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Notes on

1.63 Advanced Environmental Fluid Mechanics

Instructor: C. C. Mei, 2001

ccmei@mit.edu, 1 617 253 2994

December 1, 2002

2-5Stokes.tex

2.5 Stokes flow past a sphere

[Refs]Lamb: HydrodynamicsAcheson : Elementary Fluid Dynamics, p. 223 ff

One of the fundamental results in low Reynolds hydrodynamics is the Stokes solution forsteady flow past a small sphere. The apllicatiuon range widely form the determination ofelectron charges to the physics of aerosols.

The continuity equation reads ~q = 0 (2.5.1)

With inertia neglected, the approximate momentum equation is

0 = p

+ 2~q (2.5.2)

Physically, the presssure gradient drives the flow by overcoming viscous resistence, but doesaffect the fluid inertia significantly.

Refering to Figure 2.5 for the spherical coordinate system (r,,). Let the ambientvelocity be upward and along the polar (z) axis: (u,v,w) = (0, 0, W). Axial symmetrydemands

= 0, and ~q = (qr(r, ), q(r,), 0)

Eq. (2.5.1) becomes1

r2

r(r2qr) +

1

r

(q sin ) = 0 (2.5.3)

As in the case of rectangular coordinates, we define the stream function to satisify thecontinuity equation (2.5.3) identically

qr = 1r2 sin

, q = 1r sin

r

(2.5.4)

At infinity, the uniform velocity W along z axis can be decomposed into radial and polarcomponents

qr = W cos =1

r2 sin

, q = Wsin =

1

r sin

r, r (2.5.5)

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z

y

x

r

o

f

q

Figure 2.5.1: The spherical coordinates

The corresponding stream function at infinity follows by integration

= W2 r2 sin2 , r (2.5.6)

Using the vector identity

( ~q) = ( ~q)2~q (2.5.7)

and (2.5.1), we get

2~q = ( ~q) = ~ (2.5.8)

Taking the curl of (2.5.2) and using (2.5.8) we get

( ~) = 0 (2.5.9)

After some straightforward algebra given in the Appendix, we can show that

~q =

~e

r sin

!(2.5.10)

and

~=

~q=

~er sin

!=

~e

r sin 2r2 +

sin

r2

1

sin

!!

(2.5.11)

Now from (2.5.9)

( ~q) =

"

~er sin

!#= 0

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hence, the momentum equation (2.5.9) becomes a scalar equation for .2

r2+

sin

r2

1

sin

!!2 = 0 (2.5.12)

The boundary conditions on the sphere are

qr = 0 q = 0 on r = a (2.5.13)

The boundary conditions at is

W

2r2 sin2 (2.5.14)

Let us try a solution of the form:

(r,) = f(r)sin2 (2.5.15)

then f is governed by the equi-dimensional differential equation:

" d2dr2

2

r2#2

f = 0 (2.5.16)

whose solutions are of the form f(r) rn, It is easy to verify that n = 1, 1, 2, 4 so that

f(r) =A

r+ Br + Cr2 + Dr4

or

= sin2

A

r+ Br + Cr2 + Dr4

To satisfy (2.5.14) we set D = 0, C = W/2. To satisfy (2.5.13) we use (2.5.4) to get

qr = 0 =W

2

+A

a3+

B

a

= 0, q = 0 = WA

a3+

B

a

= 0

Hence

A =1

4W a3, B =

3

4W a

Finally the stream function is

=W

2

"r2 +

a3

2r

3ar

2

#sin2 (2.5.17)

Inside the parentheses, the first term corresponds to the uniform flow, and the second termto the doublet; together they represent an inviscid flow past a sphere. The third term iscalled the Stokeslet, representing the viscous correction.

The velocity components in the fluid are: (cf. (2.5.4) :

qr = Wcos

"1 +

a3

2r3

3a

2r

#(2.5.18)

q = W sin

"1

a3

4r3

3a

4r

#(2.5.19)

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2.5.1 Physical Deductions

1. Streamlines: With respect to the the equator along = /2, cos and qr are odd whilesin and q are even. Hence the streamlines (velocity vectors) are symmetric fore andaft.

2. Vorticity:

~= ~e

1

r

(rq)

r

1

r

qr

!~e =

3

2W a

sin

r2~e

3. Pressure : From the r-component of momentum equation

p

r=

W a

r3cos (= ( ~q))

Integrating with respect to r from r to , we get

p = p 3

2

W a

r3cos (2.5.20)

4. Stresses and strains:1

2err =

qrr

= Wcos

3a

2r2

3a3

2r4

!

On the sphere, r = a, err = 0 hence rr = 0 and

rr = p + rr = p +3

2

W

acos (2.5.21)

On the other hand

er = r

r

q

r +

1

r

qr

= 3

2

W a3

r4

sin

Hence at r = a:

r = r = er = 3

2

W

asin (2.5.22)

The resultant stress on the sphere is parallel to the z axis.

z = rr cos r sin = p cos +3

2

W

a

The constant part exerts a net drag in z direction

D = Z2

oad Z

od sin z ==

3

2

W

a4a2 = 6W a (2.5.23)

This is the celebrated Stokes formula.

A drag coefficient can be defined as

CD =D

12W

2a2=

6W a12W

2a2=

24W(2a)

=24

Red(2.5.24)

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5. Fall velocity of a particle through a fluid. Equating the drag and the buoyant weightof the eparticle

6Woa =4

3a3(s f)g

henceWo =

2

9g

a2

f

!= 217.8

a2

f

!

in cgs units. For a sand grain in water,

f=

2.5 1

1= 1.5, = 102cm2/s

Wo = 32, 670 a2cm/s (2.5.25)

To have some quantitative ideas, let us consider two sand of two sizes :

a = 102cm = 104m : Wo = 3.27cm/s;

a = 103cm = 105 = 10m, W o = 0.0327cm/s = 117cm/hr

For a water droplet in air,

f=

1

103= 103, = 0.15 cm2/sec

thenWo =

(217.8)103

0.15a2 (2.5.26)

in cgs units. If a = 103 cm = 10m, then Wo = 1.452 cm/sec.

Details of derivation

Details of (2.5.10).

r sin ~e

!= 1

r2 sin

~er ~e r sin ~er

0 0

= ~er

1

r2 sin

! ~e

1

r sin

r

!

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Details of (2.5.11).

~e

r sin = ~q

= 1r2 sin

~er r~e r sin ~er

1r2 sin

1sin

r

0

=~e

r sin

"2

r2+

sin

r2

1

sin

!#