Simple Potential flows
Dehan Yuan, Dixia Fan
November 2, 2022
1
Simple Potential flows
1. Uniform Stream ∇2 (ax + by + cz + d) = 0
1D:ϕ = U x + constantψ = U y + constant,
⃗v = (U, 0, 0)
2D:ϕ = U x + V y + constantψ = U y − V x + constant,
3D:ϕ = U x + V y + W z + constant,
⃗v = (U, V, 0)
⃗v = (U, V, W )
2. Source(Sink) flow
2D, Polar coordinates
∇2 =
p
1 ∂
∂
1 ∂2
(r ) + 2 2 , with r = x2 + y 2
r ∂r ∂r
r ∂θ
An axisymmetric solution: pϕ = a ln r + b. Verify that it satisfies
∇2 ϕ = 0, except at r =
x2 + y 2 = 0. Therefore, r=0 must be
excluded from the flow.
Define 2D source of strength m at r=0:
1
ϕ=
∂ϕ
m
m
ln r∇ϕ =
êr ⇔ vr =
, vθ = 0
2π
∂r
2πr
Net outward volume flux is:
˛
¨
¨
⃗v · n̂ds =
∇ · ⃗v ds =
∇ · ⃗v ds
S
Sε
˛
¨ 2π
⃗v · n̂ds =
vr rε dθ =
m
|{z}
|{z}
Cε
0
source strength
m
2πrε
C
If m < 0 ⇒ sink. Source m at (x0 , y0 ):
2
m p
ln (x − x0 )2 + (y − y0 )2
2π
m
m
ln r(potential function) ⇔ ψ =
θ(Stream function)
ϕ=
2π
2π
ϕ=
3D: Spherical coordinates
∇2 =
p
1 ∂ 2∂
∂ ∂
x2 + y 2 + z 2
(r
)
+
(
,
,
...),
where
r
=
r2 ∂r ∂r
∂θ ∂φ
a
A spherically symmetric solution: ϕ = + b. Verify ∇2 ϕ = 0 except
r
at r = 0.
Define a 3D source of strength m at r = 0. Then
ϕ=−
m
∂ϕ
m
⇔ vr =
=
, vθ = 0, vφ = 0
4πr
∂r
4πr2
Net outward volume flux is
!
vr dS = 4πrε2 ·
m
= m (m < 0 for a sink)
4πrε2
3. 2D point vortex
3
∇2 =
1 ∂
1 ∂2
+ 2 2
r ∂r r ∂θ
Another particular solution: ϕ = aθ + b. Verify that ∇2 ϕ = 0 except
at r = 0.
Define the potential for a point vortex of circular Γ at r = 0. Then
ϕ
1 ∂ϕ
Γ
Γ
θ ⇔ vr =
= 0, vθ =
=
2π
∂r
r ∂θ
2πr
1 ∂
ωz =
(rvθ ) = 0 except at r = 0
r ∂r
ϕ=
and
Stream function:
ψ=−
Γ
ln r
2π
Circulation:
ˆ
ˆ
⃗v · d⃗x =
C1
ˆ
⃗v · d⃗x +
C2
ˆ 2π
⃗v · d⃗x =
C1 −C2
| ˜ {z
}
S ωz dS=0
4
0
Γ
rdθ =
2πr
Γ
|{z}
vortex strength
4. Dipole(doublet flow)
A dipole is a superposition of a sink and a source with the same
strength.
2D dipole:
p
m p
[ln (x − a)2 + y 2 − ln (x + a)2 + y 2 ]
2π
p
µ
∂
lim ϕ =
ln (x − ξ)2 + y 2 |ξ=0
a→0
2π
∂ξ
|{z}
ϕ=
µ=2ma,constant
=−
µ
x
µ x
=−
2
2
2π x + y
2π r2
5
2D dipole (doublet) of moment µ at the origin oriented in the +x direction.
NOTE: dipole = µ
ϕ=−
∂
(unit source)
∂ξ
µ x cos α + y sin α
−µ cos θ cos α + sin θ sin α
=
2
2
2π
x +y
2π
r
3D dipole:
1
m
1
(p
−p
), where µ = 2ma fixed
a→0
4π
(x − a)2 + y 2 + z 2
(x + a)2 + y 2 + z 2
1
x
µ x
µ ∂
µ
p
=−
=−
|ξ=0 −
2
2
2
3/2
2
2
2
4π ∂ξ (x − ξ) + y + z
4π (x + y + z )
4π r3
ϕ = lim −
3D dipole (doublet) of moment µ at the origin oriented in the +x direction.
5. Stream and source: Rankine half-body
It is the superposition of a uniform stream of constant speed U and a
source of strength m.
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2D: ϕ = U x +
m p 2
ln x + y 2
2π
m
x
∂ϕ
=U+
2
∂x
2π x + y 2
m
u|y=0 = U +
, v|y=0 = 0 ⇒
2πx
−m
,y = 0
V⃗ = (u, v) = 0 at x = xs =
2πU
u=
For large x, u → U , and U D = m by continuity ⇒ D =
3D:ϕ = U x −
m
p
4π x2 + y 2 + z 2
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m
U
m
x
∂ϕ
=U+
2
2
∂x
4π (x + y + z 2 )3/2
m x
u|y=z=0 = U +
, v|y=z=0 = 0, w|y=z=0 = 0 ⇒
4πx |x|3
−m
V⃗ = (u, v, w) = 0 at x = xs =
,y = z = 0
4πU
u=
For large x, u → U , and U A = m by continuity ⇒ A =
m
U
6. Stream + source/sink pair: Rankine closed bodies
To have a closed body, a necessary condition is to have Σmin body = 0
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2D Rankine ovoid:
ϕ = Ux −
1
m
1
[p
−p
]
2
2
2
2
4π
(x + a) + y + z
(x − a) + y 2 + z 2
For Rankine Ovoid,
m
x+a
x−a
∂ϕ
=U+
[
−
]
2
2
2
3/2
2
∂x
4π ((x + a) + y + z )
((x − a) + y 2 + z 2 )3/2
m
1
1
u|y=z=0 = U +
[
−
]
2
4π (x + a)
(x − a)2
m (−4ax)
=U+
4π (x2 − a2 )2
m
u|y=z=0 = 0 at (x2 − a2 )2 = (
)4ax
4πU
U=
At x = 0,
u=U+
2a
m
where R = y 2 + z 2
2
4π (a + R2 )3/2
Determine radius of body R0 :
ˆ R0
2π
uRdR = m
0
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7. Stream + Dipole: circles and spheres
2D: ϕ = U x +
µx
µ
)
= cos θ(U r +
|{z}
2
2πr
2πr
x=r cos θ
The radial velocity is then
ur =
∂ϕ
µ
= cos θ(U −
)
∂r
2πr2
r
µ
. This
2πr2
is the K.B.C. for a stationary circle of radius a. Therefore, for
Setting the radial velocity vr = 0 on r = a we obtain a =
µ = 2πU a2
the potential
ϕ = cos θ(U r +
µ
)
2πr
is the solution to ideal flow past a circle of radius a.
•Flow past a circle (U, a).
10
a2
ϕ = U cos θ(r + )
r
a2
1 ∂ϕ
= −U sinθ (1 + 2 )
Vθ =
r ∂θ
r

 = 0 at θ = 0, π − stagnation points
Vθ |r=a = −2U sin θ
 = ∓2U at θ = π , 3π − maximum tangential velocity
2 2
Illustration of the points where the flow reaches maximum speed around
the circle.
3D: ϕ = U x +
µ cos θ
µ
µ
+
= U r cos θ(1 +
)
2
4π 4π r
4πr3
The radial velocity is then
vr =
∂ϕ
µ
)
= cos θ(U −
∂r
2πr3
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r
µ
. This
2πU
is the K.B.C. for a stationary sphere of radius a. Therefore, choosing
Setting the radial velocity vr = 0 on r = a we obtain a = 3
µ = 2πU a3
µ
the potential ϕ = U r cos θ(1 +
) is the solution to ideal flow past a
4πr3
sphere of radius a.
•Flow past a sphere (U, a).
ϕ = U r cos θ(1 +
a3
)
2r3
1 ∂ϕ
a3
= −U sinθ (1 + 3 )
r ∂θ
2r

 = 0 at θ = 0, π
3U
sin θ
Vθ |r=a = −
 = − 3U at θ = π
2
2
2
Vθ =
2D corner flow Velocity potential ϕ = rα cos αθ; Stream function ψ = rα sin αθ
(a) ∇2 ϕ = (
∂2
1 ∂
1 ∂2
+
+
)ϕ = 0
∂r2 r ∂r r2 ∂θ2
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(b)
∂ϕ
= αrα−1 cos αθ
∂r
1 ∂ϕ
uθ =
= −αrα−1 sin αθ
r ∂θ
so uθ = 0( or ψ = 0) on αθ = nπ, n = 0, ±1, ±2...
π 2π
i.e., on θ = θ0 = 0, , , ...(θ0 ≤ 2π)
α α
ur =
i. Interior corner flow-stagnation point origin:α > 1. For example,
α = 1, θ0 = 0, π, 2π, u = 1, v = 0
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ii. Exterior corner flow, |v| → ∞ at origin:
α<1
θ0 = 0,
π
only
α
π
Since we need θ0 ≤ 2π, we therefore require ≤ 2π, i.e., α ≥ 1/2 only.
α
For example,
α = 1/2, θ0 = 0, 2π (1/2 infinite plate, ow around a tip)
14
α = 2/3, θ0 = 0,
3π
(90o exterior corner)
2
In general the potential for a corner located at (x0 , y0 ) with an angle θ0
with respect to the positive x axis is
p
y − y0
ϕ = rα cos αθ = [ (x − x0 )2 + (y − y0 )2 ]α cos {α[tan−1 (
) − θ0 ]}
x − x0
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