turbulence 2
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Turbulence
We require that this relationship depend only on the properties of the material, not of the motion
e.g. : Newtonian fluid Relation between stress & strain rate is linear
+
=
=
i
j
j
iijijkkij
v
ijx
u
x
ueee
2
1;
3
12)(
)ijij eF=
Newtonian :( )
= ijkkijv
ij ee 1
2
Viscosity (independent of the flow,dependent, on the fluid)
(6EA)EA9:ij
(6EA)EA9:ije
Most general linear relation
=klijklij
Symmetry reduces unknowns from 81 to 27
All but (2) (1 in practice) can be eliminate for isotropic fluid Newtonian
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Turbulence
Incom ressible
Derivative of density following fluid material pt. is zero
0=Dt
D
0or0 =
=
i
i
x
uur
+= ijkkijijij eep 1
2
+
=
k
k
k
kkk
x
u
x
ue
2
1
0 incompressilbe.0
+
+=
i
j
j
iijij
x
u
x
up for incompressible.
2
+
+
=
+
=
i
j
j
i
ji
ij
ji
i
x
u
x
u
xx
P
exxDt
if =constant throughout flow ,
0= i
i
x
u
0,1 2
=
+
= iiiuuPDu
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Turbulence
Re nolds decom osition
Impossible to approach turbulence deterministically ?split flow into a mean and a random part (fluctuation part)
=~ iii
Instantaneous mean fluctuation
VVV~~ pPp +=
ijijij +=
deal with only constant density case
0
~
== R
: no densit fluctuation
( )
VV
0iii
uU
uUx
+
=+
ijij
jij
jjiix
pxx
uut
+
++
=
+++
Average these instantaneous equations to get equations governing the mean flow
( ) ( ) ( )t
UuU
tuU
tuU
t
iiiiiii
=+
=+
=+
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Turbulence
ii xx
p
=
+
( ) )()()( VijVijVij Tx
Tx
=+
: gradient of mean pressure
: divergence of mean viscous stress
( ) ( )
j
i
j
j
j
ji
j
j
j
i
j
j
i
j
j
i
j
j
i
j
j
i
j
j
ii
jj
x
uu
x
Uuu
x
U
x
UU
x
uu
x
Uu
x
uU
x
UU
x
uUuU
+
+
+
=
+
+
+
=
++
0 0
Product of fluctuating quantities can not be assumed zero, they may or may not be zero. e.g.) 02 u
Look at continuity equation
=
=
0
i
i
i
i
j
j
ii
i
x
U
xxx
=
0 mean satisfies same continuity equation as does instantaneous
at equat on oes uctuat on sat s y
0~
=
=
i
i
ux
( ) 0~ =
ii Uux
i
ix0=
i
i
x
u even fluctuation satisfy same continuity equation
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Turbulence
ibleincompress)(2 jiET ijtR
ij =
turbulent viscosity
The medium in this case is not the fluid, but the flow itself
Turbulent flow are flows !!
Keep going in spite of obvious inconsistence
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Turbulence
Can we derive a dynamical equation for Reynolds stress ?
Can we get equation for velocity fluctuation itself ?
Instantaneous
V~~~~ )(
ij
V
ij
j
ij
ij
ij
i exxx
uu
t
u ~2~where)equationStokesNavier(~
=
+
=
+
321
ij
ij
ij
ij
xuu
xuU
xUu
xUU
+
+
+
Mean
ji
jj
V
ij
ij
j
j
i uuxx
T
x
P
x
UU
t
U
+
=
+
)(
11
j
i
jx
uu
442
Recall
pPpuUu iii
+=+=
~~
Subtract mean from instantaneous to get equation for fluctuation
j
V
ij
ij
i
j
j
i
j
j
i
j
j
i
j
i
xx
P
x
uu
x
uu
x
Uu
x
uU
t
u
+
=
+
+
+
)(
11
Fluctuation
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Reynolds stress
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Turbulence
Suppose infinitesimal disturbanceRemaining equation for fluctuation is linear & Reynolds stress is gone (2nd order)
Fluctuation :V
ijiii PUuu
Uu
+
=
+
+
)(11
linearjijj xxxxt
Mean :
j
V
ij
ij
i
j
i
x
T
x
P
x
UU
t
U
+
=
+
)(
11
No Reynolds stress
Equations for infinitesimal disturbance are linear & closed (well-posed)
Equations for mean flow is just the unchanged original viscous equations
- presence of infinitesimal disturbances does not change base flow (laminar solution)
Fluctuation equation can easily be transformed into Orr-Sommerfeld equation
Superimpose disturbance
, , ...
If infinitesimal disturbance grows, where does additional energy come from?
Develop energy equation for fluctuation 22211mean uc ua on ne c nergy per un mass
To 2nd order ( multiply linearized equation by )
32122
uuuuu ii ==
iu
edisturbancmalinfinitesiofEnergyKinetic1111
)(V
ij
iii
iiiii uP
uU
uuuuUuu
+
=
+
+
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jjjj xxxxt
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Turbulence
Energy equation for turbulent fluctuation
( momentum of fluctuation ) ( velocity ) = ( kinetic energy of fluctuation )
{
volumeunitperEnergyKinetic2~~~~
momentum
== uuuu
volumeun tper
+
=
+
j
ij
j
ii
j
ij
j
i
ij
ij
ii
x
Uu
x
uu
x
uu
x
u
x
P
x
uU
t
uu
2
21
321 0 continuity
+
=
+
=
j
i
ij
i
i
j
j
i
jj
ij
x
u
xx
u
x
u
x
u
xx
e
2
2
2
22222 q
kuuuuu =++==Let2
iji
iji
j
ii
ij
Uuu
uuu
x
uu
puqU
q
+
+
=
+
022
2
2
22
Reynolds stress
orjj
j
ij
i
ij
x
eu
2
2
iii uPpuPu 1
0 continuity
=
=
j
jiji
puxxxx
term where net effect on energy is zero
2
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==
j
jj
j
j
i
ji uqxx
ux
uu22
since continuity
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Turbulence
2
2
222
2
22 iii
i
j
i
i
ix
u
x
q
x
u
x
uu
xx
u
xu
x
uu
=
=
=
( ) ( ) ( )22222 ijijijj
i
ijiji
jj
ij
i eeuxx
ueeu
xx
eu
=
=
)0( =+ ijijijij ee
Summary
+
=
+
j
i
j
i
j
i
ji
j
jj
jj
jx
u
x
u
x
Uuu
q
xuqpu
xx
qU
t
q
22
11222
222
form 1
form 1
ijij
j
i
jiijijj
jj
jee
x
Uuueuuqpu
xx
qU
t
q
22
2
1122 222
+
=
+
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Turbulence
True dissi ation
111
2
22
uuuuuu
ee ijij
0422
221
+
+
=
+
+
= Lxxxxxx ijij
Can only remove kinetic energy
Dissipation removes fluctuation kinetic energy and sends it to internal energy.
ijijeu2
.
- due to deformation work
Note difference with which acts to accelerate (increase Kinetic Energy) of neighboringu y ac on o v scous s ress.
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Turbulence
Pseudo dissipation
Commonly the dissipation in the sense that synthetic
( )( ) ( )
++=++=
22
2D
D
eeeee
x
u
x
u
ijijijijijijijijijij
j
i
j
i
0
+
+
+
=+= LL2
1
1
1
44D
xxee ijijijij
in general D
Note : D0 always
Why is the real dissipation and not Dwhen both clearly remove Kinetic Energy from flow ?
0
0
=
ij
ije
u
r
1
0
0
=
ij
ije
ur
. .
distortionbut no rotationr
rigid body rotationfluid particle has no shape changer
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Turbulence
ee 2 cannot de end on rotation directl
[ ]ijijijij
j
i
j
i eex
u
x
u +=
=D
2
Dissipation of energy is due only to distortion ( i. e. strain rates) not rotation
is real dissipation of energy
If turbulent is isotropic (almost never is) thenD
= and
115
=
ix
u
because all derivatives are interrelated
But at very high turbulent Re, the small scale turbulence is nearly isotropic local isotropy !At high turbulent Re, we shall see that dissipation occurs only at smallest scales of motion
Therefore in all turbulent flow at hi h turbulent Re 43 ~ul
= Dand15
2
1
ix
u
For turbulence modeling usually use form of Kinetic Energy equation involving D and and
treat D as if it were the dissipation.
O.K. at hi h turbulent Re
( )
jj x
q
x
22
,
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Turbulence
Taylors frozen field hypothesis
suppose I can imagine turbulence being swept a fixed probe at velocity in such awaythat the turbulence does not change as it passes the probe
cU
cU
frequency : cUf =
1x
spatial disturbance : wavelength
CU
If disturbance is convected along 1x
t
then
tUx c =
1
1
-
Taylors frozen field hypothesis
- only good when turbulence intencity is less then 10%
c
rms
U
u
- differentiate to get
- square & ave. to get
t
u1
2
1
u
uppose ave pro e o measure a xe po n1 tu
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t
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Turbulence
Using Taylor frozen field hypothesis (if applicable)
2
1
2
2
1
1 1
=
t
u
Ux
u
c
If isotro a lies or locall isotro hi h turb. Re
:
2
1
2
115
=
t
u
Uc
.
high )10( 4>=
ulRl
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Turbulence
Production
Dissi ation
= 00 VV ji
ji dVu
u
uudV = rate at which turb. KE is produced in 0V
massunitperproduction=
=0V
dV rate at which energy is dissipated in 0V
+
+
0
)(2
2
~~~
22
v
V
ijijj
j
uuudV
euuqpux
dV
r
0
2V
Since integrand is a divergence, we can apply the divergence them (since simply-connected region)
rrr
x=
vector field
( )xnn ~~~ =
+=
+
00
22
22
22
A
jijijj
V
ijijj
j
dsneuuqpueuuqpux
dV
~~~~ r
= dsnAdVArrr
snuuqup2
+
[ ] is evaluated on surface enclosing0A 0V
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0 0
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Turbulence
njj upnpunuprrr
==
0=nu if surface is at rest ( Kinematic B.C )
K.B.C 0=Dt
DF0=+
Fu
t
F r
0= nu on surface
02
2 jjnuq
on surface by kinematic B.C
ji
V
ijjiji nuneu)(2 =
ur
is identically zero on surface
- normal component due to K.B.C- tangential component due to no-slip
entire integral is exactly zero !
Therefore net effect of turbulent transport term (divergence) is zero :
on surface0)( j
V
iji nzu
ey ont ncrease or ecrease .
They only move energy around.
jpu - moves energy by action of pressure forces HLH
juq2
2- carries energy from region of higher energy to region of lower energy
i
V
ijijiueu)(
2 = - accelerate adjacent region of flow due to viscous stress
q2
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jx - s m ar to a ove
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Turbulence
Di ression on model for turbulencek- forget whole Reynolds stress eq.
- use eddy viscosity
2
kul =
LLLLk
2
3
2
2
2
2
2
~][~1
~][~][
2
2
1qk KE
e
ijijee 2= dissipation
+=
=
VUu
Euu
e
ijeji
2
xy
write & close eqn.s for &k
DP
+
=
+
jjj
kuqpu
kU
k2
1
jjj
Model transport term using a gradient-diffusion approach
- assume energy is moved down gradients in energy by these terms i.e. from high to low
Simplest model
j
jjx
kuqpu
22
1
[ transport term ]j
ex
D+
=
+
= ijeji
EECk
Ck
Uk
Eeuu
1
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e
j
e
jj xxxt
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Turbulence
Derive KE eq. for mean motion
ii PUu 1
ji
j
ij
jjj
jixxxxt
ijiji
jiijiijijj EExUuuEUUuuPU
xQ
xUQ
t 221
21
21 22 + +=+
KE eq. for mean flow
2
3
2
2
2
1 UUUUUQ ji ++==(when )
EEE ijij =2 : dissipation of KE to heat due to mean motion ( deformation )
j
iji
x
Uuu
is only term which occurs in both equs for mean and fluctuation , but with opposite sign!
Effect of this term is take energy from mean and put it into flucuation , or vice versa
iU
Almost always in engineering flow has apposite sign asjiuuj
x
0
j
iji
x
Uuu almost always 0P
mean fluctuation
j
iji
x
Uuu
PU
uu iji
is called turbulent production term
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j
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Turbulence
-
KE eq. For fluctuation
+
=
+ i
jiijij
jq
j
q uuueuuq
puu 2
1 22222
jjj
3-dimensionality ! ( tendency to isotropic )
222uuu +
2
3
2
2 uu
3217.0 uuu ==
Fluctuation mom.
j
ij
j
ij
j
ijij
jij
ij
i
x
Uuu
uux
uuexx
p
x
uUt
u
+
=
+
)2(
1
for i=1
+
=
+
j
j
j
j
j
jij
jj
jx
Uu
u
uu
x
uue
xx
p
x
uU
t
uu 111
1
111 )2(
1
or div. (no net effect on closed sys.)
j
j
j
jjj
jj
jx
ue
x
Uuu
x
upeuuu
xpu
xu
xUu
t
+
+
+
=
+
11
11
1
111
2
11
1
2
1
2
1 21
22
11
2
1
2
1
1P u
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+
jjxxx
v1
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Turbulence
j
j
j
jjj
jj
jx
ue
x
Uuu
x
upeuuu
xpu
xu
xUu
t
+
+
+
=
+
2222
2
222222
1
2222 21
22
11
2
1
2
1i=2 :
i=3 :
j
j
j
jjj
jj
jx
ue
xuu
x
upeuuu
xpu
xu
xUu
t
+
+
+
=
+
33
33
3
333
2
33
3
2
3
2
3 22222
continuity
+
=
3
3
2
2
1
1
x
u
x
u
x
u
+
=
3
3
2
2
1
1 11
x
u
x
up
x
up
The rote of the pressure-strain rate term in the eq.s for the individual compents of KE is to attempt to distribute
the energy among them, more or less equally.
This course a tendency of turbulence to return-to-isotropy when left alone (isotropy : )2
3
2
2
2
1 uuu ==
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T b l
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Turbulence
Uall derivatives of averaged quantities (except ) are identically zero
i.e.
2x( etcuuxx
u
02
1, 3
2
1
32
2
21
=
jx
ue
x
up
11
1
1
averages of products of fluctuations are not, in general ,zero even if flow is homogeneous !
write comp. of KE equs for the fluctuation.
ijij
uUu
eey
Uuv
=
=
111
20
0
P
j
j
j
j
uu
x
ue
x
up
xyx
+=
33
22
2
2
1
1
1
21
0
j
j
j
j
x
ue
y
Uuv
x
up
x
up
xe
xp
+
=
=
11
3
3
2
2
3
3
21
0
/
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Turbulence
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Turbulence
At very high turb. Re , the dissipation tends to be equally distributed between components( local isotropy of small scale )
11 Uuu
3
1
10
30
2
2
32
+=
+
=
x
up
yuv
xp
xp
3
1
10
3
3
+=
x
up
In fact it is the action of the ressure-strain rate terms which is res onsible for the local isotro ofthe small scales where the dissipation occurs
Production Dissipation in many shear flows, even where other terms are present.
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