Turbulent Flow(Final)

8/16/2019 Turbulent Flow(Final)

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A d v a n c e d F l u i d M e c h

a n i c s

Chemical & Petroleum Engineering Department

Sharif University of Technology

TURBULENT FLOW

Presented by:

Prof. D.Rashtchian


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Turbulent transport of momentum

Turbulence of random velocity fluctuations- Use statistical methods

Turbulent velocity iu~

ˆ ~ iii uU u +=

Component Velocity

gFluctuatin Mean

*************************

Fig.1

Interpret Ui as a time averaged velocity defined by:

( )∫∫ ≡+== ∞→∞→T

iiiT

T

iT

i udt uU T

dt uT

U

00

~ˆ1

lim~1

lim

( )

∫∫ =−=−==

∞→∞→

T

iii

T

i

T

i

T

i U U dt U

T

U dt u

T

u00

01

limˆ1

limˆ


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i.e. the mean value (or time average) of the fluctuating quantity is zero. Assume that

Ui the mean flow is steady (∂Ui/∂t = 0)

Note: Time averaging commutes w.r.t. differentiation.

( )i

j j

iT

i j

T

j

i

j

i

u x x

U

dt uT xdt x

u

T x

u ~~1~1~

00 ∂

∂

=∂

∂

=

∂

∂

=∂

∂

=∂

∂

∫∫

The time average of the fluctuationiû is zero, but the average of the square of the

fluctuation is not zero and the quantityi

i

U

u 2ˆ is used as a convenient measure of the

turbulent fluctuation-known as the "intensity of turbulence" and ranges from 0.01 to

0.1 for most turbulent flows.

( )2ˆiu r.m.s. velocity.

Mean K.E./unit volume =( ) ( )

nsfluctuatioflowmean

ˆ2

1ˆ

2

1 22

+

+=+= iiii uU uU KE ρ ρ


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Equations for the mean flow

Consider the momentum and continuity equations. These apply to the instantaneous velocity

in a turbulent field.

j j

i

i j

i j

x x

u

x

p

x

uu

∂∂

∂+

∂

∂−=

∂

∂ ~~1~~2

ν ρ

0~

=∂∂

i

i

x

u )1(

The equations must apply on average

ˆ~iii uU u +=

Continuity

( )0ˆ

~~1lim

0

=∂

∂=+

∂

∂=

∂

∂=

∂

∂

∫∞→ i

i

iiii

iT

i

i

T x

U uU

x x

udt

x

u

T (2)

The mean value satisfies continuity. It is the mean value of velocity that we measure and

require in applications.


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Momentum:

The equations of motion for the mean flow Ui are obtained by taking the time average of all

terms in the resulting equation.

Consider each term:

( )( ){ }

{ }

{ } (2.1) )ˆˆ(ˆˆ

ˆˆˆˆ

ˆˆ

~

~)~~(

~

~ (i)

i j

j j

i j ji ji

j

i j jii ji j

j

ii j j

j j

j

ii j

j j

i j

uu x x

U U uuU U

x

uU uU uuU U x

uU uU x x

u

uuu x x

uu

∂

∂+∂

∂=+∂

∂=

+++∂

∂=

++∂

∂=

∂

∂

−∂

∂=∂

∂

(2.2) 1

)ˆ(1~1

(ii)i

i

ii x

P p P

x x

p

∂

∂−=+

∂

∂−=

∂

∂−

ρ ρ ρ

( ) (2.3) ˆ~

(iii)222

j j

iii

j j j j

i

x x

U uU

x x x x

u

∂∂

∂=+

∂∂

∂=

∂∂

∂ν ν ν


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Hence

(3) )ˆˆ(1

2

i j

j j j

i

i j

i j uu

x x xU

x P

xU U

∂∂−∂∂∂+∂∂−=∂∂ν

ρ

Equation for mean flow has an additional term.(Drop the ^ i ji j uuuu =⇒ ˆˆ )

Term j

i j

x

uu

∂

∂ is analogous to the convective term

j

i j x

U U ∂

∂;

It represents the mean transport of fluctuating momentum by turbulent velocity

fluctuations.

If iû and jû uncorrelated i.e. 0=i juu - no turbulent momentum transfer but

experience shows that 0≠i juu - momentum transfer is a key feature of turbulentmotion.

Term )ˆˆ( i j j

uu x∂

∂thus exchanges momentum between the turbulence and the mean

flow (equation 2.1)even though the mean momentumof the turbulent velocity

fluctuations is zero ( 0~ =iu ρ ).


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Because of the decomposition iii uU u ˆ~ += , turbulent motion can be perceived as something which

produces stresses in the mean flow. For this reason, equation (3) may be rearrange so that all stresscan be put together.

)(ˆˆ ji j

i j

i

j

j

i ji

j j

i j T

xuu

x

U

x

U P

x x

U U

∂

∂=

−

∂

∂+

∂

∂+−

∂

∂=

∂

∂ ρ µ δ ρ - mean stress tensor.( τ τ ˆ~ += T )

∂

∂+

∂

∂=−+−=

i

j

j

i jii j ji ji ji

x

U

x

U uu P T µ σ ρ σ δ ; ˆˆ

(shear) (normal)

The contribution of the turbulent motion to the mean stress tensor is i jT

ji uu ρ σ −= called the

Reynolds stress tensor. Define, T ji ji σ σ +=Ω ji


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Turbulent shearing stresses

Time averaging of the equations of motion leads to the Reynolds stress tensor, i juu ˆˆ ρ .

iû and jû are the velocity fluctuations in the ji ≠ directions at one point and jiuu is a

measure of the "correlation" between the fluctuations.

Correlated variables

( )( ) ji ji j jii ji uuU U uU uU uu +=++= ˆˆ~~

If 0≠ jiuu , iû and jû are said to be correlated i.e. dependent.If 0= jiuu , uncorrelated i.e. iû and jû are independent.


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Fig2(a)

1

0

12

21

=

>

R

uu

Fig2(b)

1

0

12

21

−=

<

R

uu

Fig2(c)

1

0

12

21

≈

≈

R

uu


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A measure of the degree of correlation between 1û and 2û is obtained from:

{ }2

1

22

21

21

ˆ.ˆ

ˆˆ

uu

uu

( )2

1

0

22

021

2112

ˆ1

limˆ : ˆˆ1

lim

==′

′′= ∫∫ ∞→∞→

T

iT

i

T

iT

dt uT

uudt uu

uu

T R

21

2112uu

uu R ′′=

N.B. abbaba ≥+⇒≥− )(2

10)( 222

Hence ∫ ≤

+≤ ∞→T

T dt

uu

uu

T R

0 2

2

2

2

2

1

2

112 1

ˆˆ

ˆˆ1lim


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Pure shear flow

Consider a turbulent shear flow with U1(x2) the only non-zero velocity component.

Ω12 is the only component of the mean stress tensor, 122

112

ˆˆ uu xU ρ µ −

∂∂=Ω

Ω12 – stress in 1 direction on face, normal in 2 direction and must result from

molecular transport of momentum in the x2 direction, and turbulent transport.

Assume 02

1 >∂

∂

x

U .

A fluid particle with positive 2û is being carried by turbulence in positive x2

direction. It is coming from a region where the mean velocity is smaller i.e. is likely

to be moving downstream more slowly than its new environment. Thus 1û is negative.

Similarly negative 2û associated with positive 1û .

************************

Fig3


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{Momentum/unit volume of flow at A in 1-direction} = )ˆ(~ 111 uU u += ρ ρ

The x1-momentum is transported in the x2-direction if u1 and u2 are correlated.

{Flux of x1-momentum in x2-direction} = 211 ˆ)ˆ( uuU + ρ

{Average flux of x1-momentun in x2-direction} = 21 ˆˆ uu ρ

1û and 2û are negatively correlated: 122112 uuT T ρ σ σ −==


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Turbulent Channel Flow

The Navier-Stokes equations in rectangularcoordinates are

)(12

ji

j j j

i

i j

i j uu

x x x

U v x

P

x

U U ∂

∂−∂∂

∂+∂∂−=

∂∂

ρ

For parallel, fully developed, 2 D flow

0...0;0

0

31

32

=∴

=∂

∂=

∂

∂

==

S H L

x

U

x

U

U U

ii

0)(1

=∂∂

jiuu x

; 0)( 33

=∂

∂ uu x

i


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Hence the equations can be written in the simplified form,

)(1

02

2

uv y y

U v

x

P

∂

∂−

∂

∂+

∂

∂−= ρ

(1)

)(1

0 2v y y

P

∂

∂−

∂

∂−= ρ

(2)

At the walls 2v = 0, P = P0(x) . Hence form (2)

20 v P P

+= ρ ρ

(3)

dx

dP

x

P

x

P 00 =∂

∂=

∂

∂ (4)


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Hence (1) can be integrated from y=0 to y with 00

== y

uv

uv y

U v

y

U v

dx

dP y

y

−∂

∂−

∂

∂+−=

=0

0 )(0 ρ

At y=h, uv=0, 0/ =∂∂ yU (zero velocity gradient, no correlation)

0

0 )(=

∂

∂=−=

y

w

y

U

dx

dP h

ρ

µ

ρ ρ

τ

Defining a friction velocity u*

2

∗= uw ρ τ

Substituting in (5)

)1(2*h

yu

y

U vuv −=

∂

∂+−


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Equation (8) may be written in dimensionless form in 2 ways.

(I))1(

)/(

)/( *

*

2

* h

y

h yd

uU d

hu

v

u

uv−=+

−

R* =u* h/v. As R* becomes large, (R* is of course a Reynolds number), the

viscous stress is suppressed. Such a limit will not applied because viscous

forces must always dominate near solid boundaries.

(II)*

*

*

*

2

*

.1)/(

)/(

hu

v

v

yu

v yud

uU d

u

uv−=+−

In this case as R* becomes large the change in total stress becomes small.Defining appropriate dimensionless variables


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;*

v

yu y =+

∗

=+u

U u

; h

y=η

Then

η η

−=++− 11*2* d

du Ru

uv (11)

*2* 1 R

y

dy

du

u

uv +

−=+

+

+− (12)

Law of wall

For large R* (from 12)

12

*

=+

++−

dy

du

u

uv (13)


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The solution of this equation must be of the form,

)(2

*

+=− y g u

vu; )( +=+ y f u (law of the wall) (14)

For sufficiently small y+, turbulent stress negligible.

1=+

+

dy

du ; with u+(0)=0 (15)

u+ = y+

Core region

For large R* (from 11)

( )η −=−∗

12u

uv


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This equation gives no information, about U itself. However h and u* are the only feasible

length and velocity scales, we can write

η d

dF

h

u

dy

dU *= Where F(η ) is some function of η . (17)

Integration from the center where U=U0

)(*

0 η F u

U U =

− (18)

From equation (14),

)(*

+= y f u

U ;

+

+=

dy

ydf

v

u

dy

dU )(2

* (19)

Matching (17) & (19),

+

= ∗∗

dy

df u

d

dF

h

u

ν η

2

. ;

K dy

df y

d

dF 1==

++

η

η (20)


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.ln1

)( const K

F += η η .ln1

)( const y K

y f +=+ +

Hence

.ln1

*

0 const

K u

U U +=

−η .ln

1

*

const y

K u

U += +

Discussion

To simplify (12) to (14) requires 1* 〈〈=

+η R

y

(a)

To simplify (11) to (16) requires η η

−〈〈+

1*

1

d

du

R (b)

Matching only possible if ∞→+ y

0→η

In practice it is found that

<

>+

1.0

100

η

y are sufficient

A d v





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Now 1.0+ y 100* >∴ Rη

/100* >∴ R 1.0∴ R Experimentally

η η 5.2=+d

du

η η *

5.2

*

1

Rd

du

R=

+∴

Hence )1(*

5.2

*

1η

η η −


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Also from (20) K dy

df y

1=

++

.)ln(

1

)( const y K y f +=∴

++

(21)

Experimentally

0.1ln5.2*

0 −=−

η u

U U

0.5ln5.2

*

+= + y

u

U

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Application

1. For Engineering Purposes these

equations have been used for 1.0>η ,i.e. to describe the core region, and

also for 0→η . Note as

0→η , −∞→=+ */ uU u

2. Sometimes the Universal Velocity

profile is used.

Equn. (15) u

+

=y

+

for y

+

≤ 5

Equn. (21) u+=2.5lny

++5.0for y

+ ≥ 30.

Limits determined experimentally.

A curve fit for 5


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Example of use of turbulent velocity profiles.

momentum transfer

Friction factor 2

2

*

2

2

2

1 U

u

U

f ==

ρ

τ ω

Using the velocity defect law for flow in a tube

*

0

0 *

0

22.

1

u

U U rdy

u

U U

h

h y

y

−=

−∫=

=

π π

∫

=

= −=

1

02 }1ln5.2{

2µ

η η η π

π

d hh

r

ow y-hr = ; η η hd dyh y =→= / ; )1( η −= hr

=1)(

ηUU


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Hence ∫=

−−=− 1

0*

0 }1ln5.2){1(2)(

η

η

η η η d u

U U

Also from experimental results

0.1ln5.2**

0 +−= η u

U

u

U

0.50.1ln5.2ln5.2 ++−= + η y 0.6*ln5.2 += R

22

Re

2

**

2 f u f

v

h

v

hu R ===

6]22

Reln[5.2

*

0 += f

u

U

10

222

*

]24

5ln2

525ln5[6]22

Reln[5.2 η η η η η η η η ++−−−++= f uU

53.022

Relog07.4

110 +

= f

f

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Mass transfer : Turbulent Taylor Analysis. Proc. Royal. Soc. (1954), A223, P446, for Axial Dispersion in turbulent pipe flow.

Consider diffusion equation in rectangular coordinates for simplicity.


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Turbulence in pipe flows

Scope of Turbulence

Most flows in nature: rivers, the atmosphere

Engineer: pipe flow, packed and plate column

Pipe Flow

Laminar sublayer - viscous forces dominate, very thin

Transition region - region of damped turbulence because of nearby wall,

eddy size y.

Turbulent core - region of fully developed turbulence, eddies of size d,

velocity nearly constant.

T b l V l i i


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Turbulent Velocities

-

local downstream velocity fluctuates due to turbulent eddies .decompose

+

=

+=

velocity

eddy

velocity

localmean

velocity

us Instantane

uuu t ˆ

-

definition of u (mean velocity)

∫= T

t dt uT

u0

1

-

clearly the average of the eddy velocity is zero

0ˆ1

ˆ11

)ˆ(1

0

0 00

=

+=+=

∫

∫ ∫∫

dt uT

dt uT

udt T

dt uuT

u

T

T T T


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- the magnitude of turbulent velocities is characterized by the RMS

-

21

0

2ˆ1

=′ ∫

T

dt uT

u

(RMS fluctuating or eddy velocity.)

-

the turbulence intensity is defined by,

turbulent intensity =

u

u′(typically up to 0.1) i.e. the average eddy velocity

may be 1/10 of the mean velocity.


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Properties of turbulent flows

(with particular reference to pipe flows)

1) Irregularity local velocities fluctuate in random manner. But all

turbulent flows are irregular. E.g. smoke plume.

2) 3D Nature pipe flows are normally considered as 1 dimension in

that downstream velocity depends only on radius. However in

turbulent flows normal velocity components, though zero on average,have fluctuating components, (V ̂ and W ˆ ). These give rise to

turbulent stresses (remember the mail bag example) and are important

in turbulent energy processes. This 3D nature adds the mathematical

difficulty.


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3) Turbulence is a property of the flow not of the fluid writing Newton’slaw for a flow involving turbulent stresses.

dy

duT )( ν ν

ρ

τ +−= ; [divided by ρ ]

Where ρ

µ ν = is the kinematics viscosity . [ ]

[ ]T L 2

=T ν eddy viscosity

In laminar sub layer ν ν T

Thus T υ varies with environment and is a flow property.


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4) Mixing in turbulent flows-diffusivity

Rewrite Newton’s law in the form, explicit in shear stress.

( )dy

ud T

ρ ν ν τ )( +−=

Dimensions :[ ][ ][ ] [ ]

[ ][ ][ ]

[ ][ ] L

L

U M

T

L

T L

U M 32

2.=

i.e. (momentum flux) = (diffusivity) * (gradient of momentum / volume)

- this fundamental relation shows how transport (here of momentum) is related

to the driving force (momentum gradient).the coefficient, υ , is the momentumdiffusivity. It shows how large a flux is produced by a given gradient. Exactly

analogous laws apply for heat transfer (Fourier's Law) and mass transfer

(Flick's law).


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- Now ν ν ffT in turbulent flows.

Turbulent is a very effective mixer of momentum which accounts for

the almost constant velocity of the core region will usually be of almost

constant temperature and composition.

But T ν ν ff in laminar sublayer .


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In laminar flow it is the molecular motion which transports momentum.(Remember mail bag example). Hence lower rates of transport for a given

driving force. Alternatively if we consider heat transfer from the wall to

bulk, heat conduction across the laminar sub layer dominates the process

(Heat transfer comes later).

- Pictorially

Eddy gives rise to normal velocity V ̂ .This transports x directional momentum

in the y direction gives rise to a

momentum flux, or shear, .τ


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5) Dissipative Nature of Turbulence.

- Turbulence comprises eddies of all sizes.

- The largest eddies are as big as the flow field. They extract energy from the

flow but are not efficient at dissipating energy. In the absence of an energy

source, however, turbulence dies away .

- There is an energy cascade from the large eddies, through eddies o

progressively smaller size until a lower limit is reached. This lower limit is

controlled by viscous dissipation of energy and Kinematics viscousity and the

rate of energy supply are the important quantities. Based on dimensional analysis

this lower limit of eddy size is given by:

434

3

Re−=

′

=d ud

ν η

Where = size of small eddies; =ν kinematics viscosityd = size of largest eddies; u' = RMS turbulent velocity


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( )

=

d v

si

6

( )

=×=

d d

d

v

sii

12

4

68 3

2

( )

=×=

d d

d

v

s

iii

24

16

6

64

32

∝ v s

volume surfaceratendissipatio


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High Reynolds number phenomenon

- Express Newton's Law of viscosity in dimensionless form.

+

+

+ −=

−=−== dy

du

d y

d u

ud

dydu

d ud

u Re1

Re122 ρ

µ ρ τ τ

Reynolds's number arises in dimensionless form of Newton's Law.- Similarity: compare two flows in similar geometries(same shape but

different size)i.e. flows exhibiting geometrical similarity. Suppose

Reynolds numbers of each flow are the same though d,u, ρ and µ of

each flow may be individually different. Then as a consequence of theabove equation each flow will have the same dimensionless distribution

of stress and velocity gradient as a function of position,


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Stress and velocity gradient as a function of position, provided each has the same

Reynolds number;i.e. u+ = f(y+) kinetic similarity

τ+ = g(y+) dynamic similarity.

The consequence is that friction factor (dimensionless wall shear stress) can be

considered a unique function of Re.Consider a cylindrical element of diameter d and length of δx

Viscous forces ∝ xd dr

duδ π µ .

Interia forces ∝ dt du xd e δ π 4

2

Re≡∝dt

dr d

cesViscousFor

ces InteriaFor

µ

ρ

High Re-interia forces dominate → Turbulent flowLow Re-viscous forces dominate → Laminar flow

Summery Notes on Turbulence


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Summery Notes on Turbulence

• Most flows are turbulent both in nature and engineering.

•

A turbulent pipe flow can be divided into three regions:

a) Laminar sublayer - no eddies.

b) Transition region – damped eddies (size y)

c) Turbulent core – undamped eddies (size d)

•

Turbulent velocities:

uuut ˆ+= {Instantaneous = local mean + fluctuant}

∫=T

t dt uT

u0

1 {T is a time long enough to include many eddies}

2

1

0

2 ]ˆ1

[ ∫=′T

dt uT

u {RMS velocity characterizes turbulence}

wvu ′≅′≅′ {Turbulence is homogenous}1.0/ ≅′ uu {Turbulence intensity}

• Newton's Law in turbulent flows. It is tempting to write

dy

dut )(/ ν ν ρ τ +−=

−

−

Turbulence for ityVis Eddy

olecula for ityVis Kinematic

t cos

cos

ν

ν


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ν is a fluid property and constant.

νt is a flow property and depends on environment (eddy size)

•

Rewrite above equation as•

dy

ud vv t

)()( ρ

τ +−= dimensions22

][

][

][

][

T

L

L

M

(momentum flux)=(momentum diffusivity)(gradient of mom/vol)

Large ν implies rapid mixing. Diffusivity has dim. [L]2/[T]

νT >> ν : Turbulent flows are rapidly mixed due to eddies.

• Energy in turbulent flows: turbulent dissipates considerable energy.Large eddies take energy from mean floe, but are not efficient indispersing energy. Small eddies do dissipate energy efficiently. There

is a transfer of energy to the small eddies, which appears as heat due tofrictional effects.

Smallest eddy size, η, si given by ( dimensional analysis)

4

3

)(

−′=

v

d u

d

η {η is also a good estimate of laminar sub-layer thickness}


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h a n i c s



• Rynolds Number arises in Newton's Law in dimensionless form

+

++ −=−=−==

dy

du

d yd

uud

d udy

du

uu Re

1

)/(

)/(222 ρ

µ

ρ

µ

ρ

τ τ

It may be interpreted as the ratio (interia forces / viscous forces).

• Large Re implies dominance of interia forces which promote turbulence.Small Re will dominance of friction (viscous) forces gives laminar flows.

Similarity (Consider different flows of same Reynolds Number) If we have

geometric similarity (e.g. two different pipe flows) then we will havekinematic similarity (same du

+/dy

+) and dynamic similarity (same τ

+).

Result

f = τw

+= f(Re)

Documents

Turbulent Flow(Final)