2015 Lecture 16 Viscous Flow

7/25/2019 2015 Lecture 16 Viscous Flow

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Laboratory for Aero & Hydrodynamics

Advanced Fluid Mechanics

Lecture 11: Laminar flow

K&C: 9.1-9.8, 9.10

KC&D: 8.1-8.7

dr. Ren Delfos & dr. Daniel Tam

Slides by dr.ir. Christian Poelma

Fall 2015


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Osborne Reynolds (1883) "An experimental investigation of the circumstances which determine whether the motion of water in parallelchannels shall be direct or sinuous and of the law of resistance in parallel channels"

Laminar flows are generally associated with flowsin which viscous effects are dominant over inertial effects


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Overview

Previously, we looked at inviscid flow ( irrotational/potential flow)

For some cases, we can no longer ignore the viscous term

The viscous term can be interpreted as a vorticity diffusion term.

For some cases, we can still analytically solve the full Navier-Stokes equations:

- Fully-developed flow

- Pressure driven, Couette flow, Hagen-Poiseuille

- Taylor-Couette flow

- Self-similar solutions: Stokes' first problem

- Stokes' second problem


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Introduction

Previously, we looked at inviscidflow

In the absence of viscosity, flow fields remain irrotational potential flow

For many problems, viscous forces cannot be ignored:

in proximity of surfaces (no-slip versus slip as seen in inviscid flow)

low Reynolds number

This means that we need to solve the full Navier-Stokes equations, includingthe viscous term:

D uDt

=gp+ 2u

D u

Dt=gp+ 2u

w ' t '

+u ' w 'x '

+v ' w 'y '

+ w ' w 'z '

=p 'z '

gl

U2+

Ul(2 w '

x '2+

2 w '

y '2+

2 w '

z '2 )


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Interpretation of the viscous term

If we take the curl of the N.S. Equation, we find the vorticity equation (here 2D):

Du

Dt=gp 2u

D

Dt [+u]= 2 curl

vorticitiy only in x-y plane (z);

pressure and g disappear (curl of grad) vortex stretching term disappears (2D)

DT

Dt= 2 THeat equation: = k/Cp : thermal diffusivity

Both kin. viscosity (momentum diffusivity) and therm. diffusivity have units m 2/ s

The viscous term smears out vorticity (or velocity gradients)

K&C 9.2; Fig 9.11

t(orx )

no viscosity:jump in velocity at wall,i.e. line/sheet of vorticity

vorticity diffusesaway from wall:fluid becomes rotational


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Solving the Navier-Stokes equations...

The full Navier Stokes equations contains the non-linear advection (uu),

which prevents us solving it analytically (in general):

D u

Dt=

u t

u u=p 2u

K&C 9.4

Solutions are available for cases where the advection term is zero:

u u=uj uixj

=0

When is this the case?


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Fully-developed flow

Visualized Flow (J. Soc. of Mech. Eng. / Pergamon)


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K&C 9.4

uu uux

v uy

w uz

Consider the flow from a large reservoir into a channel (2D):

This flow is the result of a pressure difference between reservoir and channel: dp/dx


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At the entrance, the velocity profile is flat:u(y) = const. The no-slip condition (viscosity!) leads to a deceleration of fluid near the wall For a given y-location near the wall, uvaries withx: u/x0 This implies that there must be a small vertical velocity (continuity!)- Core of flow accelerates (again: due to continuity) Both terms of advective term non-zero (we only consider the momentum equation in thex

1direction)

K&C 9.4

uu uux

v uy

w uz

Consider the flow from a large reservoir into a channel (2D):

ux

vy

=0

Viscous region (boundary layer)

and inviscid region (core)


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After some length(*), the boundary layers merge and an equilibrium is reached- Core no longer accelerates (cannot go to infinity) The velocity profile does not change anymore Now, the /xterm becomes zero: fully developed flow;

Also: vterm becomes zero, so advection term vanishes! (Another interpretation: streamlines are parallel)

K&C 9.4*) This length depends on the Reynolds number : L/D ~ 0.06 Re (see K&C, page 789)

uu uux

v uy

w uz

Consider the flow from a large reservoir into a channel (2D)

u

x

v

y=0


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Fully-developed 2D channel flow

K&C 9.4

For the fully-developed case in 2D, the Navier-Stokes

equations simplify to:

0=1

px

2u

y2

0=1

p

y

2D: /z = 0Steady: /t = 0Fully-dev: /x= 0

x

y

Dui

Dt=

pxi

2 uixj

2

p/y= 0 implies thatpis not a function of y, only ofx,

from the xequation, it then follows that ucan only be a function y,

which means that the two terms in the xequation are constants

So the pressure drop per unit of length is constant over the channel

drives flow


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K&C 9.4NB:

For the fully-developed case in 2D, the Navier-Stokes

equations simplify to:

0=1

p

x+

2

u

y2

Du i

Dt=

pxi

+ 2uixj

2

Integrating this equation twice, we obtain:

0=y

2

2px

+ u+ Ay+B

The constants can be found from the boundary conditions:

u=0 at y=0 0=0

2

2px

+ 0+ A0+B B=0

u=0 at y=2 b 0=(2 b)2

2

p

x+ 0+A2b +0 A=

b

p

x

0=y

2

2px

+ u+b

px

y

u(y)=y

dp

dx

(by

2

)

Plane Poiseuille flow


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K&C 9.4NB:

u y=y

dp

dx

by

2

Q

0

2b

u y dy=[by

2

2

y3

6

dp

dx]

0

2b

=2b

3

4b3

3

dp

dx=

2 b3

3

dp

dx

u y=by

dp

dx

y2

2dp

dx

flow rate per unit length [m2/s]

VQ

2b=

b2

3 dp

dxaverage velocity


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u y=y

dp

dx

by

2

Q

0

2b

u y dy=[by

2

2

y3

6

dp

dx]

0

2b

=2b

3

4b3

3

dp

dx=

2 b3

3

dp

dx

u y=by

dp

dx

y2

2dp

dx

flow rate per unit length [m2/s]

VQ

2b=

b2

3 dp

dxaverage velocity

ij =p ij 2 eij = dudy = b 2y2 dpdx =by dpdx u

PL2b ZPR2b Z=2L z W=PLPRL2b Z2Z =

dpdx

b

PL

PR

Force balance of pressure and wall shear stress:

K&C 9.4NB:

U


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Couette flow

K&C 9.4NB:

0=

y2

2

p

x uAyB

U

We can find another fundamental flow by applying a different set of boundaryconditions:

- assume that the top wall is moving with a velocity U- assume that there is no longer a driving pressure (dp/dx= 0)

U


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Couette flow

K&C 9.4

0=

y2

2

p

x uAyB

U

We can find another fundamental flow by applying a different set of boundaryconditions:

- assume that the top wall is moving with a velocity U- assume that there is no longer a driving pressure (dp/dx= 0)

0= uAyB

u=0 at y=0 0=0A0B=0 B=0

u=U at y=2b 0= U2bA A= U2b

0= u U2b

y u=Uy

2b=

U2b


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Parallel, laminar channel flows

K&C 9.4

As the non-linear advection term is zero, the problem becomes linear. This

means that we can find solutions by combining the pressure-driven andwall-driven cases:


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Steady pipe flow (a.k.a. Hagen-Poiseuille)

K&C 9.5

The same procedure can be done for a circular geometry, using cylindrical coords.:

0=p r

0=px

r

d

dr rdudru=

r2a2

4dp

dx , Q=0

a

2 r u d r = a4

8dp

dx

this law wasdetermined empirically

by Poiseuille (mid 19thcentury)

NB: Poiseuille found that Q ~ a4p/L;(vel. profile & viscosity unknown at time)


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Circular Couette flow

K&C 9.6

The flow field between two rotating cylinders for low Reynolds numbers:

0= d

dr

[

1

r

d

dr r u

]

u

2

r=

1

dp

dr

Note that the pressureincreases with r

u=1R1 at r=R1u=2R2 at r=R2

Boundary conditions:

u= 1

1R1/R22 [21R1R2

2

]rR12

r12


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Circular Couette flow

K&C 9.6

The flow field between two rotating cylinders for low Reynolds numbers:

u= 1

1R1/R22 [21R1R2

2

]rR12

r12

2=0,R2=

flow around arotating cylinderin an infinite fluid:

our previousirrotational vortex

viscous, but irrotational (!)

1=0,R1=0

flow inside arotating cylinder

our previoussolid body rotationresult!


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Impulsively-started plate

K&C 9.7

The previous cases were all stationary. Let's consider the flow

field caused by an infinite plate at rest that suddenly startsmoving (with a velocity U) this is Stokes' first problem.

x, U

Flow

u t

=px

2 u

y2

0=py

[note the similiarity to the Couette flow case, but nowwe have a time-dependent term and other BCs]

Flow field is not a function of x (infinite plate), so:du/dx+ dv/dy = 0 dv/dt = 0 v= 0 (plate only moves inx-dir)

The pressure gradient must be zero (plate is infinitely long, invariant for shift inx), so:

u t=

2u

y 2 NB:

u y , 0 =0u 0, t =Uu , t=0

Initial conditions: fluid at rest, Surface moves at velocity U for t>0

The velocity at infinity remains unaffected by the motion of the wall

with the following boundary conditions:


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K&C 9.7

We can simplify the problem by realizing that U

only occurs in one of the BCs. We introduce the newdependent variable u' = u/U;(i.e. we make the velocity dimensionless):

x, U

Flow

u ' t

=2 u '

y 2

u 'y , 0=0u '0, t =1u ' ,t=0

From dimensional analysis, we find that the solution must be of the form

u '=u

U=fy , t ,

NB: the left-hand side is dimensionless, so right-hand side must be too!

u '=uU

=F( y t)=F()

This is the only possible combination and of course reciproke, square,etc.; we write it here so we wil end up with u= f(y/ (...))

with = y

2 t (the factor 2 is added for convenience);

The yis made non-dimensional with , t. is called a similarity variable,

Fis called a self-similar solution because one dependant variablescales with another dependant variable.

' 2 '


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K&C 9.7

To solve it, we substitute our new parameters... x, UFlow

u '=uU

=F( y

2 t)=F() u t

= UF()

t =U

F() t

=Ud F

d t

=Ud F

d y

4 t3/2

t = y

4 t3/2 =

2 t

=Ud F

d 2 t

= y2 t

u t

=

u=u ' U=U F

u ' t

= u '

y 2

I l i l d l u ' 2

u '


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K&C 9.7

To solve it, we substitute our new parameters... x, UFlow

u '=uU

=F( y

2 t)=F() uy

=UF()

y

=Ud F

d y

uy

=Ud F

d 1

2 t

y =

1

2 t

2 u

y2=

y ( U2 t F )

2 u

y2=

U

4 td

2F

d 2

u t

=2u

y 2 U

d F

d 2 t

= U

4 td

2F

d 2 2

d F

d =

d2F

d 2

We have reduced our partialdiff. equation to an ordinary2ndorder dif. equation; Also:only 2 BCs left/needed.

u y , 0 =0

u 0, t =Uu , t=0

F=0

F0=1F=0

u t

= u

y 2

=Ud F

d 2 t

u t

=

I l i l t t d l t


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K&C 9.7

We can solve this 2ndorder ODE: x, U

Flow

2 d F

d =

d2F

d 2

2 G=d Gd

d F

d =G

2C=ln G

A e2=G

A e2=

d F

d

F=A0

e'2

d '+ B

[we substitute dF/d= G]

[separation of variables]2 d =d G

G

[integrate]

[exponent of both sides, rewrite integration constant]

[substitute original term for G]

[integrate]

I l i el t ted l te


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K&C 9.7

We can solve this 2ndorder ODE: x, U

Flow

F=A0

n

e

2

d B

F=0F0=1F=0F()=A

0

e2

d +1=0 A=2

F0=A0

0

e

2

d B=1 B=1

F()=1 20 e2

d

error function erf()

u

U=1erf( y

2 t)

y

t

2 d F

d =

d2F

d 2

Impulsively started plate


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K&C 9.7

x, Uu

U=1erf( y2 t)

Two observations:

1. The initial condition at the wall is a vorticity sheet;

no additional vorticity is concentrated, it only diffuses away from the wall.

2. If we choose u/U= 0.05 to define the thickness () of the layer in which the velocity has penetrated, we find from the solution that this occurs at = 1.38.

2.76 t

2.76 x

U27.6

x

U (d /2)27.6

x

U

D U

47.6

x

D

x

D0.03R e

Inlet length can be estimated by finding out how longit takes for to reach the middle of a channel (D/2)

NB: exact value of constant depends on criteria

Diffusion of a vortex sheet


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Diffusion of a vortex sheet

K&C 9.8 (This is identical to the previous example)

u=U erf

y

2 t


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Two-dimensional image of an axisymmetric water jet, obtained by the laser-induced fluorescence technique. (From R. R. Prasadand K. R. Sreenivasan, Measurement and interpretation of fractal dimension of the scalar interface in turbulent flows,Phys. Fluids A, 2:792807, 1990)

http://www.eng.fsu.edu/~shih/succeed/jet/meanjetwidth.jpg

Another exampleof a self-similar flow:

the mean velocity inthe far-field of a turbulent jet

Oscillating plate


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Oscillating plate

K&C 9.10

Stokes' second problem describes the flow as a result of an oscillating wall.

Note that this solution is NOT self-similar!

u t

=2u

y 2

u 0, t=Ucos t

u( , t)=bounded

u=Uey / cos ( ty /2 )

2 = penetration depth

amplitude periodic phase

solve using Ansatz u

U=ei tf(y)

Example: Pulsatile flow =R

= Wo = Womersley-number


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Example: Pulsatile flow

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2015 Lecture 16 Viscous Flow