into the plaster. If the bed is painted with a white latex paint,
streamlines can be recorded by carefully placing potassium permanganate crystals on
the bed. A record of the streamlines remains as dark stains on the paint.
2.2.4 Basic Three-Dimensional Irrotational Flows
Except for the vortex, all our two-dimensional irrotational flows have three-dimensional
counterparts that qualitatively are much like their two-dimensional counterparts. (An
example of a three-dimensional vortex is a smoke ring. Mathematical representation of
such a three-dimensional phenomenon is more complicated than in the case of two-
dimensional flows.) Here, we simply list these counterparts. The analysis proceeds as
in the two-dimensional case.
Uniform stream
The velocity potential for a uniform stream is
uniform stream
=xU
x
+yU
y
+zU
z
=r ·U (2.2.22)
with a velocity field
v = = U
Surfaces of constant are planes perpendicular to U.
When U has only a component in the z direction, a Stokes stream function can be
found in the form
uniform stream
=05U
z
R
2
sin
2
=05U
z
r
2
(2.2.23)
Point source or sink (point monopole)
The velocity potential for a point source of strength m at r
0
is
source
=
−m
4
r −r
0
=
−m
4
x −x
0
2
+y −y
0
2
+z −z
0
2
(2.2.24)
Here, m is the volume discharge from the source, with continuity satisfied everywhere
except at r
0
.Ifm is positive, represents a source. If negative, it represents a sink.
Irrotationality is satisfied everywhere.
Surfaces of constant are concentric spheres centered at r
0
. The velocity is directed
along the radius of these spheres and dies out like the reciprocal of the distance squared
to satisfy continuity. The velocity is given by
v = =
mr −r
0
4
r −r
0
3
(2.2.25)
2.2 Irrotational Flows and the Velocity Potential 59
When the source lies on the z-axis, a Stokes stream function can be found in the form
source
=
−mz −z
0
4
r
2
+z −z
0
2
(2.2.26)
Point doublet (point dipole)
The velocity potential for a doublet can again be found by differentiating the potential
for a source, giving
doublet
=B·
source of strength 4
=
B·r −r
0
r −r
0
3
(2.2.27)
with velocity components
v =
B
r −r
0
3
−
3r −r
0
B·r −r
0
r −r
0
5
(2.2.28)
The constant surfaces are no longer simple geometries like circles and spheres.
When the doublet lies on the z-axis and additionally B points parallel to the z-axis,
a Stokes stream function for a doublet can be found in the form
doublet
=
−B
z
r
2
r
2
+z −z
0
2
3/2
(2.2.29)
These results are summarized in Table 2.2.1.
2.2.5 Superposition and the Method of Images
In a number of simple cases, the solution for flow past a given body shape can be
obtained by recognizing an analogy between potential flow and geometrical optics, since
the Laplace equation also governs the passage of light waves. Boundaries can be thought
of as mirrors, with images of the fundamental solutions appearing at appropriate points
to generate the “mirror” boundary.
Line source near a plane wall
Suppose we have a source of strength m a distance b from a plane wall. According
to the method of images, the plane wall can be regarded as a mirror. As the source
“looks” into the mirror, it sees an image source of the same strength at distance b
behind the mirror (see Figure 2.2.6). A mirror interchanges right and left. This does not
affect the source, but will affect signs of vortices and doublets. The potential for the
two-dimensional case is, with the x-axis acting as the wall,
=
m
2
ln
x
2
+y −b
2
+ln
x
2
+y +b
2
(2.2.30)
with velocity components
v
x
=
m
2
x
x
2
+y −b
2
+
x
x
2
+y +b
2
v
y
=
m
2
y −b
x
2
+y −b
2
+
y +b
x
2
+y +b
2
(2.2.31)
Note that v
y
vanishes on y = 0, satisfying the necessary boundary condition that the
velocity normal to the stationary wall be zero.
60 Inviscid Irrotational Flows
5
3
1
–1
–3
–5
–6 –4 –2 0 2 4 6
Source at (1, 0) near a wall at (x, 0)
Figure 2.2.6 Source and its image by a wall—streamlines
This solution can be used as a simple model of physical problems such as water
intakes or pollution sources near a straight coastline. Additional sources for more
complicated flows can be easily added.
Note that if the source is in a corner, both walls act as mirrors, and there will
be three image sources plus the original source. The third image comes about because
the mirrors extend to plus and minus infinity, so the two images that you might think
would be sufficient in turn have their own image. The preceding procedure can be easily
extended to vortices and doublets near a wall and to three-dimensional examples. It
can also be used for acoustic sources. Some loudspeaker enclosures are designed to be
placed in corners or near walls to increase their apparent power output by the image
speakers.
Example 2.2.1 Point source near walls
A point source of strength m is located at the point (1, 2, 3). There is a wall at x = 0
and a second wall at y =0. What is the velocity potential for this flow?
Solution. From equation (2.2.24), the given source has a velocity potential
0
=−
m
4
r −r
0
where r
0
=i +2j +3k
To generate the first wall, an image source of strength m at (−1, 2, 3) is needed. This
will have a velocity potential
1
=−
m
4
r−r
1
, where r
1
=−i+2j +3k. When the second
wall is added, both the original source and the image at (−1, 2, 3) will have images,
with a combined velocity potential
2
+
3
=−
m
4
r −r
2
−
m
4
r −r
3
where r
2
=i −2j +3k and r
3
=−i−2j +3k
Thus, our velocity potential is the sum of the potentials of four sources, each of strength
m, and located at the points (1, 2, 3), (−1, 2, 3), (1, −2, 3), and (−1 −2, 3). The total
potential is
2.2 Irrotational Flows and the Velocity Potential 61
total
=
0
+
1
+
2
+
3
=−
m
4
r −r
0
−
m
4
r −r
1
−
m
4
r −r
2
−
m
4
r −r
3
Notice that if there were a third wall at z =0, eight sources would be needed.
2.2.6 Vortices Near Walls
An interesting variation of the source near a wall is that of a vortex near a wall. The
image will have a circulation in the reverse direction of the original vortex because left
and right are interchanged in a reflection (see Figure 2.2.7). For a wall at x = 0 the
velocity potential for the original vortex plus its image is
=
2
tan
−1
y −b
x
−tan
−1
y +b
x
(2.2.32)
with velocity components
v
x
=
x
=
2
−y −b
x
2
+y −b
2
+
y +b
x
2
+y +b
2
v
y
=
y
=
2
x
x
2
+y −b
2
+
x
x
2
+y +b
2
(2.2.33)
The velocity at (0, b) due to the image vortex is /4b 0. This is called the induced
velocity. The vortex at (0, b) will tend to travel with the induced velocity, carrying its
image with it, since the induced velocity at the image will be the same value.
Wall
Figure 2.2.7 Vortex and its image by a wall—streamlines
62 Inviscid Irrotational Flows
The stream function to accompany equation (2.2.31) is
=
2
ln
x
2
+y −b
2
+ln
x
2
+y +b
2
=
4
ln
x
2
+y −b
2
x
2
+y +b
2
(2.2.34)
Therefore, lines of constant are given by
x
2
+y −b
2
=Cx
2
+y +b
2
(2.2.35)
where C is a constant related to the value of the stream function by = /4n C.
Expanding and rearranging, equation (4.3.33) becomes
x
2
+
y −
b1+C
1−C
2
=
4Cb
2
1−C
2
Thus, the streamlines are circles of radius r = 2b
C/1−C centered at (0, d), where
d = b
1+C
1−C
.
Since the streamlines in this example are circles, a physical realization of this flow
is a single vortex in a cup of radius r, the vortex being a distance b from the center of
the cup. The vortex will travel around the cup on a circular path of radius r at a speed
of /4b.
Still another realization of a vortex flow is a vortex pair near a wall, the vortices
being of equal but opposite circulation. Take them to be a distance a from the wall and
separated by a distance 2b (Figure 2.2.8). To generate the wall, it is necessary to add a
vortex image pair as well. The total velocity potential is then
=
2
tan
−1
y −a
x −b
−tan
−1
y −a
x +b
−tan
−1
y +a
x −b
+tan
−1
y +a
x +b
(2.2.36)
b
a
wall
Figure 2.2.8 Trajectories of a vortex pair and its images by a wall
2.2 Irrotational Flows and the Velocity Potential 63
with stream function
=
4
lnx −b
2
+y −a
2
−lnx +b
2
+y −a
2
−lnx −b
2
+y +a
2
+lnx +b
2
+y +a
2
(2.2.37)
and with velocity components
v
x
=
x
=
2
−y −a
x −b
2
+y −a
2
+
y −a
x +b
2
+y −a
2
+
y +a
x −b
2
+y +a
2
−
y +a
x +b
2
+y +a
2
v
y
=
y
=
2
x −b
x −b
2
+y −a
2
−
x +b
x +b
2
+y −a
2
−
x −b
x −b
2
+y +a
2
+
x +b
x +b
2
+y +a
2
(2.2.38)
The induced velocity at b a is
V
induced
=
4
1
a
−
a
a
2
+b
2
i +
−1
b
+
b
a
2
+b
2
j
Therefore, the equations of motion for this vortex are
db
dt
=
4
1
a
−
a
a
2
+b
2
da
dt
=
4
−1
b
+
b
a
2
+b
2
(2.2.39)
With appropriate sign changes, similar equations hold for the other three vortices.
The path the vortices travel can be found by rearranging (2.2.39) in the form
4
dt
a
2
+b
2
=
a
b
2
db =−
b
a
2
da
The variables a and b can be separated and the resulting equation integrated, giving as
the path the vortex travels
1
b
2
=
1
b
2
0
+
1
a
2
0
−
1
a
2
(2.2.40)
a
0
b
0
being the initial position of the vortex.
This model of a traveling vortex pair is useful in describing the spreading of the
vortex pair left by the wing tips of an aircraft on takeoff. In this case, the wall represents
the ground. It is also responsible for the ground effect—the increased lift an airplane
experiences on takeoff due to the proximity of the ground.
Example 2.2.2 A vortex near a plane wall
A wall is located at x =0. A vortex with circulation 20 meter
2
/second is placed 1 meter
above the wall. What is the velocity potential, and at what speed does the vortex move?
Solution. For convenience, take the instantaneous position of the vortex to be
(0, 1, 0). Then the velocity potential for the original vortex is
orig
= 10 tan
−1
y−1
x
.