Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 3.26. Consider the steady Cartesian velocity field
u=−Ay
x2+y2
( )
β
,+Ax
x2+y2
( )
β
,0
$
%
&
&
‘
(
)
)
.
a) Determine the streamline that passes through
x=(xo,yo,0)
b) Compute Rij for this velocity field.
c) For A > 0, explain the sense of rotation (i.e. clockwise or counter clockwise) for fluid elements
for
β
< 1,
β
= 1, and
β
> 1.
Solution 3.26. a) Use the definition of a streamline:
dy
dx =v
u=Ax x2+y2
( )
β
−Ay x2+y2
( )
β
=−x
y
. Use the
two ends of this extended equality to find:
ydy =−xdx
, and integrate the resulting differential
relationship to get:
y22=−x22+const
. Evaluate the constant using the required condition:
x2+y2=xo
2+yo
2
. This is a circle with radius
xo
2+yo
2
. Therefore the streamlines are circles.
b) The rotation tensor is:
, where the
second equality comes from putting the specified velocity field into the definition of Rij with
u=(u,v,w)=(u1,u2,u3)=ui
as usual. Evaluating the derivatives produces:
Rij =A
x2+y2
( )
β
+1
0−x2−y2+2
β
y2−x2+y2−2
β
x2
( )
0
x2+y2−2
β
x2− −x2−y2+2
β
y2
( )
0 0
0 0 0
“
#
$
$
%
$
$
&
‘
$
$
(
$
$
=2A(1−
β
)
x2+y2
( )
β
0−1 0
1 0 0
0 0 0
“
#
$
%
$
&
‘
$
(
$
c) The following answers are based on A > 0. For
β
< 1, fluid particles rotate counter clockwise
(positive
ω
z). For
β
= 1, fluid particles do not rotate. For
β
> 1, fluid particles rotate clockwise
(negative
ω
z). Interestingly, the streamlines are the same (circular!) for all three possibilities.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 3.27. Using indicial notation (and no vector identities) show that the acceleration a of a
fluid particle is given by:
a=
∂
u
∂
t+∇1
2u2
( )
+ω×u
where ω is the vorticity.
Solution 3.27. The acceleration of a fluid particle is
a=DuDt ≡ ∂u∂t+u⋅ ∇
( )
u
. Thus, the task
is to prove
u⋅ ∇
( )
u=∇1
2u2
( )
+ω×u
. Start from the advective acceleration written in index
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 3.28. Starting from (3.29), show that the maximum u
θ
in a Gaussian vortex occurs
when
1+2(r2
σ
2)=exp(r2
σ
2)
. Verify that this implies r ≈ 1.12091
σ
.
Solution 3.28. Differentiate the u
θ
equation from (3.29) with respect to r and set this derivative
equal to zero.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 3.29. Using (3.35) in two dimensions with F = 1, show that the time-rate-of-change of
the area of the parallelogram shown is hl(d
θ
/dt)cos
θ
when
θ
depends on time while h and l are
constants.
Solution 3.29. With F = 1, the volume integral of ∂F/∂t on the right side of (3.35) is zero, and
the integrands are simplified, so (3.35) simplifies to:
Here we note that the left and right sides move identically, so b will be the same. However, n
points in opposite directions on these two sides, so the contributions from these two sides cancel.
The x-coordinates of points on the parallelogram’s top side are hcos
θ
≤ x(t) ≤ l + hcos
θ
. The y–
coordinate of the parallelogram’s top side is y(t) = hsin
θ
. Time differentiate the location of any
x!
y!
l!
h!
θ
(t)!
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 3.30. Using (3.35) in two dimensions with F = 1, show that the time-rate-of-change of
the area of the triangle shown is
1
2b dh dt
( )
when h depends on time and b is constant.
Solution 3.30. With F = 1, the volume integral of ∂F/∂t on the right side of (3.35) is zero, and
the integrands are simplified, so (3.35) simplifies to:
d
dt
dV
V*(t)
∫= + b⋅ndA
A*(t)
∫
.
(Here the volume integration is two dimensional, and produces the triangle’s area, hb/2, which
can be time differentiated, (d/dt)(hb/2) to reach the desired result. However, this is not the
intended solution path for this problem.)
When h is time-dependent, the triangle has two moving sides (a top side of length b, and
a hypotenuse of length [h2 + b2]1/2). Thus, the simplified version of (3.35) reduces to:
d
dt dV
V*(t)
∫
= +
top si de
∫hypotenuse
∫
{ }
b⋅ndA
.
The x-coordinates of the triangle’s top side are 0 ≤ x ≤ b. The y-coordinate of the triangle’s top
side is y(t) = h(t). Time differentiate the location of any point on the triangles’s top side to find b:
b = (dx/dt, dy/dt) = (0, dh/dt).
And, on the triangle’s top side, n is ey, so
b⋅n=dh dt
,
The x–y coordinates of the triangle’s hypotenuse fall on the line y = (h/b)x for 0 ≤ x ≤ b.
The motion of points on the hypotenuse can be described by a purely vertical velocity. So, for
any constant x-location, differentiate the equation of this line to find b:
b = (dx/dt, dy/dt) = (0, (x/b)dh/dt).
Tangent and normal vectors to an x-y curve are (1, dy/dx) and (dy/dx, –1), respectively. Thus, the
outward unit normal on the hypotenuse of the triangle is:
n=h b,−1
( )
(h/b)2+1
=(h,−b)
h2+b2
, so
b⋅n=−x dh dt
( )
h2+b2
.
Thus, the simplified version of (3.35) can be written:
d
dt dV
V*(t)
∫=b⋅n
top si de
∫dA +b⋅n
hypotenuse
∫dA =dh
dt
0
b
∫dx −x dh dt
( )
h2+b2
0
b
∫1+(h/b)2dx
$
%&
‘
,
where, in two dimensions, dA is just dx on the top side and dA is a path length element along the
hypotenuse, ds = [1 + (dy/dx)2]1/2dx. The factor in [,]-brackets is this path length element in terms
of dx. Perform the integrations to find:
d
dt
dV
V*(t)
∫=bdh
dt −1
b
dh
dt
x
2
2
#
$
%&
‘
(
0
b
=b
2
dh
dt
,
and this is the desired result.
x!
y!
b!
h(t)!
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 3.31. Using (3.35) in two dimensions with F = 1, show that the time-rate-of-change of
the area of the ellipse shown is
π
b da dt
( )
when a depends on time and b is constant.
Solution 3.31. With F = 1, the volume integral of ∂F/∂t on the right side of (3.35) is zero, and
the integrands are simplified, so (3.35) simplifies to:
d
dt
dV
V*(t)
∫= + b⋅ndA
A*(t)
∫
.
(Here the volume integration is two dimensional, and produces the ellipse’s area,
π
ab, which can
be time differentiated, (d/dt)(
π
ab) to reach the desired result. However, this is not the intended
solution path for this problem.)
When a is time-dependent, the contour that defines the ellipse, (y/b)2 + (x/a)2 = 1, is also
time dependent. However, the symmetry of the ellipse allows the analysis to completed in the
first quadrant alone and then multiplied by 4. Thus, the simplified version of (3.35) reduces to:
d
dt dV
V*(t)
∫=4b⋅n
first qua drant
∫dA
.
The equation for the ellipse in the first quadrant is:
x= +a1−y b
( )
2
or
y= +b1−x a
( )
2
.
The motion of points on this curve can be described by a purely horizontal velocity when a
varies but b is constant. So, for any constant y-location, differentiate the first equation to find b:
b=dx
dt , 0
!
“
#$
%
&=da
dt 1−y b
( )
2, 0
!
“
#$
%
&=x
a
da
dt , 0
!
“
#$
%
&
,
where the final equality comes from changing the independent variable from y to x.
Tangent and normal vectors to an x-y curve are (1, dy/dx) and (–dy/dx, 1), respectively.
Thus, the outward unit normal on the ellipse in the first quadrant is:
n=−dy /dx,1
( )
(dy /dx)2+1
, so
b⋅n=x(−dy /dx)
a(dy /dx)2+1
da
dt
.
Thus, the simplified version of (3.35) can be written:
d
dt
dV
V*(t)
∫=4b⋅n
first qua drant
∫dA =4x(−dy /dx)
a(dy /dx)2+1
da
dt
0
a
∫1+(dy /dx)2dx
$
%&
‘
,
where, in two dimensions, dA is a path length element, ds = [1 + (dy/dx)2]1/2dx, along the first-
quadrant portion of the ellipse. This path length element is the factor in [,]-brackets above.
x!
y!
b!
a(t)!
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 3.32. For the following time-dependent volumes V*(t) and smooth single-valued
integrand functions F, choose an appropriate coordinate system and show that
d dt
( )
FdV
V*(t)
∫
obtained from (3.30) is equal to that obtained from (3.35).
a) V*(t) = L1(t)L2L3 is a rectangular solid defined by 0 ≤ xi ≤ Li, where L1 depends on time while
L2 and L3 are constants, and the integrand function F(x1,t) depends only on the first coordinate
and time.
b) V*(t) = (
π
/4)d2(t)L is a cylinder defined by 0 ≤ R ≤ d(t)/2 and 0 ≤ z ≤ L, where the cylinder’s
diameter d depends on time while its length L is constant, and the integrand function F(R,t)
depends only on the distance from the cylinder’s axis and time.
c) V*(t) = (
π
/6)D3(t) is a sphere defined by 0 ≤ r ≤ D(t)/2 where the sphere’s diameter D depends
on time, and the integrand function F(r,t) depends only on the radial distance from the center of
the sphere and time.
Solution 3.32. The two equations are (3.30)
d
dt
F(x,t)dx
x=a(t)
x=b(t)
∫=∂F
∂t
dx
a
b
∫+db
dt
F b,t
( )
−da
dt
F a,t
( )
,
and (3.35)
d
dt
F(x,t)dV
∫=
∂
F(x,t)
∂
t
dV
∫+F(x,t)b⋅ndA
∫
.
Now evaluate the right side terms from (3.35).
d
dt
F(x,t)dV
V*(t)
∫=
∂
F(x1,t)
∂
t
x3=0
L3
∫
x2=0
L2
∫dx1dx2dx3
x1=0
L1(t)
∫+F(L1,t)
x3=0
L3
∫
x2=0
L2
∫dL1
dt e1
$
%
& ‘
(
)
⋅e1dx2dx3
=L2L3
∂
F(x1,t)
∂
t
dx1
x1=0
L1(t)
∫+L2L3
dL1
dt
F(L1,t)
Setting the left and right side terms equal produces
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Now start from (3.35) using a control volume that encloses the cylinder. In this case the only
control surface that moves lies at R = d/2, has outward normal n = eR, and moves with velocity b
= (d(d/2)/dt)eR. First evaluate the left side term from (3.35).
d
dt
F(x,t)dV
V*(t)
∫=d
dt
F(R,t)
ϕ
=0
2
π
∫
R=0
d/ 2
∫Rd
ϕ
dRdz
z=0
L
∫=2
π
Ld
dt
F(R,t)RdR
0
d/ 2
∫
.
Now evaluate the right side terms from (3.35).
d
dt
F(x,t)dV
V*(t)
∫=
∂
F(R,t)
∂
t
ϕ
=0
2
π
∫
R=0
d/ 2
∫Rd
ϕ
dRdz
z=0
L
∫+F(d/2,t)
ϕ
=0
2
π
∫
z=0
L
∫d(d/2)
dt eR
&
‘
( )
*
+
⋅eR
d
2
d
ϕ
dz
=2
π
L
∂
F(R,t)
∂
t
RdR
R=0
d/ 2
∫+2
π
Ld(d/2)
dt
F(d/2,t)d
2
Setting the left and right side terms equal produces
2
π
Ld
dt
F(R,t)RdR
0
d/ 2
∫=2
π
L
∂
F(R,t)
∂
t
RdR
R=0
d/ 2
∫+2
π
Ld(d/2)
dt
F(d/2,t)d
2
, (b2)
which is identical to (b1).
c) Use spherical coordinates with the origin at r = 0. The sphere expands symmetrically so the
volume element is dV = 4
π
r2dr. The volume integral proceeds from r = 0 (= a) to r = D(t)/2 (= b)
so (3.30) implies:
d
dt
F(x,t)dV
V*(t)
∫=4
π
d
dt
F(r,t)r2dr
r=0
D(t)/ 2
∫= 4
π
∂F
∂t
r2dr
0
D/ 2
∫+4
π
d D 2
( )
dt
F D /2,t
( )
D
2
%
&
‘ (
)
*
2
. (c1)
Now start from (3.35) using a control volume that encloses the sphere. In this case the moving
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 3.33. Starting from (3.35), set F = 1 and derive (3.14) when b = u and V*(t) =
δ
V
→0
.
Solution 3.33. With F = 1, b = u, and V*(t) =
δ
V with surface
δ
A, (3.35) becomes:
d
dt
dV
δ
V
∫=0+u⋅ndA
δ
A
∫
.
The first integral is merely
δ
V. Use Gauss’ divergence theorem on the second term to convert it
to volume integral.
d
dt
δ
V
( )
=∇ ⋅ udV
δ
V
∫
.
As
δ
V
→0
the integral reduces to a product of
δ
V and the integrand evaluated at the center point
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 3.34. For a smooth single valued function F(x) that only depends on space and an
arbitrarily-shaped control volume that moves with velocity b(t) that only depends on time, show
that
d dt
( )
F(x)dV
V*(t)
∫=b⋅ ∇F(x)dV
V*(t)
∫
( )
.
Solution 3.34. Start from Reynolds transport theorem (3.35):
d
dt
F(x,t)dV
V*(t)
∫=
∂
F(x,t)
∂
t
dV
V*(t)
∫+F(x,t)b⋅ndA
A*(t)
∫
.
When F does not depend on time, the first term on the right drops out.
d
dt
F(x,t)dV
V*(t)
∫=F(x,t)b⋅ndA
A*(t)
∫
.
When b does not depend on location, it can be taken outside the surface integral.
d
dt
F(x,t)dV
V*(t)
∫=b⋅F(x,t)ndA
A*(t)
∫
$
%
&
‘
(
)
.
Apply Gauss’ theorem to the integral in large parentheses, to reach the desired form:
d
dt
F(x,t)dV
V*(t)
∫=b⋅ ∇F(x,t)dA
V*(t)
∫
%
&
‘
(
)
*
.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
3.35. Show that (3.35) reduces to (3.5) when V*(t) =
δ
V
→0
and the control surface velocity b is
equal to the fluid velocity u(x,t).
Solution 3.35. When V*(t) =
δ
V with surface
δ
A,
δ
V is small, and b = u,
δ
V represents a fluid
particle. Under these conditions (3.35) becomes:
d
dt
F(x,t)dV
δ
V
∫=
∂
F(x,t)
∂
t
dV
δ
V
∫+F(x,t)u⋅ndA
δ
A
∫
,
and the time derivative is evaluated following
δ
V. Use Gauss’ divergence theorem on the final
term to convert it to a volume integral,
F(x,t)u⋅ndA
δ
A
∫=∇ ⋅ F(x,t)u
( )
δ
V
∫dV
,
so that (3.35) becomes:
d
dt
F(x,t)dV
δ
V
∫=
∂
F(x,t)
∂
t+∇ ⋅ F(x,t)u
( )
‘
(
)
*
+
,
dV
δ
V
∫=
∂
F(x,t)
∂
t+F(x,t)∇ ⋅ u+u⋅ ∇
( )
F(x,t)
‘
(
)
*
+
,
dV
δ
V
∫
,
where the second equality follows from expanding the divergence of the product Fu.
As
δ
V
→0
the various integrals reduce to a product of
δ
V and the integrand evaluated at
the center point of
δ
V. Divide both sides of the prior equation by
δ
V and take the limit as
δ
V
→0
to find:
lim
δ
V→0
1
δ
V
d
dt
F(x,t)dV
δ
V
∫=lim
δ
V→0
1
δ
V
∂
F(x,t)
∂
t+F(x,t)∇ ⋅ u+u⋅ ∇
( )
F(x,t)
(
)
*
+
,
–
dV
δ
V
∫
,
lim
δ
V→0
1
δ
V
d
dt
F(x,t)
δ
V+…
[ ]
=lim
δ
V→0
1
δ
V
∂
F(x,t)
∂
t+F(x,t)∇ ⋅ u+u⋅ ∇
( )
F(x,t)
‘
(
) *
+
,
δ
V+…
–
.
/
0
1
2
, or
d
dt
F(x,t)+F(x,t)lim
δ
V→0
1
δ
V
d
dt
δ
V
( )
=
∂
F(x,t)
∂
t+F(x,t)∇ ⋅ u+u⋅ ∇
( )
F(x,t)
,
where the product rule for derivative has been used on product F
δ
V in [,]-braces on the left.
From (3.14) or Exercise 3.33:
lim
δ
V→0
1
δ
V
d
dt
δ
V
( )
=∇ ⋅ u
, so the second terms on both sides of the
last equation are equal and may be subtracted out leaving:
d
dt
F(x,t)=
∂
F(x,t)
∂
t+u⋅ ∇
( )
F(x,t)
,
and this is (3.5) when the identification
DDt ≡d dt
is made.