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

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Advanced Fluid Mechanics
Chapter 41
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 nondimensional 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 nondimensionalization)
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
unsteady part convective part
inertia forces pressureforces
viscousforces

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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.

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Advanced Fluid Mechanics
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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
Since Vv2 = grad (divV
v
) cuel curlVv
0 = curl curlVv
0=Vv
on r= a
0

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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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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

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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)

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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)

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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
radius a.
(2) Stokes formula valid only for Re

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(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. 338 (a) Cylinder data

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4.3 The Hydrodynamic Theory of Lubrication (White 39.7, p.187190)
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

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Since v

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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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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)
and is readily obtained.
(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

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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