Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
where
τ
rr = 2
µ
(∂ur/∂r) is the normal viscous stress, and the sign of the surface tension term is set
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 4.71. Redo the dimensionless scaling leading to (4.101) by choosing a generic viscous
stress,
µ
U/l, and then a generic hydrostatic pressure,
ρ
gl, to make p − p∞ dimensionless. Interpret
the revised dimensionless coefficients that appear in the scaled momentum equation, and relate
them to St, Re, and Fr.
Solution 4.71. (i) Viscous scaling of the pressure.
Start from
ρ∂
u
∂
t+u⋅ ∇
( )
u
&
‘
( )
*
+ =−∇p+
ρ
g+
µ
∇2u
, and render it dimensionless using the
dimensionless variables:
xi
∗=xil
,
t∗=Ωt
,
,
p∗=(p−p∞)l
µ
U
, and
gj
∗=gjg
.
The result is:
UΩ
∂
u*
∂
t*+U2
lu*⋅ ∇*
( )
u*=−1
ρ
l
µ
U
l∇*p*+gg*+
µ
U
ρ
l2∇*2u*
,
which can be simplied by dividing all terms by
µ
U/
ρ
l2 and using
ν
=
µ
/
ρ
:
Ωl
U
Ul
ν
$
%
& ‘
(
)
∂
u*
∂
t*+Ul
ν
$
%
& ‘
(
) u*⋅ ∇*
( )
u*=−∇*p*+gl
U2
Ul
ν
$
%
& ‘
(
)
g*+∇*2u*
.
Here the coefficients are the Strouhal number-Reynolds number product, the Reynolds number,
and the product of the inverse square of the Froude number and the Reynolds number. This
scaling is useful for low Reynolds number flow where it naturally produces a pressure gradient-
viscous force balance.
(ii) Hydrostatic scaling of the pressure.
Start from
ρ∂
u
∂
t+u⋅ ∇
( )
u
&
‘
( )
*
+ =−∇p+
ρ
g+
µ
∇2u
, and render it dimensionless using the
dimensionless variables:
xi
∗=xil
,
t∗=Ωt
,
,
p∗=(p−p∞)
ρ
gl
, and
gj
∗=gjg
.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 4.72. A solid sphere of mass m and diameter D is released from rest and falls through
an incompressible viscous fluid with density
ρ
and viscosity
µ
under the action of gravity g.
When the z coordinate increases downward, the vertical component of Newton’s second law for
the sphere is:
mduz
dt = +mg −FB−FD
, where uz is positive downward, FB is the buoyancy force on
the sphere, and FD is the fluid-dynamic drag force on the sphere. Here, with uz > 0, FD opposes
the sphere’s downward motion. At first the sphere is moving slowly so its Reynolds number is
low, but
ReD=
ρ
uzD
µ
increases with time as the sphere’s velocity increases. To account for
this variation in ReD, the sphere’s coefficient of drag may be approximated as:
CD≅1
2+24 ReD
.
For the following items, provide answers in terms of m,
ρ
,
µ
, g and D; do not use z, uz, FB, or FD.
a) Assume the sphere’s vertical equation of motion will be solved by a computer after being put
into dimensionless form. Therefore, use the information provided and the definition
t*≡
ρ
gtD
µ
to show that this equation may be rewritten:
dReD
dt*=AReD
2+BReD+C
, and determine the
coefficients A, B, and C.
b) Solve the part a) equation for ReD analytically in terms of A, B, and C for a sphere that is
initially at rest.
c) Undo the dimensionless scaling to determine the terminal velocity of the sphere from the part
c) answer as
t→ ∞
.
Solution 4.72. a) The buoyant and drag forces on the sphere will be:
FB=
ρπ
6D3g
and
FD=1
2
ρ
uz
2
π
4
D2CD=1
2
ρ
uz
2
π
4
D21
2+24
ReD
!
“
#$
%
&
.
Thus, the approximate equation for the sphere’s velocity is:
D!
µ
,
ρ
!
z!
g!
m!
vz!
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
m
µ
ρ
D
dReD
dt =m−
ρπ
6D3
“
#
$%
&
‘g−
π
8
ρµ
2ReD
21
2+24
ReD
“
#
$%
&
‘
.
Switch to the dimensionless time t* using the chain rule for differentiation and
t*=
ρ
gtD
µ
:
m
µ
ρ
D
dReD
dt =m
µ
ρ
D
dReD
dt*
dt*
dt
!
“
#$
%
&=m
µ
ρ
D
dReD
dt*
ρ
gD
µ
!
“
#$
%
&
=mg dReD
dt*=m−
ρπ
6D3
!
“
#$
%
&g−
π
8
ρµ
2ReD
21
2+24
ReD
!
“
#$
%
&.
Divide both sides of the final equality by mg, and rearrange the right side:
dReD
dt*=−
π
16
µ
2
ρ
mg
“
#
$%
&
‘ReD
2−3
πµ
2
ρ
mg
“
#
$%
&
‘ReD+1−
π
6
ρ
D3
m
“
#
$%
&
‘
.
Thus:
A=−
π
16
µ
2
ρ
mg
“
#
$%
&
‘
,
B=−3
πµ
2
ρ
mg
“
#
$%
&
‘
, and
C=1−
π
6
ρ
D3
m
.
b) Assuming C is positive, the scaled equation found for part a) can be separated and integrated:
dReD
AReD
2+BReD+C
∫=dt*
∫
, or
t∗+const.=−2
B2−4AC
tanh−12AReD+B
B2−4AC
#
$
%
&
‘
(
.
This integrated result can be algebraically inverted:
ReD=1
2A−B+B2−4AC tanh −1
2
B2−4AC t∗+const.
( )
#
$
%&
‘
(
)
*
+
,
–
.
.
The constant can be evaluated with the initial condition ReD = 0 at t = 0 to find:
ReD=1
2A−B+B2−4AC tanh −B2−4AC
2
t∗+tanh−1B
B2−4AC
#
$
%&
‘
(
)
*
+
+
,
–
.
.
/
0
1
2
1
3
4
1
5
1
,
which is a formal solution to the problem.
c) Terminal velocity will occur when dReD/dt* = 0. Use this fact and the result of part a) to find:
0=−
π
16
µ
2
ρ
mg
“
#
$%
&
‘ReT
2−3
πµ
2
ρ
mg
“
#
$%
&
‘ReT+1−
π
6
ρ
D3
m
“
#
$%
&
‘
,
where ReT is the Reynolds number at terminal velocity. Divide by the leading coefficient:
0=ReT
2+48 ReT−16
π
ρ
mg
µ
2
“
#
$%
&
‘1−
π
6
ρ
D3
m
“
#
$%
&
‘
.
Thus, using the quadratic formula and choosing the appropriate sign to ensure ReT > 0, yields:
ReT=1
2−48 +482−4(1) −16
π
ρ
mg
µ
2
“
#
$%
&
‘1−
π
6
ρ
D3
m
“
#
$%
&
‘
“
#
$%
&
‘
“
#
$
$
%
&
‘
‘=−24 +242+16
π
ρ
mg
µ
2
“
#
$%
&
‘1−
π
6
ρ
D3
m
“
#
$%
&
‘
.
so:
(uz)T=
µ
−B
“
#
$%
&
‘
2
−C
“
#
$
$
%
&
‘
‘=
µ
ρ
mg
2
“
$%
‘1−
π
ρ
D3
“
$%
‘
“
$
$
%
‘
‘
.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 4.73. a) From (4.101), what is the dimensional differential momentum equation for
steady incompressible viscous flow as
Re → ∞
when g = 0.
b) Repeat part a) for
Re →0
. Does this equation include the pressure gradient?
c) Given that pressure gradients are important for fluid mechanics at low Re, revise the pressure
scaling in (4.100) to obtain a more satisfactory low-Re limit for (4.39b) with g = 0.
Solution 4.73. Equation (4.101) with g = 0 is a scaled equation that involves the Strouhal
number St = Ωl/U, and the Reynolds number Re =
ρ
Ul/
µ
.
St
∂
u∗
∂
t∗+u∗⋅ ∇*
( )
u∗=−∇*p∗+1
Re ∇∗2u∗
.
a) When
Re → ∞
, the viscous terms drop out. In dimensional terms, the result is the Euler
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 4.74. a) Simplify (4.45) for motion of a constant-density inviscid fluid observed in a
frame of reference that does not translate but does rotate at a constant rate Ω = Ωez.
b) Use length, velocity, acceleration, rotation, and density scales of L, U, g, Ω, and
ρ
to
determine the dimensionless parameters for this flow when g = –gez and x = (x, y, z). (Hint:
subtract out the static pressure distribution.)
c) The Rossby number, Ro, in this situation is U/ΩL. What are the simplified equations of
motion for horizontal constant-density inviscid flow, u = (u, v, 0), observed in the rotating frame
of reference when Ro << 1.
Solution 4.74. a) For no translation, a constant rotation rate, and inviscid flow, (4.45) simplifies
to:
ρ
∂u
∂t+(u⋅ ∇)u
$
%
&‘
(
)=−∇p+
ρ
g−
ρ
2Ω×u+Ω×Ω×x
( )
–
./
0
,
where the primes have been dropped. When Ω = Ωez, g = –gez, and x = (x, y, z), several terms in
the equation can be evaluated:
∂u
∂t+(u⋅ ∇)u=−1
ρ
∇p−gez−2−Ωvex+Ωuey
( )
+Ω2xex+yey
( )
,
where u = (u, v, w) and whole equation has been divided by
ρ
.
b) The primary point of difficulty here is figuring out how to scale the pressure. One way to
begin is to set u = 0 in the momentum equation, to determine the equation for ps, the static
pressure distribution.
0=−1
ρ
∇ps−gez+Ω2xex+yey
( )
Subtract this equation from the full momentum equation to find:
∂u
∂t+(u⋅ ∇)u=−1
ρ
∇(p−ps)−2−Ωvex+Ωuey
( )
.
Now continue the scaling task using p – ps. For this equation, use the following definitions of
dimensionless variables:
xi
∗=xiL
,
t∗=Ωt
,
uj
∗=ujU
, and
p∗=(p−ps)
ρ
ΩUL
.
This scaling for the pressure should apply when both fluid motion and rotation of the coordinate
frame play a role. Inserting these into the revised version of (4.45) produces:
ΩU∂u∗
∂t∗+U2
L(u∗⋅ ∇∗)u∗=−
ρ
ΩUL
ρ
L∇∗p∗−2ΩU−v∗ex+u∗ey
( )
.
Divide the whole equation by ΩU to find:
∂u∗
∂t∗+U
ΩL(u∗⋅ ∇∗)u∗=−∇∗p∗−2−v∗ex+u∗ey
( )
.
Thus, the dimensionless parameter that governs this flow is U/ΩL.
c) When U/ΩL = Ro << 1, the non-linear advective accleration terms may be dropped, and this
leaves:
∂u
∂t=−1
ρ
∂p
∂x+2Ωv
and
∂v
∂t=−1
ρ
∂p
∂y−2Ωu
.
These equations are commonly used in the study of large-scale atmospheric flows (see Ch. 13).
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 4.75. From Figure 4.23, it can be seen that CD ∝ 1/Re at small Reynolds numbers and
that CD is approximately constant at large Reynolds numbers. Redo the dimensional analysis
leading to (4.99) to verify these observations when:
a) Re is low and fluid inertia is unimportant so
ρ
is no longer a parameter.
b) Re is high and the drag force is dominated by fore-aft pressure differences on the sphere and
µ
is no longer a parameter.
Solution 4.75. Overall there are 5 parameters D = drag force, U, d,
ρ
, and
µ
.
a) When
ρ
is not a parameter, the dimensional analysis proceeds a follows. The parameter &
units matrix is:
D U d
µ
M 1 0 0 1
where the 2, 4, and
π
have been absorbed into the undetermined constant.
b) When
µ
is not a parameter, the dimensional analysis proceeds a follows. The parameter &
units matrix is:
D U d
ρ
M 1 0 0 1
L 1 1 1 -3
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 4.76. Suppose that the power to drive a propeller of an airplane depends on d (diameter
of the propeller), U (free-stream velocity),
ω
(angular velocity of propeller), c (velocity of
sound),
ρ
(density of fluid), and
µ
(viscosity). Find the dimensionless groups. In your opinion,
which of these are the most important and should be duplicated in a model testing?
Solution 4.76. Start with dimensional analysis. The using P = power, the parameter & units
matrix is:
P U d
ω
ρ µ
c
M 1 0 0 0 1 1 0
L 2 1 1 0 -3 -1 1
At low speeds the Mach number should not matter so, but the Reynolds number will
determine boundary layer characteristics, and the Strouhal number will set the effective angle of
attack along the propeller’s blades; therefore:
P
ρ
U3d2=f
ρ
Ud
µ
,
ω
d
U
$
%
&
‘
(
)
for
U
c<< 1
.
At higher speeds where the impact of compressibility is felt, vicous effects may be irrelevant, so:
P
ρ
U3d2=f
ω
d
U,U
c
$
%
& ‘
(
)
for
U
c
near unity or higher.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 4.77. A 1/25 scale model of a submarine is being tested in a wind tunnel in which p =
200 kPa and T = 300 K. If the prototype speed is 30 km/hr, what should be the free-stream
velocity in the wind tunnel? What is the drag ratio? Assume that the submarine would not
operate near the free surface of the ocean.
Solution 4.77. Since submarine are not designed for extended operations on the sea surface,
gravity is unimportant. Therefore the coefficient of drag depends only on the Reynolds number
which should be duplicated in model test. Equating Reynolds numbers between the model
(subscript “m“) and the full-size submarine (subscript “s“) produces:
UmLm
ν
m
=UsLs
ν
s
, or
Um=Us
Ls
Lm
ν
m
ν
s
.
Using the numbers from the problem statement leads to:
ρ
m=pmRmTm=200,000Pa (287m2s−2K)(300K)=2.32kgm−3
ν
m=
µ
m
ρ
m
=1.85 ×10−5kgm−1s−1
2.32kgm−3=7.97 ×10−6m2s−1
, and
Um=30kph
( )
25
( )
7.97 ×10−6m2/s
10−6m2/s=5,980kph
!
This speed is not reasonable because it is so high, and because it corresponds to a Mach number
of nearly 5 in air while the Mach number of the submarine in water will be ~0.006. Thus, all the
critical dimensionless numbers cannot be matched, and this exercise illustrates a leading
difficulty of scale model testing: Reynolds numbers typically cannot be matched without
introducing other (possibly significant) problems.
If the coefficient of drag were the same in both cases (it will not be!), then the force ratio
would be:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 4.78. The volume flow rate Q from a centrifugal blower depends on its rotation rate Ω,
its diameter d, the pressure rise it works against Δp, and the density
ρ
and viscosity
µ
of the
working fluid.
a) Develop a dimensionless scaling law for Q in terms of the other parameters.
b) Simplify the result of part c) for high Reynolds number pumping where
µ
is no longer a
parameter.
c) For d = 0.10 m and
ρ
= 1.2 kg/m3, plot the measured centrifugal blower performance data
from the table in dimensionless form to determine if your result for part b) is a useful
simplification. Here RPM is revolutions per minute, Q is in liter/s, and Δp is in kPa.
RPM = 5000 8000 11,000
Q ΔP Q ΔP Q ΔP
0.3 0.54 0.5 1.4 0.9 2.6
1.1 0.51 2.0 1.3 4.1 2.2
1.5 0.48 3.3 1.1 6.2 1.8
2.8 0.37 5.0 0.84 7.7 1.3
3.8 0.24 6.5 0.49 9.5 0.81
d) What maximum pressure rise would you predict for a geometrically similar blower having
twice the diameter if it were spun at 6,500 RPM?
Solution 4.78. a) As in Example 4.17, the six parameters Q, Ω, d, Δp,
ρ
and
µ
lead to three
dimensionless groups. These are:
The flow coefficient =
Q
, The head coefficient =
Δp
, and Re =
ρ
Ωd2
.
where Φ is an undetermined function.
c) The task here is to plot the data provided in dimensionless form. Converting RPM to radians
per second and liters per second to (meters)3 per second, and then moving to dimensionless
parameters leads to the following dimensionless tabulations:
Ω = 523.6 rad/s 837.8 rad/s 1152 rad/s
Coefficients: flow head flow head flow head
0.000573 0.1641 0.000597 0.1662 0.000781 0.1633
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
The data provided in the tabulations collapse well eventhough the Reynolds number varies by
more than a factor of two; Re increases in direct proportion to the rotational speed. Thus, the
conclusion from this plot is that the assumption leading to the part b) result is a justified
0″
0.02″
0.04″
0.06″
0.08″
0.1″
0.12″
0.14″
0.16″
0.18″
0″ 0.002″ 0.004″ 0.006″ 0.008″ 0.01″
5000″RPM”
8000″RPM”
11000″RPM”
Δp
ρ
Ω2d2
Q
Ωd3
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 4.79. A set of small-scale tank-draining experiments are performed to predict the liquid
depth, h, as a function of time t for the draining process of a large cylindrical tank that is
geometrically similar to the small-experiment tanks. The relevant parameters are gravity g,
initial tank depth H, tank diameter D, orifice diameter d, and the density and viscosity of the
liquid,
ρ
and
µ
, respectively.
a) Determine a general relationship between h and the other parameters.
b) Using the following small-scale experiment results, determine whether or not the liquid’s
viscosity is an important parameter.
H = 8 cm, D = 24 cm, d = 8 mm H = 16 cm, D = 48 cm, d = 1.6 cm
h (cm) t (s) h (cm) t (s)
8.0 0.00 16.0 0.00
6.8 1.00 13.3 1.50
5.0 2.00 9.5 3.00
3.0 3.00 5.3 4.50
1.2 4.00 1.8 6.00
0.0 5.30 0.0 7.50
c) Using the small-scale-experiment results above, predict how long it takes to completely drain
the liquid from a large tank having H = 10 m, D = 30 m, and d = 1.0 m.
Solution 4.79. a) The required relationship can be obtained from dimensional analysis. The
parameter & units matrix is:
h t g H D d
ρ
µ
M 0 0 0 0 0 0 1 1
L 1 0 1 1 1 1 -3 -1
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling