Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 9.1. a) Write out the three components of (9.1) in x–y–z Cartesian coordinates.
b) Set u = (u(y), 0, 0), and show that the x– and y-momentum equations reduce to:
0=−1
ρ
∂
p
∂
x+
ν
d2u
dy2
, and
0=−1
ρ
∂
p
∂
y
.
Solution 9.1. a) Equation (9.1) is the constant-viscosity Navier-Stokes’ momentum equation for
incompressible flow:
Du
Dt =−1
ρ
∇p+
ν
∇2u
,
where
ν
is the kinematic viscosity of the flow. Using u = (u, v, w) and
∇=
∂ ∂
x,
∂ ∂
y,
∂ ∂
z
( )
, the
three components of this equation become:
x:
∂
u
∂
t+u
∂
u
∂
x+v
∂
u
∂
y+w
∂
u
∂
z=−1
ρ
∂
p
∂
x+
ν∂
2u
∂
x2+
∂
2u
∂
y2+
∂
2u
∂
z2
&
‘
(
)
*
+
,
y:
∂
v
∂
t+u
∂
v
∂
x+v
∂
v
∂
y+w
∂
v
∂
z=−1
ρ
∂
p
∂
y+
ν∂
2v
∂
x2+
∂
2v
∂
y2+
∂
2v
∂
z2
&
(
)
+
, and
And, when u depends only on y, then ∂u/∂t = ∂u/∂x = ∂u/∂z = 0 so the part a) equations simplify
further:
x:
0=−1
ρ
∂
p
∂
x+
ν∂
2u
∂
y2
&
‘
(
)
*
+
,
y:
0=−1
ρ
∂
p
∂
y
, and
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 9.2. For steady pressure driven flow between parallel plates (see Figure 8.3), there are 7
parameters: u(y), U, y, h,
ρ
,
µ
, and dp/dx. Determine a dimensionless scaling law for u(y), and
rewrite the flow-field solution (8.5) in dimensionless form.
Solution 9.2. The parameters are: u(y), U, y, h,
ρ
,
µ
, and dp/dx. First, create the parameter
matrix:
u U y h
ρ
µ dp/dx
––––––––––––––––––––––––––––––––––––––––
Mass: 0 0 0 0 1 1 1
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 9.3. An incompressible viscous liquid with density
ρ
fills the gap between two large
smooth parallel walls that are both stationary. The upper and lower walls are located at x2 = ±h,
respectively. An additive in the liquid causes its viscosity to vary in the x2 direction. Here the
flow is driven by a constant non-zero pressure gradient:
∂
p
∂
x1=const.
a) Assume steady flow, ignore the body force, set
u=u1(x2),0,0
( )
and use
∂ρ
∂
t+
∂
∂
xi
ρ
ui
( )
=0
,
ρ∂
uj
∂
t+
ρ
ui
∂
uj
∂
xi
=−
∂
p
∂
xj
+
ρ
gj+
∂
∂
xi
µ∂
ui
∂
xj
+
∂
uj
∂
xi
%
&
‘
‘
(
)
*
*
+
,
–
–
.
/
0
0
+
∂
∂
xj
µ
v−2
3
µ
%
&
‘ (
)
*
∂
ui
∂
xi
+
,
–
.
/
0
to determine u1(x2) when
µ
=
µ
o1+
γ
x2h
( )
2
( )
.
b) What shear stress is felt on the lower wall?
c) What is the volume flow rate (per unit depth into the page) in the gap when
γ
= 0?
d) If –1 <
γ
< 0, will the volume flux be higher or lower than the case when
γ
= 0?
Solution 9.3. a) The continuity equation is satisfied by the form of the velocity field. The j =1–
component of momentum equation simplifies to:
0=−
∂
p
∂
x1
( )
+
∂ ∂
x2
( )
µ∂
u1
∂
x2
( )
[ ]
.
Integrate once with
∂
p
∂
x1=const.
to find:
µ∂
u1
∂
x2
( )
=
∂
p
∂
x1
( )
x2+C
. Divide by
µ
and
integrate again:
u1=1
µ
∂
p
∂
x1
x2+C
#
$
%
&
‘
(
dx2
∫=
∂
p
∂
x1
( )
x2+C
µ
o1+
γ
x2h
( )
2
( )
dx2
∫
=h2
2
γµ
o
∂
p
∂
x1
ln 1+
γ
x2
h
#
$
% &
‘
(
2
#
$
%
%
&
‘
(
( +Ch
µ
o
γ
tan−1x2
γ
h
#
$
%
&
‘
( +D.
The boundary conditions,
u1(±h)=0
, determine the values of the constants: C = 0, and
D=−h22
γµ
o
( )
∂
p
∂
x1
( )
ln 1+
γ
( )
, thus:
u1(x2)=−h2
2
γµ
o
∂
p
∂
x1
ln 1+
γ
x2h
( )
2
1+
γ
%
&
‘
‘
(
)
*
*
.
b) From the solution of part a) with C = 0:
τ
w=
µ∂
u1
∂
x2
( )
y=−h=−h
∂
p
∂
x1
( )
c) When
γ
= 0, the flow profile is parabolic:
q=u1(x2)dx2
−h
+h
∫=−h2
2
µ
o
∂
p
∂
x1
1−x2
2
h2
%
&
‘
(
)
*
dx2
−h
+h
∫=−2h3
3
µ
o
∂
p
∂
x1
d) The volume flux will be higher because the viscosity will be reduced at the wall. Manipulation
of the near-wall viscosity with additives is sometimes used in long piping systems to reduce
pumping power requirements.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 9.4. An incompressible viscous liquid with density
ρ
fills the gap between two large
smooth parallel plates. The upper plate at x2 = h moves in the positive x1-direction at speed U.
The lower plate at x2 = 0 is stationary. An additive in the liquid causes its viscosity to vary in the
x2 direction.
a) Assume steady flow, ignore the body force, set
u=u1(x2),0,0
( )
and
∂
p
∂
x1=0
, and use the
equations specified in Exercise 8.3 to determine u1(x2) when
µ
=
µ
o1+
γ
x2h
( )
.
b) What shear stress is felt on the lower plate?
c) Are there any physical limits on
γ
? If, so specify them.
Solution 9.4. a) For
u=u1(x2),0,0
( )
, no body force, and
∂
p
∂
x1=0
in steady incompressible
flow, the continuity equation is automatically satisfied, and the momentum equation for j = 1
simplifies to:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 9.5. Planar Couette flow is generated by placing a viscous fluid between two infinite
parallel plates and moving one plate (say, the upper one) at a velocity U with respect to the other
one. The plates are a distance h apart. Two immiscible viscous liquids are placed between the
plates as shown in the diagram. The lower fluid layer has thickness d. Solve for the velocity
distributions in the two fluids.
Solution 9.5. For steady viscous flow between infinite parallel plates, the fluid velocity will be
unidirectional: u = (u, 0, 0). For this problem, no pressure gradient is specified so assume it to
be zero. Thus, the horizontal (x1-direction) momentum equation reduces to:
∂τ
21
∂
x2=
∂ ∂
x2
( )
µ∂
u
∂
x2
( )
=
µ∂
2u
∂
y2=0
,
where the last equality follows when the viscosity is constant and x2 = y. Here, the viscosity is
assumed constant within each fluid. This means that the flow profile in each fluid will be piece-
wise linear:
u(y)=A1+B1y for 0 ≤y≤d
A2+B2y for d≤y≤h
#
$
%
&
‘
(
,
where the A’s & B’s are constants and ‘1’ implies the upper fluid layer with viscosity
µ
1, and ‘2’
implies the lower fluid layer with viscosity
µ
2. The four constants can be determined from the
four boundary conditions:
i) u(0) = 0 (match the speed of the lower boundary)
ii) u(h) = U (match the speed of the upper boundary)
iii) u(d–) = u(d+), and (match flow speeds at the internal fluid-fluid interface)
iv)
τ
(d–) =
τ
(d+) (match shear stress at the internal fluid-fluid interface)
where
τ
is the shear stress in the fluid. These four boundary conditions imply:
A2 = 0,
A
1+B
1h=U
,
A2+B2d=A
1+B
1d
, and
µ
2B2=
µ
1B
1
Use the first two equations to eliminate A1 and A2 from the second two equations to find:
B2d=U−B
1(h−d)
, and
µ
2B2=
µ
1B
1
.
Eliminate B2 and solve for B1:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 9.6. Consider plane Poiseuille flow of a non-Newtonian power-law fluid using the
coordinate system and geometry shown in Figure 9.3. Here the fluid’s constitutive relationship is
given by (4.37):
τ
xy =m∂u∂y
( )
n
, where m is the power law coefficient, and n is the power law
exponent.
a) Determine the velocity profile u(y) in the lower half of the channel, 0 ≤ y ≤ h/2, using the
boundary condition u(0) = 0.
b) Given that the maximum velocity occurs at y = h/2 and that the flow profile is symmetric
about this location, plot
u(y)u(h/2)
vs.
y(h/2)
for n = 2 (a shear thickening fluid), n = 1 (a
Newtonian fluid), and n = 0.4 (a shear thinning fluid).
c) Explain in physical terms why the shear-thinning velocity profile is the bluntest.
Solution 9.6. In plane Poiseuille flow, the upper channel wall does not move. In the lower half of
the channel, the ∂u/∂y > 0, so fractional powers are readily managed.
a) For fully developed constant density flow (so that ∂p/∂x is independent of x) the horizontal
momentum equation reduces to:
0=−∂p
∂x+∂
τ
xy
∂y
.
The vertical momentum equation is simply, ∂p/∂y = 0, so ∂p/∂x does not depend on x or y. Thus,
this horizontal momentum equation can be integrated to find:
y∂p
∂x=
τ
xy +C
,
where C is a constant. However, when y = h/2,
τ
xy = 0 so C = (h/2)∂p/∂x, therefore:
h
2−y
“
#
$%
&
‘−∂p
∂x
“
#
$%
&
‘=
τ
xy =m∂u
∂y
“
#
$%
&
‘
n
, or
∂u
∂y=−1
m
∂p
∂x
#
$
%&
‘
(
1nh
2−y
#
$
%&
‘
(
1n
,
and both equations are written this way because –∂p/∂x is positive. The second equation can be
integrated to find:
u(y)=−1
m
∂p
∂x
#
$
%&
‘
(
1n1
1+1n
h
2−y
#
$
%&
‘
(
1+1n
+D
,
where D is another constant of integration. The horizontal velocity must go to zero on the lower
channel wall at y = 0, so
n+1
“
#$
%
&h
2
2m
∂x
“
#$
%
&
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
where
η
=
y(h/2)
. The three curves for n = 2, 1, and 0.4 are plotted here:
u(
η
)/u(
η
= 1)
c) From the part a) solution, the shear stress is:
τ
xy =−∂p
∂x
h
2−y
#
$
%&
‘
(
.
The highest magnitude shear stress occurs at the upper and lower wall of the channel. In a shear
thinning fluid, the local viscosity decreases when the shear stress increases. Thus, to maintain
this shear stress profile, the velocity gradient magnitude must increase in regions where the shear
stress magnitude is also large. This leads to a blunter profile for a shear thinning fluid compared
to a Newtonian (or a shear thickening) fluid.
0.00#
0.20#
0.40#
0.60#
0.80#
1.00#
1.20#
1.40#
1.60#
1.80#
2.00#
0.000# 0.200# 0.400# 0.600# 0.800# 1.000#
n=2#
n=1#
n=0.4#
η
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 9.7. Consider the laminar flow of a fluid layer falling down a plane inclined at an angle
θ
with respect to the horizontal. If h is the thickness of the layer in the fully developed stage,
show that the velocity distribution is
u(y)=g2
ν
( )
h2−y2
( )
sin
θ
, where the x-axis points in the
direction of flow along the free surface, and the y-axis points toward the plane. Show that the
volume flow rate per unit width is
Q=gh33
ν
( )
sin
θ
, and that and the frictional stress on the
wall is
τ
w =
ρ
ghsin
θ
.
Solution 9.7. In the given coordinates, there is a
component of gravity, gsin
θ
, acting in the x-direction, but
the presence of atmospheric pressure on the liquid surface
Integrating twice produces:
u(y)=−gsin
θ
2
ν
y2+Ay +B
,
where A and B are constants. The boundary condition on the liquid surface (y = 0) is zero shear
stress (
µ
∂u/∂y = 0), and this allows A to be evaluated:
0=
∂
u
∂
y
#
$
%
&
‘
(
y=0
=−gsin
θ
ν
y+A
#
$
%
&
‘
(
y=0
=A
.
The boundary condition, on the solid surface (y = h) is zero velocity (u = 0), and this allows B to
be evaluated:
0=u(h)=−gsin
θ
2
ν
h2+B
, or
B=gsin
θ
2
ν
h2
.
so the velocity profile is:
u(y)=−gsin
θ
2
ν
y2−h2
( )
.
The volume flow rate per unit width is:
Q=u(y)dy
0
h
∫=−gsin
θ
2
ν
y2−h2
( )
dy
0
h
∫=−gsin
θ
2
ν
h3
3−h3
&
‘
(
)
*
+ =gsin
θ
3
ν
h3
.
The magnitude of the shear stress is:
τ
=
µ∂
u
∂
y=
µ
gsin
θ
ν
&
‘
( )
*
+
y=
ρ
gy sin
θ
,
and the maximum shear stress,
τ
w =
ρ
ghsin
θ
, occurs at the solid surface.
y
x
h
stress
distribution
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 9.8. In two-dimensional (x,y)-coordinates, the Navier-Stokes equations for the fluid
velocity,
u=(u,v)
, in a constant-viscosity constant-density flow are:
∂u∂x+∂v∂y=0
,
∂u
∂t+u∂u
∂x+v∂u
∂y=−1
ρ
∂p
∂x+
ν
∂2u
∂x2+∂2u
∂y2
#
$
%&
‘
(
, and
∂v
∂t+u∂v
∂x+v∂v
∂y=−1
ρ
∂p
∂y+
ν
∂2v
∂x2+∂2v
∂y2
#
$
%&
‘
(
.
a) Cross differentiate and sum the two momentum equations to reach the following equation for
ω
z=∂v∂x− ∂u∂y
, the vorticity normal to the x–y plane:
∂
ω
z
∂t+u∂
ω
z
∂x+v∂
ω
z
∂y=
ν
∂2
ω
z
∂x2+∂2
ω
z
∂y2
“
#
$%
&
‘
.
b) The simplest nontrivial solution of this equation is uniform shear or solid body rotation (
ω
z =
constant). The next simplest solution is a linear function of the independent coordinates:
ω
z = ax
+ by, where a and b are constants. Starting from this vorticity field, derive the following velocity
field:
u=−b
2
(ax +by)2
a2+b2+c
“
#
$%
&
‘
and
v=a
2
(ax +by)2
a2+b2+c
!
“
#$
%
&
.
where c is an undetermined constant.
c) For the part b) flow, sketch the streamlines. State any assumptions you make about a, b, and c.
d) For the part b) flow when a = 0, b > 0, and
u=(Uo, 0)
at the origin of coordinates with Uo > 0,
sketch the velocity profile along a line x = constant, and determine
∇p
.
Solution 9.8. a) Apply –∂/∂y to the x-direction momentum equation, and ∂/∂x to the y-direction
momentum equation to reach:
−∂
∂t
∂u
∂y
#
$
%&
‘
(−∂u
∂y
∂u
∂x−u∂
∂x
∂u
∂y
#
$
%&
‘
(−∂v
∂y
∂u
∂y−v∂
∂y
∂u
∂y
#
$
%&
‘
(=1
ρ
∂2p
∂y∂x−
ν
∂2
∂x2+∂2
∂y2
#
$
%&
‘
(∂u
∂y
#
$
%&
‘
(
, and
∂
∂t
∂v
∂x
“
#
$%
&
‘+∂u
∂x
∂v
∂x+u∂
∂x
∂v
∂x
“
#
$%
&
‘+∂v
∂x
∂v
∂y+v∂
∂x
∂v
∂y
“
#
$%
&
‘=−1
ρ
∂2p
∂x∂y+
ν
∂2
∂x2+∂2
∂y2
“
#
$%
&
‘∂v
∂x
“
#
$%
&
‘
.
Add these two equations together noting that pressure terms cancel, and that the second and
fourth terms in each equation sum to zero because of the continuity equation,
∂u∂x+∂v∂y=0
.
ω
z=∂v∂x− ∂u∂y
∂
ω
z
∂t+u∂
ω
z
∂x+v∂
ω
z
∂y=
ν
∂2
ω
z
∂x2+∂2
ω
z
∂y2
“
#
$%
&
‘
.
b) To find the velocity field start with:
ω
z = ax + by = ∂v/∂x – ∂u/∂y. (1)
The part a) equation implies: ua + vb = 0 or v = –(a/b)u. (2)
The continuity equation is also needed: ∂u/∂x + ∂v/∂y = 0. (3)
Use (2) to eliminate v, from (1) and (3) to find:
ax +by =−a
b
∂u
∂x−∂u
∂y
, and
∂u
∂x−a
b
∂u
∂y=0
.
Combine these twice, first to eliminate ∂u/∂y, then to eliminate ∂u/∂x to reach:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
For consistency, the unknown functions must be: f(x) = a2x2 + const. and g(y) = b2y2 + const. so
u=−b
2
(ax +by)2
a2+b2+c
“
#
$%
&
‘
, and then from (2)
v=−a
bu=a
2
(ax +by)2
a2+b2+c
“
#
$%
&
‘
.
where c is another undetermined constant.
c) dysl/dxsl = v/u = –a/b. Thus, if a and b are both
positive constants, the streamlines are negatively-
slopped straight lines. In this case, u < 0 and v > 0,
so the arrows point upward and to the left.
d) Evaluate the constants to find:
u=Uo−b2
( )
y2
and v = 0, so the
horizontal velocity profile is parabolic. To get the pressure gradient,
evaluate the 2D Navier-Stokes equations using the part b) velocity field:
0=−1
∂p
ν
(−b)
, and
0=−1
∂p
, so
∇p=−
µ
b, 0
( )
.
!
x!
y !
b!
a!
Uo!
y !
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 9.9. Consider circular Couette flow as described by (9.10) in the limit of a thin gap
between the cylinders. Use the definitions:
R=(R
1+R2) 2
,
h=R2−R
1
,
Ω=(Ω1+Ω2) 2
,
ΔΩ =Ω2− Ω1
, and
R=R+y
to complete the following items.
a) Show that
u
ϕ
(y)≅ ΩR+ΔΩR y h
( )
when
y<< R
,
h<< R
, and all terms of order
y R
and
h R
or higher have been dropped. This is the lubrication approximation for circular Couette
flow.
b) Compute the shear stress
µ
du
ϕ
dy
( )
from the part a) flow field and put it terms of
µ
, R1, R2,
Ω1, and Ω2.
c) Compute the exact shear stress from (9.10) using:
τ
R
ϕ
=2
µ
R
2
∂
∂
R
u
ϕ
R
!
“
#$
%
&+1
2R
∂
uR
∂ϕ
‘
(
)*
+
,
, and
evaluate it at R = R1, and R = R2.
d) What values of
h R
lead to shear stress errors of 1%, 3% and 10%.
Solution 9.9. Here,
R
1=R−h/ 2
and
R2=R+h/ 2
, so
R
1
2=R
2−Rh +h2/ 4
and
R2
2=R
2+Rh +h2/ 4
. Plus,
R2
2−R
1
2=2Rh
. Use these exact relationships in (9.10) to reach:
u
ϕ
(y)=1
2RhΩ2R2+Rh+h2
4
“
#
$%
&
‘−Ω1R2−Rh+h2
4
“
#
$%
&
‘
)
*
+,
–
.R+y
( )
− ΔΩ
R2−Rh+h2
4
“
#
$%
&
‘R2+Rh+h2
4
“
#
$%
&
‘
R+y
0
1
2
2
3
2
2
4
5
2
2
6
2
2
.
Simplify terms within the large {,}-brackets.
u
ϕ
(y)=1
2Rh R +y
( )
ΔΩR2+2ΩRh +ΔΩ h2
4
#
$
%&
‘
(R+y
( )
2− ΔΩ R4−R2h2
2+h4
16
*
+
,
,
–
.
/
/
0
1
2
3
2
4
5
2
6
2
.
Perform all the multiplications within the large {,}-brackets and continue with this
simplification:
u
ϕ
(y)=1
2Rh R +y
( )
ΔΩR4+ΔΩ2R3y+ΔΩR2y2+2ΩR3h+4ΩR2hy +2ΩRhy2+
ΔΩ h2
4R2+ΔΩ h2
2Ry +ΔΩ h2
4y2− ΔΩR4+ΔΩ R2h2
2− ΔΩ h4
16
$
%
&
‘
&
(
)
&
*
&
.
Cancel equal and opposite terms, and group terms than involve the average,
Ω=(Ω1+Ω2) 2
,
ΔΩ =Ω2− Ω1
Expand
1+y R
( )
−1
in a power series for
y<< R
:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
u
ϕ
(y)=ΩR+y
( )
+ΔΩR1−y
R+y2
R2+…
$
%
&‘
(
)y
h+1
2Rh
3h2
4+y2
$
%
&‘
(
)+hy
4R2+1
2R3
hy2
4−h3
16
$
%
&‘
(
)
*
+
,–
.
/
.
Keeping only linear terms in
y R
and
h R
leads to the final series:
τ
R
ϕ
(R2)=2
µ
Ω2− Ω1
[ ]
R2
2−R
1
2R
1
2=
µ
Ω2− Ω1
[ ]
Rh
R−h
2
#
$
%&
‘
(
2
≅
µ
Ω2− Ω1
[ ]
R
h1−h
R
#
$
%&
‘
(
.
To first order, the average shear stress is:
τ
=1
2
τ
R
ϕ
(R
1)+
τ
R
ϕ
(R2)
( )
=2
µ
Ω2− Ω1
[ ]
R2
2−R
1
2R2
2=
µ
Ω2− Ω1
[ ]
h
R
,
which matches the result from the lubrication approximation.
d) To first order, the difference of either exact shear stress from this average shear stress is:
Δ
τ
τ
=
τ
R
ϕ
(R
1or 2)−
τ
τ
= ± h
R
.
Thus, shear stress errors of shear stress errors of 1%, 3% and 10% occur when
h R
= 0.01, 0.03,
and 0.10.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 9.10. Room temperature water drains through a round vertical tube with diameter d.
The length of the tube is L. The pressure at the tube’s inlet and outlet is atmospheric, the flow is
steady, and L >> d.
a) Using dimensional analysis write a physical law for the mass flow rate
˙
m
through the tube.
b) Assume that the velocity profile in the tube is independent of the vertical coordinate,
determine a formula for
˙
m
, and put it in dimensionless form.
c) What is the change in
˙
m
if the temperature is raised and the water’s viscosity drops by a factor
of two?
Solution 9.10. a) This part is just dimensional analysis. The solution parameter is
˙
m
, and
boundary condition & material parameters are:
ρ
= density of the fluid, µ = viscosity of the fluid,
g = acceleration of gravity, d = tube diameter, and L = tube length. The parameter matrix is:
˙
m
ρ µ g d L
–––––––––––––––––––––––––––––––––––
Mass: 1 1 1 0 0 0
Length: 0 -3 -1 1 1 1
Time: -1 0 -1 -2 0 0
where the z-axis points upward (opposite gravity), R is the radial coordinate, and uz is the vertical
velocity. This is the equation for Poiseuille flow with the pressure gradient replaced by
ρ
g.
Noting that downward velocity produces positive
˙
m
, the results in Section 8.2 imply:
˙
m =
ρ
Q=
ρπ
8
µ
d
2
$
%
& ‘
(
)
4
ρ
g=
π
128
ρ
2gd4
µ
or
˙
m
µ
d=
π
128
ρ
2gd3
µ
2→ Π1=
π
128 Π2
.
c) The flow rate is inversely proportional to µ and µ = µ(T). Since the ratio
µ
room temp
µ
=2
, the
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 9.11. Consider steady laminar flow through the annular space formed by two coaxial
tubes aligned with the z-axis. The flow is along the axis of the tubes and is maintained by a
pressure gradient dp/dz. Show that the axial velocity at any radius R is
uz(R)=1
4
µ
dp
dz R2−a2−b2−a2
ln b a
( )
ln R
a
#
$
%
&
‘
(
,
where a is the radius of the inner tube and b is the radius of the outer tube. Find the radius at
which the maximum velocity is reached, the volume flow rate, and the stress distribution.
Solution 9.11. Use cylindrical coordinates, and assume
uz = uz(R), where R is the radial coordinate. The axial
R
∂
uz
∂
R=R2
2
µ
∂
p
∂
z+A
, and
uz(R)=R2
4
µ
∂
p
∂
z+Alog R+B
.
The two boundary conditions are u = 0 at R = a, b and these allow the two constants A and B to
be determined. Alternatively evaluating the above formula for uz(R) at R = a and b produces.
uz(a)=0=a2
4
µ
∂
p
∂
z+Alog a+B
, and
uz(b)=b2
4
µ
∂
p
∂
z+Alogb+B
.
These two equations can be solved to find:
A=−1
4
µ
∂
p
∂
z
$
%
& ‘
(
) b2−a2
ln(b/a)
, and
B=1
4
µ
∂
p
∂
z
#
$
% &
‘
( b2−a2
ln(b/a)log(a)−a2
*
+
,
–
.
/
,
so that:
u(R)=1
4
µ
∂
p
∂
z
#
$
% &
‘
( R2−a2−b2−a2
ln(b/a)log R a
( )
*
+
,
–
.
/
.
The shear stress is:
τ
=
µ∂
u
∂
R=R
2
µ
∂
p
∂
z−1
4
µ
∂
p
∂
z
“
#
$%
&
‘b2−a2
Rln(b/a)
.
The maximum velocity occurs where
τ
= 0, and this condition implies:
0=Rmax
2
µ
∂
p
∂
z−1
4
µ
∂
p
∂
z
$
%
& ‘
(
) b2−a2
Rmax ln(b/a)
or
Rmax =b2−a2
2ln(b/a)
.
The volume flow rate Q is:
Q=2
π
u(R)RdR
a
b
∫=
π
8
µ
∂
p
∂
z
%
&
‘ (
)
* b2−a2
( )
2
ln(b/a)+a4−b4
,
–
.
.
/
0
1
1
.
b
Rmax
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 9.12. A long round wire with radius a is pulled at a steady speed U, along the axis of a
long round tube of radius b that is filled with a viscous fluid. Assuming laminar fully-developed
axial flow with
∂
p
∂
z=0
in cylindrical coordinates
(R,
ϕ
,z)
with u = (0, 0, w(R)), determine
w(R) assuming constant fluid density
ρ
and viscosity
µ
with no body force.
a) What force per unit length of the wire is needed to maintain the motion of the wire?
b) Explain what happens to w(R) when
b→ ∞
. Is this situation physically meaningful? What
additional term(s) from the equations of motion need to be retained to correct this situation?
Solution 9.12. The steady axisymmetric no-swirl equations are:
1
R
∂
∂
R
RuR
( )
+
∂
w
∂
z=0
,
uR
∂
uR
∂
R+w
∂
uR
∂
z=−1
ρ
∂
p
∂
R+
µ
ρ
1
R
∂
∂
RR
∂
uR
∂
R
%
&
‘ (
)
* +
∂
2uR
∂
z2−uR
R2
%
&
‘
(
)
*
,
0=−1
ρ
R
∂
p
∂ϕ
, and
uR
∂
w
∂
R+w
∂
w
∂
z=−1
ρ
∂
p
∂
z+
µ
ρ
1
R
∂
∂
RR
∂
w
∂
R
%
&
‘ (
)
* +
∂
2w
∂
z2
%
&
‘
(
)
*
.
Setting u = (0, 0, w(R)) automatically satisfies the cons. of mass equation. With
∂
p
∂
z=0
, the
Simultaneous solution of these two algebraic equations produces:
w(R)=U
ln b a
( )
ln b R
( )
a) The viscous shear stress on the wire,
τ
w, will act to retard its motion. The viscous force per
unit length of wire will act in the z-direction on the wire and will be
2
π
a
τ
w=2
π
a
µ∂
w
∂
R
%
&
‘ (
)
*
R=a
=2
π
a
µ
U
ln b a
( )
1
b R −b
R2
%
&
‘ (
)
*
%
&
‘
(
)
*
R=a
=−2
πµ
U
ln b a
( )
.
Thus, the force necessary to maintain the motion will be:
= + 2
πµ
Uezln b a
( )
b) When
b→ ∞
, the solution indicates that the force necessary to keep the wire moving will
approach zero. This is not physically meaningful for any finite wire length and time duration
because the fluid is viscous. The problem(s) with this limiting situation can be corrected by
including the unsteady terms,
∂ ∂
t
, or terms involving uR and
∂ ∂
z
in the conservation equations.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 9.13. Consider steady unidirectional incompressible viscous flow in Cartesian
coordinates, u = v = 0 with w = w(x,y) without body forces.
a) Starting from the steady version of (8.1), derive a single equation for w assuming that
∂
p
∂
z
is
non-zero and constant.
b) Guess w(x,y) for a tube with elliptical cross section
x a
( )
2+y b
( )
2=1
.
c) Determine w(x,y) in for a tube of rectangular cross section: –a/2 ≤ x ≤ +a/2, –b/2 ≤ y ≤ +b/2.
(Hint: find particular (a polynomial) and homogeneous (a Fourier series) solutions for w.)
Solution 9.13. a) Start with the full z-direction momentum equation in Cartesian coordinates:
∂
w
∂
t+u
∂
w
∂
x+v
∂
w
∂
y+w
∂
w
∂
z=−1
ρ
∂
p
∂
z+
ν∂
2w
∂
x2+
∂
2w
∂
y2+
∂
2w
∂
z2
&
‘
(
)
*
+
.
For steady flow with u = v = 0 and w = w(x,y), a lot terms that drop out of the above equation,
flow formula when a = b.
c) The solution for a rectangular cross section is a bit more involved. The equation for w in part
a) is inhomogeneous, so its solution will be made up of a particular solution and a homogeneous
solution: w = wp + wh. Here, the particular solution must satisfy the governing equation, but it
need not satisfy the boundary conditions. The constants in the homogeneous solution can be
used to fix up the solution at the tube walls. Therefore, start by looking for a one-variable
particular solution, i.e. wp = wp(x), and plug this into the field equation:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 9.14. A long vertical cylinder of radius b rotates with angular velocity Ω concentrically
outside a smaller stationary cylinder of radius a. The annular space is filled with fluid of
viscosity
µ
. Show that the steady velocity distribution is:
u
ϕ
=R2−a2
b2−a2
b2Ω
R
, and that the torque
exerted on either cylinder, per unit length, equals 4
πµ
Ωa2b2/(b2 − a2).
Solution 9.14. Use cylindrical coordinates, and start from equation (8.10),
u
ϕ
(R)=1
R2
2−R
1
2Ω2R2
2− Ω1R
1
2
[ ]
R− Ω2− Ω1
[ ]
R
1
2R2
2
R
%
&
‘
(
)
*
,
and set R1 = a, R2 = b, Ω2 = Ω, and Ω1 = 0 to find:
u
ϕ
(R)=1
b2−a2Ωb2−0
[ ]
R− Ω − 0
[ ]
a2b2
R
%
&
‘
(
)
* =Ωb2
b2−a2
R2−a2
R=R2−a2
b2−a2
b2Ω
R
.
From Appendix B the shear stress is
τ
R
ϕ
=
µ
R
∂
∂
R
u
ϕ
R
%
&
‘
(
)
* =
µ
Rb2Ω
b2−a2
∂
∂
R
1−a2
R2
%
&
‘
(
)
* =2
µ
a2b2Ω
b2−a2
1
R2
%
&
‘ (
)
*
.
If the axial length of the cylinders is h, then
torque on the inner cylinder = (
τ
R
ϕ
)R = a(area)(radius) =
2
µ
a2b2Ω
b2−a2
1
a2
$
%
& ‘
(
) 2
π
ah
( )
a=4
πµ
a2b2hΩ
b2−a2
,
torque on the outer cylinder = (
τ
R
ϕ
)R = b(area)(radius) =
2
µ
a2b2Ω
b2−a2
1
b2
$
%
& ‘
(
) 2
π
bh
( )
b=4
πµ
a2b2hΩ
b2−a2
.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 9.15. Consider a solid cylinder of radius a, steadily rotating at angular speed Ω in an
infinite viscous fluid. The steady solution is irrotational: u
θ
= Ωa2/R. Show that the work done by
the external agent in maintaining the flow (namely, the value of 2
π
Ru
θτ
r
θ
at R = a) equals the
viscous dissipation rate of fluid kinetic energy in the flow field.
Solution 9.15. Using the given velocity field, the shear stress is:
τ
R
ϕ
=
µ
R
∂
∂
R
u
ϕ
R
%
&
‘
(
)
* =
µ
Ωa2R
∂
∂
R
1
R2
%
&
‘ (
)
* =−2
µ
Ωa21
R2
.
The work done per unit height =
2
π
a
τ
R
ϕ
u
ϕ
{ }
R=a=2
π
a⋅2
µ
Ω⋅ Ωa=4
πµ
a2Ω2
.
From (4.58) the viscous dissipation rate of kinetic energy per unit volume for an
incompressible flow is
ρε
=2
µ
Sij Sij
, where
ε
is the viscous dissipation of kinetic energy per unit
mass. For the given flow field there is only one non-zero independent strain component:
SR
ϕ
=S
ϕ
R=R
2
∂
∂
R
u
ϕ
R
$
%
&
‘
(
) =Ωa2
2
R
∂
∂
R
1
R2
$
%
& ‘
(
) =−Ωa21
R2
.
Therefore:
ρε
=2
µ
Sij Sij =2
µ
SR
ϕ
2+S
ϕ
R
2
( )
=4
µ
Ω2a4
R4
,
so the kinetic energy dissipation rate per unit height is:
ρε
a
∞
∫2
π
RdR =8
πµ
Ω2a41
R3dR
a
∞
∫=4
πµ
Ω2a2
,
which equals the work done turning the cylinder.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 9.16. Redo the lubrication-theory scaling provided in Section 9.3 for the situation when
there is an imposed external time scale
τ
so that the appropriate dimensionless time is t* = t/
τ
,
instead of that shown in (9.14). Show that this leads to the additional requirement
ρ
h2/
µτ
<< 1
for the validity of (9.17a). Interpret this new requirement in terms of the viscous diffusion length
ντ
. Is the unsteady acceleration term needed to analyze the effect of a 100 Hz oscillation
imposed on a 0.1-mm film of 30-weight oil (
ν
≈ 4 x10–4 m2/s)?
Solution 9.16. Use the dimensionless scaling provided by (9.14) except replace that used for the
time variable with t* = t/
τ
. The remaining members of (9.14),
x* = x/L , y* = y/h = y/
ε
L, u* = u/U , v* = v/
ε
U , and p* = p/Pa, where
ε
= h/L,
are unchanged. This mildly revised scaling does not influence the continuity equation. However,
it does impact the unsteady term in both momentum equations, but it is only necessary to
consider one. The horizontal momentum equation is the more interesting equation. Substituting
in the dimensionless variables leads to:
ρ
U
τ
∂
u∗
∂
t∗+
ρ
U2
L
“
#
$%
&
‘u∗
∂
u∗
∂
x∗+
ερ
U2
ε
L
“
#
$%
&
‘v∗
∂
u∗
∂
y∗=−P
a
L
“
#
$%
&
‘
∂
p∗
∂
x∗+
µ
U
L2
“
#
$%
&
‘
∂
2u∗
∂
x∗2+
µ
U
ε
2L2
“
#
$%
&
‘
∂
2u∗
∂
y∗2
.
Dividing by
µ
U/
ε
2L2, produces:
ρε
2L2
µτ
∂
u∗
∂
t∗+
ε
2Re
( )
u∗
∂
u∗
∂
x∗+
ε
2Re
( )
v∗
∂
u∗
∂
y∗=−1
Λ
$
%
&‘
(
)
∂
p∗
∂
x∗+
ε
2
∂
2u∗
∂
x∗2+
∂
2u∗
∂
y∗2
.
The coefficient of the first term is different than that given in (9.16a), and it must be small for the
validity of (9.17a):
ρε
2L2
µτ
=
ρ
h2
µτ
=h2
ντ
<<1
or h << [
ντ
]1/2.
where
ν
=
µ
/
ρ
. This new requirement states that the diffusion distance, [
ντ
]1/2, associated with
the time scale of flow unsteadiness
τ
must much larger than the gap distance h.
For a 100 Hz oscillation (t ~ 10 ms) imposed on a 0.1 mm gap containing 30-weight oil
with
ν
≈ 4 x10–4 m2/s:
h2
ντ
=(10−4m)2
(4 ×10−4m2s−1)(0.01s)=0.0025
.
This is much less than unity, so the unsteady term is not needed in an elementary lubrication
analysis of this situation. However, if the gap were 10 times larger (1 mm) this conclusion would
be different.