KUNDU Fluid Mechanics 2 Episode 4 pdf

law, but is derived from the momentum equation for inviscid flows, namcly, the Euler equation 4.46: at ’ a x j axi p ilxi where we have assumed that gravity g = -Vgz is the only bod

are small compared to the speed of sound and in which the tcrnperature differences in the flow are small This is discussed in Section 18 It is shown there that, undcr these restrictions, heating due to the viscous dissipation term is negligible in Eq (4.66),

and that the term - p ( V u) can be combined with the left-hand side of Eq (4.66) to give ([or a perfect gas)

DT pc*- = -v -q

where K k / p C , is thc thermul dissivity, stated in m2/s and which is the same as

that of the momentum diffusivity u

The viscous heating term t may be negligible in the thcrmal energy equa- tion (4.G6), but not in the mechanical energy cquation (4.62) In fact, there must be a sink of mechanical energy so that a steady state can be maintained in the prescnce of the various types of forcing

15 Second IAW os Tlii?rmodynamic.s: Enhvpy Produclion

The second law of thermodynamics esscntially says that real phenomena can only proceed in a direction in which the “disordcr” of an isolatcd system incrcases Disor- der of a systcm is a measure of the degree of unifonnir?; of macroscopic properties in the system, which is the same as the d e p of randomness in the molecular arrangc- nients that gcnerate thesc properties In this conncction, disordcr, uniformity, and randomness havc essentially the same rncaning For analogy, a tray containing rcd balls on one side and white balls on the othcr has more order than in an arrangement

in which the balls arc mixed togcthcr A real phenomenon must thereforc proceed in a direction in which such orderly arrangements dccrease because of “mixing.” Consider two possiblc states of an isolated fluid system, onc in which there are nonuniformities

of temperaturc and velocity and the other in which thcse propertics are uniform Both

or these statcs have the same internal cnergy Can the system spontaneously go from the state in which its properties are uniform to one in which they are nonuniform? The

second law asserts that it cannot, based on cxperience Natural proccsses, therefore, tend to causc mixing duc to transport of heat, momentum, and mass

A consequcnce of the swond law is that therc must exist aproperty called enrmpy,

which is related to other thcrmodynamic propertics of the mcdium In addition, thc second law says that the entropy of an isolated systcm can only increase; entropy is thercfore a measure of disordcr or randomness of a system Lct S be the cntropy pcr unit mass It is shown in Chapter 1, Scction 8 that the changc or entropy is related to

Using Fourier's law of heat conduction, this becomes

The first term on the right-hand side, which has the form (heat gain)/T, is the cntropy gain due to reversible heat transfer because this term does not involve heat conduc- tivity The last two terms, which are proportional to the square of temperature and velocity gradicnts, represcnt the entmpy production due to hcat conduction and vis- cous generation of heat The second law of thermodynamics requks that the entropy production due to irreversible phenomena should be positive, so that

An explicit appeal to the second law of thermodynamics is therefore not required in most analyscs of fluid flows bccause it has already been satisfied by laking positivc values for the molecular coeflicicnts of viscosity and thermal conductivity

If the flow is inviscid and nonheat: conducting, entropy is preservcd along the particle palhs

16 BernouIIi hipalion

Various conservation laws for mass, momentum, energy, and entropy wcre presented

in the preceding sections The well-known Bernoulli (4.46) equation is not a separate

law, but is derived from the momentum equation for inviscid flows, namcly, the Euler

equation (4.46):

at ’ a x j axi p ilxi

where we have assumed that gravity g = -V(gz) is the only body force The advective

acceleration can be expressed in t e r n of vorticity as follows:

w h m we have used r;j = -&ijhOk (sce Eq 3.23), and used the customary notation

q2 = uz I = twice kinetic encrgy

Then the Euler equation becomes

= (u x 0 ) i (4.71)

Now assume that p is a function of p only A flow in which p = p ( p ) is called

a barotmpic.fk,w, of which isothetmal and isentropic ( p / p Y = constant) flows arc special cascs For such a flow we can writc

where Eq (4.74) has been used The preceding equation is identical to Eq (4.72)

Using Q (4.72), the Eulcr equation (4.71.) becomes

Bui a [ I 2 ST ]

-+- -q + - + + z = ( u x o ) ;

at axi 2

Defining the Bernoulli function

I i q 2 + 1 + gz = constant along streamlines and vortex lines

Bernouli equations are integrals of the conservation laws and have wide applicability

as shown by the examples that follow Important deductions can be made from the preceding equation by considering two special cases, namely a steady flow (rotational

or irrotational) and an unsteady irrotational flow These are described in what follows

which is called Bemulli’s e4ualion If, in addition, the flow is irrotational (o = 0), then Eq (4.72) shows that

P

Fikwre 4.18 Flow over a solid objwl Flow outside thc boundary layer is irrolalional

It may be shown that a sufficient condition for the existence of the surfaces con- taining streamlines and vortex lines is that the flow be barotropic Tncidentally, thesc

are called Lamb surfaces in honor of the distinguished English applied mathematician and hydrodynamicist, Horace Lamb Tn a general, that is, nonbaroh-opjc Row, a path composed of streanilinc and vortex line segments can be drawn between any two points in a flow field Thcn Eq (4.78) is valid with the proviso that the integral be evaluated on the specific path chosen As written, Eq (4.78) requires the restTictions that the flow be stcady, inviscid, and have only gravity (or other conservative) body forces acting upon it Tmtational flows are studied in Chapter 6 We shall note only the

important.point here that, in a nonmtating frame of reference, bamtropic irrotational flows rcmain irrotational irviscous dTects are negligible Considcr the flow around a solid object, say an airfoil (Figure 4.18) The flow is irrotational at all points outside the thin viscous layer closc to the surface of the body This is bccause a particle P

on a streamline outside the viscous layer started from some point S, where the flow

is uniform and consequently irrotational The Bernoulli equation (4.79) is therefore satisfied everywhere outsidc the viscous layer in this example

Unsteady Irrotational Flow

An unsteady form oPBernoulli’s equation can be derived only if the flow is irrotational For hotational flows thc velocity vector can be written as the gradient of a scalar potential Cp (called velocity potential):

114 c7unmrryacionLuurn

where the integrating function F ( r ) is independent of location This form of the Bernoulli equation will be used in studying irrotational wave motions in Chapter 7

Energy Bernoulli Equation

Return to Eq (4.65) in the steady state with neither heat conduction nor viscous stresses Then t i j = -psii and Eq (4.65) becomes

If the body force per unit mass gi is conservative, say gravity, then +i = -(a/axi)(gz),

which is the gradient of a scalar potential In addition, from mass conservation,

a(pui)/axi = 0 and thus

(4.82)

From Eq (1.13) h = e + p / p Eq (4.82) now states that gradients of B’ = h +

q2/2+gz must be normal to the local streamline direction ui Then B’ = h +q2/2+

gz is a constant on streamlines We showed in the previous section that inviscid, non-heat conducting flows are isentropic (S is conserved along particle paths), and in

Eq (1.1 8) we had the relation d p / p = dh when S = constant Thus the path integral

d p / p becomes a function h of the endpoints only if, in the momentum Bernoulli equation, both hcat conduction and viscous stresses may be neglected This latter form h m the energy equation becomes very useful for high-speed gas flows to show the interplay between kinetic energy and internal energy or enthalpy or temperature along a streamline

17 Applications of Bernoulli’s k$ualion

Application of Bernoulli’s equation will now be illustrated for some simple flows

Pitot %be

Consider first a simple device to measure the local velocity in a fluid stream by inserting a narrow bent tube (Figure 4.19) This is called apiror rube, after the French mathematician Henry Pitot (1 695-177 1 ), who used a bent glass tube to measure the velocity of the river Seine Consider two points 1 and 2 at the same level, point 1 being

away from the tube and point 2 being immediately in front of the open end where the

fluid velocity is m Friction is negligible along a streamline through 1 and 2, so that

Bernoulli’s equation (4.78) gives

from which the velocity is found to be

itot tube

P

E’igure 4.19 Pilot tuhe for rncasuring vclocily in a duct

Prcssures at thc two points are found from thc hydrostatic balance

PI = pghl and p2 = pgh2

so that [he velocity can bc found from

Because it is assumcd that thc fluid density is very much greater than that of the atmosphcre to which the tubes are exposed, the pressures at the tops of the two fluid columns are assumed to be thc same Thcy will actually differ by plumg(h2 - h l )

Use of the hydrostatic approximation abovc station 1 is valid when the streamlines arc straight and parallel betwccn station 1 and thc upper wall In working out this problem, the fluid dcnsity also has been laken to be a constant

Thc pressurc p2 measured by a pitot tubc is called “stagnation pressure:’ which

is larger than the local static pressure Evcn when there is no pitot tubc to meaqure thc stagnation pressure, it is customary to refcr LO the local valuc of thc quantity

( p + p u 2 / 2 ) as thc local stagnafiun pressure, defined as the pressure that would bc

reached i l h e local flow is imgined to slow down to zcro velocity frictionlessly The

quanlity pu2/2 is s o m e h c s called thc dynumic pm.wure; stagnation pressure is tbc

sum of static and dynamic pressures

Orifice in a lhnk

As another application or Bernoulli’s equalion, consider the flow though an orifice

or opcning in a lank (Figure 4.20) The flow is slightly unsteady due to lowering 01

A

A

Distribution of

(p -p,,J at orifice Figure 4.20 Flow through a sharp-edgcd orificc Pressure has thc almosphcric value cvcrynherc m s s

seaion CC, its dishbution across orifice AA is indicated

the water level in the tank, but this effect is small if the tank area is large as compared

to the orifice area Viscous effects are negligible everywhere away from the walls of

the tank All streamlines can be traced back to the free surface in the tank, where they have the same value of thc Bernoulli constant B = y2/2 + p / p + gz I1 .follows that

the flow is irrotational, and B is constant throughout the flow

We want to apply Bernoulli’s equation between a point at the free surface in the tank and a point in the jet However, the conditions right at the opening (section

A in Figure 4.20) are not simple because the pressure is not uniform across the jet Although pressure has the atmospheric value everywhere on the free surface of the jet (neglecting small surface tension effects), it is not equal to the atmospheric pressure inside the jet at this section The streamlines at the orifice are curved, which requires that pressure must vary across the width of the jet in order to balance the centrifugal forcc The pressure distribution across the orifice (section A) is shown in Figure 4.20

However, the streamlines in the jet become parallel at a short distance away from the orifice (section C in Figure 4.20), whcre the jet area is smaller than the orifice area The pressure across section C is u n i f m and equal lo the atmospheric value because

it has that value at the surface of the jet

Application of Bernoulli’s equation between a point on the free surface in the

tank and a point at C gives

from which the jet velocity is found as

u = J2gh,

Figure 4.21 Flow through a munded oriBcc

which simply states that the loss of potcnlial energy equals the gain of kinetic energy The mass dow rate is

rit = pA,u = PA&&,

where A, is the area of the jet at C For orifices having a sharp edge, A, has been round to bc %62% of thc orifice area

If the orifice happens to have a well-rounded opening (Figure 4.21), thcn h e jet does not contract The streamlines right at the exit are then parallel, and the pressure

at the cxit is uniform and equal to the atmosphcric pressure Consequently the mass flow rate is simply p A m , where A equals the orifice area

18 Houwinesq Approximation

For flows satisfying certain conditions, Boussinesq in 1903 suggested that the density changes in thc fluid can be neglected except in the gravity term where p is multiplicd

by g This approximation also treats the othcrpperties of the fluid (such asp, k, C p )

as constants A formal jusNication, and the conditions under which the Boussinesq approximation holds, is givcn in Spiegel and Veronis (1960) Here we shall discuss the basis OF the approximation in a somewhat intuitive manner and examinc the resulting simplifications of the equations of motion

Continuity Equation

The Boussinesq approximation replaces the continuity equation

by the incompressible form

v - u = o

(4.83)

(4.84)

However, this does not mcan that the density is regarded as constant along the direction

of motion, but simply that the magnilude of p - ’ ( D p / D t ) is small in comparison to the magnitudes of the velocity gradients in V u We can immediately think of several situations where the density variations cannot be neglected as such The first situation

is a steady flow with large Mach numbcrs (defined as U / c , where U is a typical measure of the flow speed and c is the speed of sound in the medium) At large Mach

numbers the comprcssibility effects are large, because the large pressure changes

cause large density changes Jt is shown in Chapter 16 that compressibility effects

are negligiblc in flows in which the Mach numbcr is <0.3 A typical value of c for

air at ordinary temperatures is 350m/s, so that the assumption is good for speeds

< 1.00 m/s For water c = 1470 m/s, but the speeds normally achievable in liquids are much smaller than this value and therefore the incompressibility assumption is very good in liquids

A second situation in which the compressibility effects m impartant is unsteady

flows The wavcs would propagate at infinite speed if thc density variations are

neglected

A third situation in which the compressibility effects are important occurs when the vertical scale of the flow is so large that the hydrostatic pressure variations cause large changes in density In a hydrostatic field the vertical scale in which thc density changes become important is of order c2/g - 10 km for air (This length agrees with the e-folding height R T / g of an “isothermal atmospherc,” because c2 = y RT; see

Chapter 1, Section 10.) The Boussinesq approximation therefore requires that the vertical scale of the flow be L << c2/g

In the three situations mentioned the medium is regarded as “compressible,” in which the density depends strongly on pressure Now suppose the compressibility effects are small, so that the density changes are caused by temperature changes

alone, as in a thermal convection problem .In this case the Boussinesq approximation

applies whcn the temperature variations in the flow are small Assume that p changes with T according to

P

where a = -p-’(ap/aT), is the thermal expansion coefficient Far a perfect gas

a = 1 / T - 3 x K-l and for typical liquids a - 5 x I O4 K-’ With a temper-

ature difference in Lhc fluid of 10 “C, thc varialion of density can be only a few percent a1 most 1.t turns out that p-’(Dp/Df) can also bc no larger than a few percent of the

velocity gradients in V u To see this, assume that the flow field is characterized by

a len@h scale L, a velocity scale U , and a tempcrature scale 6 1 By this we mean

18 Ilouw%eq Appmmhalion 119

that the velocity varies by U and the temperature varies by ST, in a distance of order

L The ratio of the magnitudes of the two terms in the continuity equation is

which allows us to replace continuity equation (4.83) by its incompressible

form (4.84)

Momentum Equation

Because of the incompressible continuity equation V u = 0, the stress tensor is

givcn by Eq (4.41) From Eq (4.43, the equation of motion is then

Du

Dt

p- = - v p + p g + p v 2 u (4.85)

Consider a hypothetical static reference state in which the density is po everywhere

and the pressure is po(t), so that Vpo = f i g Subtracting this state from Eq (4.85)

and writing p = po + p’ and p = po + pl, we obtain

where 11 = p/po The ratio p’/po appears in both the inertia and the buoyancy terms

For small values of p’/po, the density variations generate only a small correction to

the inertia term and can be neglected However, the buoyancy term p’glpo is very

important and cannot be neglected For example, it is these density variations that

drive tbe convective motion when a layer of fluid is heated The magnitude of p’g/po

is therefore of the same order as the vertical acceleration awlat or the viscous term

uV2 w We conclude that the density variations are negligible the momentum equation,

except when p is multiplied by g

Heat Equation

From Q (4.66), h e thermal energy equation is

Although the continuity equation is approximately V u = 0, an important point is

that the volume expansion term p(V u) is not negligible compared to other dom-

inant terms of Eq (4.87); only for incompressible liquids is p(V u) negligible in

Eq (4.87) We have

Assuming a perfect gas, for which p = p R T , C, - C, = R and (Y = 1 / T , the

foregoing estimate becomes

where we used e = C,T for a pedcct gas Note that we would have gotten C, (instead

of C,) on the left-hand side of Eq (4.88) if we had dropped V u in Eq (4.87)

Now we show that the heating due to viscous dissipation of energy is negligi- ble under the restrictions underlying the Boussincsq approximation Comparing the magnitudes of viscous heating with thc left-hand si& of Eq (4.88), we obtain

In typical situations this is extremely small (-

Fourier’s law of heat conduction

Neglecting 4, and assuming

Summary: The Boussinesq approximation applies if th Mach number of th flow

is small, propagation of sound or shock waves is not considered, the vertical scale of the flow is not too large, and the temperature differences in the fluid are small Then

the density can be treated as a constant in both the continuity and the momentum

equations, except in the gravity term Properties of the fluid such as p, it, and C,

are also assumed constant in this appi-oximation Omitting Coriolis forces, the set of equations corresponding LO the Boussinesq approximation is

where the z-axis is taken upward Thc constant pa is a reference density correspond- ing to 8 refercnce temperaturc TO, which can be taken to be the mean temperaturc

in the flow or the temperature at a boundary Applications of the Boussincsq set can be found in several places throughout the book, for example, in the problems of wave propagation in a density-stratificd medium, thermal instability, turbulence in a stratified medium, and gcophysical fluid dynamics

19 Boundary Condidions

The differential equations wc have derived for the conservation laws are subject to boundary conditions in order to properly formulate any problem Specifically, the Navier-Stokes equations me of a form that requires the vclocity vector to be given on all surfaccs bounding thc flow domain

If we arc solving for an external flow, that is, a flow ovcr some body, we must spccify the velocity vector and the thermodynamic state on il closed distant surface

On a solid boundary or at the interface betwccn two immiscible liquids, conditions may be derived from the thrcc basic conservation laws as follows

In Figure 4.22, a "pillbox" is drawn through the interrace surface separating medium 1 (fluid) From medium 2 (solid or liquid immiscible with fluid 1) Here dAl

and dA1 are elements of the end face areas in medium 1 and medium 2: rcspectively, locally tangent to the interfacc, and separatcd from each other by a distance 1 Now apply the conservation laws to the volume &fined by the pillbox Next, let 1 + 0:

keeping AI and A2 in the different media As 1 + 0, all volumc integrals + 0 and the

integral over the side area, which is proportional to 1, tends to zero as well Define a

unit vector n, normal to the interface at thc pillbox and pointed into medium 1 Mass conservation gives plul - n = p2u2 n at each point on thc interfacc as the end face area becomes small

If medium 2 is a solid, then u2 = 0 there Tf medium 1 and medium 2 are immiscible liquids, no mass flows across thc boundary surrace In cjther case, u1 - n =

0 on the boundary Thc same proccdure applied to the integral form of the momentum

cquatim (4.16) gives the result that the forcdarea on the surface, ni ti,i is continuous across the interface if surface tension is neglected If surface tension is includcd, a jump in pressure in the direction normal to the interfacc must be added; see Chapter 1, Section 6

Applying the integral form of energy conservation (4.64) to a pillbox of infinites-

imal height 1 gives the rcsult niyi is continuous across thc interface, or explicity,

kl (aTl/an) = k2(aT2/an) at the interrace surface The hcat flux must be continuous

at the interfacc; it cannot store heat

Figure 4.22 Interhccc bclwcen two mcdia; evaluation or boundary conditions

lhrcims

1 Let a one-dimemional velocity field be u = u ( x , t ) , with u = 0 and UJ = 0

The dendy varies a$ p = po(2 - cos wt) Find an expression for u(x, t) if

u(0, t) = u

2 Tn Section 3 we derived the continuity equation (4.8) by starting from the

integral form of the law of conservation of mass for a j x e d region Derive Eq (4.8)

by starting From an inkgal form for a material volume [Hint: Formulate the principle

for a material volume and then use Eq (4.5).]

3 Consider consemtion of angular momentum derived from the angular

momentum principle by the word statement: Rate of increase of angular momen-

tum in volume V = net influx of angular momentum across the bounding surface

A of V + torqucs due to surface forces + Lorques due to body forces Here, the only

torques are due to the same forces that appear in (linear) momentum conservation The

possibilities for body torques and couple stresses havc been ncglected The torques

due to thc surface forces are manipulated as follows The torquc about a point 0 due

to the element of surface f m tmkdA, is SEijkXjtmkdAmr where x is the position vector from 0 to thc element dA Using Gauss’ theorem, we write this as a volume integral,

where wc have used axj/axm = Sjm The second term is Sv x x V - t d V and

combines with the remaining terms in thc conservation of angular momentum to give

sv x x (Lincar Momcntum: Eq (4.17)) dV = 1, Eijkrjk d V Since the left-hand

side = 0 for any VOlUmC v, WC conclude that & j j k t k j = 0, which leads to t i j = t j i

4 Near Ihe end of Scction 7 we derived the equation of motion (4.15) by starting

from an intcgral €orm for a material volumc Derive Eq (4.15) by starting from the

integral statemcnt for ajixed region, given by Eq (4.22)

5 Verify Lhc validity of lhc second form of thc viscous dissipation given in

EQ (4.60) [Hint: Complcte the square and use S i j d i j = Sii = 3.1

6 A rcctangular tank is placed on wheels and is given a constant horizontal

acceleration a Show that, at steady state, the anglc made by the free surface with the

horizontal is givcn by lan 0 = a / g

7 A jet of water wilh adiameter of 8 cm and a speed of 25 m/s impinges normally

on a large stationary flat plate Find the €orce required to hold the platc stationary

l5w7!iM# 123

Compare the avcrage pressurc on the plate with h e stagnation pressure if the plate is

20 times the area of the jet

8 Show that the thrust dcveloped by a stationary rocket motor is F = p A U 2 +

A ( p - pah), where patm is the atmospheric pmsure, and p, p, A, and I/ are, respec-

tively, the pressure, density, area, and velocity of the fluid at thc nozzle exit

9 Consider the prqcllcr of an airplane moving with a velocity U1 Takc a

reference frame in which the air is moving and the propeller [disk] is stationary Then

the effect of the propeller is to accelerate the fluid from the upstream value UI to

the downstream value UZ > U t Assuming incompressibility show that the thrust

developed by the propeller is given by

F = - ( U 22 - u 2 1 ) -

where A is thc projected arca of the propellcr and p is the density (assumed constant)

Show &so that the velocity of the fluid at the plane of the propeller is the average value

U = ( U I + U2)/2 [Hint: The flow can be idcalized by a pressure jump, of magnitude

Ap = F / A right at the location of the propeller Also apply Bernoulli’s equation

between a section far upstream and a section immediately upstream of the propeller

Also apply the Bernoulli equation between a section immediately downstream of h e

propeller and a section far downstream This will show that Ap = p(U,’ - U ; ) / 2 ]

10 A hemispherical vessel of radius R ha5 a small rounded orifice of area A at

the bottom Show that the time required to lower the level from hl to h2 is given by

1 1 Consider an incompressible planar Couette flow, which is the flow between

two parallel plates separated by a distance b The upper plate is moving parallel to

itself at speed U , and the lower plate is stationary Let the x-axis lie on the lower plate

All flow fields are independent of x Show that the pressure distribution is hydrostatic

and that the solution of the Navier-Stokes equation is

UY

u ( y ) = -

b

Writc the expressions for the stress and strain rate tensors, and show that the viscous

dissiparion per unit volume is

Take a rectangular control volume for which the two horizontal surfaces coincide

with the walls and the two vertical surfaces are perpendicular to the flow Evaluate

every term of energy equation (4.63) forthis control volume, and show that tbe balance

is between the viscous dissipation and the work done in moving the upper surface

12 The components of a mass flow vector p u are p u = 4 x 2 y , p u = x y z ,

pw = yz2 Compute the net outflow through the closed surface formed by the planes

= p U 2 / b 2

x = 0, x = 1, 4’ = 0, y = 1, z = 0, z = 1

(a) Tntegrate ovcr the closed sudace

(b) Tntegrate over the volume bounded by that surface

13 Prove that the velocity field given by ur = 0, ug = k / ( 2 n r ) can have only two possible values of the circulation They are (a) r = 0 for any path not enclosing the origin, and (b) r = k for any path enclosing the origin

14 Water flows through apipe in a gravitalional field as shown in the accompany- ing figure Ncglect the effects of viscosity and surface tension Solve the appropriate conservation equations for the variation of Lhc cross-sectional area of the fluid column

A ( z ) after the water has left the pipe at z = 0 The velocity of the fluid at z = 0 is uniform at uo and the cross-sectional area is Ao

15 Redo the solution for the ''orifice in a tank" problem allowing for the fact that

in Fig 4.20, h = h ( t ) How long does the tank take to empty?

llileruzhm Cikd

Aris, R (1962) Vectors, Tensors and the Basic Equa/ion.s of Fluid Mechanics, Englcwood ClilYs, NJ:

PrcnticeHall (Thc basic equationsof motion and thc various rormsof thc Rcynolds wansport thcorem

are derivd wd discussd.)

Batchelor, G K (1967) An Znrwduc/ion to Fluid Dynamics London: Cdmhridgc Univenily Press (This

contains an cxcellent and authoritative treatmcnt of the basic equations.)

Holton, J R (1979) An fntmducrion to Dynamic iUefeorology, Ncw York Acdcmic Prcss

Pedlosky, I (1 987) Geophysical Fluid Dynamics, Ncw York Springer-Verlag

Spiegel, E A and G Vcronis (1960) On thc Boussinesq approximation for a compmssible fluid Asfro-

Stommcl H M and D W Moore (1989) An Introduction to the Corio1i.s Force New York: Columbia 'Itucsdcll, C A (1952) Stokes' principle of viscosity JoumZ oJRationol Mechanics u d Analysi.s 1:

physical Journul131: 442447

Univemity Press

228-231

Supplernim?al Reading

Chandrdwkhar, S ( 1961) Hydwxlyinmic und Hydmmagnetic Stabilify, London: Oxford Universiv h s s

(This is a p o d sourcc to learn thc basic quaticins in a hrielwd simplc way.)

Chapter 5

Vorticity Dynmics

I Irifnniuchn 125

2 Ihrikx ruuI h e x I?mmf? 134

luhs 126

3 Hole of KwxMit!y in 1htdona.l and Rrmc 136

Irrntntioruil hrficw 126

Solid-Rodg HOUI~~OII 127

Irmtnrionalhrtex 127 7 h t m d o n of Vorhw 141

Diecussion 130 8 fi)rlcrL.%l 144

4 Ke1nin:X (,'irculnhn 'I7mm-m 1 30 h'xmiws 145

1)kcmsion of KelvZs 'I'hcni.c:rri 132 I,ihmim (,'ikd 146

Helrii holm \irks 'I'hcorans 134 .Supplemenhi1 f i x d r g 147

5 VorLki& Kqu&n in n ,Yonmlahg 6 Vorticiy Equalion in n Rotaling I Memiing of (w V ) u 139

M ~ 1 1 h ~ of 2 ( 8 V)U 140

1 Inhduciion

Motion in circular streamlines is called vortex motion The presence of closed stream-

lines does not necessarily mean that the fluid particles are rotating about their own

centers, and we may have rotational as well as irrotational vortices depending on

whether the fluid parficles have vorticity or not The two basic vortex flows are the

solid-body rotation

(5.1)

ug = p r , I

l l

and the irrotational vortex

I

ug = -

2 a r

These are discussed in Chapter 3, Section 1 1, where also, the angular velocity in the

solid-body rotation w a denoted by 00 = w / 2 Morcover, the vorticity of an element is

everywhere equal to w for the solid-body rotation represented by Eq (5.1), so that the

circulation around any contour is w times the area enclosed by the contour In contrast,

the flow represented by Eq (5.2) is irrotational everywhere except at the origin, where

the vorticity is iniinite All the vorticity of this flow is therefore concentrated on a

line coinciding with the vortex axis Circulation around any circuit not enclosing the

125

origin is therefore zero, and that enclosing the origin is r An irrotational vortex is

therefore called a line vortex Some aspects a€ the dynamics of flows with vorticity

are examined in this chapter

2 hrhx JJines and V o r h 71dbes

A vortex line is a curve in the fluid such that its tangent at any point gives the direction

of the local vorticity A vortex line is therefon: related to the vorticity vector the same way a streamline is related to the velocity vector If w,, wJ, and w, are the Cartesian

components of the vorticity vector o, then the orientation of a vortex line satisfies the

Vortex lines passing through any closcd curve form a tubular sudace, which

is called a vortex tube Just as streamlincs bound a streamtube, a group of vortcx

lines bound a vortex tube (Figure 5.1) The circulation around a narrow vortex tube

is dI' = o d A , which is similar to the expression for the rate of flow d Q = u d A through a n m w skamtube The strength of a vortex nrhe is defined as thc circulation

around a closed circuit taken on the surface of thc tube and embracing it just once

From Stokes' theorem it follows that the strength of a vortex tube is equal to the mcan vorticity times its cross-sectional area

3 Rule of f i m o d y in Rotational and Irmlutional Vortices

The role of viscosity in the two basic types of vortex flows, namely thc solid-body rota- tion and the irrotational vortex, is examined in this section Assuming incompressible

Slreamiubc Vorlcx lube

Agnre 5.1 Analogy bclween strcmtube and lube

3 R d e n/ hwiily in Ilotkdbinul and Imiluiiunal k i k s

flow, we shall see that in one of these flows the viscous k m in the momentum equa-

tion drop out, although the viscous stress and dissipation of energy are nonzero The

two flows are examined separately in what follows

Solid-Body Rotation

As discussed in Chapter 3, fluid elements in a solid-body rotation do not deform

Because viscous stresses are proportional to deformation rate, they am zero in this

flow This can bc demonstratcd by using the expression for viscous stress in polar

127

coordinates:

where we have substituted ue = o r / 2 and ur = 0 We can therefore apply the inviscid

Eulcr equations, which in polar coordinates simplify to

which are paraboloids of revolution (Figure 5.2)

The important point to note is that viscous stresses are absent in this flow (The

viscous stresses, howevcr, are important during the transient period of iniriuting the

motion, say by steadily rotating a tank containing a viscous fluid at rest.) Tn terms of

velocity, Eq (5.5) can be written as

1 2

PZ - 3 ~ ~+ 0 P ~ Z Z 2 = P I - ~ P U ; , + pgel,

which shows that the Bernoulli function B = u i / 2 + g r + p / p is not constant €or

points on different streamlines This is expected of inviscid rotational flows

Figure 5 2 Constant pressurc surhces in a solid-body mtnlion gencmled in a rotating kink containing

liquid

the viscous stress is

I au, a u P r

9 0 = P [ ;= + r c ($)I = -2’

which is nonzero everywhere This is because fluid elements do undergo deformation

in such a Row, as discussed in Chapter 3 However, the interesting point is that the nef

viscous.force on an element again vanishcs, just as in the case of solid body rotation

In an incompressible flow, the net viscous force per unit volume is related to vorticity

which is zcro for irrotational flows The viscous forces on the surfaces 01 an element

cancel out, leaving a zero resultant The equutions of motion therefore reduce to the inviscid Euler equation.s, although viscous stresses are izonzem everywhere The

pressure distribution can therefore be found from the inviscid sct (5.4), giving

where we have used ug = r / ( k r ) Tnlegration between any two points gives

which implies

PI 4 1 h 4 2

- + - +gz1 = - + - +szz

idealized by ug = w R 2 / 2 r ; hcre we have chosen the value of circulation such that U O

is continuous at r = R (see Figure 3.16b) The strength of such a vortex is given by

r = (vorticity)(core m a ) = nwR2

One way of gcnerating an irrotational vortex is by rotating a solid circular cylinder

in an infinite viscous fluid (see Figure 9.7) It is shown in Chapter 9, Section 6 that the stcady solution of the NavicrStokes equations satisfying the no-slip boundary condilion (ue = w R / 2 at r = R) is

where R is the radius of the cylindcr and w / 2 is its constant angular vclocity; sec Eq

(9.1 5) This flow does not havc any singularity in the cntire field and is irrotational everywhere Viscous stresses are present, and the resulting viscous dissipadon of kinetic encrgy is exactly compensated by the work done at thc surface of the cylinder However, there is no net viscous force at any point in the steady state

Discussion

The examples given in this scction suggest that irrotationulity does not imply the ahsence ofviscous stresses In fact, they must always be present in irrotational flows

of real Ruids, simply because the fluid elements deform in such a flow However the

net viscous force vanishes if o = 0, as can be seen in Eq (5.6) We have also givcn

an example, namely that of solid-body rotation, in which there is uni$otm vorticity

and no viscous stress at all However, this is the only example in which rotation can

take place without viscous effects, because Eq (5.6) implies that the net force is zero

in arotational flow if o is miform everywhere Except for this example, fluid rotation

is accomplished by viscous effects Indeed, we shall see later in this chapter that viscosity is a primary agency for vorticity generation

4 Kklvin'x Circulation Ti?uwn?rn

Several theorems of vortex motion in an inviscid fluid were published by Helmholtz

in 1858 He discovered these by analogy with electrodynamics Inspired by this work, Kelvin in 1868 introduced the idea of circulation and proved the following theorem:

In an inviscid, bumtropicflow with conservative body forces, the Circulation around

a closed curve moving with thefluid remuins comtant with time, if the motion is

observed € o m a nonrotating frame The theorem can be restated in simple terms as follows: At an instant of time take any closed contour C and locate the new position

of C by following the motion of all of its fluid elements Kelvin's circulation theorem

states that the circulations around the two locations of C are the same In other words,

D r

1 D t = O * 1 (5.7)

where D / D r has been used to emphasize that the circulation is calculated around a material contour moving with the fluid

To prove Kelvin's theorem, the rate of change of circulation is found as

where dx is the separation between two points on C (Figure 5.4) Using the momentum equation