Elementary Fluid Dynamics by Acheson
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Elementary Fluid Dynamics D.J.Acheson, Clarendon Press (90) Notes by CKWong
1 Introduction Sources: D.J.Acheson, “Elementary Fluid Dynamics”, Clarendon Press (90) R.E.Meyer, ”Introduction to Mathematical Fluid Dynamics”, Wiley (71). Experiment: Aerofoil & Starting Vortex Applications: Waves (§3.1) Flow Instability (§9.1) Hydraulic Jumps (§3.10) Smoke Ring Interactions (§5.4) Atmospheric Jet Stream (§9.8) Quantum Vortices (§5.8) Sperm Movement (§7.5) Spindown of Stirred Tea (§8.5)
1.1
Preliminary
Flow velocity:
u u x, t u, v , w
(1.1)
Steady Flow:
u 0 t
(1.2)
2-D Flow:
u u x, y , t , v x, y , t ,0
(1.3)
2-D Steady Flow: u u x, y , v x, y ,0
(1.4)
Streamline: Curve with tangent equal to u everywhere. Let curve be: x x ( s ) , y y ( s ) , z z ( s ) Then
dx / ds dy / ds dz / ds u v w
(1.5)
Streak photographs: illuminated polystyrene beads of neutral buoyancy.
Material derivative, or, rate of change “following the fluid”: Df d rc f [ x ( t ), y ( t ), z ( t ), t ] Dt dt fvv
f f dx f dy f dz t x dt y dt z dt
f f f f u v w t x y z
f ( u ) f t
(1.6)
Acceleration: Du u ( u )u Dt t Uniform rotation: u ( y , x,0)
(1.7)
u 0 t
Steady flow: So that
Du u u Dt
y x y , x,0 x y
2 x, y ,0
For steady flow: Df f u f u s f u Dt s where s ss is the tangent vector to a streamline.
u f
0 means f is constant along a streamline but it can vary between different
streamlines. Df 0 means f is constant for a fluid particle but it can differ between different Dt
particles. A material (dyed) volume is a collection of designated fluid particles.
It may deform during the course of motion.
1.2 Ideal Fluids An ideal fluid is Incompressible:
ρ = constant, or
D 0, Dt
which implies u 0 .
Inviscid (no viscosity): pn . Mass of a volume V fixed in space & bounded by surface S is: m dV V
Rate of change is dm d dV dV V t dt dt V
Net rate of mass leaving the volume is
S
u ndS u dV V
Conservation of mass means dV u dV V t V
(A)
Actually, one can show that eq.(A) is true even if V denotes a volume moving along with the fluid. [ see R.E.Meyer, Chap 1 ]. More generally, the convection theorem ( Meyer, p.10) states that D Df fdV f u dV V V Dt Dt
Going back to eq.(A), since it is true for all V, we have u 0 t
which is called the equation of continuity. Using
u u u and
D u Dt t
We can writing the equation of continuity as D u 0 Dt
For an incompressible fluid, ρ = constant, or u 0
D 0 , so that Dt
(incompressible condition)
Force on dyed blob in inviscid fluid: pndS pdV S
V
which means the force density is p .
1.2.1 Euler’s Equation Newton’s law applied to a fluid volume: D udV p g dV V Dt V
Using the convection theorem, the left side is
D u D Du u u u u u dV dV V V Dt Dt Dt V
Du dV Dt
where the last equality was obtained with the help of the eq. of continuity. We thus obtain Euler’s eqs.:
Du p g Dt
(1.12)
where g is the gravitational acceleration. Writing g we have
Du u u p Dt t
(1.13)
or p u u t Using
u u u u
1 u 2 2
we have p 1 u u u u 2 t 2
(1.14)
1.2.2 Bernoulli Streamline Theorem For steady flow, (1.14) becomes
u u H where p
1 u2 2
H
(1.15)
Using u u u 0
we have
u H 0
(1.16)
ie, H = const along streamline of an ideal fluid in steady flow. Irrotational flow means u 0
(1.17)
Bernoulli Streamline Theorem for irrotational flow: H = const everywhere in a steady irrotational flow of an ideal fluid.
1.3
Irrotational Flow
Vorticity: 2-D flow:
ω u
(1.18)
u [u ( x, y , t ), v ( x, y , t ),0]
ω
i
j
k
x u
y v
v u 0,0, 0,0, z x y 0
v u x y
(1.19)
Consider 2 mutually perpendicular line segments AB & AC (fig.1.3): v B v A v ( x x, y , t ) v ( x, y , t )
v x = CCW angular velocity of AB x
u C u A u ( x, y y , t ) u ( x, y , t )
u y = CW angular velocity AC y
Hence 1 1 v u = average angular velocity of AB & AC. 2 2 x y Note: ω donotes local ‘spin’ & has nothing to do with global rotations. (Fig.1.6) Eg. Shear flow (fig.1.4)
u y ,0,0
(1.20)
has no global rotation but v u 0 x y Eg. Line vortex flow θ kˆ u r
(1.21)
has
ω u
r
θ r
z
r
θ r
z
1 r r ur
ru
1 z r r uz 0
k
0 z 0
but is obviously rotating as a whole. Eg. Flow
θ u r ˆ spin has same rate as global rotation. Eg. Rankine Vortex (fig.1.7) r u a 2 r ur u z 0
1.4
for
ra ra
(1.23)
Vorticity Equation
Euler’s eq. (1.14): u u u H t
can be written as u ω u H t both sides gives
ω ω u 0 t
(1.24)
Using
a b b a a b a b b a we have
ω u u ω ω u ω u u ω u
ω
ω u
where u 0
for incompressible fluid
ω u 0
for any vector u.
Thus (1.24) becomes ω ω u ω u 0 t
or Dω ω u Dt
(1.25)
which is called the vorticity equation. For 2-D flow u u x, y , t , v x, y , t ,0
ω 0,0,
we have
ω u
u0 z
so that Dω 0 Dt
(1.27)
Thus, for a 2-D flow of ideal fluid under conservative forces, ω is conserved for each fluid element. For steady flow, we have ω u 0 so that ω is conserved along each streamline. Thus, steady 2-D flow pass an aerofoil is irrotational since 0 for x .
1.5 Steady Flow Past Fixed Wing Lift of wing in steady flow is due to (fig.1.9; also demonstrated by Wright program) pupper p
plower p
Flow is irrotational. Bernoulli theorem means p Thus lift means uupper ulower .
1 2 u is constant everywhere. 2
1.5.1 Circulation Circulation Г around closed curve C is u dx
(1.31)
C
Stokes theorem:
C
u dx u ndS
(1.32)
S
ω ndS S
where C is the boundary of S. For the irrotational flow past an aerofoil, 0 for any C not enclosing the aerofoil. 0 for any C that encloses the aerofoil
(surface of foil is another boundary of S). Furthermore, for fig.1.8, lift means 0 .
1.5.2 Kutta-Joukowski Hypothesis For wing with sharp trailing edge, u can be finite everywhere only for one value of circulation K . Otherwise, u at the sharp edge. That K describes the actual flow is the Kutta-Joukowski hypothesis. For thin, symmetric wing with small α, (see Chap 4) K UL sin (1.34) where U u
L = length of wing
1.5.3 Lift For ideal flow, Drag = 0 Lift L U For K , we have
(1.35)
L U 2 L sin
(1.36)
For large angles, 1.36 overestimates badly (fig.1.11).
1.6 Conclusion Near wing tip, flow is not 2-D & can’t be irrotational. This leads to trailing vortices & hence drag. (fig.1.12) Generation of Γ is a response to starting vortices (Chap 5), which in turn requires viscosity.
2 Viscous Flow Source: D.J.Acheson, “Elementary Fluid Dynamics”, Chapter 2, Clarendon Press (90)
2.1 Introduction Theoretically, inviscid, and in particular, irrotational (potential) flow is easiest to deal with. Naively, we may expect it to describe flows with low viscosity. In practice, we do find the theory useful in describing phenomena such as water waves, sound waves, and slow flow past streamlined objects. For flow past a general object, the down-stream side becomes turbulent. The root cause of this may be traced to the viscosity related no-slip boundary condition which a potential flow usually cannot satisfy. Thus, flows from the inviscid theory are valid only if they can be made to reconcile with the no-slip condition within a sufficiently thin transitional unseparated boundary layer. If the inviscid theory predicts an increase of pressure along the flow direction on the boundary, separation of boundary layer is expected and turbulence results. (see chapters 4 & 8). On the other hand, theory of very viscous fluid is very successful (see chapter 7). Newtonian fluid: du dy
(2.1)
where τ is the shear stress, and μ is the coefficient of viscosity. Navier-Stokes Eqs (for incompressible Newtonian fluids): u 1 (2.3) u u p 2u g t u 0
See Landau for a derivation. The kinematic viscosity is defined as
(2.2)
No-slip condition: at stationary boundary. u0 Reynolds Number: R
UL
(2.4)
where U and L are the characteristic flow velocity and dimension of the system, respectively. And ν is the kinematic viscosity. Using estimates
L
u O U
u O U
on the Navier-Stokes eqs. (2.3), we have
L u O U L u u
Inertia term:
2 O U
2
Viscous term:
2
(2.5)
so that U2 inertia term L O R O U viscous term L2
(2.6)
High R flows: Corresponds to flows with small viscosity. Under favorable circumstances, eg. flow past streamlined bodies, it corresponds to the flow of ideal fluid except for a thin boundary layer (required for the no-slip condition) and a narrow trailing wake. Typical thickness of such a boundary layer is
L
O R
1
2
(2.7)
In general, high R flows are unstable & easily become turbulent. Low R flows: R 0.01 . Flow is well-ordered & reversible.
2.2 Navier-Stokes Eqs. Consider the Euler eq D u Du u u p Dt Dt Using D u u u u u u u u Dt t
u uu t
we can re-write it as u uu p 0 t or u 0 (A) t where the momentum flux density tensor П is ij ui u j p ij
Eq(A) is just the “eq of continuity for momentum” which expresses the conservation of momentum. Since we expect the principle of the conservation of momentum to be valid even in viscous fluids, eq(A) should apply there with a suitable modification of П. We thus set ij ui u j ij
where the stress tensor ij is related to the viscous stress tensor 'ij by
ij p ij 'ij [Note: Acheson used Tij instead of ij (see chap 6) ] What this means is that the generalization of the Euler’s eq to viscous fluids can be done by replacing p with p ' so that, for example,
Du p Dt
becomes
Du p ' Dt
or u u u p ' t
(B)
Next, we need to determine the form of ' . Now, viscous effects are due to “friction” between adjacent fluid elements moving ui with different velocities. Hence ' should depend on but not on u itself. x j Furthermore, there’s no friction if the fluid rotates as a whole with uniform angular Ω r velocity Ω so that u or ui ijk j xk . Now: ui u ijk j km ijm j mji j m xm xi
and ui iji j 0 xi
Hence, the choice ui u j u b k ij ' ji x j xi xk
'ij a
where a, b are constants, will satisfy the above requirements. A more common form is ui u j 2 uk u k ij x j xi 3 xk ij xk
'ij
(C)
where the positive constants
a
2 3
2 3
b ab
are called coefficients of viscosity. (Landau used η instead of μ) This also defines the Newtonian fluid. [cf eq(2.1)]. Using (C), (B) becomes
ui u p ' ji uj i t x j xi x j
p xi x j
u u j 2 u u k ij k ij i xk x j xi 3 xk
2 p 3
uk x j xk
ui u j x j xi
xi
ui 1 u p k xi x j x j 3 xk
or in vector form: u u u p ' t 1 p 3
1 p 3
2 u u
2 u u
(D)
For incompressible fluids, we have the Navier-Stokes eqs.: u u u p 2u t u 0
(2.3)
Note also that for incompressible fluids, ui u j x j xi
'ij [cf (2.1)]
2.3 Simple Flows 2.3.1 Plane Parallel Shear Flow Plane parallel shear flow is defined as u u y , t ,0,0
(2.8)
Thus u
u 0 x
(incompressible)
Navier-Stokes eqs becomes
u 1 p 2u 2 t x y p p 0 y z
(2.9)
The last eq means p and hence
p dp is a function of x dx
x, t
only.
Hence, the right and left hand side of
1 p u 2u 2 x t y is a function of
x, t
and
y, t , respectively. Thus, either can only be function of t
only.
2.3.2 Impulsively Moved Boundary Let fluid lies at rest in region 0 y at t 0 . At t 0 , boundary at y 0 is set in motion in x-direction with constant speed U. No-slip condition means u U ,0,0 at boundary for t 0 . In general, we expect u u y , t ,0,0 so that it is a plane parallel shear flow. According to the last section,
p f t so that p f t x C . x
Let there be no externally applied pressure gradient, ie., p at x are equal. This means we must have f t 0 and p C .
Eq(2.9) thus becomes
u 2u 2 t y which is simply the 1-D diffusion eq.
(2.12)
This must be solved with the initial condition
u y ,0 0
for
y0
and boundary conditions
u 0, t U
for t 0
u , t 0
for t 0
Now, (2.12) is invariant under the scale transformation t 2t y y where α is a constant. This means we can attempt a similarity solution:
u f
where
y
t
so that y 1 t y t 2 t3 u y f' f ' f ' t t 2t 2 t3 u 1 f' f' y y t
2u 1 1 f '' f '' 2 y y t t df where f ' . Hence (2.12) becomes d f '
2t
f ''
1 t
or f '' 1 f' 2 so that 1 ln f ' 2 const 4
or f ' Be
2 4
(2.13)
whence
f A B e
s2
4
0
ds
where A, B are constants. Using
0
for t 0 or y for y 0
the boundary & initial conditions become
f 0
f 0 U
and
Hence
0 A B e
s2
4
0
1
ds A B
UA
whence 1 f U 1
e
0
s2
4
ds
(2.14)
or 1 u y, t U 1 [see fig 2.8]
y
0
t
e
s2
4
ds
The vorticity is
u U e y
2
4
y U e y t
2
4 t
(2.15)
which indicates a diffusion with standard deviation
2 viscous diffusion time O L
(2.16)
where L is a distance.
2.3.3 Flow Down Inclined Plane With reference to fig. 2.9, the Navier-Stokes eqs u 1 u u p 2u g t becomes
2 t . In other words,
2 u 1 p 2 u v u 2 2 u g sin t x y x y x 2 v 1 p 2 u v v 2 2 v g cos t x y y y x where u u, v,0 . The no-slip condition means u 0 for y 0 . Hence u must vary with y if it’s not identically 0. The simplest solution possible is therefore a steady flow of the form u u y , v y ,0
The incompressibility condition u v 0 x y becomes v 0 y so that v const . However, v 0 0 so that v 0 for all y.
The steady state Navier-Stokes eqs. thus become
0
1 p d 2u 2 g sin x dy
0
1 p g cos y
(2.17)
The 2nd eq gives
p gy cos f x
(A)
Since all streamlines are in the x direction, the free surface must be y h where h is a constant. For a flat interface, p must be equal on both sides and the tangential stress must vanish. Let the atmospheric pressure be p0 , we have
p p0
and
du 0 dy
at
yh
Putting these into (A) gives
p0 gh cos f x so that (A) becomes p gy cos p0 gh cos or
p p0 g h y cos whence p 0 x
and the remaining Navier-Stokes eq reduces to
0
d 2u g sin dy 2
which gives du g sin yA dy u
g sin 2 y Ay B 2
Putting in the boundary conditions at u0 y0 and
du 0 dy
at
yh
we have 0B 0
g sin
h A
so that
u
g sin y 2 hy 2
(2.19)
The volume flux per unit length in the z-direction is
(2.18)
h
g sin
0
Q udy
y2 g sin h 3 h 2 g sin 3 hy dy h h 0 2 6 2 3 h
2.3.4 Another Example Consider the impulsively moved plane boundary problem with an added stationary plane boundary a distance h above it. (see fig 2.10) The problem is to solve u u y , t ,0,0 from
u 2u 2 t y
(2.20)
with the initial condition
u y ,0 0
for h y 0
and boundary conditions
u 0, t U
and
u h, t 0
for t 0
The form of (2.20) clearly suggests the use of the method of separation of variables. Let
u Y y T t (2.20) becomes
Y
dT d 2Y T 2 dt dy
or
dT d 2Y Tdt Ydy 2
const
Thus, for 0 , T Ae t Y Be y Ce y where A, B, C are constants and
For 0 , we have TA
Y B' y C'
where B ', C ' are constants. Hence, u y , t a0 y b0 e t a e y b e y
(A)
0
is a general solution of (2.20). The next task is to adjust the constants to satisfy the initial & boundary conditions: u y ,0 a0 y b0 a e y b e y 0
for h y 0
u 0, t b0 e t a b U
for t 0
u h, t a0h b0 e t a e h b e h 0
for t 0
0
0
0
The last 2 eqs can be satisfied only if the time dependent sums vanish identically. Hence, we have b0 U a0h b0 0
so that
a0
U h
and e a b 0 t
0
e a e t
0
h
b e h 0
for t 0
Since the e t terms are independent, each term in these sums must vanish individually so that a b 0 a e h b e h 0
which gives b a e h e h 2sinh h 0 so that
h n i and n 2 2 2 0 h Eq(A) becomes 2
U u y, t y U e h n 1
where An 2ia u y ,0
0
n 2 2 t h2
n An sin y h
are to be determined by the initial condition, namely,
U n y U An sin y 0 h h n 1
for h y 0
Using
sin nx sin mxdx 2
nm
0
we have
h
h
0
U n dy y U sin y An 0 2 h h
or h 2 y n An U dy 1 sin y h 0 h h n 2 z U dz 1 sin z 0 n n
2U n
where
z
n y h
n
1 cos z n z cos z sin z 0
2U n
so that, finally n y 2 2 t 1 n u y , t U 1 e h sin y h n h n 1 2 2
(2.21)
2.4 Flow with Circular Streamlines In cylindrical coordinates
r, , z , (see Appendix A.5)
u ur , u , uz e r ur e u e z uz e r e1 cos e 2 sin e r e
e e r
e e1 sin e 2 cos
e z e3
(all other partials vanishes)
(2.23)
f e r e ez f r z r u f ur u uz f r z r V
V
V V rVr z rr r z
er
re
ez
1 r r Vr
rV
z Vz
2 2 f r 2 2 2 f z rr r r 2
To write the Navier-Stokes eqs u 1 u u p 2u t in cylindrical coordinates, we 1st consider the term
u u :
u e rur ur u e r e r u ur ur ur u u z e r e r u ur r z r ur u
e r e r u ur r
ur u e e r u u r r
u e u u u e e u u u ur u uz e e u u r z r u u
e e u u r
u 2 e r e u u r
u e zuz uz u e z e z u uz
e z u uz so that ur u u 2 e e r u u r e r e u u e z u uz u u r r
u 2 uu e r u ur e u u r e z u uz r r Next, we turn to 2 u :
2 2 2 e r ur r e r ur 2 2 z 2 rr r r 2 2 er r 2 ur 2 2 e r ur r rr r z Now
2 e r ur eu u e r r 2 r r 2e r e r ur 2 ur ur 2 er 2 2 ur 2 ur e r u r 2 e er 2 so that
2 1 2 e r ur e r r 2 ur 2 r rr r z
ur 2 ur e u 2 e e r r r 2
ur e u 2 e r r u u e r 2ur 2r 2e 2 r r r
e r 2 ur
1 r2
Similarly
2 e u eu u e 2
2 e e u 2u u 2 e 2 2
e u 2e r
u 2u e 2
so that
2 2 e u e r 2 u 2 2 e u r rr r z 2
2 1 u 2u e r u e u 2 e e r 2 r2 2 rr r z e 2u
1 u e u 2e r 2 r
Finally, with
2 e z uz e z 2u z we have u u 1 u 2u e r 2ur 2r 2e 2 r e 2u 2 e u 2e r e z 2uz r r r u 2 u u 1 e r 2ur 2r 2 e 2u 2 2 r 2 u e z 2uz r r r r
Hence
ur u2 1 p u 2 u u ur 2ur 2r 2 t r r r r u uu 1 p u 1 u u r 2u 2 2 r 2 u t r r r r uz 1 p u uz 2uz t z where u f ur u uz f r z r
2 2 f r 2 2 2 f z rr r r 2
The incompressibility condition is simply u
u u rur z 0 rr r z
2.4.1 Differential Eq. Consider a 2-D circular flow
(2.22)
u u r, t e
(2.27)
The incompressibility condition u
u u rur z 0 rr r z
is automatically satisfied. The Navier-Stokes eqs (2.22) becomes ur : u : uz :
u 2 1 p r r
u 1 p u r t r rr r 1 p 0 z
Now, u is a function of
r, t
1 2 u r
only.
The u eq can therefore be written as p f r, t
Integrating, we have, since p is independent of z:
p r, , t f r, t C r, t where C is another arbitrary function of
r, t .
Now, p must be single valued. This means
p r, , t p r, 2n , t This is possible only if f 0 , ie,
p is a function of
r, t
The Navier-Stokes eqs can be further simplify to
u u r, t e
p p r, t
u 2 1 p r r
ur :
u :
u u r t rr r
1 2 u r
(2.30)
only.
2.4.2 Steady Flow Between Cylinders For steady flow, (2.30) becomes d du r u 0 dr dr Setting y ln r , it becomes r
d du dy dy
u 0
so that u Ae y Be y Ar
B r
(2.31)
If the fluid occupies region r1 r r2 between 2 cylinders rotating with angular velocities 1 and 2 , respectively, the no-slip condition gives r11 Ar1
B r1
r2 2 Ar2
B r2
Solving for A and B gives A B
r121 r2 2 2 r12 r2 2 1 2 r 2r 2 21 2 2 1 2 1 1 2 r2 r1 2 r1 r2
(2.32)
This formula will be of use in the study of Taylor vortices. (see Fig 9.8)
2.4.3 Spin-Down Let fluid occupies the interior of an infinitely long cylinder of radius a. Initially, the whole system is rotating with uniform angular velocity Ω so that u r for r a t0 The cylinder is suddenly stopped at t 0 . We thus need to solve u u 1 r 2 u t rr r r with the above initial condition as well as the (no-slip) boundary condition u 0 at r a for all t 0 . Using the separation of variables method:
u r, t R r T t we have dT d dR 1 r R Tdt R rdr r r 2 so that for 0 ,
T t e t R r J1 r
where J n is the Bessel function defined by d d 2 n 2 r k 2 J n kr 0 r rdr dr
[see M.Abramowitz, I.A.Stegun, “Handbook of Mathematical Functions”] Hence u r, t c e t J 1 r
To satisfy the no-slip condition at r a , we need
a n
where n are roots of J 1 , ie., J 1 n 0 .
Therefore n a
2
and n t a 2
u r, t cn e n 1
r J 1 n a
The initial condition u r for r a
t0
means r r cn J 1 n a n 1 Using the orthogonality of the Bessel functions:
a
2 a2 r r J J rdr J 1 n nm 0 a n a m 2
where n is the nth root of J , we have a
2 a2 r J 1 n r 2dr cn J 2 n 2 a 0
Using 1
J ar r
1
dr
0
1 J 1 a a
( see I.S.Gradshteyn, I.M.Ryzhik, “Table of Integrals, Series, and Products”, formula 6.561.5) we have a 3
1
n
2 a2 J 2 n 2
J 2 n cn
so that
cn
2 a n J 2 n
whereupon
2
n t 2 a r u r, t e a J 1 n a n 1 n J 2 n
Finally, using J 1 x J 1 x
2 J x x
we have J 0 n J 2 n
2
n
J 1 n 0
( n are roots of J 1 )
so that
2
n t 2 a r u r, t e a J 1 n a n 1 n J 0 n
(2.33)
This equation overestimate the spin down time of a cup of tea by an order of magnitude since there’s no account for the effect of the bottom (see chapter 5).
2.4.4 Viscous Decay of Line Vortex A line vortex u
0 e 2 r
(2.34)
The vorticity singularity at r 0 is expected to diffuse in a viscous fluid. Consider the circulation of a circular path C of radius r, 2
r, t u dr u r, t rd 2 ru r, t C
0
(2.35)
Eq(2.30) u u 1 r 2 u t rr r r can be rewritten in terms of as follows: u
1 2 r
u 1 t 2 r t
u 1 r 2
1 1 2 r r r
2u 1 2 2 1 2 2 r 2 2 r 3 r r r r 2
u r rr r
2 u 1 u 2 r r r
1 2
1 2
2 2 1 2 1 1 1 2 3 2 2 r r r r 2 r r r r r 1 1 1 2 r 3 r 2 r r r 2
so that 1 1 2 r t 2
or
1 1 1 2 1 3 2 2 3 r r r r 2 r r
1 2 2 t r r r
(2.36)
For the line vortex decay, the initial condition is
r,0 0 The requirement of u be finite at r 0 afterwards means
0, t 0
for t 0
Eq(2.36) is invariant under the same scale transformation as (2.12), namely, r r t 2t where α is a constant. We therefore seek a similarity solution: r r, t f where t As in the case of (2.13), we have r 1 t r t 2 t3 r f' f ' f ' t t 2t 2 t3 1 f' f' r r t
2 1 1 f '' f '' 2 r r t t so that (2.36) becomes f '
1 1 1 f ' f '' 2t t t r
or f '
2
1
f ' f ''
df ' 1 d f ' 2 ln f ' ln or
f'
Ce
2 4
2 4
C
f C e
2 4
d ae
2 4
b
Hence ae
r2
4 t
b
Putting in the initial & boundary conditions, we have 0 b 0ab
so that
a b 0
Hence r 0 e
2
4 t
1
(2.37)
The circulation thus diffuses with a standard deviation of 2 t . (see fig.2.12) For very small r, ie., r 2 t , we have 0
r2 4 t
so that u 0
r 8 t
(2.38)
which corresponds to uniform rotation with angular velocity
0
8 t
.
2.5 Convection & Diffusion of Vorticity Applying the same technique that begets vorticity eq (1.25) to the Navier-Stokes eq u u u p 2u t we obtain the viscous vorticity eq.
Dω ω u 2 ω Dt
For 2-D flow u u x, y , t , v x, y , t ,0
ω 0,0, we have (cf 1.27),
(2.39)
Dω 2 ω Dt
or
2 2 u 2 2 t y x Obviously, the term
u
ω
(2.40)
describes the vorticity convection while 2ω its
diffusion.
2.5.1 2-D Flow Near Stagnation Point Consider the irrotational flow pattern near a stagnation point as shown in fig.2.13 with u x (2.41) v y where 0 is a constant. Obviously, the no-slip condition is not satisfied at the boundary y 0 . However, the mainstream flow speed | u | | x | increases in the flow direction along the boundary. According to the Bernoulli’s theorem, the mainstream pressure decreases in the flow direction along the boundary. Therefore, we expect a thin, unseparated boundary layer with thickness Using R
UL
L2 , we have O
L
. O 1 R
.
The no-slip condition creates a vortex sheet at the boundary which diffuses into the fluid. The existence of an unseparated boundary layer implies the convective effects are sufficient to balance this diffusion and confine it within the layer. (Do Ex.2.14)
2.5.2 High R Flow Past Flat Plate
3. Waves Ref: D.J.Acheson, “Elementary Fluid Dynamics”, Clarendon Press (90)
3.1 Introduction Consider a simple harmonic surface wave:
A cos kx t
(3.1)
Wave speed is dispersive:
c
k
g k
(3.2)
so that waves with longer wavelengths 2
k
travel faster.
The group velocity is defined as cg
d dk
(3.3)
For a dispersion given by (3.2), we have
d 1 g 1 gk c (3.5) dk 2 k 2 Hence, individual wave crests travel twice as fast as the group so that they continually appear at the back of the group & disappear at the front. cg
Dispersion is also responsible for wake pattern behind moving boat. Stationary wave patterns of flow past object contains both up & down stream disturbances because of surface tension effects. Waves at interface between 2 fluids caused by buoyancy effects have speed
c
g 1 2 k 1 2
(3.6)
where 1 and 2 are the densities of the fluids. A variant of this is internal gravity waves that travel in a fluid with density that varies with height (eg. the atmosphere). Sound waves are due to compressibility and have speed
a0
p0 0
(3.7)
which is non-dispersive.
3.2 Surface Waves on Deep Water 2-D water waves: u [u( x , y , t ), v ( x , y , t ),0] Irrotational motion: i u x u
j y v
k 0 z 0
v u 0 x y Thus, there exists a velocity potential φ : u v x y Incompressibility: u 0
2 2 0 x 2 y 2 Let wave on free surface be described by y ( x , t )
3.2.1 Condition at Free Surface Surface condition: Fluid particles on surface remain there. Let
F x, y , t y x, t the surface condition means that F x, x, t , t 0
ie, F 0 on surface. Thus
DF F u F 0 Dt t
Using F t t
on y x, t
F x x
(3.17)
F 1 y
(3.17) becomes
u v 0 t x
on y x, t
(3.18)
3.2.2 Bernoulli’s Eq. for Unsteady Irrotational Flow Consider the Euler eq. for incompressible, irrotational flow: p 1 u u 2 where t 2 Writing u and integrate, we have p 1 2 u G t t 2
gy
(3.19)
which is called the Bernoulli’s eq. for unsteady irrotational flow. Note that the arbitrary function G is doesn’t affect the value of u. The incompressibility condition in Eq(3.19) can be removed as follows. (see Landau) Consider the specific (per unit mass) enthalpy h, 1 dh Tds vdp Tds dp
we have, for adiabatic processes ( ds 0 ), 1 p h
Therefore, (3.19) becomes 1 h u2 G t t 2
3.2.3 Pressure Condition at Free Surface For inviscid fluid, there is no surface tension. Hence, the condition at a free surface is simply p p0
on
y x, t
where p0 is the (usually constant) atmospheric pressure. The unsteady Bernoulli eq thus becomes, with a proper choice of G, 1 2 p on y x, t (3.20) u v 2 g G t 0 0 t 2 0
3.2.4 Linearization of Surface Conditions To 1st order in u, v, and η, (3.20) becomes g 0 t
on
y x, t
Now x, y , t t
y
x, y , t t
2 x, y , t t 2 y 0
y 0
is already of order u, the true 1st order approximation of (3.20) is t
Since
g 0 t
on
y0
on
y0
on
y0
(3.22)
Similarly, for (3.18):
v 0 t
Using v
, we have y
0 t y
(3.21)
3.2.5 Dispersion Relation Eliminating η from (3.21-22) gives
1 2 0 g t 2 y which is clearly separable. Setting
x, y , t f y x, t we have
on
y0
y 2 t 2
f ' 0 x, t y 0
f 0 y 0
2 t 2
so that
1 2 f 0 2 f ' 0 0 g t or 2 2 0 t 2
where
g
f' f
y 0
On the other hand, incompressibility means satisfies the Laplace eq. inside the fluid for all t:
2 2 0 x 2 y 2 ie 2 f f '' 0 x 2 Since f
is a function of y only, we must have
f '' k 2 const 2 x f 2
which, combining with the t eq., gives 1 2 1 2 2 t 2 k 2 x 2 Thus, both the time and spatial components of Φ are harmonic oscillators while together, they obey the 1-D wave eq. with propagation speed c
k
.
A general solution is of the form
x, t C sin kx t C sin kx t where C, δ are arbitrary constants. If we consider a simple wave propagating in the x direction, we have
x, t C sin kx t
Now, f itself satisfies f '' k 2 f 0 so that
f y B exp ky D exp ky
(3.A)
Since the wave should die out towards the bottom, we must have
f y 0
(3.B)
so that D 0 and
f y B exp ky whereupon
g
f' f
gk
(3.26)
y 0
giving a wave speed
c
k
g k
The potential itself is
x, y , t C exp ky sin kx t
(3.25)
From (3.22), we have 1 x, t C cos kx t A cos kx t g t y 0 g
(3.23)
where A
g
C
k
C
3.2.6 Meaning of Small Amplitude From (3.23,5), we have u
Ck exp ky cos kx t A exp ky cos kx t x
v
Ck exp ky sin kx t A exp ky sin kx t y
Ck sin kx t Ak sin kx t x g
2 C sin kx t Ck sin kx t A exp ky sin kx t t g Thus, the linearization condition that leads to (3.21) u
v x
means 1 x
ie. A
1 k 2
the wave amplitude of the surface wave is much small than its wavelength λ. For (3.22), we need u 2 v 2 g which means A2 2 gA ie A
g
2
1 k 2
3.2.7 Particle Path In the Eulerian picture, the equations
u A exp ky cos kx t v A exp ky sin kx t give the velocity at each point
x, y
at time t.
In the Lagrangian picture, the corresponding velocity of a fluid particle whose trajectory is x t , y t is described by
dx t A exp ky t cos kx t t dt dy t A exp ky t sin kx t t dt which can be integrated to give the particle path. It is straightforward to prove that the ansatz
x t x A exp ky sin kx t y t y A exp ky cos kx t is a solution to 1st order in A. It describes a circular motion of radius A exp ky about the average position
x, y .
3.3 Dispersion Given a dispersion relation
k
(3.28)
the group velocity is defined as cg
d dk
(3.29)
Properties:
1.
Waves of a single wavelength travels with phase velocity c
2.
Whenever superpositions of nearby wavelengths are substantial, the packet travels with the group velocity cg
3.
k
d . This is true whether the packet is dk
isolated or just a Fourier component of a complicated excitation. Energy is transferred at group velocity. (see Ex.3.12)
3.3.1 Motion of a Wave Packet A general disturbance can be written as
.
x, t a k exp i kx t dk
(3.30)
where a k is the Fourier amplitude and it is understood that only real quantities are physically meaningful. When a k is narrowly centered at k0 , the disturbance is called a wave packet.
Writing
k 0 k ' cg 0
(3.31)
where
0 k0 ,
k k0 k ' ,
cg 0
d dk
(3.32) k k0
we have kx t k0 k ' x 0 k ' cg 0 t k0 x 0t k ' x cg 0t
so that (3.30) becomes
x, t exp i k0 x 0t a k0 k ' exp ik ' x cg 0t dk '
3.3.2 Gaussian Packet Consider a Gaussian wave packet with standard deviation σ:
k k0 2 1 a k exp 2 2 The integral in (3.33) becomes
I
1 2
k '2 exp 2 ik ' x cg 0t dk '
2 2 1 1 dk ' exp x cg 0t exp k ' i x c t g 0 2 2 2
(3.33)
2 exp x cg 0t 2
so that
x, t exp i k0 x 0t
x c t 2 2
g0
which shows a wave packet with an envelope moving with velocity cg 0 .
3.3.3 Large Time Response to Localized Disturbance After a sufficiently long time, different Fourier components of a local disturbance will be greatly dispersed. What’s left is a sinusoidal wavetrain with both k and ω slowly varying functions of
x, t .
g so that waves with longer wavelengths k travel faster, one expects the wavelength λ to increase with x. (see fig 3.8) For surface waves on deep water, c
Let the slowly varying wave train be
x, t A x, t exp i x, t
(3.34)
where x, t is a phase function. The local wavenumber k and frequency ω are defined by k
x
t
(3.35)
Now k 2 2 t tx xt x or k 0 t x
Hence
(3.36)
k d k 0 t dk x
or cg k 0 x t
(3.37)
ie., k is constant for an observer moving with velocity cg . This can also be seen from the fact that k f x cg t
(3.38)
satisfies (3.37) and is obviously constant for x cg t const .
3.3.4 Surface Waves on Deep Water gk
(3.39)
1 g 2 k After a long time, any local disturbance becomes locally sinusoidal with wavenumber k in the neighborhood of distance cg
x cg t
(3.40)
from the initial disturbance region so that g gt k 2 4cg 4x
2
(3.41)
gt 2x
and hence gt 2 gt 2 x 4 x t 2x which admits a solution gt 2 4x where ε is a constant. Eq(3.34) thus becomes gt 2 4x
x, t A x, t exp i
(3.42)
3.4a Laplace’s Formula Source: Landau Consider 1st the interface between 2 inviscid fluids. Let the principal radii of curvature at a point on the interface be R1 and R2 . ( R 0 if it is drawn into fluid 1) The corresponding line elements there would be dl1 R1d1 and dl2 R2d 2 where dθ denotes the angle subtended by the line element. A surface element there is dS dl1dl2 . Let the interface be moved by an infinitesimal amount r . Both R1 and R2 would be changed by an amount r nˆ , where nˆ the interface normal pointing from fluid 1 to fluid 2. Line element dl1 would become
R r n d R r n dlR
1
1
1
1
1
1 dl1 1 r n R1
and similarly for dl2 The surface element dS then becomes
1 1 dl1 1 r n dl2 1 r n R1 R2 The change induced in dS is therefore
1 1 r ndS R R 2 1 This will change the surface energy by
1 1 r ndS R R 2 1
where α is the surface tension coefficient. On the other hand, the work required to be done against the pressure difference is
p S
2
ˆ p1 r ndS
In equilibrium, the total change of energy must be zero so that we have 1
R
1
1 p2 p1 r ndS 0 R2
ie.
1 1 p1 p2 R1 R2
For viscous fluids, this is easily generalized to
1 1 n R1 R2
2 1 n
where the stress tensor is defined by pI '
3.4b
Curves
Source: T.M.Apostol, “Calculus”, vol I. Consider a curve C t with parameter t in an n-dimensional space. If C is the path of a particle, it is usually denoted in vector notation as r t . The tangent (velocity) of C is defined by v
dr vvˆ dt
where v v is the speed (magnitude of v) and vˆ is the unit tangent. Assuming Euclidean metric for all vectors, the path length can be written as
s t v t dt t
0
so that v
ds dt
The principal normal is defined by
n
dvˆ dt
or dvˆ d v 1 dv v dv 1 dv 2 a v nnˆ dt dt v v dt v dt v dt where nˆ is the unit normal and a the acceleration. A more familiar way to write this is n
a
dv d vvˆ dvˆ dv v vˆ dt dt dt dt
Consider the rate of vˆ with respect to s: d v dt d v 1 n n n nˆ ds ds dt v v where n 1 dvˆ v v dt is called the curvature. The radius of curvature is R
1
3.4b.1 Circle Consider curve r t a cos t , a sin t which describes a circular path of radius a in the x-y plane. The tangent (velocity) is v t a sin t , a cos t . The unit tangent is vˆ t sin t ,cos t . The speed is
v a .
The path length is s t adt at . t
0
The normal is The unit normal is
n t
dvˆ cos t , sin t . dt
nˆ t cos t , sin t
d v 1 1 1 n cos t ,sin t nˆ ds v a a 1 The curvature is . a
The radius of curvature is
1
a.
3.4b.2 Graph Consider a graph of f t vs t. The foregoing analysis can be applied here using r t t , f t
Hence The tangent (velocity) is
v t 1, f '
The unit tangent is vˆ t
1
The speed is
where
f '
df . dt
1, f ' .
1 f '2
v 1 f '2 .
The path length is s t 1 f '2 dt . The normal is
n t
dvˆ d 1 1, f ' . dt dt 1 f '2
f ' f '' f '2 f '' f '' , 3 3 2 2 2 2 1 f '2 1 f ' 1 f '
The unit normal is
f ''
1 f '2
nˆ t
3
2
f ',1
1 1 f '2
f ',1
d v 1 f '' f '' n f ',1 nˆ 2 3 2 2 2 ds v 1 f ' 1 f '
The curvature is
1 f '' . 3 2 2 R 1 f ' 1 1 f ' R 2
The radius of curvature is
3
f ''
2
.
3.4 Surface Tension & Capillary Waves For our 2-D wave problem for an inviscid fluid, the boundary condition on the free surface when surface tension effects are included is (see 3.4a and b) 1 2 2 ( neglected) R x x where α is the coefficient of surface tension, p0 the (constant) atmospheric pressure 2
p0 p
and R the radius of curvature of the surface ( R 0 if it is drawn inside the fluid). Eq(3.22) thus generalized to
2 g 0 t x 2
on
y0
(3.44)
while the other eqs remain unchanged, ie. on y 0 0 t y
(3.21)
2 2 0 x 2 y 2 Eliminating η gives
2 3 g 0 t 2 y x 2y
on
y0
Setting
x, y , t f y x, t gives
2 2 f 0 2 f ' 0 g 0 t x 2 while the Laplace eq. still gives
2 f '' k 2 const 2 x f with solution f y B exp ky as before.
(A)
Hence (A) becomes g 2 2 2 2 k g k 0 k k 2 x 2 t 2 t 2 and the wave speed is c k
g k
(3.46)
so that
kc k 3
gk
(3.45)
g d cg dk 2 k 3 gk 3k 2
(3.47)
For large k (short wavelengths), we have c
k
cg
3 k 3 c 2 2
Such waves are called capillary waves or ripples. In contrast with the gravity waves, they travel faster if their wavelengths are shorter. Furthermore, the group velocity is greater than the corresponding phase velocity. Thus, wave crests in a wave packet travel backwards. In general, we have capillary-gravity waves. (see fig 3.11 for dispersion) Salient points are: 4g
1
4
at k
g .
1.
there is a minimum phase velocity cmin
2.
there is both a capillary & a gravity wave for each phase velocity.
3.5 Finite Depth Effects The effect of a finite depth h is to change the boundary condition satisfied by
f y B exp ky D exp ky to
(3.A)
v h
df dy
0
(3.C)
y h
ie
B exp kh D exp kh 0 D exp 2kh B f y B exp ky exp 2kh ky
Be kh exp k y h exp k y h 2 Be kh cosh k y h
so that (3.25) is replaced by
x, y , t C cosh k y h sin kx t
(3.D)
The dispersion relation (3.26) becomes
g
f' f
g y 0
k sinh kh gk tanh kh cosh kh
(3.51)
so that the phase speed is (see fig 3.12)
g tanh kh k k which, for h large, becomes c
g k as it should. For small h, we have c
g kh gh k which is non-dispersive. c
(3.52)
(3.53)
(3.54)
3.6a Thermodynamics of Ideal Gas An ideal gas is defined by 2 eqs. Pv RT N Ak BT
u cv 0T where v is the molar volume, u the molar energy. The molar heat capacity at constant volume cv 0 is a function of T only. (Note: specific heat is heat capacity per unit mass) R 8.3144
Joules universal gas constant mole Kelvin
N A 6.023 1023 Avogardro's number
k B 1.381 1023
Joules Boltzmann's constant Kelvin
From 1st law: du Tds Pdv or ds
du P dv T T
which, for an ideal gas, becomes ds
cv 0 R dT dv T v
and integrated to
v cv 0 dT R ln T0 T v0
s s0
T
The T integral cannot be done until the functional form of cv 0 is known. Assuming cv 0 to be a constant, we have cv 0 R T v s s0 R ln T0 v0
Next, consider the molar enthalpy h u Pv which, for an ideal gas can be written as h c p 0T
where the molar heat capcity at constant pressure c p 0 is a function of T. Thus c p 0T cv 0T RT
so that c p 0 cv 0 R
or
1
R cv 0
where
cp0 cv 0
Hence, for cv 0 independent of T: 1 T v 1 T 1 v R s s0 R ln ln T0 v0 1 T0 v0
For an adiabatic process, s 0 so that Tv 1 const Other forms of this are Pv P const
3.6 Sound Wave For an inviscid, compressible fluid, we have u u u p t
Euler’s eq.:
(3.55)
Eq. of continuity:
u 0 t
(3.56)
Note that an implicit assumption for inviscid fluid is that all processes are adiabatic. (see Landau, §2). Let the fluid be an ideal gas. Adiabatic processes satisfy c p0 p const cv 0 or D p 0 Dt
3.6.1 Small Amplitude Waves Let the unperturbed state be a homogeneous one at rest. For small perturbations, we can write
(3.57)
u u1
p p0 p1
0 1
(3.58)
where subscript 0 denotes unperturbed quantities and 1 the (small) perturbations. Since the unperturbed state is homogeneous, p0 0 is independent of position. For adiabatic perturbations, p remains p0 0 for each fluid element and is therefore also independent of position. Hence
p0 p1 0 1
p0 0
ie p1 1 1 1 p0 0
1
which, to 1st order of perturbation, gives
p1 1 p 1 1 1 1 1 1 p0 0 p0 0 ie p1 1 0 p0 0
which, in terms of the sound velocity (see later),
a0
p0 0
(3.59)
can be written as p1 a0 2 1
(3.60)
To 1st order of perturbation, the Euler eq (3.55) becomes
0
u1 p1 t
(3.61)
and the eq of continuity (3.56), 1 0 u1 0 t
Eliminating u1 from (3.61-62), we have
(3.62)
2 u1 2 1 t t which, with the help of (3.60), becomes 2 p1 0
2 p1
1 2 p1 0 a0 2 t 2
2 1
1 2 1 0 a0 2 t 2
(3.63)
or
Thus, both the pressure and density variation are governed by the well–known wave equation.
2 1 2 2 2 0 c t whose general solution is of the form ξ x,t x vt so that i vi t t i i 2 2 v v v v i j i j t 2 i j i j
j ij xi xi j j i 2 2 xi xi i i
p1 p1 x vt
with
v a0
so that the perturbation, called sound, travels with speed a0 .
3.6.2 Spherical Waves For spherically symmetric waves, r, t . The wave eq. becomes
1 2 r r 2 r r Setting
2 1 2 2 0 c t
(3.65)
1 r
r, t h r, t we have 1 1 h r r r r 2
2 1 1 1 2 2 2 2 h 2 h 2 2 r r r r r r r r r r 3
1 2 2 2 1 2 h r r 2 r r r 2 r r r r 2 so that (3.65) simplifies to a 1-D wave eq. for h: 2h 1 2h 0 r 2 c 2 t 2 with solution
h r, t h r ct h r ct so that 1 (3.66) r and h obviously denote outgoing and incoming waves, respectively.
r, t h r ct h r ct The h
The radiation condition simply sets h 0 .
3.7 Supersonic Flow Past a Thin Aerofoil Consider a 2-D inviscid, compressible, adiabatic flow past an aerofoil of the form
u U u1 , v1 ,0
p p0 p1
0 1
(3.67)
As in §3.6.1, we consider only relations up to 1st order in perturbed quantities. The adiabatic relation is the same as in §3.6.1: p1 a0 2 1
(3.68)
The Euler eqs. U u1 v1 U u1 p0 p1 x y x t
0 1
U u1 v1 v1 p0 p1 x y y t
0 1 simplify to
p U u1 1 x x t p 0 U v1 1 x y t
0
which, for steady flow, become
0U
u1 p 1 x x
0U
v1 p 1 x y
Combining these 2 eqs gives
0U
2u1 2 p1 2v 0U 21 xy xy x
0U
u1 v1 0 x y x
ie.
which says the vorticity
u1 v1 is independent of x. y x
On the other hand, 0 for the uniform flow far away from the aerofoil. Hence, 0 everywhere. The flow is irrotational (potential) so that we can write (3.69) u1 v1 x y The Euler eqs thus become p1 0 0U x x p1 0 0U y x
which means p1 0U
C x
where C is a constant. Since p1 0 when u1
Similarly, the eq of continuity u 0 t
(3.70) 0 , we must have C 0 . x
ie
0 1 0 1 U u1 0 1 v1 0 t x y becomes, to 1st order perturbation, u v 1 U 1 0 1 1 0 t x x y which, for steady state, simplifies to u v 1 0 1 1 0 x x y which, in terms of , becomes U
U
2 2 1 0 2 2 0 x y x
Putting in the adiabatic condition (3.68), we have
2 2 U p1 20 0 2 a0 2 x y x which, by (3.70), becomes
0U 2 2
2 2 20 0 2 x 2 y x
a0 2
ie
1 M x y 0 2
2
2
2
2
(3.71)
where M
U a0
is called the Mach number for the flow.
3.8
Internal Gravity Waves
Incompressible, inviscid, fluids with stratified density variation can support internal gravity waves which are driven by buoyancy forces. One example is salty water with vertical salt concentration variation. Consider the 2-D case where u u x, y , t , v x, y , t ,0
In the static equilibrium state, we assume
0 y
with
0 ' 0
so that the density increases as one goes down toward the bottom. The corresponding pressure distribution is given by the steady state Euler eqs., 0
p0 x
0
p0 0 g y
so that by a suitable choice of the origin, we have
p0 y 0 gy
Since the fluid is incompressible & inviscid, the relevant eqs are u u p g t u 0
(3.80)
u u 0 t t
which, in our choice of coordinates, becomes p p 0 1 u1 v1 u1 0 1 x y x t
p0 p1 u1 v1 v1 0 1 g x y y t
0 1
u1 v1 0 x y t u1 x v1 y 0 1 0 For 1st order perturbations, these simplify to
0
u1 p 1 t x
0
v1 p 1 1 g t y
u1 v1 0 x y 1 v1 0 0 t y
(3.81)
Consider a Fourier mode of the form a x, y , t A y exp i kx t
(3.82)
where a can be u1 , v1 , p1 , or 1 . Eq(3.81) becomes i 0U1 ikP1 i 0V1 ikU1
dP1 R1 g dy
dV1 0 dy
i R1 V1
(3.83)
d 0 0 dy
Using ' to denote
d , the 3rd, 1st & last eqs give dy
i U1 V1 ' k P1 R1
0 k
U1 i
0 k2
V1 '
V1 0 ' i
so that the 2nd become i 0V1 i
k
2
V 0V1 ' ' ig 1 0 '
which simplifies to 1 ' V1 2 0 'V1 ' 0V1 '' g 0 2 V1 k 0 0 or
V1 ''
0 ' ' V1 ' k 2 g 0 2 1V1 0 0 0
2 N2 2 N V1 '' V1 ' k 2 1V1 0 g
where the buoyancy number N is defined by g ( 0 ' 0 ) N 2 0 ' 0
0
(3.84)
The simplest case is when 0 is exponential, ie y H
0 y A exp so that N2
g H
is a constant. So are the coefficients Eq(3.84), the solution of which is simply
V1 B exp y where ξ satisfies
2
N2 N2 k 2 2 1 0 g
2
1 g k 2 2 1 0 H H
or
ie.
1 2H
1 2 g il 1 1 4 Hk 2 H 2 H
where
l
1 g 4 Hk 2 2 H 1 2H
(3.84A)
The solution (3.82) thus become 1 v1 x, y , t B exp il y i kx t 2 H y B exp exp i kx ly t 2H
(3.86)
which is a plane wave with wavevector k k , l ,0 and amplitude that decays exponentially as it rises. V1 V y 0 ' i 1 exp , we have i H H B y 1 x , y , t i exp exp i kx ly t H 2H
Using R1
(3.88)
Note that 1 is usually more accessible to experimental observation. Solving for ω from (3.84A), we get the dispersion relation: 1 l2H H 2 4 Hk 2 k 2 g
2
gk 2 H
g N 2k 2 1 1 l2H k 2 l2 1 k2 l2 H 2 2 2 4H 4H 2 4 Hk k
which is clearly anisotropic. If the wavelength 2 2 k k2 l2 is small compared with the scale height H, (3.87) simplifies to N 2k 2 2 2 k l Nk 2 2 k l which gives a group velocity 2
(3.89)
c g k , , k l m
1 N 2 2 k l
Nl
k 2 l 2 k
l 2
l2 k
3
k k , l, m
k2
k
2
l
ˆ k k
so that cg c p 0
,
k
2
l2
3
,0
l , k ,0
l , k ,0
Now, the phase velocity is cp
2 3
kl
(3.90)
k k2 l2
k , l ,0
(3.91)
(3.87)
ie.,
c g c p . (see fig.3.13)
3.9a Partial Differential Eqs. Source: F.B.Hildebrand, “Advanced Calculus for Applications”, 2nd ed, Chapter 8, Prentice Hall (1976) nth order ODE: n independent arbitrary constants. linear ODE: General solution = linear combination of n independent functions. Non- linear ODE: may exist singular solution with no arbitrary constants. nth order PDE: General solution = arbitrary functions of specific functions (arguments). In case of specific boundary conditions, it’s usually easier to determine a set of particular solutions & combine them to satisfy the BCs.
3.9a1 Quasi-Linear Eq. of 1st Order Most general quasi-linear eq. of 1st order: z z P x, y , z Q x, y , z R x, y , z x y
(15)
Linear case: P x, y
z z Q x, y R1 x, y z R2 x, y x y
(16)
Let
u x, y , z c
(17)
defines a solution (integral surface) of (15). Treating z as z x, y , the chain rule gives: u u z 0 x z x or
u u z 0 y z y
(18)
u z y u y z
u z x u x z
which, when substituted into (15), gives u u u P Q R 0 x y z
(19)
(20)
Introducing a vector V P, Q, R in the
x, y , z
V u 0
space, (20) becomes (21)
Since u is normal to the integral surface (17), V is tangent to a curve on the latter. Conversely, a curve with tangent parallel to V everywhere must be on the integral surface. Such curves are called characteristic curves of the differential eq. Any curve can be denoted by the position vector r r s where s is the arc length. The unit tangent to the curve is dr dx dy dz i j k ds ds ds ds
If it is a characteristic curve, we must have P
dx ds
Q
dy ds
R
where μ can be any function of
dz ds
(22)
x, y , z .
Eq(22) can be written as dx dy dz P Q R
(23)
Let the independent solutions to (23) be
u1 x, y , z c1
u2 x, y , z c2
(24)
where c1 , c2 are constants. Any surface specified as
or
F u1 , u2 0
(25a)
u2 f u1
(25b)
will be an integral surface of (15).
3.9a2 Initial Conditions For the linear case z z P x, y Q x, y R1 x, y z R2 x, y x y
(35)
One of the eqs in (23), namely, dx dy P Q is an ordinary differential eq. Let its solution be
u1 x, y c1
(35a)
which is called a characteristic cylinder. (The intersection of this cylinder with the xy plane is called a characteristic base curve.) Eq(35a) can be used to express y in terms of x and c1 so that another eq of (23), say, dx dz P R1 z R2
ie. dz R1 R z 2 dx P P
is just an ODE involving z x . The solution to this can be written as
u2 z1 x, c1 2 x, c1 c2 or
u2 z 1 x, y 2 x, y c2 The general solution to (35) is therefore z
x, y 1 f u1 x, y 2 1 x, y 1 x , y
or z s1 x, y f s2 x, y s3 x, y
(36)
For a 1st order ODE, the arbitrary integration constant can be determined by requiring
the integral curve to go through a specific point in the xy plane. For a 1st order PDE, the arbitrary integration function in (25) can be determined by requiring the integral surface to include a specific curve in the xyz space. Let this curve be given by the intersection of 2 surfaces
1 x, y , z 0
2 x, y , z 0
(42)
and the independent solutions to (23) are
u1 x, y , z c1
u2 x, y , z c2
(43)
The elimination of x, y , z then results in an eq of the form F c1 , c2 0 . The required integral surface is then simply F u1 , u2 0 .
3.9a3
Characteristics of Linear 1st Order Eqs.
Any curve curve in the xyz space can be specified as C:
x x ,
y y ,
z z
(90)
The projection of this onto the xy plane is C0 :
x x ,
y y ,
z0
(91)
The prescribed curve (42) can alternatively be given as a specification of z on C0 in the form of (90). The question is whether the PDE has an integral surface that actually contains C. Now, on C0 , everything, including z, is a known function of λ. Hence dz z dx z dy d x d y d
(94)
Also, the PDE itself becomes an eq of independent variable λ. z z (93) P Q R1 z R2 x y These 2 eqs can be considered as simultaneous eqs for The condition for the existence of a unique solution is
z z and on C0 . x y
P dx d
Q dy dx Q 0 dy P d d d
(95)
ie. dx dy P Q
on C0
(96)
In other words, a unique solution that satisfies the boundary condition prescribed on a curve C0 exists only if C0 is no where tangent to a characteristic base curve. Now, in case C0 is itself a characteristic base curve, the determinant in (95) vanishes identically: P dx d
Q dy 0 d
(96a)
There’ll either be no or infinitely many solutions to (93) & (94). For the latter case, we need, according to Cramer’s rule, R1 z R2 dz d
Q dy 0 d
or
P
R1 z R2
dx d
dz d
0
(96b)
This means a curve C that satisfies eqs (96a,b) must be a characteristic curve. Indeed, this provides a way to find characteristic curves for more complicated PDEs.
3.9
Finite-Amplitude Waves in Shallow Water
Let the bottom of the fluid be at y 0 ; its free surface, y h x, t . Let h0 be some typical value of h x, t , and L be a typical horizontal length scale. The shallow water approximation means h0 L (3.92) The full 2-D eqs for ideal fluid are u u u 1 p u v t x y x v v v 1 p u v g t x y y
(3.93) (3.94)
u v 0 x y
(3.95)
The shallow water approximation means (see §3.9.1 for justification) Dv g Dt
so that (3.94) simplifies to 1 p 0 g y which can be integrated immediately to give
p gy f x, t The integration constant f x, t can be determined by the boundary condition that we have atmospheric pressure p0 at the free surface y h x, t . Hence p p0 g y h x, t
With this, (3.93) becomes Du h g Dt x
which means the change of u for each fluid element is independent of y. Hence, if u is independent of y at some time, it will be so for all time. In which case, (3.93) simplifies to u u h u g t x x
(3.96)
and (3.95) can be integrated to give v
u y f x, t x
The integration constant f x, t can be determined by the boundary condition at the bottom, ie., at v0 so that f 0 and v
u y x
y0
(3.96a)
As shown in §3.2.1, the kinematic condition at the free surface is h h u v t x
on y h x, t
(3.18)
Combining with (3.96a), we have h h u u h 0 t x x
on y h x, t
(3.97)
Eqs(3.96-7) are known as the shallow water equations.
3.9.1 Shallow Water Equations The shallow water eqs are (see Acheson §3.9) ut uu x ghx 0 (3.96) hux ht uhx 0 (3.97) where subscripts denote partial derivatives. Let u, h be specified on curve C0 . We have u ' ut t ' u x x ' h ' ht t ' hx x ' where ' denotes derivative with respect to λ. Rewriting these as a set of simultaneous eqs for ut , u x , ht , hx , we have
1 u 0 h t ' x' 0 0
g ut 0 1 u u x 0 0 0 ht u ' t ' x ' hx h ' 0
The determinant of coefficients is
h 1 u u 0 g D x' 0 0 t' h 1 u 0 t' x' 0 t' x' x ' x ' ut ' t ' u x ' ut ' ght ' x '2 2 x ' t ' u t '2 gh u 2
Setting D 0 gives
x ' ut ' u 2t '2 t '2 gh u 2
x ' t ' u gh
dx (a) u gh dt which give us the characteristic base curves C0 .
According to Cramer’s rule, the condition for infinitely many solutions for ux is 1
0
0
g
0 0 1 u 0 t' u' 0 0 0 h' t' x' ie
0 1 u 0 0 g u ' 0 0 t ' 0 1 u u ' x ' uu ' t ' t ' gh ' 0 h' t' x' h' t' x' h' x ' t 'u g u' dx dh ug dt du
(b)
Combining (a) and (b) gives gh g
dh du
ie du g
dh 2d gh h
u 2 gh const
(c)
which are satisfied on the characteristic base curves. Following Acheson, we define
c gh
(3.98)
so that the characteristic base curves (a) become
dx uc dt
(3.103)
on which u 2c const . Consider the functions and u 2c
u 2c
so that u
1 2
h
1 2 1 2 c g 16 g
ht
1 t t 8g
c
1 4
and
The shallow water eq (3.96) thus become 1 1 1 t t x x x x 0 2 4 8
which simplifies to
t t
1 3 x 3 x x x 0 4
4 t t 3 x 3 x 0
(d)
Similarly, (3.97) becomes 1 1 1 2 x x t t x x 0 32 g 8g 16 g
x x 4 t t 2 x x 0 4 t t 3 x 3 x x x 0 4 t t 3 x 3 x 0
(d)+(e):
4t 3 x 0 1 3 0 t 4 x
(e)
t u c x u 2c 0
4t 3 x 0
(d)-(e):
1 t 4 3 x 0 t u c x u 2c 0
3.9.2
(3.99)
(3.100)
Flow Caused By Dam Break
Let static 2-D water be contained between 0 y h0 for x 0 by a dam at x 0 . (see fig.3.15a) The problem is to find the water flow after the dam breaks at t 0 . Consider 1st the initial conditions. For x 0 , t 0 , we have u 0 , h h0 , so that c c0 gh0 . The characteristic base curves dx uc dt
(3.103)
become dx c0 dt
ie., x c0t x0
(3.104a)
which are straight lines that intercept the x axis at x0 . (see fig.3.15c) After the dam breaks, these characteristic base curves will extend continuously, wherever possible, into the t 0 region. On such extensions, u 2c 2c0 . Now, some points may be reachable by extensions of both types of base curves. (Point P in fig.3.15c) Both eqs u 2c 2c0 must then be satisfied, which means u0
and
c c0
ie., the water there are not yet disturbed and the characteristic base curves are still straightlines given by (3.104a). As is obvious from fig.3.15d, the upper boundary of this undisturbed region is the line x c0t .
Beyond the line x c0t , a point (eg., Q in fig.3.15.d) can be reached only by the extension of characteristic base curves of the ‘positive’ type u 2c 2c0
dx u c on which dt
(3.104)
The other, ‘negative’ type, characteristic base curve that pass through Q , ie. dx uc dt
(3.104b)
requires (3.104c) u 2c k where k 2c0 is a constant along the ‘negative’ base curve (3.104b). The solution of (3.104) & (3.104c) is u c0
k 2
and
c
1 k c0 2 2
(3.105a)
which are both constants on the ‘negative’ base curve (3.104b) which itself becomes a straight line dx 1 3 c0 k dt 2 2
ie. 1 3 c0 k t x0 2 2 where x0 is the value of x at t 0 . x
Now, different ‘negative’ base curves are characterized by different k so that they have different slopes. To avoid crossing of such curves, we need x0 0 so that the ‘negative’ base curves all eminate from the origin: x
1 3 c0 k t 2 2
(3.105b)
The ‘positive’ base curves are given by dx 1 1 3c0 k dt 2 2
(3.105c)
which are not straightlines since k is not constant on such curves. From (3.105b), we have k
2 x c0 2 3 t
so that the ‘positive’ base curve (3.105c) becomes dx 1 1 x 1 x 3c0 c0 2 4c0 dt 2 3 t 3 t
while (3.105a) becomes u
2 c0 3
x t
1 c 2c0 3
and
x t
(3.105,6)
Since c gh 0 , these relations are valid only for x 2c0t Using (3.106) we have h
c2 1 x 2 gh0 g 9g t
2
for
c0t x 2c0t
At x 2c0t , we have h 0 , which should remain so for x 2c0t on physical grounds. (see fig.3.15b).
3.9.3 Non-Linear Wave Distortion The dam break flow may be taken as a smoothing out of the initial discontinuity in
h x, t through non-linear mechanism.
Let us return to the shallow water eqs t u c x u 2c 0 t u c x u 2c 0
(3.99) (3.100)
Consider the region c0t x 2c0t where u 2c 2c0 . Eq (3.99) is automatically satisfied. Eliminating u from (3.100) gives t 2c0 3c x c 0 which can be rewritten as z z z 0 t x
(3.109)
z 3c 2c0
(3.108)
with
The characteristics eqs of (3.109) are (see §3.9a1)
dt dx dz 1 z 0
so that z c1 and dx z dt
ie.
x zt c2
The general solution is therefore
z F x zt
(3.110)
where F is an arbitrary differentiable function. Note that (3.110) is an implicit solution. To show that (3.110) is indeed a solution of (3.109), we need to evaluate the various partials of (3.110) with x and t taken as the independent variables. Setting x zt the partials on (3.110) are z dF z z tF ' t t d t z dF z 1 t F ' x x d x
where F '
dF . d
Solving for the partials of z, we have z zF ' t 1 tF '
(3.111)
z F' x 1 tF '
(3.112)
which obviously satisfies (3.109) identically for arbitrary differentiable F. Now, (3.110) is a wave of z that travels with velocity –z. Assuming positive amplitude ( z 0 ), the wave travels to the left with speed that is proportional to its amplitude, thus providing the mechanism for smoothing out the discontinuity in the dam break problem. Consider now the eq
z z z 0 t x
(3.113)
Applying the foregoing analysis, we see that its solution is
z F x zt
(3.114)
which describes a wave of z that travels with velocity z. Assuming positive amplitude ( z 0 ), the wave travels to the right with speed that is proportional to its amplitude, thus steeping the wave profile as shown in fig.3.16. From the analog of (3.112): F ' X z x 1 tF ' X
we see that
X x zt
(3.115)
z goes to infinity at time tc defined by x
1 tc F ' x ztc 0
(3.116)
In case of multiple solutions, only the minimum value will be physically meaningful since the wave will collapse after that. (see fig.3.16c)
3.9.4 The Formation of a Bore Let fluid be contained by a vertical plate in the region x 0 , 0 y h0 for t 0 . At t 0 , the plate at x 0 begins to move in the x direction with speed U t , where α is a constant. (see fig.3.17) Once again, we need to solve the shallow water eqs (3.96-7) or (3.99-100). The treatment is analogous to that of the dam break problem so we’ll simply highlight the salient points. For t 0 , the fluid occupies x 0 with u 0 and h h0 , c c0 gh0 . The corresponding characteristic base curves x c0t x0 (3.104a) are straight lines in the 4th quadrant that intercept the x axis at x0 . (see fig.3.18) The undisturbed region ( u 0 and c c0 ) is found by extending the above straight lines into the 1st quadrant. The upper boundary of this is obviously the line x c0t . (see fig.3.18)
As is clear from fig.3.18, only ‘negative’ type of base curves
dx u c can be dt
extended across the boundary x c0t . On such base curves, u 2c 2c0 (3.117) so that (3.100) is automatically satisfied. Eliminate c from (3.99) gives 3 t 2 u c0 x u 0 whose characteristics are given by
(3.118)
dt dx du 1 3uc 0 0 2 so that u c1 and 3 x c1 c0 t c2 2 The general solution of (3.118) is therefore dx 3 u c0 dt 2
ie.
3 u F x u c0 t 2
(3.119)
Now, we expect u to have a maximum at the plate and decrease to zero at x c0t . Eq(3.118) is then of the profile steepening type (3.113) discussed in §3.9.3. The actual form of F is of course determined by the boundary conditions. 1 At time t, the vertical plate is at x t 2 moving with speed u t . 2
Eq(3.119) becomes 1 2
3 2
t F t 2 t c0 t F t 2 c0t with
0 F 0
Writing
t 2 c0t we have
t0
(3.120)
t
1 c0 c0 2 4 2
so that (3.120) becomes F
1 c0 c0 2 4 2
0
(3.120)
where the + root is chosen so that F 0 0 .
3 Eq(3.119) is evaluated by putting x u c0 t in (3.120) so that 2
u
1 3 c0 c0 2 4 x u c0 t 2 2
Solving fo u, we have
2u c0
2
3 c0 2 4 x u c0 t 2
4u 2 4c0 6 t u 4 x c0t 0 3 u 2 c0 t u x c0t 0 2 so that 2 1 3 3 u c0 t c0 t 4 x c0t 2 2 2
(3.121)
where the + root is chosen to make u 0 at the undisturbed boundary x c0t . Eq(3.117) can be used to find c: 2 u 1 3 3 c c0 c0 c0 t c0 t 4 x c0t 2 4 2 2
Hence
c x 2
2 3 c0 t 4 x c0t 2
At the boundary x c0t ,
1
2
(3.122)
c 3 x 2 c0 t 2
which explodes at time tc
2c0 3
(3.123)
At this moment, the vertical plate is at position 1 1 x tc 2 c0tc 2 3
whereupon (3.122) becomes
1 1 8 4 4 x c0tc c0 c0tc c0 4 4 3 3 which means the water level at the gate at time tc is c c0
h
1 2 16 c h0 g 9
3.10 Hydraulic Jumps & Shock Waves 3.10.1 Hydraulic Jumps A bore (§3.9.4) that is stationary to the observer is called a hydraulic jump. Both are shallow water phenomena which exhibit an abrupt jump of water level. One way to produce it is to turn on the kitchen tap. Water in the sink splays outward radially in a thin layer which exhibits a sudden increase in height at a certain radius that depends on the flow rate. With reference to fig.3.19 for notations, we see that 1. Conservation of mass means U1h1 U 2h2 (3.124) 2.
Conservation of momentum means 1 2 1 gh1 h1U12 gh2 2 h2U 2 2 2 2
(3.125)
where terms proportional to g come from pressure, and those to U from convective contributions to the stress tensor. 3.
Conservation of energy means turbulence at the jump will dissipate energy in the
form of heat. The energy loss at the jump is (Ex.3.20) gU1 3 (3.128) h2 h1 4h2 where h2 h1
(3.129)
Introduce now the Froude number U U F gh c p
(3.126)
where c p gh is the phase speed of water waves in shallow water. It plays a similar role as the Mach number in compressible flow. Flows with F 1 ( F 1 ) are called sub- (super-) critical. Now, eliminating U 2 using (3.124), (3.125) becomes 1 2 1 h 2U 2 gh1 h1U12 gh2 2 1 1 2 2 h2 so that U1 2
gh2 h2 2 h12 2h1 h2 h1
gh2 h2 h1 2h1
Interchanging indices 1 2 gives gh h h U 22 1 2 1 2h2 Hence, the corresponding Froude numbers are
F1
h h h U1 2 12 2 h1 gh1
F2
h h h U2 1 12 2 h2 gh2
(3.127)
Since h2 h1 , we have F1 1
and
F2 1
(3.130)
ie., the jump changes a supercritical flow to a subcritical one.
3.10.2
Unsteady 1-D Gas Dynamics
Consider the sound wave problem (§3.6) without the small amplitude approximation. The relevant eqs are u u u p t
Euler eq.:
(3.55)
Eq. of continuity:
u 0 t
(3.56)
Adiabatic (homentropic) condition:
D p 0 Dt
Using the Leibniz rule, (3.57) becomes Dp D 1 0 Dt Dt
or Dp p D D a2 Dt Dt Dt
(3.133a)
where a2
p
(3.133)
Taking the material derivative of (3.133) gives 2a
1 Dp p D Da 2 Dt Dt Dt
a 2 a 2 D Dt
1
a 2 D Dt
Rewriting (3.56) as D u Dt
(3.133b) becomes 2
Da 1 a u Dt
[from (3.133a)]
(3.133b)
(3.57)
or 2 a u a a u 0 1 t
(3.133c)
Now, if the gas is initially at rest, (3.57) becomes p const so that, for example, p p 1 0 ie p
p a 2
Also, analogous to (3.133b), we have
2aa 1
a2
so that p
2a a 1
The Euler eq. then becomes u 2a u u a 0 t 1
(3.133d)
For 1-D gas flow with u u x, t ,0,0 , (3.133c,d) simplify to [cf. Ex.3.22] 2 a a u 0 u a 1 t x x
u u 2a a u 0 t x 1 x
(3.133e)
These are similar to, but not the same as, the shallow water eqs(3.96-7). The characteristics are obtained by considering u 1 0 2a 1 x' t ' 0 0
0 1 0 t'
a 1 a 0 2 t ax 0 u u a ' t 0 ux u ' x '
This set of eqs is indeterminate if 2 conditions hold:
I. 1
u
0
a 1 2
2a 1 1 t' x' 0 0 0 t' 0
0
u 0 x'
ie.,
2a u 0 1 u 1 2a x' 0 0 t' 1 1 0 t' x' 0 t'
a 1 2 u x'
x ' x ' ut ' t ' u x ' ut ' a 2t ' x ' ut ' a 2t '2 0 2
which requires
x ' u a t ' dx ua dt
(3.133f)
II. 1
0
0
0
0
1
t' a' 0 0 u' t'
a 1 2 u 0 0 x'
ie. 0
1
u
0
0
a' 0 0 t' 0 1 u' t' x' u' t' a ' x ' ut ' t ' u '
a 1 2 u x'
a 1 2
a ' x ' ut ' t ' u '
a 1 0 2
so that du 2 dx u da a 1 dt Combining with (3.133f), we have du 2 da 1 ie. u
2a c 1
To summarize. The characteristic base curves are dx ua dt
(3.133f)
on which u
2a c 1
(3.133g)
where c are constants. These can be used to simplify (3.133e) as follows. [cf. §3.9.1] Consider the functions 2a b 1
ub
ub
and
so that u a
1 2
1 2
b
b
1 4
1 2
The 1st eq in (3.133e) thus become 1 1 1 x x 0 t t x x 2 2 8
which simplifies to 4 t t 1 3 x 3 1 x 0
(d)
Similarly, the 2nd one becomes 1 1 1 x x 0 t t x x 2 2 8 4 t t 1 3 x 3 1 x 0
(d)+(e):
(e)
4t 1 3 x 0
Now:
1 3 1 u
2a 2a 3 u 1 1
4 u a so that we have
t u a x 0 2a t u a x u 1 0 Similarly, (d)-(e):
(3.131)
4t 3 1 x 0
Now:
3 1 3 u
2a 2a 1 u 1 1
4 u a so that we have
t u a x 0 2a t u a x u 1 0
(3.132)
Consider the case of a piston moving with speed U t into a long tube containing gas at rest with sound speed a0
p0 . 0
The situation is analogous to the formation of a bore (§3.9.4). For t 0 , the gas occupies x 0 with
u0
a a0
and
The characteristic base curves are straightlines in the 4th quadrant: dx 1 ie. a0 t t0 x dt a0 where t0 is the intercept with the t axis. The + type lines must have t0 0 . The line t
1 x a0
ie.,
x a0 t
thus form an upper boundary for them. For the – type lines, only those with t0 0 can be extended across this boundary. Thus, inside the bore region, we have, along the – type characteristic base curves 2a 2a (3.134) u 0 1 1 and (3.132) is automatically satisfied. For the + type characteristic base curves, eliminating a using (3.134) turns (3.131) into 1 t 2 1 u a0 x u 0
(3.135)
The characteristics of this are given by dt dx du 1 1 1 u a0 0 2 The general solution of (3.135) is therefore 1 u F x 1 u a0 t (3.136) 2 where F is an arbitrary differentiable function. The function F is determined by the boundary condition at the piston: u t
at
1 x t2 2
ie., 1 2
t F t a0 t Setting
t0
1
t a0 t 2 we have t
1
a 0
so that 1
a 0
a0 2 2
a0 2 2 F
where the + root was chosen to make F 0 0 . Thus
u
1 1 a0 a0 2 2 x 1 u a0 t 2
Solving for u, we have
1 1 u a0 t 2
2u 2 2 a0u a0 2 a0 2 2 x
u 2 2a0 1 t u 2 x a0t 0 u
2 1 2a0 1 t 2a0 1 t 8 x a0t 2
where the + root is chosen to make u 0 at x a0t . Hence u 2 x 2a 1 t 2 8 x a t 0 0
Now, (3.136) is of the wave steepening type,
u will eventually become singular, x
which happens when 2a0 1 t 8 x a0t 0 2
At x a0t , this occurs at time 2 a0 tc 1 whence (3.135) breaks down and a shock propagates down the tube.
3.10.3
Normal Shock Waves
A shock normal to the flow is similar to a hydraulic jump (§3.10.1). With reference to fig.3.20 [cf. fig.3.19], we have the Rankine-Hugoniot equations: Conservation of mass: 1U1 2U 2 p1 1U12 p2 2U 2 2
Conservation of momentum:
(3.137)
1 2 a12 1 a2 U1 U 22 2 2 1 2 1
Conservation of energy:
[The last is different from the hydraulic jump case which substains an energy loss across the jump.] Furthermore, the 2nd law of thermodynamics S2 S1 0 means that log p2 2 log p11
ie., p2 2
p11
or
p2 2 p1 1
(3.138)
We leave as an exercise to show that 1. U 2 a0 U1 , ie., M 2 1 M 1 2.
p2 p1 , 2 1
3.11 Viscous Shocks & Solitary Waves 3.11.1 Weak Viscous Shocks For a weak viscous shock propagating in x direction into a gas at rest, the velocity
u x, t satisfies Burger’s equation 2 2u 1 1 u a0 u 2 3 x x t 2
(3.140)
where
if both μ and are small. 0
For 0 , (3.140) reverts to (3.135). Consider the ansatz
u f x Vt
(3.141)
such that
f U1
f 0
Now: u Vf ' t u f' x where df f ' d
2u f '' x 2
x Vt
Eq(3.140) becomes 2 1 V 1 f a0 f ' f '' 3 2 which, using df ' df df ' df ' f '' f' d d df df becomes 2 1 df ' V 1 f a0 df 3 2 ie 2 1 f ' 1 f 2 V a0 f Const 3 4
or 4 a0 V 8 f ' f2 f C 3 1 1
where C is a constant.
Hence 3 1 8
df 4 a0 V f2 f C 1
In order to the formula
ax
2
dx 1 x p ln bx c a p q x q
where p, q are the roots of ax 2 bx c , we set 4 a0 V a 1 cC b 1 p, q
1 b b 2 4C 2
so that 3 1 1 f p 2 ln 8 f q b 4C Now, the points require f p, q . The condition f 0 thus require either
b b2 4C 0
or
b b2 4C 0
Only the 1st eq has a solution with C 0 . Therefore, we have 3 1 1 f b ln 8 b f
The condition f U1 then means b U1 so that 3 1 1 f U1 ln 8 U1 f which can be solved for u f to give
3 1 U1 U1 u exp u 8
( u U1 )
ie., u
U1 U1 3 1 U1 1 exp x Vt 1 exp 8
(3.142)
where the shock thickness is given by 8 3 1 U1
(3.144)
Thus, u decays from U1 to 0 within a distance of order
.
Finally, the condition on b means 4 a0 V U1 1 Solving for the shock speed V gives 1 V a0 U1 1 4
3.11.2
(3.143)
Solitary Waves in Shallow Water
We now look for a nonlinear shallow water wave of permanent form whose steepening effects are caused by weak dispersion. Let
0 be typical magnitude of
, the vertical displacement of water surface.
h0 be that for h, the depth of water. L
be that for horizontal length scale.
0
If h0
and
h0 2 L2
are small & of the same order of magnitude, then
(3.145) satisfies the Korteweg- de Vries
equation: 3 3c0 1 2 c0 c0h0 0 t 2h0 x 6 x 3
where the last term describes dispersive effects. We seek a traveling solution of the form
f x Vt f
(3.146)
with boundary conditions
f ( n ) 0
n 0,1, 2,
(3.145a)
Thus, (3.146) becomes
3c 1 Vf ' c0 0 f f ' c0h0 2 f ''' 0 2h0 6 or
c0 V f '
3c0 2 1 f ' c0h0 2 f ''' 0 4h0 6
which, on integrating, becomes 3c 1 c0 V f 0 f 2 c0h02 f '' C 4h0 6
(3.146a)
where C is a constant. Putting in the boundary conditions (3.145a) gives C 0 . Writing d 1 2 f' df 2 (3.146a) becomes f ''
1 3c 1 c0h0 2d f '2 c0 V f 0 f 2 df 6 4h0 2 1 1 c c0h0 2 f '2 c0 V f 2 0 f 3 C 12 2 4h0
(3.146b)
where C is another constant. Putting in the boundary conditions (3.145a) gives C 0 . Rearranging terms in (3.146b) gives V 1 3 2 h0 f ' 2h0 1 3 c0
f f 2 a f f 2
(3.146c)
where
V a 2h0 1 c0
(3.146d)
Taking the square root of (3.146c):
f 'f
3 a f h03
so that 3 h03 f
df
a f
C
which, upon the use of the formula
x
dx 1 ln a bx a
a bx a 2 a bx tanh 1 a a bx a a
gives
3 2 a f tanh 1 3 h0 a a where C 0 according to the boundary conditions (3.145a). Solving for f gives
a f 3a tanh a 4h03 3a 3a f a 1 tanh 2 a sech 2 3 4h0 4h03
ie.,
3a x Vt 3 4h0
a sech 2
(3.147)
Eq(3.146d) can also be rewritten as
a V c0 1 2h0
(3.148)
Since sech ranges between 0 and 1, a is the maximum of the wave height 0 . [see fig.3.23]. Owing to the assumptions (3.145), these eqs are valid only for a h0 . According to (3.148), waves with larger amplitude a travels with greater velocity V. Thus, a higher wave can overtake a lower one. What seems astonishing is that after
such a collision, both waves emerge unscathed. [see fig.3.24] Such waves are called solitons.
4. Classical Aerofoil Theory Source: D.J.Acheson, “Elementary Fluid Dynamics”, Chapter 4, Clarendon Press (90)
4.1 Introduction 4.2 Velocity Potential & Stream Function 4.2.1 Velocity Potential For irrotational flow, u 0 so that one can define a velocity potential at point P by P
( P ) u dx
(4.2)
u
(4.3)
O
where O is some reference point. In a simply connected region, is path independent & hence single- valued. In a multiply connected region, can be path dependent & hence multi- valued. The circulation around a closed contour C is u dx dx C C
C
(4.4)
where C is the change of in going around C.
Examples: 1.
Uniform flow u U ,0,0 has Ux const .
2.
Irrotational flow u x, y ,0 has stagnation point at the origin. To solve
x x
y y
0 z
we see that the 3rd eq means is a function of x and y only. Next, we integrate the 1st eq to get 1 2
x2 f ( y) Taking
of this & equating the result to the 2nd eq gives y
df y dy which gives 1 f y2 C 2
C const
so that
x2 y2 C 1 2
which is single valued so that 0 for any circuit. 3.
Line vortex flow u
k e r
is irrotational except at origin. Since u is not defined at origin, any domain that covers the origin is multiply connected. In cylindrical coordinates, θ z ˆ ˆ rˆ r r z so that we need to solve 0 r z
and
k r r
This gives k C which is multi-valued as expected. Circulation for a circuit that goes round the origin n times is 2 nk
4.2.2 Stream Function For 2-D incompressible irrotational flow, we can write u u, v , x y so that u v u 0 x y
(4.5)
(4.6)
is automatically satisfied. Now, u v (4.7) 0 y x y x x y so that ψ is constant along a streamline. ψ is called a stream function since streamlines are simply curves with const .
u
Another way to write (4.5) is u k
i
j
x 0
y 0
k , z y x
(4.8)
In cylindrical coordinates, we have θ k θ ˆ ˆ rˆ r 1 ˆ u k rˆ r r z r r 0 0 or ur
r
u
r
(4.9)
which obviously satisfies the incompressibility condition trivially u 0 rur rr r
(4.10)
4.3
Complex Potential
Consider a 2-D incompressible flow. u v 0 x y we have u
y
v
x
(4.11)
If it is also irrotational, u v 0 y x we have u
x
v
y
(4.11a)
Thus, and satisfy the Cauchy- Riemann conditions x y
y x
(4.11b)
so that they are the real & imaginary parts of an analytic function, ie., (4.12) w i where w z , with z x iy , is known as the complex potential.
By definition f z z f z df lim dz z0 z df is independent of the path z approaches 0. dz Thus, if f a ib is analytic, we have
We say f is analytic at z if
a df a b b i i i dz x x y y for the 2 cases z x and z iy . Another useful property for f to be analytic at z is that it has a Taylor expansion there. Since w is analytic, we have
dw i u iv dz x x
(4.15)
The flow speed is therefore q u2 v2
dw dz
(4.16)
Either of the real & imaginary parts of an analytic function satisfies the Laplace eq. individually. Hence
2 2 0 x 2 y 2
(4.13)
2 2 0 x 2 y 2
(4.14)
4.3.1 Uniform Flow At Angle Here u U cos
v U sin
so dw u iv U cos i sin Ue i dz
and, integrating gives w Uze i
(4.17)
4.3.2 Line Vortex u
e 2 r
(4.18)
From §2.4.4, we have
2
(4.19)
From (4.9), we have ur
0 r
u
so that
f r
ln r 2
r 2 r
to x-Axis
Hence w i
i ln r i ln r i i ln z 2 2 2
(4.20)
where z rei
This can immediately generalized to a line vortex at z z0 : w i
ln z z0 2
(4.21)
4.3.3 Flow Near Stagnation Point The Taylor expansion of w near z0 is w z w z0 z z0 w ' z0
1 2 z z0 w '' z0 2
The constant term w z0 is inconsequential. If z0 is a stagnation point, w ' z0 0 . Hence, the flow near a stagnation point is determined, to the lowest order, by the quadratic term. Writing
w '' z0 ei
and
z z0 z1
we have w const
1 2 i z1 e 2
1 const z2 2 2
where z2 z1e
i 2
Dropping the constant term & subscripts, the complex potential near a stagnation point is therefore 1 1 w z 2 x 2 y 2 2ixy 2 2
where α is real.
(4.22)
The velocity potential is
Re w x 2 y 2 1 2
so the equipotential lines are rectangular hyperbolae. The stream function is Im w xy
(4.24)
so the streamlines (constant ψ) are rectangular hyperbolae perpendicular to the equipotentials [see fig.4.1] The corresponding flow is dw u Re Re z x dz dw v Im Im z y dz
(4.23)
4.4 Method of Images Let there be a rigid plane wall at x 0 . at r d ,0 [see fig.4.2].
Consider a line vortex of strength
The boundary condition
u 0, y 0
x
or
0 x 0
can be satisfied by replacing the wall with a mirror image of the vortex. As in electrostatics, the image should be at r d ,0 with strength .
According to (4.21), the complex potential is therefore w i
ln z d i ln z d 2 2
(4.25)
ie.,
i i
zd ln 2 z d
i
2
zd ln z d i
(4.26)
where zd z d i e zd zd
Hence
zd ln 2 zd
so that the streamlines are
zd ln const 2 zd
or simply zd R zd
(4.27)
where R is a constant. Now, (4.27) can be written as
x d
2
2 y 2 R 2 x d y 2
or, on collecting terms:
1 R x 2
2
d 2 y 2 2 1 R 2 dx 0
1 R2 x 2 dx d 2 y 2 0 2 1 R 2
2 2 2 2 2R 1 R2 2 2 1 R d y d 1 d x 2 2 2 1 R 1 R 1 R
1 R 2 2R which is a circle centered at z d . d ,0 with radius a 2 2 1 R 1 R
Such circles are known as coaxial circles. The distances between the center of a coaxial circle and those of the vortices at x d are 1 R2 2d R 2 c d 1 1 R2 1 R2 1
so that
2
2d 2 c c R a2 2 1 R
(4.27a)
Consider now the flow inside a circular cylinder z a due to a line vortex of strength
at z c a,0 . The boundary condition of no flow in the normal
direction is satisfied if z a is a streamline. This is easily accomplished by placing an image vortex of strength at a point which according to (4.27a) is
a2 z a,0 . The corresponding complex potential is [cf. (4.25)]: c a2 w i ln z c i ln z 2 2 c
(4.28)
4.4.1 Milne-Thomson’s Circle Theorem Consider w f ( z ) where all singularities of f lie in z a . Then a2 w f z f z
(4.29)
is the complex potential of a flow with 1. 2.
same singularities as f z in z a .
z a as a streamline.
Proof: 1.
Since all singularities of f z
a2 are in z a , those of f and hence z
a2 a2 a2 a ; ie., a z . Therefore, f is regular z a and f are in z z z
w has the same singularities as f z there.
2.
On z a , we have z aei
(
real)
2
a aei z z so that
w z f z f z 2 Re w z Hence,
0
on
z a
which means z a is a streamline.
4.5
Irrotational Flow Past A Circular Cylinder
IR flow in x direction, with speed U at infinity, past fixed cylinder z a . Complex potential for unperturbed flow is:
f z Uz (singular only at infinity).
Applying the circle theorem (§4.4.1), we have
a2 a2 f U Uz z z
a2 a2 f U z z
so
a2 wz U z z
(4.31)
is the complex potential with z a as a streamline.
Any superposition with a line vortex of strength Г is also a solution with z a as a streamline so that a more general form of (4.31) is
a2 wz U z i ln z z 2
(4.32)
Consider first (4.31). Writing z rei we have
a2 w z U rei e i i r so that
a2 cos r
(4.33)
a2 U r sin r
(4.34)
U r
Hence
a2 ur U 1 2 cos r r u
a2 U 1 2 sin r r
(4.35)
The flow is symmetric fore & aft (see fig4.4a) At r a , u 2U sin 0 for 0, , ie, there is slip everywhere except for 2 points. Define the slip velocity as us u 0 at r a . The arc length along the top of the cylinder from the forward stagnation point is
s a . Eq(4.35) then gives (with r a ), s s us 2U sin 2U sin a a
(4.37)
dus 2U s cos ds a a
The slip velocity therefore rises from 0 at front stagnation point to a maximum at
2 , then falls to 0 at rear stagnation point. For finite Г, we deal with (4.32), which becomes
a2 w z U rei e i i ln r i r 2 so that
U r
a2 cos r 2
a2 U r sin ln r r 2 and
a2 ur U 1 2 cos r r u
a2 U 1 2 sin r r 2 r
(4.38)
For aerofoil problems, one usually deal with 0 ; in which case, let B
0 2 Ua
(4.39)
Now we search for the stagnation points at which u 0 . On r a , (4.38) gives ur 0 us u 2U sin
U 2sin B 2 a
Thus, for B 2 , there are 2 stagnation points at [see fig.4.4.b],
sin 1
B 2
and
sin 1
B 2
(4.39a)
For B 2 , these 2 stagnation points coalesce into 1 at sin 1 1
2
or
3 . 2
For B 2 , there is no real solution to (4.39a), ie., the stagnation points, if exist, are off the cylinder. This means ur 0 must now be satisfied by setting cos 0 in
the 1st eq in (4.38). We put
3 so that our result may converge properly to that 2
of the B 2 case. The 2nd eq in (4.38) thus becomes
a2 a 2 Ba u U 1 2 U 1 2 0 r 2 r r r ie., r 2 Bar a 2 0 with solutions
r
a B B2 4 2
However, the – root gives r 0 at B 2 so that it must be discarded. Therefore, there is only 1 stagnation point at r
a B B2 4 , 2
3 2
(4.40)
Since we have irrotational steady flow, the Bernoulli theorem applies and we have p 1 2 everywhere. u const 2 Note: if the flow is not irrotational, the above still applies to the surface of the cylinder since it is a streamline. On the cylinder, r a so that u 2 ur 2 u 2 u 2 2U sin 2 a 4U 2 sin 2 2U sin const a The Bernoulli eq. thus becomes p U const 2U 2 sin 2 sin a
2
on
ra
(4.40a)
The net force F on the cylinder is
ˆ a F pndS
2
0
prˆ d
where rˆ i cos j sin Now, (4.40a) is symmetric fore & aft, ie., invariant under transformation .
Since cos changes sign under the same transform, the x-component of F vanishes. For the y-component, we have
Fy a
2
0
p sin d
2 U a const 2U 2 sin 2 sin sin d 0 a (4.41) U
where only the 3rd term in the integrand contributes. Thus, we have no drag but an upward lift L U for
0 , as mentioned in
§1.6.3. If the oncoming stream makes an angle α with the x-axis, the unperturbed complex potential is given by (4.17) as w Uze i . Eqs(4.31,2) are accordingly modified to
i a 2 i w z U ze e z a2 w z U ze i ei i ln z z 2
(4.42)
Fig.4.4 should likewise be turned anticlockwise through angle α to describe the present situation.
4.6
Conformal Mapping
Let w z be the complex potential of some 2-D irrotational flow in the z-plane with w i .
Consider an analytic function
Z f z
(4.43)
with inverse
z F Z
(4.44)
which is also analytic. Then W Z w F Z w z
is analytic.
(4.45)
Let Z X iY
(4.46)
and write
W Z X , Y i X , Y
(4.47)
where X, Y , Φ, Ψ are real. Since W Z is analytic, the Cauchy-Riemann conditions give u* X , Y
X Y
(4.48) Y X where u* and v* are the flow velocity components in the Z-plane. v* X , Y
Thus, dW i u* iv* dZ X X
Similarly, dw u iv dz
Writing
dw dW dz dZ dZ dz
(4.49)
we have
u* iv*
u iv f ' z
(4.50)
where, from (4.43), we’ve used Z ' f ' . Thus, if we want the mapping to preserve the uniform flow at infinity, we must have
f ' z 1.
as
z
Consider a point z0 with Z 0 f z0 . Let f ( n ) z0 be the 1st non-vanishing derivative at z0 .
The Taylor expansion of Z around z0 gives
Z z0 z Z 0
z
n
n!
f n z0 O z
n 1
so that
Z Z z0 z Z 0
z
n
f
n!
n
z0 O z
n 1
(4.50a)
Any complex function f can be written as
f f exp i arg f For the product of 2 complex functions, fg fg exp i arg f arg g
so that
arg fg arg f arg g Applying this to (4.50a) gives arg Z n arg z arg f
n
z0
where we’ve dropped the higher order terms and made use of the fact that the arg of a real function vanishes. Consider now 2 such expansions around z0 , namely, arg Z1 n arg z1 arg f
n
arg Z 2 n arg z2 arg f
n
z0 z0
Subtracting them leads to arg Z 2 arg Z1 n arg z2 arg z1
(4.51)
Hence, for n 1 , the angle between small line segments is preserved under the mapping. Such mappings are called conformal. The Joukowski transformation is given by c2 Z f z z z
(4.52)
Now, c2 2c 2 f ' 1 2 f '' 3 z z so that at z c , we have n 2 . The inverse of (4.52), ie., f 1 Z z , is obtained by solving for z from z 2 Zz c 2 0 which gives
z
1 Z Z 2 4c 2 2
so that f 1 has 2 branches with branch points at
Z 2c . As is the usual practice,
we take the line joining these banch points to be the branch cut. Furthermore, to ensure zZ
as
Z
we choose the + branch so that z
4.7
1 Z Z 2 4c 2 2
(4.53)
Irrotational Flow Past An Elliptical Cylinder
We now apply the Joukowski transformation Z z
c2 z
(4.52)
to the circle z aei with 0 c a .
Thus c 2 i Z X iY ae e a i
c2 c2 a cos i a sin a a ie.,
c2 X a cos a Eliminating
c2 Y a sin a
gives
X2 c2 a a
2
Y2 c2 a a
2
1
which is an ellipse with semi-axis a
(4.54)
c2 . [see fig.4.5.b]. a
The complex potential of a flow passing an elliptical cylinder is therefore obtained by substituting the inverse z
1 Z Z 2 4c 2 2
(4.53)
of (4.52) into the complex potential
a2 w z U ze i ei i ln z z 2 of a flow past a circular cylinder. Here
(4.42) is the angle that the uniform flow makes
with the x-axis. Using
1 2 Z Z 2 4c 2 z Z Z 2 4c 2 2c 2 we have 1 Z Z 2 4c 2 i W Z U Z Z 2 4c 2 e i a 2 e 2 2 2 c i
4.8
1 ln Z Z 2 4c 2 2 2
(4.55)
Irrotational Flow Past A Finite Flat Plate
If we choose c a so that (4.52) becomes Z z
a2 z
(4.56)
The ellipse
c2 X a cos a collapses to X 2a cos
c2 Y a sin a Y 0
which describes a flat plate of length 4a on the X-axis. From
i a 2 i w z U ze e i ln z z 2
(4.42)
we have
i a 2 i dw U e 2 e i dz z 2 z From (4.56), we have dZ a2 1 2 dz z Therefore
dw dW u* iv* dz dZ dZ dz
a2 a2 U e i 2 ei i 1 2 z z 2 z
1
(4.57)
Using z
1 Z Z 2 4c 2 2
(4.53)
eq(4.57) can be rewritten as a function of Z. At the ends of the plate where Z 2a or z a , the flow speed given in (4.57) usually becomes infinite. [see, eg., fig.4.6a] One exception is when the numerator of (4.57) also vanishes, ie., U e i ei i
0 2 a
at
z a
or 2 aiU e i ei 4 aU sin
Thus, to remove the singularity at the trailing edge ( z a ), the circulation must be 4 aU sin
(4.58) in which case, (4.57) is evaluated by the L’Hospital rule to be 2Ua 2ei 2a 2 u* iv* lim i 3 z a 2 z 2 z 3 z
2Uei 2U sin 2 i a a a
1
1
Uei iU sin U cos
so that u* U cos
v* 0
which means the flow leaves the trailing edge horizontally. Note that the singularity at the leading edge ( z a ) is still present. [see fig.4.6.b]
4.9
Flow Past A Symmetric Aerofoil
A mapping of a circular cylinder to an aerofoil is obtained by applying a2 Z z z
(4.56)
to a circle centered at z ,0 , of radius r a , where 0 . Note that the circle goes through the point z a,0 but encloses z a,0 . [see fig.4.7a]
Points on the circle are given by
z a ei where
(4.59a)
is the phase angle [see fig.4.7a].
Putting it into (4.56) gives Z a ei
a2 a ei
(4.59)
The positions of the trailing / leading edges are giving by ei 1 , so that Z a
a2 a
2a a2 a 2 a 2
Turning to the potentials, we must 1st adapt
i a 2 i w z U ze e i ln z z 2
(4.42)
to the present situation. This is easily done by the substitution z z a a whereupon (4.42) becomes 2 a i i wz U z e e i ln z z 2
Doing the same to
dw a2 U e i 2 ei i dz z 2 z gives i a 2 i dw U e e i 2 z dz z
Hence,
dw dW u* iv* dz dZ dZ dz i a 2 i a 2 1 U e 1 2 e i z z 2 z
(4.60)
The most important difference from the flat plate case is that the singularity at z a is inconsequential since it now resides inside the aerofoil. The othe
singularity at the trailing edge, z a , is eliminated as before by setting the numerator to zero, ie., U e i ei i 0 2 a which gives
4 a U sin
(4.61)
When a , the aerofoil given by (4.59) is thin & symmetric. Keeping only 1st order terms in
, (4.59) becomes
Z a ei ae i 1 1 e i a
a ei a e i e 2i 2a cos 1 cos 2 i 2sin sin 2 The length of the aerofoil is L Re Z 0 Re Z 2a 2 2a 2 4a
The thickness of the foil is T Im Z Im Z 2 2sin sin 2
The extremum point satisfies dT 2 2 cos 2 cos 2 0 d ie., cos 2 cos2 1 cos which gives cos
1 1 1 1 8 1 4 2
or 0 2 3
0 T 3 2 3 2 3 3
Hence, the maximum thickness is 3 3 .
(4.61a)
Using (4.61a) and dropping the
term, (4.61) becomes
LU sin
4.10 Blasius’s Theorem Consider a steady irrotational 2-D flow about a fixed body with boundary C. Let w be the complex potential of the flow, Fx , Fy be components of net force on body, then 2
1 dw Fx iFy i dz C dz 2
(4.62)
This is known as the Blasius’s theorem. Proof: Reminder: a complex number z x iy is equivalent to a 2-D vector
x, y .
Consider an infinitesimal segment z on the boundary C. We can write
z z ei ei s where s is the length, and
the angle the segment made with the x-axis.
As s 0 , the segment, which points in the direction
cos ,sin , becomes
tangent to C. The (outward) normal to C is therefore in the direction
sin , cos .
This means the pressure force on the segment is p sin , cos s . [see fig.4.8] In other words,
Fx sin p s
Fy cos p s
so that
Fx i Fy p sin i cos s pie i s
(4.62a)
For inviscid fluid, C is a streamline so that
u u, v q cos ,sin
on C
where q u u 2 v 2 is the speed.
Hence, dw u iv q cos i sin qe i dz
on C.
(4.62b)
Since the flow is steady & irrotational, the Bernoulli theorem applies so that p
1 2 q k 2
where
k const
Eq(4.62a) thus becomes 1
Fx i Fy q 2 k ie i s 2 which, with the help of (4.62b) to eliminate q, becomes 1 dw 2 2i Fx i Fy e k ie i s 2 dz 2
1 dw i i i e s ike s 2 dz Now e i s z x i y
so that 2
1 dw Fx i Fy i ei s ik x i y 2 dz 2
1 dw i z ik z 2 dz Integrating over C then gives the theorem (4.62) since the integration of the 2nd term depends only on the end points and hence vanishes over any closed contour. We leave it as an exercise (Ex.4.5) to show that the moment about the origin is 2 1 dw Re z dz C dz 2
(4.63)
4.10.1 Uniform Flow Past A Circular Cylinder This problem was already treated in §4.5 so it serves as a check on the Blasius’s theorem. The complex potential is
a2 wz U z i ln z z 2
(4.32)
so that
a2 dw U 1 2 i dz z 2 z Applying the theorem gives 2
1 dw Fx iFy i dz C 2 dz dw 2 Res dz U i i U
Hence Fx 0
Fy U
(4.64)
as before.
4.10.2
Uniform Flow Past An Elliptical Cylinder
The complex potential W Z is assumed to be obtained by conformal mapping from the circular cylinder result w z as described in §4.7. We shall calculate the torque using 2 1 dW Re Z dZ ellipse dZ 2
(4.63)
In terms of the circular cylinder quantities: c2 Z z z
(4.52)
dw dW dz dZ dZ dz
dZ
c2 dZ dz 1 2 dz dz z
we have 2
1
2
dW dw dZ dZ dz dZ dz dz Eq(4.63) becomes 2 1 dw dz Re z a Z dz dZ dz 2
For 0 ,
a2 w z U ze i ei z
(4.42)
so that
dw a2 U e i 2 ei dz z 1 2
U 2 Re I
(4.65a)
where c 2 i a 2 i I z z a z e z 2 e
z z a
2
2
c 2 z 2e i a 2ei
z
2
c2 z3
1
c2 1 z 2 dz 2
dz
so the poles are at z 0, c . Since the pole at z 0 is of 3rd order, direct evaluation of the residue there will involve 2nd derivatives of the other parts of the integrand. This is rather tedious. We thus choose an alternative way & evaluate the residue by Laurent’s expansion. Using 1
1 1 c2 1 c2 c4 1 1 2 4 2 2 2 2 2 z c c z c z z
(z c)
the integrand of I becomes
1 2 1 c2 c4 2 4 2 i 2 2 4 2 i 2 z c z e 2a z a e 3 1 2 4 c z z z The coefficient of z 1 is
a2 1 2 2i 2 2 2 i 2 c e 2a c e 2 2 e 2i c c Therefore
a2 a2 2 i I 4 i 2 e 4 i 2 cos 2 4 sin 2 c c so that (4.65a) becomes 2U 2c 2 sin 2
(4.65)
which is clockwise & tends to turn the ellipse broadside to the stream.
4.11 The Kutta-Joukowski Lift Theorem Consider a steady irrotational 2-D flow past a body with contour boundary C. Let the flow at infinity be uniform with u U ,0 , and the circulation around the body be
. Then
F 0, U
(4.66)
This is the Kutta-Joukowski Lift Theorem. Proof: 1st, choose the coordinates so that the body lies entirely inside a circle z R . Assuming the absence of any singularities, the flow
dw is analytic for z R , dz
dw R dw inclusive of the infinity where lim has a Taylor series in U . Hence, z dz dz z so that dw a a U 1 22 dz z z
(4.67)
Consider now the Blasius’s theorem 2
1 dw Fx iFy i dz C 2 dz
(4.62)
dw is analytic for z R , the contour integral is path independent there. dz
Since
Putting in (4.67) gives 2
1 a a Fx iFy i U 1 22 dz C' 2 z z 1 i 2 i 2Ua1 2 2Ua1
(4.68)
where C’ satisfies z R .
Now, (4.67) implies dw dz 2 ia1 C ' dz
Thus
2 a1 wC ' i C '
where
F C
(4.68a)
denotes the difference of the value of F going once around C.
By means of analytic continuation, (4.68a) can be extended to C so that 2 a1 wC i C
Now, since C is a streamline, const there; hence, 2 a1 C u dl C
(4.69)
Eq(4.68) thus becomes Fx iFy i U
which is just the theorem.
4.12 Lift: The Deflection of Airstream We’ve explained the lift of an aerofoil in terms of the pressure difference due to the negative circulation around the aerofoil (K-J theorem). Another way to look at it is to say that the aerofoil imparts a downward momentum to
the oncoming stream & thereby, according to Newton’s law of action-reaction, attains an upward lift. For a single aerofoil, the deflection of the airstream is not apparent since it becomes undetectible at infinity. One thus needs to study the flow past an infinite stack of airfoils for such effects.
5. Vortex Motion 5.1
Kelvin’s Circulation Theorem
5.2
Persistence Of Irrotational Flow
5.3
Helmholtz Vortex Theorem
5.4
Vortex Rings
5.5
Axisymmetric Flow
5.6
Motion Of A Vortex Pair
5.7
Vortices In Flow Past A Circular Cylinder
5.8
Instability Of Vortex Patterns
5.9
Steady Viscous Vortex Maintained By A
Secondary Flow 5.10 Viscous Vortices: Prandtl-Batchelor Theorem
5.1 Kelvin’s Circulation Theorem Theorem Circulation is conserved along particle path for an ideal fluid subject to conservative forces.
Proof Let the circulation be
C t
u dx
(5.1)
To avoid confusion, let us denote derivatives of coordinates by δ and reserve the symbol d for time derivatives. Hence [see Ex.5.2 for a more formal approach],
C t
u x
d du d x x u C t dt C t dt dt
(5.1a)
Now u
d x dx 1 u u u u2 dt dt 2
so that
u C t
d x 1 1 u 2 u 2 C 0 dt 2 C t 2
since u is single valued. Hence, (5.1a) becomes d du x C t dt dt
For an ideal fluid under conservative forces f , the Euler eq. becomes p du u u u dt t where
p
Hence d Ct x C 0 dt
since ξ is single valued.
Notes on the Theorem 1.
The theorem only applies to circuits that ‘move’ with the fluid; not to stationary
2.
ones. The theorem applies whenever we can write du dt
(a)
Hence, the listed assumptions on the fluid can be relaxed as long as (a) is valid. 3.
Similarly, the theorem applies wherever (a) holds on a circuit C; whether it holds for other regions of the fluid is immaterial.
4.
The theorem does not require the fluid domain to be simply connected since we did not invoke the Stokes theorem.
Generation of Lift on Aerofoil As mentioned in §1.6 and Chapter 4, the uplift of an aerofoil requires a negative circulation Γ around the aerofoil. If the aircraft is initially at rest, 0 . If the air is an ideal fluid, Γ will always be zero according to Kelvin’s circulation theorem. Hence, no flight is possible. If air is viscous, a wake of separated boundary layers and vortices will always be present downstream of the aircraft. In particular, if the starting vortex is positive, its shedding will give the aerofoil a negative Γ. [see Fig.5.2]
Encarsia Formosa This tiny wasp employs the Weis-Fogh mechanism of lift generation that works even for inviscid fluid. [see Fig.5.3 & caption for explanation]
5.2
Persistence Of Irrotational Flow
Cauchy-Lagrange Theorem Consider an ideal fluid under conservative forces. Any portion of fluid that is irrotational at some time will remain so afterwards.
Proof Suppose ω u is not identically 0 for that portion V of fluid at some later time. Then, by Stoke’s theorem, there exists a circuit C bounding the surface S of V such that ω n u dx u ndS dS 0 C
S
S
However, this violates Kelvin’s theorem since 0 at some time before. QED.
Comments As discussed in §1.5, the vorticity eq. Dω ω u Dt
(1.25)
simplifies to Dω 0 Dt
(1.27)
for 2-D flows, so that the above result is rather obvious. The usefulness of the Cauchy-Lagrange theorem is that it applies also to 3-D flows. For an irrotational flow, we can write (5.3) u where is single-valued if the flow region is simply connected. If the flow is incompressible, ie., u 0 2 0 [See Ex.5.23-29 for applications in this respect]
(5.4)
5.3
Helmholtz Vortex Theorem
A vortex line is, at one instance, a curve x s x s , y s , z s
whose tangent at each point is parallel to the vorticity vector ω x , y , z u
(5.5)
ie.,
dx
dy ds
x
ds
y
dz
ds
z
or
dx
x
dy
y
dz
z
(5.5a)
A vortex surface is a surface to which ω is tangent at each point. [fig.5.4b] The set of all vortex lines that pass through a closed curve in space forms the boundary of a vortex tube. [See fig.5.4a]
Theorems Consider an ideal fluid under conservative forces so that Kelvin’s circulation theorem applies. The Helmholtz vortex theorems state that 1. Vortex lines, & hence vortex tubes, move with the fluid. 2.
The strength of a vortex tube, as defined by, ω ndS S
(5.6)
is time independent and the same for all cross section of the tube.
Proof of (1) Consider a circuit C that lies, at time t 0 , completely on a vortex surface S. Let C be bounded by a simply connected subsurface S* of S.
The circulation around C is, with the help of Stoke’s theorem, ω n u dx u ndS dS C
S*
S*
which is simply the strength of a vortex tube that goes through C. Since S* is a vortex surface, ω n 0 there. Hence 0
on C
Now, let C t be the circuit composed of, at time t, the same fluid particles as C at t 0 . [In short, C moves with the fluid flow to C t at time t.]
Let t the circulation of C t . According to Kelvin’s theorem, t 0 for all t. Since C is arbitrary, this means ωn 0
on S t
where S t is the surface composed of, at time t, the same fluid particles as S at t 0 . [In short, S moves with the fluid flow to S t at time t.]
Thus, S t is also a vortex surface. Another way to say it is that a vortex surface moves with the flow. A vortex line can be considered as the intersection of 2 vortex surfaces. Since both surfaces moves with the flow, so is their intersection, the vortex line.
Proof of (2) Consider the volume V inside a vortex tube bounded by 2 of its cross sections [see fig.5.4a]. Denote the section of the vortex tube surface involved by ST ; the cross sections by S1 and S2 ; their bounding circuits, C1 and C2 , respectively. The Gauss theorem gives
ST S1 S2
ω ndS ωdV V
Now, since ST is a vortex surface, ω n 0 everywhere on it so
ω ndS 0 ST
Next,
ω u 0 so
ωdV 0 V
Hence, ω ndS 0 S1 S2 Noting that n here are outward normals of the surface bounding V, this becomes 1 2 where i is the strength of the tube at Si , as defined by (5.6). QED.
Thin Vortex Tube Let the cross section S of a thin vortex tube be small enough so that
is
practically constant over it. Thus, S . Since the surface of the tube is a vortex surface, it moves with the flow. Thus, the volume of fluid inside the tube and bounded by 2 cross sections will remain thus bounded during the flow. Since the flow is incompressible, this volume is conserved. Therefore, stretching the tube will decrease S . Since is a constant, must increase to compensate. To summarize, stretching a vortex tube will intensify local vorticity. [see fig.5.5 for application to a tornado; fig.5.6 for that to spin down of a cup of tea].
5.3.1 Vorticity Equation The Helmholtz theorems were originally proved starting with the vorticity equation Dω ω u Dt
(5.7)
Some of the conclusions drawn from the Helmholtz theorems are also obtainable directly through (5.7) One such example is the local intensification of vorticity due to the stretching of a vortex tube. Let the vortex lines be mostly in the z-direction so that ω k and Dω u Dt z
(5.8)
The z-component of which is D w Dt z
Now, stretching the fluid in the z direction makes
w D 0 so that 0 as z Dt
promised. Another simple example is that of 2-D flow. Vortex tubes are all straight & parallel to the z-axis and w 0 . There is no stretching of tubes and D 0 Dt
(5.9)
Still another example is the axisymmetric flow, which, in terms of cylindrical coordinates
R, , z , is of the form
u uR R, z , t e R u z R, z , t e z
Since u is independent of The vorticity is
(5.10)
, the streamlines all lie in planes const .
ω
eR
Re
ez
1 R R uR
0
uR uz e z z R uz
(5.11)
The vortex lines are rings centered around the z axis. Vortex tubes are surfaces of doughnuts similarly centered. According to the 1st Helmholtz theorem, they move with the fluid, which means they can only expand or contract about the z axis. Incompressibility of the fluid means the volume enclosed by each tube is conserved. Let the radius of a thin tube be R, its cross section S , so that its volume is 2 R S . As R varies in time with the flow, S also changes to maintain a constant volume. However, according to the 2nd Helmholtz theorem, S const so that const R In more precise language, we write D 0 Dt R
(5.12)
which can be deduced more formally from (5.7) using (5.10). When, in axisymmetric flow, an isolated vortex tube surrounded by irrotational flow is called a vortex ring. [see fig.5.7].
5.3.2 Proof of eq(5.12) Using
u uR e R u z e z u R, z , t ω e
where
uR uz R, z , t z R
we have
ω u
uR e R u z e z uR e R R
where we’ve used e R e Thus, the vorticity eq (5.7) becomes Dω u R e Dt R
Writing Dω D D e Re Dt Dt Dt R D D Re Re Dt R R Dt
Now D R e u Re Dt
uR u z Re z R u R e
so that Dω D Re u R e u R e Dt Dt R R R
ie., D 0 Dt R
5.4 Collision Leapfrog
Vortex Rings
5.5
Axisymmetric Flow
5.5.1 The Stokes Stream Function 5.5.2 Irrotational Flow Past A Sphere 5.5.3 Hill’s Spherical Vortex
5.5.1 The Stokes Stream Function For an incompressible 2-D flow,
u e z
(4.8)
ensures that the incompressibility condition u 0 is automatically satisfied while the stream function x, y , t is constant along streamlines, ie.,
u
0
(4.7)
The question is whether an analogous form u ' e , with ' ' R, z, t , exists for axisymmetric incompressible flows. Now
u ' uR
u uz ' R R z
uR uz ' z R
Using eR u
1 R R 0
Re
ez
' ' ' eR ez z z R R R ' 0
we have ' ' ' ' ' ' ' z R R R z R z so that ' is not generally constant along streamlines.
u '
Now, the offending term originates from the curl of ' in the expression of u. It is easily removed by setting ' u e R
so that R
(5.13)
eR 1 R R 0
Re
ez eR ez z z R 0
whereupon
u
0 z R R z
ie., the Stokes stream function is constant along streamlines.
In spherical coordinates er 1 A 2 r sin r Ar
r, , , re rA
r sin e r sin A
The analog (5.13) of is obviously u e r sin so that
u
(5.14)
er
re
r sin e
1 r sin r 0
0
2
1 er e r sin r r
while r, , t is constant along streamlines:
u u r r r sin u ur r r 1 1 0 r sin r r r r
u ur
(5.15)
5.5.2 Irrotational Flow Past A Sphere In a steady flow, the vorticity eq. for axisymmetric flow, D 0 Dt R
(5.12)
reduces to
u
0 r sin where R r sin .
Thus
r sin
(5.16)
is constant along a streamline.
Consider a uniform inviscid flow past a rigid sphere with surface r a . If there are no closed streamlines, all streamlines will start and end at the infinities, where 0 . Hence, according to (5.16), 0 everywhere and the flow is irrotaional. Now, for an axisymmetric flow in spherical coordinates:
u ur r, , t e r u r, , t e er 1 ω u 2 r sin r ur
re ru
r sin e 0
1 u ru r e e r r
where 1
u
u
u
1 u
r ru r r r r r r
(5.17)
In terms of the Stokes stream function Ψ defined by 1 u e er e r r sin r sin r we have
(5.14-5)
1
1 2
1 1
r sin r 2 r 2 sin
(5.18)
Thus, for our irrotational flow past a sphere, we need to solve
2 sin 1 2 0 r 2 r sin with boundary conditions ur 0 on r a
ra
(5.19)
[impenetrable wall]
(5.20a)
and ur U cos , u U sin
as r
which, by (5.15), means 1 U cos , r sin
1 U sin r sin r
r 2U sin cos ,
rU sin 2 r
2
as r
or as
r
(5.20b)
which, upon integration, becomes
1 2 r U sin 2 C1 r 2
1 2 as r r U sin 2 C2 2 In order for these 2 expressions to be compatible, the arbitrary functions C1 , C2 must
be constants that can be conveniently set to zero. Hence
1 2 r U sin 2 2
as
r
(5.20)
Eq (5.19), as well as the boundary conditions (5.20,a,b), suggests that we can look for a separable solution, ie.,
r, R r whereupon, (5.19) becomes 2 ODEs d 2R R0 dr 2 r 2 d 1 d sin d sin d where α is a constant.
(5.21a) (5.21b)
In order to satisfy (5.20), we have R r
1 2 rU 2
sin 2
(5.21c)
Substituting into (5.21b), we have L.H .S . 2sin
d cos 2sin 2 d
R.H .S . sin 2 so that (5.21b) is satisfied only if 2 ; whereupon, (5.21a) becomes
d 2R 2 R0 dr 2 r 2 Since this is homogeneous, we look for a power solution
(5.21d)
R rn which turns (5.21d) into
n n 1 2 0 with solutions n 2 , or 1 so that R Ar 2
B r
with A, B constants.
From (5.21c), we have 1 A U 2
so that B 1 r, Ur 2 sin 2 r 2
Finally, to satisfy (5.20a), we have 1 1 1 2 B 2B 2 Ur 2sin cos U 3 cos r sin r sin 2 r r so that at r a ur
2
2B U 3 cos 0 a which gives 1 B Ua 3 2
so that
1 2 a3 2 r, U r sin 2 r
for r a
(5.21)
a3 ur U 1 3 cos r 1 U sin a3 a3 u 2r 2 U 1 3 sin r sin r 2r r 2r The streamlines const are shown in Fig.5.12a. The slip velocity on the sphere surface is
ur r a 0 3 u r a U sin 2
(5.22)
Thus, the fore & aft points 0, are stagnation points. Maxima are at
2
.
According to the Bernoulli theorem, this implies a severe adverse pressure gradient over the downstream side of the sphere. ( 0 boundary layer is expected.
2
). A large wake with separated
5.5.3 Hill’s Spherical Vortex In §5.5.2, we solved the steady axisymmetric flow outside a solid sphere r a . Now, let the surface r a to be an imaginary one in the fluid. We shall show that there is a flow pattern, called the Hill’s spherical vortex, inside the sphere that can match up at r a with the one given in §5.5.2. In other words, one can find vortices embedded in a steady, uniform flow. [see Fig.5.12] The problem is to solve
u
0 r sin with the boundary conditions
for
ra
(5.16)
ur r a 0 3 u r a U sin 2
Now, (5.16) means that
r sin
(5.22)
is a constant along each streamline, a property
shared by the Stokes stream function Ψ. Hence, (5.16) implies
r sin
c
where c is an arbitrary function. Using the spherical coordinate relation (5.18) between ω and Ψ, we have
1
1 2
1 1
c r sin r sin r 2 r 2 sin or
2 sin 1 2 2 2 c r sin r 2 r sin Now, in terms of , (5.22) become 1 0 2 a sin r a
(5.23)
1 3 U sin a sin r r a 2
(5.24)
Now, the easiest way to match (5.24) is to try the same type of ansatz as in the r a region, namely,
r, g r sin 2
for r a
Plugging into (5.23) gives sin 2
d 2g g 2 2 sin 2 c r 2 sin 2 2 dr r
or 1 d 2g g 2 4 c 2 2 r dr r
(5.24a)
Since the R.H.S. is independent of θ, we must have c const . The homogeneous eq to (5.24a) the same as the radial eq. for r a . Thus, the homogeneous solution is g Ar 2
B r
To find the particular solution, we again try a power solution g rn so that (5.24a) becomes
n n 1 2 r n 4 c which gives n4
and
c 10
Hence, the general solution for (5.24a) is g Ar 2
B c 4 r r 10
so that B c r, Ar 2 r 4 sin 2 r 10
Now, to keep Ψ regular at r 0 , we must set B 0 so that c r, Ar 2 r 4 sin 2 10
and c 2 2 c 4 Ar r sin cos 2 A r cos 10 10 1 4c 3 2 1 2 u 2 Ar r sin 2 A cr sin r sin 10 5
ur
2 r sin 2
Matching the conditions (5.24) at r a requires c 2 A a 2 cos 0 10
and 1 3 2 A ca 2 sin U sin 5 2
ie., A
c 2 a 0 10
1 3 A ca 2 U 5 4
with solution 3 A U 4
c
15U 2a 2
so that
3 r2 r, Ur 2 1 2 sin 2 4 a
for r a
(5.25)
The corresponding streamlines, which are all closed, can be found in Fig.5.12b. The circulation around each streamline differs for different streamlines. The maximum is the streamline that goes around a full hemispherical cross section. By Stokes theorem, max
a
0
0
dS d drr
hemi
Using
cr sin
15U r sin 2a 2
we have
max
15U 2a 2
0
a
d drr 2 sin 0
a3 15U 2 5Ua 2a 2 3
Thus, a Hill spherical vortex will travel through stationary fluid with a speed U
max 5a
5.6
Motion Of A Vortex Pair
A line vortex is defined as u
e 2 r
Consider a vortex pair of strength and a distance 2d apart as shown in Fig.5.13a. Each will induce a downward flow / 4 d at the instantaneous position of the other. Thus, the pair, as well as the accompanying streamline pattern, moves downward at that speed. In a reference frame moving with the vortices, an additional uniform upward flow / 4 d is experienced. The full complex potential is therefore w
iz i i ln z d ln z d 4 d 2 2
(5.26)
When calculating the velocity of one of the vortices, the contribution of that vortex to w must be subtracted. Thus, the velocity U ,V of the vortex at z d is given by
U iV
d iz i ln z d dz 4 d 2 z d i i 0 4 d 2 z d z d
(5.27)
ie., it’s stationary as stated before. Separating the real and imaginary parts in (5.26), we have w
y zd x i ln 4 d 2 z d 4 d 2
where zd z d i e zd zd
Thus, the stream function is
Now,
x zd 2 ln 4 d z d
z d x iy d z d x iy d
x d iy x d iy x d y 2 zd zd x d iy x d iy x d 2 y 2 2
2
so that
4
x x d 2 y 2 ln 2 2 d x d y
The corresponding streamlines are sketched in Fig.5.13b.
5.7
Vortices In Flow Past A Circular Cylinder
Consider a circular cylinder of radius a initially at rest in a fluid of kinematic viscosity ν. Let it be suddenly set in uniform motion with speed U perpendicular to its axis so that the Reynolds number R
2aU
(5.28)
is about 200.
5.7.1 The Initial Phase: Almost Irrotational Flow 5.7.2 Flow At Later Stage: The von Karman Vortex Street 5.7.3 A Simple Model
5.7.1 The Initial Phase: Almost Irrotational Flow If the fluid were inviscid, our 2-D flow would be governed by D 0 Dt
(5.9)
Since the fluid is initially at rest, 0 . According to (5.9), it will remain so for all subsequent time so the flow will always be irrotational. Consider now our viscous fluid. During a very short initial phase before the cylinder has moved a distance comparable to its radius, the flow is predominantly irrotational (see fig.4.4a). There is intense vorticity in the rapidly thickening boundary layer. However, there is not yet time for separation to occur so the mainstream is still free of vorticity. At this stage, results of the (inviscid) irrotational flow give the velocity at the edge of the boundary layer. Positions at which reversed flow 1st occur are those where the velocity decreases most rapidly with distance along the boundary in the flow direction. In case of a circular cylinder, this happens at the rear stagnation point. (see Fig.5.14a)
5.7.2 Flow At Later Stage: The von Karman Vortex Street Once reversed flow is initiated, the flow begins to differ substantially from the irrotational results. 1st, the 2 attached eddies behind the cylinder grow in size (Fig.5.14b). Then the flow becomes asymmetric about the centerline (Fig.5.14c). Finally, it settles into an unsteady but highly structured flow in which vortices are shed alternatively from the 2 sides of the cylinder, giving rise to the von Karman vortex street (Fig.5.14d,e). Note that the wake is entirely turbulent for R 2000 while for R 30 , the 2 eddies remain symmetrically attached.
5.7.3 Von Karman Vortex Street: A Simple Model We now model a fully formed vortex street (Fig.5.14d,e) by 2 sets of line vortices as shown in fig.5.15. Members of the lower set of vortices, all of strength Г, are at z na , with n 0, 1, 2,. 1 Those of the upper set, of strength , are at z n a ib . 2
As in §5.6, we assume each line vortex to move with the local velocity due to the combined flow of everything other than itself. Consider any vortex. Only the x-component U of velocity due to vortices in the other row is finite. All else vanishes since contributions from other vortices cancel in pairs. By symmetry, U is the same (and points to the left for 0 ) for all vortices. Since the complex potential w z due to a line vortex at z0 is w z
i ln z z0 2
(4.21)
the total complex potential due to the lower line of vortices is w z
i 2
i 2
i 2
i 2
i 2
ln z na
n
1 ln z ln z na ln z na n n 1 1 ln z ln z na ln z na n n 1 2 2 2 ln z ln z n a n 1 z2 ln z ln 1 ln n 2a 2 2 2 n 1 n a n 1
i z 2 ln z 1 C 2 n 1 n 2a 2
where C
i 2
ln n a 2
2
n 1
is a constant and can be dropped. Using the identity (see Carrier et al) z2 sin z z 1 2 n n 1
we have w
i z ln sin const 2 a
(5.29)
Hence dw i z cot dz 2a a
1 For an upper row vortex at z n a ib , we have 2 dw i 1 b i b b cot n i tan i tanh dz 2a 2 a 2a 2a a a
Thus, the whole street moves with velocity b u tanh ,0 a 2a
5.8
Instability Of Vortex Patterns
Consider 2 line vortices of equal strength Г and distance 2a apart. Each vortex induces, at the center of the other, a flow of speed
perdendicular 4 a
to the line joining them. Hence, both rotates about the mid-point of that line with angular velocity / 4 a 2 . Next, consider n line vortices of strength Г placed symmetrically on a circle of radius a. (see Fig.5.16) The angular separation between adjacent vortices is distance between the mth neighbors is therefore d m 2a sin with m
m 2
2 . The n
, where m
2 m n
n , and the floor a of a is the largest integer that is equal or less than a. 2
Note that all neighbors occur in pairs except for the
n th one, which is unique, when 2
n is even. Consider a vortex and the pair of its mth neighbors. The line joining a vortex to its
m
mth neighbor makes an angle
2
with the radius that passes through the vortex.
The flow at the vortex, induced by that neighbor, then makes an angle with the radius.
m 2
m n
By symmetry, the radial components of the flows induced by a pair
of the neighbors cancels each other. The tangential component is equal to 2 sin m 2 d m 2 2 a For n even, there is only one
n th neighbor situated at the other end of the diameter 2
that goes through the vortex. The induced flow is again tangential with magnitude . 4 a
For n odd, there are
n 1 / 2
pairs of neighbors so that the total flow is
n 1 . n 1 4 a 2 2 a
For n even, there are n / 2 pairs and a unique
n th neighbor. The total flow is 2
n n 1 4 a 2 2 a 4 a In both case, the vortex rotates with angular velocity n 1
4 a 2
(5.32)
Stability of a dynamical system can be investigated by the introduction of a small perturbation. If the state of the system remains close to the unperturbed one, it is stable. Otherwise, it is unstable. There are cases that do not quite fit in this scheme of classification. For example, some systems are stable under infinitesimal perturbation but unstable under finite perturbation. Others can be stable under any instantaneous perturbation but unstable under a prolonged one. These cases are sometimes called metastable. An introduction to the subject can be found in chapter 9. Here, we list, without proof, some results of interest concerning the stability of vortex patterns. 1.
The double row vortex pattern in fig.5.15, §5.7.3, is unstable except for the case cosh
2.
b a
2
ie.,
b 0.281 a
(5.31)
The equally spaced vortices on a circle pattern is stable for n 7 , unstable for n 7 and metastable for n 7 . This is observed in liquid helium.
Finite Patches of Concentrated Vorticity An elliptical path of uniform vorticity ω rotates with angular velocity ad (5.33) 2 a b where a and b are the semi-axes of the ellipse. This motion is stable for
1 b 3. 3 a
A circular array of n patches of vorticity is unstable even for n 7 , if the patches are big enough. The critical patch size decreases with increasing n.
For the finite patch version of the vortex street, there is again only 1 spacing ratio for which the pattern is stable. This ration is close to the von Karman value and only weakly dependent on the patch size. A patch of 2-D vorticity distribution v u x y is still governed by the vorticity eq. D u v 0 Dt t x y For incompressible flow, we also have u v 0 x y
(5.9)
(5.35)
Now, (5.9) implies ω is conserved for a fluid element while (5.35) implies the cross-section of that element in the x-y plane is conserved. Thus
dS const
(5.36)
where the integration is over the whole flow plane. There are other relationships of this kind:
xdS const
ydS const
(5.37)
Vortex Merging Let 2 circular patches of uniform and equal vorticity, each of radius R, have centers a distance d apart at t 0 . For
d 3.5 , the patches end up deformed and rotating about a common center. R
For
d 3.5 , the patches quickly merge. To satisfy the conservation laws, they do R
this by wrapping around each other with a layer of irrotational flow between them. When the vorticity are of opposite sign, a vortex pair may be formed.
5.9
Steady Viscous Vortex Maintained By A
Secondary Flow In the presence of viscosity, the vorticity eq. becomes ω ω u ω ω 2 (5.38) u t The processes controlled by the various terms are roughly: ω convection of vortex lines. u :
ω u :
intensification of vorticity (when vortex lines are stretched).
2 ω :
diffusion of vorticity.
Burgers Vortex One known situation that involves all 3 processes and still admits an exact solution to the Navier Stokes eqs is known as the Burgers vortex. Basically, it is a line vortex whose viscous decay (§2.4.4) is countered by a secondary flow which 1.
sweeps the vorticity back towards the axis.
2.
intensifies the vorticity by stretching fluid element along the axis direction.
The result is a steady vortex of the form 1 ur r 2
u
2 1 e r / 4 2 r
where 0 and Г are constants. The velocity profile is shown in fig.5.18. The vorticity
uz z
(5.39)
ω
1 r
er
re
ez
r 1 r 2
z
2 1 e r / 4 2
z
r 2 / 4 e ez 4
is concentrated within a core of radius of order
(5.40)
.
In (5.39), the secondary flow, as parametrized by α, and the rotary flow, as parametrized by Г are not coupled since both parameters can vary independently. For a real flow, the secondary and rotary flows are usually coupled due to the presence of rigid boundaries.
5.10 Viscous Vortices: Prandtl-Batchelor Theorem For a 2-D steady inviscid flow, the vorticity eq. reduces to
u 0
ω k
where
ie., ω is constant along a streamline so that we can write
(5.41)
since the stream function Ψ is, by definition, constant along a streamline. If, as in fig.1.8, all flow lines can be traced to the infinities where the flow is uniform, we can deduce immediately that 0 everywhere. If closed streamlines are present, no simple conclusion can in general be drawn for inviscid flows. For steady viscous flow, the Navier-Stokes eq become p u u 2u
Using u u
1 u 2 u u 2
u u 2u we have 1 2 p u u u u u 2 or 1 2
u u u Integrating over a closed contour C, we have
u u 2
p
u dx u u dx C
C
If C is a streamline, dx is parallel to u so that the right hand side vanishes. Thus
C
u dx 0
if
0
ie.,
ω dx 0
(5.42)
C
For our 2-D flow, i ω x 0
j y 0
k , ,0 z y x
d , ,0 d y x d d u, v,0 u d d
so that (5.42) becomes d C u dx 0 d
(5.43)
or d C 0 d where C.
d can be taken outside the integral because ψ is constant on the streamline d
In case of a finite circulation C , we have d 0 d
which is the Prandtl-Batchelor theorem: in a steady 2-D viscous flow, the vorticity is constant throughout any region of closed streamline in the limit 0 .
6. The Navier-Stokes Equations 6.1
Introduction
6.2
The Stress Tensor
6.a Navier-Stokes Eqs.
6.B
Convection Theorem
6.2
The Stress Tensor
Let x be the position vector of some point fixed in the fluid, and S a small geometrical surface element, with unit normal n, that passes through x. Let the force exerted on this surface by the fluid on the side pointed by n be (6.1) t S where t is called the stress vector. For an inviscid fluid, we have
t p x, t n so that the stress aligns with the unit normal. For viscous fluid, we expect t to have non-vanishing components tangent to the surface. This is described by the stress tensor T whose Cartesian component Tij is defined as the i-component of stress on a surface with unit normal n e j .
Thus, for a surface with unit normal n n j e j , the stress vector is ti Tij n j
or
t T n
For a more fastidious proof, see Acheson.
(6.3)
6.3
Cauchy’s Equation Of Motion
Convection Theorem Ref:
R.E.Meyer, “Introduction to Mathematical Fluid Dynamics”, Wiley (71)
Theorem 3.1 If f x, t C 1 t , D Dt
z
t
fdV
z FGH t
Corollary 3 Let 1 (t t1 ) , D Dt
z
t
fdV
t
z
IJ K
Df f v dV Dt
1
fdV
z
1
Proof t H t 0 So that
z
t
fdV
z
0
g( a, t ) J ( a, t )dV0
where g a, t f x a, t , t a x (a,0)
J det
Thus
xi a j
fv ndS
(3.5)
(3.6)
D Dt
z
t
fdV
t
z z z FGH z FGH z FGH
0
0
t
t
gJ dV0 0 t
b g Df J I J f J dV Dt t K Df J I f J dV Dt Jt K Df I f vJ dV K Dt
gJdV0
0
J J v t
if
Now: J ik
xi x Akn n Ank an ai
where Akn are the cofactors (Caution: our Aij is Meyer’s Aji ), Akn ( ) k n kn
where kn is the determinant of the minor matrix obtained by striking out the kth x row & nth column of the matrix i . a j
Writing the ith row of J as a vector ri with components
ri j , we have
det r1 , r2 , r3 det r1 , r2 , r3 det r1 , r2 , r3 det r1 , r2 , r3 A1k r1 k A2 k r2 k A3k r3 k where ∂ is the differential operator, ie., x J Ajk j ak [see Determinant.doc for a better derivation] Hence:
v J x v x v Akn k Akn k Akn k i J ik k J v t t an xi an xi an For the corollary:
QED
Df f f v fv Dt t
b g
so that
z
1
b g
fv dV
z
1
b fvg ndS
6.4
A Newtonian Viscous Fluid: The
Navier-Stokes Equations
6.a Navier-Stokes Eqs. Consider the Euler eq D u Du u u p Dt Dt Using D u u u u u u u u Dt t
u uu t
we can re-write it as u uu p 0 t or u 0 (A) t where the momentum flux density tensor П is ij ui u j p ij
Eq(A) is just the “eq of continuity for momentum” which expresses the conservation of momentum. Since we expect the principle of the conservation of momentum to be valid even in viscous fluids, eq(A) should apply there with a suitable modification of П. We thus set ij ui u j ij
where the stress tensor ij is related to the viscous stress tensor 'ij by
ij p ij 'ij [Note: Acheson used Tij instead of ij (see chap 6) ] What this means is that the generalization of the Euler’s eq to viscous fluids can be done by replacing p with p ' so that, for example,
Du p Dt
becomes
Du p ' Dt
or u u u p ' t
(B)
Next, we need to determine the form of ' . Now, viscous effects are due to “friction” between adjacent fluid elements moving ui with different velocities. Hence ' should depend on but not on u itself. x j Furthermore, there’s no friction if the fluid rotates as a whole with uniform angular Ω r velocity Ω so that u or ui ijk j xk . Now: ui u ijk j km ijm j mji j m xm xi
and ui iji j 0 xi
Hence, the choice ui u j u b k ij ' ji x j xi xk
'ij a
where a, b are constants, will satisfy the above requirements. A more common form is ui u j 2 uk u k ij x j xi 3 xk ij xk
'ij
(C)
where the positive constants
a
2 3
2 3
b ab
are called coefficients of viscosity. (Landau used η instead of μ) This also defines the Newtonian fluid. [cf eq(2.1)]. Using (C), (B) becomes
ui u p ' ji uj i t x j xi x j
p xi x j
u u j 2 u u k ij k ij i xk x j xi 3 xk
2 p 3
uk x j xk
ui u j x j xi
xi
ui 1 u p k xi x j x j 3 xk
or in vector form: u u u p ' t 1 p 3
1 p 3
2 u u
2 u u
(D)
For incompressible fluids, we have the Navier-Stokes eqs.: u u u p 2u t u 0
Note also that for incompressible fluids, ui u j x j xi
'ij [cf (2.1)]
(2.3)
6.B
Convection Theorem
Ref:
R.E.Meyer, “Introduction to Mathematical Fluid Dynamics”, Wiley (71)
Theorem 3.1 If f x, t C 1 t , D Dt
z
t
fdV
z FGH t
Corollary 3 Let 1 (t t1 ) , D Dt
z
t
fdV
t
IJ K
Df f v dV Dt
z
1
fdV
z
1
(3.5)
fv ndS
(3.6)
Proof t H t 0 So that
z
t
fdV
z
0
g( a, t ) J ( a, t )dV0
where g a, t f x a, t , t a x (a,0)
J det
Thus D Dt
z
t
xi a j
fdV
t
z z z FGH
0
0
gJ dV0 0 t
b g Df J I J f J dV Dt t K
gJdV0
0
z FGH z FGH
t
t
IJ K Df I f vJ dV K Dt
Df J f dV Dt Jt J J v t
if
Now: J ik
xi x Akn n Ank an ai
where Akn are the cofactors (Caution: our Aij is Meyer’s Aji ), Akn ( ) k n kn
where kn is the determinant of the minor matrix obtained by striking out the kth x row & nth column of the matrix i . a j
Writing the ith row of J as a vector ri with components
ri j , we have
det r1 , r2 , r3 det r1 , r2 , r3 det r1 , r2 , r3 det r1 , r2 , r3 A1k r1 k A2 k r2 k A3k r3 k where ∂ is the differential operator, ie., x J Ajk j ak [see Determinant.doc for a better derivation] Hence:
v J x v x v Akn k Akn k Akn k i J ik k J v t t an xi an xi an For the corollary: Df f f v fv Dt t
b g
so that
z
1
b g
fv dV
z
1
b fvg ndS
QED
7. Very Viscous Flow 7.1
Introduction
7.2
Low R Flow Past A Sphere
7.3
Corner Eddies
7.4
Uniqueness & Reversibility of Slow Flows
7.5
Swimming At Low R
7.6
Flow in a Thin Film
7.7
Flow in a Hele-Shaw Cell
7.8
Adhesive Problem
7.9
Thin Film Flow Down a Slope
7.10 Lubrication Theory
7.1
Introduction
For a steady viscous flow, the Navier-Stokes eqs become 1 (7.1) u u p 2u g
The term ‘very viscous flow’ is used to denote situations in which the convective
u u
(inertia) term
is negligible, whereupon (7.1) is linearized.
There are 2 rather different ways to achieve this, as described in the following.
Slow Flow Equations This happens when R
UL
1
(7.2)
Since
U2 u u O L
U 2 L
2 u O
we have u u O UL O R 1 2 u Furthermore, if gravity is negligible, (7.1) simplifies to (7.3) 0 p 2u which, together with the incompressible condition u 0
(7.3a)
are known as the slow flow eqs.
Thin Film Equations This applies to motion of thin films of fluid. If the thickness of the film is sufficiently small, the velocity gradient across it may be enhanced to a level where the convective term is overcome by the viscous term, even though R is quite large in the conventional sense.
The governing eqs, called the thin-film eqs, willl be derived in §7.6.
7.2
Low R Flow Past A Sphere
We now seek a solution to the slow flow eqs (7.3) for the uniform flow past a sphere of radius a.
7.2.1 Equation for the Stokes Stream Function 7.2.2 Boundary Conditions 7.2.3 Solutions 7.2.4 Drag 7.2.5 Further Considerations 7.2.6 Navier-Stokes Eqs For Ψ 7.2.7 Matched Asymptotic Expansions
7.2.1 Equation for the Stokes Stream Function Let the uniform flow be in the z direction. The flow is axisymmetric about the z-axis, so that in terms of spherical coordinates
r, , , it takes the form u ur r, , u r , ,0
The incompressibility condition u 0 is u e r sin so that
u
(5.14)
er
re
r sin e
1 r sin r 0
0
2
1 er e r sin r r
(5.15)(7.6)
and
u
1 r sin 2
er
re
r sin e
r
1 sin r
1 r sin r
1 1 2 1 1 e r sin r 2 r 2 sin e
1 E 2 r sin
where
E2
2 sin 1 2 r 2 r sin
2 Q2 r 2 r 2
0
and Q 2 sin
1 sin
Hence er 1 u 2 r sin r 0
1 r sin 2
re 0
r sin e E 2
2 e E e r E 2 r r
Using
u u 2u 2u eq(7.3) can be written as
p u
e r E 2 e r E 2 r sin r 2
ie., p 2 E 2 r r sin 1 p 2 E r r sin r
Eliminating p gives
1 1 1 2 2 E E 2 0 2 2 r sin sin r or
2 sin 1 2 r 2 r 2 sin E 0 E 2 E 2 0 2
2 sin 1 r 2 r 2 sin 0
(7.7)
7.2.2 Boundary Conditions The no-slip boundary condition means 1 ur a , 2 0 a sin r a
u a,
1 0 a sin r r a
ie.,
0 r a r r a
(BC1)
As the flow becomes uniform at infinity, ie., u Ue z for r we have
ur , U cos
u , U sin
so that 1 U cos r r sin
lim
1 U sin r r sin r
lim
2
(7.7a)
which is possible only if Ψ is of the form
r 2 f whereupon (7.7a) become 1 f ' U cos sin
and
2 f U sin sin
ie., 1 f U sin 2 2
1 2 r U sin 2 2
for r
(BC1a)
Finally, u 0 on the surface of the sphere so that every curve on it is a streamline with zero flow velocity. Thus, Ψ is a constant on r a and, for convenience, we can set
a, 0
(BC1b)
7.2.3 Solutions In view of the form of (7.7) & (BC1), a simple ansatz could be
f r sin 2 To see if it works, we calculate: Q 2 sin
1 cos 2 2 f sin sin
d 2 2 2 2 Q2 2 2 E 2 2 f ''sin 2 2 2 f sin r r r r dr 2
2
2 Q2 E E 2 2 r r 2
2
2 Q 2 d 2 2 2 2 2 2 2 f sin r dr r r
d 2 2 2 2 2 r dr
f sin 2
so that (7.7) is satisfied if 2
d2 2 dr 2 r 2 f 0
(7.7b)
with the boundary conditions
f a f ' a 0
lim f r
1 2 rU 2
(BC2)
(BC2) suggests the ansatz f r so that
d2 2 2 2 2 f 1 2 r dr r 2
d2 2 4 dr 2 r 2 f 1 2 2 3 2 r
Thus, (7.7b) is satisfied if 1 2 2 3 2 0
which means
1 2 0
2 3 2 0
or
so that 2
1
1 1 8 2 1
or
4
1
5 25 16 2 1
A general solution to (7.7b) is therefore f
A Br Cr 2 Dr 4 r
where A, B, C, D are constants. Putting in (BC2), we have 0
A Ba Ca 2 Da 4 a
0
A B 2Ca 4 Da 3 2 a
D0
1 C U 2
Substituting the last 2 into the first 2 gives A 1 Ba Ua 2 a 2
A B Ua a2
which gives A
1 1 3 1 3 1 Ua Ua 2 2 4
B
1 1 2 3 1 Ua Ua 2a 2 4
Hence
f
U a3 3ar 2r 2 4 r
(7.8a)
U a3 2 2 3ar 2r sin 4 r
See Fig.7.2 for the corresponding streamlines.
(7.8)
7.2.4 Drag Now, from (7.8a),
d2 2 3Ua U a3 3a f 2 2 0 2 2 2 2 2 2 3 r 4 r r dr 2r so that E 2
3Ua 2 sin 2r
3Ua E 2 sin cos r 3Ua E 2 2 sin 2 r 2r
and p 3Ua 3 cos r r
(7.8a)
1 p 3Ua 3 sin r 2r
(7.8b)
Integrating (7.8a) gives p
3Ua cos c 2r 2
Taking the partial with respect to θ: p 3Ua 2 sin c ' 2r
which, on comparison with (7.8b), gives c' 0 ie., c const . Thus, (7.8c) becomes p
3Ua cos p 2r 2
where p p r .
Since the unit normal of the spherical surface is n er
(7.8c)
the stress vector t T n can be written in spherical coordinates as t tr , t , t Trr , T r , T r
where, according to eq(A.44) Trr p 2
T r r
ur r
u r r
ur u u 1 ur r r r r
Tr 0
Now, from
U a3 2 2 3ar 2r sin 4 r
(7.8)
we have
ur
1 U a 3 2 3ar 2r 2 cos 2 r sin 2r r
1 U a3 u 2 3a 4r sin r sin r 4r r so that, on r a , ur u 0 as expected. Further differentiation gives
ur U a 3 a 3Ua a 2 3 4 3 2 cos 2 1 2 cos r 2 r r 2r r ur U a3 2 3ar 2r 2 sin 2r r u U a3 a 3Ua a 2 3 4 3 2 sin 2 1 2 sin r 4 r r 4r r which, on r a , become
ur ur 0 r
u 3U sin r 2a
Hence, on the surface of the sphere, the stress vector is tr p t
3U cos p 2a
3U sin 2a
t 0
The net force on the sphere should, by symmetry, in the direction of the uniform stream (z-axis). The corresponding component of t being t z tr cos t sin
3U 3U cos2 p cos sin 2 2a 2a
3U p cos 2a
The drag on the sphere is therefore 2 1
D
t a d cos d 2
z
0 1
2 a
1
2
t d cos z
1
1
3U 2 a 2 p cos d cos 2a 1 3U 2 a 2 6 aU a
(7.9)
This formula was confirmed experimentally by measuring the terminal velocity U T of a steel ball dropped into a pot of glycerine. Thus, the viscous drag exactly balances the buoyancy reduced pull of the gravity so that 6U T a
4 3 a shere fluid g 3
7.2.5 Further Considerations The above theory was due to Stokes. It was found to be flawed since, for example, its application to the 2-D case fails in the sense that the solutions it provides cannot satisfy all the required boundary conditions (see Ex.7.4).
The trouble lies in the fact that the non-linear term
u u
dropped to obtain the
slow flow eqs (7.3) is not negligible in the transitional region where the deflection due to the sphere begins to die out and the flow is more or less, but not quite, uniform. One way to handle the problem is by means of the mathched asymptotic expansions due to I. Proudman and J.R.A. Pearson. [J.Fluid.Mech. 2, 237-62 (1957)].
7.2.6 Navier-Stokes Eqs for Ψ
7.2.6 Navier-Stokes Eqs for Ψ We now re-write the full steady state Navier-Stokes eqs 1 u u p 2u
in terms of the Stokes stream function Ψ. Using u u
1 u 2 u u 2
u u 2u 2u we have 1 p u 2 u u u 2 1
Using u e
1 E 2 r sin
we have er u u ur
e u
0
0
e r er
e u 1 E 2 r sin
u ur E 2 e E 2 r sin r sin
1 2 1 2 E e 3 2 E 2 r sin r r sin 2
1 1 2 er e 2 2 E r r sin r
From §7.2.1, we have u
Hence
1 r sin 2
2 2 e r E e r r E
1 p u2 2 1 1 2 2 e r sin r sin E 2 e r sin r r r 1
ie., 1 1 2 1 2 sin p u 2 2 E r 2 r sin r 1 1 2 1 1 r sin E 2 p u 2 2 r 2 r r sin r
Eliminating p
1 2 u by cross differentiation gives 2
2 1 1 2 1 r 2 sin 2 r sin E r r sin 2 r r sin r E
Terms proportional to ν were already worked out in §7.2.1. Thus
sin
1 2 1 2 r 2 sin 2 r E r r 2 sin 2 E 1 1 2 E 2 2 2 E 2 2 r r sin r r sin
E 2 E 2
1 2 cos 1 2 2 E 2 3 r r sin sin 1 2 1 2 2 E sin r 3 r 2 r
1 r sin 2 2
r 2 cot
2 2 E r r
ie.,
E 2 E 2
1 r sin 2
r 2 cot
2 2 E r r
Shifting to dimensionless variables, we set r'
r a
u'
u U
'
Ua 2
so that E 2 E 2 E '2 E '2 '
U a2
1 2 1 ' 2 U2 E E ' ' r 2 r r '2 r ' a
(7.11a)
Substituting these into (7.11a) and then leave out the primes for clarity, we have 2 2 r 2 cot r r E where the Reynolds number is defined by E 2 E 2
R
Ua
R r sin 2
(7.12)
(7.11)
7.2.7 Matched Asymptotic Expansions The method is rather involved. Here, we are contended with only a summary of the salient results. Interested readers should consult the original paper: I. Proudman and J.R.A. Pearson. [J.Fluid.Mech. 2, 237-62 (1957)], where they obtained solution to the Navier-Stokes eqs (7.11) in 2 parts.
Near the sphere They found
1 2 r 1 sin 2 4
1 3 1 1 3 1 R 2 R 2 2 cos r 8 r r 8
(7.13)
which is just the sum of the O 1 and O R terms in an asymptotic expansion for Ψ that is exact in the limit R 0 for fixed r. If we take only the O 1 terms, we have
1 1 2 r 1 2 sin 2 4 r
1 2 1 2 2 r 3r sin 4 r
which is just Stokes solution (7.8) in dimensionless quantities. The O R term improves the accuracy of the solution so that, for example, the drag (7.9) becomes 3 D 6Ua 1 R 8
Far from the sphere With r* Rr
(7.15)
they found that
1 r*2 31 1 sin 2 1 cos 1 exp r* 1 cos 2 2R 2R 2
(7.14)
which is just the sum of the O R 2 and O R 1 terms in an asymptotic expansion for Ψ that is exact in the limit R 0 for fixed r* . Thus, by ‘far away’ from the sphere, we mean a distance r of order R 1 or greater as R0.
Matching Solutions (7.13) & (7.14) are matched in the following sense. Rewriting (7.13) in terms of r* , we have 2
1 r* 2 1 sin 4 R 3 R 3 R R2 1 R 2 R 2 2 cos r* 8 r* r* 8
Keeping only terms of O R 2 and O R 1 , 1r * 1 4 R
2
3 1 2 2 1 cos R sin r* 4
1 r*2 3 3 r*2 2 2 2 1 cos sin 4 R 4 r* R
1 r*2 3 3 2 1 cos R sin 2 2 4 R 4 r*
(7.15a)
Rewriting (7.14) in terms of r, we have
1 2 2 31 1 r sin 1 cos 1 exp rR 1 cos 2 2R 2
Keeping only the O 1 and O R terms,
1 2 2 31 1 1 2 r sin 1 cos rR 1 cos r 2 R 2 1 cos 2 2R 8 2
1 2 2 3 3 r sin r sin 2 r 2 R 1 cos sin 2 2 4 16
1 2 2 3 3 r sin 2 R 1 cos 4 r 4
(7.15b)
In view of (7.15), we see that (7.15a) & (7.15b) are identical. This is what we mean by saying that (7.13) & (7.14) are matched.
7.3
Corner Eddies
In a 2-D slow flow past a symmetric obstacle, eddies may occur symmetrically for and aft of the body (see Fig.7.1a). If the shape of the obstacle is a circular arc, eddie-free flow is still possible if the corner angle is larger than C 146.3 . Below C , an infinite sequence of eddies of alternating senses occurs (see Fig.7.4). Here, we follow an elementary approach to the problem due to Moffatt.
7.3.1 Governing Equations 7.3.2 Solutions 7.3.3 Boundary Conditions 7.3.4 Discussions
7.3.1 Governing Equations We begin with the stream function representation v u x y ie.,
u k
or
ui ij 3
x j
so that u k
or
u i ijk kl 3
2 x j xl
il j 3 i 3 jl
2 x j xl
2 2 i3 x3xl x j x j
Since ψ is a function of x and y only, we have
u i i 3
2 i 3 2 x j x j
or u e 3 2
Consider now the slow flow equation p 2u
(7.3)
Taking the curl gives 0 2u 2 u e 3 2 2
ie., 2 2 0
(7.16)
In terms of cylindrical coordinates, r
(7.17)
2 1 1 2 r 2 r r r 2 2 0
(7.18)
ur
1 r
u
2 1 1 2 r 2 r r r 2 2 so that (7.16) becomes 2
2
7.3.2 Solutions Seeking for a separable solution of (7.18), we write
g r f It is easy to see that separation can be achieved if d2 f const f m 2 f (7.18a) 2 d where the particular form of the constant is chosen for later convenience. Substituting into (7.18) gives 2
d 2 1 d m2 dr 2 r dr r 2 g 0
(7.18b)
which is homogeneous in r. A natural ansatz is therefore g r Now
d 2 1 d m2 2 2 dr 2 r dr r 2 r 1 m r 2
d 2 1 d m2 2 2 2 dr 2 r dr r 2 r 1 m 2 3 2 m r
2 m 2 2 m 2 r 2 2
so that (7.18b) is satisfied if or m 2 m Now, a general solution to (7.18a) is f c1 cos m c2 sin m For a given λ, the general solution to (7.18) is therefore
r A cos B sin C cos 2 D sin 2 where A, B, C, D are constants.
(7.18c)
7.3.3 Boundary Conditions The system of interest is shown in Fig.7.4. We shall put the origin at the vertex with the x-axis bisecting the corner angle. The no-slip condition is therefore ur u 0 on where 2α is the angle of the corner. Applying them on the solution (7.18c) gives u :
A cos B sin C cos 2 D sin 2 0 A cos B sin C cos 2 D sin 2 0
ur :
A sin B cos 2 C sin 2 2 D cos 2 0
A sin B cos 2 C sin 2 2 D cos 2 0
They are easily decoupled to give
A cos C cos 2 0
(7.19a)
A sin 2 C sin 2 0
(7.19b)
B sin D sin 2 0
(7.19c)
B cos 2 D cos 2 0
(7.19d)
Eqs(7.19a,b) are consistent only if cos
sin
cos 2
2 sin 2
0
ie.,
2 cos sin 2 cos 2 sin 0
which can be simplified as
tan 2 tan 2
sin 2 sin 2 cos cos 2
sin cos 2 2 sin 2 cos sin cos 2 sin 2 cos 2sin 2 cos
(7.19e)
sin 2 sin 2 1 sin 2
1 sin 2 sin 2 1 where we’ve used the trigonometric identities
sin A cos B cos A sin B sin A B 2sin A cos B sin A B sin A B Setting
x 2 1 we have x sin 2 sin x 2 sin 2 sin x 2 x
(7.19)
Now, eqs(19.c,d) can be obtained from eqs(19.a,b) by interchanging sine and cosine. Hence, the counterpart of (7.19e) is
cos sin 2 cos 2 sin 2 cos 2 sin
which can be simplified to sin 2 sin 2 1 sin 2
1 sin 2 sin 2 1 sin 2 sin x 2 x
(7.19’)
Obviously, (7.19) & (7.19’) cannot be satisfied simultaneously. Thus, we have 2 independent solutions, namely, 1.
r A cos C cos 2 with
2.
sin 2 sin x 2 x
(7.19)
r B sin D sin 2 with
sin 2 sin x 2 x
(7.19’)
7.3.4 Discussions For a given corner subtending an angle 2α, we can solve eq(7.19) or eq(7.19’) to get x and hence x 2 Both ur and u , as calculated from (7.17), vary with r as
1
r 1 r x / 2 . Thus, for x real and positive (λ real and greater than 1), u 0 at r 0 Furthermore, for a fixed θ, neither ur nor u changes sign as r varies. The flow is therefore of the simple form with no eddies, as shown in Fig.7.3a. For λ complex, ie., of the form with p, q real p iq the measured value of any physical quantity is defined to be the real part of the corresponding complex function. For a fixed θ, both ur and u are of the form Re cr 1 Re cr p 1iq Re cr p 1eiq ln r
where c is some complex constant (different ones for ur and u , of course), and we’ve used r iq exp ln r iq exp iq ln r
Writing c c ei , where ε is real, we have Re c r p 1e
i q ln r
c r p 1 cos q ln r
which changes sign twice with r whenever q ln r changes by 2π. Note that each sign change indicates the presence of a vortex (eddie). Since lim ln r , the sign changes occur with increasing rapidity as one r 0
approaches the vertex. Thus, an infinite sequence of eddies is indicated. Now, eq(7.19) is usually solved by graphical method as shown in fig.7.5. Thus, the soloution x0 is simply the ordinate of the intersection of the curve y
sin x sin 2 and the straight line y . x 2
As is clear from fig.7.5, no intersect is possible if
sin 2 0.217 2
ie., sin 2 0.217 2
Solving sin 2 C 0.217 2 C gives 2 C 146.3
Eddies occur if 2 2 0 .
7.4
Uniqueness & Reversibility of Slow Flows
Theorem Consider viscous fluid in region V with closed surface S. There is at most one solution to the slow flow eqs (7.3) which satisfies a given boundary condition
u uB x
on S.
Proof Let there be another solution u* that satisfies the same boundary condition. Define the ‘difference flow’ by v u* u and the ‘difference pressure’ by P p* p The linearity of the slow flow eqs u 0 p 2u implies P 2 v
v 0
with the boundary conditions v0
on S.
In component form, these eqs become P 2vi xi x j x j
vi 0 xi
Multiplying the 1st with vi , we have vi
Using
P 2vi vi xi x j x j
vi P P v P vi P i vi xi xi xi xi we have Pvi 2vi vi xi x j x j
Integrating over V and applying the divergence theorem gives
Pvi dS vi S
V
2vi dV x j x j
The left hand side vanishes since v 0 on S. Together with x j
vi 2vi v v i i vi vi x j x j x j x j x j
we get 0 V x j
vi vi vi dV vi x j x j x j
Applying the divergence theorem to the 1st term shows that it vanishes on account of the boundary condition on S. Thus 2
v v v 0 i i dV i dV x j x j i , j V x j V
Since the integrands are all non-negative, we must have vi 0 for all i, j at all x V x j which means v const for all x V . Since v 0 on S, we have v 0 everywhere so that u* u . QED.
Reversibility Let u1 x , together with the attendant pressure field p1 x , be the solution to the slow flow eqs p 2u
u 0
with boundary conditions u x u B x on S.
Next, we change the boundary condition to u x u B x on S. By inspection, we see that u1 x , together with a pressure field p1 x c , c being a constant, is a solution that satisfies the new boundary condition. Furthermore, according to the uniqueness theorem, it is the only such solution. Thus, we see that for flows governed by the slow flow eqs, Reversed boundary conditions leads to reversed flow. One example of this is the concentric cylinder experiment shown in Fig.2.6. Note that for real fluids, reversal of flow is only partial since 1.
The slow flow eqs themselves are only approximations.
2.
Due to thermal and mechanical fluctuations, no boundary condition can be exactly reversed in practice.
7.5
Swimming At Low R
Consider a battery powered mechanical fish consisting of a cylindrical body and a plane tail which flaps as shown in Fig.7.6a. It swims easily in water but makes no progress in syrup. The problem it encounters in the latter case is caused by the reversibility of flow for low Reynolds numbers as discussed in §7.4. Swimming in low R situations thus requires motions that is not time-reversible. This can be achieved by replacing the plane tail with a rotating helical coil. Spermatozoa also used this mechanism, sending helical waves down their tails.
7.5.1 Swimming of a Thin Flexible Sheet 7.5.2 Basic Eqs 7.5.3 Dimensionless Form 7.5.4 Iterative Equations 7.5.5 1st order Solution 7.5.6 2nd order Solution
7.5.1 Swimming of a Thin Flexible Sheet A simple model for ciliary propulsion is a thin extensible sheet which flexes according to
ys a sin k x t
xs x where
xs , y s
(7.21)
denote the coordinates of a particle on the sheet (see Fig.7.7).
Thus, a wave travels along the x direction with speed c
k
while the particles on the sheet move only in the y direction with velocity dy s a cos k x t dt
This motion is not time-reversible since ‘running the film backward’ entails the wave moving in the opposite direction. We shall show that, for a / small, 2 / k being the wavelength, the flow induced by this oscillatory flexing contains a steady component 2
a (7.22) U 2 c in the x-direction. Thus, the sheet swims to the left in a fluid that is otherwise at rest. 2
7.5.2 Basic Eqs As in §7.3, we work in the stream function representation with (7.23) v u x y so that u e 3 2
and the slow flow eq becomes 2 2 0
which, in Cartesian coordinates, is simply 2
2 2 x 2 y 2 0
(7.24)
The boundary condition on the sheet is simply u u s . In terms of ψ, we have us
y
vs
0 y ys
x
a cos k x t y ys
on y ys a sin k x t .
(7.25)
7.5.3 Dimensionless Form Without loss of generality, we shall solve eq(7.24) only at t 0 . Solutions at other times can be obtained by replacing kx with k x t . To further facilitate manipulations, we introduce the dimensionless quantities:
'
y ' ky
x ' kx
k a
(7.26)
On substituting into eqs(7.24-25) and then dropping the primes for clarity, we have, 2
(7.24):
2 2 x 2 y 2 0
(7.25):
us
y
vs on y ys sin x .
(7.27)
0 y ys
x
cos x y ys
(7.28)
7.5.4 Iterative Equations For ε small, the boundary conditions (7.28) can be approximated by a Taylor series expansion: y x
y ys
y ys
y
sin x y 0
x
sin x y 0
2 y 2
0 y 0
2 yx
cos x
(7.30)
y 0
Similarly, we write ψ as a power series in ε,
1 2 2 3
(7.31)
where the n ’s are independent of ε. On substituting into (7.27) and using the fact that the coefficient of each power of ε must vanish, we have 2
2 2 x 2 y 2 n 0
for all n.
By successive differentiations on (7.31), we have n y n
y 0
n y n 1x
n 1 y n
y 0
y 0
n 1 y n 1x
n 2 y n
2 y 0
y 0
n 2 y n 1x
n 3 y n
y 0
2 y 0
n 3 y n 1x
y 0
On substituting into the 1st boundary conditions in (7.30), we have
1 y
y 0
2 sin x 21 y
2 y
2 y 0
2 2 y 2 y 0
3 y
y 0
2 3 y 2 y 0 2
y 0
m 1 m 1 sin x m m! y
m 2 y m y 0
m 3 y m y 0 2
y 0
0
Setting the coefficients of each power of ε to zero, we have
1 y 2 y
0 y 0
sin x y 0
2 1 y 2
0 y 0
m y
y 0
sin n x n 1 m n n! y n 1
y 0
sin m 1 x m 1 m 1 ! y m
0 y 0
Doing the same to the 2nd boundary conditions in (7.30), we have 1 cos x x y 0 2 x
sin x y 0
2 1 yx
0 y 0
m x
y 0
sin n x n 1 m n n ! y n x
y 0
sin m 1 x m 1 m 1 ! y m1x
0 y 0
To summarize, to get m , we must solve 2
2 2 2 m 0 y 2 x
with the boundary conditions m y
y 0
sin n x n 1 m n n! y n 1
y 0
sin m 1 x m 1 m 1 ! y m
0 y 0
m x
y 0
sin n x n 1 m n n ! y n x
y 0
sin m 1 x m 1 m 1 ! y m1x
which requires previous knowledge of 1 ,, m 1 .
0 y 0
7.5.4 Solution
7.5.5 1st order Solution The 1st order solution 1 satisfies 2
2 2 2 1 0 y 2 x
(7.32)
with boundary conditions
1 y
1 x
0 y 0
cos x y 0
A separable solution
1 X x Y y together with the boundary conditions suggest X to be sinusoidal so that we set X '' 2 X so that X c1 cos x c2 sin x
In order to satisfy the 2nd boundary condition, we need
Y 0 1
X sin x
so that X '' X
and 2
d2 1 Y 0 dy 2
Setting Y e y gives
1
2 2
0
which has 2 pairs of double roots 1 The general solution is therefore
Y A By e y C Dy e y If u is to be finite as y , we must have CD0
With Y 0 1 , we have
Y 1 By e y
Finally,
1 y
0 means y 0
1 By B e y sin x 0
at y 0
ie., B 1 so that
1 1 y e y sin x
(7.34)
1 1 y 1 e y sin x ye y sin x y
(7.34a)
7.5.6 2nd order Solution The 2nd order solution 2 satisfies 2
2 2 2 2 0 y 2 x
(7.33)
with boundary conditions 2 y 2 x
sin x y 0
sin x y 0
2 1 y 2 2 1 yx
0 y 0
0 y 0
Using the 1st order result
1 1 y e y sin x
(7.34)
we have 1 1 y 1 e y sin x ye y sin x y
2 1 y 1 e y sin x 2 y 2 1 y 2
sin x y 0
2 1 ye y cos x xy 2 1 xy
0 y 0
so that the boundary conditions become
2 y
sin 2 x y 0
A separable solution
2 X x Y y
2 x
0 y 0
(7.35)
with X c1 cos x c2 sin x
Y A By e y
(7.35a)
obviously cannot satisfy (7.35) To proceed, we rewrite the 1st eq. in (7.35) as
2 y
y 0
1 1 cos 2 x 2
so that the cosine term is of the manageable type (7.35a). On the other hand, the constant term requires 2 to be of the form
2 X xY y f y
(7.35b)
so that the boundary conditions become X x Y ' 0 f 0
1 1 cos 2 x 2
Y 0 sin 2 x 0 which requires 1 X x cos 2 x 2
Y ' 0 1
Y 0 0
f ' 0
1 2
(7.35c)
Plugging into (7.35a) gives
Y 0 A 0 Y ' 0 A B B 1 so that Y ye y Furthermore, for (7.35b) to remain a solution, f itself must be a solution of (7.33), ie., f (4) 0 so that f is of the form f Ay 3 By 2 Cy D
and f ' 3 Ay 2 2 By C Now, u is proportional to f’ so that to keep the former finite at y , we need A B 0
Plugging in the boundary condition gives f 'C
1 2
so that, finally, we have f
1 y 2
where D, which is of no physical significance, was dropped for convenience. Putting everything together, we have 1 2
2 ye 2 y cos 2 x
1 y 2
(7.36)
so that 2 1 1 2 y 1 e 2 y cos 2 x y 2 2 Together with 1 1 y 1 e y sin x ye y sin x y
(7.34a)
we have 1 2 y y y 1 1 ye y sin x y e 2 y cos 2 x 2 2
(7.37)
Reverting to unprimed quantities, this becomes 1 1 1 kye ky sin kx ky e 2 ky cos 2kx a y 2 2 or u
y
1 1 akye ky sin kx a ky e 2 ky cos 2kx 2 2
1 1 ye ky sin kx c 2 ky e 2 ky cos 2kx 2 2 Generalizing to arbitrary time, we have 1 1 u ye ky sin kx t c 2 ky e 2 ky cos 2 kx t 2 2 which has a steady component U
1 2 c 2
(7.22)
(7.38)
7.6
Flow in a Thin Film
7.6.1 Basic Conditions 7.6.2 Basic Equations
7.6.1 Basic Conditions Consider a viscous fluid in steady flow between 2 rigid boundaries z0
and
z h x, y
Let U and L be typical flow speed and length scale in the horizontal ( ) direction. The condition for a thin film flow is (7.39) hL Now, the no-slip condition must be satisfied at z 0 and z h . As z increases from 0 to h, u 1st increases to order U, then drops back to 0, thus u U O z h Meanwhile,
u also experiences a sign change somewhere mid-stream so that z
2 u U O 2 2 z h
U By comparison, the horizontal gradients u and 2u , which are of O and L U O 2 , respectively, are seen to be much weaker owing to (7.39). L
Although the magnitude of u we z is not specified, the no-slip condition still applies, so that its horizontal gradients are also expected to be much weaker than the vertical one. Thus, the viscous term can be approximated as 2u u 2 z 2
From the incompressibility condition, u v w w u 0 x y z z we see that
w U O z L
and hence Uh w O O U u L
Order of magnitude estimates also give u u w
U Uh U z L Lh L
Uh h u U ,U , U 1,1, L L so that
u u
U2 h 1,1, L L
2u U h 2 1,1, 2 z h L
2u Thus, u u may be neglected when compared to 2 if z U2 U 2 L h
ie.,
Uh 2 1 L
or 2
UL h 1 L
(7.40)
which is the 2nd condition for a thin film flow. [the 1st is (7.39)]
7.6.2 Basic Equations Under the conditions eqs(7.39,40) and in the absence of external forces, the steady state Navier-Stokes eqs simplify to
0
1
p
2u z 2
u 0
i.e., p 2u 2 x z
p 2w 2 z z
p 2v 2 y z
u v w 0 x y z
As shown in the last section, w and
(7.41)
2w are both smaller by a factor h / L than z 2
their horizontal counterparts. According to (7.41), so is
p . z
Setting p to be independent of z, the 1st 2 eqs in (7.41) can be integrated to give u 1 p v 1 p z A zC z x z y u
1 p 2 z Az B 2 x
v
1 p 2 z Cz D 2 y
where A, B, C, D are functions of x, y only. From 2 u U O 2 2 z h
eq(7.41) implies p p U O 2 x y h
so that UL p O 2 h
Consider now the stress tensor
(7.44)
(7.42)
ui u j x j xi
ij p ij
(7.43)
The largest term in the parenthesis is u U UL O O 2 p z h h Hence
ij p ij
(7.45)
which means the thin-film flow behaves like an inviscid flow in this aspect.
7.7
Flow in a Hele-Shaw Cell
A Hele-Shaw cell flow is a thin film flow between 2 parallel plane boundaries. Usually, fluid flow is driven by horizontal pressure gradient to past by cylindrical objects in the gap as shown in Fig.7.8. Applying the no-slip condition on z 0 and z h to (7.42) gives 1 p 2 0B 0 h Ah 2 x 0
0D
1 p 2 h Ch 2 y
so that A
1 p h 2 x
u
1 p h z z 2 x
v
1 p h z z 2 y
C
1 p h 2 y
and
(7.46)
Note that p v y p u x and hence the direction of flow and the streamline patterns, are all independent of z.
Eliminating p in (7.46) by cross differentiation, we have u v (7.47) 0 y x so that at a given z, the flow is similar to a 2-D irrotational flow. However, if we calculate the circulation around any closed curve C, we find udx vdy C
p 1 p z h z dx dy 2 x y C
1 z h z C p dx 2
1 z h z p C 2
0
(7.48)
where we’ve used the fact that p is a physical quantity and hence must be single-valued. Now, (7.48) is valid irrregardless of the global topology of the fluid domain. This behavior is markedly different from that of true 2-D flows for which can take on finite values. Thus, if we place a flat plate at an angle of attack to the oncoming stream, the streamline patterns will be as shown in Fig.4.6a but never as in Fig.4.6b. Another example is the flow into a rectangular opening as shown in Fig.7.9b, which is to be contrasted with that of high R flow shown in Fig.7.9a.
7.8
Adhesive Problem
It is well known that a rather large force is required to pull a disc away from a rigid plane if the two are separated by a thin film of viscous fluid. Here we study the problem in terms of the steady state thin film flow eqs and attribute all the time dependence of the flow to the changing boundary conditions. Justification of this assumption will be left as an exercise. Let the radius of the disk be a and the height of the fluid be h t . [see Fig.7.10] By symmetry of the setup, we expect the (unsteady) flow to be of the axisymmetric:
u ur r , z , t e r u z r , z , t e z In terms of cylindrical coordinates, the relevant thin film eqs are (cf.7.41) p 2u 2r r z 0
2uz z 2
1 u rur z 0 r r z
Using the fact that p is independent of z, we can integrate the 1st eq. and get ur 1 p z A z r ur
1 p 2 z Az B 2 r
where A, B are functions of r & t only. Sticking in the no-slip boundary condition at z 0 and z h t , we have 0B
0
1 p 2 h Ah 2 r
so that ur
1 p h z z 2 r
Thus
ur 1 2 p h z z r 2 r 2 and the incompressibility condition becomes
1 2 p 1 p u h z z h z z z 0 2 2 r 2 r r z
i.e.,
uz 1 2 p 1 p h z z z 2 r 2 r r which integrates to
uz
1 2 p 1 p 1 2 1 3 hz z C 2 r 2 r r 2 3
No-slip conditions at z 0 then gives
1 2 p 1 p 1 2 1 3 uz hz z 2 r 2 r r 2 3
The boundary condition at the moving disk is uz
dh dt
at z h t
so that
dh 1 2 p 1 p 3 h dt 12 r 2 r r which can be rearranged to give
2 p 1 p 1 p 12 dh r r 2 r r r r r h 3 dt Integrating, we get r
p 6 dh 2 r D r h 3 dt
p
3 dh 2 r D ln r E h 3 dt
where D, E are functions of t only.
(7.49)
To avoid a singularity at r 0 , we set D 0 . With the boundary condition p p0 at r a , p0 being the atmospheric pressure, we have p
3 dh 2 r a 2 p0 h 3 dt
The force exerted on the disk by the fluid is therefore
F
2
0
p p rdrd a
0
0
6 dh a 2 r a 2 rdr h 3 dt 0
6 dh 1 4 1 4 a a h 3 dt 4 2
3 dh 4 a 2h 3 dt
(7.50)
Thus, if one tries to pull up the disk, i.e.,
dh 0 , then F 0 , so that the pull by the dt
fluid is downward. Furthermore, the h 3 dependence makes F rather large for small h.
7.9
Thin Film Flow Down a Slope
Consider the 2-D problem of a layer of viscous fluid spreading down a slop under the action of gravity. [See Fig.7.11 for the definitions of the coordinate system and various quantities.]
As in §7.8, we neglect
u and attribute all time dependencies to that of the t
boundaries. In the presence of gravity, the thin film flow eqs become
0
1 p 2u 2 g sin x z
0
1 p g cos z
(7.52)
The 2nd eq can be integrated to give
p gz cos f x, t
(7.52a)
Consider the free surface
z h x, t Its unit tangent and normal vectors in the x-z plane are (see §3.4b.2)
vˆ
h 1, 2 x h 1 x 1
nˆ
h , 2 x h 1 x
Under the thin film approximation (L is horizontal length scale) hL we have h h 1 x L
so that
vˆ 1, 0 xˆ
nˆ 0, 1 zˆ
Consider now the stress tensor
1
1
vi v j x j xi
ij p ij
The stress vector at the surface is t nˆ zˆ u u w t1 13 z z x t 3 33 p 2
w p z
[see (7.45)]
Hence. the continuity of the stress across an interface gives u 0 z p p0
(7.53) (7.53a)
on z h x, t .
Putting (7.53a) into (7.52a) gives p0 gh cos f so that
p g h z cos p0 p h g cos x x
The 1st eq. in (7.52) thus becomes
2u h g cos g sin 2 z x
(7.54)
Now, h h O 1 x L
so that unless α is very small, eq(7.54) can be approximated by 2u 2 g sin z Integrating, we get
u g z sin A z 1 2 z g sin Az B 2 where A, B are functions of x, t . u
(7.54a)
Applying (7.53) at z h , we have 0 h
g
sin A
Applying the no-slip condition at z 0 , we have 0B Hence, (7.54a) becomes u
z 1 h z g sin 2
(7.55)
u z h g sin x x
so that the incompressibility condition gives w z h g sin z x
which can be integrated to give w
z 2 h g sin A 2 x
Putting in the no-slip condition at z 0 , we have A0 so that w
z 2 h g sin 2 x
Putting this into the surface wave condition, w
h h u t x
at
zh
[see eq(3.18)]
we have
h h h2 u g sin 0 t x 2
Now, (7.55) gives
at
zh
(7.55a)
h2 g sin at 2 so that (7.55a) becomes u
zh
h h h 2 g sin 0 t x
(7.56)
The characteristics of this obey (see §3.9a3) dt dx dh 2 1 h g sin 0 so that h c1 dx g sin c12 dt
x c12
g sin
t c2
Hence, the general solution of (7.56) is g sin 2 h f x h t where f is an arbitrary function.
(7.56a)
Now, (7.56a) is of the wave-steepening type discussed in §3.9.3. Thus, taking its partial, we have h 2 g sin h 1 ht f ' x x
1
f' 2 g sin
tff '
which blows up at t tC , where 1
2 g sin
tff 't t
0
C
In practice, surface tension effects set in to prevent instability of the type depicted in Fig.3.16c. The wave profile at the front therefore remains in the form of a ‘nose’ shown in Fig.3.16b and Fig.7.11. According to Huppert, beyond tC , h is described by the similarity solution h
g sin
x t
(7.57)
[see H.E.Huppert, J.Fluid Mech 173, 557-94 (1986)]
The position xN of the front edge is obtained from the constant volume condition xN
hdx A
(7.58)
0
where A is the area of the cross section of the fluid in the x-z plane. Putting in (7.57), we have
A
xN
tg sin
xdx
0
2 3/ 2 xN tg sin 3
or 1/ 3
9 A2 g sin xN t 4
(7.59)
7.10 Lubrication Theory When a solid body slides over another one, the frictional resistance it experiences is usually comparable in magnitude to the normal force it exerts on the other. However, if there is a thin film of fluid between them, the frictional resistance can be reduced drastically. This is called lubrication. The main reason for this is that, in thin film flows, the ratio between the magnitudes of the tangential and normal stresses is of the order [see §7.6]
U UL
h h 1 L h2
7.10.1
Slider Bearing
7.10.2
Flow Between Eccentric Rotating Cylinders
7.10.1
Slider Bearing
Consider the 2-D system shown in Fig.7.12. The rigid lower boundary at z 0 moves with velocity U past a stationary block of length L. The space between them is filled with viscous fluid, and the pressures at both ends of the bearing are equal to the atmospheric value p0 . Working with the 2-D version of the thin film flow eqs (7.41), we can begin at the solution 1 dp 2 (7.42) u z Az B 2 dx where p is a function of x only. The boundary conditions are u U
at z 0 at z h u0 which, when plugged into (7.42), gives UB 0
1 dp 2 h Ah U 2 dx
so that u
1 dp z h z z U 1 2 dx h
1 dp U z h z h 2 dx
(7.60)
The integral form of the incompressibility condition is that the mass flux Q is the same across all cross-sections of the film. In terms of (7.60), we have h
Q udz 0
1 dp U z h z dz 2 dx h 0 h
1 dp 1 1 3 U h 2 dx 2 3 h
1 2 1 h 2
1 dp 3 U h h 12 dx 2
(7.61)
Rearranging terms gives dp 12 U 3 h Q dx h 2
which integrates to
12 U h Q dx 3 h 2 0 x
p p0
x x 1 1 6 U 2 dx 2Q 3 dx h 0 0h
(7.61a)
where p p0 at x 0 . Now, we also have p p0 at x L , so that L L 1 1 0 6 U 2 dx 2Q 3 dx h 0 0h
ie., 6 0 h 2 Udx L
Q
12 0 h3 dx L
L
1
U h 2
2
dx
0 L
(7.62)
1 0 h3 dx
In the special case of a plane slider bearing, h vary linearly from h1 at x 0 to h2 at x L , ie., h x h2 h1
x h1 L
Using the formulae dx
a bx
2
3
dx
a bx we have
1 b a bx 1 2b a bx
2
dx
h
2
dx
h
3
1 h2 h1 h2 h1 h1 L L
L2 h2 h1 h1L h2 h1 x
L h2 h1 h 1
2
x
h2 h1 h2 h1 h1 L L
x
2
L3 2 h2 h1 h1L h2 h1 x
2
L 2 h2 h1 h 2
so that
1 1 dx L L 2 0 h h2 h1 h2 h1 h2h1 L
L
dx
h 0
3
1 L 1 L 2 2 2 2 h2 h1 h2 h1 h2 h1 h2 h1
whereupon eq(7.62) gives U h2h1 Q 2 h2 h1 Also, Eq(7.61a) becomes
p p0 UL 1 1 h2h1 L 1 1 U 2 2 6 h2 h1 h h1 h2 h1 h2 h1 h h1
UL 1 1 h2h1 1 1 2 2 h2 h1 h h1 h2 h1 h h1
ULh2h1 1 1 1 1 1 1 2 2 2 2 h2 h1 h h1 h2 h1 h h1
1 1 1 1 1 2 h h2 h1 h1h2 h
ULh2h1 h22 h12
UL h h2 h1 h 2 h1h2 2 2 h h h 1
2 2
UL h h1 h h2 h h12 h 2 2 2
(7.63)
For h2 h h1 , the last expression is positive so that p p0 and we have an upward lift. Hence, lubrication occurs if h decreases in the direction of flow as shown in Fig.7.12.
7.10.2
Flow Between Eccentric Rotating Cylinders
Consider the setup shown in Fig.7.13. Viscous fluid fills the narrow gap between a fixed outer cylinder of radius
r a 1 , and an inner, offset, cylinder of radius a which rotates with peripheral velocity U. This serves as a simple model for an axle rotating in its housing.
Geometry Let the centers of the outer and inner cylinders be at x 0,0 and x ,0 , respectively. The quantity h then provides a convenient measure of the ‘height’ of the fluid.
Simple trigonmetric maneuver gives
h a 1 where
2 2 a 2 2a cos
sin
sin a
Now, ε and hence η are small. Keeping only 1st order terms in them, we have sin sin a a
sin 1
cos cos
cos cos sin sin cos cos sin sin
cos
a
sin 2
a 1 cos a 1 cos a a
and
h a cos a 1 cos a where
(7.64)
offset of centers eccentricity a difference of radii
The minimum of h is at 0 :
h 0 a 1 0 Hence, 0 1 . For 0 , the cylinders are coaxial. For 1 , they are in contact at 0 .
Solution For small gaps, curvature effects may be neglected so that the results for the plane slider bearing are applicable here with straightforward adaptations such as x a , u u , etc. Thus, eq(7.60) becomes 1 dp U u z h z (7.65) h 2 a d where z is the distance across the gap, measured along the radial direction. Similarly, eq(7.61) becomes 1 dp 3 U Q h h 12 a d 2
(7.66)
Using
h a 1 cos
(7.64)
the counterparts to the integrals in (7.62) are 2
I2 a 0
d 1 2 2 h a
d
1 cos 0
2
2
d 1 I3 a 3 2 3 h a 0
2
d
1 cos
3
0
These can be evaluated by contour integral as follows. 1st, let z ei so that
1 i 1 1 e e i z 2 2 z
cos
dz iei d izd whereupon 1 dz 4 I2 2 2 a C ia 2 2 1 iz 1 z z 2
I3
1
a 2 3 C
dz 1 iz 1 z z 2
3
C
zdz
2 z 1 z
8
2
2
z 2dz
ia 2 3 3 2 C 2 z 1
z
3
where C is the unit circle centered at the origin. Now, the poles are the roots of z2 1
2
z0
i.e.,
z
1
1
1
1 1
1 2
Since 0 1 , they are both real. 2
Furthermore, z z 1 so that only one of them is inside C. Obviously, z z , so that only the pole at z that contributes to the integrals. With the help of z z
2
z z
we have
2
1 2
I2
4
zdz
ia 2
I3
z z z z 2
C
2
d
2 dz z z z z
4
z
2 i 2
ia 2
1 2z 2 a z z z z 3 8 2
2
2
8 z z
a 2 2 z z
3
2
a 1 2 2
3/ 2
z 2dz
8
3 3 ia 2 3 3 C z z z z
1 d2 z2 2 3 3 2 i 2 ia 2 dz z z 3 zz 8
d 2z 3z 2 a 2 3 3 dz z z 3 z z 4 z z 8
d z z 2 z 2 3 3 a dz z z 4 z z
2z 2z 4 z z 2 z 4 5 a z z z z
8
8
2 3
16 a 2 3
3
3
z z
5
z2 z2 2 z z 2 z
16 a 2 3
3
z z
5
16 a 2 3 3 z z
2
2a 2 3 1
2 5/ 2
5
z
2
4 z z z2
z z 2 2 z z
4 2 2
2 2
a 2 3 1 2
5/ 2
Thus, (7.62) becomes UI Q 2 2I3 a 2 3 1 2 U 2 2 a 2 1 2 3/ 2 2 2
5/ 2
1 2 Ua 2 2
(7.67)
dp is an even function of θ since h is. d Hence, p, and hence p cos , are odd functions of θ.
From (7.66), we see that
The horizontal force on the inner cylinder Fx p cos dS S
therefore vanishes. The calculation of the vertical force is too involved to be included here. We merely list the result: F
12U
2 1 2 2 2
Of particular interest is the factor 1 2
(7.68)
1/ 2
which makes
1. In other words, by moving horizontally towards the outer cylinder, the inner one can substain arbitrarily large vertical loads. F as
Let us consider the flow at . From (7.64), we see that
h a 1 From (7.67), we have
at
Q 1 U h 2 2
at
We can therefore write (7.66) as dp U Q 12 a 2 3 d h h U 1 1 2 h 2 2
12 a
U 1 2 12 a 2 h 2 2 at which implies an adverse pressure and hence reversed flow in that neighborhood. 0
Differentiating (7.65), we have 1 dp u 1 dp U z h z z 2 a d h 2 a d
1 dp U h 2z 2 a d h
6 Uh U Q h 2z 3 h 2 h
6U h3
h 1 2 U a h 2z 2 2 h 2
Hence
u z
z h
6U h2
h 1 2 U a 2 2 h 2
6U 2 h
6U h2
h 1 2 a 2 2 3 1 1 2 a 1 cos a 2 2 3
2Ua 3 3 2 1 cos h2 2 2
2Ua 1 4 2 cos 2 2 h 2
2Ua 1 4 2 cos h2 2 2
(7.69)
For this to be negative, we need 1 4 2 cos 0 2 2
(7.69a)
Now, reversed flow, if occurs, is expected around where cos 1 or cos 1 . Rewriting (7.69a) to reflect this, we have 1 4 2 1 2 2
u z
0 z h
so that no reversed flow occurs. Rewrite the criterion as 0 3 4 2 2 1 1 2 3 1
(7.69b)
The critical values of λ are the roots of the polynomial on the right side, i.e.,
1
1 3 13 2
However, since 0 1 , the only valid value is
1 3 13 0.30 2
Writing
2 3 1 we see that near so that , eq(7.69b) becomes
0 13 13 so that to satisfy (7.69b), we must have 0 , i.e., Conversely, if
(7.70)
there will exist a range of θ such that
u z
0 so that reversed flow is expected. z h
This feature, though of theoretical interest, is usually overwhelmed by other complications in practice.
8. Boundary Layers 8.1
Prandtl’s Paper
8.2
Steady 2-D Boundary Layer Eq.
8.3
Boundary Layer On Flat Plate
8.4
High R Flow In Converging Channel
8.5
Rotating Flows Controlled By Boundary
Layers 8.6
Boundary Layers Separation
8.1
Prandtl’s Paper
8.2
Steady 2-D Boundary Layer Eq.
We now derive the eq. for a steady, incompressible, 2-D boundary layer adjacent to a rigid wall at y 0 . The no-slip condition for a stationary boundary means at (8.2) uv0 y0 which differs from the inviscid flow boundary condition only by the additional demand of u 0 . Our main concern is therefore to see how u merges from 0 to the inviscid flow value. For a thin boundary layer, we expect u u y x
(8.4a)
To get an order of magnitude estimates, let U 0 be some typical value of u. Let the distance for u to change by an amount U 0 be of order L in the x-direction, and δ in the y-direction. Hence, (8.4a) means U0
U0 L
ie.,
L
(8.4)
The Navier-Stokes eqs. are
2u 2u u u 1 p u v 2 2 x y x y x
u
2v 2v v v 1 p v 2 2 x y y y x
u v 0 x y The 3rd
eq. in (8.5) means
v u U 0 y x L
(8.5)
so that v is of order
U 0 . Hence v u . L
The other 2 eqs. in (8.5) then imply p p x y which means we may write p p x in the boundary layer.
Similarly, from
2u U 0 2 x 2 L
2u U 0 y 2 2
we deduce
2u 2u 2 y 2 x
These estimates can be used to simplify (8.5) into the boundary layer eqs.,
u
u u 1 dp 2u v 2 x y dx y
u v 0 x y
(8.1) (8.2)
Eq(8.1) provides the means to estimate the magnitude of the layer thickness δ. Now, U 0 2 U 0 U 0 U 0 2 L.H .S . L L L Assuming viscosity effect to be significant within the layer, we have U 02 U 20 L ie.,
L
U0L
1 R
where R is the Reynolds number. Hence, thin boundary layer needs large R.
By interpreting x as the direction tangent, y as that normal, to the boundary, (8.1-2) are applicable to curved boundaries. What’s worth mentioning is that although p p , we still have p p x . x y Let U x be the slip velocity at y 0 for the corresponding inviscid flow. For a thin boundary layer, we expect u U x at the “edge” y , where, according to the Bernoulli theorem, p
1 U 2 const along a streamline. Hence, 2
dp dU U dx dx
(8.8)
so that a rise in U is accompanied by a drop in p and vice versa. To ensure the boundary layer merges smoothly into the mainstream inviscid flow, we impose the boundary condition on (8.1-2) y y u U x as or
(8.9)
8.2.1 Example Consider the problem
u 0 0 ,
u '' u ' 1 ;
u 1 2
(8.10)
where 0 is a small constant. Integrating, we have y
du ' ln 1 u ' C1 1 u'
or, 1 u ' C2 e y /
so that u y C2 e y / dy y C3e y / C4 Putting in the boundary conditions, we have 0 C3 C4 2 1 C3e 1/ C4
which is easily solved to give C3
1 e
1/
C4
and
1
1 e
1/
1
so that u y
e y / 1 e 1/ 1
(8.11)
Now, e 1/ 1 . Also e y / 1 unless y . Hence, u as given by (8.11) may be splitted into 2 parts: 1. Mainstream part: uM y 1 for y . 2.
Boundary layer part:
uBL 1 e y /
Note that uM lim u 0
with
y fixed.
for
0 y .
uBL lim u
with
0
y /
fixed.
and
lim uBL lim uM
y /
y 0
which is equivalent to (8.9).
Alternative Approach Another approach which is more akin to that taken in fluid dynamic problems is to take advantage of the smallness of ε at the earliest stage. Thus, (8.10) simplifies to uo ' 1 with solution uo y c There is now only 1 arbitrary constant so that only 1 of the 2 boundary conditions can be satisfied. On making uo 1 2 , we obtain an ‘outer’ solution uo y 1 What we did so far is similar to obtaining the mainstream inviscid flow: The problem is simplified by invoking the smallness of a parameter to drop the highest derivatives. However, this lowering the order of the system reduces the number of arbitrary constants in the solution so that not all boundary conditions can be satisfied. Thus, an ‘inner’ solution, or boundary layer, is needed to satisfy the other boundary conditions. To accentuate the relatively rapid change of u inside the boundary layer, we shift to the variable Y
y
so that (8.10) becomes d 2u du dY 2 dY On invoking 1 , we have the ‘inner’ (boundary layer) eq., d 2ui dui 0 dY 2 dY
which, on integrating, gives du Y ln i C1 dY
ie., dui C2 e Y dY
so that ui C3e Y C4
On making this satisfy the ‘no-slip’ boundary condition u 0 0 get 0 C3 C4 so that ui C3 e Y 1
Matching the inner & outer solutions with
lim ui lim uo
Y
y 0
we get C3 1 so that the full solution is y 1 y u as 0 for fixed y / 1 e y / as found before.
at
y 0 , we
8.3
Boundary Layer On Flat Plate
If the fluid is inviscid, a uniform stream approaching a flat plate at zero incidence angle suffers no deflection. Hence U x const . According to (8.8),
dp 0 so that dx
the boundary layer eqs (8.1-2) reduce to
u
u u 2u v 2 x y y
(8.12)
u v 0 x y
(8.13)
For a semi-infinite plate that extends from x 0 to x , it seems natural to seek a similarity solution: y u with (8.14) h g x U and g is to be determined. 1st of all, (8.13) can be satisfied if we introduce the stream function ψ so that (8.15) v u x y With the help of (8.14), we have
x, y u x, y dy k x Uh
dy d k x d
Ug x h d k x If the plate is a streamline, ψ is constant on it. For convenience, we can set
( x, y 0) x, 0 0 and get rid of k x by writing
x, Ug x h d Ug x f 0
where
(8.16)
f h d
so that
f 0 0
0
Thus,
u
1 df df Ug x U y g x d d
(8.17)
Using y dg dg 2 x g dx g dx we have dg dg df v U f g g dx d x x dx df dg U f (8.18) d dx Similarly,
u d 2 f dg d 2 f U U x x d 2 g dx d 2 u d 2 f U d 2 f U y y d 2 g d 2
(8.18a)
2u U d 3 f U d 3 f y 2 g y d 3 g 2 d 3 so that (8.12) becomes U
df dg d 2 f df dg U d 2 f U d3 f U U f d g dx d 2 d dx g d 2 g 2 d 3
or
Ug
dg d 2 f d3 f f dx d 2 d 3
(8.18b)
which is self-consistent only if g
dg c1 dx
d3 f d2 f c Uf 0 1 d 3 d 2 where c1 is a constant.
Integrating the 1st eq. gives 1 2 g c1 x c2 2 where c2 is another constant.
(8.18c)
Now, eqs(8.18a) show that the partials of u are all proportional to g 1 , which blows up as g 0 . The only place such extremes can be tolerated is at the leading edge of the plate where x 0 . Hence, we set c2 0 in (8.18c). To summarize, we now have g 2c1 x
dg c1 dx 2x
U 2c1 x f u U
df d
y 2c1 x (8.19)
df c1 v U f d 2 x
d 3 f c1U d 2 f f 0 d 3 d 2
(8.17,18)
(8.20)
Note that the value of η at the origin x y 0 depends on the path one chooses to take the limit. Thus, if we approach along the y-axis but 0 if we approach along any other straight line y mx on which m
x . 2c1
Since (8.20) is a 3rd order ODE, there’ll be 3 arbitrary integration constants (besides c1 ) in a general solution. Their determination require 3 boundary conditions. From (8.16), we already have
f 0 0 . Next, as the flow approaches the leading edge of the plate along the x-axis ( 0 ), we expect u 0 at 0 so that (8.17) gives
f ' 0
df 0 d 0
Finally, far away from the plate ( y or ) the flow should be u U ,0 . Hence, (8.17) gives
f ' 1
while (8.18) gives lim f f ' 0
Finally, inspection of the eqs shows that c1 serves only as a scale for x so that its role is more or less redundant for a similarity solution. For convenience, we can set it to c1
U
so that (8.20) becomes
d3 f d2 f f 0 d 3 d 2
(8.20’)
which is free of any parameters of the problem. Solution to this problem must be obtained numerically. The results are shown in fig.8.8. Of interest is that
u 0.97 at 3 . U
The boundary layer thickness δ can be estimated from y x U
O
(8.22)
as shown in Fig.2.14. The horizontal stress on the plate is
u
v
u
xy y x y 0 y y 0 v where 0 since v x,0 0 for all x 0 . x y 0 Uisng (8.18a) and , we have
xy
U d2 f g d 2
0
Since
g 2c1 x we have
2 x U
U f '' 0 g
U to give 2 x
xy U
U d2 f 2 x d 2
U 0
U 2x
f '' 0
(8.23)
which decreases as x increases.
Plate of Length L Assuming (8.23) to hold for a plate of finite length L, the drag on it is L
D 2 xy dx 2 U 0
U 2
L
f '' 0 0
dx 2 U 2 UL f '' 0 x
where the factor 2 accounts for the equal contributions from both the upper & lower side of the plate. In terms of the Reynolds number R D 2 2 U 2 Lf '' 0 R 1/ 2 Note that D is proportional to
UL
, we have
(8.24) L . Hence
lim D 0 0
Numerically, f '' 0 0.4696 .
Both (8.24) & the velocity profile agrees with experiment unless R is very high, at which case, the boundary layer becomes unstable and turbulence ensues. The critical value of R for this onset is about 105 ~ 106 .
8.4
High R Flow In Converging Channel
We consider the high R, 2-D flow between 2 plane walls and y x tan y0 A narrow slit at the origin provides the means for the necessary intake or discharge to maintain a steady flow. For the mainstream flow, the simplest ansatz is purely radial:
u ur r , e r The incompressibility condition u
u rur 0 rr r
then becomes rur 0 r
Therefore Q (8.25a) r where Q is a constant and the – sign is inserted so that Q 0 represents inflow. ur
Next, the 2-D Euler eqs in polar coordinates are [cf. eq(2.22), §2.4]
ur u2 1 p u ur t r r u uu 1 p u u r t r r which, for our steady flow ansatz, become, u 1 p ur r r r
p p r Thus, ur is also a function of r only and, upon integration, 1 2 1 ur p c1 2 Using (8.25a), we have 1 Q 2 p ur 2 c1 2 c1 2 r
Boundary Layer on
y0
The mainstream flow, according to (8.25a), is U x
Q x
(8.25)
The boundary eqs dp dU U dx dx
u
u u 1 dp 2u v 2 x y dx y
u v 0 x y
(8.8) (8.1) (8.2)
thus become dp Q Q2 U 2 3 dx x x
u
u u Q2 2u v 3 2 x y x y
(8.26)
Following the approach of §8.3, we try a similarity solution y Q u U x h h with g x x Introducing the streamline function ψ to satisfy (8.2), we have v u x y and
x, y u x, y dy k x
Q dy h d k x x d
Q g x h d k x x Setting the wall y 0 to be a streamline, ψ is constant on it. For convenience, we
can set
( x, y 0) x, 0 0 and get rid of k x by writing
Q Q x, g x h d g x f x x 0 where
f h d
so that
f 0 0
0
Thus u
Q Q g f ' f ' y x y x
and, with y g' 2 g ' x g g we have
v
g g' g g' Q 2 f f f ' x x x g x
Qg g' g' 1 x f x f ' 2 x g g
Qg g' f x f f 2 x g
'
Similarly,
1 u 1 g' Q g' Q 2 f ' f '' 2 f ' x f '' x x g x g x u Q f '' y xg
2u Q 2 f ''' 2 y xg so that (8.26) becomes
Q Q g ' Qg g' f ' 2 f ' x f '' 2 f x f f x x g g x
Q Q2 2 f ''' 3 x xg which simplifies to
Q ' f '' xg
g' x2 f '2 1 x ff '' 1 2 f ''' g gQ
(8.26a)
This can be consistent by setting g c1 x so that (8.26a) becomes 2 f ' 1
c12Q
f '''
To make this parameter free, we set c12
Q
, where Q is used to ensure the reality
of c1 . Hence f '2 1 f '''
(8.27a)
where the +/- sign corresponds to Q positive/negative, ie., to in/out flow. One of the boundary conditions associated with (8.27a) was already found to be
f 0 0 Another one should be provided by the no-slip condition for all uv0 y0 Since v y 0 0 is automatically satisfied by (8.27), we get only one new condition
f ' 0 0 The 3rd boundary condition is obtained by merging the solution with the mainstream flow: u y
Q x
ie.,
f ' 1
Setting
F f ' eq.(8.27a) can be transformed into a 2nd order eq., F 2 1 F '' with boundary conditions
F 0 0
F 1
(8.28)
(8.29)
Summary Taking stock of progress so far, we have g
Q
x
Q y x
Q f u
Q f' x
(8.27) v
Q Qg f ' f ' 2 x x u Q 2 y x
u Q f ' f '' x x 2
Q
f ''
F f'
where the +/- sign corresponds to in/out flow. The ODE to solve is F '' 1 F 2 0
with boundary conditions
F 0 0 ;
F 1
Solutions Using F ''
1 d F '2 2 dF
the ODE becomes 1 d F '2 1 F 2 0 2 dF
which is integrated to 1 2 1 F ' F F 3 c2 2 3
From the mainstream flow (8.25a), we have
(8.29a)
dur Q dr r 2
so that dur dr
y 0
Q x2
The boundary layer quantity that should merge to this is u Q Q 2 F F ' 2 x x x Since F 1 , this means
F ' 0
Substituting back to (8.29a) gives c2
2 3
so that (8.29a) becomes 1 2 1 2 F ' F F3 0 2 3 3
which can be rearranged to give 3 2 2 F ' 2 3F F 3 2 F 1 F 2
For the outflow case, we have
3 F ' 2 F 1 F 2 Since F 0 0 , the right hand side is imaginary at 0 . Hence, there is no solution that can satisfy the required boundary conditions. For the inflow case, we need to solve
3 F ' 2 F 1 F 2 To get rid of the square root, we set G2 2 F so that 2GG ' F '
and (8.29b) becomes
(8.29b)
2
3 GG ' G 3 G 2 2
or G'
1 3 G2 6
Integrating gives dG 1 G tanh 1 c3 2 3G 6 3 3 or G 3 tanh c4 2 F 3tanh 2 c4 2 2 where the ci ’s are constants.
(8.30)
Now, (8.30) satisfies F 1 automatically. To satisfy F 0 0 , we need
c4 tanh 1
2 1.14 3
That c4 can have 2 possible values is an example of the non-uniqueness of flow at high R. [see §9.7] The flow pattern for c4 1.14 is shown in fig.8.9 and is typically observed in experiment. The flow pattern for c4 1.14 involves reversed flow close to the wall. Using the fact that F 1 1 , the layer thickness δ can be estimated from
1
Q x
ie.,
Q
x
(8.31)
which decreases as one approaches the corner.
8.5
Rotating Flows Controlled By Boundary
Layers 8.5.1 Almost Uniform Rotation: Basic Eqs. 8.5.2 Steady, Inviscid Flow 8.5.3 Ekman Boundary Layers 8.5.4 The ‘Interior’ Flow 8.5.5 Unsteady Flow: ‘Spin-Down’
8.5.1 Almost Uniform Rotation: Basic Eqs. The time rate of change of a vector Q as seen from a fixed and a rotating frame are related by dQ dQ ΩQ dt F dt R where is the angular velocity vector of the rotation. Setting Q x , Ω we have x uF uR Taking the time derivative of the whole eq. gives du F du R dx Ω dt F dt F dt F
du dx R Ω uR Ω Ω Ω x dt R dt R du R 2Ω u R Ω Ω x dt R
For fluid motion,
d D is replaced with . Thus, we have dt Dt Ω u Ω
Ω x u F u R u F u F u R u R 2 R t F t R where the 3rd and 4th terms on the right are the Coriolis and centrifugal acceleration, respectively. [For a more rigorous derivation, see R.E.Meyer, “Introduction to Mathematical Fluid Mechanics”, Chap 5, Wiley (71)] At a given instance of time, quantities in the 2 frames are related by simple coordinate transformations. Hence, provided the 2 frames are kept in the same orientation, we have
Q F Q R Dropping the subscript R for clarity, the Navier-Stokes eqs in the rotating frame are simply
Ω u Ω Ω x u u 1 2 u u 2 p t u 0
Using
a b c d a c b d a d b c we have
Ω x Ω x Ω 2x 2 Ω x
2
Using
a b a b b a a b b a we have 1 2 a a a a a 2
[cf.
1 2 f f f ] 2
Ω x Ω x Ω x Ω x Ω Ω x 2 Ω x Ω x 2 Ω x Ω 2
x 2 2 x x 2x x 2 x x 2x where we have used x x i ijk k ijk kj ijj 0 x j
x Ω x i j i j ij i x j
x x x i x j i x j ij xi x j Thus Ω x Ω 2 x 2 Ω x 2
2
2Ω2x 2 Ω x Ω 2Ω x Ω or
(8.34)
1 2 Ω Ω x Ω x 2
(8.35)
Thus, for const , (8.34) becomes Ω u u u 1 2 u u 2 p* t
(8.34a)
where 1 2 p* p Ω x 2
(8.36)
which contains the centrifugal forces, is called the reduced pressure. With the understanding that only reduced pressures will be used in the following, we can drop the subscript * for the sake of clarity. Let U be typical value of u , L be typical length scale of flow. We are interested in cases where the flow u is small compared to the rotation of the system, ie.,
U 1 L
(8.36a)
Now,
u u
U2 O L
Ω u O U Eq.(8.36a) therefore implies
u u
U O 1 Ωu L
Eq(8.34a) thus simplifies to u 1 2Ω u p 2u t u 0
(8.37) (8.38)
The relevant Reynolds number is R
L2
[not
UL
]
and the flow will be considered inviscid if R 1 .
8.5.2 Steady, Inviscid Flow Let the coordinate system attached to the rotating frame be
x, y , z
with the z-axis
parallel to the axis of rotation, so that Ω 0,0, . With u u, v, w , we have
i j k Ω u 0 0 v i u j v, u,0 u v w
For a steady, inviscid flow, Eq(8.37) becomes 1 p (8.39) 2 v x 2 u 0
1 p y
1 p z
while (8.38) becomes u v w 0 x y z
(8.40) (8.41)
(8.42)
From (8.41), we see that p is independent of z. So are u and v by way of (8.39-40). Eq(8.42) then put w also independent of z. Thus, u is independent of z, which is known as the Taylor-Proudman theorem.
8.5.3 Ekman Boundary Layers Consider the steady flow between 2 boundaries that rotate with slightly different angular velocities. In the rotating frame attached to 1 of the boundaries, the fluid flow is obviously small so that the almost uniform rotation eqs. (8.37-38) apply. We assume the flow consists of 2 components. One is the inviscid interior (mainstream, high R) flow that obeys eqs(8.39-42). The other is the so called Ekman boundary layer flow for which the viscous term
2u can be approximated by 2 u .
Let the boundaries be planes perpendicular to the z-axis. The angular velocities of the boundaries at z 0 and z L are Ω and 1 , respectively. As stated earlier, 1 . Obviously, 2u u 2 z 2
(8.43a)
Consider 1st the boundary layer on z 0 . Borrowing the results from §8.5.2, and the help of (8.43a), eqs. (8.37-38) become
2 v 2 u 0
1 p 2u 2 x z
(8.43)
1 p 2v 2 y z
(8.44)
1 p 2w 2 z z
(8.45)
u v w 0 x y z
(8.46)
The arguments presented in §8.2 can be applied here by the transformations
x x, y
and
Thus, (8.4a) becomes
yz
u u u , z x y
v v v , z x y
(8.4a’)
To get an order of magnitude estimates, let U 0 be some typical value of u and v. Let the distance for u or v to change by an amount U 0 be of order L in the x-y plane, and δ in the z-direction. Hence, (8.4a’) means U0
U0 L
ie.,
L
(8.4)
Eq. in (8.46) means w u v U 0 z x y L so that w is of order
U 0 . Hence w u , v . L
Eq(8.43-5) then imply p p p or z x y so that we may write p p x, y in the boundary layer. Hence,
p p and remain close to their inviscid ‘interior’ values, which obey x y
2 v I 2 u I
1 pI x
(8.39)
1 pI y
(8.40)
The boundary layer eqs(8.43-4) can therefore be written as 2u 2 v v I 2 z
(8.47)
2v 2 u u I 2 z
(8.48)
8.47 i 8.48
gives
2 v iu vI iuI
2 u iv z 2
2 i iv u iv I uI
Setting
f u iv uI ivI we have
2 f 2 i f z 2
(8.48a)
Consider the ansatz f ea z where a is a constant. Eq(8.48a) is satisfied if
a 2 2 i or
a i
2
2
ei / 4
2 1 i cos i sin 4 4
Setting
z* z
[ i ei / 2 ]
the general solution to (8.48a) is f Ae 1i z* Be1i z*
where A and B are arbitrary functions of x and y. Hence u iv uI ivI f
uI ivI Ae 1i z* Be1i z*
To match the interior flow, we need f 0 as z which requires B 0 , so that
u iv uI ivI Ae
1i z*
The no-slip condition u v 0 at the wall z 0 gives 0 uI ivI A so that
u iv uI ivI 1 e
1i z*
uI iv I 1 e z* cos z* i sin z*
Equating real and imaginary parts, we have u uI 1 e z* cos z* v I e z* sin z* v v I 1 e z* cos z* uI e z* sin z*
or
u uI e z* uI cos z* vI sin z*
(8.49)
v vI e z* vI cos z* uI sin z*
(8.50)
Taking the partials u uI v u e z* I cos z* I sin z* x x x x
v v v I u e z* I cos z* I sin z* y y y y and putting them into the incompressibility condition (8.46), we have w u v z x y u v u v I I 1 e z* cos z* I I e z* sin z* x y y x From (8.39-40), we have
vI 1 2 pI y 2 xy uI 1 2 pI x 2 xy so that uI vI 0 x y and
u v w I I e z* sin z* z y x or w uI v I z* e sin z* z* y x
With w z 0 0 , we have *
uI
vI * z* w e sin z*dz* y x 0 z
At the edge of the Ekman layer where z* , we have
uI
v wE I e z* sin z*dz* y x 0 Now,
I e z* sin z*dz* 0
1 i 1 z Im e z iz dz Im e i 1 0
1 1 Im 0 i 1 2
so that wE
1 uI v I 2 y x
(8.51)
1 I 2 where I is the z-component of the vorticity of the interior flow.
If the boundary is rotating with angular velocity B relative to the rotating frame, (8.51) is generalized to wE
1
I B 2
(8.52)
Similarly, if the upper boundary at z L rotates with angular velocity T relative to the rotating frame, we have
wE
1 T I 2
(8.53)
8.5.4 The ‘Interior’ Flow According to §8.5.2, uI , vI , and wI are all independent of z. This means wE at the top and bottom of the interior flow must match. Eqs(8.52-3) thus imply 1 1 I B T I 2 2
(8.54)
or
I T B In the case of Fig.8.10, B 0 , T so that
I
vI uI x y
or, in terms of cylindrical coordinates er 1 ωI r r 0 ez
re ru I
r, , z ,
ez z 0
1 d ru I e z r dr
where we’ve used the fact
u I 0. z
Integrating, we have ru I
1 r 2 c1 2
or 1 c r 1 2 r is to be regular at r 0 , we must set the constant c1 0 , so that
u I
If u I
u I
1 r 2
Thus, the angular velocity of the fluid is the average of those of the boundaries. In other words, the motion of the fluid is entirely controlled by the boundary layers.
We now obtain the rest of the solution. Using (8.51), we have
uzI
1 1 I 2 2
The incompressibility condition in cylindrical coordinates 1 1 u u rurI I zI 0 r r r z
then reduces to 1 rurI 0 r r
which gives rurI c2 However, if urI is to be regular at r 0 , we must set the constant c2 0 , so that urI 0 The secondary flow is therefore purely in the z direction. (see fig.8.10).
8.5.5 Unsteady Flow: ‘Spin-Down’ Consider the situation depicted in fig.8.11. Initially, the fluid, together with its boundaries at z L , rotate about the z-axis with angular velocity 1 . At t 0 , the angular velocities of both boundaries are reduced to Ω. Obviously, the fluid will eventually ‘spin-down’ to the same angular velocity. What interests us is the relevant time-scale. Physically, we expect the spin down to begin with the formation of Ekman layers on both boundaries. At this point, the interior inviscid flow still rotates essentially with angular velocity 1 . As is the case in §8.5.4, uI and vI are independent of z but not wI .
In fact, according to eqs(8.52-3), wI 0 ( wI 0 ) near the lower
(upper) Ekman layer. Thus, as time goes by, the Ekman layers extend towards the interior flow, which is the essence of the spin-down process. The almost uniform rotation eqs(8.37-8) for the inviscid interior flow are uI 1 pI (8.56) 2 v I t x vI 1 pI 2 u I t y
(8.57)
wI 1 pI t z
(8.57a)
uI v I wI 0 x y z
(8.46)
Eliminating p from (8.56-7) gives uI v I t y x
v I uI 2 y x 0 which, with the help of (8.46), becomes uI v I wI 2 0 t y x z
(8.58)
As discussed earlier,
wI 0 so that the vorticity in the interior decreases with time: z
I uI vI 0 t t y x in agreement of the Helmholtz vortex theorem (§5.3). Now, uI and vI are independent of z so that (8.58) can be integrated over the vertical span of the interior fluid to give I (8.58a) 2 wI 0 t where L is the height of the interior fluid and wI is the difference of wI at its top L
& bottom. By eq(8.51),
1 wI I 2
top lower on of the Ekman layer bottom upper
Hence, (8.58a) becomes L
I 2 I 0 t
(8.60)
with solution
I A exp 2
t Ae t / T L
where A is an arbitrary function of T
x, y
L
(8.61)
2
is called the spin down time. Applying the initial condition I 2 at we have 2 A and
I 2 e t / T
and
t0
(8.59)
In terms of cylindrical coordinates
ωI
er
re
ez
1 r r 0
ru I
z 0
ez
r, , z
and assuming axisymmetry, we have
1 d ru I 2 e t / T e z r dr
which is easily integrated to give ru I r 2 e t / T c1
Regularity at r 0 then sets c1 0 so that u I r e t / T
(8.62)
which decribes the main ‘dissipation’ process of the excess rotation. The other components of u I are obtained from (8.58) and (8.55) as urI
r t / L e L
uzI 2
z t / L e L
(8.63)
[cf. the last part of §8.5.4] The streamlines of this secondary flow are described by
dz dz u 2z dt zI dr dr urI r dt which integrates to ln z 2 ln r const or zr 2 const (see Fig.8.11).
8.6
Boundary Layers Separation
9.
Instability
9.1
The Reynolds Experiment
9.2a Interface Waves 9.2
Kelvin-Helmholtz Instability
9.3
Thermal Convection
9.4
Centrifugal Instability
9.5
Instability of Parallel Shear Flow
9.6
General Theorem on Stability of Viscous
Flow 9.7
Uniqueness & Non-Uniqueness Of Steady
Viscous Flow 9.8
Instability, Chaos, And Turbulence
9.9
Instability At Very Low Reynolds Number
9.1
The Reynolds Experiment
9.2
Kelvin-Helmholtz Instability
Consider a deep layer of inviscid fluid of density 2 moving with uniform speed U over another stationary deep layer of density 1 2 . [see Fig.9.4] Let the interface be y x, t . A small traveling wave disturbance there can be described by
x, t Aei kx t
(9.2)
with the understanding that only the real part of any complex quantity has physical meaning. The dispersion relation is (detailed derivation is given in §9.2a),
1 2 kU 2 k 2U 2 1 2 1 2 k g 1 2 k 2
(9.3)
where α is the coefficient of surface tension. This can be written as R iI where
R
kU 2 1 2
I
1 k 2U 2 1 2 1 2 k g 1 2 k 2 1 2
Thus, ω is complex if I is real, ie., k 2U 2 1 2 1 2 k g 1 2 k 2 0
or U 2 1 2 g 2 k 1 2 k 1
(9.4)
in which case, the dispersion indicates an exponential growth with time of any disturbance. This is known as the Kelvin-Helmholtz instability. The minimum flow U C that can induce such an instability is given by
g U C 2 1 2 min 1 2 k k 1 2 k
Setting f k
g 1 2 k k
with
k 0
we have df g 2 1 2 dk k
The extremum of f are therefore at kC
g 1 2
with f k C 2 g 1 2
which is obviously a minimum. Hence U C 2 1 2 2 g 1 2 1 2
(9.6)
which says that both gravity & surface tension serves to stabilize the system by raising the value of U C . Kelvin-Helmholtz instability can also occur in a continuously stratified fluid with d 0 0 dy The buoyancy frequency [see §3.8] g d 0 N2 0 dy
(9.7)
then serves as a measure of the stabilizing effects of the density distribution. We leave it as an exercise (Ex.9.2) to show that the system becomes unstable only if the Richardson number
J
N2
dU / dy
2
(9.8)
is less than
1 somewhere in the flow. 4
Such instabilities are observed in the atmosphere, sometimes in the form of ‘clear air turbulence’ and sometimes marked by distinctive cloud patterns.
9.2a Interface Waves We now generalize the surface waves results in §§3.2-4 to waves at the interface of 2 fluids. All quantities related to the upper and lower fluids are labeled 2 and 1, respectively. Mechanical equilibrium in the unperturbed state then requires 1 2 . The interface is at y x, t .
For ease of reference, we list below the basic eqs governing surface waves. on y x, t (3.18) u 0 t x y p 1 2 u gy G t t 2
(3.19)
2 2 0 x 2 y 2
(3.24)
Eqs(3.18,19) are linearized as: on y 0 t y p gy G t t
(3.21) (3.19a)
with the boundary condition p p0 on y and setting G t
p0
Eq(3.19a) becomes g 0 t
on
y0
For the interface waves, eqs(3.21,19a,24) become 1 2 on y 0 t y t y
(3.22)
(3.21a)
1 p1 gy G1 t t 1
(3.19b)
2 p2 gy G2 t t 2
(3.19c)
21 21 0 x 2 y 2
(3.24a)
22 22 0 x 2 y 2
(3.24b)
The boundary conditions are
1 y
0 y
p1 p2 p , and from (3.21a), 1 2 y y
2 y
0
(A1)
y
on y
(A2)
on y 0
(A3)
Consider the ansatz
Aei k x t
(3.23)
1 f1 y ei k x t
2 f 2 y ei k x t
To satisfy (3.24a,b) and (A1), we need f1 '' k 2 f1 0
f 2 '' k 2 f 2 0
with the boundary conditions
f1 0
f2 0
f1 c1e k y
f 2 c2 e k y
Thus
where c1 , c2 are constants. The interface condition (A3) then demands c2 c2 C so that (B) becomes
1 Cek y i k x t
2 Ce k y i k x t
Eq(3.21) gives kC i A
(3.21b)
(B)
iC gA ei k x t G1
p
iC gA ei k x t G2
1
Since G1 , G2 are functions of t only, they must vanish identically. Consistency then demands p Pe
i k x t
where P is a constant, so that (C) become P P iC gA iC gA
1
2
Eliminating P gives
iC 1 2 gA 1 2 0 Eq(3.21b,c) are compatible only if
i
k
i 1 2 g 1 2
0
ie.,
kg 1 2 2 1 2 0 or
1 2 1 2
2 kg
1 2 1 2
kg
The phase velocity is therefore
vp
k
g 1 2 k 1 2
and the group velocity:
vg
d 1 g 1 2 1 vp dk 2 k 1 2 2
(3.21c)
p
2
(C)
9.2a.1
Surface Tension
The effect of the surface tension at the interface is to modify the interface condition (A2) to 2 y x, t p2 p1 2 at x where α is the coefficient of surface tension. Thus, eq(C) is modified into p iC gA ei k x t G1 1
1
and
iC gA ei k x t G2
1
2
p k Ae 2
1
As before, G1 G2 0 and
p1 Pei k x t so that iC gA
P
1
and
k 2 P iC g A 2 2 Eliminating P gives iC 1 2 A g 1 2 k 2 0
which is compatible with (3.21b) only if
k
i
i 1 2 g 1 2 k 2
0
ie.,
k 3 kg 1 2 2 1 2 0 or
2
2 k 3 kg 1 1 2 1 2
i k x t
9.2a.2
Uniform Flow
We now consider the case where the flow of the upper fluid tends to a uniform speed U sufficiently far away from the interface. To emphasize this fact, we write u1 u1 , v1 1 , 1 x y where
u 2 U u2 , v2 U 2 , 2 x y
u1 y v1 y u2 y v2 y 0
The linearized version of eqs(3.18) is 1 t y 2 U t x y
on y 0
(3.21’)
which combines to give an interface condition 1 2 on y 0 U y y x to supplement 2 p2 p1 2 x
at
y x, t
The linearized version of eqs(19,24) are 1 p1 gy G1 t t 1
(3.19’)
2 p2 1 2 U U 2 gy G2 t t 2 2 x
(3.19’’)
21 21 0 x 2 y 2
(3.24a)
22 22 0 x 2 y 2
(3.24b)
Consider the ansatz
Aei k x t
1 f1 y ei k x t
(3.23)
2 f 2 y ei k x t
(B)
The solutions to (3.24a,b) are still
1 C1ek y i k x t
2 C2e k y i k x t
Eq(3.21’) gives i A kC1 0
i kU A kC2 0
(3.21b’)
Eq(3.19’,’’) become
iC1 gA ei k x t G1
p1
1
1 2 1 i k x t U p k 2 Ae 2 2 1
iC2 ikUC2 gA ei k x t G2
Since G1 , G2 are functions of t only, we must have G1 0
and
1 G2 U 2 2
Consistency also demands
p1 Pei k x t where P is a constant, so that (C’) become P iC1 gA
1
k 2 P i kU C2 g A 2 2 Eliminating P gives i1C1 i kU 2C2 g 1 2 k 2 A 0
which becomes, with the help of the 1st eq in (3.21b’),
2 1 2 i kU 2C2 g 1 2 k A0 k Eq(3.21b’,c’) are compatible only if k i kU 2
i kU g 1 2 k 2
2 1 0 k
ie., kg 1 2 k 3 2 1 kU 2 0 2
(3.21c’)
(C’)
or
1 2 2 2kU 2 k 2U 2 2 kg 1 2 k 3 0
[cf. Ex.3.6]
so that
1 1 2
kU k 2U 2 2 k 2U 2 kg k 3 2 2 1 2 2 1 2
1 1 2
kU k 2U 2 k g k 2 2 1 2 1 2 1 2
or
1 2 kU 2 k 2U 2 1 2 1 2 k g 1 2 k 2
(9.3)
9.3
Thermal Convection
Consider a viscous fluid resting between 2 horizontal rigid boundaries at z 0 and z d . A temperature difference T is maintained between the boundaries, with the lower one being hotter. If the density of the fluid decreases with rising temperature, the fluid becomes top-heavy. Nonetheless, if T is increased from 0 slowly by small steps, the fluid can remain stable up to a critical value whereupon an organized cellular motion sets in. (see fig.9.6)
9.3.1 Linear Stability Theory 9.3.2 Stability to Finite-Amplitude Disturbances 9.3.3 Experimental Results
9.3.1 Linear Stability Theory For small temperature changes, the fluid density ρ at temperature T can be approximated by
1 T T
(9.9)
where α is the volume coefficient of thermal expansion, and the density at temperature T . Since the change in ρ is usually slight, the fluid is still approximately incompressible: D 0 Dt
and
u 0
(9.10)
Assuming the viscosity to be temperature independent, the conservation of momentum still takes the form of the Navier-Stokes eqs.
Du p 2u g Dt
(9.11)
Thermal conduction in a fluid is a rather complicated process. For an introductory discussion, see Landau & Lifshitz, “Fluid Mechanics”, chapter V. Here, we simply generalize the elementary heat diffusion eq. T 2T t
to include convective effects, ie., DT 2T Dt
(9.12)
Here, κ is the thermal diffusivity of the fluid. It is related to the thermal conductivity χ by
cp
where c p the specific heat at constant pressure.
What (9.12) describes is a situation where the change of the amount of heat inside a fluid element is due solely to the exchange of heat by conduction with the surrounding fluid. Neglected are all other energy sources (eg., work done on fluid element by stresses exerted by the surrounding fluid) and sinks (eg., energy dissipation due to
viscosity). In the unperturbed state of rest, the temperature T0 z must satisfy the steady state, no flow version of (9.12), namely d 2T0 0 dz 2 of which solution is
(9.12a)
T0 z c1 z c2
Using the boundary conditions (see Fig.9.6)
T0 0 Tl we have Tl c2
T0 d Tl T
and
and
Tl T c1d c2
which is easily solved to give T0 z
z T Tl d
(9.13)
The corresponding density distribution eq(9.9) thus becomes
0 z 1 T0 z T
(9.14)
with the accompanying hydrostatic pressure is given by the static, stationary version of (9.11): 0
dp0 0 z g dz
(9.15)
Consider a slight perturbation to the system such that
T T0 z T1
0 z 1
p p0 z p1
u u1 u1 , v1 , w1
(9.16)
where the perturbations, distinguished by the subscript 1, are assumed small and functions of
x, y , z , t .
Eqs(9.9) thus become
0 1 1 T0 T1 T which, with the help of (9.14) becomes 1 T1
(9.17)
Eqs(9.10) is now 1 u1 1 0 t u1 0
(9.18)
Eq(9.11) is
0 1
Du1 p0 p1 0 1 2u1 g Dt
which, with the help of (9.15) becomes
0 1
Du1 p1 0 1 2u1 1g Dt
which in term can be linearized to
0
u1 p1 0 2u1 1g t
(9.19)
Finally, eq(9.12) becomes D T0 T1 2 T0 T1 Dt
which, with the help of eqs(9.12a,13), becomes w1
dT0 DT1 2T1 dz Dt
which in term can be linearized to w1T0 '
T1 2T1 t
(9.20)
where, according to (9.13) T0 '
dT0 T const dz d
For small density changes, 0 z in eq(9.19) may be approximated by the constant
. Eqs(9.17-20) then become a set of PDE’s with constant coefficients. We now aim to obtain an equation involing w1 alone.
To begin, we eliminate p1 in (9.19) by taking the curl of the whole equation: 2 u1 1g 1 g t Using (9.17) to replace 1 , we have
2 u1 T1 g t Taking the curl again and using the vector identities
a a 2a a b b a a b a b b a we have 2 2 2 u1 g T1 g T1 t where we’ve also used (9.18).
(9.20a)
Since g 0,0, g , the z-component of (9.20a) is
2T1 2 2 2 w1 g 2 g T1 z t 2T 2T g 21 21 y x
(9.21)
Rewriting (9.20) as 2 T1 w1T0 ' t T1 can be eliminated from eq(9.21) by operating on the whole equation with 2 , so that t
2 2 2 2 2 2 w1 g 2 2 T1 y t t t x 2 2 gT0 ' 2 2 w1 y x
Since derivatives of x and y are always in the combination 2
(9.22)
2 2 , we see x 2 y 2
that w1 is isotropic in the x-y plane. groups: t,
x, y , and z.
The independent variables thus fall into 3
A separable ansatz thus takes the form
w1 W z f x, y g t
(9.23)
Now, a successful separation means that we can write
DF
(9.23a) cF F where F is any one of the separated functions, cF a constant, and D an ordinary differential operator that involves only functions and derivatives of the independent variables of F. The complicated form of (9.22) means that the most general form of separation is not easily achieved. However, the situation will be greatly simplified if the operators
D for 2 of the separated functions take the simplest possible forms. Since the primary changes are in the z direction, we should simplify f and g so that 1 2 (9.24) f a 2 f 1 dg s g dt
(9.24a)
where the peculiar form of the proportionality constant for f is chosen to conform with Acheson’s notation, as well as for future convenience. Note that (9.24a) can be solved immediately to give
g t c1e s t
One useful property of eq(9.23a) is
D w1
Let D
w1 D F cF w1 F
(9.23b)
d .. dz
With the help of eqs(9.24, 24a, 23b), we have, 2 w1 D 2 a 2 w1
so that eq(9.22) becomes
2 2 2 w1 t t
s D 2 a 2 s D 2 a 2 D 2 a 2 w1 gT0 ' a 2 w1
Cancelling fg from both sides gives s D 2 a 2 s D 2 a 2 D 2 a 2 W gT0 ' a 2W
(9.25)
Eq(9.25) is a 6th order ODE. Its general solution thus contains 6 arbitrary constants. However, since the equation is homogeneous, only 5 of them are independent. (The 6th, taken as the overall constant multiplication factor, is immaterial). Since the boundary conditions always appear in pairs (1 each for the upper & lower boundary), the integration constants and boundary conditions can never be matched entirely by themselves. As will be shown later, there are 6 boundary conditions. We therefore have an eigenvalue problem whereby a solution satisfying all the boundary conditions exists only for some special values of the parameters in (9.25). The no-slip condition gives us u1 v1 w1 0
(BC1)
u1 v1 w1 0 t t t
(BC2)
n u1 n v1 n w1 n 0 x n x x n
(BC3)
n u1 n v1 n w1 n 0 y n y y n for all n, at z 0, d .
(BC4)
Since the boundaries are kept at fixed temperatures, we have
T1 nT1 nT1 n n 0 t x y for all n, at z 0, d . T1
(BC5)
We must now translate these conditions into ones on W so that they can be applied to (9.25). (BC1) together with (9.23) gives us
W 0
at z 0, d .
(BCa)
(BC3,4), with n 1 , and the incompressibility condition (9.18) gives us at z 0, d . (BCb) DW 0 (BC5), with n 2 , gives
2 2 2 T1 0 y 2 x
on z 0, d
so that (9.21) becomes 2 2 w1 0 t
on z 0, d
ie., s D 2 a 2 D 2 a 2 W 0
at z 0, d
On expansion, we get D 4 2 a 2 s D 2 a 2 W 0
at z 0, d
which, on applying (BCa), becomes D 4 2 a 2 s D 2 W 0
at z 0, d
(BCc)
Eqs(BCa-c) thus provide 6 boundary conditions to (9.25). Now, (9.25) is an ODE with constant coefficients. Its solution is therefore of the form e z which, on substituting into (9.25), gives
s 2 a 2 s 2 a 2 2 a 2 gT0 ' a 2
(9.25a)
For given values of a and s, this is a 3rd degree polynomial of 2 . Let the roots be
i2 , i 1, 2,3 .
The general solution to (9.25) is W Ai e i z Bi e i z 3
i 1
As mentioned before, only 5 of the 6 constants Ai , Bi are independent.
Ai , Bi , s
The determination of
so that the boundary conditions as well as (9.25a)
are satisfied is a straightforward but tedious task which is not particularly illuminating. For illustrative purposes, it may be more fruitful to attack another related problem where the boundary conditions are simplified to at z 0, d W D 2W D 4W 0 which is easily seen to be satisfied by the ansatz W sin
N z d
(BC)
N 1, 2,3,
N The secular equation for the eigenvalue s is simply (9.25a) with 2 . d Setting 2
N a a a d (9.25a) becomes 2 *
2
2
2
2
s a*2 s a*2 a*2 gT0 ' a 2
or s 2 a*2 s a*4 gT0 '
a2 0 a*2
with solution
1 s a*2 2
2
2
a2 4 a 4 a* gT0 ' 2 a* 4 *
Using T0 '
T d
we have s
1 2 a* 2
a*4 4 g
T a 2 d a*2
Hence, for T 0 , the factor inside the square root is always positive so that s is always real. Furthermore, if
T a 2 2 a*4 (9.26a) 2 d a* we have s 0 , so that the perturbation grows exponentially with time and the
2
a*4 4 g
system is unstable. Eq(9.26a) can be simplified as
g
T a 2 a*4 2 d a*
g T a*6 1 2 N 2 2 2 2 a d a a d 2
3
(9.27b)
Now, a is not a parameter related to the system. It is simply some unknown constant related to the horizontal length scale via eq(9.24). Hence, (9.27b) is satisfied as soon as the left side is greater than the minimum of the right side with respect to both a and N. The minimum with respect to N can be found by inspection to be at N 1 . That for a satisfies 3
2
2 2 3 2 3 a 2 2 2 a 2 2 2a 0 a d a d
or
2 2 a 2 3a 2 0 d a2
2
2d 2 so that the instability criterion is 3
g T 2d 2 3 2 27 4 2 2 d 2d 4d 4 which, in terms of the Rayleigh number
g Td 3 R
(9.27)
is simply
R
27 4 4
For the boundary conditions (BCa-c), the instability criterion is
R 1708 which corresponds to a 3.1/ d .
9.3.2 Stability to Finite-Amplitude Disturbances We have just shown that a critical value of T exists, above which disturbances grow exponentially with time so that even infinitesimal ones may become finite as t . A related question is whether a critical value of T also exists, below which any disturbances die out eventually. In our idealized system of thermal convection, the answer is yes and the 2 critical values are equal. Thus, for R 1708 , all disturbances subside eventually and the system will return to the unperturbed state of rest.
9.3.3 Experimental Results Criteria for the onset of instability as calculated from the linear theory are usually in good agreement with experiment. However, after the system become unstable, the predicted exponential growth of the disturbances cannot be substained indefinitely since non-linear effects will eventually become substantial enough to halt the growth. The system then reaches a steady state which cannot be described by the linear theory. With respect to the thermal convection discussed earlier, this means only the critical value R c of the Rayleigh number agrees with observation. Other interesting observations that are beyond the reach of the linear theory include: 1. Convection patterns such as the 2-D rolls and hexagonal cells in the x-y plane (see fig.9.7). 2.
Shifts of the unperturbed steady state convection to other, often time- dependent, patterns as one continues to increase R to values well beyond R c .
3.
Effects due to variation of surface tension in case of a free upper surface. (Benard instabilities).
9.4
Centrifugal Instability
Consider 2 concentric, rotating, circular cylinders with the gap between them filled with a viscous fluid. Let the inner and outer cylinders, together with their associated quantities, be labeled 1 and 2, respectively. As discussed in §2.4.2, the steady flow u Ar
B r
(2.31) (9.28)
with A
r121 r2 2 2 r12 r2 2
B
r12 r2 2 1 2 r2 2 r12
(2.32) (9.29)
is an exact solution to the Navier-Stokes eqs that satisfies the no-slip boundary condition. If the cylinders rotates in the same sense, instability occurs if 1 c , where c is some critical value that depends on 2 , r1 , r2 , and ν. [see eq(9.41)]. The ensuing flow consists of counter-rotating Taylor vortices superimposed on the main rotary flow. [see Fig.9.8]
9.4.1 Linear Stability Theory 9.4.2 Inviscid Theory: the Rayleigh Criterion 9.4.3 Experiments
9.4.1 Linear Stability Theory Since the flow is axisymmetric, all quantities related to the unperturbed, steady floware functions of r only. Small quantities describing the perturbation will be distinguished by a prime. They are assumed to be functions of
r, z and t.
For example, we write u ur ',U u ', uz ' p p0 p '
(9.30)
The Navier-Stokes eqs. in cylindrical coordinates were derived in §2.4:
ur u2 1 p u 2 u u ur 2ur 2r 2 t r r r r u uu 1 p u 1 u u r 2u 2 2 r 2 u t r r r r uz 1 p u uz 2uz t z where u f ur u uz f r z r
2 2 f r 2 2 2 f z rr r r 2
The incompressibility condition is: u
u u rur z 0 rr r z
The unperturbed steady flow satisfies
u0 U r e
p0 p0 r
U 2 1 dp0 r dr
d dU r rdr dr
Note that
1 2 U 0 r
(2.30)
(2.22)
u 0 U
r
so that u0 ur u0 u u0 uz 0 Using (2.30) to cancel out the unperturbed parts, eq(2.22) become ur ' 1 1 p ' u ' 2 u ' ur ' 2U u ' u '2 ur ' r2 t r r r
u ' U u ' u ' dU 1 2 ur ' u ' u ' r u ' 2 u ' t dr r r uz ' 1 p ' 2u ' u ' uz ' z t z and u
u ' rur ' z 0 rr z
where 2 2 2 r 2 z 2
The linearized version is ur ' 2U u ' 1 p ' u ' 2 ur ' r2 t r r r u ' U dU ur ' t r dr
1 2 u ' 2 u ' r
uz ' 1 p ' 2u ' z t z
(9.31)
and u
u ' rur ' z 0 rr z
To simplify the mathematics, we assume the gap between the cylinder to be small, ie., d r2 r1 r1 Now u ' 2ur ' O r2 d
so that
u ' ur ' O r2 2 r r1
ur ' r2 and similarly for u ' . 2 ur '
Eq(9.31) thus simplify to 2U u ' 1 p ' 2 ur ' r r t dU U 2 u ' ur ' 0 r t dr 1 p ' 2 u z ' z t
(9.33a) (9.33b) (9.33c)
Using
u ' u ' ur ' u ' O r O r r r r d r1 the incompressibility condition becomes ur ' uz ' 0 r z
(9.33d)
Finally, from (9.28), we see that dU U B B A 2 A 2 2A dr r r r
so that (9.33b) is simplified to 2 u ' 2 Aur ' 0 t
(9.33b’)
Eliminating p ' between (9.33a,c) gives 2 ur ' uz ' 2U u ' 0 r r z t z Further eliminating uz ' using (9.33d), we have 2 2 2 2 ur ' ur ' 2U u ' 0 2 r 2 r z 2 t z
or 2 2 2u ' 2U u ' 0 (9.34) r r z 2 t Still further eliminating u ' using (9.33b’), we have
2 2 2u ' 4 AU ur ' 0 r r z 2 t 2
(9.38a)
which is a 6th order linear PDE for ur ' alone.
Now, the coefficients of (9.38a) are all constants except for the term
U . r
Thus, for 1 2 , we can write U 1 1 2 r 2
so that all the coefficients in (9.38a) are constants. In which case, we have 2 2 2u ' 4 A ur ' 0 r z 2 t 2
(9.38b)
We must now establish and then satisfy the relevant boundary conditions. No-slip condition obviously requires ur ' u ' uz ' 0 n ur ' n ur ' 0 n z n
(BC1) (BC2)
n u ' n u ' 0 n z n n uz ' n uz ' 0 n z n for all n on the surfaces r r1 , r2 . That is, only partials with respect to r can be finite. The incompressibility condition then implies ur ' 0 r
on r r1 , r2
(BC3)
From (9.34), we have 2 2 ur ' 0 t Since 2 2 2 r 2 z 2 we have
on r r1 , r2
(BC4)
2 2 2 2 4 4 4 2 2 2 2 z 2 r 2 z 2 r 4 z 2r 2 z 4 r
so that (BC4) becomes 2 4 4 2 4 ur ' 0 2 r 2z 2 r t r
on r r1 , r2
(BC4a)
Our task is to find a solution to (9.38b) that satisfy (BC1,2,3,4a). Eq(9.38b) is obviously separable. Since the partials with respect to z are all of even order, a periodic solution can be supported. Thus, we consider the ansatz
ur ' ur r e s t cos nz
(9.35)
so that ur ' sur ' t
2 ur ' n 2 ur ' z 2
2 ur ' e s t cos nz D 2ur 2 r
where D
d . dr
2u ' e s t cos nz D 2 n 2 u r r
Thus, (9.38b) becomes 2 s D 2 n 2 D 2 n 2 ur 4 An 2ur 0
while the boundary conditions at r r1 , r2 simplify to ur 0 (BC1,2): (BC3): Dur 0 (BC4a):
sD 2 D 4 2n 2 D 2 ur 0
ie.,
4 s 2 2 D 2 n D ur 0
(9.38)
(BC5) (BC6)
(BC7)
Since (9.38) is homogeneous and of even order, there is an odd number (=5) of independent integration constants. The number of boundary conditions are even (=6). Therefore, it is an eigenvalue problem for which a parameter, eg. s, can take on only special values.
In general, s can be complex. Let s sR i , where both sR and ω are real. The time dependence of ur ' is then e sR t ei t . Instability occurs if sR 0 . The threshold to instability is therefore at sR 0 . For non-oscillatory evolution, 0 , so that the threshold is at s 0 . The equations decribing this marginal state are
2 D 2 n 2 ur 4 An 2ur 0 3
(9.38):
(9.38a)
subject to boundary conditions at r r1 , r2 : (BC5): ur 0 (BC6): Dur 0
D
(BC7):
4
2 n 2 D 2 ur 0
(BC7a)
Shifting to a more natural coordinate defined by r r1 r r1 x r2 r1 d we have D
dx d 1 d dr dx d dx
Eq(9.38a) thus becomes 3
1 d2 2 2 n 2 ur 4 An 2ur 0 d dx 2
which can be rearranged to give 3
d2 4 An 2d 6 2 nd u ur 0 2 r 2 dx
On setting T
a nd
4 Ad 4
2
it becomes 3
d2 2 2 2 a ur Ta ur 0 dx
(9.38b)
subject to boundary conditions at x 0,1 : (BC5): ur 0 dur (BC6): 0 dx 2 d4 2 d 2 a u 0 4 2 r dx dx
(BC7a):
(9.40)
Using, A
r121 r2 2 2 r12 r2 2
(2.32)
the Taylor number becomes 4d 3 r121 r2 2 2 2d 3 r121 r2 2 2 (9.41) 2 r1 r2 2 r1 where the last equality made use of the narrow gap appoximation r1 r2 . T
For a given a, eq(9.38b) and the boundary conditions (9.40) constitute an eigen problem with eigenvalues T a . Let
Tl a min T a
for given a.
The threshold of instability corresponds to the minimum of Tl ranging over all a. Remarkably, eq(9.38b) & (9.40) take the same form as the thermal instability eqs (9.25-6) with s 0 . Hence, the threshold for centrifugal instability is also (9.42) T 1708
Likewise, the critical value of n is approximately
3.1 . d
The streamlines of the secondary flow are shown in Fig.9.8. Since the period of the flow in the z direction is H
n
2 , the height of each cell is n
d 3.1
Although the final steps of our derivation invoke the assumption 1 2 , eqs(9.41-2) turns out to quite adequate whenever 1 , 2 are of the same sign.
Note that for 1 , 2 0 , eq(9.41-2) implies the necessary condition for instability is that r 2 decreases sufficiently quickly with r.
9.4.2 Inviscid Theory: the Rayleigh Criterion
9.4.3 Experiments
9.5
Instability of Parallel Shear Flow
9.5.1 The Inviscid Theory 9.5.2 The Viscous Theory 9.5.3 Experimental Results
9.5.1 The Inviscid Theory Consider the 2-D, inviscid, incompressible, flow between 2 flat plates at y L . (see Fig.9.10). The basic eqs are the Euler eqs u u u 1 p u v t x y x
(E1)
v v v 1 p u v t x y y
(E2)
and the incompressibility condition u v 0 x y
(C1)
Consider the parallel shear flow u0 U y ,0
(9.46)
where U is an arbitrary function of y. Substituting (9.46) into the basic eqs, we have 1 p (E1): 0 x (E2):
0
(C1):
00
1 p y
Hence, u0 is a solution to the basic eqs with p p0 const . Consider a general and presumably small 2-D disturbance. All quantities related to the disturbance will be distinguished by the subscript 1. They are in general functions of
x, y , t .
For example, u u 0 u1 ' U y u1 , v1
p p0 p1
The basic eqs becomes dU u1 u1 u 1 p1 U u1 1 v1 t x x dy y
(E3)
v1 v v 1 p1 U u1 1 v1 1 t x y y
(E4)
u1 v1 0 x y
(C2)
of which the 1st two can be linearized to u1 u 1 p1 U 1 U ' v1 t x x v1 v 1 p1 U 1 t x y
(E5) (E6)
where a prime denotes derivative with respect to y. Since all coefficients in these eqs are independent of t and x, the natural ansatz is f1 x, y , t f y ei kxt (9.47) where f can be u, v, or p. Thus (E5):
1 i kU u U ' v ikp
(E7)
(E6):
1 i kU v p '
(E8)
(C2):
iku v ' 0
(C3)
Eliminating p from (E7,8) gives
ikU ' u i kU u ' U '' v U ' v ' k kU v Further eliminating of u through (C3) gives
1 U ' v ' kU v '' U '' v U ' v ' k kU v k
which can be simplified to 1 kU v '' U '' k kU v 0 k
or kU '' v '' k2 v 0 kU
(9.48)
Boundary conditions to (9.48) are the impenetrable wall condition
v 0
y L
at
(9.49)
Eq(9.48) is therefore an eigenvalue problem for .
The following trick is due to Lord Rayleigh (1880). Let the complex conjugate of v be denoted by v * . L
dyv 9.48 *
gives
L
L
L
* v v ''dy
L
2 kU '' k 2 dy 0 kU
v
L
(9.50)
Now, integration by part gives, with the help of (9.49), L
* * L v v ''dy v v ' L
L
L
L
L
L
* v ' v 'dy
2 v ' dy
so that (9.50) becomes L
2 v ' dy
L
L
2 kU '' k 2 dy 0 kU
v
L
(9.50a)
Writing R iI the imaginary part of (9.50a) is L
2
I k v L
U ''
kU
2
dy 0
(9.51)
For I 0 , this means L
2
v
L
U ''
kU
2
dy 0
which is possible only if U '' changes sign somewhere in the interval. Since instability results from exponential growth with time, which in turn requires I 0 , this gives us the Rayleigh’s inflection point theorem, which says: A necessary condition for the linear instability of an inviscid shear flow U y is that U '' y changes sign somewhere in the flow.
9.5.2 The Viscous Theory Consider now the case where the fluid is viscous. The discussion in §9.5.1 is modified as follows. The basic eqs are now the Navier-Stokes eqs instead of the Euler eqs.
2u 2u u u u 1 p u v 2 2 t x y x y x
(NS1)
2v 2v v v v 1 p u v 2 2 t x y y y x
(NS2)
and the incompressibility condition u v 0 x y
(C1)
With the same ansatz as (9.47), the counterpart of eqs(E7,8) are 1 (NS7) i kU u U ' v ikp k 2u u ''
1 i kU v p ' k 2v v ''
(NS8)
while (C3) remains unchanged: iku v ' 0
NS 7 '
(C3)
gives
1 ikU ' u i kU u ' U '' v U ' v ' ikp ' k 2u ' u '''
(NS7a)
ik NS 8 gives 1 k kU v ikp ' ik k 2v v ''
NS 7a NS 8a
gives
ikU ' u i kU u ' U '' v U ' v ' k kU v k 2u ' u ''' ik k 2v v ''
(NS8a)
Eliminating u through (C3) gives
1 U ' v ' kU v '' U '' v U ' v ' k kU v k
i ikv '' v '''' ik k 2v v '' k which can be simplified to 1 kU v '' U '' k kU v k i v '''' 2k 2v '' k 4v k
(NS9)
Introduce the stream function
y ei kxt so that u1 and
y
u '
v1
x
v ik
Eq(NS9) becomes i kU '' ik U '' k kU '''' 2k 2 '' k 4
which can be rearranged to give i '''' 2k 2 '' k 4 kU '' k U '' k kU kU '' k 2 kU ''
In terms of ψ, the no-slip boundary conditions are at y L ' 0
(9.53)
(9.54)
We leave it as an exercise to show that for a plane Poiseuille flow defined as
y2 U y U max 1 2 L
(9.55)
(see Ex.2.3)
solutions of (9.53-4) leads to a curve of marginal stability shown in fig.9.11. Of particular interest is the fact that instability occurs for a band of wavenumbers k (slashed region in fig.9.11) if R
U max L
5772
(9.56)
In contrast, the flow (9.55) is stable for all wavenumbers k in an inviscid flow. Thus, viscosity plays a dual role. According to (9.56), it is stabilizing since it raises the value of threshold value of U max to reach instability. On the other hand, it is de-stabilizing because the flow would have been stable if viscosity is absent altogether.
9.5.3 Experimental Results Criterion (9.56) for the Poiseuille flow has been confirmed experimentally. However, it applies only to systems with extremely low level of background turbulence. Furthermore, non-linear effects are significant so that the criterion applies only when the amplitudes of the disturbances are sufficiently low.
9.6
General Theorem on Stability of Viscous
Flow Theorem: Consider an incompressible viscous fluid occupying region V t that is enclosed within a sphere of diameter L. Let u x,t be a bounded solution to the Navier-Stokes eqs in V t satisfying the boundary condition u u B x, t on the boundary S t of V t . Thus, there exists uM such that for all x V t at all t.
uM u x, t
Let u* x,t be another solution satisfying the same boundary condition but different initial conditions at t 0 . The “difference flow” v u* u
(9.59)
thus satisfies v0
on
S t
(9.60)
The kinetic energy E of v defined as E
1 v 2dV 2 V t
(9.61)
satisfies 3 2 2 t E E0 exp uM2 L2
where E0 is its initial value.
(9.57)
Thus if R
uM L
3
(9.58)
then as t and the flow is unstable. E0
Before the actual proof of the theorem, we 1st establish the following relation: v v 1 dE i v j ui i x dt V t x j j
2
dV
9.6.1 Proof of Eq(9.62) 9.6.2 Proof of the Theorem
(9.62)
9.6.1 Proof of Eq(9.62) Both u and u* are solutions of the Navier-Stokes eqs., therefore u 1 u u p 2u t u* 1 u* u* p* 2u* t Subtracting, we have v 1 u* u* u u p* p 2 v t Now,
u* u* u v u v u u v u u* v so that by setting 1 P p* p
we have v v u u* v P 2 v t
or, in index notations, vi u v P 2vi v j i u* j i t x j x j xi x j x j
Multiplying by vi and sum over i gives 1 2 ui 1 2 P 2vi v v v u v v v *j i i i j t 2 x j x j 2 xi x j x j
(A)
We now try to put as many as possible the spatial derivatives in divergence form. Thus u v v vi v j ui vi v j j ui v j i ui vi j x j x j x j x j u v vi v j j ui v j i where x j x j
v j x j
v 0
so that vi v j
u j x j
v vi v j ui ui v j i x j x j
(B)
Similarly u v 2 v 2 v 2u* j u* j v 2 * j u* j x j x j x j x j
u* j
where
x j
v* 0
P v P vi P vi P i vi xi xi xi xi
x j
(C)
(D)
vi vi vi vi vi vi x j x j x j x j x j
so that vi
vi x j x j x j
vi vi vi vi x j x j x j
(E)
Puuting (C-E) into (A), we have 1 2 v t 2 x j
1 2 v v v v vi v j ui v u* j vi P vi i ui v j i i i 2 x j x j x j x j
Upon integrating with
dV
the divergence term vanishes since
V t
1 2 v vi v j ui v u* j vi P vi i dV x j 2 x j V t
1 2 vi v v u v u v P v ni dS i i j i 2 * j i x j S t
v u 2 v u
S t
0
Hence
1
j i
i *j
P
vi vi n j dS x j
since v 0 on S t
1 2 vi vi vi v dV u v t 2 i j x j x j x j dV V t V t
The convective (Reynold’s) theorem gives
d 1 2 1 2 1 2 v dV v u ndS v dV dt V t 2 t 2 2 V t S t
1 2 v dV t 2 V t
since v 0 on S t
Thus
d 1 2 vi vi vi v dV u v i j x j x j x j dV dt Vt 2 V t which is (9.62).
9.6.2 Proof of the Theorem From 2
vi Aij 0 x j
for any Aij
we have 2
vi v 2 2 Aij i Aij 0 x j x j
(9.63)
which holds even when, as will be assumed hereafter, summation over repeated indices is implied.
Choosing Aij
ui v j
, we have
2
vi ui v j vi ui v j 2 0 x x j j 2
or 2 v 1 v ui v j i ui v j i x j 2 2 x j
2
On substituting into (9.62), we have dE dt 2
2 2 2 vi ui v j dV x V t j
Using
u v dV u
2 V t
2
i
j
2 M
v dV u
2 V t
2 j
2 M
E
we have 2 dE 1 2 2 vi uM E dV dt 2 V t x j
(9.64)
On the other hand, setting Aij h j vi in (9.63) gives 2
vi 2 v 2h j vi i h j vi x j x j
which, upon using h v h j vi2 2h j vi i vi2 j x j x j x j becomes 2
vi h 2 h j vi2 vi2 j h j vi x j x j x j
On integrating over V t , the divergence term vanishes since v 0 on its boundary
S t . Hence 2
vi 2 h j 2 dV v h v x j i x j j i dV V t V t
h h v dV 2
2
(9.64a)
V t
Now, (9.64a) holds for arbitrary h. If we manage to find an h such that h h2 C 0
(9.65)
for all x in V t for all t, where C is some real positive constant, (9.64a) becomes 2
vi 2C 2 E dV C v dV x j V t V t which, when substituted into (9.64), gives dE 1 2 uM C 2 E dt
Integrating, we have t E E0 exp uM2 C 2 which is our theorem eq(9.57) with
(9.66)
C
3 2 L2
(9.66a)
Our task is therefore 2-fold. 1st, we must show that such an h do exist and find the maximum C it affords. 2nd, we need to find the optimal h that can maximize C among all h’s. At this point, the dimension of the system enters into consideration. As stated in the theorem, we shall assume V t always lies inside a sphere, center at the origin, of diameter L. Hence, all points of the system satisfy x L .
The easiest construction is perhaps of the type
h h r er so that h h2
1 d 2 r h h2 2 r dr
Setting h r ar , where a is a constant, we have where r
f h h 2 3a a 2 r 2
Since f 2 a 2 r 0 r
for all r,
f is a monotonically decreasing function of r. Its minimum is therefore at r
L where 2
a 2 L2 4 The value of a that maximize this is given by f 3a
1 3 aL2 0 2
ie., a so that
6 L2
L . 2
h r
6 r L2
f h h2
6 6 3 2 r2 2 L L
and C f
r
L 2
6 6 9 3 2 2 L 4 L
which is only slightly less than the value (9.66a) given in the theorem. Eq(9.66a) is achieved with h r
L
tan
r L
so that L dh r for r sec 2 0 2 dr L L Thus, h is monotonically increasing with minimum at r 0 . 2
Furthermore f
dh h 2 h2 dr r
r 2 r r sec 2 tan tan 2 L rL L L L L 2 r 2 tan L L r L 2
L
2
r 2L tan 1 r L 2 2 1 tan x x L 2
The minimum of f is given by
2 2 tan x sec 2 x 0 2 x x
which implies sin x cos x x 0 ie., 2 x sin 2 x
whose solution is x0
or
r0
where
x
r L
At which point, (9.66a) is achieved and the theorem is proved.
9.7
Uniqueness & Non-Uniqueness Of Steady
Viscous Flow Theorem Consider a fixed region V of fluid that is enclosed in a sphere of diameter L. Let u and u* be 2 steady solutions of the Navier-Stokes eqs in V with the same value
u B x on the boundary of V. Let uM be an upper bound to u in V. If R
uM L
3
(9.68)
the 2 flows are identical, ie., u u* In other words, bounded steady viscous flow with R 5 is unique.
Proof The proposition of this theorem is just the steady state version of the theorem in §9.6. Thus, the t limit of (9.57) is satisfied, ie., 2 3 2 2 t E lim E0 exp uM 0 t L2
where (9.68) ensured the coefficient of t is negative. QED.
9.7.1 An Example of Non-Uniqueness of Steady Flow 9.7.2 Hysteresis
9.7.1 An Example of Non-Uniqueness of Steady Flow We now consider some variants to the Taylor experiment of §9.4 involving viscous fluid occupies the gap r1 r r2 between 2 coaxial cylinders. The differences we introduce are: 1. 2 0 . 2.
Fluids are also bounded by stationary plane walls at z 0, L with L adjustable. (see fig.9.13).
The system is characterized by 3 dimensionless numbers: r 1. Radius ratio 1 . r2 2.
Reynolds number
3.
Aspect ratio
R
1r1d
where
d r2 r1 .
L . d
Benjamin & Mullin (1982) had performed measurements on such an apparatus with r1 (9.71) 12.61 R 359 0.6 r2 The salient points were: 1. 2.
Some 20 different stable steady flows were observed. On theoretical grounds, they inferred the existence of 19 other steady flows that
3.
were unstable and hence unobservable. All observed flows were of an axisymmetric cellular nature as shown in fig.9.8.
4.
Different flows are distinguished by different number of cells and/or different
5.
sense of rotation within each cell. Appearance of a particular flow pattern depends on the way the boundary
6.
conditions (9.71) were achieved from an initial state of rest. If R 359 were achieved by small steps from 0, the same flow consisting 12 cells was always observed.
For more details, see T.B.Benjamin, T.Mullin, J.Fluid Mech 121, 219-30 (1982).
9.7.2 Hysteresis Measurements on very short cylinders, with 4 , indicated that transitions between 2-cell & 4-cell modes were decribed by a state diagram shown in fig.9.14. Such diagrams can be studied using catastrophic theory, a good reference of which is J.M.T.Thompson, “Instabilities & Catastrophes in Science & Engineering”, Wiley (1982). Folds in the surface implies non-uniqueness of solutions in parts of the R plane; hence hysteresis. The middle sheet of the fold corresponds to unstable solutions which are not observable.
9.8
Instability, Chaos, And Turbulence
Consider again the Taylor vortex apparatus with 1 2 and a temperature difference T T2 T1 0 maintained between the cylinders. This may serve as a crude model for the atmosphere as a rotating fluid under differential heating. Ref:
R.Hide, Q.J.R.Met.Soc. 103,1-28 (1977).
If Ω is sufficiently small, a weak differential rotation is observed (see Fig.9.16a). As Ω is increased in small steps past a critical value C that depends on T , baroclinic instabilities set in which amplify non-axisymmetric waves. Further increasing Ω increases the amplitude of these waves leading to a meandering jet structure reminiscent of atmospheric jet streams (see Fig.9.16b). Amplitude, shape, and wavenumber of this jet can be either steady or oscillatory. At still higher values of Ω, complicated, aperiodic fluctuations set in and the system become chaotic (see Fig.16c).
9.9
Instability At Very Low Reynolds Number
The theorem of §9.6 guarantees stability for flows of sufficiently low R. However, it applies only if u is prescribed on some, possibly varying, boundary. For flows with free boundaries, instabilities are possible for arbitrarily small R.
Viscous Fingering in Hele-Shaw Cell. A Hele-Shaw cell (see §7.7) is formed by pressing 2 flat sheets of transparent plastics together with syrup filling the small gap (~2mm) in between. A hole is drilled on the top sheet for the insertion of a syringe. Air is then injected by the syringe. In principle, one might expect the air to displace the syrup in a symmetrical manner, so that the air-syrup interface is circular. However, such an interface is found to be unstable. Ripples soon form and develop into fingers as shown in Fig.9.20. Such behavior is observed whenever a more viscous fluid is displaced by a less viscous one. It is called the Saffman- Taylor instability.
Buckling of Viscous Jets A falling jet of viscous fluid is approximately symmetrical about a vertical axis if the height H is below a critical value H C . (see Fig.9.21a). If H H C , the jet becomes unstable and buckles (see Fig.9.21b). What makes it interesting is that, in contrast with the other instabilities discussed so far, it occurs only if R is less than some critical value.
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