1
Lecture Notes on Fluid Dynamics
(1.63J/2.21J)
by Chiang C. Mei, MIT
2-6oseen.tex
[Ref] Lamb : Hydrodynamics
2.6
2.6.1
Oseen’s improvement for slow flow past a cylinder
Oseen’s criticism of Stokes’ approximation
Is Stokes approximtion always good for small Reynolds number?
The exact Navier Stokes equations read
∇ · q~ = 0
(2.6.1)
~q · ∇~q = −∇p + ν∇2 q~.
(2.6.2)
Let us estimate the error of Stokes approximation where the velocity field is given by
3a
a3
3a
a3
, qz = −W sin θ 1 − 3 −
.
qy = W cos θ 1 + 3 −
2r
2r
4r
4r
!
!
(2.6.3)
In the far field r/a 1, the viscous stress is dominated by the last term
a3
∇ q~ = O 3 .
r
2
!
The inertia term is dominated by
∂~q
a2
W
∼O 2 .
∂z
r
!
Hence the error is their ratio
∂~
q
W ∂z
r
=O
,
2
ν∇ ~q
a
which becomes unbounded for r/a 1. Thus inertia cannot be ignored in the far field.
The difficulty is severe for the two-dimensional flow past a cylinder. By taking the curl,
Stokes equation gives
∇2 ζ~ = 0.
Since the body is a source of vorticity, ζ~ would become unbounded logarithmically for large
r/a. This is certainly unphysical and is known as Stokes’ paradox.
2
In view of the increasing importance of inertia in the far field, Oseen suggests that the
linear approximation of the inertia term, which is of dominant importance in the far field,
and not in the near field, be added. Let ~q = W ~k + q~0 the Oseen equations are:
∇ · q~0 = 0
W
(2.6.4)
∇p
∂ q~0
=−
+ ν∇2 q~0 .
∂z
ρ
(2.6.5)
The added linear inertia term is of dominant importance in the far field and not in the near
field.
We shall demonstrate the Oseen approximation for the two-dimensional case of a circular
cylinder.
2.6.2
Oseen’s theory for a circular cylinder
Because of (2.6.4), the pressure is still harmonic
∇2 p = 0.
(2.6.6)
Now the velocity can be expressed as the sum of a potential part ∇φ and a solenoidal part
q~00 , where
q~0 = ∇φ + q~00
(2.6.7)
It follows that
∇2 φ = 0
(2.6.8)
and vorticity is only associated with q~00 . Substituting Eq. () into Eq. () and setting
p = p∞ = −ρW
∂φ
∂z
(2.6.9)
we get
W
∂ q~00
= ν∇2 q~00 ,
∂z
(2.6.10)
∂ ζ~
~
= ν∇2 ζ,
∂z
(2.6.11)
which implies
W
where ζ~ = ∇ × q~00 . To solve for q~00 let us introduce a vector potential ~kσ
ζ~ = ∇ × (~kσ).
Now
∂σ ~ 2
∇ × ζ~ = ∇ × (∇ × ~kσ) = ∇(∇ · (~kσ)) − ~k∇2 σ = ∇
− k∇ σ.
∂z
(2.6.12)
3
On the other hand
W ∂ q~00
∇ × ζ~ = ∇ × (∇ × q~00 ) = ∇(∇ · q~00 ) − ∇2 q~00 = −∇2 q~00 = −
.
ν ∂z
after using (2.6.10). It follows by equating the two preceding equations,
−
W ∂ q~00
∂σ ~ 2
=∇
− k∇ σ.
ν ∂z
∂z
(2.6.13)
Now (2.6.12) does not define σ uniquely. Let us impose a condition
∂σ
= ν∇2 σ.
∂z
(2.6.14)
W~
W ~00
q = ∇σ −
k σ,
ν
ν
(2.6.15)
W
Then we can integrate to get
−
hence,
ν
q~00 = ~k σ − ∇σ.
W
In polar coordinates the velocity components are
(2.6.16)
ν ∂σ
W ∂r
(2.6.17)
ν 1 ∂σ
.
W r ∂θ
(2.6.18)
∂φ
ν ∂σ
+ σ cos θ −
∂r
W ∂r
(2.6.19)
qr00 = σ cos θ −
qθ00 = −σ sin θ −
The total disturbance is
qr0 =
1 ∂φ
ν 1 ∂σ
− σ sin θ −
.
r ∂θ
W r ∂θ
It is convenient to rearrange (2.6.14) by introducing
qθ0 =
(2.6.20)
σ(y, z) = eW z/2ν σ̄(y, z)
so that
W
2ν
We now introduce dimensionless variables
∇2 σ̄ −
φ = W a Φ,
σ = W Σ,
2
(2.6.21)
σ̄ = 0,
σ̄ = W Σ̄,
~
q~0 = W Q,
(2.6.22)
r = aR.
(2.6.23)
Then the governing equatios are
∇2 Φ = 0
(2.6.24)
4
and
(∇2 − 2 )Φ = 0
=
Wa
2ν
(2.6.25)
The velocity components are
1 ∂Φ
∂Φ
+ σ cos θ −
∂R
2 ∂R
1 ∂Φ
1 1 ∂Φ
=
− φ sin θ −
.
R ∂R
2 R ∂θ
Qr =
Qθ
(2.6.26)
The boundary conditions are
QR + cos θ = 0 R = 1
(2.6.27)
Qθ − sin θ = 0 R = 1.
(2.6.28)
The general solution to (2.6.24) and (2.6.25) is
Φ = A0 ln R + A1
cos θ
R
(2.6.29)
and
σ = C0 K0 (R)
(2.6.30)
We shall determine the coefficients A0 A1 and C0 approximately for small = Re/2.
On the cylinder R = 1, we use the approximation for R 1
1
K0 (R) ∼
= − σ + ln
2
1
∼ − σ + ln
=
2
R
2 R 2
I0 (R) +
+···
2
2
R
+ ...
2
(2.6.31)
σ ∼
= C0 eR cos θ K0 (R)
1
R
∼
+ · · ·)
= −C0 (1 + R cos θ + · · ·)(σ + ln
2
2
Details of calculation for small R are given for convenience
∂Φ
A0 A1
=
− 2 cos θ
∂R
R
R
1 ∂Φ
A1
=−
sin θ |1
R ∂θ
R
R
cos θ
2
R
∼
−σ sin θ |1 = +C0 (1 + R cos θ + · · ·) σ + ln
sin θ
2
σ cos θ |1 ∼
= −C0 (1 + R cos θ + · · ·) σ + ln
(2.6.32)
5
1
R
1 ∂σ ∼
C0 cos θ(1 + R cos θ) σ + ln
+···
= +
2 ∂R
2
2
1
1
+ Co (1 + R cos θ + · · ·) −
+···
2
R
1 1
1 1 ∂σ ∼
−
C0 (−R sin θ)(1 + cos θ) σ + ln
= +
2 R ∂θ
2 R
2
−
On the cyolinder R = 1 condition on the radial velocity requires that
QR = − cos θ = A0 − A1 cos θ − C0 (1 + R cos θ) σ + ln
1
−C0 cos θ(1 + cos θ) σ + ln
+···
2
2
1
1
+···
− C0 (1 + R cos θ) −
2
R
−
R
2
cos θ
Neglecting O( ln ) and equating coefficients of constant terms an cos θ separately,
C0
=0
2
A0 +
(2.6.33)
1
C0
−A1 − 1 −
C0 σ + ln
+
= −1
2
2
2
From the condition Qθ = sin θ, we get
+A1 + 1 −
C0
2
The final results are :
C0 =
1
2
A0 = +
1
−2
− σ + ln
and
A1 =
σ + ln
1
1
2
2
− σ + ln
2
(2.6.35)
.
− σ + ln
1
2
1
2
=0
2
2
(2.6.34)
(2.6.36)
(2.6.37)
1
= C0
4
(2.6.38)
As a physical deduction, we leave it as an exercise to show that
drag force/length =
1
2
4πµW
− σ − ln
(2.6.39)
2
so that the drag coefficient is
CD = 1
2
D
ρW 2 2a
=
2π
1
2
1
− σ − ln
2
(2.6.40)
6
In the far wake R 1
1
π −R
e
1−
+···
2R
8R
(2.6.41)
π −R
π −R(1−cos θ)
e
= C0
e
2R
2R
(2.6.42)
K0 (R) ∼
=
r
the rotational part is
C0 e
R cos θ
r
r
There is significant contribution only in the region θ 1. Since 1 − cos θ ∼
=
π −R(1−cos θ) Q0
e
+
2R
R r
r
π
A
π −Y 2 /2X A0
2 /2
0
−Rθ
∼
e
+
= C0
e
+
.
= C0
2X
R
2X
R
QR = C 0
θ2
,
2
r
(2.6.43)
the wake is therefore parabolic in shape. In the far field, the potential part s emits fluid
isotropically as a source at the discharge rate (2πA0 ). On the other hand the rotational wake
is a fluid sink with the influx rate
√
Z ∞ r
π −Rθ2 /2
2C0 π Z ∞ −η2
πC0
e
R dθ =
e dη =
= −2πA0
(2.6.44)
2C0
2R
0
0
in view of (2.6.33). Mass is conserved.