Lecture 19-20: Natural convection in a plane layer.

Principles of linear theory of hydrodynamic stability

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z

x

kGrTvpvvtv

TPr

TvtT

1

0div v

Governing equations:

T=0

T=A

h=1

We will distinguish two cases:

A =-1 – layer heated from aboveA =1 – layer heated from below

Quiescent state

2

1. Quiescent state: 00 t

v ,

AzczcTT

zTkTkTkTkGrTp

2100

000000

0

000

Linear stability of a quiescent state

3

ppp

TT

v

0

0

Let us analyse the time evolution of a small perturbation of the quiescence

The linearised equations for a perturbation read

kGrvptv

PrTv

t1

0

0div v

v0 =0, p0 and T0 is the basic state;v, θ and p’ is a small perturbation

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Grvzp

tv

vxp

tv

z

z

x

x

PrAv

t z

1

0

zv

xv

zx

For 2D flow, we may introduce the stream-function defined as

xv

zv

zx

,

The continuity equation is satisfied automatically 0

22

xzxzzv

xv zx

Taking x- and z-projections of the Navier-Stokes equation gives

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The Navier-Stokes and heat transfer equation can be re-written as

Grxz

pxt

zxp

zt

PrxA

t1

z-derivative of the first equation minus x-derivative of the second equations gives

xGr

t

2

PrxA

t1

Boundary conditions:

00 zx

vv ,

0For temperature, at the upper and lower plate:

For velocity, at the upper and lower plate(rigid walls):

00

xz

,or, in terms of stream-function

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iri

tittir

expexpexp Time dependence of a perturbation:

In general,

If λr>0 then the perturbation will exponentially grow (with the rate λr).If λr<0 then the perturbation will exponentially decay (with the rate λr).

If λi ≠ 0 then the growth (or decay) is oscillatory.If λi = 0 then the growth (or decay) is monotonic.

Basic idea of the stability analysis: (i)Seek a solution in the form of the normal modes. (ii)Find λ that satisfy the equations (we may find several discrete values of λ or even continuous spectrum of λ).(iii)If at least one λ has positive real part then the considered basic state is unstable.

We will analyse stability of the quiescent state only in respect to normal modes:

ikxtztzxikxtztzx exp,,,exp,,

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For this problem it can be shown that perturbations develop monotonically, i.e. λi = 0.

ikGrkkk iv 422 2

21k

PrikA

Substitution of the normal modes gives Boundary conditions:

000 ,,

Next, system (*) together with the above boundary conditions can be solved numerically.

For the case, when the upper and lower plates are both free surfaces (which is not a good assumption as both plates cannot be free surfaces), the solution can be obtained analytically.

(*)

At a free boundary:

000 ,,

000

,,xv

zv

v zx

xzz

In terms of the stream-function, for normal perturbations:

8

z sin0

Functions and satisfy the boundary conditions. Let us substitute these functions into system (*):

z sin0

00

4224

0

22 2 ikGrkkkr

0

22

00

1 kPr

ikAr

or

000

22222 ikGrkkr

01

0

22

0

k

PrikA

r

We obtained the homogeneous system of linear equations,

0

0

022021

012011

aa

aa

This system has a non-trivial (non-zero) solution if

or 00

0

2221

1211

aa

aa

02221

1211

aa

aadet

(**)

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The last condition written for equations (**) is

01 22

22222

kPr

ikA

ikGrkk

r

r

det

or

01 2222222

AGrkk

Prkk

rr

or

011

122

2

222222

AGr

kk

kPr

kPr rr

This quadratic equation can be written in the following form:

AGrk

kk

Prc

kPr

bcbrr

22

2

222

222

1

01

10

,

10

A basic state is unstable if λr>0. For this, we need

0440 222 ccbbcbbr

This equation will have two solutions defined by the formula:

242 cbb

r

01

22

2

222

AGrk

kk

Pr

or

Finally,

3

2

322

,gL

PrGrRak

kARa

For the layer heated from above, A=1. The instability may occur if

2

322

kk

Ra

But Ra>0, this condition is never satisfied. Hence, the layer heated from above is hydrodynamically stable. Fluid will remain at the quiescent state.

-- Rayleigh number

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For the layer heated from below, A=-1. The instability may occur if

2

322

kk

Ra

Ra

unstable

stable

neutral curve, λr=0(stability curve)

kkc

Rac

Let us determine Rac. Condition of minimum:

2222

0 .d

d c

kk

Ra

6584

27 4

cc

kRaRa

Quiescence becomes unstable for the layer heated from below if the temperature difference between the plates is high enough for Ra>Rac.

Convective rolls with dimensions of will be observed. c

kh

2

ck2

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For the case of rigid-rigid boundaries, the stability diagram is very similar but

For the free-rigid boundaries,

1708123 cc

Rak ,.

1101682 cc

Rak ,.

Cloud streets

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Horizontal convective rolls producing cloud streets (lower left portion of the image) over the Bering Sea

Simple schematic of the production of cloud streets by horizontal convective rolls

Good pictures:http://www.meteorologynews.com/2009/10/29/cloud-streets-photographed-over-gulf-of-mexico/

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Remarks:

(i) The method used to define the instability threshold is universal. This method can be used for finding the thresholds of instability of any solution for any partial differential equations.

(ii) Next, we can take the convective rolls as a basic state; represent all physical quantities as sums of a basic state with small disturbances; linearise the equations; and determine the conditions when the found rolls become unstable.

(iii) At the first instability threshold, the state of quiescence is replaced by convective rolls with a typical horizontal size kc. Passing the next instability threshold, the convective motion will represent the combination of the rolls of two different sizes. And so on. At large Gr~105, convective motion becomes turbulent: superposition of the rolls with the sizes determined by continuous spectrum of k.

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John William Strutt, 3rd Baron Rayleigh (12 November 1842 – 30 June 1919) was an English physicist who, with William Ramsay, discovered the element argon, an achievement for which he earned the Nobel Prize in Physics in 1904. He also discovered the phenomenon now called Rayleigh scattering, explaining why the sky is blue, and others.