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Chapter 2 Hydrodynamics Irrotational Flows and Flux In general, flows can be descibed as laminar flows in which the fluid acts as though it is comprised of orderly la yers or plates. Laminar flows can be represented visually using streamlines. On the other hand, turbulent flows are very c haotic and cannot by visually r epresented by streamlines. In this chapter we investigate the geometrical properties of lanminar irrotaional flows, orderly flows that have no circulation. We will c over steady state flows in which the flows or fields remain steady and constant in time and may be represente d by streamlines. The particles that compose the system may move around and accelerate in moving from one part of a system to another, but the field patterns themselves rema in constant in time. At the heart of classical field theory lies two very powerful theorems: the Uniqueness theorem and Helmholtz's theorem. We will see that two properties called fl ux and circulation uniquely determine any vector field. 2 . 1 Characterizing Vector Fields. One of the incredible things about vector fields is that all the different types of field p atterns that one can dream up follow from two very simple fie ld patterns. The first type is known as a diverging field where the field can be thought of as diverging from a source of field, or conversely, converging to a sink or drain of field. source Figure 2.1 A source produces a diverging field. Classical Fluids, Chapter 2 -21-

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• Chapter 2Hydrodynamics

Irrotational Flows and FluxIn general, flows can be descibed as laminar flows in which the fluid acts as though it is comprisedof orderly layers or plates. Laminar flows can be represented visually using streamlines. On theother hand, turbulent flows are very chaotic and cannot by visually represented by streamlines. Inthis chapter we investigate the geometrical properties of lanminar irrotaional flows, orderly flowsthat have no circulation. We will cover steady state flows in which the flows or fields remainsteady and constant in time and may be represented by streamlines. The particles that compose thesystem may move around and accelerate in moving from one part of a system to another, but thefield patterns themselves remain constant in time. At the heart of classical field theory lies two verypowerful theorems: the Uniqueness theorem and Helmholtz's theorem. We will see that twoproperties called flux and circulation uniquely determine any vector field.2 .1 Characterizing Vector Fields.

One of the incredible things about vector fields is that all the different types of field patterns thatone can dream up follow from two very simple field patterns. The first type is known as adiverging field where the field can be thought of as diverging from a source of field, or conversely,converging to a sink or drain of field.

source

Figure 2.1A source produces a diverging field.

Classical Fluids, Chapter 2 -2 1-

• In Figure 2.1 the field lines for a single source is shown. This is the classic diverging fieldpattern. The second type of pattern is known as a circulating or solenoidal field. A whirlpool isthe classic example,

Figure 2.2A whirlpool is an example of a circulating field.

and is shown in Figure 2.2. Complicated fields can be produced from a combination of circulatingand diverging fields. For instance, your bath tub drain is a sink that a produces a diverging flow,but the residual angular momentum from filling the tub causes the water to swirl around the drainas it moves in towards the drain forming a more complicated whirlpool.There are some subtleties. Just because a field curves, does not mean that the field has acirculating component.

Figure 2.3A purely diverging field produced by two sources and one sink.

Classical Fluids, Chapter 2 -2 2-

• Figure 2.3 shows a purely diverging field produced by sources and sinks. What distinguishes thisfield from a circulating field is that although the field swirls on one side of a given source it tendsto swirl the opposite way on the other side. A circulating field will cause the field to swirl in thesame direction on both sides with one exception we will discuss later, however, the definingdistinction that labels this field as purely diverging is that as the fields leave (or enter) the sources(sinks) the lines are perpendicular to the boundaries of the sources. In short note that for divergingflows, the field lines end on sources. Anywhere field lines end, the flow will be diverging. Incontrast, a pure circulating field always has the field lines circulate around and close onthemselves. If the field lines form closed loops, then the field is a pure circulating field.The next step is to take this heuristic classification scheme and make it rigorous. We will do thisby introducing two quantities: the flux and the circulation. After we have defined these quantities,we will formally state the uniqueness theorem, and follow it with an application to fluids.2 .2 The Uniqueness Theorem & Helmholtz's Theorem.

We will now just state the uniqueness theorem. It is very important and lays the mathematicalfoundation for the rest of fluids and electricity and magnetism.The Uniqueness Theorem

A vector field F will be uniquely determined by specifying its flux andcirculation.

flux = F dA

circulation = F d (2 .1)

Once the flux and circulations are specified, then the vector field may be found by solving bothequations (2.1) simultaneously. The flux is expressed in terms of a surface integral and thecirculation is a path integral. Often either the flux or the circulation will be zero and the solutionsimplifies readily. One case important case is for a conservative field which requires the circulationto be zero. In the case of a force field, the work done is W = F d. For a conservative force,the work done around a closed path must be zero: F d = 0.Another theorem that is closely related to the uniqueness theorem is Helmholtz's theorem.Helmholtz's theorem is used to define scalar and vector potentials.

Helmholtz's Theorem

Any vector field F may be written as the sum of a pure polar vector and apure axial vector.

The polar part leads to a flux, and the axial part leads to circulation.Classical Fluids, Chapter 2 -2 3-

• Polar and axial vectors behave differently. Axial vectors are typically the result of a cross productand we have seen that the resultant of the cross product is a vector; however, the cross productdoes not commute. Are there any other properties of the cross product that are unusual? As youmight guess the answer is yes. We are going to look at the way vectors behave under reflection.This will tell us something very important about vectors.Vectors formed by cross products are called axial vectors of which angular momentum and torqueare examples. Normal vectors like position and momentum are called polar vectors. Consider atoy top with constant angular momentum L given by

L = r p. (2.2)For a small chunk of the top m, both r and p rotate with the top.

rp p

rL

Figure 2.4For a constant angular momentum L the position and momentum

vectors continually change direction.

If L is constant, it cannot lie in the same plane as r and p because everything in that plane isswirling around nothing in the plane of rotation is constant. The only constant direction is alongthe symmetry axis of the top, and L lies along this axis. Hence, L is called an axial vector. As inthis example, axial vectors in general are used to describe the rotational properties of a system.Axial vectors behave differently than polar vector upon refection. To see this consider the mirrorimage of a polar vector say a displacement vector r. A mirror doesn't really flip left and right,i.e., if a displacement is made in the plane parallel to the mirror, the reflected displacement vectorr' retains its direction, but if the displacement is made perpendicular to the mirror, then thereflected displacement vector is flipped

r'r

r'

r

Classical Fluids, Chapter 2 -2 4-a) b )

• Figure 2.5a) The image of a displacement parallel to the mirror is not filliped. b) The imageof a displacement perpendicular to the mirror is flipped.

as shown in Figure 2.5. Hence, a mirror actually flips things that are perpendicular to the mirror.Contrast the polar vector to the axial vector case. Remember, an axial vector is given by the crossproduct of two polar vectors so to see how an axial vector behaves when reflected we must lookfirst at the constituent polar vectors.

rp

r'

p'

L

L'

r

p

r'

p'

L'

L

a) b )Figure 2.6

a) When the angular momentum of the spinning disk is parallel to the mirror, the imageof the disk spins in the opposite direction so its angular momentum flips. b) Whenthe angular momentum of the disk is perpendicular to the mirror , the image rotates inthe same sense so the angular momentum of the image does not flip.

As an example, we will use angular momentum. When the angular momentum is parallel to themirror, both of the polar vectors cannot be parallel to the mirror. For the case shown inFigure 2.7, the r is taken parallel to the mirror, so the p vector will be perpendicular, that is, themirror image of the rotating disk appears to be spinning in the opposite direction.L' = r (- p) = - L (2.3)

Thus, its angular momentum must point in the opposite direction. On the other hand when L isperpendicular to the mirror, then both r and p lie in the plane of the mirror and neither will beflipped. Then the image of the disk rotates in the same direction leaving the direction of L 'unchanged. Thus, the axial vectors and polar vectors behave exactly opposite to one another underreflections. It is important to differentiate between the two because the two types of vectors behavedifferently and manifest different vector properties. It was Hermann von Helmholtz who firstproved that the simplest form in which any vector field can be written is the sum of a polar part andan axial part, and this important theorem bears his name.2 .3 Flux

Consider a region of space with volume where a fluid is flowing through the volume, coming inClassical Fluids, Chapter 2 -2 5-

one side and out the other. Of importance is the amount of fluid, mass, within the region . It

• should be clear that the amount of fluid within the volume is related to the velocity distribution onthe boundary of the volume. For instance, if more fluid is flowing into than is flowing out, thenthe mass will increase in time. When we have more fluid flowing into a volume than out, we saythat there is a net negative flux, and conversely, when there is more fluid passing out of the volumethan into, we have a net positive flux. The two are illustrated in figure 3.4

(a) (b)Figure 2.8

(a) A net negative flux. (b) A net positive flux.

From Figure 2.8 we can see that if there are more field lines leaving a volume than entering, thenthere is a net positive flux. Thus, we see that in any continuous medium where we can invent avector field, the concept of flux is fundamental to describing what is happening. It's rather likerunning a grocery store. If you don't know what's arriving relative to what's leaving, you're notlikely to succeed.To define flux, nothing could be more natural than to say it is the net amount of mass passingthrough the surface per unit time.

flux = dmdt passing through a surface

(2.4)This should depend on how the velocity field is distributed over the surface. Suppose we have auniform flow through the surface S

Sx

AFigure 2.9

A flux through a simple surface S.

as shown in Figure 2.9 where the fluid flows a distance x in a time t. In the time t, the fluidwill sweep out a volume Ax where A is the surface area of the surface S. Multiplying anddividing the flux by this volume

m

tm

VA x

tS=

so that m

tvA

S= (2.5)

where v = x/t is the speed of the fluid.Classical Fluids, Chapter 2 -2 6-Let's make the situation a bit more realistic. Suppose that the surface is tipped relative to the flow

• so that the unit normal to surface makes an angle with the velocity

S

Figure 2.10The surface S is tipped relative to the flow so the unit normal to the surface makes

an angle with v.

as in Figure 2.10. Obviously the flux must have decreased. Furthermore, only the part of thesurface that is perpendicular to the flow will have a flux through it. This means we want theoverlap between the unit normal to the surface and the velocity which suggests using the dotproduct flux = v A. (2.6)

SFigure 2.11

Only the part of the surface perpendicular to the flow contributes to the flux.

The quantity v occurs so often that it is expedient to give it a name;J = v current density (2.7)

J is called the mass current density or simply the current density. J has units ofJ mass

time area[ ] =

.The flux becomes flux = J A. (2.8)

Suppose that our surface were not a simple plane but enclosed a volume. Surfaces that completelyenclose a volume are called closed surfaces. The surface might also not be rectangular.dA

Figure 2.12

Classical Fluids, Chapter 2 -2 7-

An infinitesimal patch (surface element) of a closed surface.

• Then we would have to divide the surface into infinitesimal patches dA as in Figure 2.12, calculatethe flux for each dA, and then sum all the dA over the whole surface. Note that the direction of dAis normal to the surface. Following the above procedure the flux becomes

dmdt

v dAS s

= r r mass flux (2.9)

which is the net mass flux through a volume bounded by a closed surface S, and dA is normal to Sand by convention always points out of the volume.Example 2.1An incompressible fluid is flowing with a uniform velocity v through a cube of side . a) Supposeat some moment that the flow has only reached the first half of the cube

(a) (b)Figure 2.13

a) A flow beginning through a cube. b) The flow is now steady.

as in Figure 2.13. What is the mass flux through the cube? b) Suppose now that the flow hasbecome fully developed through the cube as in Figure 2.13(b). What is the mass flux through thecube?Solution: a) There is a negative flux through the face perpendicular to the flow. There is no fluxthrough any of the other faces. Through the face where the flow is entering the volume,

J = J cos = - Jwhere is exactly opposite to the flow as in Figure 2.14 so that the angle between v and is 180and cos 180 = -1.

v180

Figure 2.14A negative flux.

Because J is uniform it may be pulled out of the integral. flux = S J dA = - J A = - J 2Classical Fluids, Chapter 2 -2 8-

• where A is the area of only the face the fluid is flowing through. A = 2. Note that the flux isnegative because it flows into the cube.b) Every field line that enters the left hand surface also leaves through the right hand surface so thenet flux is zero. Mathematically:

S J dA = - J2 + J2 = 0Note that since the field lines are parallel to the top, bottom, front, and back sides, there is no fluxthrough those sides, i.e., J dA = 0.In Example 2.1 there seems to be a connection between the number of field lines entering andleaving and the flux. Graphically we interpret the magnitude or strength of the field to be given bythe density of the field lines which would have units of (no. of lines/sq. cm) since we aremultiplying the field strength by the area (sq. cm) to get the flux, the flux is proportional to the netnumber of field lines passing through the surface. The flux must thus be connected with thesource of the vector field, i.e., the only way to have a net number of field lines pass through thesurface is to have field lines end inside the surface.A non zero flux through a closed surface exists only if the surfacecontains a source of the field, i.e., field lines end within the volumeenclosed by the surface.

This property can be seen by comparing the cases shown in Figure 2.13(a) which has a flux and(b) which has no flux. We can use this property to our advantage as shown by Example 2.2.Example 2.2A hemisphere of radius R is oriented in a uniform flow as shown in Figure 2.15 with the circulardisk perpendicular to the flow. The fluid is incompressible with a density . Calculate the fluxthrough the spherical part of the hemisphere only.

disk sphere

JFigure 2.15

Uniform flow through a hemispherical volume.

Solution: Because of the changing dot product J dA over the spherical part of the surface, the fluxis somewhat difficult to calculate; however, the flux through the circular disk is very easy tocalculate. J is uniform over the disk so it can be pulled out of the integral, and the dot productgives a -1 because is opposite to J.Classical Fluids, Chapter 2 -2 9-

• input flux = circle J dA = - J dAinput flux = - J r2

Since every field line that enters the surface also passes out through the surface, the net flux mustbe zero net flux = input flux + output flux = 0,and we can solve for output fluxoutput flux =- input flux = + Jr2

Thus, we have used the properties of flux to avoid calculating a complicated surface integral.Example 2.3The mass flux of water through a circular wire hoop or radius 0.5 m is 700 kg/s. If the plane ofthe wire hoop is tilted at an angle of 30 with respect to the flow and if the flow is uniform, find thespeed of the flow.

J

plane of the hoop

3060

Figure 2.16Flux through a wire hoop tilted at 30 with the current density J.

Solution: The mass flux is dm/dt = J dA (note we are not using a closed surface). Before wecan evaluate the integral, we must calculate the dot productJ dA = JdAcos

where is the angle between J and . Note that if the plane of the flow makes an angle of 30 withJ, then the normal must make an angle of 60 with J. Since J is uniform over the surface, we canpull J out of the flux integral yieldingdm/dt = J A cos 60

from which we can solve for the velocity using J = v.v

dm dtr

=

= ( ) ( ) /

cos . cos 2 260700

0 5 60 kg/s

1000 kg/m m3 = 1.8 m/s

Note that since water is so dense, even a very low flow rate of 1.8 m/s will create quite a fluxthrough a surface one meter in diameter.

Classical Fluids, Chapter 2 -3 0-

• 2 .4 Conservation of Mass

From many of the discussions we have had, it should be dawning upon you that flux is naturallylinked to conserved quantities. After all, if a quantity such as mass or charge is conserved, itmeans that it may be neither created nor destroyed. So if the quantity is flowing in a steady state,what passes in through any closed surface must also pass out, thus, providing a zero flux. Thedays when you can describe a conserved quantity merely by saying that it may neither be creatednor destroyed are numbered. Conservation implies far more, and it is time to glimpse the powerthat these principles actually give us.To understand the conservation laws, you must have a firm understanding of flux. Previously inmechanics, when we have said that a quantity is conserved, we have meant that it remainsconstant. This is fine for closed isolated systems where mass and energy cannot be exchangedwith any other system, but what about open systems, such as, a volume V where fluid can enterand leave? What does it mean to say that mass is conserved? Here we extend the idea ofconservation to mean that the amount of mass need not be constant; however, mass can be neithercreated or destroyed so that the only way the amount of mass in V can change is if there is a flowof mass through the surface bounding V, that is, the only way to change the mass in a givenvolume is if there is a flux through the surface bounding the volume!Let m represent the amount of mass inside V at any given moment. If mass cannot be either creatednor destroyed, then the only way to change the amount of mass inside a volume to move the massthrough the surface bounding the volume, i.e., create a flux. The mass in the volume will increaseif we create a negative flux (move mass into the volume).

dmdt

dmdtvolume bounding surface

= (2.10)Conservation of mass is a statement that the rate at which the mass within a volume changes isequal and opposite to the rate that mass leaves through the surface of the volume where the surfaceS completely bounds the volume V with no leaks. The right-hand side of equation (2.10) is justthe mass flux. Giving,

dmdt

v dAvolume S

= r r Continuity equation (2.11)Equation (2.11) expresses conservation of mass in its global form and is known as the continuityequation. The meaning of the minus sign is that a negative mass flux will increase the amount ofmass in the volume, and a positive flux will decrease the amount of mass in the volume.The continuity equation can be relatively easy to solve for steady flows that do not change in time.

Steady StateWhen a flow has become steady and does not change in time, the systemis said to have reached a steady state.

When a system is in a steady state, the mass within any given volume must remain steady in timeand not change. This means the continuity equation for a steady state simplifies toClassical Fluids, Chapter 2 -3 1-

Figure 2.17

The flux through the closed grey surface must be zero. As the pipe narrows, the fluid must speedup to keep the flux constant. Note that the field lines are much denser in the narrow section of thepipe indicating a larger velocity. Thus, an understanding of flux allows us to easily see that thespeed of the fluid must be greater in the narrow section of the pipe before we've even done anymathematics. Mathematically, J dA = 0 - 1v1A1 + 2v2A2 = 0

Classical Fluids, Chapter 2 -3 2-and using = constant:

• v1A1 = v2A2 Continuity for an (2.13)incompressible fluidv

AA

v21

21= v1 = 4 v1.

Thus, since A1 > A2, v2 > v1.From this example we can also gain insight into why the flux through A1 must be equal andopposite to the flux through A2. Let the flux through A1 be greater than the flux through A2. Thenfluid would build up inside the surface, and this build up cannot be maintained in a steady statebecause pressure effects would either cause the flow into the surface to diminish or the pipe wouldburst. An example of a case where more fluid flows in than out is a balloon being blown up. It isobvious that if a steady state is attempted and the flow rate into the balloon is kept constant, theballoon would soon pop. A popping balloon is anything but a steady state!The continuity equation allows us to add to our understanding the possible ways of creating flowsin a fluid. Flux is well-suited for investigating the source of the field. Note that the currentdensity is comprised of two quantities: the density, and the velocity.

J = vBoth the density and the velocity can create a flux. In order to have a net flux there must be aninhomogeniety in the flow, either more flowing out than in, or vice versa. Suppose first that thevelocity v varies with position, and the density is constant. Such a case is shown in Figure 2.18where the fluid on the left is at rest, and the fluid on the right accelerates from rest creating apositive flux through the grey box.

Fluid atrest here v

Figure 2.18A flux in J is produced by a variation in velocity with position.

Note that we could not maintain this flow for long because we would run out of fluid in the greyregion. This example is a time-dependent flow which is short lived.Consider next an example which can be maintained in a steady state in which the flux in J is causedpurely by a gradient in the density. Such an example occurs at a pipe with a section containingrefrigeration equipment which cools the fluid flowing in the pipe. As we enter the grey regionshown in Figure 2.19, the temperature drops and the fluid becomes denser. If the flow is steady-

Classical Fluids, Chapter 2 -3 3-

state, there must be the same flux through the left hand side as the right hand side.

• JHot Coldbecomingdenser

Figure 2.19The fluid becomes denser in the region of the gradient in the density.

But how can the flux be the same? If the density is going up, shouldn't flux go up? Let the area ofthe pipe be A. If the input flux must match the output flux, then1v1A = 2v2A

v v2 12

1=

(2.14)

Thus, this is a weird case where the net flux in J is zero, but there is a flux in v. The example isinstructive for more than flow in a pipe. Think of a freeway as rush hour approaches. If you takea section without entrance or exit ramps, there are no sources or sinks for cars. Cars are conservedso the influx to the section must equal the out flux as in equation (2.14); hence, as the density ofcars goes up (maybe cause by a couple of slow drivers), the velocity of the traffic flow goes down.Example 2.5Consider a 3-D region containing an incompressible fluid without gravity. A spherical source atthe center supplies fluid at a constant rate and causes a radial flow. If the velocity of the flow is 3m/s at r = 6 m from the center, what is the velocity at an arbitrary radius r from the center?Solution:A steady state flow requires the mass flux at r = 5 m to be the same at any other radius if mass isconserved:

v1A1 = v2A2The area perpendicular to the flow is the surface area of a sphere is 4r2:

vr

rv2

12

2 1= = 12 m /sr3

2

The velocity field falls-off as an inverse square field. Do you know any other inverse squarefields?It should no longer surprise you that the equations we have just defined apply to more than justconservation of mass in a fluid. The continuity equation applies to any conserved quantity. Theenergy of a system is also conserved so it too must obey a continuity equation.Classical Fluids, Chapter 2 -3 4-

• 2 .5 Conservation of Energy & Bernoulli's Principle.

One of the most fantastic developments is the similarity in form among all the conservation laws.As you will soon see, this similarity in form lends a formal simplicity and elegance to fieldtheories. We have talked about conservation of mass and yet there is nothing that we have saidwhich is peculiar to mass. Indeed, the only assumptions we have made are that the mass inquestion must have the ability to flow and that it is conserved, i.e., it cannot be created ordestroyed. These assumptions also fit energy, but rather than go through another lengthyderivation, let's draw upon our experience to guess the equation for energy conservation.Recall that for mass we defined a mass density

= M/V .We start in the same place with the energy U by defining an energy density u

u = U/V energy density (2.15)We also need a current density which for mass we defined to be

J = v.Similarly, we can define an energy current density S

S = uv energy current (2.16)densitywhere v is the velocity at which energy flows in the system. Note that S has units ofenergy/(area time). As with J, we can form an energy flux; however, the energy flux is nothingmysterious because the amount of energy passing through a surface per unit time is simply thepower passing through the surface.

Power = S dA (2.17)Recalling the mass continuity equation

dmdt

v dAvolume S

= r r (2.11)We expect that if energy is conserved the following continuity equation holds

dUdt

S dAvolume S

= r r (2.18)Energy, however, is not exactly like mass. In general, energy is not conserved. Nonconservativeforces such as friction can act in a fluid. Friction does play a very important role in fluids, andlater we will discuss some of its effects. For the moment, there are many important effects that weClassical Fluids, Chapter 2 -3 5-

first wish to study unencumbered by the technical difficulties friction introduces. In addition, there

• might also be work being done by forces applied to the volume by external forces.

dUdt

S dA dWdtvolume S

ext= + r r (2.19)

Such a case exists in a fluid where a volume of the fluid will experience a force due to the fluidaround it. This force is transmitted by a pressure gradient. We can modify equation (4-24) toaccount for this force by including the rate at which the surrounding fluid does work on thevolume.

dWdt

F v= r r (2.20)

dUdt

S dA F vvolume S

= + r r r r Continuity of Energy (2.21) in a fluid.

We will explore only the first rudimentary solution to equation (2.21).Consider an incompressible fluid in motion under the influence of gravity. In the following modelwe will make three assumptions:

1) The flow has reached a steady state.2) The fluid is incompressible, = const.3) All the forces are conservative - no friction.4) The only forces we will consider are those due to gravity and pressure gradients.The first assumption requires that the amount of energy in any volume must remain constant, i.e.,steady.

dUdt Volume

= 0 (2.22)Equation (2.22) has two contributions: one due to flux of energy, and the other due to rate atwhich the surrounding fluid does work.

- S S dA + F v = 0 (2.23)To evaluate Equation (2.23) consider a flux tube, that is, a tube whose walls are formed bystreamlines, as shown in Figure 2.20. The end caps to this tube are perpendicular to the velocity,and we will consider the radius of the tube to be small enough so that the velocity is uniform over

Classical Fluids, Chapter 2 -3 6-

the end caps.

• AA

12

Figure 2.20A Flux tube in the fluid.

There is a negative flux through surface 1 and a positive flux through surface 2.Recall that S = uv where u is the energy density of the fluid. The energy can come in three forms:kinetic, potential, and internal energy related to temperature and chemical potential. The internalenergy we'll assume it is approximately constant so we can ignore it. Then

u = 12 v2 + gh (2.24)where gh is the gravitational potential energy per unit volume, and the rate of energy flux is then

S S dA = - ( 12 v12 + gh1) v1A1 + ( 12 v22 + gh2) v2A2 (2.25)The rate at which work is done by the fluid is figured from the pressure times the area and notingthat at surface 1 the force exerted on the flux tube by the surrounding fluid is along v and against atsurface 2, as shown in Figure 2.21.

F v = P1A1v1 - P2A2v2. (2.26)

F = P A1 1 1

F = P A22v

Figure 2.21The force exerted on surface 1 is along v, and the force exerted on surface 2 is

opposite to v.

Substituting the last two equations into the energy continuity equation (2.23)Classical Fluids, Chapter 2 -3 7-

• + ( 12 v12 + gh1) v1A1 - ( 12 v22 + gh2) v2A2 + P1A1v1 - P2A2v2 = 0.Rearranging terms

( 12 v12 + gh1 + P1) v1A1 = ( 12 v22 + gh2 + P2 ) v2A2 (2.27)We can simplify this further by using the continuity of mass equation which by the steady stateassumption means that what flows in must also flow out for an incompressible fluid.

v1A1 = v2A2Equation (2.27) becomes

12 v2 + gh + P = Const. Bernoulli's (2.28)EquationWe can gain a feeling for what Bernoulli's equation tells us by applying it to some simpleexamples.Example 2.6An incompressible fluid has a speed v in a level section of pipe of radius r1 where the pressure isP. The pipe narrows to a radius r2. Find the pressure in the narrow section of pipe.Solution: Since the pipe is level, the height y is constant which simplifies Bernoulli's equation:

12 v2 + P = 12 v22 + P2. (2.29)Note that there are two unknowns, v2 & P2 so Bernoulli's equation alone is not enough to solve theproblem. For a steady state to exist, the continuity equation determines the velocity of the flow.The input mass flux must equal the output mass flux. The continuity equation gives

Av = A2v2

v

r

rv2

12

22= o .

Thus, the velocity of the flow is totally determined by how much fluid must fit through the narrowsection of pipe. The speed of the flow through the narrow sections is greater. Now we can solvefor P2 from equation (2.29).

Classical Fluids, Chapter 2 -3 8- P P v r

r2

2 12

22

12

1= +

o o

• Since r1 > r2, P2 < P1 So although the velocity is greatest in the narrow section, thepressure is least there. This example forms the basis for a device known as a venturi where apartial vacuum is created by a fast moving flow.The result of Example 2.6 is true in general and is known as Bernoulli's principle.

Bernoulli's Principle

The speed of the flow will be greatest where the pressure is least.

At first glance Bernoulli's principle may seem to contradict common sense but remember that theforce on a fluid always points towards low pressure so that a fluid will accelerate towards lowpressure, and thus, will be moving faster as it gets closer to the low pressure region.From Bernoulli's principle we can explain many seemingly magical effects. For instance a favoritedisplay by stores having a sale on vacuum cleaners is to place a beach ball in the output streamfrom the vacuum cleaner. Common sense dictates that the ball would immediately be pushed outof the stream and go flying across the store, but it doesn't. The ball magically floats in the stream.This is not accomplished by hidden wires or other slights of hand. The salesmen are usingBernoulli's principle to keep the ball in the stream. Due to friction the flow from the vacuumcleaner will be faster at the center than on the edges of the flow. As the ball starts to move from thecenter of the stream as in Figure 2.22, the flow is faster on the center side of the ball.

LowHigh

fastslowspeed

pressure(a) (b)

Figure 2.22(a) A ball centered in the flow has equal pressures on each side. (b) As the ball movesoff center, the flow is faster toward the center. The low pressure on the center side ofthe ball places a restoring force on the ball keeping it in the flow.

By Bernoulli's principle the pressure must be lower on the center side of the ball. There is thus anatural restoring force that tends to keep the ball in the stream. So the vacuum salesman doesn'thave to do anything special; she merely sets the ball in the air stream and Bernoulli's principalnaturally keeps the ball there. Least you think this is magic, notice one other important fact fromClassical Fluids, Chapter 2 -3 9-the flow in Figure 2.22. The air on the center side of the ball in (b) is flowing faster and causes a

• bending of the flow to the left in the wake of the ball. Since the ball has deflected the air to the leftthe air places an equal and opposite force on the ball pushing it to the right by Newton's third law.The shape of an airplane wing, which is called an air foil , is also determined by Bernoulli'sprinciple and conservation of mass. Consider the air foil shown in Figure 2.23. Because of itsasymmetrical shape, the volume of fluid above the center line (shown as a dashed line) of the airfoil is constricted more than the volume below the center line.

(a)

Figure 2.23(a) The velocity above the airfoil must be greater because of the constricted volumeabove the center line (dashed). (b) Flow past a symmetrical airfoil and an attack angle.

As we have seen in two examples, conservation of mass requires that where the area perpendicularto the flow becomes constricted the flow speed must increase. Therefore, the speed of the fluidwill be greater above the airfoil than below, and thus, the air pressure will be less above thanbelow. The resultant pressure gradient across the air foil creates a lifting force. Be careful! Sinceair is highly compressible and the flows around an actual wing become turbulent, Bernoulli'sequation cannot be expected to yield good results when applied to an airplane wing. Also,airplanes do not rely upon the shape of the wing alone for lift. The wing is set at an attack angle sothat it actually bends the flow coming off the tail downwards even further than that shown in figure4.10. Since the air experiences a downwards force, by Newton's third law, the wing experiencesan upwards force.As a final example, have you ever wondered why the roofs of buildings always blow off in highwind storms? Consider the flow shown in Figure 2.24. The velocity of the flow inside therectangular building is zero so there is high pressure inside. The flow velocity above the roof isvery high, so by Bernoulli's principle, the pressure is low. This gives rise to a lifting force

F P P Aoutside inside roof= ( ) (2.30)

Classical Fluids, Chapter 2 -4 0-

• Figure 2.25Velocity of flow over a rectangular building. The velocity on the top is very high.

The roof is very literally sucked off the building. It would be better to open the windows and letthe air flow through the building thereby decreasing the pressure inside the building.There are many more examples we could use, and some of these, as in the case of a baseball'sinfamous curve ball, will be treated later, and others, such as a frisbie, will be left as homeworkproblems. We have however, not by any means exhausted the applications of Bernoulli'sprinciple. The following example of a siphon illustrates the diverse places Bernoulli's principlecan be used.Example 2.7

What is the greatest height h1 in Figure 2.26 that the siphon will lift water?

hh

A

B

Ch

1

2

3

Figure 4.11A simple siphon for draining the fluid from a tank.

Solution: A siphon is used to drain tanks that don't have outlets. The weight of the fluid in thetube over h pulls the fluid up over the hump. Once a flow is established, it will continue until allthe fluid is drained. At first it might seem wise to apply Bernoulli's principle to points A and B;however, we don't know the pressure at point A. It will not be P + gh3 because the fluid has asubstantial velocity at point A which will reduce the pressure below that for a static fluid. At pointClassical Fluids, Chapter 2 -4 1-

C the fluid enters air so it must approximately be at atmospheric pressure P. We can then apply

• Bernoulli's equation to points B and C. Setting the coordinate system origin at point C,PB + 12 vB2 + g(h1 + h2 + h3) = P + 12 vC2 + 0.

Because the flow is steady, conservation of mass determines the speed of the flow in the tube.AvB = AvC vB = vC.

Solving for h1: h P P

gh hB1 2 3=

+( )oNote that our solution depends upon the difference in pressure between points C and B. Pressurecan never be negative so the largest value this difference can have is when PB = 0 .

h P

gh h1 2 3max = +( )o

2 .6 Divergence

The road to this point has been a very theoretical, and we have come a long way, but there is onemore refinement that we can make that will increase our understanding even further. InFigure 2.27 we have one surface S that encloses a source of flow, while surface S' does not. Fromour previous discussions it should be immediately apparent that S' has zero flux through it becauseevery field line that enters also leaves. S, on the other hand has many lines leaving so it has a netflux through it. We must conclude that there is something different about the volume bounded byS than the volume bounded by S'.

S

S'

Figure 6.1S bounds a source and has a finite flux.

S' does not bound a source and has zero flux.

If we want to get to the heart of the matter, then we need to find a way to mathematically get towhat is going on inside the volumes and not be content to just deal with the surfaces, that is, wewish to mathematically explore the structure of the sources themselves. One way to do this is totake the surface S and shrink it until it is inside the source, until it contains only a smallClassical Fluids, Chapter 2 -4 2-infinitesimal piece of the source. If we do this, we will get to the heart of what a diverging field

• really is, and so justifiably, we'll define this to be the divergence of the field. Suppose we have aflow J. The divergence can technically be defined to be

div JV

J dA

VS

r

r r

=

lim

0. (2.31)

where S bounds the volume V and we take the limit as the volume goes to zero. Equation (2.31) isa coordinate-free definition of the divergence. It gets across the local nature of the divergence;however, it is not a practical means to calculate the divergence which is found by considering aflow through a cubical surface shown in Figure 2.28. The center of the cube is at the point (x, y,z), and has side lengths x, y, z so the volume is V = xyz.

J

xx0 x + x

x

Figure 2.28A cube has a flux passing through the faces perpendicular to the x axis.

The right hand side has a greater positive flux than the left hand side has negative flux. The netflux is J dA = Jx(x + x, y ,z) yz - Jx(x, y, z) yz (2.32)

where Jx(x + x,y,z) means the x component of J evaluated at the point (x + x, y, z). Dividing bythe volume and taking the limit as V goes to zero:

div Jx y z

J x x y z J x y zx y z

x xr=

+( ) ( )lim

0 .

Note that the yz divides out leaving

div Jx

J x x J xx

x xr=

+( ) ( )lim

0 . (2.33)

Classical Fluids, Chapter 2 -4 3-What's left is nothing less than the definition of the partial derivative of Jx with respect to x. We

• could also repeat this for flows through the y and z faces and obtain similar results that can besummarized as div J J

x

Jy

Jz

x y zr= + +

, (2.34)

which we can write in a coordinate free manner by taking the dot product of del on Jdiv J = J . divergence (2.35)

We will take equation (2.36) as the definition of the divergence. Recall the reason we areintroducing the divergence is that we want to explore the volume where the flux is originating.We can interpret the divergence from a wonderful theorem first proved by Gauss and known asGauss's Theorem (as opposed to Gauss's Law) or as simply the divergence theorem. Thedivergence theorem states,The divergence at first looks to be a horrible beast, full of vector nasties. But in actuality equation(2.31) allows us to easily attach physical significance to the divergence because the divergence isnothing but a flux density. The only way to have a non zero flux is to have field lines end withinthe surface S. This means that within the volume V where the field lines end, there will be adivergence. Thus, any volume in which field lines end will contain a divergence as illustrated inFigure 2.29.

V V'Figure 6.3Only regions where the field lines end will have a divergence. There is no divergenceanywhere in the boxed region V . There is a positive divergence in the boxed regionV ' .

The divergence is a local property of a vector field, but Flux is a global property. The divergencebeing a density has meaning at a given point in space, but flux exists only over a surface. Notethat the divergence forms a scalar field which we can interpret as the flux density.2 .7 The Divergence And The Continuity Equation.

We can gain a different perspective on the continuity of mass by writing the continuity equation inits local form. To accomplish this task divide both sides of equation (2.11) by the volume and takethe limit as V 0

Classical Fluids, Chapter 2 -4 4-

• lim limV

ddt

m

VV

J dA

VS

=

0 0

r r

(2.37)

The left hand side gives the density which can be a function of position as well as time, so thederivative must become a partial derivation. The right hand side is the definition of the divergence,hence:t J=

r . Continuity (6-10)local form

Recall that divergence only occurs in a region where field lines end. This means that for anyconserved quantity, the only way to change the density is to have field lines end. This connectionof divergence with the ends of field lines actually forms a very powerful visual tool. The onlyregions where the density is changing is where field lines of mass current density J end. The Jfield depicted in Figure 6.4 has two regions of positive divergence, 1 and 2, where the densitymust be decreasing because fluid is flowing out of these regions, and one region, 3, where thedivergence is negative and the density is increasing because fluid is flowing into the region.

1

2

3

Figure 2.30A field with three regions of divergence

Classical Fluids, Chapter 2 -4 5-

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