Elementary Fluid Dynamics
D.J.Acheson, Clarendon Press (90)
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: uu x
,t u v w, ,
(1.1) Steady Flow: u t 0 (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( ), zz s( ) Then dx ds u dy ds v dz ds w / / / (1.5)
Material derivative, or, rate of change “following the fluid”: Df Dt d dt f x t y t z t t [ ( ), ( ), ( ), ] rc 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 f t u (1.6) Acceleration: D Dt t u u u u ( ) (1.7) Uniform rotation: u ( y, x, 0) Steady flow: 0 t u So that
D Dt u u u
, , 0
y x y x x y
2 , , 0 x y For steady flow:
Df f f f Dt s u u s uwhere sss is the tangent vector to a streamline.
u
f 0 means f is constant along a streamline but it can vary between different streamlines.0
Df
Dt means f is constant for a fluid particle but it can differ between different
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:
V m
dV Rate of change is V V dm d dV dV dt dt t
Net rate of mass leaving the volume is
S
dS V
dV
u n
u
Conservation of mass means
V t dV V dV
u (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
V V D Df fdV f dV Dt Dt
uGoing back to eq.(A), since it is true for all V, we have
0 t uwhich is called the equation of continuity.
Using
u u u
D Dt t
u
We can writing the equation of continuity as
0
D
Dt u
For an incompressible fluid, ρ = constant, or D 0
Dt
, so that
0
u (incompressible condition) Force on dyed blob in inviscid fluid:
S p dS V pdV
n
which means the force density is p.
1.2.1 Euler’s Equation
Newton’s law applied to a fluid volume:
V V
D
dV p dV
Dt
u
gUsing the convection theorem, the left side is
V V D D D dV dV Dt Dt Dt
u
u u u u u u V D dV Dt
uwhere the last equality was obtained with the help of the eq. of continuity.
We thus obtain Euler’s eqs.:
D
p Dt
u g (1.12)
where g is the gravitational acceleration.
Writing g (1.13) we have D p Dt t
u u uor p t
u u Using
1 2 2 u u u u u we have
1 2 2 p t
u u u u (1.14)1.2.2 Bernoulli Streamline Theorem
For steady flow, (1.14) becomes
u
u H where 2 1 2 p H
u (1.15) Using
0 u u u we have
u
H 0 (1.16)ie, H = const along streamline of an ideal fluid in steady flow.
Irrotational flow means
0
u (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: ω u (1.18)
[ ( , , ), ( , , ), 0]u x y t v x y t u
0, 0, 0, 0, 0 v u x y z x y u v
i j k ω v u x y
(1.19)Consider 2 mutually perpendicular line segments AB & AC (fig.1.3):
( , , ) ( , , ) B A v v v v x x y t v x y t x x = CCW angular velocity of AB ( , , ) ( , , ) C A u u u u x y y t u x y t y y
= CW angular velocity AC Hence 1 1 2 2 v u x y
= average angular velocity of AB & AC.
Note: ω donotes local ‘spin’ & has nothing to do with global rotations. (Fig.1.6)
Eg. Shear flow (fig.1.4)
y, 0, 0
u (1.20)
has no global rotation but
0
v u x y
Eg. Line vortex flow
ˆ k r u θ (1.21) has 1 1 0 0 0 r z r r r r z r r z u ru u k
r θ z r θ z ω u but is obviously rotating as a whole.
ˆ
r
u
θ
spin has same rateas global rotation. Eg. Rankine Vortex (fig.1.7)
2 r u a r for r a r a (1.23) 0 r z u u
1.4
Vorticity Equation
Euler’s eq. (1.14):
H t u u u can be written as H t u ω u both sides gives
0 t ω ω u (1.24) Using
a b b a a b a b b a we have
ω u u ω ω uω u u ω
u ω ω u where 0 u for incompressible fluid
0 ω u for any vector u.
Thus (1.24) becomes
0 t ω u ω ω uor
D Dt ω ω u (1.25)which is called the vorticity equation.
For 2-D flow
, ,
, , ,
, 0 u x y t v x y t u ω
0, 0,
we have
0 z ω u u so that 0 D Dt ω (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
u
0 ωso 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)
upper
p p plower p
Flow is irrotational.
Bernoulli theorem means 1 2 2
p u is constant everywhere. Thus lift means uupper ulower.
1.5.1 Circulation
Circulation Г around closed curve C is
C d
u x (1.31) Stokes theorem:
C d S dS
u x
u n
(1.32) S dS
ω nwhere 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) sin K
UL
(1.34) where U u L = length of wing1.5.3 Lift
For ideal flow, Drag = 0
Lift L U (1.35)
2 sin U L
L
(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
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):
1 2 p t
u u u u g (2.3) 0 uSee Landau for a derivation.
The kinematic viscosity is defined as
No-slip condition: 0 u at stationary boundary. Reynolds Number: UL R (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
O U u O
U L uon the Navier-Stokes eqs. (2.3), we have
Inertia term:
O
U2 L u u Viscous term: 2 O
U 2 L u (2.5) so that
2 2 inertia term viscous term U L O O R U L (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
1 2 O R L
(2.7)In general, high R flows are unstable & easily become turbulent.
Low R flows:
0.01
R .
2.2 Navier-Stokes Eqs.
Consider the Euler eq
D D p Dt Dt
u u u u Using
D Dt t
u u u u u u u u
t
u uu we can re-write it as
0 p t
u uu or
0 t
u (A)where the momentum flux density tensor П is
ij u ui j pij
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 u ui 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,
D
p Dt
u
' D p Dt u or ' p t
u u u (B)Next, we need to determine the form of '.
Now, viscous effects are due to “friction” between adjacent fluid elements moving
with different velocities. Hence ' should depend on i j
u x
but not on u itself.
Furthermore, there’s no friction if the fluid rotates as a whole with uniform angular
velocity Ω so that u
Ω r
or ui ijkjxk. Now:
i m
ijk j km ijm j mji j
m i u u x
x and 0 i iji j i u x
Hence, the choice
' i j k ' ij ij ji j i k u u u a b x x x
where a, b are constants, will satisfy the above requirements.
A more common form is
2 ' 3 j i k k ij ij ij j i k k u u u u x x x x
(C)where the positive constants
a
2 2
3 b 3a b
are called coefficients of viscosity. (Landau used η instead of μ) This also defines the Newtonian fluid. [cf eq(2.1)].
'ji i i j j i j u u p u t x x x
2 3 j i k k ij ij i j j i k k u p u u u x x
x x x
x
2 3 j k i i k j j i u u u p x
x
x x x 1 3 k i i k j j u u p x
x
x x or in vector form: ' p t
u u u 2 1 3 p
u u
2 1 3 p
u u (D)For incompressible fluids, we have the Navier-Stokes eqs.:
2 p t
u u u u (2.3) 0 uNote also that for incompressible fluids,
' i j ij j i u u x x
[cf (2.1)]2.3 Simple Flows
2.3.1 Plane Parallel Shear Flow
Plane parallel shear flow is defined as
, , 0, 0u y t
Thus 0 u x u (incompressible)
Navier-Stokes eqs becomes
2 2 1 u p u t
x
y 0 p p y z (2.9)The last eq means p and hence p dp
x dx
is a function of
x t, only.Hence, the right and left hand side of
2 2 1 p u u 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 t0.
At t0, boundary at y0 is set in motion in x-direction with constant speed U. No-slip condition means u
U, 0, 0
at boundary for t0.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
x
so that p f t x
C.Let there be no externally applied pressure gradient, ie., p at x are equal. This means we must have f t
0 and pC.Eq(2.9) thus becomes
2 2 u u t
y (2.12)This must be solved with the initial condition
, 0 0u y for y0 and boundary conditions
0,u t U for t0
, 0u t for t0
Now, (2.12) is invariant under the scale transformation
yy t2t
where α is a constant.
This means we can attempt a similarity solution:
u f
where y t
(2.13) so that 3 2 y t t
1 y t
3 ' ' ' 2 2 u y f f f t t t t
1 ' ' u f f y y t
2 2 1 1 '' '' u f f y y t t
where f ' df d
. Hence (2.12) becomes 1 ' '' 2 f f t t or '' 1 ' 2 f f
so that 2 1 ln ' 4 f const or 2 4 ' f Be whence
2 4 0 s f
A B
e dswhere A, B are constants.
Using
for t0 or y
0
for y0
the boundary & initial conditions become
0 f and f
0 U Hence 2 4 0 1 0 A B e s ds A B
U A whence
2 4 0 1 1 s f
U e ds
(2.14) or
2 4 0 1 , 1 y s t u y t U e ds
[see fig 2.8] The vorticity is 2 2 4 4 y t u U U e e y y t
(2.15)which indicates a diffusion with standard deviation 2 t
. In other words, viscous diffusion time O
L2 (2.16)where L is a distance.
2.3.3 Flow Down Inclined Plane
With reference to fig. 2.9, the Navier-Stokes eqs
1 2 p t
u u u u g becomes2 2 2 2 1 sin u p u v u u g t x y
x
x y
2 2 2 2 1 cos v p u v v v g t x y
y
x y
where u
u v, , 0
.The no-slip condition means u0 for y0. Hence u must vary with y if it’s not identically 0.
The simplest solution possible is therefore a steady flow of the form
, , 0u y v y
u
The incompressibility condition
0 u v x y becomes 0 v y so that vconst.
However, v
0 0 so that v0 for all y. The steady state Navier-Stokes eqs. thus become2 2 1 0 p d u gsin x
dy
1 0 p gcos y
(2.17) The 2ndeq gives
cos p
gy
f x (A)Since all streamlines are in the x direction, the free surface must be yh 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 p , we have0
0
p p and du 0
dy
at yh (2.18)Putting these into (A) gives
0 cos
p
gh
f xso that (A) becomes
0 cos cos p
gy
p
gh
or
0 cos pp
g hy
whence 0 p x and the remaining Navier-Stokes eq reduces to
2 2 0 d u gsin dy
which gives sin du g y A dy
2 sin 2 g u y Ay B Putting in the boundary conditions 0 u at y0 and 0 du dy
at yh we have 0B sin 0 g h A so that 2 sin 2 g y u
hy
(2.19)2 3 2
3
0 0
sin sin sin
2 6 2 3 h g h y g h h g Q udy
hy dy
h
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 from2 2 u u t
y (2.20)with the initial condition
, 0 0u y for h y 0 and boundary conditions
0,u t U and u h t
, 0 for t0The form of (2.20) clearly suggests the use of the method of separation of variables. Let
uY y T t (2.20) becomes 2 2 dT d Y Y T dt
dy or 2 2 dT d Y Tdt
Ydy
const Thus, for 0, t T Ae Y BeyCeywhere A, B, C are constants and
For 0, we have T A Y B y' C'where B C', ' are constants. Hence,
0 0
0 , t y y u y t a y b e a e b e
(A) is a general solution of (2.20).The next task is to adjust the constants to satisfy the initial & boundary conditions:
0 0
0 , 0 y y 0 u y a y b a e b e
for h y 0
0
0 0, t u t b e a b U
for t0
0 0
0 , t h h 0 u h t a h b e a e b e
for t0The last 2 eqs can be satisfied only if the time dependent sums vanish identically. Hence, we have 0 b U 0 0 0 a hb so that a0 U h and
0 0 t e a b
0 0 t h h e a e b e
for t0Since the et terms are independent, each term in these sums must vanish individually so that 0 a b 0 h h a e b e which gives b a 2 sinh 0 h h e e
h so that h n i and 2 2 2 2 0 n h
Eq(A) becomes
2 2 2 1 , sin n t h n n U n u y t y U e A y h h
where An 2ia
0
are to be determined by the initial condition, namely,
1 , 0 nsin 0 n U n u y y U A y h h
for h y 0 Using 0 sin sin 2 nm nx mxdx
we have 0 sin 2 0 h n U n dy y U y A h h h
or 0 2 1 sin h n y n A U dy y h h h
0 2 1 sin n z U dz z n n
where z n y h
0 2 1cos cos sin
n U z z z z n n
2U n so that, finally
2 2 2 1 2 1 , 1 sin n t h n y n u y t U e y h n h
(2.21)2.4 Flow with Circular Streamlines
In cylindrical coordinates
r, ,
z
, (see Appendix A.5)
u u ur, , z
rur u zuz u e e e 1cos 2sin r
e e e e e1sin
e2cos
ez e3 r e e r er z f f r r
z e e e r z f u u u f r r
z u
z r V V rV r r r z V 1 r z r z r r r z V rV V
e e e V 2 2 2 2 2 2 f r f r r r r
z To write the Navier-Stokes eqs
1 2 p t
u u u uin cylindrical coordinates, we 1stconsider the term
u
u:
u
erur
ur
u
ere ur
ur
r r z r r r u u u u u r r
z e e u
r r r r u u u r e e u
r r r u u u r e e u
u
e u
u
u
e e
u
u
r z u u u u u r r z
e e u
u u u r e e u
2 r u u r e e u
u
ezuz
uz
u
ezez
u
uz
z uz e u so that
r
2
r r r z z u u u u u u r r u u e e u e e u e u
2
r
r r z z u u u u u u r r e u e u e u Next, we turn to2u :
2 2
2 2 2 2 rur r rur r r r r
z e e
2 2 2 2 2 r r ur rur r r r z r
e e Now
2 2 r r r r r r u u u
e e e 2 2 2 2 2 r r r r r r u u u
e e e 2 2 2 r r r r r u u u
e e e so that
2 2 2 2 2 2 1 2 r r r r r r r r r u u u r u u r r r z r
e e e e e 2 2 1 2 r r r r r u u u r
e e e 2 2 2 2 r r r r u u u r r
e e Similarly
2 2 u u u
e e e 2 2 2 2 2 u u u
e e e 2 2 2 r u u u
e e eso that
2 2
2 2 2 2 u r u u r r r z r
e e e 2 2 2 2 2 1 2 r u u r u u r r r z r
e e e e 2 2 1 2 r u u u r
e e e Finally, with
2 2 zuz z uz e e we have 2 2 2 2 2 2 2 1 2 2 r r r r r z z u u u u u u u r r r
u e e e e e e 2 2 2 2 2 2 2 2 1 2 r r r r z z u u u u u u u r r r r
e e e Hence
2 2 2 2 1 2 r r r r u u p u u u u t r r r r
u
2 2 2 1 1 2 r r u u u p u u u u t r r r r
u (2.22)
1 2 z z z u p u u t
z
u where r z f u u u f r r
z u 2 2 2 2 2 2 f r f r r r r
z The incompressibility condition is simply
z 0 r u u ru r r r z u2.4.1 Differential Eq.
,u r t
u e (2.27)
The incompressibility condition
z 0 r u u ru r r r z u is automatically satisfied.The Navier-Stokes eqs (2.22) becomes
r u : 2 1 u p r r
u : u 1 p r u 12u t r r r r r
z u : 0 1 p z
Now, u is a function of
r t, 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 twhere C is another arbitrary function of
r t, . Now, p must be single valued. This means
, ,
, 2 ,
p r
t p r
n
tThis is possible only if f 0, ie, p is a function of
r t, only. The Navier-Stokes eqs can be further simplify to
, u r t u e p p r t
, r u : 2 1 u p r r
u : u r u 12u t r r r r
(2.30)2.4.2 Steady Flow Between Cylinders
For steady flow, (2.30) becomes
0 d du r r u dr dr Setting ylnr, it becomes 0 d du u dy dy so that y y B u Ae Be Ar 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 1 1 1 1 B r Ar r 2 2 2 2 B r Ar r
Solving for A and B gives
2 2 1 1 2 2 2 2 1 2 r r A r r
2 2 1 2 1 2 1 2 2 2 2 1 2 2 1 2 1 1 r r B r r r r (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 ra t0 The cylinder is suddenly stopped at t0. We thus need to solve
2 1 u u r u t r r r r
with the above initial condition as well as the (no-slip) boundary condition 0
u at ra for all t0.
,
u r t R r T t we have 2 1 dT d dR r R Tdt R rdr r r
so that for 0,
t T t e
1 R r J
r
where Jn is the Bessel function defined by
2 2 2 n 0 d d n r k J kr rdr dr r [see M.Abramowitz, I.A.Stegun, “Handbook of Mathematical Functions”] Hence
, 1 t u r t c e J r
To satisfy the no-slip condition at ra, we need
n
a
where
n are roots of J , ie.,1 J1
n 0.Therefore 2 n a
and
2 1 1 , n t a n n n r u r t c e J a
The initial condition
u r for ra t0 means 1 1 n n n r r c J a
2 2 1 0 2 a n m nm n r r a J J rdr J a a
where
n is the nth root of J, we have
2 2 2 1 2 0 2 a n n n r a J r dr c J a
Using
1 1 1 0 1 J ar r dr J a a
( see I.S.Gradshteyn, I.M.Ryzhik, “Table of Integrals, Series, and Products”, formula 6.561.5) we have
2
2 3 2 2 1 2 n n n n a a J
c J
so that
2 2 n n n a c J
whereupon
2 1 1 2 2 , n t a n n n n a r u r t e J J a
Finally, using
1 1 2 J x J x J x x we have
0 2 1 2 0 n n n n J
J
J
(
n are roots of J )1 so that
2 1 1 0 2 , n t a n n n n a r u r t e J J a
(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 0 2 r u e (2.34)The vorticity singularity at r0 is expected to diffuse in a viscous fluid. Consider the circulation of a circular path C of radius r,
2
0 , , 2 , C r t d u r t rd
ru r t
u r
(2.35) Eq(2.30) 2 1 u u r u t r r r r
can be rewritten in terms of as follows: 1 2 u r 1 2 u t r t 2 1 1 1 2 u r r r r
2 2 2 3 2 2 1 2 2 1 2 u r r r r r r
2 2 1 u u u r r r r r r r 2 3 2 2 2 1 2 2 1 1 1 1 2
r r r r r 2
r r r r 2 3 2 2 1 1 1 1 2
r r r r r so that 2 3 2 2 3 1 1 1 1 1 1 2
r t
2
r r r r r 2
r or2 2 1 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 r0 afterwards means
0,t 0 for t0
Eq(2.36) is invariant under the same scale transformation as (2.12), namely,
r
r t2twhere α is a constant. We therefore seek a similarity solution:
r t, f
where r t
As in the case of (2.13), we have
3 2 r t t
1 r t
3 ' ' ' 2 2 r f f f t t t t
1 ' ' f f r r t
2 2 1 1 '' '' f f r r t t
so that (2.36) becomes 1 1 1 ' ' '' 2 f f f t r t t
or 1 ' ' '' 2 f
f f
' 1 ' 2 df d f
2 ln ' ln 4 f
C or 2 4 ' f Ce
2 2 4 4 f C e d ae b
Hence 2 4 r t ae b Putting in the initial & boundary conditions, we have
0 b 0 a b so that a b 0 Hence 2 4 0 1 r t e (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 have2 0 4 r t
so that 0 8 r u 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
2 p t
u u u uwe obtain the viscous vorticity eq.
2 D Dt ω ω u ω (2.39) For 2-D flow
, ,
, , ,
, 0 u x y t v x y t u
0, 0,
ω we have (cf 1.27),2 D Dt ω ω or
22 22 t x y
u (2.40)Obviously, the term
u
ω
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 v y (2.41)where 0 is a constant.
Obviously, the no-slip condition is not satisfied at the boundary y0.
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 O 1
L R . Using 2 UL L R
, we have
O
.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)
3. Waves
Ref: D.J.Acheson, “Elementary Fluid Dynamics”, Clarendon Press (90)
3.1 Introduction
Consider a simple harmonic surface wave:
cos
A kx t
(3.1)Wave speed is dispersive:
g c
k k
(3.2)
so that waves with longer wavelengths 2
k
travel faster. The group velocity is defined asg d c dk (3.3)
For a dispersion given by (3.2), we have
1 1 2 2 g d g c gk c dk k (3.5)
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.
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
1 2 1 2 g c k
(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).
0 0 0 p a
(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: u i j k x y z u v 0 0 v x u y 0
Thus, there exists a velocity potential
φ
:u x v y
Incompressibility: u 0 2 2 2 2 0
x yLet 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 tthe surface condition means that
, , , 0
F x x t t
