Fluid Dynamics 3  Solutions to Sheet 6
1.
(i) (a) The potential of a uniform stream is
φ
=
Uz
, and the result follows from
z/r
= cos
θ
.
(b) The potential of a dipole
μ
=
μ
ˆ
z
is
μ
· ∇
1
r
=

μ
·
x
r
3
=

μz
r
3
=

μ
cos
θ
r
2
.
(ii) The Laplacian gives (a)
4
φ
=
U
cos
θ
r
2
∂r
2
∂r

U
r
sin
θ
∂
sin
2
θ
∂θ
=
2
U
cos
θ
r

2
U
cos
θ
r
= 0
.
and (b)
4
φ
=
2
μ
cos
θ
r
2
∂r

1
∂r
+
μ
r
4
sin
θ
∂
sin
2
θ
∂θ
=

2
μ
cos
θ
r
4
+
2
μ
cos
θ
r
4
= 0
.
(iii) The velocity components are (a)
u
r
=
∂φ
∂r
=
U
cos
θ,
u
θ
=
1
r
∂φ
∂θ
=

Ur
sin
θ,
and (b)
u
r
=
∂φ
∂r
=
2
μ
cos
θ
r
3
,
u
θ
=
1
r
∂φ
∂θ
=
μ
sin
θ
r
3
.
(iv)
n
· ∇
φ
=
u
r
, and
u
r

r
=
R
=
U
cos
θ
+
2
A
cos
θ
R
3
,
so
A
=
UR
3
/
2 does the trick.
2.
(i) In three dimensions,
4
(1
/r
) = 0. Thus
4
φ
=
A
· ∇4
1
r
= 0
.
Explicit calculation gives
φ
=
A
· ∇
1
r
=

A
·
r
r
3
.
(ii) The boundary condition is
u
·
n
=
U
·
n
.
Now
u
i
=
∂
i
φ
=

∂
i
A
j
r
j
r
3
=

A
i
r
3
+
3
A
j
r
j
r
i
r
5
,
and hence
u
·
n

r
=
R
=

A
·
n
R
3
+
3
A
·
n
R
3
=
2
A
·
n
R
3
.
Thus using the boundary condition we have
U
= 2
A
/R
3
and
φ
=

R
3
2
r
2
U
·
n
.
(iii) Since the centre of the sphere moves according
to
x
=
U
t
, the above potential has the form
φ
(
x
, t
) =
φ
(
x

U
t
), and thus
∂φ
∂t
=

U
∇
φ
=

R
3
2
r
3
3(
U
·
n
)
2

U
2
.
This means one has to use the timedependent
form of Bernoulli’s equation
u
2
2
+
p
ρ
=
p
atm
ρ

∂φ
∂t
.
Now
u
=
∇
φ
=
R
3
2
r
3
[3
n
(
U
·
n
)

U
]
,
and consequently
u
2
=
R
6
4
r
6
3(
U
·
n
)
2
+
U
2
.
Thus the pressure distribution around the sur
face is
p

p
atm

r
=
R
=
ρU
2
8
(
9 cos
2
θ

5
)
,
where
θ
is the angle between
U
and
n
. This is
the same as found in the lecture for a sphere
1
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 Fall '11
 Eggers
 Trigraph, SEPTA Regional Rail, Boundary value problem, φ

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