Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
4.42. Air, water, and petroleum products are important engineering fluids and can usually be
treated as Newtonian fluids. Consider the following materials and try to classify them as:
Newtonian fluid, non-Newtonian fluid, or solid. State the reasons for your choices and note the
temperature range where you believe your answers are correct. Simple impact, tensile, and shear
experiments in your kitchen or bathroom are recommended. Test and discuss at least five items.
a) toothpaste d) glass g) hot oatmeal j) silly putty
b) peanut butter e) honey h) creamy salad dressing
c) shampoo f) mozzarella cheese i) ice cream
Solution 4.42. One simple way to conduct
rheological tests is with two ordinary plastic
cups that are meant to stack inside each
other. Coat the outside of one of the cups
with liquid or material of interest. Then slide this cup into the second one to make a crude
a torsional spring connecting the two cups for very small angular deflections. In addition, rising
air bubbles in shampoo have pointed tails and this indicates there is memory within the fluid.
Moving bubbles in Newtonian fluids do not have pointed tails.
j) Silly putty can be considered a visco-elastic solid (when formed into a ball, the ball bounces)
or a non-Newtonian liquid (when a ball of silly putty is left alone on a flat surface, gravity
compresses it and it begins to flow horizontally).
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 4.43. The equations for conservation of mass and momentum for a viscous Newtonian
fluid are (4.7) and (4.39a) when the viscosities are constant.
a) Simplify these equations and write them out in primitive form for steady constant-density flow
in two dimensions where
ui=u1(x1,x2),u2(x1,x2),0
( )
,
p=p(x1,x2)
, and gj = 0.
b) Determine
p=p(x1,x2)
when
u1=Cx1
and
u2=−Cx2
, where C is a positive constant.
Solution 4.43. a) There are a number of simplifications to make. For steady flow
∂ ∂
t→0
; for
constant density flow
; for 2D flow
∂ ∂
x3→0
and u3 = 0; and gj = 0 implies g1 = g2 = 0.
Thus, the continuity equation is simplified as follows:
∂ρ
∂
t+
∂
∂
xi
ρ
ui
( )
=0→
steady
∂
∂
xi
ρ
ui
( )
=0→
exp and
ρ∂
ui
∂
xi
+ui
∂ρ
∂
xi
=0→
ρ
=const≠0
∂
ui
∂
xi
=0 →
2D
∂
u1
∂
x1
+
∂
u2
∂
x2
=0
.
Following a similar procedure and using this result, the simplified momentum equation is:
ρ
ui
∂
uj
∂
xi
=−
∂
p
∂
xj
+
µ∂
2uj
∂
xi
2
.
The components of this equation are obtained by setting j = 1 and 2:
ρ
u1
∂
u1
∂
x1
+
ρ
u2
∂
u1
∂
x2
=−
∂
p
∂
x1
+
µ∂
2u1
∂
x1
2+
∂
2u1
∂
x2
2
%
&
‘
(
)
*
, and
ρ
u1
∂
u2
∂
x1
+
ρ
u2
∂
u2
∂
x2
=−
∂
p
∂
x2
+
µ∂
2u2
∂
x1
2+
∂
2u2
∂
x2
2
%
&
‘
(
)
*
.
b) With
u1=Cx1
and
u2=−Cx2
, the simplified continuity equation is satisfied. The two
momentum equations become:
ρ
C2x1=−
∂
p
∂
x1
, and
ρ
C2x2=−
∂
p
∂
x2
.
Integrate both to find:
p=−
ρ
C2x1
22+f(x2)
, and
p=−
ρ
C2x2
22+g(x1)
, where f and g are
single-variable functions. These two equations for p are consistent when:
f(x2)=−C2x2
22+const
, and
g(x1)=−C2x1
22+const
. Thus,
p(x1,x2)=const −C2x1
2+x2
2
( )
2
,
where the constant is readily determined as the pressure at the origin of coordinates; thus
p(x1,x2)=p(0,0) −C2x1
2+x2
2
( )
2
.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 4.44. Simplify the planar Navier-Stokes momentum equations (given in Example 4.9)
for incompressible flow, constant viscosity, and conservative body forces. Cross differentiate
these equations and eliminate the pressure to find a single equation for
ω
z = ∂v/∂x – ∂u/∂y. What
process(es) might lead to the changes in
ω
z for fluid elements in this flow?
Solution 4.44. The starting-point equations are:
ρ∂
u
∂
t+u
∂
u
∂
x+v
∂
u
∂
y
!
“
#$
%
&=−
∂
p
∂
x+
ρ
gx+
∂
∂
x
2
µ∂
u
∂
x+
µυ
−2
3
µ
!
“
#$
%
&
∂
u
∂
x+
∂
v
∂
y
!
“
#$
%
&
(
)
*+
,
–+
∂
∂
y
µ∂
v
∂
x+
∂
u
∂
y
!
“
#$
%
&
(
)
*+
,
–
, and
ρ∂
v
∂
t+u
∂
v
∂
x+v
∂
v
∂
y
!
“
#$
%
&=−
∂
p
∂
y+
ρ
gy+
∂
∂
x
µ∂
u
∂
y+
∂
v
∂
x
!
“
#$
%
&
(
)
*+
,
–+
∂
∂
y
2
µ∂
v
∂
y+
µυ
−2
3
µ
!
“
#$
%
&
∂
u
∂
x+
∂
v
∂
y
!
“
#$
%
&
(
)
*+
,
–
In two-dimensional incompressible flow, ∂u/∂x + ∂v/∂y = 0. Thus, these equations simplify to:
ρ∂
u
∂
t+u
∂
u
∂
x+v
∂
u
∂
y
!
“
#$
%
&=−
∂
p
∂
x+
ρ
gx+
∂
∂
x
2
µ∂
u
∂
x
(
)
*+
,
–+
∂
∂
y
µ∂
v
∂
x+
∂
u
∂
y
!
“
#$
%
&
(
)
*+
,
–
, and
ρ∂
v
∂
t+u
∂
v
∂
x+v
∂
v
∂
y
!
“
#$
%
&=−
∂
p
∂
y+
ρ
gy+
∂
∂
x
µ∂
u
∂
y+
∂
v
∂
x
!
“
#$
%
&
(
)
*+
,
–+
∂
∂
y
2
µ∂
v
∂
y
(
)
*+
,
–
When the body force is conservative,
g=−∇Φ
, where Φ if the body-force potential (a scalar
function). When the viscosity is constant the terms on the right may be rearranged and simplified
with the incompressibility requirement, ∂u/∂x + ∂v/∂y = 0:
ρ∂
u
∂
t+u
∂
u
∂
x+v
∂
u
∂
y
!
“
#$
%
&=−
∂
p
∂
x−
ρ
∂Φ
∂x+
µ∂
2u
∂
x2+
µ∂
∂
x
∂
u
∂
x+
∂
v
∂
y
!
“
#$
%
&+
µ∂
2u
∂
y2
=−
∂
p
∂
x−
ρ
∂Φ
∂x+
µ∂
2u
∂
x2+
∂
2u
∂
y2
!
“
#$
%
&,
ρ∂
v
∂
t+u
∂
v
∂
x+v
∂
v
∂
y
!
“
#$
%
&=−
∂
p
∂
y−
ρ
∂Φ
∂y+
µ∂
2v
∂
x2+
µ∂
∂
y
∂
u
∂
x+
∂
v
∂
y
!
“
#$
%
&+
µ∂
2v
∂
y2
=−
∂
p
∂
y−
ρ
∂Φ
∂y+
µ∂
2v
∂
x2+
∂
2v
∂
y2
!
“
#$
%
&.
Now divide each equation by
ρ
, apply –∂/∂y to the first equation, apply ∂/∂x to the second
equation and add the equations to together. This causes the body force terms to cancel out.
∂
∂x
∂
u
∂
t+u
∂
v
∂
x+v
∂
v
∂
y
“
#
$%
&
‘−∂
∂y
∂
v
∂
t+u
∂
u
∂
x+v
∂
u
∂
y
“
#
$%
&
‘
=∂
∂x−1
ρ
∂p
∂y
“
#
$%
&
‘−∂
∂y−1
ρ
∂p
∂x
“
#
$%
&
‘+
µ
∂
∂x
1
ρ
∂
2v
∂
x2+
∂
2v
∂
y2
“
#
$%
&
‘
)
*
+,
–
.−
µ
∂
∂y
1
ρ
∂
2u
∂
x2+
∂
2u
∂
y2
“
#
$%
&
‘
)
*
+,
–
..
Expand derivatives on both sides of the equation and use the incompressible flow requirement
and definition
ω
z = ∂v/∂x – ∂u/∂y to consolidate terms.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 4.45. Starting from (4.7) and (4.39b), derive a Poisson equation for the pressure, p, by
taking the divergence of the constant-density momentum equation. [In other words, find an
equation where
∂
2p
∂
xj
2
appears by itself on the left side and other terms not involving p appear
on the right side]. What role does the viscosity
µ
play in determining the pressure in constant
density flow?
Solution 4.45. For constant density flow the continuity and momentum equations are:
and
∂
uj
∂
t+ui
∂
uj
∂
xi
=−1
ρ
∂
p
∂
xj
+gj+
µ
ρ
∂
2uj
∂
xi
2
.
Apply
∂ ∂
xj
to the momentum equation and swap the order of differentiation in the first and last
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 4.46. Prove the equality of the two ends of (4.40) without leaving index notation or
using vector identities.
Solution 4.46. Equation (4.40) holds for incompressible flow where ∂ui/∂xi = 0. Therefore, start
from the first equality of (4.40), and subtract
µ
(∂/∂xj) (∂ui/∂xi) = 0 to start building something that
looks like a cross product.
µ
∇2u
( )
j=
µ∂
2uj
∂
xi
2=
µ∂
2uj
∂
xi
2−
µ∂
2ui
∂
xj
∂
xi
=−
µ∂
∂
xi
∂
ui
∂
xj
−
∂
uj
∂
xi
%
&
‘
‘
(
)
*
*
The terms in large parentheses on the right are the rotation tensor Rij from (3.13). Thus,
µ
∇2u
( )
j=−
µ∂
∂
xi
−
ε
ijk
ω
k
( )
=
µε
ijk
∂
∂
xi
ω
k=−
µε
jik
∂
∂
xi
ω
k=−
µ
∇ × ω
( )
j
,
where (3.15), Rij = –
ε
ijk
ω
k has been used.
Alternatively, start with the curl of the vorticity and expand it to show both cross
products in terms of the alternating tensor:
−
µ
∇ × ω
( )
j=−
µε
jkl
∂
∂
xk
ω
l=−
µε
jkl
∂
∂
xk
ε
lmn
∂
∂
xm
un
(
)
*
+
,
– =−
µε
jkl
ε
lmn
∂
2un
∂
xk
∂
xm
.
Now use epsilon-delta relation (2.19) for the product
ε
ijk
ε
lmn, to find:
−
µ
∇ × ω
( )
j=−
µδ
jm
δ
kn −
δ
jn
δ
km
( )
∂
2un
∂
xk
∂
xm
=−
µ∂
2uk
∂
xk
∂
xj
−
∂
2uj
∂
xk
∂
xk
(
)
*
*
+
,
–
– =
µ∂
2uj
∂
xk
2=
µ
∇2u
( )
j
.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 4.47. The viscous compressible fluid conservation equations for mass and momentum
are (4.7) and (4.38). Simplify these equations for constant-density constant-viscosity flow and
where the body force has a potential,
gj=−
∂
Φ
∂
xj
. Assume the velocity field can be found
from
uj=
∂φ ∂
xj
, where the scalar function
φ
depends on space and time. What are the
simplified conservation of mass and momentum equations for
φ
?
Solution 4.47. For constant density, the continuity equation simplifies:
0=
∂ρ
∂
t+
∂
∂
xi
ρ
ui
( )
=
∂ρ
∂
t+ui
∂ρ
∂
xi
+
ρ∂
ui
∂
xi
=
ρ∂
ui
∂
xi
, or
.
For the momentum equation with constant density and constant viscosities, the viscous stress
Note that
∂φ
∂
xi
∂
∂
xi
∂φ
∂
xj
=
∂φ
∂
xi
∂
∂
xj
∂φ
∂
xi
=1
2
∂
∂
xj
∂φ
∂
xi
$
%
&
‘
(
)
2
, rearrange terms, and insert
gj=−
∂
Φ
∂
xj
:
∂
∂
xj
ρ∂φ
∂
t+
ρ
2
∂φ
∂
xi
%
&
‘
(
)
*
2
%
&
‘
‘
(
)
*
* =−
∂
∂
xj
p+
ρ
Φ
( )
+
µ∂
∂
xj
∂
2
φ
∂
xi
2
.
The final term is zero from the continuity equation, put everything on one side, pull out
∂ ∂
xj
,
and divide by
ρ
:
∂
∂
xj
∂φ
∂
t+1
2
∂φ
∂
xi
$
%
&
‘
(
)
2
+p
ρ
+Φ
$
%
&
&
‘
(
)
) =0
. Thus, the simplified equations are
∂
2
φ
∂
xi
2=0
, and
∂φ
∂
t+1
2
∂φ
∂
xi
$
%
&
‘
(
)
2
+p
ρ
+Φ=f(t)
,
which is the Laplace equation and an unsteady Bernoulli equation. Here, f(t) is an undetermined
function of time that may be set by boundary conditions on
φ
and p. These are the equations for
unsteady potential flow, and can be used to solve fluid flow problems when: i) the fluid’s density
is constant, ii) the fluid’s viscosity is constant, iii) the flow is irrotational, and iv) only a no
through-flow boundary condition is required on solid surfaces. The no slip boundary condition
cannot be satisfied (in general) with these equations.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 4.48. The viscous compressible fluid conservation equations for mass and momentum
are (4.7) and (4.38).
a) In Cartesian coordinates (x,y,z) with
g=gx, 0, 0
( )
, simplify these equations for unsteady one-
dimensional unidirectional flow where:
ρ
=
ρ
(x,t)
and
u=u(x,t), 0, 0
( )
.
b) If the flow is also incompressible, show that the fluid velocity depends only on time, i.e.,
u(x,t)=U(t)
, and show that the equations found for part a) reduce to
∂ρ
∂
t+u
∂ρ
∂
x=0
, and
ρ∂
u
∂
t=−
∂
p
∂
x+
ρ
gx
.
c) If
ρ
=
ρ
o(x)
at t = 0, and
u=U(0) =Uo
at t = 0, determine implicit solutions for
ρ
=
ρ
(x,t)
and
U(t)
in terms of x, t,
ρ
o(x)
, Uo,
∂
p
∂
x
, and gx.
Solution 4.48. a) For one-dimensional flow:
∂ ∂
y=
∂ ∂
z=0=v=w
, so the continuity equation
reduces to:
∂ρ
∂
t+u
∂ρ
∂
x+
ρ∂
u
∂
x=0
.
Similarly, the Navier-Stokes momentum equation in the x-direction becomes:
ρ∂
u
∂
t+
ρ
u
∂
u
∂
x=−
∂
p
∂
x+
ρ
gx+
µυ
+4
3
µ
( )
∂
2u
∂
x2
.
b) Incompressible flow means
∇ ⋅ u=0
. Here this implies:
∂
u
∂
x=0
, thus the equations for part
a) simplify to:
∂ρ
∂
t+u
∂ρ
∂
x=0
, and
ρ∂
u
∂
t=−
∂
p
∂
x+
ρ
gx
.
c) For
∂
u
∂
x=0
,
u(x,t)=U(t)
. Therefore, start with what’s left of the continuity equation and
look for a method-of-characteristics solution, i.e. set
ρ
=
ρ
x(t),t
( )
so that
d
ρ
x(t),t
( )
dt =
∂ρ
∂
t+dx
dt
$
%
& ‘
(
)
∂ρ
∂
x=0
,
where the characteristic curves are specified by:
dx dt =u(x,t)=U(t)
, or
x=Udt +xo
o
t
∫
. The
u(x,t)=U(t)
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 4.49. a) Derive the following equation for the velocity potential for irrotational inviscid
compressible flow in the absence of a body force:
∂
2
φ
∂
t2+
∂
∂
t∇
φ
2
( )
+1
2∇
φ
⋅ ∇ ∇
φ
2
( )
−c2∇2
φ
=0
where
, as usual. Start from the Euler equation (4.41), use the continuity equation,
assume that the flow is isentropic so that p depends only on
ρ
, and denote
∂
p
∂ρ
( )
s=c2
.
b) What limit does c → ∞ imply?
c) What limit does
∇
φ
→ 0 imply?
Solution 4.49. a) The path to the final equation may not be obvious, so the steps listed here may
not necessarily appear in logical sequence at first glance. Start with
∂
u
∂
t+(u⋅ ∇)u=−1
ρ
∇p
, and
use the vector identity (B3.9) for the nonlinear advective acceleration to find:
∂
u
∂
t+∇u2
2
$
%
&
&
‘
(
)
) +1
ρ
∇p=−ω×u=0
where the final equality holds in irrotational flow. Take the time derivative of this equation to
find:
∂
2u
∂
t2+
∂
∂
t∇u2
2
$
%
&
&
‘
(
)
) +
∂
∂
t
1
ρ
∇p
$
%
&
‘
(
) =0
(1)
Here we assume that the flow isentropic so that the density is only a function of the pressure,
ρ
=
ρ
(p)
; this means that the final term of (1) can be rewritten:
∂
∂
t
1
ρ
∇p
%
&
‘
(
)
* =−1
ρ
2
d
ρ
dp
∂
p
∂
t∇p+1
ρ
∇
∂
p
∂
t=∇1
ρ
%
&
‘
(
)
*
∂
p
∂
t+1
ρ
∇
∂
p
∂
t=∇1
ρ
∂
p
∂
t
%
&
‘
(
)
*
.
Thus with
∇
φ
=u
, (1) becomes:
∇
∂
2
φ
∂
t2+1
2
∂
∂
t∇
φ
2+1
ρ
∂
p
∂
t
&
‘
(
)
*
+ =0
.
This means the quantity in parentheses is at most a function of time. Thus a suitable redefinition
of
φ
will produce:
∂
2
φ
∂
t2+1
2
∂
∂
t∇
φ
2+1
ρ
∂
p
∂
t=0
(2)
Save this result, and start again with the Euler equation, but this time take a dot product with u:
∂
∂
t
u2
2
#
$
%
%
&
‘
(
( +u⋅ ∇ u2
2
#
$
%
%
&
‘
(
( +1
ρ
u⋅ ∇p=0
Insert
into the first two terms of this equation and add it to (2)
∂
2
φ
∂
t2+
∂
∂
t∇
φ
2+∇
φ
⋅ ∇ 1
2∇
φ
2
&
‘
( )
*
+ +1
ρ
∂
p
∂
t+1
ρ
u⋅ ∇p=0
Now use the isentropic relationship,
dp =c2d
ρ
, to eliminate p.
∂
2
φ
∂
t2+
∂
∂
t∇
φ
2+∇
φ
⋅ ∇ 1
2∇
φ
2
&
‘
( )
*
+ +c21
ρ
∂ρ
∂
t+1
ρ
u⋅ ∇
ρ
–
.
/
0
1
2 =0
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 4.50. Derive (4.43) from (4.42).
Solution 4.50. The entire statement of equation (4.42) is:
u=dx
dt =dX
dt +d“
x
dt =U+d
dt “
x
1“
e
1+“
x
2“
e
2+“
x
3“
e
3
( )
=U+d“
x
1
dt “
e
1+d“
x
2
dt “
e
2+d“
x
3
dt “
e
3+“
x
1
d“
e
1
dt +“
x
2
d“
e
2
dt +“
x
3
d“
e
3
dt =U+“
u +Ωד
x
,
Thus, we see that
d“
x dt =“
u +Ωד
x
. Now time differentiate u, to find:
du
dt ≡a=dU
dt +d#
u
dt +d
dt Ω×#
x
( )
=dU
dt +d
dt #
u
1#
e
1+#
u
2#
e
2+#
u
3#
e
3
( )
+dΩ
dt ×#
x +Ω×d#
x
dt
=dU
dt +d#
u
1
dt #
e
1+d#
u
2
dt #
e
2+d#
u
3
dt #
e
3+#
u
1
d#
e
1
dt +#
u
2
d#
e
2
dt +#
u
3
d#
e
3
dt +dΩ
dt ×#
x +Ω×#
u +Ω×#
x
( )
Here the various terms written in component form may be identified. The second through fourth
terms are the fluid particle acceleration, a´, observed in the non-inertial frame of reference. The
fifth through seventh terms, which involve the time derivatives of the unit vectors, can be written
in terms of a cross product:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 4.51. Observations of the velocity u´ of an incompressible viscous fluid are made in a
frame of reference rotating steadily at rate Ω = (0, 0, Ωz). The pressure at the origin is po and g =
–gez.
a) In Cartesian coordinates with u´ = (U, V, W) = a constant, find p(x,y,z).
b) In cylindrical coordinates with u´ = –ΩzRe
ϕ
, determine p(R,
ϕ
,z). Guess the result if you can.
Solution 4.51. The pressure and velocity of an incompressible viscous fluid in a rotating
coordinate system both appear in (4.45).
ρ
!
D!
u
Dt
“
#
$%
&
‘
O!
1!
2!
3
=−!
∇p+
ρ
g−dU
dt −2Ω×!
u−dΩ
dt ×!
x−Ω×Ω×!
x
( )
,
–
./
0
1+
µ
!
∇2!
u
a) With U = 0, Ωz = const., u´ = const., g = –gez, , this equation reduces to:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 4.52. For many atmospheric flows, rotation of the earth is important. The momentum
equation for inviscid flow in a frame of reference rotating at a constant rate
Ω
is:
∂
u
∂
t+(u⋅ ∇)u=−∇Φ − 1
ρ
( )
∇p−2Ω×u−Ω×(Ω×x)
For steady two-dimensional horizontal flow,
u=(u,v, 0)
, with Φ = gz and
ρ
=
ρ
(z), show that the
streamlines are parallel to constant pressure lines when the fluid particle acceleration is
dominated by the Coriolis acceleration
(u⋅ ∇)u<< 2Ω×u
, and when the local pressure gradient
dominates the centripetal acceleration
Ω×(Ω×x)<< ∇p
ρ
. [This seemingly strange result
governs just about all large-scale weather phenomena like hurricanes and other storms, and it
allows weather forecasts to be made based on surface pressure measurements alone.]
Hints. 1. If Y(x) defines a streamline contour, then
dY dx =v u
is the streamline slope.
2. Write out all three components of the momentum equation and build the ratio v/u.
3. Using hint 1., the pressure increment along a streamline is:
dp =
∂
p
∂
x
( )
dx +
∂
p
∂
y
( )
dY
Solution 4.52. Since the rotation rate is steady, choose the coordinate system so that the z-axis is
parallel to Ω: i.e.
Ω=(0,0,Ωz)
. When the flow is steady,
Ω×(Ω×x)<< ∇p
ρ
.
(u⋅ ∇)u<< 2Ω×u
, the viscous terms are neglected, and Φ = gz with
ρ
=
ρ
(z), the momentum
equation becomes:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 4.53. Show that (4.55) can be derived from (4.7), (4.53), and (4.54).
Solution 4.53. Start with (4.53),
∂
∂
t
ρ
e+1
2uj
2
!
“
#$
%
&
‘
(
)*
+
,+
∂
∂
xi
ρ
e+1
2uj
2
!
“
#$
%
&ui
‘
(
)*
+
,=
ρ
giui+
∂
∂
xi
Tijuj
( )
−
∂
qi
∂
xi
,
and work on the left side terms first:
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 4.54. Multiply (4.22) by uj and sum over j to derive (4.56).
Solution 4.54. The goal here is to develop the partial differential equation that governs the
kinetic (or mechanical) energy,
1 2
( )
ρ
uj
2
, in a fluid flow. Start from (4.22), multiply this by uj
and consider the implied sum over j to find:
uj
∂
∂
t
ρ
uj
( )
+uj
∂
∂
xi
ρ
uiuj
( )
=
ρ
ujgj+uj
∂
∂
xi
Tij
( )
.
Expand and regroup the terms on the left side,
uj
∂
∂
t
ρ
uj
( )
+uj
∂
∂
xi
ρ
uiuj
( )
=
ρ
uj
∂
uj
∂
t+ujuj
∂ρ
∂
t+
∂
∂
xi
ρ
ui
( )
$
%
&
‘
(
)
+
ρ
ujui
∂
uj
∂
xi
.
Recognizing that the contents of the [,]-brackets are zero because of (4.7), and simplify the terms
on the left to find:
ρ∂
∂
t
uj
2
2
!
“
#
#
$
%
&
&+
ρ
ui
∂
∂
xi
uj
2
2
!
“
#
#
$
%
&
&=
ρ
ujgj+uj
∂
∂
xi
Tij
( )
.
Now substitute for the stress tensor from (4.27):
ρ∂
∂
t
uj
2
2
!
“
#
#
$
%
&
&+
ρ
ui
∂
∂
xi
uj
2
2
!
“
#
#
$
%
&
&=
ρ
ujgj+uj
∂
∂
xi
−p
δ
ij +
τ
ij
( )
,
use the definition of D/Dt and utilize the index-exchange property of
δ
ij, to reach
ρ
D
Dt
uj
2
2
!
“
#
#
$
%
&
&=
ρ
ujgj−uj
∂
p
∂
xj
+uj
∂τ
ij
∂
xi
,
which is (4.56).
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 4.55. Starting from
ε
=1
ρ
( )
τ
ij Sij
, derive the right most expression in (4.58).
Solution. 4.55. First determine
τ
ij from (4.27) and (4.37),
τ
ij = 2
µ
Sij −1
3Smm
δ
ij
“
#
$%
&
‘+
µυ
Smm
δ
ij
,
then substitute this into
ε
=1
ρ
( )
τ
ij Sij
to find:
ε
≡1
ρτ
ij Sij =1
ρ
2
µ
Sij −1
3Smm
δ
ij
#
$
%&
‘
(+
µυ
Smm
δ
ij
#
$
%&
‘
(Sij
=2
µ
ρ
Sij −1
3Smm
δ
ij
#
$
%&
‘
(Sij +
µυ
ρ
Smm
δ
ij Sij
=2
µ
ρ
Sij −1
3Smm
δ
ij
#
$
%&
‘
(Sij −1
3Smm
δ
ij +1
3Smm
δ
ij
#
$
%&
‘
(+
µυ
ρ
Smm Sjj
=2
µ
ρ
Sij −1
3Smm
δ
ij
#
$
%&
‘
(
2
+2
µ
ρ
Sij −1
3Smm
δ
ij
#
$
%&
‘
(1
3Smm
δ
ij +
µυ
ρ
Smm Sjj
=2
µ
ρ
Sij −1
3Smm
δ
ij
#
$
%&
‘
(
2
+
µυ
ρ
Smm Sjj +2
µ
ρ
Sij
δ
ij −1
3Smm
δ
ij
δ
ij
#
$
%&
‘
(Smm
3
.
The first two terms on the last line look promising since Sjj = Smm = (∂um/∂xm)2. The third term on
the last line is readily determined to be zero since
δ
ij
δ
ij =
δ
jj = 3. Therefore the contents of the last
set of parentheses is:
Sij
δ
ij −1
3Smm
δ
ij
δ
ij =Sjj −1
3Smm (3) =0
.
So,
ε
=2
µ
ρ
S
ij
−1
3S
mm
δ
ij
&
‘
( )
*
+
2
+
µυ
ρ
S
mm
S
jj
=2
ν
S
ij
−1
3
∂
u
m
∂
x
m
δ
ij
&
‘
(
)
*
+
2
+
µυ
ρ
∂
u
m
∂
x
m
&
‘
(
)
*
+
2
,
where
ν
≡
µρ
.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 4.56. For many gases and liquids (and solids too!), the following equations are valid:
q=−k∇T
(Fourier’s law of heat conduction, k = thermal conductivity, T = temperature),
e = eo + cvT (e = internal energy per unit mass, cv = specific heat at constant volume), and
h = ho + cpT (h = enthalpy per unit mass, cp = specific heat at constant pressure),
where eo and ho are constants, and cv and cp are also constants. Start with the energy equation
ρ∂
e
∂
t+
ρ
ui
∂
e
∂
xi
=−p
∂
ui
∂
xi
+
τ
ij Sij −
∂
qi
∂
xi
for each of the following items.
a) Derive an equation for T involving uj, k,
ρ
, and cv for incompressible flow when
τ
ij = 0.
b) Derive an equation for T involving uj, k,
ρ
, and cp for flow with p = const. and
τ
ij = 0.
c) Provide a physical explanation why the answers to a) and b) are different.
Solution 4.56. a) Put
q=−k∇T
, e = eo + cvT, and
τ
ij = 0 into the energy equation to find:
ρ
cv
∂
T
∂
t+ui
∂
T
∂
xi
!
“
#$
%
&=−p
∂
ui
∂
xi
+
∂
∂
xi
k
∂
T
∂
xi
!
“
#$
%
&
.
For incompressible flow
∂
ui
∂
xi=0
, so the final equation is
ρ
cv
DT
∂
k
∂
T
!
#$
&
q=−k∇T
.
Substitute the final term into the energy equation:
ρ
De
Dt =−
ρ
D p
ρ
( )
Dt +
∂
∂
xi
k
∂
T
∂
xi
%
&
‘
(
)
*
. Collect like
terms and recall that
h=e+p
ρ
to find:
ρ
Dh
Dt =
∂
∂
xi
k
∂
T
∂
xi
$
&
‘
)
. Now use h = ho + cpT to get
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 4.57. Derive the following alternative form of (4.60):
ρ
cp
DT
Dt =
α
TDp
Dt +
ρε
+
∂
∂
xi
k
∂
T
∂
xi
!
“
#$
%
&
, where
ε
is given by (4.58) and
α
is the thermal expansion
coefficient defined in (1.26). [Hint:
dh =
∂
h
∂
T
( )
pdT +
∂
h
∂
p
( )
Tdp
]
Solution 4.57. Start from (4.60),
ρ
De
Dt =−p
∂
um
∂
xm
+2
µ
Sij −1
3
∂
um
∂
xm
δ
ij
&
‘
(
)
*
+
2
+
µυ
∂
um
∂
xm
&
‘
(
)
*
+
2
+
∂
∂
xi
k
∂
T
∂
xi
&
‘
(
)
*
+
,
and use (4.58) so simplify the appearance of the middle terms on the right:
Now work on the thermodynamic variables. Use the hint to make the choice of T and p as
independent variables, h = h(T,p), so that:
dh =
∂
h
∂
T
!
“
#$
%
&
p
dT +
∂
h
∂
p
!
“
#$
%
&
T
dp =cpdT +
∂
h
∂
p
!
“
#$
%
&
T
dp
. (†)
However, the natural variables for h are s and p, h = h(s,p), so that
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 4.58. Show that (4.68) is true without abandoning index notation or using vector
identities.
Fluid Mechanics, 6th Ed. Kundu, Cohen, and Dowling
Exercise 4.59. Consider an incompressible planar Couette flow, which is the flow between two
parallel plates separated by a distance b. The upper plate is moving parallel to itself at speed U,
and the lower plate is stationary. Let the x-axis lie on the lower plate. The pressure and velocity
fields are independent of x, and fluid has uniform density and viscosity.
a) Show that the pressure distribution is hydrostatic and that the solution of the Navier–Stokes
equation is u(y) = Uy/b.
b) Write the expressions for the stress and strain rate tensors, and show that the viscous kinetic-
energy dissipation per unit volume is
µ
U2/b2.
c) Evaluate the kinetic energy equation (4.56) within a rectangular control volume for which the
two horizontal surfaces coincide with the walls and the two vertical surfaces are perpendicular to
the flow and show that the viscous dissipation and the work done in moving the upper surface
are equal.
Solution 4.59. In this flow is one dimensional, u = (u(y), 0), and ∂u/∂x = 0. This means that the
continuity equation is satisfied, and that the flow is incompressible. Thus, the x (horizontal) and
y (vertical) components of the constant-viscosity Navier-Stokes’ momentum equation are:
∂
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−g−
ν∂
2v
∂
x2+
∂
2v
∂
y2
&
‘
(
)
*
+
.
Here, v is zero, ∂u/∂x = ∂u/∂t = ∂p/∂x = 0, so these equations simplify to:
0=
∂
2u
∂
y2
, and
0=−1
ρ
∂
p
∂
y−g
.
a) The simplified vertical mometum equation can be rearranged to: ∂p/∂y = –
ρ
g, which is the
equation of hydrostatics and is readily integrated to find p = po –
ρ
gy, when the density is
constant. The simplified horizontal momentum equation can be integrate twice to find u = Ay +
B, and the boundary conditions on the lower plate (u = 0 on y = 0) and on the upper plate (u = U
on y = b), allow the constants A (= 0) and B (= U/b) to be determined. The final velocity result is:
u(y) = Uy b.
b) Strain rate:
Sij =1
2
∂
ui
∂
xj
+
∂
uj
∂
xi
#
$
%
%
&
‘
(
( =
∂
u
∂
x1
2
∂
u
∂
y+
∂
v
∂
x
( )
1
2
∂
v
∂
x+
∂
u
∂
y
( )
∂
v
∂
y
#
$
%
&
‘
( =
0U2b
U2b0
#
$
%
&
‘
(
For incompressible flow, the stress tensor is:
Tij =−p
δ
ij +2
µ
Sij =−p
µ
U b
µ
U b −p
“
#
$
$
%
&
‘
‘
The viscous dissipation per unit volume is:
ρε
=2
µ
Sij Sij =2
µ
S12
2+S21
2
[ ]
=2
µ
U24b2+U24b2
[ ]
=
µ
U2b2
c) The steady-flow kinetic energy equation,
ρ
ui
∂
∂
xi
1
2uj
2
!
“
#$
%
&=
ρ
giu+uj
∂
∂
xi
Tij
( )
, can be rewrittten:
U
y
x
b