ES330 Assignment 10 Solutions
Chapters 8 and 9
Due Date: Thursday December 2, 2010
I
Problem 1
Consider the steady, twodimensional, incompressible velocity field,
~
V
= (
u, v
) = (
ax
+
b
)
ˆ
i
+(

ay
+
c
)
ˆ
j
, where
a
,
b
, and
c
are constants. Calculate the pressure as a function on
x
and
y
.
Solution
To find the pressure, we need to solve the 2D Navier Stokes Equations. First, we will list
the assumptions given in the problem statement.
Assumptions
1. Steady and Laminar(
∂
∂t
≡
0)
2. Incompressible, Newtonian Flow (
ρ
≈
constant and
μ
≈
constant)
3. We know
∂u
∂x
=
a
,
∂v
∂y
=

a
, and
∂u
∂y
=
∂v
∂x
=
∂
2
u
∂x
2
=
∂
2
u
∂y
2
=
∂
2
v
∂x
2
=
∂
2
v
∂y
2
= 0.
xComponent
ρ
∂u
∂t
+
u
∂u
∂x
+
v
∂u
∂y
=

∂P
∂x
+
ρg
x
+
μ
"
∂
2
u
∂x
2
+
∂
2
u
∂y
2
#
(1)
∂P
∂x
=

ρu
∂u
∂x
=

ρ
(
ax
+
b
)(
a
)
P
=

ρ
Z
(
a
2
x
+
ab
)
dx
=

ρ
a
2
x
2
2
+
abx
+
f
(
y
) +
C
yComponent
ρ
∂v
∂t
+
u
∂v
∂x
+
v
∂v
∂y
=

∂P
∂y
+
ρg
y
+
μ
"
∂
2
v
∂x
2
+
∂
2
v
∂y
2
#
(2)
∂P
∂y
=

ρv
∂v
∂y
=

ρ
(

ay
+
c
)(

a
)
P
=
ρ
Z
(

a
2
y
+
ac
)
dy
=
ρ

a
2
y
2
2
+
acy
+
f
(
x
) +
C
C is some arbitrary scalar. We can see that
f
(
y
) =
ρ
h

a
2
y
2
2
+
acy
i
and
f
(
x
) =

ρ
h
a
2
x
2
2
+
abx
i
,
therefore:
P
=

ρa
ax
2
2
+
bx
+
ay
2
2

cy
+
C
1
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II
Problem 2
Consider steady, incompressible, laminar flow of a Newtonian fluid in an infinitely long round pipe
annulus of inner radius
R
i
and outer radius
R
o
.
Ignore effects of gravity.
A constant negative
pressure gradient
∂P/∂x
is applied in the
x

direction,
∂P
∂x
=
P
2

P
1
x
2

x
1
, where
x
1
and
x
2
are arbitrary
locations along the
x

axis, and
P
1
and
P
2
are the pressures at those two locations. The pressure
gradient may be caused by the pump and/or gravity. Note that we adopt a modified cylindrical
coordinate system here with
x
instead of
z
for the axial component, namely, (
r, θ, x
) and (
u
r
, u
θ
, u
x
).
Derive an expression for the velocity field in the annular space in the pipe.
Solution
To solve for the velocity field, we will need to solve the Navier Stokes and Continuity
equations in cylindrical coordinates.
Here are the assumptions we will use to help us
reduce the equations to a solvable form.
Assumptions
1. Steady and Laminar(
∂
∂t
≡
0)
2. Incompressible, Newtonian Flow (
ρ
≈
constant and
μ
≈
constant)
3. Axisymmetric (
∂
∂θ
≡
0,
u
θ
≡
0)
4. Negligible End EffectsNo boundary layer effects because infinitely long (
∂
∂x
≡
0)
5. Boundary Conditions:
 Due to No Slip
u
x

r
=
R
i
=
u
x

r
=
R
o
= 0
 Flow can’t penetrate a solid wall
u
r

r
=
R
i
=
u
r

r
=
R
o
= 0
Continuity
1
r
∂
(
ru
r
)
∂r
+
1
r
∂u
θ
∂θ
+
∂u
z
∂z
= 0
(3)
This tells us that
u
r
does not depend on
r
, and therefore, since is doesn’t depend on
θ
(Assumption 3) or
x
(Assumption 4), it must be a constant. Also, since we know
u
r
= 0
at the walls and it is constant, it must be zero everywhere. Therefore
u
r
≡
0 everywhere.
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 Spring '11
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 Fluid Dynamics, Trigraph, dx, ∂x

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