Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 5.1. A closed cylindrical tank 4 m high and 2 m in diameter contains water to a depth
of 3 m. When the cylinder is rotated at a constant angular velocity of 40 rad/s, show that nearly
0.71 m2 of the bottom surface of the tank is uncovered. [Hint: The free surface is in the form of a
paraboloid of revolution. For a point on the free surface, let h be the height above the
(imaginary) vertex of the paraboloid and r be the local radius of the paraboloid. From Section
5.1,
h=
ω
0
2r22g
, where
ω
0 is the angular velocity of the tank. Apply this equation to the two
points where the paraboloid cuts the top and bottom surfaces of the tank.]
Solution 5.1. Denote the local radius of the paraboloid of revolution as r(h), and let h = 0 at
vertex of this parabaloid. The goal of this problem is to determine
π
r2(h1), where h1 coincides
with the bottom of the tank.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 5.2. A tornado can be idealized as a Rankine vortex with a core of diameter 30 m. The
gauge pressure at a radius of 15 m is −2000 N/m2 (i.e.,, the absolute pressure is 2000 N/m2 below
atmospheric).
(a) Show that the circulation around any circuit surrounding the core is 5485 m2/s. [Hint: Apply
the Bernoulli equation between infinity and the edge of the core.]
(b) Such a tornado is moving at a linear speed of 25 m/s relative to the ground. Find the time
required for the gauge pressure to drop from −500 to −2000 N/m2. Neglect compressibility
effects and assume an air temperature of 25°C. (Note that the tornado causes a sudden decrease
of the local atmospheric pressure. The damage to structures is often caused by the resulting
excess pressure on the inside of the walls, which can cause a house to explode.)
Solution 5.2. At 25°C and one atmosphere, air density is
ρ
air = (101.3 kPa)/(287m2s–2K–1)(298K)
= 1.18 kgm–3.
a) The flow is irrotational outside the Rankine vortex core, so the steady constant density
Bernoulli equation implies at any distance r from the center of vortex core:
1
2
ρ
airU∞
2+p∞=1
2
ρ
airU2(r)+p(r)
, or
U(r)=2( p∞−p(r))
ρ
air
[ ]
1 2
,
where the “∞” subscript refers to conditions very far from the vortex (U∞ ≈ 0). From equation
(3.28), U(r) = Γ/2
π
r, so where r = rc = the core radius:
Γ=2
π
rc2( p∞−pc)
ρ
air
[ ]
1 2 =2
π
(15) 2(2000) 1.18
[ ]
1 2 =5485m2s−1
.
b) The radial distance at which the gauge pressure is –500 Pa can be determined from the results
developed for part a) for U(r):
U(r) = Γ/2
π
r =
=2( p∞−p(r))
ρ
air
[ ]
1 2
, or
r=Γ2
π
( )
2( p∞−p(r))
ρ
air
[ ]
−1 2
.
Evaluate to find r, and divide by (r – rc) by 25 m/s to determine the time for the gauge pressure
to drop from −500Pa to −2000Pa.
r=Γ2
π
( )
2( p∞−p(r))
ρ
air
[ ]
−1 2 =5485m2s−12
π
( )
2(500Pa) 1.18kgm−3
[ ]
−1 2 =30.0m
, so
time = (30m – 15m)/25ms–1 = 0.60 s.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 5.3. The velocity field of a flow in cylindrical coordinates (R,
ϕ
, z) is u = (uR, u
ϕ
, uz) =
(0, aRz, 0) where a is a constant.
(a) Show that the vorticity components are ω = (
ω
R,
ωϕ
,
ω
z) = (–aR, 0, 2az).
(b) Verify that
∇ ⋅ ω=0
.
(c) Sketch the streamlines and vortex lines in an (R,z)-plane. Show that the vortex lines are given
by zR2 = constant.
Solution 5.3. a) The vorticity components in cylindrical coordinates are provided in Appendix B:
ω
R=1
R
∂
uz
∂ϕ
−
∂
u
ϕ
∂
z=0 – aR =−aR
ωϕ
=
∂
uR
∂
z−
∂
uz
∂
R=0–0=0
, and
ω
z=1
R
∂
(Ru
ϕ
)
∂
R−1
R
∂
uR
∂ϕ
=2az – 0 =2az
.
b)
∇ ⋅
ω
=1
R
∂
(R
ω
R)
∂
R+1
R
∂ωϕ
∂ϕ
+
∂ω
z
∂
z=1
R
∂
(−aR2)
∂
R+1
R
∂
(0)
∂ϕ
+
∂
(2az)
∂
z=−2a+2a=0
c) The only non-zero velocity component is u
ϕ
, so the streamlines are circles centered on the z
axis. Vortex lines are given by:
dR
ω
R
=dz
ω
z
→dR
−aR =dz
2az → − ln(R)=1
2ln(z)+const.
Exponentiate the last expression to find:
1R=const.z
, or R2z = const.
Thus the vortex lines asymptote to the plane z = 0 and to the z-axis.
R
z
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 5.4. Starting from the flow field of an ideal vortex (5.2), compute the viscous stresses
τ
rr,
τ
r
θ
, and
τθθ
, and show that the net viscous force on a fluid element, (∂
τ
ij/∂xi), is zero.
Solution 5.4. In two dimensions, the ideal vortex flow field is u =(ur, u
θ
) = (0, Γ/2
π
r). This flow
field is incompressible since
∇ ⋅ u=1r
( )
∂
u
θ∂θ
( )
=0
. Therefore, the viscous stress listings for
plane polar coordinates in Appendix B apply:
τ
rr =2
µ
Srr =2
µ∂
ur
∂
r=0
,
τθθ
=2
µ
S
θθ
=2
µ
1
r
∂
u
θ
∂θ
+ur
r
!
“
#$
%
&=0
, and
τ
r
θ
=2
µ
Sr
θ
=
µ
r
∂
∂
r
u
θ
r
!
“
#$
%
&+1
r
∂
ur
∂θ
!
“
#$
%
&=
µ
rΓ
2
π
r3
!
“
#$
%
&(−2) =−
µ
Γ
π
r2
.
For incompressible flow (∂ui/∂xi = 0) with constant viscosity:
∂τ
ij
∂
xi
=2
µ∂
∂
xi
Sij =
µ∂
∂
xi
∂
ui
∂
xj
+
∂
uj
∂
xi
!
“
#
#
$
%
&
&=
µ∂
∂
xj
∂
ui
∂
xi
+
µ∂
2uj
∂
xi
2=
µ∂
2uj
∂
xi
2=∇2u
.
Using the velocity field given above, and remembering to differentiate unit vectors, this
becomes:
∇2u=1
r
∂
∂
r
r
∂
∂
r
u
θ
e
θ
( )
%
&
‘ (
)
* +1
r2
∂
2
∂θ
2u
θ
e
θ
( )
=1
r
∂
∂
r
re
θ
∂
u
θ
∂
r
%
&
‘ (
)
* +1
r2
∂
∂θ
−u
θ
er
( )
=e
θ
1
r
∂
u
θ
∂
r+e
θ
∂
2u
θ
∂
r2−u
θ
r2e
θ
=−e
θ
µ
Γ
π
−1
r3+2
r3−1
r3
%
&
‘ (
)
* =0.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 5.5. Consider the situation depicted in Figure 5.6. Use a Cartesian coordinate system
with a horizontal x-axis that puts the barrier at x = 0, a vertical y-axis that puts the bottom of the
container at y = 0 and the top of the container at y = H, and a z-axis that points out of the page.
Show that, at the instant the barrier is removed, the rate of baroclinic vorticity production at the
interface between the two fluids is:
D
ω
z
Dt =2(
ρ
2−
ρ
1)g
(
ρ
2+
ρ
1)
δ
,
where the thickness of the density transition layer just after barrier removal is
δ
<< H, and the
density in this thin interface layer is assumed to be (
ρ
1 +
ρ
2)/2. If necessary, also assume that
fluid pressures match at y = H/2 just after barrier removal, and that the width of the container
into the page is b. State any additional assumptions that you make.
Solution 5.5. At the instant the barrier is removed hydrostatic pressure
will persist in each liquid. Let this pressure be p2 for the fluid on the
where u is the element’s horizontal velocity, and b is the dimension of the container into the
page. Because the pressure forces are hydrostatic, there is no vertical pressure force imbalance so
the vertical component of Newton’s second law for the same fluid element is:
ρ
2+
ρ
1
2
#
$
% &
‘
( b
δ
dy
( )
dv
dt =0
.
Thus, since v is initially zero along the interface, it remains that way until the fluid starts moving.
Returning to the horizontal equation, canceling common factors, and inserting the
relationships for p1 and p2 produces:
ρ
2+
ρ
1
2
#
$
% &
‘
(
δ
( )
du
dt =po−
ρ
2g y −H2
( )
−po+
ρ
1g y −H2
( )
=
ρ
2−
ρ
1
( )
g−y+H2
( )
.
Simplify and solve for du/dt to find:
du
dt =2
ρ
2−
ρ
1
( )
g
ρ
2+
ρ
1
( )
δ
−y+H2
( )
.
In this circumstance the z-component of vorticity is
ω
z = (∂v/∂x) – (∂u/∂y). At the instant of
interest, v = dv/dt = 0, so
d
dt
ω
z=d
dt −
∂
u
∂
y
%
&
‘
(
)
* =−
∂
∂
y
du
dt
%
&
‘ (
)
* =2
ρ
2−
ρ
1
( )
g
ρ
2+
ρ
1
( )
δ
.
Thus, when the heavier fluid is on the left, positive vorticity directed along the z-axis is produced
when the barrier is removed.
y = H
dy
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 5.6. At t = 0 a constant-strength z-directed vortex sheet is created in the x–z plane (y =
0) in an infinite pool of a fluid with kinematic viscosity
ν
, that is ω(y,0) = ez
γδ
(y). The symmetry
of the initial condition suggests that ω =
ω
zez and that
ω
z will only depend on y and t. Determine
ω(y,t) for t > 0 via the following steps.
a) Determine a dimensionless scaling law for
ω
z in terms of
γ
,
ν
, y, and t.
b) Simplify the general vorticity equation (5.13) to a linear field equation for
ω
z for this situation.
c) Based on the fact that the field equation is linear, simplify the result of part a) by requiring
ω
z
to be proportional to
γ
, plug the simplified dimensionless scaling law into the equation
determined for part b), and solve this equation to find the undetermined function to reach:
ω
z(y,t)=
γ
2
πν
texp −y2
4
ν
t
‘
(
)
*
+
,
Solution 5.6. a) This part of the exercise is dimensional analysis. The parameters are:
ω
z,
γ
,
ν
, y,
and t. Use these to create the parameter matrix:
ω
z
γ
ν
y t
––––––––––––––––––––––––––––
Mass: 0 0 0 0 0
Length: 0 1 2 1 0
Here ω = (0, 0,
ω
z), so this equation simplifies to:
∂ω
z
∂
t+u
∂ω
z
∂
x+v
∂ω
z
∂
y+w
∂ω
z
∂
z=
ω
z
∂
w
∂
z+
ν∂
2
ω
z
∂
x2+
∂
2
ω
z
∂
y2+
∂
2
ω
z
∂
z2
%
&
‘
(
)
*
.
where u = (u, v, w). However, the exercise statement suggests that
ω
z only depends on y and t. In
this case the vorticity equation simplifies further.
∂ω
z
∂
t+v
∂ω
z
∂
y=
ω
z
∂
w
∂
z+
ν∂
2
ω
z
∂
y2
%
&
‘
(
)
*
.
The boundary conditions on
ω
z are independent of x and z. Therefore, all the dependent field
variables should all be independent of these directions. This implies ∂u/∂x = ∂w/∂z = 0, so the
incompressible flow continuity equation becomes ∂v/∂y = 0 which implies v = constant. A
constant velocity can be removed from the field equations via a Galilean transformation to a
moving coordinate system. Thus, v can be chosen equal to zero without loss of generality, so the
middle terms in the vorticity equation drop out completely leaving:
∂ω
z
∂
t=
ν∂
2
ω
z
∂
y2
%
&
‘
(
)
*
.
c) For
ω
z to be the solution of a linear equation it must be proportional to
γ
the initial field
strength. Therefore set:
ω
z = (
γ
/[
ν
t]1/2)F(y/[
ν
t]1/2), where F is merely another undetermined
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
function. Compute the derivatives that compose the field equation using
η
= y/[
ν
t]1/2 as the
argument of F.
This last equation can be rewritten, and integrated once to find:
d
d
η
#
F (
η
)+1
2
η
F(
η
)
$
%
& ‘
(
) =0
→
“
F (
η
)+1
2
η
F(
η
)=C
,
where C is a constant. The resulting first-order differential equation has an integrating factor of
exp{+
η
2/4} that leads to:
d
d
η
e+
η
24F(
η
)
( )
=Ce+
η
24
, so
F(
η
)=De−
η
24+e−
η
24C e+
ξ
24
0
η
∫d
ξ
.
where D is another constant. The initial condition,
ω
z = 0 for y > 0 at t = 0, requires C = 0. This
leaves:
ω
z(y,t)=D
ν
texp −y2
4
ν
t
%
&
‘
(
)
*
.
To match the initial spatial profile,
ω
z(y,0) =
γδ
(y), integrate in y and take the limit as
t→0
.
γδ
(y)dy
−∞
+∞
∫=
γ
=lim
t→0
ω
z(y,t)dy
−∞
+∞
∫
)
*
+ ,
–
. =lim
t→0
D
ν
texp −y2
4
ν
t
0
1
2
3
4
5
dy
−∞
+∞
∫
)
*
+ ,
–
. =2Dlim
t→0exp −
ζ
2
{ }
d
ζ
−∞
+∞
∫
)
*
+ ,
–
. =2D
π
.
Thus,
D=
γ
2
π
, so the final solution is:
ω
z(y,t)=
γ
2
πν
texp −y2
4
ν
t
‘
(
)
*
+
,
.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 5.7. a) Starting from the continuity and Euler equations for an inviscid compressible
fluid,
∂ρ ∂
t+∇ ⋅
ρ
u
( )
=0
and
ρ
DuDt
( )
=−∇p+
ρ
g
, derive the Vazsonyi equation:
D
Dt
ω
ρ
$
%
&
‘
(
) =ω
ρ
$
%
&
‘
(
)
⋅ ∇u+1
ρ
3∇
ρ
× ∇p
,
when the body force is conservative:
g=−∇Φ
. This equation shows that ω/
ρ
in a compressible
flow plays the nearly same dynamic role as ω in an incompressible flow [see (5.26) with Ω = 0
and
ν
= 0].
b) Show that the final term in the Vazsonyi equation may also be written:
1
ρ
( )
∇T× ∇s
.
c) Simplify the Vazsonyi equation for barotropic flow.
Solution 5.7. a) Start by taking the curl of the Euler equation:
∇ ×
∂
u
∂
t+(u⋅ ∇)u=−1
ρ
∇p+g
(
)
*
+
,
– →
∂
ω
∂
t+∇ × (u⋅ ∇)u=−∇ ×1
ρ
∇p
0
1
2
3
4
5 +∇ × g
.
The second term can modified by using the vector identity:
(u⋅ ∇)u=ω×u+1
2∇u2
,
b) Start from the thermodynamic property relationship: de = Tds – pd(1/
ρ
). Therefore:
∇e=T∇s−p∇1
ρ
( )
, and
∇ × ∇e=0=∇ × T∇s
( )
− ∇ × p∇1
ρ
( )
( )
=∇T× ∇s− ∇p× ∇ 1
ρ
( )
,
since
∇ × ∇s=∇ × ∇ 1
ρ
( )
=0
. Thus,
1
ρ
3
( )
∇p× ∇
ρ
=1
ρ
( )
∇T× ∇s
.
c) In barotropic flow,
∇p
and
∇
ρ
will be parallel, so the Vazsonyi equation becomes
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 5.8. Starting from the unsteady momentum equation for a compressible fluid with
constant viscosities:
ρ
Du
Dt +∇p=
ρ
g+
µ
∇2u+
µυ
+1
3
µ
( )
∇ ∇ ⋅ u
( )
,
show that
∂
u
∂
t+ω×u=T∇s− ∇ h+1
2u2+Φ
( )
−
µ
ρ
∇ × ω+1
ρµυ
+4
3
µ
( )
∇ ∇ ⋅ u
( )
where T = temperature, h = enthalpy per unit mass, s = entropy per unit mass, and the body force
is conservative:
g=−∇Φ
. This is the viscous Crocco-Vazsonyi equation. Simplify this equation
for steady inviscid non-heat-conducting flow to find the Bernoulli equation (4.78)
h+1
2u2+Φ
= constant along a streamline,
which is valid when the flow is rotational and nonisothermal.
Solution 5.8. Divide the given momentum equation by
ρ
and use a vector identity (B3.9 with u =
F) for the advective acceleration term,
(u⋅ ∇)u=ω×u+1
2∇u2
,
Du
Dt +1
ρ
∇p=
∂
u
∂
t+ω×u+1
2∇u2+1
ρ
∇p=g+
µ
ρ
∇2u+1
ρµυ
+1
3
µ
( )
∇ ∇ ⋅ u
( )
.
Using the second equality,
g=−∇Φ
, and the vector identity (B3.13) for the Laplacian of a
vector,
∇2u=∇(∇ ⋅ u)− ∇ × (∇ × u)=∇(∇ ⋅ u)− ∇ × ω
, allows the last equation to be rewritten:
∂
u
∂
t+ω×u+1
2∇u2+1
ρ
∇p=−∇Φ+
µ
ρ
∇(∇ ⋅ u)− ∇ × ω
( )
+1
ρµυ
+1
3
µ
( )
∇ ∇ ⋅ u
( )
.
Collect and combine terms to see how close this is to the target equation.
∂
u
∂
t+ω×u=−1
ρ
∇p− ∇ 1
2u2+Φ
( )
−
µ
ρ
∇ × ω+1
ρµυ
+4
3
µ
( )
∇ ∇ ⋅ u
( )
.
Comparing this equation with the target equation shows that the term involving the
thermodynamic variable must be modified. Use the property relationship, dh = Tds + (1/
ρ
)dp, to
deduce:
∇h=T∇s+1
ρ
( )
∇p→ − 1
ρ
( )
∇p=T∇s− ∇h
.
Substituting this into the last equation produces:
∂
u
∂
t+ω×u=T∇s− ∇ h+1
2u2+Φ
( )
−
µ
ρ
∇ × ω+1
ρµυ
+4
3
µ
( )
∇ ∇ ⋅ u
( )
,
which is the requisite result.
Steady flow implies (∂/∂t)(anything) = 0. Inviscid implies
µ
=
µυ
= 0. Inviscid
(frictionless) non-heat conducting flows are isentropic, so
∇s=0
. Thus, the Crocco-Vazsonyi
equation simplifies to:
ω×u=−∇ h+1
2u2+Φ
( )
.
Take the dot-product of this equation with the streamline direction,
es≡u u
, noting that
ω×u
is perpendicular to es.
es⋅ω×u
( )
=0=−es⋅ ∇ h+1
2u2+Φ
( )
→
∂
∂
s
h+1
2u2+Φ
( )
=0
,
where s is the path length along a streamline. Thus,
h+1
2u2+Φ
= constant along a streamline.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 5.9. a) Solve
∇2G(x,#
x )=
δ
(x−#
x )
for G(x,x´) in a uniform unbounded three-
dimensional domain, where
δ
(x – x´) =
δ
(x – x´)
δ
(y – y´)
δ
(z – z´) is the three dimensional Dirac
delta function.
b) Use the result of part a) to show that:
φ
(x)=−1
4
π
q(%
x )
x−%
x
all %
x
∫d3%
x
is the solution of the Poisson
equation
∇2
φ
(x)=q(x)
in a uniform unbounded three-dimensional domain.
Solution 5.9. a) 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 G becomes:
∇XYZ
2G=
δ
(X)=
δ
(x−%
x )
.
Integrate this equation inside a sphere of radius r:
∇XYZ
2G dV
sphere
∫∫∫ =∇XYZG⋅ndA
spherical surface
∫∫ =
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 unity.
The dot product in the middle portion of the above equation simplifies to ∂G/∂r because
n = er on the spherical surface and
er⋅ ∇XYZ =
∂ ∂
r
. Plus, in an unbounded uniform environment,
there are no preferred directions so G = G(r) alone (no angular dependence). Thus, the integrated
field equation simplifies to
∂
G
∂
r
#
$
% &
‘
(
r2sin
θ
d
θ
d
ϕ
=0
2
π
∫
θ
=0
π
∫
ϕ
=r2
∂
G
∂
r
sin
θ
d
θ
d
ϕ
=0
2
π
∫
θ
=0
π
∫
ϕ
=4
π
r2
∂
G
∂
r=1
,
which implies
∂
G
∂
r=1
4
π
r2
or
G=−1
4
π
r=−1
4
π
x−$
x
.
b) Work in the unshifted coordinate system, and start with the given information about the
Green’s function,
∇2G(x,#
x )=
δ
(x−#
x )
, and multiply both sides by
q(“
x )
to get:
∇2q(#
x )G(x,#
x )
[ ]
=q(#
x )
δ
(x−#
x )
. (1)
The q-function can slide inside the differential operator,
∇2
, because
q(“
x )
depends on x´ while
φ
(x)=q(x)
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 5.10. Start with the equations of motion in the rotating coordinates, and prove (5.29)
Kelvin’s circulation theorem for the absolute vorticity. Assume that the flow is inviscid and
barotropic and that the body forces are conservative. Explain the result physically.
Solution 5.10. The general idea is to redo the ordinary derivation of Kelvin’s theorem in a
steadily rotating coordinate system where the inviscid momentum equation is:
ρ
Du
Dt =−∇p+
ρ
ge−2Ω×u
[ ]
,
($)
Here all velocities are observed in the rotating frame so the primes used in Section 4.7 have been
Consider the product of vectors in the integrand
−2Ω×u
( )
⋅dx=−2
ε
ijkΩjukdxi=2Ωj
ε
jik dxiuk=−2Ω⋅u×dx
( )
.
Now define n to be the unit vector perpendicular to the plane defined by u and dx, and let
u⊥
be
the component of u perpendicular to dx so that:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 5.11. In (R,
ϕ
,z) cylindrical coordinates, consider the radial velocity uR = −R–1(∂
ψ
/∂z),
and the axial velocity uz = R–1(∂
ψ
/∂R) determined from the axisymmetric stream function
ψ
(R,z)=Aa4
10
R2
a2
#
$
%
&
‘
( 1−R2
a2−z2
a2
#
$
%
&
‘
(
where A is a constant. This flow is known as Hill’s spherical
vortex.
a) For R2 + z2 ≤ a2, sketch the streamlines of this flow in a plane that contains the z-axis. What
does a represent?
b) Determine u = uR(R,z)eR + uz(R,z)ez
c) Given
ωϕ
=
∂
uR
∂
z
( )
−
∂
uz
∂
R
( )
, show that ω = ARe
ϕ
in this flow and that this vorticity field is
a solution of the vorticity equation (5.13).
d) Does this flow include stretching of vortex lines?
Solution 5.11. a) The zero streamline is given by R = 0
and by R2 + z2 = a2.
For R2 + z2 ≤ a2, the others appear as shown to the right.
c) Compute the components of the vorticity. Here, u
ϕ
= 0 and neither uR
or uz depend on
ϕ
, so
ω
R=1
R
∂
uz
∂ϕ
−
∂
u
ϕ
∂
z=0
,
ωϕ
=
∂
uR
∂
z−
∂
uz
∂
R=AR
5−Aa2
5−4R
a2
&
‘
( )
*
+ =AR
, and
ω
z=1
R
∂
(Ru
ϕ
)
∂
R−1
R
∂
uR
∂ϕ
=0
.
Therefore, ω = ARe
ϕ
in this flow. The vorticity equation is:
∂
ω
∂
t+u⋅ ∇
( )
ω=ω⋅ ∇
( )
u+
ν
∇2ω
.
The flow is steady so the ∂/∂t term is zero. For ω =
ωϕ
e
ϕ
= ARe
ϕ
, the remaining terms are:
advection term:
u⋅ ∇
( )
ωϕ
e
ϕ
( )
=e
ϕ
uR
∂ωϕ
∂
R=AuRe
ϕ
,
vortex-stretching term:
ωϕ
e
ϕ
⋅ ∇
( )
u
( )
=
ωϕ
R
∂
∂ϕ
uReR+uzez
( )
=AR
R
uR
∂
eR
∂ϕ
=AuRe
ϕ
,
and viscous term:
ν
∇2
ωϕ
e
ϕ
( )
=
∂
2
∂
R2+1
R
∂
∂
R+1
R2
∂
2
∂ϕ
2
‘
(
)
*
+
,
ωϕ
e
ϕ
( )
=e
ϕ
∂
2
ωϕ
∂
R2+e
ϕ
1
R
∂ωϕ
∂
R+
ωϕ
1
R2
∂
2e
ϕ
∂ϕ
2
=e
ϕ
∂
2
ωϕ
∂
R2+1
R
∂ωϕ
∂
R−
ωϕ
R2
&
‘
(
)
*
+ =e
ϕ
0+1
R
A−AR
R2
&
‘
( )
*
+ =0
So, the vorticity equation satisfied by a balance of the advection and vortex stretching terms.
d) Yes, this flow includes stretching of vortex lines.
z
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 5.12. In (R,
ϕ
,z) cylindrical coordinates, consider the flow field uR = –
α
R/2, u
ϕ
= 0, and
uz =
α
z.
a) Compute the strain rate components SRR, Szz, and SRz. What sign of
α
causes fluid elements to
elongate in the z-direction? Is this flow incompressible?
b) Show that it is possible for a steady vortex (a Burgers’ vortex) to exist in this flow field by
adding u
ϕ
= (Γ/2πR)[1 – exp(–
α
R2/4
ν
)] to uR and uz from part a) and then determining a pressure
field p(R,z) that together with u = (uR, u
ϕ
, uz) solves the Navier-Stokes momentum equation for a
fluid with constant density
ρ
and kinematic viscosity
ν
.
c) Determine the vorticity in the Burgers’ vortex flow of part b).
d) Explain how the vorticity distribution can be steady when
α
≠ 0 and fluid elements are
stretched or compressed.
e) Interpret what is happening in this flow when
α
> 0 and when
α
< 0.
Solution 5.12. a) Using the strain rates from Appendix B,
SRR =
∂
uR
∂
R=−
α
2
,
Szz =
∂
uz
∂
z=
α
, and
SRz =1
2
∂
uR
∂
z+
∂
uz
∂
R
#
$
% &
‘
( =0
.
Therefore, when
α
> 0, fluid elements elongate in the z-direction. The divergence of the velocity
field is:
1R
( )
∂
(RuR)
∂
R
( )
+
∂
uz
∂
z=1R
( )
−2
α
R2
( )
+
α
=0
, so the flow is incompressible.
b) Here: uR = –
α
R/2, u
ϕ
= (Γ/2πR)[1 – exp(–
α
R2/4
ν
)], and uz =
α
z. None of these velocities
depends on the angle
ϕ
. Consider the three components of the Navier-Stokes momentum
equation:
(i) Steady R-direction equation with ∂/∂
ϕ
= 0, from Appendix B:
uR
∂
uR
∂
R+uz
∂
uR
∂
z−u
ϕ
2
R=−1
ρ
∂
p
∂
R+
ν
1
R
∂
∂RR∂uR
∂R
‘
(
) *
+
, +∂2uR
∂z2−uR
R2
‘
(
)
*
+
,
.
The various terms involving the velocity are:
uR
∂
uR
∂
R=−
α
R
2−
α
2
%
&
‘ (
)
* =
α
2R
4
,
uz
∂
uR
∂
z=0
,
1
R
∂
∂R
R∂uR
∂R
#
$
% &
‘
( =−
α
2R
,
∂2uR
∂z2=0
, and
−uR
R2=
α
2R
.
So, the R-direction equation becomes:
α
2R
4−Γ2
4
π
2R31−exp −
α
R2
4
ν
‘
(
)
*
+
,
–
.
/
0
1
2
2
=−1
ρ
∂
p
∂
R
,
which can formally be integrated
p(R,z)=−
ρα
2
%
R
4−Γ
2
4
π
2
%
R
3
1−exp −
α
%
R
2
4
ν
)
*
+
,
–
.
/
0
1
2
3
4
2
5
6
7
8
7
9
:
7
;
7
0
R
∫
d%
R +f(z)
.
where f(z) is an unknown function.
(ii) Steady
ϕ
-direction equation with ∂/∂
ϕ
= 0, from Appendix B, is
uR
∂
u
ϕ
∂
R+uz
∂
u
ϕ
∂
z+uRu
ϕ
R=
ν
1
R
∂
∂R
R∂u
ϕ
∂R
%
&
‘
(
)
* +∂2u
ϕ
∂z2−u
ϕ
R2
%
&
‘
(
)
* =
ν
∂
∂R
1
R
∂(Ru
ϕ
)
∂R
%
&
‘
(
)
* +∂2u
ϕ
∂z2
%
&
‘
(
)
*
.
The various terms are:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Multiply by 4
π
/Γ
α
, and collect like terms:
1
R1−1
( )
−1
R1−1
( )
exp −
α
R2
4
ν
%
&
‘
(
)
* −
α
R
2
ν
exp −
α
R2
4
ν
%
&
‘
(
)
* =−
α
R
2
ν
exp −
α
R2
4
ν
%
&
‘
(
)
*
;
so the
ϕ
-direction momentum equation is satisfied.
(iii) Steady z-direction equation with ∂/∂
ϕ
= 0, from Appendix B:
uR
∂
uz
∂
R+uz
∂
uz
∂
z=−1
ρ
∂
p
∂
z+
ν
1
R
∂
∂RR∂uz
∂R
&
‘
( )
*
+ +∂2uz
∂z2
&
‘
(
)
*
+
.
The various terms are:
uR
∂
uz
∂
R=0
,
uz
∂
uz
∂
z=
α
2z
,
1
R
∂
∂R
R∂uz
∂R
#
$
% &
‘
( =0
, and
∂2uz
∂z2=0.
So, the z-direction equation becomes:
α
2z=−1
ρ
∂
p
∂
z
, or
p(z,R)=−
ρα
2z22+g(R)
,
and this determines the unknown function,
f(z)=−
ρα
2z22
, in the final equation of (i), so the
pressure is:
p(R,z)−p(0,0) =−
ρα
2
2z2+R2
4
%
&
‘
(
)
* +
ρ
Γ2
4
π
2–
R 31−exp −
α
–
R 2
4
ν
%
&
‘
(
)
*
/
0
1
2
3
4
2
5
6
7
8
7
9
:
7
;
7
0
R
∫d–
R
.
c) There is only one non-zero component of the vorticity:
ω
z=1
R
∂
∂
R
Ru
ϕ
( )
−1
R
∂
uR
∂ϕ
= + Γ
α
4
πν
exp −
α
R2
4
ν
*
+
,
–
.
/
.
d) When
α
> 0, the vorticity distribution can be steady because the strain field concentrates the
vorticity at precisely the correct rate to balance the vorticity’s outward diffusion.
e) When
α
> 0, then flow comes inward toward the z-axis along z = 0 and turns vertically upward
and downward along the +z and –z axis, respectively. Here the vorticity is concentrated near the