SESM3004 Fluid Mechanics
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Transcript SESM3004 Fluid Mechanics
Lecture 19-20:
Natural convection in a plane layer.
Principles of linear theory of
hydrodynamic stability
z
Governing equations:
T=A
h=1
T=0
x
We will distinguish two cases:
v
v v p v GrT k
t
T
1
v T
T
t
Pr
div v 0
A =-1 – layer heated from above
A =1 – layer heated from below
1
Quiescent state
1. Quiescent state:
p GrT k 0
0
0
t
T k 0 T k 0
v 0,
0
0
0
T k T z
0
0
T 0 T c z c Az
0
0
1
2
2
Linear stability of a quiescent
state
Let us analyse the time evolution of a small perturbation of the quiescence
v
T T
v0 =0, p0 and T0 is the basic state;
v, θ and p’ is a small perturbation
p p p
0
0
The linearised equations for a perturbation read
v
p v Grk
t
1
v T0
t
Pr
div v 0
3
Taking x- and z-projections of the Navier-Stokes equation gives
v
p
v
t
x
v
p
v Gr
t
z
1
Av
t
Pr
x
x
z
z
z
v v
0
x z
x
z
For 2D flow, we may introduce the stream-function defined as
v
x
, v
z
x
z
The continuity equation is satisfied
automatically
v x v z 2 2
0
x
z xz xz
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The Navier-Stokes and heat transfer equation can be re-written as
p
t z
x
z
p
Gr
t x
z
x
1
A
t
x Pr
z-derivative of the first equation minus x-derivative of the second
equations gives
Gr
t
x
1
A
t
x Pr
For temperature, at the
Boundary
upper and lower plate:
conditions:
2
0
v 0, v 0
For velocity, at the upper
and lower plate
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or, in terms of
0,
0
(rigid walls):
stream-function
z
x
x
z
We will analyse stability of the quiescent state only in respect to
normal modes:
x , z , t z expt ikx, x , z , t z expt ikx
In general, i
r
i
Time dependence of a perturbation: expt exp t expi t
r
i
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.
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Substitution of the normal modes gives
k 2k k ikGr
2
(*)
Boundary
conditions:
ikA
iv
2
4
1
k
Pr
0, 0, 0
2
For this problem it can be shown that perturbations develop
monotonically, i.e. λi = 0.
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: v 0, v v 0, 0
z x
In terms of the stream-function,
0, 0, 0
for normal perturbations:
x
z
z
xz
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Functions sin z and sin z satisfy the boundary
conditions.
Let us substitute these functions into system (*):
0
0
k 2k k ikGr
2
2
4
r
ikA
r
2
2
4
0
0
0
0
1
k
Pr
2
0
2
0
or
(**)
2
r
k k
2
2
2
ikGr 0
2
0
0
1
ikA k 0
Pr
2
2
r
0
0
We obtained the homogeneous system of linear equations,
a a 0
11
0
12
0
a a 0
21
0
22
0
or
a
a
11
21
a
a
12
22
0
0
0
This system has a non-trivial (non-zero) solution if
a a
det
0
a
a
11
12
21
22
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The last condition written for equations (**) is
k k
det
ikA
2
2
2
r
or
k
2
2
ikGr
1
k
Pr
2
2
2
r
k 1 k k AGr 0
2
2
2
r
or
r
2
r
2
2
Pr
1
1
1 k k
Pr
Pr
2
2
0
2
2
2
r
2
k
AGr 0
k
2
2
2
This quadratic equation can be written in the following form:
b c 0, b 1
2
r
r
1
k
Pr
2
1
k
c
Pr
2
2
2
0
k
AGr
k
2
2
2
2
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This equation will have two solutions defined by the formula:
b b 4c
2
2
r
A basic state is unstable if λr>0. For this, we need
0
r
b b 4c
b b 4c c 0
2
2
2
or
1
k
Pr
2
Finally,
2
2
k
AGr 0
k
2
2
ARa
2
2
k
k2
,
2 3
gL3 -- Rayleigh
Ra PrGr
number
For the layer heated from above, A=1. The instability may occur if
k
Ra
2
2
k
3
2
But Ra>0, this condition is never satisfied. Hence, the layer heated from
above is hydrodynamically stable. Fluid will remain at the quiescent 10
state.
For the layer heated from below, A=-1. The instability may occur if
Ra
2
k
k2
2 3
Let us determine Rac.
Condition of minimum:
unstable
Ra
d Ra
0k
2.22
dk
2
neutral curve,
λr=0
(stability curve)
c
27
Ra Ra k
658
4
Rac
4
c
c
stable
kc
k
Quiescence becomes unstable for the layer heated
from below if the temperature difference between the
plates is high enough for Ra>Rac.
2
Convective rolls with dimensions of h
will be
k
observed.
c
2 k
c
11
For the case of rigid-rigid boundaries, the stability diagram is very
similar but k 3.12, Ra 1708
c
c
For the free-rigid boundaries, k 2.68, Ra 1101
c
c
12
Good pictures:
http://www.meteorologyn
ews.com/2009/10/29/clo
ud-streetsphotographed-over-gulfof-mexico/
Cloud streets
Simple schematic of the production of
cloud streets by horizontal convective
rolls
Horizontal convective rolls producing
cloud streets (lower left portion of the
image) over the Bering Sea
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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.
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