Trapo SS14 IIa (1)

p. II.1/1

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Chapter II.1 – Flow Phenomena – Laminar Flow

Chapter II : Flow Phenomena

● Transport of momentum● Resistance due to boundaries or obstacles in the flow field

laminar means: the fluid elements move in (almost) parallel streamlines with different velocities leading to a velocity gradient

Two kinds of flow: laminar ↔ turbulent

II.1 : Laminar Flow

x

yt0

t1v

x(y)

( )xyx

dv y

dyt h h r= - =viscosity νwith in Pa.sspecial case -

incompressible fluids:

( ( ))xyx

d v y

dy

rt = -ν

First index in t: direction of gradient

Second index in t: direction of flow

p. II.1/2

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Examples for relations between shear stress and velocity gradient:

linear linear with offset power law

h … (Newtonian) t0 … yield stress n … power law index

viscosity K … viscosity for 1 s-1

First we consider a few very important cases of flows with simple relationships

p. II.1/3

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II.1.a : Stationary laminar flow between two horizontal plates

“1” p

1

“2” p

2

xy

W

L

dL … lengthd … thicknessW … width

The balance equation reads in this case:

( ), , , ,

0;0;

( )( ) ( )x

V in x in V out x out x

d vV v v F

dt

rf r f r

==

= - +å constant flow and cross section

stationarity

The equation is reduced to a mere force balance: 0xF =åThis force balance reads for the test volume with coordinates (0<x'<L, 0<y'<y, 0<z'<W) as follows:

1 20 ( ) ( ) ( )x yxF p Wy p Wy LWt= = - -åvolume forces surface force

and thus: 1 2( )yx

p py y

Lt

-=

p. II.1/4

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1 2( )xdv y p py

dy Lh-

= -For Newtonian liquids one gets:

The solution of this equation with boundary conditions: ( ) 02

forx

dv y y= = ±

reads:

In particular one has for the maximum velocity:

21 2

max (0)8x

p p dv v

L h-

= =

for the mean velocity:

/ 2 21 2

max

/ 2

1 2( )

12 3

d

x

d

p p dv v y dy v

d L h

+

-

- æ ö= = =ç ÷è øò

and the velocity gradient at the wall: 1 2

/ 2

( )

2x

y d

dv y p p d

dy L h=

-= -

221 2 2

( ) 18x

p p d yv y

L dh

æ ö- æ ö= -ç ÷ç ÷ç ÷è øè ø

p. II.1/5

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II.1.b : Stationary laminar flow through a horizontal circular tube

The balance equation reduces again to a force balance, which reads for the test volume with coordinates (x

1<x'<x

2, 0<r'<r) as follows::

2 21 2 2 10 ( ) ( ) (2 ( ))x rxF p r p r r x xp p t p= = - - -å

and thus:1 2

2 1

( )2rx

p p rr

x xt -

=-

R … radiusr … radial coordinatex … axial coordinate

Rr

x2-x

1

r

x

For Newtonian liquids one gets: 1 2

1 2

( )

2xdv r p p r

dr x x h-

=-

p. II.1/6

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The solution of this equation with boundary conditions: ( ) 0 forxv y y R= =

reads:

and for the mean velocity:

21 2

max22 10

1 1( ) 2

8 2

R

x

p p Rv v r r dr v

R x xp

p h- æ ö= = =ç ÷- è øò

221 2

2 1

( ) 14x

p p R rv r

x x Rh

æ ö- æ ö= -ç ÷ç ÷ç ÷- è øè ø

For the velocity gradient at the wall:

1 2

1 2

( )

2x

r R

dv r p p R

dr x x h=

-= -

-

In particular one has for themaximum velocity:

21 2

max2 1 4

p p Rv

x x h-

=-

and for the volume flux (Hagen-Poisseuille):4

2 1 2

2 18V

p p RR v

x x

pf ph

-= =

-

p. II.1/7

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II.1.c : Stationary laminar flow through a horizontal annulus

The balance equation reduces again to a force balance, which reads for the test volume with coordinates (0<x'<L, r<r'<r+dr) as follows:

1 20 (2 ) (2 ) (2 ' ) (2 ' )x rx rxr r drF p r dr p r dr r L r Lp p t p t p

+= = - + -å

pressure forces shear forces

Since one has: ( ( ) ) ( ( ) ) ( ( ) )rx rx rxr dr rd r r r r r r drt t t

+é ù= -ë û

one ends at: 1 2( ( ) )rx

p pd r r r dr

Lt -

=

First integration leads to:1 2 1( )2rx

p p Cr r

L rt -

= +

with some integration constant C1

p. II.1/8

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For Newtonian liquids the second integration leads to:

The shear stress is zero and the velocity is maximum at radial position

( )( )

2

1

2ln

abr bba

-=

21 2 12

1( ) ln( )

4x

p p Cv r r r C

L h h-

= + +

( ) 0xv r r a r b= = =for andwith a second integration constant. Using as boundary conditions the “No flow condition” at the two walls:one gets the following velocity distribution:

( )2221 2

1( ) 1 ln

4 ln( )x

ap p b r rbv r

bL b ba

h

ì ü-- ï ïæ ö æ ö= - +í ýç ÷ ç ÷è ø è øï ï

î þ

p. II.1/9

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II.1.d : Flow caused by moving surfaces

In contrast to II.1.c we replace the pressure forces by an axial movement of the inner cylinder.The balance equation reduces again to a force balance, which reads for the test volume with coordinates (0<x'<L, r<r'<r+dr) as follows:

0 (2 ' ) (2 ' )x rx rxr r drF r L r Lt p t p

+= = + -å

only shear forces

Thus one gets: 1

( ( ) )0 ( )rx

rx

d r rr r C

dr

tt= ® =

Second integration for Newtonian liquids leads to: 12( ) ln( )x

Cv r r C

h= - +

0( ) 0 ( )x xv r r b v r v r a= = = =for and forwith a second integration constant. Using as boundary conditions the conditions at the two walls:one gets the following velocity distribution:

p. II.1/10

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The shear stress distribution is given by:

( )0 1

( )ln

rx

vr

bra

ht =

In the limiting case b/a=1+x with small x one obtains the following approximation:

( )( ) ( )2

0 0 0 02 2

1 1 2( )

1ln 1 1rx

v v v va b a ar

br r r b a b a r b abaa

h h h ht +

= » = =- + --+ -

For r = a,b the limiting expressions of the figure above are obtained. In the denominator of the last expression a2 can also be replaced by b2 .

( )( )0

ln( )

lnx

brv r vba

=

p. II.1/11

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II.1.e : Flow caused by moving surfaces (rotating inner tube)

In contrast to II.1.d we replace the axial movement of the inner cylinder by a rotation with rotational speed n in s-1.

Instead of a force balance we have now a balance equation for the torque (rotational moments), which reads for the test volume with coordinates (0<x'<L, r<r'<r+dr) as follows:

0 '.(2 ' ) '.(2 ' )r rr r drM r r L r r Lq q qp t p t

+= = + -å

lever . force

Thus one gets:2

21

( ( ) )0 ( )r

r

d r rr r C

drq

qt

t= ® =

In cylinder coordinates one has for the Newtonian law: r

vdr

rdr

q

qt h

æ öç ÷è ø=

Which leads finally to:

with a second integration constant C2.

1 2( )2

C Cv r r

rq h h= -

p. II.1/12

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2 2 2

2 2( )

na b rv r

b a rq-

=-

0( ) ( ) 0v r v na r a v r r bq q= = = = =for and for

Using as boundary conditions the conditions at the two walls:

one gets the following velocity distribution:

And for the shear stress distribution:2 2

2 2 2

2( )r

na br

b a rJht =

-

For the torque one has of course:

( )

( )

2 2

2 2

2 2

2 2

( ) ( ) 2 ( ) 4

( ) ( ) 2 ( ) 4

r

r

a bM a a F a a aL a L n

b a

a bM b b F b b bL b L n

b a

q q J

q q J

p t p h

p t p h

ü= = = ïï- =ý

ï= = = ï- þ

p. II.1/13

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II.1.f : Flow through pipes with other cross sections

In II.1.a flow through parallel plates and in II.1.b flow through circular pipes have been considered. For elliptic, triangular and square cross sections also analytical solutions exist.

The results can be summarized in generalizing the result from II.1.a

where d is a characteristic thickness and B a characteristic width of the pipe.

The following figure gives the flow coefficient M0 for stationary flow of

Newtonian liquids in various cross-sections as functions of the ratio d/B :

for the volumetric flow rate:

3

12V

p Bdv Bd

Lf

hD

= =

by introduction of a flow coefficient M0

to:3

012V

p BdM

Lf

hD=

p. II.1/14

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p. II.1/15

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II.1.g : Non-stationary flow

Next we consider the problem, that for a Newtonian fluid at rest occupying the half-infinite space (y >0) the confining wall (at y =0) starts to move at constant velocity v

0 for t >0.

which reads: ( )xyx yxy y dy

d vdx dy dz dx dz dx dz

dt

rt t

+= -

We start with a micro-momentum balance for the volume with coordinates in

[ ] [ ] [ ], , ,x x dx y y dy z z dz+ ´ + ´ +

Using yx yx yxy dy yd dyt t t

+é ù= -ë û

one ends up with:

2

2x xv v

t y

¶ ¶=

¶ ¶ν with the kinematic viscosity

p. II.1/16

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The solution of this partial differential equation with initial and boundary conditions

reads:

This is one of the most important solutions for diffusion equations, which will be discussed more in chapter III and IV. Here only one important feature will be discussed. The velocity distribution near the wall can be approximated by:

with the error function defined as:2

0

2erf( )

zuz e du

p-= ò

0( , ) 1 erf4

x

yv y t v

t

é ùæ ö= -ê úç ÷è øë ûν

0( 0, 0) 0 ( 0, 0)x xv y t v y t v> = = = > =and

0 0( , ) (0, ) 1xx

v yv y t v t y v

y tp¶ æ ö» - = -ç ÷¶ è øν

The broken lines in the left figure represent this linear approximation. One thus defines the penetration depth

( )peny t tp= ν

For the wall shear stress one gets:

0( ) (0, )xw

v vt t

y t

ht h

p¶

= =¶ ν

Trapo SS14 IIa (1)

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Transcript of Trapo SS14 IIa (1)