Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 5.13. A vortex ring of radius a and strength Γ lies in the x–y plane as shown in the
figure.
a) Use the Biot-Savart law (5.13) to reach the following formula for the induced velocity along
the x-axis:
u(x)=−Γez
4
π
(xcos
ϕ
−a)ad
ϕ
x2−2ax cos
ϕ
+a2
#
$%
&
3 2
ϕ
=0
2
π
∫
.
b) What is u(0), the induced velocity at the origin of coordinates?
c) What is u(x) to leading order in a/x when x >> a?
Solution 5.13. a) Start with:
u(x)=Γ
4
π
e
ϕ
0
2
π
∫×(x−%
x)
x−%
x3ad
ϕ
,
where e
ϕ
= – exsin
ϕ
+ eycos
ϕ
, x = xex, x´ = aeR = a(excos
ϕ
+ eysin
ϕ
),
e
ϕ
×ex=−ezcos
ϕ
,
z!
ϕ”
x!
Γ!y!
a!
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 5.14. An ideal line vortex parallel to the z-axis of strength Γ intersects the x–y plane at x
= 0 and y = h. Two solid walls are located at y = 0 and y = H > 0. Use the method of images for
the following.
a) Based on symmetry arguments, determine the horizontal velocity u of the vortex when h =
H/2.
b) Show that for 0 < h < H the horizontal velocity of the vortex is:
u(0,h)=Γ
4
π
h1−21
(nH /h)2−1
n=1
∞
∑
‘
(
)
*
+
,
,
and evaluate the sum when h = H/2 to verify your answer to part a).
Solution 5.14. There are two flat solid walls at y = 0
and y = H. The combined effect of each requires many
image vortices.
a) When the vortex is located at x = (0, H/2), the
induced velocities from the two closest image vortices
Now consider the first image vortex above the upper wall. It must be located at y = 2H – h and it
must have a strength of –Γ to mimic the effect of the wall at y = H. Based on the drawing, this
induced velocity will be in the negative x-direction, so
u(0,h)=Γ
2
π
1
2h−1
2H−2h+…
%
&
‘
(
)
*
However, the first upper image vortex spoils the lower wall boundary condition, so another
image vortex must be added at y = –2H + h and it must have a strength of +Γ to mimic the effect
of the wall at y = 0. Based on the drawing, this induced velocity will be in the negative x–
direction, so
u(0,h)=Γ
2
π
1
2h−1
2H−2h−1
2H+…
%
&
‘
(
)
*
But, now another vortex with strength +Γ needs to added at y = 2H + h to preserve the upper wall
boundary condition, so
u(0,h)=Γ
2
π
1
2h−1
2H−2h−1
2H+1
2H+…
%
&
‘
(
)
*
.
Continuing this construction leads to:
u(0,h)=Γ
2
π
1
2h−1
2H−2h−1
2H+1
2H+1
2H+2h−1
4H−2h−1
4H+1
4H+1
4H+2h…
%
&
‘
(
)
*
,
t
t
t
t
y = H
y = 2H
y = 3H
y
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
which can be simplified by removing terms that are equal and opposite:
u(0,h)=Γ
2
π
1
2h−1
2H−2h+1
2H+2h−1
4H−2h+1
4H+2h−1
6H−2h+1
6H+2h…
%
&
‘
(
)
*
.
The first term is unique while all the others follow in pairs, so
u(0,h)=Γ
2
π
1
2h+−1
2nH −2h+1
2nH +2h
%
&
‘ (
)
*
n=1
∞
∑
–
.
/
0
1
2 =Γ
4
π
h1−21
n2H h
( )
2−1
n=1
∞
∑
–
.
/
/
0
1
2
2
.
where the second equality follows from algebraic manipulations within the big parentheses.
When h = H/2, this formula becomes:
u(0,H/2) =Γ
4
π
h1−21
4n2−1
n=1
∞
∑
‘
(
)
*
+
,
.
The sum is 1/2, and this can be found by looking in an appropriate mathematical reference (see
Gradshteyn, I. S., and Ryzhik, I. M., Tables of Integrals, Series, and Products [Academic Press,
New York, 1980], p. 8), or by considering the terms in the sum:
1
4n2−1
n=1
∞
∑=1
2
1
2n−1−1
2n+1
%
&
‘ (
)
*
n=1
∞
∑=1
2
1−1
3
%
&
‘ (
)
* +1
3−1
5
%
&
‘ (
)
* +1
5−1
7
%
&
‘ (
)
* +…
+
,
–
.
/
0
=1
2
1+−1
3+1
3
%
&
‘ (
)
* +−1
5+1
5
%
&
‘ (
)
* +−1
7+1
7
%
&
‘ (
)
* +…
+
,
–
.
/
0 =1
2
.
The pair-cancellation of terms is the mathematical signature of the symmetry discussed in part
a), and series of this type are sometimes known as telescoping series. With either evaluation
approach, the net induced velocity is
u(0,H/2) =Γ
4
π
h1−21
2
%
&
‘ (
)
*
+
,
–
.
/
0 =0
,
which matches the part a) result.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 5.15. The axis of an infinite solid circular cylinder with radius a coincides with the z–
axis. The cylinder is stationary and immersed in an incompressible inviscid fluid, and the net
circulation around it is zero. An ideal line vortex parallel to the cylinder with circulation Γ passes
through the x–y plane at x = L > a and y = 0. Here two image vortices are needed to satisfy the
boundary condition on the cylinder’s surface. If one of these is located at x = y = 0 and has
strength Γ. Determine the strength and location of the second image vortex.
Solution 5.15. If the net circulation around the
cylinder is zero, then the second image vortex must
have strength –Γ. Therefore, the fluid velocity u at any
fluid velocity at the cylinder-surface point (xs, ys).
Using Cartesian unit vectors and the diagram to the right, with the circulation directions of the
vortices shown, leads to:
u(xs,ys)=Γ
2
π
a−ys
a
ex+xs
a
ey
%
&
‘ (
)
* +Γ
2
π
(L−xs)2+ys
2−ys
(L−xs)2+ys
2
ex−(L−xs)
(L−xs)2+ys
2
ey
%
&
‘
‘
(
)
*
*
+Γ
2
π
(xs−+
x )2+(ys−+
y )2
ys−+
y
(xs−+
x )2+(ys−+
y )2
ex−xs−+
x
(xs−+
x )2+(ys−+
y )2
ey
%
&
‘
‘
(
)
*
*
This can be simplified to:
+Γ
2
π
a
a2−(L−xs)ys
(L−xs)2+ys
2−(xs−&
(xs−&
x )2+(ys−&
y )2
(
)
+
, =0.
After dividing out common factors and cancelling equal and opposite terms this reduces to:
−Lys
(L−xs)2+ys
2+#
x ys−xs#
y
(xs−#
x )2+(ys−#
y )2=0
,
which must be true for all values of (xs, ys). So, at xs = a and ys = 0, this equation becomes:
(xs,ys)
y
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 5.16. Consider the interaction of two vortex rings of equal strength and similar sense of
rotation. Argue that they go through each other, as described near the end of section 5.6.
Solution 5.16. Consider two vortex rings with a similar sense of
rotation as shown in the figure to the right. The radii and speeds of
the two vortices are equal. The motion at A is the resultant of VB,
VC, and VD, while the motion at C is the resultant of VA, VB, and
VD. Comparing the velocity components on A and C, it is clear that
the net resultant is an enlargement of the vortex on the right, and a
AC
VB
VD
VA
VD
VC
VB
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 5.17. A constant density irrotational flow in a rectangular torus has a circulation Γ and
volumetric flow rate Q. The inner radius is r1, the outer radius is r2, and the height is h. Compute
the total kinetic energy of this flow in terms of only
ρ
, Γ, and Q.
Solution 5.17. For an irrotational vortical flow with
circulation Γ. The velocity in the angular direction will be:
u
θ
=Γ
2
π
r
.
The volumetric flow rate will be:
Q=u
θ
dzdr
0
h
∫
r
1
r2
∫=Γ
2
π
rdzdr
0
h
∫
r
1
r2
∫=hΓ
2
π
ln r2
r
1
&
‘
(
)
*
+
.
The total kinetic energy, KE, of the fluid in the in the torus
will be:
KE =1
2
ρ
u
θ
2
0
2
π
∫dzrdrd
θ
0
h
∫
r
1
r2
∫=
ρ
Γ2
4
π
1
rdzdr
0
h
∫
r
1
r2
∫=
ρ
Γ2h
4
π
ln r2
r
1
‘
(
)
*
+
, =1
2
ρ
ΓQ
.
h
r1r2
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 5.18. Consider a cylindrical tank of radius R filled with a viscous fluid spinning
steadily about its axis with constant angular velocity Ω. Assume that the flow is in a steady state.
(a) Find
ω⋅ndA
A
∫
where A is a horizontal plane surface through the fluid normal to the axis of
rotation and bounded by the wall of the tank, and n is the normal on A.
(b) The tank then stops spinning. Find again the value of
ω⋅ndA
A
∫
.
Solution 5.18. a) At steady state the flow inside the tank will be in solid body rotation, and the
vorticity will be twice the rotation rate: ω = 2Ω = constant. Thus,
ω⋅ndA
A
∫=2Ω
π
R2
.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 5.19. Using Figure 5.13, prove (5.32) assuming that: (i) the two vortices travel in
circles, (ii) each vortex’s speed along it’s circular trajectory is constant, and (iii) the period of the
motion is the same for both vortices.
Solution 5.19. When the two vortices travel in circles, with radii h1 and h2, at constant speeds
with the same period T, the circumference of each circular trajectory divided by the induced
speed of the vortex on that trajectory must match:
2
π
h
1
Γ22
π
h=T=2
π
h2
Γ12
π
h
,
where the term on the left applies to the vortex on the left (1), and the term on the right applies to
the vortex on the right (2). This relationship can be simplified to:
h
1Γ2=h2Γ1
. (#)
Now use h = h1 + h2 to substitute for h2 using h2 = h – h1, and solve for h1 to find:
h
1
Γ2
=h−h
1
Γ1
, or
h
1
1
Γ2
+1
Γ1
“
#
$%
&
‘=h
Γ1
, which implies
h
1=h
Γ1
1
Γ2
+1
Γ1
“
#
$%
&
‘
−1
=Γ2h
Γ2+Γ1
.
The relationship (#) then implies:
h2=Γ
1h
Γ
1+Γ2
for the other vortex.
Thus, G is located at the place where the diagonal line connecting the tips of the induced velocity
vectors crosses the line segment that connects the vortices. So, G is halfway between the two
vortices when they are of equal strength, but it lies closer to the stronger vortex when the vortices
are of unequal strength. Here G is also the “center of vorticity”. For example, when G is located
at the origin of coordinates:
center of vorticity ≡rnΓn
n=1
2
∑
=(−h
1, 0)Γ1+(h2, 0)Γ2=−h
1Γ1+h2Γ2, 0
( )
=−Γ2hΓ1+Γ1hΓ2
Γ1+Γ2
, 0
%
&
‘(
)
*=(0, 0).
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 5.20. Consider two-dimensional steady flow in the x–y plane outside of a long circular
cylinder of radius a that is centered on and rotating about the z-axis at a constant angular rate of
Ωz. Show that the fluid velocity on the x-axis is u(x,0) = (Ωza2/x)ey for x > a when the cylinder is
replaced by
a) a circular vortex sheet of radius a with strength
γ
= Ωza, and
b) a circular region of uniform vorticity ω = 2Ωzez with radius a.
c) Describe the flow for x2 + y2 < a2 for parts a) and b).
Solution 5.20. a) For a circular vortex sheet, an
element of the circle of length ad
θ
located at angle
θ
Separate the components and consider them individually.
u(x,0) =Ωza2ex
2
π
asin
θ
d
θ
x2+a2−2ax cos
θ
−
π
π
∫+Ωza2ey
2
π
(x−acos
θ
)d
θ
x2+a2−2ax cos
θ
−
π
π
∫
.
The x-component of u is zero because it is specified as an integration of an odd function on an
even interval. The remaining y-component integration can be rewritten to reach a tabulated
integral form, and then evaluated:
u(x,0) =Ωza2ey
2
π
1
2x1+x2−a2
x2+a2−2ax cos
θ
&
‘
(
)
*
+
d
θ
−
π
π
∫
=Ωza2ey
2
π
1
2x
θ
+2(x2−a2)
x2+a2
( )
2−4a2x2
tan−1x2+a2
( )
2−4a2x2
x2+a2−2ax tan
θ
2
&
‘
(
(
(
)
*
+
+
+
&
‘
(
(
(
)
*
+
+
+
−
π
+
π
=Ωza2ey
2
π
1
2x
θ
+2tan−1x2−a2
(x−a)2tan
θ
2
&
‘
(
)
*
+
&
‘
(
)
*
+
−
π
+
π
=Ωza2ey
2
π
1
2x2
π
+2
π
2−2−
π
2
&
‘
( )
*
+
&
‘
( )
*
+ =Ωza2ey
x.
[Thus,
u(x,0) =(x−acos
θ
)d
θ
x2+a2−2ax cos
θ
−
π
+
π
∫=2
π
x
and is independent of a!]
b) For a uniform distribution of vorticity within the circle specified by x2 + y2 < a2, an element of
area dA = rd
θ
dr will have a circulation dΓ =
ω
zdA = 2Ωzrd
θ
dr. The distance between the vortex
!
(acos!,asin!)
y
du
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 5.21. An ideal line vortex in a half space filled with an inviscid constant-density fluid
has circulation Γ, lies parallel to the z-axis, and passes through the x–y plane at x = 0 and y = h.
The plane defined by y = 0 is a solid surface.
a) Use the method of images to find u(x,y) for y > 0 and show that the fluid velocity on y = 0 is
u(x,0) =
Γhex
π
(x2+h2)
[ ]
.
b) Show that u(0,y) is unchanged for y > 0 if the image vortex is replaced by a vortex sheet of
strength
γ
(x)=−u(x,0)
on y = 0.
c) (If you have the patience) Repeat part b) for u(x,y) when y > 0.
Solution 5.21. a) If the actual vortex is at (0, h) and has circulation Γ, then an image vortex of
with circulation –Γ at (0, –h) produces a no-through-flow boundary condition on y = 0. In
Cartesian coordinates, the velocity u at any location (x, y) is the sum of the two induced
velocities:
u(x,y)=Γ
2
π
x2+(y−h)2
−(y−h)ex+xey
x2+(y−h)2
%
&
‘
‘
(
)
*
* +Γ
2
π
x2+(y+h)2
(y+h)ex−xey
x2+(y+h)2
%
&
‘
‘
(
)
*
*
,
and this form can be simplified to:
u(x,y)=Γ
2
π
−(y−h)ex+xey
x2+(y−h)2+(y+h)ex−xey
x2+(y+h)2
%
&
‘
(
)
*
. (†)
When evaluated on y = 0, the velocity is:
u(x,0) =Γ
2
π
hex+xey
x2+h2+hex−xey
x2+h2
%
&
‘
(
)
* =Γhex
π
(x2+h2)
.
b) First of all, determine u(0,y) from the results of part a)
u(0, y)=Γ
2
π
−ex
y−h+ex
y+h
%
&
‘
(
)
*
. (%)
For a flat vortex sheet lying on y = 0, an element of the length dx´ located at x´ will have a
circulation of dΓ =
γ
(x´)dx´. The distance between this vortex element and the location (x, y) is
(x−#
x )2+y2
. The magnitude and direction of the induced velocity increment du at (x, y) are
dΓ2
π
(x−%
x )2+y2
( )
and
−yex+(x−#
x )ey
(x−#
x )2+y2
.
When the image vortex is replaced by a vortex sheet, u(x,y) for y > 0 will be the sum of the
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling