Fluid Dynamics 3  Solutions to Sheet 4
1. (i) For the given flow field,
u
θ
=
f
(
r
) in cylindrical
polars. Thus, applying the formula from chapter 0,
∇ ·
u
=
1
r
∂u
θ
∂θ
= 0
.
(ii) We have to solve the steady Euler equation
(
∇ ·
u
)
u
=
−∇
p/ρ.
According to chapter 0, the convective term on the
left is
(
∇ ·
u
)
u
=
−
u
2
θ
r
ˆ
r
=
−
f
2
r
ˆ
r
.
Thus
p
can only depend on
r
, so that
∇
p
=
p
′
(
r
)ˆ
r
,
and the required equation follows.
(iii) According to chapter 0,
∇ ×
u
=
ˆ
z
r
∂rf
∂r
,
which vanishes if
f
=
A/r
. From (ii), we have
p
′
(
r
) =
ρA
2
r
3
,
and so
p
(
r
) =
−
ρA
2
2
r
2
+
p
atm
.
2.
(i) Conservation of mass is volume flux in equals
volume flux out, so
Ud
=
U
1
b
1
+
U
2
b
2
(ii) Use the momentum integral theorem.
Need
to choose a control surface  this should be a
boundary where we know stuff about the solu
tion.
So follow example in lectures and take
it to be around the flow cutting the inflow
and outflows far away from the impingement.
Let
S
be the contour
ABCDEF
in the figure:
Deal with pressure term first:
A
B
C
D
E
F
integraldisplay
S
p
n
d
S
=
integraldisplay
S
(
p
−
p
atm
)
n
d
S
+
integraldisplay
S
p
atm
n
d
S
The last term is zero by the divergence theo
rem (see lectures). The first term is zero along
all parts of
S
apart from the section
DE
. The
other term is
integraldisplay
S
ρ
u
(
u
.
n
)d
S
which is zero on
AF
,
DE
,
BC
since
u
.
n
= 0
(there is no flow across these boundaries). So
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 Fall '11
 Eggers
 Fluid Dynamics, Dot Product, Patm

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