Conservation of Energy

Control-Volume Approach

T

he third fundamental law to be applied tofluid-flow analyses is thefirst law of thermodynamics. An integral expression for the conservation of energy applied to a control volume will be developed from thefirst law of thermodynamics, and examples of the application of the integral expression will be shown.

▶ 6.1

INTEGRAL RELATION FOR THE CONSERVATION OF ENERGY The first law of thermodynamics may be stated as follows:

If a system is carried through a cycle, the total heat added to the system from its surroundings is proportional to the work done by the system on its surroundings.

Note that this law is written for a specific group of particles—those comprising the defined system. The procedure will then be similar to that used in Chapter 5—that is, recasting this statement into a form applicable to a control volume that contains different fluid particles at different times. The statement of thefirst law of thermodynamics involves only scalar quantities, however, and thus, unlike the momentum equations considered in Chapter 5, the equations resulting from the first law of thermodynamics will be scalar in form.

The statement of the first law given above may be written in equation form as I

dQˆ1 J I

dW (6-1)

where the symbolH

refers to a“cyclic integral”or the integral of the quantity evaluated over a cycle. The symbols dQ and dW represent differential heat transfer and work done, respectively. The differential operator, d, is used as both heat transfer and work are path functions and the evaluation of integrals of this type requires a knowledge of the path. The more familiar differential operator, d, is used with a “point”function. Thermodynamic properties are, by definition, point functions, and the integrals of such functions may be evaluated without a knowledge of the path by which the change in the property occurs between the initial and final states.¹The quantity Jis the so-called “mechanical equivalent of heat,”numerically equal to 778.17 ft lb/Btu in engineering units. In the SI

1For a more complete discussion of properties, point functions and path fuctions, the reader is referred to G. N.

Hatsopoulos and J. H. Keenan,Principles of General Thermodynamics. Wiley, New York, 1965, p. 14.

65

system,Jˆ1 N m/J. This factor will not be written henceforth, and the student is reminded that all equations must be dimen- sionally homogeneous.

We now consider a general thermo- dynamic cycle, as shown in Figure 6.1.

The cycle a occurs between points 1 and 2 by the paths indicated. Utilizing equation (6-1), we may write, for cyclea,

Z 2 1adQ‡

Z 1 2adQˆ

Z 2 1a dW‡

Z 1 2adW

(6-2a)

A new cycle between points 1 and 2 is postulated as follows: the path between points 1 and 2 is identical to that considered previously; however, the cycle is completed by pathbbetween points 2 and 1, which is any path other thanabetween these points. Again equation (6-1) allows us to write

Z 2 1adQ‡

Z 1 2b dQˆ

Z 2 1adW‡

Z 1

2b dW (6-2b)

Subtracting equation (6-2b) from equation (6-2a) gives Z 1

2adQ Z 1

2b dQˆ Z 1

2adW Z 1

2b dW which may be written

Z 1

2a…dQ dW† ˆ Z 1

2b…dQ dW† (6-3)

As each side of equation (6-3) represents the integrand evaluated between the same two points but along different paths, it follows that the quantity,dQ dW, is equal to a point function or a property. This property is designateddE, the total energy of the system. An alternate expression for the first law of thermodynamics may be written

dQ dW ˆdE (6-4)

The signs ofdQanddW were specified in the original statement of thefirst law;dQis positive when heat is added to the system,dWis positive when work is done by the system.

For a system undergoing a process occurring in time intervaldt, equation (6-4) may be written as

dQ dt dW

dt ˆdE

dt (6-5)

Consider now, as in Chapter 5, a general control volumefixed in inertial space located in afluid-flowfield, as shown in Figure 6.2. The system under consideration, designated by dashed lines, occupies the control volume at timet, and its position is also shown after a period of time Dthas elapsed.

P

a a

b 1

2

Figure 6.1 Reversible and irreversible thermodynamic cycles.

In thisfigure, region I is occupied by the system at timet, region II is occupied by the system att‡Dt, and region III is common to the system both attand att‡Dt.

At timet‡Dt the total energy of the system may be expressed as Ej_t‡DtˆE_IIj_t‡Dt‡E_IIIj_t‡Dt

and at time t

EjtˆEIjt‡EIIIjt

Subtracting the second expression from thefirst and dividing by the elapsed time interval, Dt, we have

Ej_t‡Dt Ej_t

Dt ˆE_IIIj_t‡Dt‡E_IIj_t‡Dt E_IIIj_t E_Ij_t Dt

Rearranging and taking the limit as Dt®0 gives

Dlimt®0

Ejt‡Dt Ejt

Dt ˆ lim

Dt®0

EIIIjt‡Dt EIIIjt

Dt ‡ lim

Dt®0

EIIjt‡Dt EIjt

Dt (6-6)

Evaluating the limit of the left-hand side, we have

Dlimt®0

Ejt‡Dt Ejt

Dt ˆdE dt

which corresponds to the right-hand side of thefirst-law expression, equation (6-5).

On the right-hand side of equation (6-6) thefirst limit becomes

Dlimt®0

EIIIjt‡Dt EIIIjt

Dt ˆdE_III dt

which is the rate of change of the total energy of the system, as the volume occupied by the system as Dt®0 is the control volume under consideration.

The second limit on the right of equation (6-6)

Dlimt®0

EIIjt‡Dt EIjt

Dt

represents the net rate of energy leaving across the control surface in the time intervalDt.

I III II

Boundary of system at time t Stationary

control volume

Streamlines at time t

Boundary of system at time t + t

Figure 6.2 Relation between a system and a control volume in afluid-flowfield.

6.1 Integral Relation for the Conservation of Energy ◀ 67

Having given physical meaning to each of the terms in equation (6-6), we may now recast thefirst law of thermodynamics into a form applicable to a control volume expressed by the following word equation:

rate of addition of heat to control

volume from its surroundings 8>

><

>>

:

9>

>=

>>

;

rate of work done by control volume on its surroundings 8<

:

9=

;ˆ

rate of energy out of control volume due to fluid flow 8>

><

>>

:

9>

>=

>>

;

rate of energy into control volume due

to fluid flow 8<

:

9=

;‡

rate of accumulation of energy within

control volume 8<

:

9=

;

(6-7)

Equation (6-7) will now be applied to the general control volume shown in Figure 6.3.

The rates of heat addition to and work done by the control volume will be expressed as dQ/dtand dW/dt.

Consider now the small areadAon the control surface. The rate of energy leaving the control volume throughdA may be expressed as

rate of energy effluxˆe…ru†…dAcosq†

The product…ru†…dAcosq†is the rate of mass efflux from the control volume through dA, as discussed in the previous chapters. The quantityeis the specific energy or the energy per unit mass. The specific energy includes the potential energy,gy, due to the position of thefluid continuum in the gravitationalfield; the kinetic energy of thefluid,u²/2, due to its velocity; and the internal energy, u, of thefluid due to its thermal state.

The quantitydAcosqrepresents the area,dA, projected normal to the velocity vector,v.

Theta (q) is the angle betweenvand the outwardly directed normal vector,n. We may now write

e…ru†…dAcosq† ˆerdA‰jvjjnjŠcosqˆer…v

^n†dA

which we observe to be similar in form to the expressions previously obtained for mass and momentum. The integral of this quantity over the control surface

Z Z

c:s:er…v

^n†dA

represents the netefflux of energyfrom the control volume. The sign of the scalar product, v

n, accounts both for efflux and for influx of mass across the control surface as considered

v n dA

Streamlines at time t

Figure 6.3 Fluidflow through a control volume.

previously. Thus, the first two terms on the right-hand side of equation (6-7) may be evaluated as

rate of energy out of control

volume 8<

:

9=

;

rate of energy into control

volume 8<

:

9=

;ˆ Z Z

c:s:er…v

^n†dA

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

@

@t Z Z Z

c:v:erdV Equation (6-7) may now be written as

dQ dt

dW dt ˆ

Z Z

c:s:er…v

^n†dA‡_@^@_t^{Z Z Z}_c:v:^{erdV (6-8)}

Afinal form for thefirst-law expression may be obtained after further consideration of the work-rate or power term,dW/dt.

There are three types of work included in the work-rate term. Thefirst is the shaft work,W_s, which is that done by the control volume on its surroundings that could cause a shaft to rotate or accomplish the raising of a weight through a distance. A second kind of work done isflow work,W_s, which is that done on the surroundings to overcome normal stresses on the control surface where there is fluid flow. The third type of work is designated shear work,W_t, which is performed on the surroundings to overcome shear stresses at the control surface.

Examining our control volume for flow and shear work rates, we have, as shown in Figure 6.4, another effect on the elemental portion of control surface,dA. VectorSis the force intensity (stress) having componentssiiandtijin the directions normal and tangential to the surface, respectively. In terms ofS, the force ondAisSdA, and the rate of work done by the fluidflowing throughdAisSdA

^v.

The net rate of work done by the control volume on its surroundings due to the presence of Sis

Z Z

c:s:v

^SdA

where the negative sign arises from the fact that the force per unit areaon the surroundings is S.

dA Sv

n Figure 6.4 Flow and shear work for a general control volume.

6.1 Integral Relation for the Conservation of Energy ◀ 69

Thefirst-law expression, equation (6-8), may now be written as dQ

dt dWs

dt ‡ Z Z

c:s:v

^{SdAˆZ Z}_c:s:^er…v

^n†dA‡_@^@_t^{Z Z Z}_c:v:^{erdv (6-9)}

wheredWs=dtis the shaft work rate.

Writing the normal stress components ofSassiin, we obtain, for the net rate of work done in overcoming normal stress,

Z Z

c:s:v

^SdA

normal

ˆ Z Z

c:s:v

^{siindAˆZ Z}

c:s:sii…v

^n†dA

The remaining part of the work to be evaluated is the part necessary to overcome shearing stresses. This portion of the required work rate,dW_t=dt, is transformed into a form that is unavailable to do mechanical work. This term, representing a loss of mechanical energy, is included in the derivative form given above, and its analysis is included in Example 3, to follow. The work rate now becomes

dW dt ˆdW_s

dt ‡dW_s dt ‡dW_t

dt ˆdW_s dt

Z Z

c:s:sii…v

^n†dA‡dW_dt^t

Substituting into equation (6-9), we have dQ

dt dW_s

dt ‡ Z Z

c:s:sii…v

^n†dA‡dW_dt^{t ˆZ Z}_c:s:^er…v

^n†dA‡_@^@_t^{Z Z Z}_c:v:^erdv

The term involving normal stress must now be presented in a more usable form. A complete expression forsiiis stated in Chapter 9. For the present, we may say simply that the normal stress term is the sum of pressure effects and viscous effects. Just as with shear work, the work done to overcome the viscous portion of the normal stress is unavailable to do mechanical work. We shall thus combine the work associated with the viscous portion of the normal stress with the shear work to give a single term, dW_m=dt, the work rate accomplished in overcoming viscous effects at the control surface. The subscript,m, is used to make this distinction.

The remaining part of the normal stress term that associated with pressure may be written in slightly different form if we recall that the bulk stress,sii, is the negative of the thermodynamic pressure,P. The shear andflow work terms may now be written as follows:

Z Z

c:s:sii…v

^{n†dA dW}_dt^{tˆ Z Z}_c:s:^P…v

^{n†dA dW}_dt^m

Combining this equation with the one written previously and rearranging slightly will yield thefinal form of thefirst-law expression:

dQ dt

dWs

dt ˆ Z Z

c:s:

e‡P r

r…v

^n†dA‡_@^@_t^{Z Z Z}_c

:v:

erdv‡dW_m

dt (6-10)

Equations (6-10), (4-1), and (5-4) constitute the basic relations for the analysis offluid flow via the control-volume approach. A thorough understanding of these three equations and a mastery of their application places at the disposal of the student very powerful means of analyzing many commonly encountered problems influidflow.

The use of the overall energy balance will be illustrated in the following example problems.

▶ 6.2

APPLICATIONS OF THE INTEGRAL EXPRESSION

Example 1

As afirst example, let us choose a control volume as shown in Figure 6.5 under the conditions of steadyfluidflow and no frictional losses.

For the specified conditions the overall energy expression, equation (6-10), becomes dQ

dt dWS

dt ˆ Z Z

c:s:r e‡P r

v

ⁿ

… †dA‡@

@t Z Z Z

c:v: erdV

0 steady flow

‡dW_m dt

Considering now the surface integral, we recognize the productr(v

^n)dAto be the massflow rate with the sign of this product indicating whether massflow is into or out of the control volume, dependent upon the sense of (v

n). The factor by which the mass-flow rate is multiplied,e‡P/r, represents the types of energy that may enter or leave the control volume per mass offluid.

The specific total energy,e, may be expanded to include the kinetic, potential, and internal energy contributions, so that e‡P

rˆgy‡v² 2 ‡u‡P

r

As mass enters the control volume only at section (1) and leaves at section (2), the surface integral becomes Z Z

c:s:r e‡P r

…v

^{n†dAˆ v}₂^{22‡gy2‡u2‡P}_r²

2

…r2v2A2† v²1

2‡gy1‡u1‡P1

r1

…r1v1A1†

The energy expression for this example now becomes dQ

dt dWs

dt ˆ v²2

2‡gy2‡u2‡P2

r2

…r2v2A2† v²1

2‡gy1‡u1‡P1

r1

…r1v1A1†

In Chapter 4, the mass balance for this same situation was found to be _

mˆr1v1A1ˆr1v2A2

If each term in the above expression is now divided by the massflow rate, we have q W_s

_

m ˆ v²2

2‡gy2‡u2‡P2

r2

v²1

2 ‡gy1‡u1‡P1

r1

!

0

!

1

2

dt

dt

Figure 6.5 Control volume with one-dimensionalflow across boundaries.

6.2 Applications of the Integral Expression ◀ 71

or, in more familiar form

v²1

2‡gy₁‡h1‡q _ mˆv²2

2 ‡gy₂‡h2‡W_s

_ m

where the sum of the internal energy andflow energy,u‡P/r, has been replaced by the enthalpy,h, which is equal to the sum of these quantities by definitionhu‡P=r.

Example 2

As a second example, consider the situation shown in Figure 6.6. If waterflows under steady conditions in which the pump delivers 3 horsepower to thefluid,find the massflow rate if frictional losses may be neglected.

Defining the control volume as shown by the dashed lines, we may evaluate equation (6-10) term by term as follows:

dQ dt ˆ0 dWs

dt ˆ …3hp†…2545 Btu=hp h†…778 ft lbf=Btu†…h=3600 s†

ˆ1650 ft lbf=s Z Z

c:s: e‡P r

r…v

^{n†dAˆZ Z}_A

2

e‡P r

r…v

^{n†dA Z Z}_A

1

e‡P r

r…v

^n†dA

ˆ n²

2 ‡gy₂‡u2‡P2

r2

…r2n2A2† n²1

2‡gy₁‡u1‡P1

r1

…r1n1A1†

ˆ n²2 n²1

2 ‡g…y2 y1† ‡ …u2 u1† ‡P2

r2

P1

r1

…rnA†

Here it may be noted that the pressure measured at station (1) is the static pressure while the pressure measured at station (2) is measured using a pressure port that is oriented normal to the oncomingflow—that is, where the velocity has been reduced to zero. Such a pressure is designated thestagnation pressure, which is greater than the static pressure by an amount equivalent to the change in kinetic energy of theflow. The stagnation pressure is thus expressed as

PstagnationˆP0ˆPstatic‡1 2rv²

12 in. Pump

Shaft work

6 in.

1

2

6 in.

Hg

Figure 6.6 A control volume for pump analysis.

for incompressibleflow; hence the energyflux term may be rewritten as Z Z

c:s: e‡P r

r…v

^{n†dA ˆP02}_r ^{P1 v}₂^21…rvA†

ˆ

(6…1 1=13:6†in:Hg…14:7 lb=in:²†…144 in:²=ft²† …62:4 lbm=ft³†…29:92 in:Hg†

v²1

64:4…lbmft=s²lbf† )

f…62:4 lbm=ft³†…v1†…p=4 ft²†g

ˆ 6:30 v²1

64:4

49v1

… †ft lbf=s

@

@t Z Z Z

c:v:erdvˆ0 dW_m

dt ˆ0

In the evaluation of the surface integral, the choice of the control volume coincided with the location of the pressure taps at sections (1) and (2). The pressure sensed at section (1) is the static pressure, as the manometer opening is parallel to thefluid-flow direction. At section (2), however, the manometer opening is normal to theflowingfluid stream. The pressure measured by such an arrangement includes both the staticfluid pressure and the pressure resulting as afluidflowing with velocityn2is brought to rest. The sum of these two quantities is known as the impact or stagnation pressure.

The potential energy change is zero between sections (1) and (2) and as we consider theflow to be isothermal, the variation in internal energy is also zero. Hence, the surface integral reduces to the simple form indicated.

Theflow rate of water necessary for the stated conditions to exist is achieved by solving the resulting cubic equation. The solution is

v1ˆ16:59 ft=s…5:057 m=s† m_ ˆrAvˆ813 lbm=s…370 kg=s†

Example 3

A shaft is rotating at constant angular velocitywin the bearing shown in Figure 6.7.

The shaft diameter isdand the shear stress acting on the shaft ist. Find the rate at which energy must be removed from the bearing in order for the lubricating oil between the rotating shaft and the stationary bearing surface to remain at constant temperature.

The shaft is assumed to be lightly loaded and concentric with the journal (the part of the shaft in contact with the bearing). The control volume selected consists of a unit length of thefluid surrounding the shaft as shown in Figure 6.7. Thefirst law of thermodynamics for the control volume is

dQ dt

dWs

dt ˆ Z Z

c:s:r e‡P r

…v

^n†dA

‡@

@t Z Z Z

c:v:redv‡dW_m dt

D d

dt

Figure 6.7 Bearing and control volume for bearing analysis.

6.2 Applications of the Integral Expression ◀ 73

From thefigure we may observe the following:

1. Nofluid crosses the control surface.

2. No shaft work crosses the control surface.

3. Theflow is steady.

ThusdQ/dtˆdW_m=dtˆdW_t=dt. The viscous work rate must be determined. In this case all of the viscous work is done to overcome shearing stresses; thus, the viscous work isR R

c.s.t(v

^et)dA. At the outer boundary,vˆ0, and at the inner boundary,R R

c.s.t(v

^{et)dAˆ t(w}d/2)A, wheree_tindicates the sense of the shear stress,t, on the surroundings. The resulting sign is consistent with the concept of work being positive when done by a system on its surroundings. Thus,

dQ

dt ˆ twd²p 2

which is the heat transfer rate required to maintain the oil at a constant temperature.

If energy is not removed from the system, thendQ/dtˆ0, and

@

@t Z Z Z

c:v:erdvˆ dW_m dt As only the internal energy of the oil will increase with respect to time,

@

@t Z Z Z

c:v:erdvˆrp D² d² 4

dm dt

dW_m dt ˆwd²p

2 t or, with constant specific heatc

cdT

dt ˆ 2twd² r…D² d²† whereDis the outer bearing diameter.

In this example the use of the viscous-work term has been illustrated. Note that

1. The viscous-work term involves only quantities on the surface of the control volume.

2. When the velocity on the surface of the control volume is zero, the viscous-work term is zero.

▶ 6.3

THE BERNOULLI EQUATION

Under certainflow conditions, the expression of thefirst law of thermodynamics applied to a control volume reduces to an extremely useful relation known as the Bernoulli equation.

If equation (6-10) is applied to a control volume as shown in Figure 6.8, in whichflow is steady, incompressible, and inviscid, and in which no heat transfer or change in internal energy occurs, a term-by-term evaluation of equation (6-10) gives the following:

dQ dt ˆ 0 dWs

dt ˆ 0

Z Z

c:s:r e‡P r

…v

^{n†dAˆZ Z}_A

1

r e‡P r

…v

^n†dA

‡ Z Z

A₂r e‡P r

…v

^n†dA

ˆ gy₁‡v²1

2 ‡P1

r1

… r1v1A1†

‡ gy₂‡v²₂ 2 ‡P₂

r2

… r2v2A2†

@

@t Z Z Z

c:v:

erdvˆ0 The first-law expression now becomes

0ˆ gy₂‡v²₂ 2 ‡P₂

r

…rv2A2† gy₁‡v²₁ 2 ‡P₁

r1

…rv1A1† As flow is steady, the continuity equation gives

r1n1A1ˆr2n2A2

which may be divided through to give gy₁‡v²1

2 ‡P1

r ˆgy₂‡v²2

2 ‡P2

r (6-11a)

Dividing through by g, we have y₁‡v²₁

2g‡P₁

rgˆy₂‡v²₂ 2g‡P₂

rg (6-11b)

Either of the above expressions is designated the Bernoulli equation.

Note that each term in equation (6-11b) has the unit of length. The quantities are often designated“heads”due to elevation, velocity, and pressure, respectively. These terms, both individually and collectively, indicate the quantities that may be directly converted to produce mechanical energy.

Equation (6-11) may be interpreted physically to mean that the total mechanical energy is conserved for a control volume satisfying the conditions upon which this relation is based—that is, steady, incompressible, inviscid, isothermalflow, with no heat transfer or work done. These conditions may seem overly restrictive, but they are met, or approached,

2

1 Streamlines

A1

A₂

Control volume

Figure 6.8 Control volume for steady, incompressible, inviscid, isothermalflow.

6.3 The Bernoulli Equation ◀ 75