Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 7.31. A pair of equal strength ideal line vortices having axes perpendicular to the x–y
plane are located at
xa(t)=xa(t), ya(t)
( )
, and
xb(t)=xb(t), yb(t)
( )
, and move in their mutually-
induced velocity fields. The stream function for this flow is given by:
ψ
(x,y,t)=−Γ
2
π
ln x−xa(t)+ln x−xb(t)
( )
. Explicitly determine
xa(t)
and
xb(t)
given that
xa(0) =(−r
o,0)
and
xb(0) =(r
o,0)
. Switching to polar coordinates at some point in your solution
may be useful.
Solution 7.31. This problem can quickly get very tedious without clear organization. For
notational simplicity, let
(a)=x−xa(t)
( )
2+y−ya(t)
( )
2
, and
(b)=x−xb(t)
( )
2+y−yb(t)
( )
2
.
Differentiate the stream function to find:
, and
v(x,y,t)=−
∂ψ
∂
x= + Γ
2
π
x−xa
(a)+x−xb
(b)
‘
(
)
*
+
,
.
This velocity field leads to 4 nonlinear first-order differential equations for point vortex
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 7.32. Two unequal strength ideal line vortices having axes perpendicular to the x–y
plane are located at
x1(t)=x1(t), y1(t)
( )
with circulation Γ1, and
x2(t)=x2(t), y2(t)
( )
with
circulation Γ2, and move in their mutually induced velocity fields. The stream function for this
flow is given by:
ψ
(x,y,t)=− Γ12
π
( )
ln x−x1(t)− Γ22
π
( )
ln x−x2(t)
. Explicitly determine
x1(t)
and
x2(t)
in terms of Γ1, Γ2, h1, h2, and h, given
x1(0) =(h
1, 0)
,
x2(0) =(h2, 0)
, and h2 – h1
= h > 0 [Hint: choose a convenient origin of coordinates, and switch to polar coordinates after
finding the shape of the trajectories.]
Solution 7.32. This problem can quickly get very tedious without clear organization. For
notational simplicity, let
(1) =x−x1(t)
( )
2+y−y1(t)
( )
2
, and
(2) =x−x2(t)
( )
2+y−y2(t)
( )
2
.
Differentiate the stream function to find:
u=
∂ψ
∂
y=−1
2
π
Γ1
(y−y1)
(1) +Γ2
(y−y2)
(2)
#
$
%
&
‘
(
, and
v=−
∂ψ
∂
x= + 1
2
π
Γ1
(x−x1)
(1) +Γ2
(x−x2)
(2)
#
$
%
&
‘
(
.
This velocity field leads to 4 nonlinear first-order differential equations for point vortex
coordinate positions.
dx1
dt =u(x1,y1,t)=−Γ2
2
π
y1−y2
(x2−x1)2+(y2−y1)2
#
$
%
&
‘
(
,
dx2
dt =u(x2,y2,t)=−Γ1
2
π
y2−y1
(x2−x1)2+(y2−y1)2
#
$
%
&
‘
(
,
dy1
dt =v(x1,y1,t)=Γ2
2
π
x1−x2
(x2−x1)2+(y2−y1)2
#
$
%
&
‘
(
, and
dy2
dt =u(x2,y2,t)=Γ1
2
π
x2−x1
(x2−x1)2+(y2−y1)2
#
$
%
&
‘
(
.
From these, it is clear that:
2
π
Γ2
dx1
dt =−2
π
Γ1
dx2
dt =y2−y1
(x2−x1)2+(y2−y1)2
, and
−2
π
Γ2
dy1
dt =2
π
Γ1
dy2
dt =x2−x1
(x2−x1)2+(y2−y1)2
,
so:
d
dt
x1
Γ2
+x2
Γ1
“
#
$%
&
‘=0
, and
d
dt
y1
Γ2
+y2
Γ1
“
#
$%
&
‘=0
.
Thus, the contents of the parentheses are constants. The initial conditions then imply:
Γ2
dt
Γ2
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
where the third equality follows from ($). Integrating the second and final members of this
extended equality and using the initial conditions produces:
x1
2+y1
2=h1
2
. Similar steps for the
second vortex lead to:
x2
2+y2
2=h2
2
. Hence the vortices move in circles with radii h1 and h2. In
addition, the distance between them remains constant:
(x2−x1)2+(y2−y1)2=x2
21+Γ2
Γ1
#
$
%&
‘
(
2
+y2
21+Γ2
Γ1
#
$
%&
‘
(
2
=h2
21+Γ2
Γ1
#
$
%&
‘
(
2
=h2
,
where the first equality comes from substituting for x1 and y1 from ($), and the final equality
comes from the second equation of (%). Based on this, let x1(t) = h1cos
θ
(t), y1(t) = h1sin
θ
(t), x2(t)
= h2cos
θ
(t), and y2(t) = h2sin
θ
(t). With these substitutions, the distance between the vortices
remains constant (= h):
(x2−x1)2+(y2−y1)2=(h2−h1)2(cos2
θ
+sin2
θ
)=h2
,
so the only remaining task is to find
θ
(t).
Evaluate the differential equation for x1(t) in terms of
θ
(t):
dx1
dt =−h1sin
θ
d
θ
dt =−Γ2
2
π
y1−y2
(x2−x1)2+(y2−y1)2
#
$
%
&
‘
(=Γ2
2
π
hsin
θ
h2
.
Use the second and final terms to reach:
d
θ
dt =−Γ2
2
π
hh
1
=Γ2+Γ1
2
π
h2
, or
θ
(t)=Γ2+Γ1
2
π
h2
!
“
#$
%
&t
,
where the initial conditions have been used to find that the constant of integration is zero. With
this relationship, reconstruction of the whole solution is possible:
x1(t)=−hΓ2
Γ1+Γ2
cos Γ2+Γ1
2
π
h2
#
$
%&
‘
(t
)
*
+,
–
.
,
y1(t)=−hΓ2
Γ1+Γ2
sin Γ2+Γ1
2
π
h2
#
$
%&
‘
(t
)
*
+,
–
.
,
x2(t)=hΓ1
Γ1+Γ2
cos Γ2+Γ1
2
π
h2
“
#
$%
&
‘t
(
)
*+
,
–
, and
y2(t)=hΓ1
Γ1+Γ2
sin Γ2+Γ1
2
π
h2
“
#
$%
&
‘t
(
)
*+
,
–
.
Hence, the pair of vortices circles the origin with a constant angular velocity. When the two
vortices are of equal strength, this solution reverts to that for Exercise 7.31. In the limit as
Γ1→ −Γ2
(i.e. the two vortices are of equal and opposite sign),
x2−x1→const.=h
, and the two
y-coordinates become equal: y1, y2
→ − Γ2t
2
π
h
, and this matches the discussion of interacting
vortices in Section 5.6.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 7.33. Consider the unsteady potential flow of two ideal sinks located at
xa(t)=xa(t),0
( )
and
xb(t)=xb(t),0
( )
that are free to move along the x-axis in an ideal fluid that
is stationary far from the origin. Assume that each sink will move in the velocity field induced by
the other.
φ
(x,y,t)=−qs
2
π
ln x−xa(t)
( )
2+y2+ln x−xb(t)
( )
2+y2
“
#
$%
&
‘
, with qs > 0.
a) Determine
xa(t)
and
xb(t)
when
xa(0) =−L,0
( )
and
xb(0) = +L,0
( )
b) If the pressure far from the origin is p∞ and the fluid density is
ρ
, determine the pressure p at x
= y = 0 as function of p∞,
ρ
, qs, and xa(t).
Solution 7.33. a) All the action takes place in the x-direction so ignore the vertical (v) component
of velocity.
u(x,y,t)=
∂φ
∂
x
!
“
#$
%
&=−qs
2
π
x−xa(t)
x−xa(t)
( )
2+y2+x−xb(t)
x−xb(t)
( )
2+y2
!
“
#
#
$
%
&
&
For
x=xa(t)
and y = 0:
u(xa, 0, t)=−qs
xa−xb
2+y2
“
$
$
%
‘
‘
=−qs
1
“
$%
‘=dxa
, and
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 7.34. Consider the unsteady potential flow of an ideal source and sink located at
x1(t)=x1(t),0
( )
and
x2(t)=x2(t),0
( )
that are free to move along the x-axis in an ideal fluid that
is stationary far from the origin. Assume that the source and sink will move in the velocity field
induced by the other.
φ
(x,y,t)=qs
2
π
ln x−x1(t)
( )
2+y2−ln x−x2(t)
( )
2+y2
“
#
$%
&
‘
, with qs > 0.
a) Determine
x1(t)
and
x2(t)
when
x1(0) =−, 0
( )
and
x2(0) = +, 0
( )
.
b) If the pressure far from the origin is p∞ and the fluid density is
ρ
, determine the pressure p at x
= y = 0 as function of p∞,
ρ
, qs, and x1(t).
Solution 7.34. a) All the action takes place in the x-direction so ignore the vertical (v) component
of velocity.
u(x,y,t)=
∂φ
∂
x
!
“
#$
%
&=qs
2
π
x−x1(t)
x−x1(t)
( )
2+y2−x−x2(t)
x−x2(t)
( )
2+y2
!
“
#
#
$
%
&
&
For
x=x1(t)
and y = 0:
u(x1, 0, t)=qs
2
π
−x1−x2
x1−x2
( )
2+y2
“
#
$
$
%
&
‘
‘
y=0
=−qs
2
π
1
x1−x2
“
#
$%
&
‘=dx1
dt
, and
for
x=x2(t)
and y = 0:
u(x2, 0, t)=qs
x2−x1
2+y2
“
$
$
%
‘
‘
=qs
1
“
$%
‘=dx2
.
=qs
2
4
π
2
1
qs4
π
( )
2t2−2
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 7.35. Consider a free ideal line vortex oriented parallel to the z-axis in a 90° corner
defined by the solid walls
θ
= 0 and
θ
= 90°. If the vortex passes the through the plane of the
flow at (x, y), show that the vortex path is given by: x–2 + y–2 = constant. [Hint: Three image
vortices are needed at points (−x, −y), (−x, y) and (x, −y). Carefully choose the directions of
rotation of these image vortices, show that dy/dx = v/u = −y3/x3, and integrate to produce the
desired result.]
Solution 7.35. The induced velocity at point (x, y) will include
contributions from the three image vortices. If the first vortex at
(x, y) has strength +Γ, then the second vortex at (−x, y) and the
third vortex at (x, −y) will have strength –Γ. The fourth vortex at
(−x, −y) will have strength +Γ. Therefore:
12
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 7.36. In ideal flow, streamlines are defined by d
ψ
= 0, and potential lines are defined
by d
φ
= 0. Starting from these relationships, show that streamlines and potential lines are
perpendicular.
a) in plane flow where x and y are the independent spatial coordinates, and
b) in axisymmetric flow where R and z are the independent spatial coordinates.
[Hint: For any two independent coordinates x1 and x2, the unit tangent to the curve x2 = f(x1) is
t=e1+(df dx1)e2
( )
1+(df dx1)2
; thus, for a) and b) it is sufficient to show
t
( )
ψ
=const ⋅t
( )
φ
=const =0
]
Solution 7.36. a) In two-dimensional ideal flow,
d
ψ
=
∂ψ
∂
xdx +
∂ψ
∂
ydy =−vdx +udy.
Thus a curve
defined by d
ψ
= 0, has a slope
dy dx =v u
. Using the hint, the tangent vector to this curve is:
t
( )
ψ
=const =ex+(v u)ey
( )
1+(v u)2=uex+vey
( )
u2+v2
.
Similarly,
d
φ
=
∂φ
∂
xdx +
∂φ
∂
ydy =udx +vdy
, so a curve defined by d
φ
= 0, has a slope
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 7.37. Consider a three-dimensional point source of strength Q (m3/s). Use a spherical
control volume and the principle of conservation of mass to argue that the velocity components
in spherical coordinates are u
θ
= 0 and ur = Q/4πr2 and that the velocity potential and stream
function must be of the form
φ
=
φ
(r) and
ψ
=
ψ
(
θ
). Integrate the velocity, to show that
φ
=
−Q/4
π
r and
ψ
=
−
Qcos
θ
/4
π
.
Solution 7.37. For a point source of strength Q (m3/s), the tangential velocity is zero because of
symmetry. The radial velocity times the area 4πr2 equals Q so ur = Q/ 4πr2. Therefore,
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 7.38. Solve the Poisson equation
∇2
φ
=Q
δ
(x−&
x )
in a uniform unbounded three-
dimensional domain to obtain the velocity potential
φ
= –Q/4π|x – x´| for an ideal point source
located at x´.
Solution 7.38. This exercise is similar to Exercise 5.9. First apply a simple shift transformation
that places x´ at the origin of coordinates. Define these new coordinates by:
X=x−#
x
,
Y=y−#
y
,
Z=z−#
z
, and set
r=x−#
x =X2+Y2+Z2
. The gradient operator
∇XYZ
in the
shifted coordinates X = (X, Y, Z) is the same as
∇
in the unshifted coordinates (x, y, z), so the
field equation for
φ
becomes:
∇XYZ
2
φ
=Q
δ
(X)=Q
δ
(x−&
x )
.
Integrate this equation inside a sphere of radius r:
∇XYZ
2
φ
dV
sphere
∫∫∫ =∇XYZ
φ
⋅ndA
spherical surface
∫∫ =Q
δ
(X)
x=−r2−y2−z2
x=−+r2−y2−z2
∫
y=−r2−z2
y= + r2−z2
∫
z=−r
z= +r
∫dXdYdZ
,
where the first equality follows from Gauss’ divergence theorem, and the triple integral on the
right side includes the location X = 0 (aka x = x´) so a contribution is collected from the three-
dimensional delta function. Thus, the right side of this equation is Q (times unity).
The dot product in the middle portion of the above equation simplifies to ∂
φ
/∂r because n
= er on the spherical surface and
er⋅ ∇XYZ =
∂ ∂
r
. Plus, in an unbounded uniform environment,
there are no preferred directions so
φ
=
φ
(r) alone (no angular dependence). Thus, the integrated
field equation simplifies to
∂φ
∂
r
$
%
& ‘
(
)
r2sin
θ
d
θ
d
ϕ
=0
2
π
∫
θ
=0
π
∫
ϕ
=r2
∂φ
∂
r
sin
θ
d
θ
d
ϕ
=0
2
π
∫
θ
=0
π
∫
ϕ
=4
π
r2
∂φ
∂
r=Q
,
which implies
∂φ
∂
r=Q
4
π
r2
or
φ
=−Q
4
π
r=−Q
4
π
x−%
x
.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 7.39. Using
(R,
ϕ
,z)
-cylindrical coordinates, consider steady three-dimensional
potential flow for a point source of strength Q at the origin in a free stream flowing along the z–
axis at speed U:
φ
(R,
ϕ
,z)=Uz −Q
4
π
R2+z2
.
a) Sketch the streamlines for this flow in any R–z half-plane.
b) Find the coordinates of the stagnation point that occurs in this flow.
c) Determine the pressure gradient,
∇p
, at the stagnation point found in part b).
d) If R = a(z) defines the stream surface that encloses the fluid that emerges from the source,
determine a(z) for
z→+∞
.
e) Use Stokes’ stream function to determine an equation for a(z) that is valid for any value of z.
f) Use the control-volume momentum equation,
ρ
u u ⋅n
( )
dS =
S
∫−pndS +F
S
∫
where n is the
outward normal from the control volume, to determine the force F applied to the point source to
hold it stationary.
g) If the fluid expelled from the source is replaced by a solid body having the same shape, what
is the drag on the front of this body?
Solution 7.39. a) A simple sketch
appears to the right.
b) From symmetry, the stagnation point
will lie on R = 0, thus only the axial
velocity needs to be considered to find
R
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 7.40. In (R,
ϕ
, z) cylindrical coordinates, the three-dimensional potential for a point
source at (0,0,s) is given by:
φ
=−Q4
π
( )
R2+(z−s)2
[ ]
−1 2
.
a) By combining a source of strength +Q at (0,0,–b), a sink of strength –Q at (0,0,+b), and a
uniform stream with velocity Uez, derive the potential (7.85) for flow around a sphere of radius a
by taking the limit as Q → ∞, and b → 0, such that
d=−2bQez
= –2πa3Uez = constant. Put your
final answer in spherical coordinates in terms of U, r,
θ
, and a.
b) Repeat part a) for the Stokes stream function starting from
ψ
=−Q4
π
( )
(z−s)R2+(z−s)2
[ ]
−1 2
.
Solution 7.40. In cylindrical coordinates, the three-dimensional potential for a uniform stream of
U=(0,0,U)
, a point source of strength +Q at
x=(0,0,−b)
, and a point-sink of strength –Q at
x=(0,0,+b)
is:
φ
=Uz −Q
4πR2+(z+b)2+Q
4πR2+(z−b)
.
Expand the square roots for b → 0, letting
r=R2+z2
d=−2bQez
U=(0,0,U)
x=(0,0,+b)
r=R2+z2
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 7.41. a) Determine the locus of points in uniform ideal flow past a circular cylinder of
radius a without circulation where the velocity perturbation produced by the presence of the
cylinder is 1% of the free stream value.
b) Repeat for uniform ideal flow past a sphere.
c) Explain the physical reason(s) for the differences between the answers for a) and b).
Solution 7.41. a) From (7.34), the velocity components for ideal flow about a cylinder with
radius a are:
ur=U1−a2
r2
#
$
%
&
‘
(
cos
θ
, and
u
θ
=−U1+a2
r2
$
%
&
‘
(
)
sin
θ
.
Clearly, the velocity magnitude when the cylinder is absent = U. Therefore, the magnitude of the
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 7.42. Using the figure for Exercise 7.29
with
A3→0
and
r→ ∞
, expand the three-
dimensional potential for a stationary arbitrary-
shape closed body in inverse powers of the
distance r and prove that ideal flow theory predicts
zero drag on the body.
Solution 7.42. The figure is reproduced here and a
portion of this solution follows that given for
Exercise 7.29. However, this time the geometry is
three-dimensional and the figure must be
interpreted as a slice through three dimensional control volume so that A1 is a spherical surface.
In ideal flow, the only surface forces are pressure forces. Therefore, the hydrodynamic
force F on the body is defined by:
F=−p−p∞
( )
n2dA
A2
∫=p−p∞
( )
ndA
A2
∫
, (a)
where the free stream velocity is U = Uex, and n is the outward normal on the total control
volume; thus it points into the body on surface A2 so the usual minus sign is missing from the
final equality in (a) because n = –n2. Note that the constant p∞ can be included in the pressure
integration because
p∞ndA
closed surface
∫=0
.
The starting point for the derivation is the integral form of the steady ideal flow
momentum equation applied to the entire clam-shell control volume:
ρ
u(u⋅n)dA
A1+A2+A3
∫=−p−p∞
( )
ndA
A1+A2+A3
∫
with the implied limit that the tube denoted by A3 goes to zero. When this limit is taken, the net
contribution of A3 is zero because the surface area of A3 goes to zero. This leaves:
ρ
u(u⋅n)dA
A1
∫+
ρ
u(u⋅n)dA
A2
∫=
ρ
u(u⋅n)dA
A1
∫+0=−p−p∞
( )
ndA
A1+A2
∫
,
where the first equality follows because
u⋅n
= 0 on the solid surface A2. Now substitute in (a)
for the pressure integration over A2, and rearrange:
ρ
u(u⋅n)dA
A1
∫=−p−p∞
( )
ndA
A1
∫−F
or
F=−p−p∞
( )
ndA
A1
∫−
ρ
u(u⋅n)dA
A1
∫
.
Thus, the force on the body can be obtained from integrals over a spherical surface (A1) that is
distant from the body.
The pressure can be written in term of the velocity using the Bernoulli equation:
p∞+1
2
ρ
U2=p+1
2
ρ
u2=p+1
2
ρ
u⋅u
( )
,
and this allows the pressure integral to be written:
−p−p∞
( )
ndA
A1
∫=−1
2
ρ
U2−u⋅u
( )
ndA
A1
∫=1
2
ρ
u⋅u
( )
ndA
A1
∫
,
where the final equality follows because U2 is a constant so
U2ndA
closed surface
∫=0
. So, the force on the
body can be obtained entirely from considerations of the velocity field:
F=
ρ
1
2u⋅u
( )
n−u(u⋅n)
( )
dA
A1
∫
.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 7.43. Consider steady ideal flow over a hemisphere of constant radius a lying on the y–z
plane. For the spherical coordinate system shown, the potential for this flow is:
φ
(r,
θ
,
ϕ
)=Ur(1+a32r3)cos
θ
where U is the flow velocity far from the hemisphere. Assume
gravity acts downward along the x-axis. Ignore fluid viscosity in this problem.
a) Determine all three components of the fluid velocity on the surface of the hemisphere, r = a, in
spherical polar coordinates:
(ur,u
θ
,u
ϕ
)=∇
φ
=
∂φ
∂
r,1
r
∂φ
∂θ
,1
rsin
θ
∂φ
∂ϕ
‘
(
)
*
+
,
.
b) Determine the pressure, p, on r = a.
c) Determine the hydrodynamic force, Rx, on
the hemisphere assuming stagnation pressure
is felt everywhere underneath the hemisphere.
[Hints:
er⋅ex=sin
θ
cos
ϕ
,
sin2
θ
d
θ
=
π
2
0
π
∫
,
and
sin4
θ
d
θ
=3
π
8
0
π
∫
].
d) For the conditions of part c) what density
ρ
h
must the hemisphere have to remain on the
surface.
Solution 7.43. a) Differentiate
φ
(r,
θ
,
ϕ
)=Ur 1+a32r3
( )
cos
θ
and evaluate on r = a.
ur=
∂φ
∂
r=U1−a3
r3
%
&
‘
(
)
*
cos
θ
–>
ur(r=a)=U1−a3
a3
#
$
%
&
‘
(
cos
θ
=0
u
θ
=1
r
∂φ
∂θ
=−U1+a3
2r3
&
(
)
+
sin
θ
–>
u
θ
(r=a)=−3U
2sin
θ
θ
!
ϕ
!
r!
z!
x!
y!