Control-Volume Approach

T

he second of the fundamental physical laws upon whichfluid-flow analyses are based is Newton’s second law of motion. Starting with Newton’s second law, we shall develop integral relations for linear and angular momentum. Applications of these expressions to physical situations will be considered.

▶ 5.1

INTEGRAL RELATION FOR LINEAR MOMENTUM Newton’s second law of motion may be stated as follows:

The time rate of change of momentum of a system is equal to the net force acting on the system and takes place in the direction of the net force.

We note at the outset two very important parts of this statement: first, this law refers to a specific system, and second, it includes direction as well as magnitude and is therefore a vector expression. In order to use this law, it will be necessary to recast its statement into a form applicable to control volume that contains different fluid particles (i.e., a different system) when examined at different times.

In Figure 5.1, observe the control volume located in afluid-flow field. The system considered is the material occupying the control volume at timet, and its position is shown both at timetand at time t‡Dt.

Referring to thefigure, we see that

I III

Boundary of system at time t

Streamlines at time t

Boundary of system at time t + t Stationary control volume

II

Figure 5.1 Relation between a system and a control volume in a fluid-flowfield.

44

Region I is occupied by the system only at timet.

Region II is occupied by the system att‡Dt.

Region III is common to the system both at tand att‡Dt.

Writing Newton’s second law for such a situation, we have

∑Fˆ d

dt…mv† ˆ d

dtP (5-1)

where the symbolsF,m, andvhave their usual meanings andPrepresents the total linear momentum of the system.

At timet‡Dtthe linear momentum of the system that now occupies regions II and III may be expressed as

Pj_t‡Dtˆ jP_IIj_t‡Dt‡ jP_IIIj_t‡Dt and at time twe have

Pj_tˆPIj_t‡PIIIj_t

Subtracting the second of these expressions from thefirst and dividing by the time interval Dtgives

Pjt‡Dt Pjt

Dt ˆPIIjt‡Dt‡PIIIjt‡Dt PIjt PIIIjt

Dt

We may rearrange the right-hand side of this expression and take the limit of the resulting equation to get

Dlimt®0

Pj_t‡Dt Pj_t Dt ˆ lim

Dt®0

P_IIIj_t‡Dt P_IIIj_t Dt ‡ lim

Dt®0

P_IIj_t‡Dt P_Ij_t

Dt (5-2)

Considering each of the limiting processes separately, we have, for the left-hand side,

Dlimt®0

Pjt‡Dt Pjt

Dt ˆ d dtP

which is the form specified in the statement of Newton’s second law, equation (5-1).

The first limit on the right-hand side of equation (5-2) may be evaluated as

Dlimt®0

PIIIjt‡Dt PIIIjt

Dt ˆ d dtPIII

This we see to be the rate of change of linear momentum of the control volume itself, since, as Dt®0, region III becomes the control volume.

The next limiting process

Dlimt®0

P_IIj_t‡Dt P_Ij_t Dt

expresses the net rate of momentum efflux across the control surface during the time interval Dt. AsDtapproaches zero, regions I and II become coincident with the control-volume surface.

Considering the physical meaning of each of the limits in equation (5-2) and Newton’s second law, equation (5-1), we may write the following word equation for the conservation of linear momentum with respect to a control volume:

5.1 Integral Relation for Linear Momentum ◀ 45

sum of forces acting

on control volume 8>

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>>

:

9>

>=

>>

;ˆ

rate of momentum out of control

volume 8>

><

>>

:

9>

>=

>>

;

rate of momentum into control

volume 8>

><

>>

:

9>

>=

>>

;

|fflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl{zfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflfflffl}

net rate of momentum efflux from control volume

‡

rate of accumulation of momentum within control

volume 8>

>>

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>>

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9>

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>=

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; (5-3)

We shall now apply equation (5-3) to a general control volume located in afluid-flow field as shown in Figure 5.2 and evaluate the various terms.

The total force acting on the control volume consists both of surface forces due to interactions between the control-volumefluid, and its surroundings through direct contact, and of body forces resulting from the location of the control volume in a forcefield. The gravitationalfield and its resultant force are the most common examples of this latter type.

We will designate the total force acting on the control volume as∑F.

If the small areadA on the control surface is considered, we may write rate of momentum effluxˆv…rv†…dAcosq†

Observe that the product…rv†…dAcosq†is the rate of mass efflux from the control volume throughdA, as discussed in Chapter 4. Recall further thatdAcosqis the area,dA, projected in a direction normal to the velocity vector, v, whereq is the angle betweenv and the outwardly directed normal vector,n. We may then multiply the rate of mass efflux byvto give the rate of momentum efflux through dA.From vector algebra this product may be written as

v…rv†…dAcosq† ˆv…rdA†‰jvj jnjcosqŠ

The term in square brackets is the scalar or dot product,v

n, and the momentum efflux term becomes

rv…v

^n†dA

Integrating this quantity over the entire control surface, we have Z Z

c:s:vr…v

^n†dA

which is thenet momentum effluxfrom the control volume.

In its integral form the momentum flux term stated above includes the rate of momentum entering the control volume as well as that leaving. If mass is entering the control volume, the sign of the productv

ⁿis negative, and the associated momentumflux is an input. Conversely, a positive sign of the productv

ⁿis associated with a momentum

v n dA

Streamlines at time t

Figure 5.2 Fluidflow through a control volume.

efflux from the control volume. Thus, thefirst two terms on the right-hand side of equation (5-3) may be written

rate of momentum out of control

volume 8<

:

9=

;

rate of momentum into control

volume 8<

:

9=

;ˆ Z Z

c:s:vr…v

^n†dA

The rate of accumulation of linear momentum within the control volume may be expressed as

@

@t Z Z Z

c:v:vrdV

and the overall linear-momentum balance for a control volume becomes

∑Fˆ Z Z

c:s:vr…v

^n†dA‡_@^@_t^{Z Z Z}

c:v:rvdV (5-4)

This extremely important relation is often referred to influid mechanics as themomentum theorem.Note the great similarity between equations (5-4) and (4-1) in the form of the integral terms; observe, however, that equation (5-4) is a vector expression opposed to the scalar form of the overall mass balance considered in Chapter 4. In rectangular coordinates the single- vector equation, (5-4), may be written as three scalar equations:

∑Fxˆ Z Z

c:s:vxr…v_?n†dA‡@

@t Z Z Z

c:v:rvxdV (5-5a)

∑F_yˆ Z Z

c:s:vyr…v_?n†dA‡@

@t Z Z Z

c:v:rvydV (5-5b)

∑Fzˆ Z Z

c:s:vzr…v_?n†dA‡@

@t Z Z Z

c:v:rvzdV (5-5c) When applying any or all of the above equations, it must be remembered that each term has a sign with respect to the positively definedx,y, andzdirections. The deter- mination of the sign of the surface integral should be considered with special care, as both the velocity component (vx) and the scalar product (v

n) have signs. The combination of the proper sign associated with each of these terms will give the correct sense to the integral. It should also be remembered that as equations (5-5a–c) are written for thefluid in the control volume,the forces to be employed in these equations are those acting on thefluid.

A detailed study of the example problems to follow should aid in the understanding of, and afford facility in using, the overall momentum balance.

▶ 5.2

APPLICATIONS OF THE INTEGRAL EXPRESSION FOR LINEAR MOMENTUM

In applying equation (5-4), it isfirst necessary to define the control volume that will make possible the simplest and most direct solution to the problem at hand. There are no general rules to aid in this definition, but experience in handling problems of this type will enable such a choice to be made readily.

5.2 Applications of the Integral Expression for Linear Momentum ◀ 47

Example 1

Considerfirst the problem offinding the force exerted on a reducing pipe bend resulting from a steadyflow offluid in it. A diagram of the pipe bend and the quantities significant to its analysis are shown in Figure 5.3.

Thefirst step is the definition of the control volume. One choice for the control volume, of the several available, is allfluid in the pipe at a given time. The control volume chosen in this manner is designated in Figure 5.4, showing the external forces imposed upon it. The external forces imposed on thefluid include the pressure forces at sections (1) and (2), the body force due to the weight offluid in the control volume, and the forces due to pressure and shear stress,Pwandtw, exerted on thefluid by the pipe wall. The resultant force on thefluid (due toPwandtw) by the pipe is symbolized asB, and itsxandycomponents asBxand By, respectively.

Considering thex-andy-directional component equations, (5-5a) and (5-5b), of the overall momentum balance, the external forces acting on thefluid in the control volume are

∑FxˆP1A1 P2A2cosq‡Bx

and

∑FyˆP2A2sinq W‡By

Each component of the unknown forceBis assumed to have a positive sense. The actual signs for these components, when a solution is obtained, will indicate whether or not this assumption is correct.

Evaluating the surface integral in both thexandydirections, we have Z Z

c:s:vxr…v?n†dAˆ …v2cosq†…r2v2A2† ‡ …v1†… r1v1A1† Z Z

c:s:vyr…v_?n†dAˆ … v2sinq†…p2v2A2† v₁

v₂ 1

2 W

y

x

Figure 5.3 Flow in a reducing pipe bend.

1

2 P₁

P₂ Pw

W Pw

Figure 5.4 Control volume defined by pipe surface.

The complete momentum expressions in thexandydirections are

Bx‡P1A1 P2A2cosqˆ …v2cosq†…r2v2A2† ‡v1… r1v1A1† and

By‡P2A2sinq Wˆ … v2sinq†…r2v2A2† Solving for the unknown force componentsBxandBy, we have

Bxˆv²2r2A2cosq v²1r1A1 P1A1‡P2A2cosq and

Byˆ v²2r2A2sinq P2A2sinq‡W

Recall that we were to evaluate the force exerted on the pipe rather than that on thefluid. The force sought is the reaction toB and has components equal in magnitude and opposite in sense toBxandBy. The components of the reaction force,R, exerted on the pipe are

Rxˆ v²2r2A2cosq‡v²1r1A1‡P1A1 P2A2cosq and

Ryˆv²2r2A2sinq‡P2A2sinq W

Some simplification in form may be achieved if theflow is steady. Applying equation (4-2), we have r1v1A1ˆr2v2A2ˆṁ

whereṁis the massflow rate.

Thefinal solution for the components ofRmay now be written as

Rxˆm…v̇ 1 v2cosq† ‡P1A1 P2A2cosq Ryˆmv̇ 2sinq‡P2A2sinq W

The control volume shown in Figure 5.4 for which the above solution was obtained is one possible choice. Another is depicted in Figure 5.5. This control volume is bounded simply by the straight planes cutting through the pipe at sections (1) and (2). The fact that a control volume such as this can be used indicates the versatility of this approach, that is, that the results of complicated processes occurring internally may be analyzed quite simply by considering only those quantities of transfer across the control surface.

For this control volume, thex-andy-directional momentum equations are

Bx‡P1A1 P2A2cosqˆ …v2cosq†…r2v2A2† ‡v1… v1r1A1† and

By‡P2A2sinq Wˆ … v2sinq†…r2v2A2† 1

W 2 B_y

P₁

P₂ Bx

Figure 5.5 Control volume includingfluid and pipe.

5.2 Applications of the Integral Expression for Linear Momentum ◀ 49

where the force having componentsBxandByis that exerted on the control volume by the section of pipe cut through at sections (1) and (2). The pressures at (1) and (2) in the above equations are gage pressures, as the atmospheric pressures acting on all surfaces cancel.

Note that the resulting equations for this control volume are identical to those obtained for the one defined previously. Thus, a correct solution may be obtained for each of several chosen control volumes so long as they are analyzed carefully and completely.

Example 2

As our second example of the application of the control-volume expression for linear momentum (the momentum theorem), consider the steam locomotive tender schematically illustrated in Figure 5.6, which obtains water from a trough by means of a scoop. The force on the train due to the water is to be obtained.

The logical choice for a control volume in this case is the water-tank/scoop combination. Our control-volume boundary will be selected as theinteriorof the tank and scoop. As the train is moving with a uniform velocity, there are two possible choices of coordinate systems. We may select a coordinate system eitherfixed in space or moving¹with the velocity of the train,v0. Let us first analyze the system by using a moving coordinate system.

The moving control volume is shown in Figure 5.7 with thexycoordinate system moving at velocityv0. All velocities are determined with respect to thexandyaxes.

The applicable expression is equation (5-5a)

∑Fxˆ Z Z

c:s:vxr…v?n†dA‡@

@t Z Z Z

c:v:vxrdV

In Figure 5.7,∑Fxis represented asFxand is shown in the positive sense. As the forces due to pressure and shear are to be neglected,Fxis the total force exerted on thefluid by the train and scoop. The momentumflux term is

Z Z

c:s:vxr…v_?n†dAˆr… v0†… 1†…v0†…h† …per unit length† h

v0

Figure 5.6 Schematic of locomotive tender scooping water from a trough.

h y

x Fx

v0

Figure 5.7 Moving coordinate system and control volume.

1Recall that a uniformly translating coordinate system is an inertial coordinate system, so Newton’s second law and the momentum theorem may be employed directly.

and the rate of change of momentum within the control volume is zero, as thefluid in the control volume has zero velocity in thexdirection.

Thus,

Fxˆrv²0h

This is the force exerted by the train on thefluid. The force exerted by thefluid on the train is the opposite of this, or rv²0h.

Now let us consider the same problem with a stationary coordinate system (see Figure 5.8). Employing once again the control-volume relation for linear momentum

∑Fxˆ Z Z

c:s:vxr…v

^n†dA‡_@t^{@Z Z Z}

c:v:vxrdV we obtain

Fxˆ0‡@

@t Z Z Z

c:v:vxrdV

where the momentumflux is zero, as the enteringfluid has zero velocity. There is, of course, nofluid leaving the control volume.

The terms @=@tRRR

c:v:vxrdV, as the velocity, vxˆv0ˆ constant, may be written as v0@=@tRRR

c:v:rdV or v0…@m=@t†, whereṁ is the mass offluid entering the control volume at the rate @m=@tˆrv0hso thatFxˆrv²0has before.

The student should note that in the case of a stationary coordinate system and a moving control volume, care must be exercised in the interpretation of the momentumflux

Z Z

c:s:vr…v

^n†dA

Regrouping the terms, we obtain

Z Z

c:s:

vr…v

^{n†dAZ Z}

c:s:

vdṁ

Thus, it is obvious that whilevis the velocity relative tofixed coordinates,v

ⁿis the velocity relative to the control-volume boundary.

Example 3

A jet offluid exits a nozzle and strikes a vertical plane surface as shown in Figure 5.9.

(a) Determine the force required to hold the plate stationary if the jet is composed of i. water

ii. air

h y

x

Fx v₀

Figure 5.8 Stationary coordinate system and moving control volume.

5.2 Applications of the Integral Expression for Linear Momentum ◀ 51

(b) Determine the magnitude of the restraining force for a water jet when the plate is moving to the right with a uniform velocity of 4 m/s.

The control volume to be used in this analysis is shown Figure 5.10.

The coordinates arefixed with the control volume, which, for parts (a) and (b) of this example, is stationary.

Writing thex-directional scalar form of the momentum theroem, we have

∑Fxˆ Z Z

c:s:vxr…v

^n†dA‡_@t^{@Z Z Z}_c:v:^vxrdv

Evaluation of each term in this expression yields

∑Fxˆ F Z Z

c:s:vxr…v

^{n†dAˆvjr… vjAj†}

@

@t Z Z Z

c:v:vxrdvˆ0 and the governing equation is

FˆrAjv²_j y

C.V.

x F

3 2

1

.

Figure 5.10 Control volume for Example 3.

Plate F Aj = 0.005 m²

V_j = 12 m/s

Figure 5.9 Afluid jet striking a vertical plate.

We may now introduce the appropriate numerical values and solve forF.For case (a), i. rwˆ1000 kg=m³

Fˆ …1000 kg=m³†…0:005 m²†…12 m=s†²

ˆ720 N ii. rwˆ1:206 kg=m³

Fˆ …1:206 kg=m³†…0:005 m²†…12 m=s†²

ˆ0:868 N

For case (b), the same control volume will be used. In this case, however, the control volume andthe coordinate systemare moving to the right at a velocity of 4 m/s. From the perspective of an observer moving with the control volume, the velocity of the incoming water jet is…vj v0† ˆ8 m=s.

Thex-directional component form of the momemtum theorem will yield the expression FˆrAj…vj v0†²

Substitution of appropriate numerical values yields

Fˆ …1000 kg=m³†…0:005 m²†…12 4 m=s†²

ˆ320 N

▶ 5.3

INTEGRAL RELATION FOR MOMENT OF MOMENTUM

The integral relation for the moment of momentum of a control volume is an extension of the considerations just made for linear momentum.

Starting with equation (5-1), which is a mathematical expression of Newton’s second law of motion applied to a system of particles (Figure 5.11),

∑Fˆ d

dt…mv† ˆ d

dtP (5-1)

we take the vector or “cross”product of a position vector,r, with each term and get r∑Fˆr d

dt…mv† ˆrd

dtP (5-6)

The quantity on the left-hand side of equation (5-6),r∑F, is the resultant moment,∑M, about the origin as shown in Figure 5.11, due to all forces applied to the system. Clearly,

y

z

x r

mv

Figure 5.11 A system and its displacement vectorr.

5.3 Integral Relation for Moment of Momentum ◀ 53

we may write

r∑Fˆ∑rFˆ∑M

where∑Mis, again, the total moment about the origin of all forces acting on the system.

The right-hand side of equation (5-6) is the moment of the time rate of change of linear momentum. This we can write as

rd

dtmvˆ d

dt…rmv† ˆ d

dt…rP† ˆ d dtH

Thus, this term is also the time rate of change of the moment of momentum of the system.

We shall use the symbolHto designate moment of momentum. The complete expression is now

∑Mˆ d

dtH (5-7)

As with its analogous expression for linear momentum, equation (5-1), equation (5-7) applies to a specific system. By the same limit process as that used for linear momentum, we may recast this expression into a form applicable to a control volume and achieve a word equation

sum of moments acting on control volume 8>

>>

><

>>

>>

:

9>

>>

>=

>>

>>

;

ˆ

rate of moment of momentum out of control

volume 8>

>>

><

>>

>>

:

9>

>>

>=

>>

>>

;

rate of moment of momentum into

control volume 8>

><

>>

:

9>

>=

>>

;

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net rate of efflux of moment of momentum from control volume

‡

rate of accumulation

of moment of momentum

within control volume 8>

>>

>>

>>

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:

9>

>>

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>=

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;

(5-8) Equation (5-8) may be applied to a general control volume to yield the following equation:

∑Mˆ Z Z

c:s:…rv†r…v

^n†dA‡_@^@_t^{Z Z Z}

c:v:…rv†rdV (5-9) The term on the left-hand side of equation (5-9) is the total moment of all forces acting on the control volume. The terms on the right-hand side represent the net rate of efflux of moment of momentum through the control surface and the rate of accumulation of moment of momentum within the control volume, respectively.

This single-vector equation may be expressed as three scalar equations for the orthogonal inertial coordinate directionsx,y, andzas

∑Mxˆ Z Z

c:s:…rv†xr…v

^n†dA‡_@^@_t^{Z Z Z}

c:v:…rv†xrdV (5-10a)

∑Myˆ Z Z

c:s:…rv†yr…v

^n†dA‡_@^@_t^{Z Z Z}

c:v:…rv†yrdV (5-10b) and

∑M_zˆ Z Z

c:s:…rv†_zr…v

^n†dA‡_@^@_t^{Z Z Z}

c:v:…rv†_zrdV (5-10c)

The directions associated withM_xand (rv) are those considered in mechanics in which the right-hand rule is used to determine the orientation of quantities having rotational sense.

▶ 5.4

APPLICATIONS TO PUMPS AND TURBINES

The moment-of-momentum expression is particularly applicable to two types of devices, generally classified as pumps and turbines. We shall, in this section, consider those having rotary motion only. If energy is derived from a fluid acting on a rotating device, it is designated a turbine, whereas a pump adds energy to afluid. The rotating part of a turbine is called a runner and that of a pump an impeller.

The following two examples illustrate how moment-of-momentum analysis is used to generate expressions for evaluating turbine performance. Similar approaches will be used in Chapter 14 to evaluate operating characteristics of fans and pumps.

Example 4

Let usfirst direct our attention to a type of turbine known as the Pelton wheel. Such a device is represented in Figure 5.12. In this turbine, a jet offluid, usually water, is directed from a nozzle striking a system of buckets on the periphery of the runner. The buckets are shaped so that the water is diverted in such a way as to exert a force on the runner that will, in turn, cause rotation.

Using the moment-of-momentum relation, we may determine the torque resulting from such a situation.

We must initially define our control volume. The dashed line in Figure 5.13 illustrates the control volume chosen. It encloses the entire runner and cuts the jet of water with velocityn0as shown. The control surface also cuts through the shaft on both sides of the runner.

v₀

Bottom view of bucket Figure 5.12 Pelton wheel.

P_at

v₀ y

x

Figure 5.13 Control volume for analysis of Pelton wheel.

5.4 Applications to Pumps and Turbines ◀ 55