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### Transcript of Advanced Fluid Mechanics - Chapter 04 - Very Slow Motion

• 7/28/2019 Advanced Fluid Mechanics - Chapter 04 - Very Slow Motion

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Chapter 4-1

d

U

Chapter4 Very Slow Motion

4.1 Equations of motion

Consider a constant density flow, the equations of motion are:

Continuity: 0= Vv

Momentum: VpVVt

V vvvv

2+=

+

Introduce the characteristic velocity : U

characteristic length : d

characteristic pressure : p0

characteristic time : t0

then the non-dimensional properties become

=U

VV

v

v~,

d

rr

v

v

=~

,0

~

p

pp = ,

0

~

t

tt =

and =rv

=

rd~

1v

=

d

~

= d~

(The magnitude of order 1)

Continuity: 0~~ = Vv

(continuity equation is invarant for non-dimensionalization)

Momentum:

t

V

tU

d~

~

0

v

+ VV~~~ vv

= pU

P ~~

2

0

+ VdU

~~

/

1 2 v

If we denote:

Reynolds No. =

dU

And pick up: P0 = U2 = dynamic pressureEquation becomes

t

V

tU

d~

~

0

v

+ VV~~~ vv

= p~~ + V

~~

Re

1 2 v

inertia forces pressureforces

viscousforces

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Chapter 4-2

1) If Re t

V

tU

d~

~

0

v

+ VV~~~ vv

= p~~ (4.1)

2) If Re 0 0~ 2

= V

v

(4.2)

Note that there is no balance term. We want to have a balance term. Multiply (4.11)

by Re

Ret

V

tU

d~

~

0

v

+Re VV~~~ vv

= pU

P ~~

20

Re

+ V

~~ 2 v

(0, as Re0)(i) The unsteady term coefficient:

For a oscillation body flow, w = frequency of oscillation

we can choose: t0 =

1

the first coefficient:

Re0tU

d

=ReU

d 0, asRe 0 and is not very large

Remark: 1 if there is no body oscillation, we may pickt0 = U/d

2

For a highly oscillation body, the unsteady term can

t neglected.

(ii) The pressure coefficientWe want to pick up P0 such that Re

2

0

U

P

1, and this term can be left to

balance the viscous term. Therefore

P0 =Re

2U =

/

2

dU

U

=d

U

And the momentum equation (4.1) become

0 = p~~ + V

~~ 2 v (4.3)

Summary: For a steady, constant density, slow flow (Re0)

V~~ v

= 0

0 = p~~ + V

~~ 2 v

Also name as: Slow flow, Creeping flow, or stokesflow.

• 7/28/2019 Advanced Fluid Mechanics - Chapter 04 - Very Slow Motion

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Chapter 4-3

4.2 Slow flow past a sphere

consider a steady, constant density flow withRe0.

Z

a

ruru

U

Vv

=rreuv

+euv

+ euv

0

upstream sphere u u ru

zeU

Mass: divVv

= 0

+

)sin()sin(

sin

1 22

ururrr

r =0

r

The streamfunction are takes such that the continuity equation is satisfied

automatically. Thus

r2sinur=

, rsinu=

r

or ur=

sin

1

2r

, u=rr

sin

1(4.4)

Momentum equation:

0 = Vpv2

v

) cuel curlVv

0 = -curl curlVv

0=Vv

on r= a

0

• 7/28/2019 Advanced Fluid Mechanics - Chapter 04 - Very Slow Motion

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Chapter 4-4

take curl on both side

0 = 0 -curl curl curlVv

(4.5)

v

v

= cuelVv

=sin

12

r0

sin

ruur

erere

r

r

=sin

12

r

++

rr

u

r

rueree

)(sin)0()0(

=

+

ru

r

uru

r

esub. Eqn (4.4)

=r

e

+

)

sin

1()

sin

1(

sin

12

rrrrr

rr

=r

e

+

)

sin

1(

1)

11(

sinsin

122

2

2

rrrrr

r

rr

=

+

)

sin

1(

sin

sin

22

2

rrr

e

= e

Where sin

1

r D

D differential operator 2

2

r + )

sin1(sin

2

r

Then

Curl v

=sin

12

r

00

sin

r

ererer

=sin

12

r

rerer

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Chapter 4-5

Finally, we can obtain

curl curlv

=

sin

r

eD2

Eq. (4.5)

D2 = 0

or

2

2

r+

2

2)

sin

1(

sin

r=0 (4.6)

B.CS in terms of : (Recall ur=

sin

12

r, u=

rr

sin

1)

(i) On r = a:ur= 0

= 0

u= 0 r

= 0 (4.7a)

(ii) Infinity condition V

v

= U ze = U[( cos) re + (-sin) e ]

ur=

sin12

r Ucos as r (4.8a)

u=rr

sin

1 Usin as r (4.8b)

integrate (4.8a) and (4.8b), we obtain

~ U2

2rsin

2 asr (4.7b)

Assume: (r,) = f(r)sin2, then the B.CS become

(4.7a) r= a )(' af =f(a) = 0

(4.7b) r f(r) ~ U2

2r

asr

• 7/28/2019 Advanced Fluid Mechanics - Chapter 04 - Very Slow Motion

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Chapter 4-6

Sub. Into Eq. (4.6), we get

0)2

)(2

()2

(22

2

22

22

22

2

==r

f

dr

fd

rdr

df

rdr

d

Aside: 02

22

3

33

4

44 =++++ df

dr

dfcr

dr

fdbr

dr

fdar

dr

fdr

we can assume solution to be the form off= Arn,

we will have 4 roots for n, n= 1, -1, 2, 4.

f=r

A+ Br+ Cr2 +Dr4 (4.9)

B.CS:

(1) f(r) ~2

2rU asr

compare with (4.9), we observe that we need to take C =2

U and D = 0

to satisfyf(r) ~2

2rU forr . ( The value of A, B are not important,

since they are not the highest order term, and r2 >> ras r)

Eq. (4.9)

f=r

A+ Br+

2

U r2 (4.10)

(2) f(a) = 0 a

A+ Ba +

2

U a2 = 0

)('

af = 0 2

A

a+ B + U a = 0

A = 4

1

a3

U , B = 4

3

aU

(r,) = a2Usin2

+ 2)(

2

1)(

4

3)(

4

1

a

r

a

r

r

a+ const (4.11a)

ur= Ucos

+ 3)(

2

1)(

2

31

r

a

r

a

u= -Usin

3)(4

1)(4

31 r

a

r

a(4.11 b, c)

• 7/28/2019 Advanced Fluid Mechanics - Chapter 04 - Very Slow Motion

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Chapter 4-7

The streamlines are:

= 0=0=5=

10=

5=

10=

Remark:

(1) The streamlines possess perfect forward and backward symmetry: there isno wake. It is the role of the convective acceleration terms, here neglected, to

provide the strong flow asymmetry typical of higher Reynolds number flows.

(2) The local velocity is everywhere retarded from its freestream value: there isno faster region such as occurs in potential flow.

(3) The effect of the sphere extent to enormous distance: at r= 10a, the velocityare still 10 percent below their freestream values.

(4) The streamlines and velocity are entirely independent of the fluid viscosity.

The pressure distribution is

0 = -p -curlv

or r

p

=

sin

12

r;

p

r

1=

sin

1

r

integrate the eqns with the known value of , we finally obtain

P=P2

3aU

2

cos

r

(4.12)

The shear stress in the fluid is

r =(r

uur

r+

1 ) =

rU sin

+ 3)(4

5)(431

ra

ra (4.13)

• 7/28/2019 Advanced Fluid Mechanics - Chapter 04 - Very Slow Motion

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• 7/28/2019 Advanced Fluid Mechanics - Chapter 04 - Very Slow Motion

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Chapter 4-9

Remarks:

(1) Stokes formula :D=6a Uprovides a method to determine the viscosityof a fluid by observing the terminal velocity U of a small falling ball of

(2) Stokes formula valid only for Re

• 7/28/2019 Advanced Fluid Mechanics - Chapter 04 - Very Slow Motion

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Chapter 4-10

(5) ForRe>1. Oseen use perturbation method and obtain a modified formula forCD.

CD = Re

24

(1+16

3

Re) (valid forRe < 3~5) (4.16)

Other curve fitting formula are, for example,

CD Re

24+

Re1

6

++ 0.4 (0 Re 2105 ) (4.17)

Fig. 3-38 (a) Cylinder data

• 7/28/2019 Advanced Fluid Mechanics - Chapter 04 - Very Slow Motion

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Chapter 4-11

4.3 The Hydrodynamic Theory of Lubrication (White 3-9.7, p.187-190)

Lubrication between journals and bearings are achieved by filly a thin film of oil

between then.

tyeccentricie ,

For the sake of simplification, we take a model of

y h1h

2h

x

l

blockfixed

wallmoving,U

Assume: 1 h

• 7/28/2019 Advanced Fluid Mechanics - Chapter 04 - Very Slow Motion

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Chapter 4-12

Since v

• 7/28/2019 Advanced Fluid Mechanics - Chapter 04 - Very Slow Motion

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Chapter 4-13

B.Cs:

1 y =0, u =U2 y =h, u = 03 x = 0,p = p0

4 x =L,p = p0

Note that dp/dx here is no longer constant (such as the couette flow between two

parallel walls), it must satisfy the pressure P0 at both ends. The dp/dx must be

determined in such a way as to satisfy the continuity equation in every section of the

form

Q = (udy + vdx) = )(

0

xhudy = const (4.20)

The solution of Eq. (4.19) with given B.Cs is

u = U(1h

y) )1()(

2

2

h

y

h

y

dx

dph

(4.21)

Here, dp/dx is determined by sub. (4.21) into (4.20), as

Q =2

Uh )(

12

3

dx

dph

Or

dx

dp= 12(

22h

U

3h

Q) (4.22)

integrate with B.C (p = p0 at x=0 ), we have

p = p0+ 6U x

h

dx

0 2 12Q x

h

dx

0 3 (4.23)

b1(x) b2(x)Inserting B.C ofp =p0 atx = L, we get

Q =2

1U

)(

)(

2

1

Lb

Lb=

2

1UH (4.24)

characteristic thickness H

This is an assumed assumption for the model. For a

certain segment in lubrication fluid, the pressure isnot the same on both ends

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Chapter 4-14

We may conclude the procedure of solution as follows:

(1) Known wedge shape h(x)(2) Obtain b1(L) & b2(L), as well asH& Q(3) The pressure distribution (4.23) can be rewritten as

p(x) =p0 + 6Ub1(x) 12Qb2(x) (4.25)

(4) The dp/dx, Eq. (4.22) can be written and calculated as

dx

dp=

2

6

h

U(1

h

H) (4.26)

(5) Knowing dp/dx, the velocity distribution can be found from Eq. (4.21)

Remark:

(1) pmax orpmin occurs where h =H.(2) For a straight wedge with h1 & h2 at both ends, we get

p(x) = p0 + 6U2

212

2

2

1

))((

h

hhhh

hh

L

2

1

0

/ hUL

pp

1

2

3

0 5.0 1

8.0

7.0

5.0

4.0

3.01

2=

hh

exit.near theoccursandhigherbecome

pressuremaxthedecreaseas 12

h

h

• 7/28/2019 Advanced Fluid Mechanics - Chapter 04 - Very Slow Motion

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Chapter 4-15

atm250the

,5.0with

example,aboveFor the

max

1

2

=

=

p

hh

(3) Taking from F.M. White text:Recall that stokes flow, being linear, are reversible. If we reverse the

wall in the figure to the left, that is, U