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Unformatted text preview: D Some simple ﬂows and their potentials
D.1 Two dimensional ﬂows
We will give the answer either in Cartesians or polars, depending on which is more convenient.
Reminder:
∂ψ
∂φ
∂ψ
∂φ
=
, v=
=−
where φ is the
∂x
∂y
∂y
∂x
velocity potential and ψ is the streamfunction.
(i) in Cartesians, u = (u, v, 0) and u = (ii) In polars, u = (ur , uθ , 0) and ur =
Type of ﬂow 1 ∂ψ
1 ∂φ
∂ψ
∂φ
=
, uθ =
=− .
∂r
r ∂θ
r ∂θ
∂r ﬂow ﬁeld u Horizontal
dipole,
strength µ, at r = 0 − Uz Ur cos(θ − α) Ur sin(θ − α) Uze−iα k2
(x − y 2 )
2 kxy k2
z
2 m
log r
2π mθ
2π m
ln z
2π ˆ
Γθ
2πr Line vortex, circulation Γ, at r = 0 Uy mˆ
r
2πr Line source, strength
m, at r = 0 Ux u = kx
v = −ky Stagnation point ﬂow
at origin complex pot. w ˆ
U cos αx+
ˆ
U sin αy Uniform stream at
angle α to x axis streamfunction ψ ˆ
Ux Uniform stream parallel to x axis potential φ Γθ
2π µ
z
ˆ
r
z−2 ˆ
2
2πr
r −µ cos θ
2πr − Γ
log r
2π µ sin θ
2πr − − D.2 Axisymmetric ﬂows
We use cylindrical or spherical polars, whichever is more convenient. In cylindrical coordinates, (r, θ, z ), u = (ur , uθ , uz ); in terms of the potential φ(r, z ) or the Stokes streamfunction Ψ(r, z ),
∂φ
1 ∂Ψ
∂φ
1 ∂φ
ur =
=−
, uz =
=
.
∂r
r ∂z
∂z
r ∂r
In spherical polar coordinates, (r, θ, ϕ), u = (ur , uθ , uϕ ); in terms of the potential φ(r, θ)
or the Stokes streamfunction Ψ(r, θ),
ur = 1 ∂Ψ
∂φ
=2
,
∂r
r sin θ ∂θ uθ = 1 ∂φ
1 ∂Ψ
=−
.
r ∂θ
r sin θ ∂r iΓ
ln z
2π
µ
2πz Type of ﬂow
Uniform
stream
aligned with axis
of symmetry
Stagnation point
ﬂow at origin
Point
strength
r=0 source,
m, at ﬂow ﬁeld u potential φ ˆ
Uz Uz k
r
2
uz = −kz k2
(r − 2 z 2 )
4 ur = m
ˆ
r
4πr 2 µ
z
ˆ−3 ˆ
Dipole, strength µ, −
r
z
3
4πr
r
in ˆdirection
z − m
4πr −µ cos θ
4πr 2 streamfunction Ψ coordinate system
1
Ur 2
2 − kr 2 z
2 cylindrical cylindrical m
(1 − cos θ)
4π spherical sin2 θ
4πr spherical µ ...
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This note was uploaded on 01/20/2012 for the course MATH 33200 taught by Professor Eggers during the Fall '11 term at University of Bristol.
 Fall '11
 Eggers

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