Geology Reference
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Therefore, substituting these expressions into
(
13.82
) and taking the arctangent gives, for small
values of the parameter
a
:
2
tan
1
2ar sin ™
Q
Qar sin ™
.r
2
§
D
Š
r
2
a
2
a
2
/
(13.84)
Finally, taking the limit as
a
!
0and
Q
!1
,
we obtain:
Fig. 13.9
Method for calculating the stream function of
a doublet flow at a point
P
by superposition of elemental
source and sink stream functions
K sin
™
r
§
D
(13.85)
where the parameter
K
is the
doublet strength
.
As mentioned above, the most important feature
of the basic flows in fluid dynamics is that more
complex and realistic flows can be constructed
by superposition of the elemental stream func-
tions associated with the basic flows. A more
in-depth introduction to this topic can be found
in Batchelor (
2000
). Here we mention the so-
called
Rankine half
-
body
, which results from
superposition of a uniform flow of velocity
v
0
and
a source of strength
Q
. Assuming that the source
S
is placed at the origin of the reference frame,
the stream function of this flow can be obtained
summing the functions (
13.75
)and(
13.78
):
case § does not depend from ™ by (
13.72
). In the
simplest case, the tangential component of
v
is
given by:
K
r
u
™
D
(13.79)
where
K
is a constant whose sign discriminates
between clockwise and counterclockwise vor-
tices.
Solving (
13.72
) from this velocity field gives:
§
D
K ln r
(13.80)
Finally, let us consider a doublet flow. As
illustrated in Fig.
13.9
, the stream function results
from the superposition of the individual stream
functions associated with the source and with the
sink:
Q
2
™
§
D
v
0
r sin ™
C
(13.86)
The corresponding flow is illustrated in
Fig.
13.6
and has a parabolic pattern. This model
originally was introduced to explain the shape
of the broad topographic swell associated with
the Hawaiian hotspot (Sleep
1987
,
1990
), but
has been successfully employed to represent in
general the interaction of mantle plumes with the
asthenospheric flow (e.g., Ribe and Christensen
1994
;Walkeretal.
2005
). By (
13.72
) we obtain
for the velocity components:
Q
2
.™
2
™
1
/
§
D
(13.81)
Taking the tangent of (
13.81
)gives:
tan
2 §
Q
tan ™
2
tan ™
1
1
C
tan ™
1
tan ™
2
(13.82)
D
tan.™
2
™
1
/
D
From elementary geometry considerations we
have that:
Q
2 r
I
u
™
D
v
0
sin ™
(13.87)
u
r
D
v
0
cos™
C
r sin ™
r cos™
a
I
tan ™
2
D
r sin ™
r cos™
C
a
(13.83)
tan ™
1
D
For (
r
,™)
D
(
Q
/2
v
0
, ) both components of
the velocity vanish and we have a stagnation
