Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 3.15. Consider the following Cartesian velocity field
u=A(t)f(x),g(y),h(z)
( )
where
A, f, g, and h are non-constant functions of only one independent variable.
a) Determine ∂u/∂t, and (u⋅∇)u in terms of A, f, g, and h, and their derivatives.
b) Determine A, f, g, and h when Du/Dt = 0, u = 0 at x = 0, and u is finite for t > 0.
c) For the conditions in b), determine the equation for the path line that passes through xo at time
to, and show directly that the acceleration a of the fluid particle that follows this path is zero.
Solution 3.15. a) Here there are three components of the fluid velocity; thus
∂
u
∂
t=
∂
∂
tAf ,Ag,Ah
( )
=fdA
dt ,gdA
dt ,hdA
dt
#
$
% &
‘
( =dA
dt f,g,h
( )
, and
u⋅ ∇
[ ]
u=A f ,g,h
( )
⋅
∂
∂
x,
∂
∂
y,
∂
∂
z
%
‘
(
*
+
–
.
0 Af ,Ag,Ah
( )
The first term in this equation only depends on t while the second one only depends on x, thus,
each must be equal and opposite, and constant (= C). So,
1
A2
dA
dt =−C→ − dA
A2=Cdt →1
A=C(t−to)→A=1
C(t−to)
, and
df
dx =C→f=C(x−xo)
,
where to and xo are constants of integration. Similarly, for the other two component directions:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 3.16 If a velocity field is given by u = ay and v = 0, compute the circulation around a
circle of radius ro that is centered on at the origin. Check the result by using Stokes’ theorem.
Solution 3.16. In plane polar coordinates, the vector path-length element, ds, on a circle of radius
ro is ds =
tds =e
θ
r
od
θ
. From Example 2.1, the radial and angular velocity component are:
ur = ucos
θ
+ vsin
θ
= arsin
θ
cos
θ
+ 0, and u
θ
= –usin
θ
+ vcos
θ
= –arsin2
θ
,
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 3.17. Consider a plane Couette flow of a viscous fluid confined between two flat plates
a distance b apart. At steady state the velocity distribution is u = Uy/b and v = w = 0, where the
upper plate at y = b is moving parallel to itself at speed U, and the lower plate is held stationary.
Find the rates of linear strain, the rate of shear strain, and vorticity in this flow.
Solution 3.17. Here there is only one velocity component: u = Uy/b. The strain rate tensor is:
Sij =1
2
∂
ui
∂
xj
+
∂
uj
∂
xi
#
$
%
%
&
‘
(
( →Sij =
∂
u
∂
x1
2
∂
u
∂
y+
∂
v
∂
x
( )
1
2
∂
v
∂
x+
∂
u
∂
y
( )
∂
v
∂
y
*
+
,
–
.
/
=
0U2b
U2b0
*
+
,
–
.
/
.
Thus, the linear strain rates are both zero, and the shear strain rate is U/2b.
The vorticity vector has one non-zero component:
ω
z=
∂
v
∂
x−
∂
u
∂
y=−U
b
.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 3.18. The steady two-dimensional flow field inside a sloping passage is given in (x,y)-
Cartesian coordinates by u =
(u,v)=3q4h
( )
1−y h
( )
2
( )
1, y h
( )
dh dx
( )
( )
where q is the volume
flow rate per unit length into the page, and h is the passage’s half thickness. Determine the
streamlines, vorticity, and strain rate tensor in this flow away from x = 0 when h =
α
x where
α
is
a positive constant. Sample profiles of u(x,y) vs. y are shown at two x-locations in the figure.
What are the equations of the streamlines along which the x– and y-axes are aligned with the
principal axes of the flow? What is the fluid particle rotation rate along these streamlines?
Solution 3.18. For planar flow in Cartesian coordinates, the streamlines are determined from:
dy
dx =v
u=y h
( )
dh dx
( )
1=y
α
x
( )
α
=y x
.
Separate the differentials and integrate to find: ln(y) = ln(x) + const. Exponentiate to find: y = Cx,
where C is a constant. Thus, the streamlines are straight lines through the origin of coordinates.
The vorticity
ω
z is determined from:
ω
z=∂v
∂x−∂u
∂y=∂
∂x
3q
4h
#
$
%&
‘
(1−y2
h2
#
$
%&
‘
(y
h
dh
dx
)
*
+,
–
.−∂
∂y
3q
4h
#
$
%&
‘
(1−y2
h2
#
$
%&
‘
(
)
*
+,
–
.
=3q
4
∂
∂x
1
α
x
1−y2
α
2x2
#
$
%&
‘
(y
α
x
α
)
*
+,
–
.−3q
4
∂
∂y
1
α
x
1−y2
α
2x2
#
$
%&
‘
(
)
*
+,
–
.
=3q
4−2y
α
x3+4y3
α
3x5
)
*
+,
–
.−3q
4−2y
α
3x3
)
*
+,
–
.=3qy
2
α
x3−1+2y2
α
2x2+1
α
2
)
*
+,
–
.
The strain-rate tensor Sij is computed from the following velocity derivatives:
S11 =∂u
∂x=∂
∂x
3q
4h
“
#
$%
&
‘1−y2
h2
“
#
$%
&
‘
)
*
+,
–
.=3q
4
∂
∂x
1
α
x
1−y2
α
2x2
“
#
$%
&
‘
)
*
+,
–
.=3q
4−1
α
x2+3y2
α
3x4
)
*
+,
–
.=3q
4
α
x2−1+3y2
α
2x2
)
*
+,
–
.
S22 =∂v
∂y=∂
∂y
3q
4h
“
#
$%
&
‘1−y2
h2
“
#
$%
&
‘y
h
dh
dx
)
*
+,
–
.=3q
4
∂
∂y
1
α
x
1−y2
α
2x2
“
#
$%
&
‘y
α
x
α
)
*
+,
–
.=3q
4
α
x21−3y2
α
2x2
“
#
$%
&
‘
, and
S12 =S21 =1
2
∂u
∂y+∂v
∂x
“
#
$
%
&
‘=1
2
∂
∂y
3q
4h
(
)
*+
,
–1−y2
h2
(
)
*+
,
–
/
0
12
3
4+1
2
∂
∂x
3q
4h
(
)
*+
,
–1−y2
h2
(
)
*+
,
–y
h
dh
dx
/
0
12
3
4=3qy
4
α
x3−1
α
2−1+2y2
α
2x2
/
0
12
3
4
,
where the differentiation details for S12 are available above in the calculation of
ω
z.
The principle axes occur then the off-diagonal strain rate components are zero. This
occurs when
x!
y!
h(x)!
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 3.19. For the flow field
u=U+Ω×x
, where U and Ω are constant linear- and angular-
velocity vectors, use Cartesian coordinates to a) show that Sij is zero, and b) determine Rij.
Solution 3.19. Since no simplifications are given, all the components of U = (U1, U2, U3) and Ω
= (Ω1, Ω2, Ω3) should be treated as being non-zero. In Cartesian coordinates, the velocity field is
u=U+Ω×x=U1e1+U2e2+U3e3+
e1e2e3
Ω1Ω2Ω3
∂
u1
∂
x3
∂
u2
∂
x3
∂
u3
∂
x3
,
/
+Ω2−Ω10
,
/
This result can be used to construct the strain rate tensor Sij:
Sij =1
2
∂
ui
∂
xj
+
∂
uj
∂
xi
#
%
%
&
(
( =1
2
∂
ui
∂
xj
+
∂
ui
∂
xj
#
%
%
&
(
(
T
#
%
%
&
(
( =
01
2−Ω3+Ω3
( )
1
2+Ω2− Ω2
( )
1
2+Ω3− Ω3
( )
01
2−Ω1+Ω1
( )
+
,
–
/
0
–
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 3.20. Starting with a small rectangular volume element
δ
V =
δ
x1
δ
x2
δ
x3, prove (3.14).
Solution 3.20. The volumetric strain rate for a fluid element is:
1
δ
V
D
Dt
δ
V
( )
=1
δ
x1
δ
x2
δ
x3
D
Dt
δ
x1
δ
x2
δ
x3
( )
=1
δ
x1
D
Dt
δ
x1
( )
+1
δ
x2
D
Dt
δ
x2
( )
+1
δ
x3
D
Dt
δ
x3
( )
.
From Section 3.4 in the text, the linear strain rate corresponding to elongation or contraction of a
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 3.21. Let Oxyz be a stationary frame of reference, and let the z-axis be parallel with the
fluid vorticity vector in the vicinity of O so that
ω=∇ × u=
ω
zez
in this frame of reference. Now
consider a second rotating frame of reference
O“
x “
y “
z
having the same origin that rotates about
the z-axis at angular rate Ωez. Starting from the kinematic relationship,
u=(Ωez)×x+$
u
, show
that in the vicinity of O the vorticity
“
ω =“
∇ ד
u
in the rotating frame of reference can only be
zero when 2Ω =
ω
z, where
“
∇
is the gradient operator in the primed coordinates. The following
unit vector transformation rules may be of use:
“
e
x=excos(Ωt)+eysin(Ωt)
,
“
e
y=−exsin(Ωt)+eycos(Ωt)
, and
“
e
z=ez
.
Solution 3.21. The approach here is to compute
“
ω =“
∇ ד
u
in the stationary frame of reference
and then determine the parameter choice(s) necessary for ω´ to be zero. The vector x must have a
consistent representation in either frame, so
x=xex+yey=“
x “
e
x+“
y “
e
y=“
x excos(Ωt)+eysin(Ωt)
( )
+“
y −exsin(Ωt)+eycos(Ωt)
( )
.
Equating components in the stationary frame of reference produces:
x=“
x cos(Ωt)−“
y sin(Ωt)
, and
y=“
x sin(Ωt)+“
y cos(Ωt)
.
The remaining independent variables are the same in either frame:
t=“
t
, and
z=“
z
. Thus, spatial
derivatives are related by:
∂
∂
#
x =
∂
x
∂
#
x
∂
∂
x+
∂
y
∂
#
x
∂
∂
y+
∂
z
∂
#
x
∂
∂
z+
∂
t
∂
#
x
∂
∂
t=cos(Ωt)
∂
∂
x+sin(Ωt)
∂
∂
y
,
∂
∂
#
y =
∂
x
∂
#
y
∂
∂
x+
∂
y
∂
#
y
∂
∂
y+
∂
z
∂
#
y
∂
∂
z+
∂
t
∂
#
y
∂
∂
t=−sin(Ωt)
∂
∂
x+cos(Ωt)
∂
∂
y
, and
∂
∂
#
z =
∂
x
∂
#
z
∂
∂
x+
∂
y
∂
#
z
∂
∂
y+
∂
z
∂
#
z
∂
∂
z+
∂
t
∂
#
z
∂
∂
t=
∂
∂
z
.
Using the above information, the gradient operator in the rotating coordinates can be rewritten in
terms of the stationary frame coordinates and unit vectors:
“
∇ = “
e
x
∂
∂
“
x +“
e
y
∂
∂
“
y +“
e
z
∂
∂
“
z
=excos(Ωt)+eysin(Ωt)
( )
cos(Ωt)
∂
∂
x+sin(Ωt)
∂
∂
y
$
%
&
‘
(
)
+−exsin(Ωt)+eycos(Ωt)
( )
−sin(Ωt)
∂
∂
x+cos(Ωt)
∂
∂
y
%
&
‘
(
)
*
+ez
∂
∂
z
“
∇ = excos2(Ωt)
∂
∂
x+exsin(Ωt)cos(Ωt)
∂
∂
y+eysin(Ωt)cos(Ωt)
∂
∂
x+eysin2(Ωt)
∂
∂
y
&
‘
(
)
*
+
+exsin2(Ωt)
∂
∂
x−exsin(Ωt)cos(Ωt)
∂
∂
y−eysin(Ωt)cos(Ωt)
∂
∂
x+eycos2(Ωt)
∂
∂
y
%
&
‘
(
)
* +ez
∂
∂
z
“
∇ = excos2(Ωt)
∂
∂
x+exsin(Ωt)cos(Ωt)
∂
∂
y+eysin(Ωt)cos(Ωt)
∂
∂
x+eysin2(Ωt)
∂
∂
y
&
‘
(
)
*
+
+exsin2(Ωt)
∂
∂
x−exsin(Ωt)cos(Ωt)
∂
∂
y−eysin(Ωt)cos(Ωt)
∂
∂
x+eycos2(Ωt)
∂
∂
y
%
&
‘
(
)
* +ez
∂
∂
z
“
∇ = ex
∂
∂
x+ey
∂
∂
y+ez
∂
∂
z
.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 3.22. Consider a plane-polar area element having dimensions dr and rd
θ
. For two-
dimensional flow in this plane, evaluate the right-hand side of Stokes’ theorem
ω⋅ndA
∫=u⋅ds
∫
and thereby show that the expression for vorticity in plane-polar coordinates
is:
ω
z=1
r
∂
∂
r
ru
θ
( )
−1
r
∂
ur
∂θ
.
Solution 3.22. Using the element shown with angular width d
θ
,
ω⋅ndA
∫=u⋅ds
∫→
ω
zrd
θ
dr =urdr +u
θ
+
∂
u
θ
∂
r
dr
(
)
* +
,
– r+dr
( )
d
θ
−ur+
∂
ur
∂θ
d
θ
(
)
* +
,
–
dr −u
θ
rd
θ
, or
ω
zrd
θ
dr =urdr +ru
θ
d
θ
+r
∂
u
θ
∂
r
drd
θ
+u
θ
drd
θ
−urdr −
∂
ur
∂θ
d
θ
dr −u
θ
rd
θ
+…
=r
∂
u
θ
∂
r
drd
θ
+u
θ
drd
θ
−
∂
ur
∂θ
d
θ
dr +…
where + … indicates the presence of higher order terms in dr and d
θ
. Division by rd
θ
dr and
passing to the limit where dr and d
θ
go to zero produces:
ω
z=
∂
u
θ
∂
r+1
r
u
θ
−1
r
∂
ur
∂θ
=1
r
∂
∂θ
ru
θ
( )
−
∂
ur
∂θ
&
‘
( )
*
+
.
ur
u!
u! + (!u!/!”)dr
u
r
+ (!u
r
/!
!
)d
!
dr
!
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 3.23. The velocity field of a certain flow is given by
u=2xy2+2xz2
,
v=x2y
, and
w=x2z
. Consider the fluid region inside a spherical volume x2 + y2 + z2 = a2. Verify the validity
of Gauss’ theorem
∇ ⋅ udV =u⋅ndA
A
∫∫
V
∫∫∫
by integrating over the sphere.
Solution 3.23. First compute the divergence of the velocity field.
∇ ⋅ u=
∂
u
∂
x+
∂
v
∂
y+
∂
w
∂
z=
∂
∂
x
2xy2+2xz2
( )
+
∂
∂
yx2y
( )
+
∂
∂
zx2z
( )
=2y2+2z2+2x2=2r2
.
The volume integral of
∇ ⋅ u
is:
∇ ⋅ udV =2r2
( )
4
π
r2dr =
r=0
r=a
∫
V
∫∫∫ 8
5
π
a5
.
Now work on the surface integration using spherical coordinates (see Figure 3.3d, and Appendix
B). Here,
n=er=excos
ϕ
sin
θ
+eysin
ϕ
sin
θ
+excos
θ
, so
u⋅n=ucos
ϕ
sin
θ
+vsin
ϕ
sin
θ
+wcos
θ
=2xy2+2xz2
( )
cos
ϕ
sin
θ
+x2y
( )
sin
ϕ
sin
θ
+x2z
( )
cos
θ
Unfortunately, this result is in mixed variables so convert everything to spherical polar
coordinates using
x=rcos
ϕ
sin
θ
,
y=rsin
ϕ
sin
θ
,
z=rcos
θ
.
This conversion produces:
u⋅n=2r3cos
ϕ
sin2
ϕ
sin3
θ
+cos
ϕ
sin
θ
cos2
θ
( )
cos
ϕ
sin
θ
+r3cos2
ϕ
sin
ϕ
sin3
θ
( )
sin
ϕ
sin
θ
+r3cos2
ϕ
sin2
θ
cos
θ
( )
cos
θ
=2r3sin2
ϕ
sin2
θ
+cos2
θ
( )
cos2
ϕ
sin2
θ
+r3sin2
ϕ
sin2
θ
( )
cos2
ϕ
sin2
θ
+r3cos2
θ
( )
cos2
ϕ
sin2
θ
=r33sin2
ϕ
sin2
θ
+3cos2
θ
( )
cos2
ϕ
sin2
θ
=r33
4sin2(2
ϕ
) 1−cos2
θ
( )
2+3cos2
ϕ
cos2
θ
1−cos2
θ
( )
( )
So, the surface integral produces:
u⋅ndA
A
∫∫ =a33
4sin2(2
ϕ
) 1−cos2
θ
( )
2+3cos2
ϕ
cos2
θ
1−cos2
θ
( )
( )
θ
=0
θ
=
π
∫
ϕ
=0
ϕ
=2
π
∫a2sin
θ
d
θ
d
ϕ
=a53
4sin2(2
ϕ
)d
ϕ
1−cos2
θ
( )
2
θ
=
π
∫
ϕ
=2
π
∫sin
θ
d
θ
+a53cos2
ϕ
d
ϕ
cos2
θ
1−cos2
θ
( )
θ
=
π
∫
ϕ
=2
π
∫sin
θ
d
θ
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 3.24. A flow field on the xy–plane has the velocity components u = 3x + y and v = 2x –
3y. Show that the circulation around the circle (x − 1)2 + (y − 6)2 = 4 is 4
π
.
Solution 3.24. The circle is centered at (1,6) and its radius is 2. So, if (x,y) is a point on the circle
then x = 1 + 2cos
θ
, and y = 6 + 2sin
θ
, where
θ
is the angle from the horizontal. The velocity
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 3.25. Consider solid-body rotation about the origin in two dimensions: ur = 0 and u
θ
=
ω
0r. Use a polar-coordinate element of dimension rd
θ
and dr, and verify that the circulation is
vorticity times area. (In Section 5 this was verified for a circular element surrounding the origin.)
Solution 3.25. Using the element shown with angular width d
θ
,
the circulation is around the element is the sum or four terms:
Γ=u⋅ds
∫=urdr +u
θ
+
∂
u
θ
∂
r
dr
‘
(
) *
+
, r+dr
( )
d
θ
−ur+
∂
ur
∂θ
d
θ
‘
(
) *
+
,
dr −u
θ
rd
θ
.
Now substitute in the velocity field: ur = 0 and u
θ
=
ω
0r.
Γ=
ω
0r+
ω
0dr
( )
r+dr
( )
d
θ
−
ω
0r2d
θ
=2
ω
0rdrd
θ
+
ω
0(dr)2d
θ
=(2
ω
0)rdrd
θ
,
where the final equality holds in the limit as the differential elements become small and (dr)2d
θ
is negligible compared to rdrd
θ
. Thus, since the voriticity is 2
ω
0 and rdrd
θ
is the area of the
element, the final relationship states: circulation = (vorticity) x (area).
ur
u!
u! + (!u!/!”)dr
ur + (!ur/!!)d!
dr
!