Advanced Fluid Dynamics of The Environment- Stokes Flow Past a Sphere

6 Pages·0.06 MB·English

by  Chiang C. Mei

1
Lecture Notes on Fluid Dynamics
(1.63J/2.21J)
by Chiang C. Mei, MIT
2-5Stokes.tex
2.5 Stokes (cid:13)ow past a sphere
[Refs]
Lamb: Hydrodynamics
Acheson : Elementary Fluid Dynamics, p. 223 (cid:11)
One of the fundamental results in low Reynolds hydrodynamics is the Stokes solution for
steady (cid:13)ow past a small sphere. The apllicatiuon range widely form the determination of
electron charges to the physics of aerosols.
The continuity equation reads
r(cid:1)q~ = 0 (2.5.1)
With inertia neglected, the approximate momentum equation is
rp
0 = (cid:0) +(cid:23)r2~q (2.5.2)
(cid:26)
Physically, the presssure gradient drives the (cid:13)ow by overcoming viscous resistence, but does
a(cid:11)ect the (cid:13)uid inertia signi(cid:12)cantly.
Refering to Figure 2.5 for the spherical coordinate system (r;(cid:18);(cid:30)). Let the ambient
velocity be upward and along the polar (z) axis: (u;v;w) = (0;0;W). Axial symmetry
demands
@
= 0; and q~ = (q (r;(cid:18));q (r;(cid:18));0)
r (cid:18)
@(cid:30)
Eq. (2.5.1) becomes
1 @ 1 @
(r2q )+ (q sin(cid:18)) = 0 (2.5.3)
r (cid:18)
r2@r r@(cid:18)
As in the case of rectangular coordinates, we de(cid:12)ne the stream function to satisify the
continuity equation (2.5.3) identically
1 @ 1 @
q = ; q = (cid:0) (2.5.4)
r (cid:18)
r2sin(cid:18) @(cid:18) rsin(cid:18) @r
At in(cid:12)nity, the uniform velocity W along z axis can be decomposed into radial and polar
components
1 @ 1 @
q = W cos(cid:18) = ; q = (cid:0)W sin(cid:18) = (cid:0) ; r (cid:24) 1 (2.5.5)
r (cid:18)
r2sin(cid:18) @(cid:18) rsin(cid:18) @r
2
z
r
q
o
y
f
x
Figure 2.5.1: The spherical coordinates
The corresponding stream function at in(cid:12)nity follows by integration
W
= r2sin2(cid:18); r (cid:24) 1 (2.5.6)
2
Using the vector identity
r(cid:2)(r(cid:2)~q) = r(r(cid:1)~q)(cid:0)r2~q (2.5.7)
and (2.5.1), we get
r2~q = (cid:0)r(cid:2)(r(cid:2)~q) = (cid:0)r(cid:2)(cid:16)~ (2.5.8)
Taking the curl of (2.5.2) and using (2.5.8) we get
r(cid:2)(r(cid:2)(cid:16)~) = 0 (2.5.9)
After some straightforward algebra given in the Appendix, we can show that
~e
~q = r(cid:2) (cid:30) (2.5.10)
rsin(cid:18)!
and
~e ~e @2 sin(cid:18) @ 1 @
(cid:16)~ = r(cid:2)q~ = r(cid:2)r(cid:2) (cid:30) = (cid:0) (cid:30) + (2.5.11)
rsin(cid:18)! rsin(cid:18) @r2 r2 @(cid:18) sin(cid:18) @(cid:18)!!
Now from (2.5.9)
~e
r(cid:2)r(cid:2)(r(cid:2)~q) = r(cid:2)r(cid:2) r(cid:2) r(cid:2) (cid:30) = 0
" rsin(cid:18)!#
3
hence, the momentum equation (2.5.9) becomes a scalar equation for .
@2 sin(cid:18) @ 1 @ 2
+ = 0 (2.5.12)
@r2 r2 @(cid:18) sin(cid:18)@(cid:18)!!
The boundary conditions on the sphere are
q = 0 q = 0 on r = a (2.5.13)
r (cid:18)
The boundary conditions at 1 is
W
! r2sin2(cid:18) (2.5.14)
2
Let us try a solution of the form:
(r;(cid:18)) = f(r)sin2(cid:18) (2.5.15)
then f is governed by the equi-dimensional di(cid:11)erential equation:
d2 2 2
(cid:0) f = 0 (2.5.16)
"dr2 r2#
whose solutions are of the form f(r) / rn, It is easy to verify that n = (cid:0)1;1;2;4 so that
A
f(r) = +Br+Cr2 +Dr4
r
or
A
= sin2(cid:18) +Br+Cr2 +Dr4
r
(cid:20) (cid:21)
To satisfy (2.5.14) we set D = 0;C = W=2. To satisfy (2.5.13) we use (2.5.4) to get
W A B A B
q = 0 = + + = 0; q = 0 = W (cid:0) + = 0
r (cid:18)
2 a3 a a3 a
Hence
1 3
A = Wa3; B = (cid:0) Wa
4 4
Finally the stream function is
W a3 3ar
= r2 + (cid:0) sin2(cid:18) (2.5.17)
2 " 2r 2 #
Inside the parentheses, the (cid:12)rst term corresponds to the uniform (cid:13)ow, and the second term
to the doublet; together they represent an inviscid (cid:13)ow past a sphere. The third term is
called the Stokeslet, representing the viscous correction.
The velocity components in the (cid:13)uid are: (cf. (2.5.4) :
a3 3a
q = W cos(cid:18) 1+ (cid:0) (2.5.18)
r
" 2r3 2r#
a3 3a
q = (cid:0)W sin(cid:18) 1(cid:0) (cid:0) (2.5.19)
(cid:18)
" 4r3 4r#
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2.5.1 Physical Deductions
1. Streamlines: With respect to the the equator along (cid:18) = (cid:25)=2, cos(cid:18) and q are odd while
r
sin(cid:18) and q are even. Hence the streamlines (velocity vectors) are symmetric fore and
(cid:18)
aft.
2. Vorticity:
1@(rq ) 1@q 3 sin(cid:18)
(cid:16)~ = (cid:16) ~e (cid:18) (cid:0) r ~e = (cid:0) Wa ~e
(cid:30) (cid:30) (cid:30) (cid:30)
r @r r @(cid:18) ! 2 r2
3. Pressure : From the r-component of momentum equation
@p (cid:22)Wa
= cos(cid:18)(= (cid:0)(cid:22)r(cid:2)(r(cid:2)q~))
@r r3
Integrating with respect to r from r to 1, we get
3(cid:22)Wa
p = p1 (cid:0) cos(cid:18) (2.5.20)
2 r3
4. Stresses and strains:
1 @q 3a 3a3
e = r = W cos(cid:18) (cid:0)
rr
2 @r 2r2 2r4!
On the sphere, r = a, e = 0 hence (cid:27) = 0 and
rr rr
3(cid:22)W
(cid:28)rr = (cid:0)p+(cid:27)rr = (cid:0)p1 + cos(cid:18) (2.5.21)
2 a
On the other hand
@ q 1@q 3Wa3
e = r (cid:18) + r = (cid:0) sin(cid:18)
r(cid:18)
@r r r @(cid:18) 2 r4
(cid:18) (cid:19)
Hence at r = a:
3(cid:22)W
(cid:28) = (cid:27) = (cid:22)e = (cid:0) sin(cid:18) (2.5.22)
r(cid:18) r(cid:18) r(cid:18)
2 a
The resultant stress on the sphere is parallel to the z axis.
3(cid:22)W
(cid:6)z = (cid:28)rrcos(cid:18)(cid:0)(cid:28)r(cid:18)sin(cid:18) = (cid:0)p1cos(cid:18) +
2 a
The constant part exerts a net drag in z direction
2(cid:25) (cid:25) 3(cid:22)W
D = ad(cid:30) d(cid:18)sin(cid:18)(cid:6) == 4(cid:25)a2 = 6(cid:25)(cid:22)Wa (2.5.23)
z
2 a
o o
Z Z
This is the celebrated Stokes formula.
A drag coe(cid:14)cient can be de(cid:12)ned as
D 6(cid:25)(cid:22)Wa 24 24
C = = = = (2.5.24)
D 1(cid:26)W2(cid:25)a2 1(cid:26)W2(cid:25)a2 (cid:26)W(2a) Re
2 2 (cid:22) d
5
5. Fall velocity of a particle through a (cid:13)uid. Equating the drag and the buoyant weight
of the eparticle
4(cid:25)
6(cid:25)(cid:22)W a = a3((cid:26) (cid:0)(cid:26) )g
o s f
3
hence
2 a2(cid:1)(cid:26) a2 (cid:1)(cid:26)
W = g = 217:8
o
9 (cid:23) (cid:26)f ! (cid:23) (cid:26)f !
in cgs units. For a sand grain in water,
(cid:1)(cid:26) 2:5(cid:0)1
= = 1:5; (cid:23) = 10(cid:0)2cm2=s
(cid:26) 1
f
W = 32;670 a2cm=s (2.5.25)
o
To have some quantitative ideas, let us consider two sand of two sizes :
a = 10(cid:0)2cm = 10(cid:0)4m : W = 3:27cm=s;
o
a = 10(cid:0)3cm = 10(cid:0)5 = 10(cid:22)m; W = 0:0327cm=s = 117cm=hr
o
For a water droplet in air,
(cid:1)(cid:26) 1
= = 103; (cid:23) = 0:15 cm2=sec
(cid:26) 10(cid:0)3
f
then
(217:8)103
W = a2 (2.5.26)
o
0:15
in cgs units. If a = 10(cid:0)3 cm = 10(cid:22)m, then W = 1:452 cm=sec.
o
Details of derivation
Details of (2.5.10).
~e ~e rsin(cid:18)~e
1 r (cid:18) (cid:30)
r(cid:2) ~e = @ @ @
rsin(cid:18) (cid:30)! r2sin(cid:18) (cid:12)(cid:12) @r @(cid:18) @(cid:30) (cid:12)(cid:12)
(cid:12) 0 0 (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
(cid:12) (cid:12)
1 @ 1 @
=~e (cid:0)~e
r (cid:18)
r2sin(cid:18) @(cid:18)! rsin(cid:18) @r!
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Details of (2.5.11).
~e
r(cid:2)r(cid:2) (cid:30) = r(cid:2)~q
rsin(cid:18)
~e r~e rsin(cid:18)~e
1 r (cid:18) (cid:30)
= @ @ @
r2sin(cid:18) (cid:12)(cid:12)(cid:12) 1@r @ (cid:0)1@(cid:18)@ @0(cid:30) (cid:12)(cid:12)(cid:12)
(cid:12) r2sin(cid:18) @(cid:18) sin(cid:18) @r (cid:12)
(cid:12) (cid:12)
~e (cid:12)@2(cid:30) sin(cid:18) @ 1 @ (cid:12)
= (cid:18) (cid:12) + (cid:12)
rsin(cid:18) "@r2 r2 @(cid:18) sin(cid:18) @(cid:18)!#