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How does Law of Conservation of Energy work

how to prove law of conservation of energy and how are the law of conservation of energy and mass similar and conservation of energy newton's second law
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Chapter6 Conservation of Energy: Control-Volume Approach The third fundamental law to be applied to fluid-flow analyses is the first law of thermodynamics. An integral expression for the conservation of energy applied to a control volume will be developed from the first 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. Notethatthislawiswrittenforaspecificgroupofparticles—thosecomprisingthe 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 which contains different fluid particles at different times. The statement of the first law of thermo- dynamics involves only scalar quantities however, and thus, unlike the momentum equationsconsideredinChapter5,theequationsresultingfromthefirstlawofthermo- dynamics will be scalar in form. The statement of the first law given above may be written in equation form as 1 %dQ¼ %dW (6-1) J where the symbol% refers to a ‘‘cyclic integral’’ or the integral of the quantity evalu- ated 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 1 the property occurs between the initial and final states. The quantity J is the so-called ‘‘mechanical equivalent of heat,’’ numerically equal to 778.17 ft lb/Btu in engineering units. In the SI system, J¼ 1 N m/J. This factor will not be written henceforth, and the student is reminded that all equations must be dimensionally homogeneous. 1 For 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. 6364 Chapter 6 Conservation of Energy: Control-Volume Approach We now consider a general thermo- 2 dynamic cycle, as shown in Figure 6.1. a The cycle a occurs between points 1 and 2 by the paths indicated. Utilizing equation a (6-1), we may write, for cycle a P b Z Z 2 1 dQþ dQ 1a 2a 1 (6-2a) Z Z 2 1 ¼ dWþ dW r 1a 2a Figure 6.1 Reversible and irreversible A new cycle between points 1 and 2 is thermodynamic cycles. postulated as follows: the path between points 1 and 2 is identical to that considered previously; however, the cycle is completed by path b between points 2 and 1, which is any path other than a between these points. Again equation (6-1) allows us to write Z Z Z Z 2 1 2 1 dQþ dQ¼ dWþ dW (6-2b) 1a 2b 1a 2b Subtracting equation (6-2b) from equation (6-2a) gives Z Z Z Z 1 1 1 1 dQ dQ¼ dW dW 2a 2b 2a 2b which may be written Z Z 1 1 ðdQdWÞ¼ ðdQdWÞ (6-3) 2a 2b 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 designated dE, the total energy of the system. An alternate expression for the first law of thermodynamics may be written dQdW ¼ dE (6-4) ThesignsofdQanddWwerespecifiedintheoriginalstatementofthefirstlaw;dQ is positive when heat is added to the system, dW is positive when work is done by the system. Forasystemundergoingaprocessoccurringintimeintervaldt,equation(6-4)maybe written as dQ dW dE  ¼ (6-5) dt dt dt Considernow,asinChapter5,ageneralcontrolvolumefixedininertialspacelocatedin a fluid-flow field, as shown in Figure 6.2. The system under consideration, designated by dashed lines, occupies the control volume at time t, and its position is also shown after a period of timeDt has elapsed.6.1 Integral Relation for the Conservation of Energy 65 Boundary of system at time t Streamlines Stationary at time t control volume II III I Figure 6.2 Relation between a system and a Boundary of system control volume in a fluid- at time t + ∆ t flow field. In this figure, region I is occupied by the system at time t, region II is occupied by the system at t þ Dt, and region III is common to the system both at t and at t þ Dt. At time tþDt the total energy of the system may be expressed as Ej ¼ E j þE j II III tþDt tþDt tþDt and at time t Ej ¼ Ej þE j I III t t t Subtractingthesecondexpressionfromthefirstanddividingbytheelapsedtimeinterval, Dt, we have Ej Ej E j þE j E j Ej III II III I tþDt t tþDt tþDt t t ¼ Dt Dt Rearranging and taking the limit asDt 0 gives Ej Ej E j E j E j Ej III III II I tþDt t tþDt t tþDt t lim ¼ lim þ lim (6-6) Dt Dt Dt Dt0 Dt0 Dt0 Evaluating the limit of the left-hand side, we have Ej Ej dE tþDt t lim ¼ Dt0 Dt dt which corresponds to the right-hand side of the first-law expression, equation (6-5). On the right-hand side of equation (6-6) the first limit becomes E j E j dE III III III tþDt t lim ¼ Dt0 Dt dt whichistherateofchangeofthetotalenergyofthesystem,asthevolumeoccupiedbythe system asDt 0 is the control volume under consideration. The second limit on the right of equation (6-6) E j Ej II I tþDt t lim Dt Dt0 represents the net rate of energy leaving across the control surface in the time inter- val Dt.66 Chapter 6 Conservation of Energy: Control-Volume Approach Having given physical meaning to each of the terms in equation (6-6), we may now recastthefirstlawofthermodynamicsintoaformapplicabletoacontrolvolumeexpressed by the following word equation: 8 9 8 9 8 9 rate of addition rate of energy rate of work done = = = of heat to control out of control  by control volume ¼ volume from volume due to : ; on its surroundings : ; : ; its surroundings fluid flow (6-7) 8 9 8 9 rate of energy into= rate of accumulation=  control volume due þ of energy within : ; : ; to fluid flow control volume Equation(6-7)willnowbeappliedtothegeneralcontrolvolumeshowninFigure6.3. Streamlines at time t dA q v n Figure 6.3 Fluid flow through a control volume. The rates of heat addition to and work done by the control volumewill be expressed as dQ/dt and dW/dt. Consider now the small area dA on the control surface. The rate of energy leaving the control volume through dA may be expressed as rate of energy efflux¼ e(rv)(dAcosu) The product (rv)(dA cos u) is the rate of mass efflux from the control volume through dA, as discussed in the previous chapters. The quantity e is the specific energy or the energy per unit mass. The specific energy includes the potential energy, gy, due tothepositionofthefluidcontinuuminthegravitationalfield;thekineticenergyofthe 2 fluid, v /2, due to its velocity; and the internal energy, u, of the fluid due to its thermal state. ThequantitydAcosurepresentsthearea,dA,projectednormaltothevelocityvector,v. Theta(u)istheanglebetweenvandtheoutwardlydirectednormalvector,n.Wemaynow write : e(rv)(dAcosu)¼ erdA½jvjjnj cosu¼ er(v n)dA whichweobservetobesimilarinformtotheexpressionspreviouslyobtainedformassand momentum. The integral of this quantity over the control surface ZZ : er(v n)dA c:s: representstheneteffluxofenergyfromthecontrolvolume.Thesignofthescalarproduct, : v n,accountsbothforeffluxandforinfluxofmassacrossthecontrolsurfaceasconsidered6.1 Integral Relation for the Conservation of Energy 67 previously. Thus, the first two terms on the right-hand side of equation (6-7) may be evaluated as 8 9 8 9 ZZ rate of energy= rate of energy= : out of control  into control ¼ er(v n)dA : ; : ; c:s: volume volume The rate of accumulation of energy within the control volume may be expressed as ZZZ erdV t c:v: Equation (6-7) may now be written as ZZ ZZZ dQ dW :  ¼ er(v n)dAþ erdV (6-8) dt dt t c:s: c:v: Afinalformforthefirst-lawexpressionmaybeobtainedafterfurtherconsiderationof the work-rate or power term, dW/dt. There are three types of work included in the work-rate term. The first is the shaft work, W ,whichisthatdonebythecontrolvolumeonitssurroundingsthat couldcause s a shaft torotate or accomplishtheraisingofaweightthrougha distance.A secondkind ofworkdoneisflowwork,W ,whichisthatdoneonthesurroundingstoovercomenormal s stresses on the control surface where there is fluid flow. The third type of work is designated shear work, W , which is performed on the surroundings to overcome shear t 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. Vector S is the forceintensity(stress)havingcomponentss andt inthedirectionsnormalandtangential ii ij tothesurface,respectively.IntermsofS,theforceondAisSdA,andtherateofworkdone : by the fluid flowing through dA is S dA v. S dA v q n Figure 6.4 Flow and shear work for a general control volume. Thenetrateofworkdonebythecontrolvolumeonitssurroundingsduetothepresence of S is ZZ :  v SdA c:s: wherethenegativesignarisesfromthefactthattheforceperunitareaonthesurroundings is –S.68 Chapter 6 Conservation of Energy: Control-Volume Approach The first-law expression, equation (6-8), may now be written as ZZ ZZ ZZZ dQ dW s : :  þ v SdA¼ er(v n)dAþ erdV (6-9) dt dt t c:s: c:s: c:v: where dW /dt is the shaft work rate. s WritingthenormalstresscomponentsofSass n,weobtain,forthenetrateofwork ii done in overcoming normal stress  ZZ ZZ ZZ : : : v SdA ¼ v s ndA¼ s ðv nÞdA ii ii c:s: c:s: c:s: normal 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 /dt, is transformed into t 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 ZZ dW dW dW dW dW dW s s t s t : ¼ þ þ ¼  s (v n)dAþ ii dt dt dt dt dt dt c:s: Substituting into equation (6-9), we have ZZ ZZ ZZZ dQ dW dW s t : :  þ s ðv nÞdAþ ¼ erðv nÞdAþ erdV ii dt dt dt t c:s: c:s: c:v: The term involving normal stress must now be presented in a more usable form. A completeexpressionfors isstatedinChapter9.Forthepresent,wemaysaysimplythat ii thenormalstresstermisthesumofpressureeffectsandviscouseffects.Justaswithshear work, the work done to overcome the viscous portion of the normal stress is unavailable todomechanicalwork.Weshallthuscombinetheworkassociatedwiththeviscousportion of the normal stress with the shear work to give a single term, dW /dt, the work rate m accomplishedinovercomingviscouseffectsatthecontrolsurface.Thesubscript,m,isused to make this distinction. Theremainingpartofthenormalstressterm,thatassociatedwithpressure,maybe writteninslightlydifferentformifwerecallthatthebulkstress,s ,isthenegativeofthe ii thermodynamic pressure, P. The shear and flow work terms may now be written as follows: ZZ ZZ dW dW t m : : s (v n)dA ¼ P(v n)dA ii dt dt c:s: c:s: Combiningthisequationwiththeonewrittenpreviouslyandrearrangingslightlywillyield the final form of the first-law expression: ZZ ZZZ dQ dW P dW s m :  ¼ eþ r(v n)dAþ erdVþ (6-10) dt dt r t dt c:s: c:v: Equations (6-10), (4-1), and (5-4) constitute the basic relations for the analysis of fluid 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 in fluid flow. The use of the overall energy balance will be illustrated in the following example problems. 6.2 Applications of the Integral Expression 69 6.2 APPLICATIONS OF THE INTEGRAL EXPRESSION EXAMPLE 1 As a first example, let us choose a control volume as shown in Figure 6.5 under the conditions of steady fluid flow and no frictional losses. For the specified conditions the overall energy expression, equation (6-10), becomes  ZZ ZZZ dQ dW P dW s m :  ¼ r eþ (v n)dAþ erdV þ dt dt r t dt c:s: c:v: 0 0steady flow dQ 1 dt 2 υ , A , r 1 1 1 Figure 6.5 Control volume with υ , A , r one-dimensional flow across 2 2 2 dW s boundaries. dt : Consideringnowthesurfaceintegral,werecognizetheproductr(v n)dAtobethemassflow ratewiththesignofthisproductindicatingwhethermassflowisintooroutofthecontrolvolume, : dependent upon the sense of v n. The factor by which the mass-flow rate is multiplied, eþ P/r, represents the types of energy that may enter or leave the control volume per mass of fluid. The specific total energy, e, may be expanded to include the kinetic, potential, and internal energy contributions, so that 2 P v P eþ ¼ gyþ þuþ r 2 r As mass enters the control volume only at section (1) and leaves at section (2), the surface integral becomes ZZ   2 P v P 2 2 : r eþ (v n)dA¼ þgy þu þ (r v A ) 2 2 2 2 2 r 2 r c:s: 2  2 v P 1 1  þgy þu þ (r v A ) 1 1 1 1 1 2 r 1 The energy expression for this example now becomes   2 2 v v dQ dW P P s 2 1 2 1  ¼ þgy þu þ (r v A ) þgy þu þ (r v A ) 2 2 2 2 1 1 1 1 2 1 dt dt 2 r 2 r 2 1 In Chapter 4, the mass balance for this same situation was found to be m _ ¼ r v A ¼ r v A 1 1 2 2 1 1 If each term in the above expression is now divided by the mass flow rate, we have  2 2 _ qW v P v P s 2 1 2 1 ¼ þgy þu þ  þgy þu þ 2 2 1 1 m _ 2 r 2 r 2 170 Chapter 6 Conservation of Energy: Control-Volume Approach or, in more familiar form 2 2 _ v q v W s 1 2 þgy þh þ ¼ þgy þh þ 1 1 2 2 2 m _ 2 m _ wherethesumoftheinternalenergyandflowenergy,uþP/r,hasbeenreplacedbytheenthalpy,h, which is equal to the sum of these quantities by definition h uþ P/r. EXAMPLE 2 As a second example, consider the situation shown in Figure 6.6. If water flows under steady conditionsinwhichthepumpdelivers3horsepowertothefluid,findthemassflowrateiffrictional losses may be neglected. Shaft work 2 12 in. 1 Pump 6 in. 6 in. Hg Figure 6.6 A control volume for pump analysis. Defining the control volume as shown by the dashed lines, we may evaluate equation (6-10) term by term as follows: dQ ¼ 0 dt dW s  ¼ (3hp)(2545Btu/hph)(778ftlb /Btu)(h/3600s) f dt ¼ 1650ftlb /s f ZZ ZZ ZZ  P P P : : : eþ r(v n)dA ¼ eþ r(v n)dAþ eþ r(v n)dA r r r c:s: A A 2 1  2 v P 2 2 ¼ þgy þu þ (r v A ) 2 2 2 2 2 2 r 2  2 v P 1 1  þgy þu þ (r v A ) 1 1 1 1 1 2 r 1   2 2 v v P P 2 1 2 1 ¼ þg(y y )þ(u u )þ  (rvA) 2 1 2 1 2 r r 2 1 Here it may be noted that the pressure measured at station (1) is the static pressurewhile the pressure measured at station (2) is measured using a pressure port that is oriented normal to theoncomingflow,thatis,wherethevelocityhasbeenreducedtozero.Suchapressureisdesignated the stagnation pressure, which is greater than the static pressure by an amount equivalent to the change in kinetic energy of the flow. The stagnation pressure is, thus, expressed as 1 2 P ¼ P ¼ P þ rv stagnation 0 static 26.2 Applications of the Integral Expression 71 for incompressible flow, hence the energy flux term may be rewritten as ZZ  2 P P P v 0 1 2 1 : eþ r(v n)dA¼  (rvA) r r 2 c:s: ( 2 2 2 6(11/13:6)in:Hg(14:7lb/in: )(144in: /ft ) ¼ 3 (62:4lb /ft )(29:92in:Hg) m ) 2 v 1 3 2  f(62:4lb /ft )(v )(p/4ft )g m 1 2 64:4(lb ft/s lb ) m f  2 v 1 ¼ 6:30  (49v )ftlb /s 1 f 64:4 ZZZ erdV ¼ 0 t c:v: dW m ¼ 0 dt Intheevaluationofthe surfaceintegralthechoiceofthecontrolvolumecoincidedwiththe locationofthepressuretapsatsections(1)and(2).Thepressuresensedatsection(1)isthestatic pressure,asthemanometeropeningisparalleltothefluid-flowdirection.Atsection(2),however, themanometeropening isnormal totheflowingfluidstream.Thepressure measuredby such an arrangementincludesboththestaticfluidpressureandthepressureresultingasafluidflowingwith velocityv isbroughttorest.Thesumofthesetwoquantitiesisknownastheimpactorstagnation 2 pressure. Thepotentialenergychangeiszerobetweensections(1)and(2)andasweconsidertheflowto beisothermal,thevariationininternalenergyisalsozero.Hence,thesurfaceintegralreducestothe simple form indicated. The flow rate of water necessary for the stated conditions to exist is achieved by solving the resulting cubic equation. The solution is v ¼ 16:59fps(5:057m/s) 1 m _ ¼ rAv¼ 813lb /s(370kg/s) m EXAMPLE 3 A shaft is rotating at constant angular velocity v in the bearing shown in Figure 6.7. The shaft diameter is d and the shear stress acting on the shaft is t. Find the rate at which energy must be removed from the bearing in order that the lubricating D oil between the rotating shaft and the stationary bearing d surface remains at constant temperature. The shaft is assumed to be lightly loaded and concentric with the journal.The control volume selected consists of dQ a unit length of the fluid surrounding the shaft as shown w dt in Figure 6.7. The first law of thermodynamics for the control volume is ZZ  dQ dW P s :  ¼ r eþ (v n)dA dt dt r c:s: ZZZ dW m þ redVþ Figure 6.7 Bearing and control t dt c:v: volume for bearing analysis.72 Chapter 6 Conservation of Energy: Control-Volume Approach From the figure we may observe the following: 1. No fluid crosses the control surface. 2. No shaft work crosses the control surface. 3. The flow is steady. Thus dQ/dt = dW /dt = dW /dt. The viscous work rate must be determined. In this case all of the m t RR : viscousworkisdonetoovercomeshearingstresses;thustheviscousworkis t(v e )dA.Atthe t c:s: RR : outer boundary, v¼ 0 and at the inner boundary t(v e )dA¼t(vd/2)A, where e indicates t t c:s: thesenseoftheshearstress,t,onthesurroundings.Theresultingsignisconsistentwiththeconcept of work being positive when done by a system on its surroundings. Thus, 2 dQ vd p ¼t dt 2 which is the heat transfer rate required to maintain the oil at a constant temperature. If energy is not removed from the system then dQ/dt¼ 0, and ZZZ dW m erdV¼ t dt c:v: As only the internal energy of the oil will increase with respect to time  ZZZ 2 2 2 D d dm dW d p m erdV ¼ rp ¼ ¼ v t t 4 dt dt 2 c:v: or, with constant specific heat c 2 dT 2tvd c ¼ 2 2 dt r(D d ) where D is 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. Whenthevelocityonthesurfaceofthecontrolvolumeiszero,theviscous-worktermiszero. 6.3 THE BERNOULLI EQUATION Undercertainflowconditions,theexpressionofthefirstlawofthermodynamicsapplied to a control volume reduces to an extremely useful relation known as the Bernoulli equation. Ifequation(6-10)isappliedtoacontrolvolumeasshowninFigure6.8,inwhichflow issteady,incompressible,andinviscid,andinwhichnoheattransferorchangeininternal energy occurs, a term-by-term evaluation of equation (6-10) gives the following: dQ ¼ 0 dt dW s ¼ 0 dt6.3 The Bernoulli Equation 73 A 2 Streamlines 2 Control volume 1 A 1 Figure 6.8 Control volume for steady, incompressible, inviscid, isothermal flow.   ZZ ZZ P P : : r eþ (v n)dA¼ r eþ (v n)dA r r c:s: A 1 ZZ  P : þ r eþ (v n)dA r A 2  2 v P 1 1 ¼ gy þ þ (r v A ) 1 1 1 1 2 r 1  2 v P 2 2 þ gy þ þ (r v A ) 2 2 2 2 2 r 2 ZZZ erdV ¼ 0 t c:v: The first-law expression now becomes   2 2 v P v P 2 1 2 1 0¼ gy þ þ (rv A ) gy þ þ (rv A ) 2 2 2 1 1 1 2 r 2 r As flow is steady, the continuity equation gives r v A ¼ r v A 1 1 2 2 1 2 which may be divided through to give 2 2 v v P P 1 2 1 2 gy þ þ ¼ gy þ þ (6-11a) 1 2 2 r 2 r Dividing through by g, we have 2 2 v P v P 1 2 1 2 y þ þ ¼ y þ þ (6-11b) 1 2 2g rg 2g rg Either of the above expressions is designated the Bernoulli equation. Notethateachterminequation(6-11b)hastheunitoflength.Thequantitiesareoften designated‘‘heads’’duetoelevation,velocity,andpressure,respectively.Theseterms,both individually and collectively, indicate the quantities which may be directly converted to produce mechanical energy. Equation(6-11)maybeinterpretedphysicallytomeanthatthetotalmechanicalenergy isconservedforacontrolvolumesatisfyingtheconditionsuponwhichthisrelationisbased, thatis,steady,incompressible,inviscid,isothermalflow,withnoheattransferorworkdone. These conditions may seem overly restrictive, but they are met, or approached, in many physicalsystems.Onesuchsituationofpracticalvalueisforflowintoandoutofastream74 Chapter 6 Conservation of Energy: Control-Volume Approach tube. As stream tubes may vary in size, the Bernoulli equation can actually describe the variationinelevation,velocity,andpressureheadfrompoint-to-pointinafluid-flowfield. AclassicexampleoftheapplicationoftheBernoulliequationisdepictedinFigure6.9, in which it is desired to find the velocity of the fluid exiting the tank as shown. Constant level 1 maintained in tank y 1 2 Figure 6.9 Control volume for Bernoulli u 2 equation analysis. The control volume is defined as shown by dashed lines in the figure. The upper boundaryofthecontrolvolumeisjustbelowthefluidsurface,andthuscanbeconsidered tobeatthesameheightasthefluid.Thereisfluidflowacrossthissurface,butthesurface area is large enough that the velocity of this flowing fluid may be considered negligible. Under these conditions, the proper form of the first law of thermodynamics is equation (6-11), the Bernoulli equation. Applying equation (6-11), we have 2 v P P atm atm 2 y þ ¼ þ 1 rg 2g rg from which the exiting velocity may be expressed in the familiar form pffiffiffiffiffiffiffi v ¼ 2gy 2 Asa final illustration of the useof the control-volumerelations,an exampleusingall three expressions is presented below. EXAMPLE 4 In the sudden enlargement shown below in Figure 6.10, the pressure acting at section (1) is considereduniformwithvalueP .Findthechangeininternalenergybetweenstations(1)and(2)for 1 steady,incompressibleflow.Neglectshearstressatthewallsandexpressu u intermsofv ,A , 2 1 1 1 and A . The control volume selected is indicated by the dotted line. 2 P = P d 1 u , u 1 1 A 2 A , r 1 1 Figure 6.10 Flow through a 1 2 sudden enlargement. Conservation of Mass ZZ ZZZ : r(v n)dAþ rdV ¼ 0 t c:s: c:v:6.3 The Bernoulli Equation 75 Ifweselectstation(2),aconsiderabledistancedownstreamfromthesuddenenlargement,the continuity expression, for steady, incompressible flow, becomes r v A ¼ r v A 1 1 2 2 1 2 or A 1 v ¼ v (6-12) 2 1 A 2 Momentum ZZ ZZZ : F¼ rv(v n)dAþ rvdV t c:s: c:v: and thus 2 2 P A P A ¼ rv A rv A 1 2 2 2 2 1 2 1 or  P P A 1 2 1 2 2 ¼ v v (6-13) 2 1 r A 2 Energy  ZZ ZZZ dQ dW P dW s m :  ¼ r eþ (v n)dAþ redVþ t dt r t dt c:s: c:v: Thus,   P P 1 2 e þ (rv A )¼ e þ (rv A ) 1 1 1 2 2 2 r r or, since rv A ¼ rv A , 1 1 2 2 P P 1 2 e þ ¼ e þ 1 2 r r The specific energy is 2 v e¼ þgyþu 2 Thus our energy expression becomes 2 2 v P v P 1 2 1 2 þgy þu þ ¼ þgy þu þ (6-14) 1 1 2 2 2 r 2 r Thethreecontrol-volumeexpressionsmaynowbecombinedtoevaluateu u .Fromequation 2 1 (6-14), we have 2 2 v v P P 1 2 1 2 u u ¼ þ þg(y y ) (6-14a) 2 1 1 2 r 2 Substitutingequation(6-13)for(P P )/randequation(6-12)forv andnotingthaty ¼ y , 1 2 2 1 2 we obtain   2 2 2 2 A A v v A 2 1 2 1 1 1 1 u u ¼ v v þ  2 1 1 1 A A 2 2 A 2 2 2 "   2 2 2 2 v A A v A 1 1 1 1 1 ¼ 12 þ ¼ 1 (6-15) 2 A A 2 A 2 2 276 Chapter 6 Conservation of Energy: Control-Volume Approach Equation (6-15) shows that the internal energy increases in a sudden enlargement. The temperature change corresponding to this change in internal energy is insignificant, but from equation (6-14a) it can be seen that the change in total head,  2 2 P v P v 1 2 1 2 þ þgy  þ þgy 1 2 r 2 r 2 isequaltotheinternalenergychange.Accordingly,theinternalenergychangeinanincompressible flowisdesignatedastheheadloss,h ,andtheenergyequationforsteady,adiabatic,incompressible L flow in a stream tube is written as 2 2 P v P v 1 2 1 2 þ þy ¼ h þ þ þy (6-16) 1 L 2 rg 2g rg 2g Note the similarity to equation (6-11). 6.4 CLOSURE In this chapter the first law of thermodynamics, the third of the fundamental relations uponwhich fluid-flow analyses arebased, hasbeen used to developan integral expression for the conservation of energy with respect to a control volume. The resulting expression, equation (6-10), is,in conjunction with equations (4–1) and (5–4),oneof thefundamental expressions for the control-volume analysis of fluid-flow problems. AspecialcaseoftheintegralexpressionfortheconservationofenergyistheBernoulli equation, equation (6-11). Although simple in form and use, this expression has broad application to physical situations. PROBLEMS 3 6.1 The velocity profile in the annular control volume of 6.5 Duringtheflowof200ft /sofwaterthroughthehydraulic Example 3 is approximately linear, varying from a velocity of turbineshown,thepressureindicatedbygageAis12psig.What zero at the outer boundary to a value of vd/2 at the inner should gage B read if the turbine is delivering 600 hp at 82% boundary. Develop an expression for the fluid velocity, v(r), efficiency?GageBisdesignedtomeasurethetotalpressure,that 2 where r is the distance from the center of the shaft. is, P + rv /2 for an incompressible fluid. 3 6.2 Sea water, r¼ 1025kg/m , flows through a pump at A 3 0:21m /s. The pump inlet is 0.25 m in diameter. At the inlet Turbine thepressureis–0.15mofmercury.Thepumpoutlet,0.152min diameter,is1.8mabovetheinlet.Theoutletpressureis175kPa. D = 2 ft 15 ft Iftheinletandexittemperatureareequal,howmuchpowerdoes the pump add to the fluid? B 3 6.3 Air at 708F, Flows into a 10-ft reservoir at a velocity of 110 fps. If the reservoir pressure is 14 psig and the reservoir temperature 708F, find the rate of temperature increase in the reservoir. Assume the incoming air is at reservoir pressure and flows through a 8-in.-diameter pipe. 6.4 Water flows through a 2-in.-diameter horizontal pipe at a flow rate of 35 gal/min. The heat transfer to the pipe can be 6.6 During the test of a centrifugal pump, a Bourdon pres- neglected,andfrictionalforcesresultinapressuredropof10psi. sure gage just outside the casing of the 12-in.-diameter suction What is the temperature change of the water? pipereads6psig(i.e.,vacuum).Onthe10-in.-diameterdischargeProblems 77 pipeanothergagereads40psig.Thedischargepipeis5ftabovethe 6.12 Thepressurizedtankshownhasacircularcrosssectionof suctionpipe.Thedischargeofwaterthroughthepumpismeasured 6ftindiameter.Oilisdrainedthroughanozzle2in.indiameter 3 to be 4 ft /s. Compute the horsepower input of the test pump. in the side of the tank. Assuming that the air pressure is maintained constant, how long does it take to lower the oil 6.7 A fan draws air from the atmosphere through a 0.30-m- surfaceinthetankby2ft?Thespecificgravityoftheoilinthe diameter round duct that has a smoothly rounded entrance. A tank is 0.85 and that of mercury is 13.6. differentialmanometerconnectedtoanopeninginthewallofthe ductshowsavacuumpressureof2.5cmofwater.Thedensityof 3 airis1.22kg/m .Determinethevolumerateofairflowintheduct P – P = 4 in. Hg atm incubicfeetpersecond.Whatisthehorsepoweroutputofthefan? 5 ft Fan Oil 0.30 m 2 in. 6 ft 6.13 An automobile is driving into a 45-mph headwind at 40mph.Ifthebarometerreads29in.Hgandthetemperatureis 2.5 cm 408F,whatisthepressureatapointontheautowherethewind 6.8 Find the change in temperature between stations (1) and velocity is 120 fps with respect to the auto? (2)intermsofthequantitiesA ,A ,v ,v ,c ,andu.Theinternal 1 3 1 3 v 6.14 Water is discharged from a 1.0-cm-diameter nozzle energyisgivenbyc T.ThefluidiswaterandT ¼ T ,P ¼ P . v 1 3 1 3 that is inclined at a 308 angle above the horizontal. If the jet strikes the ground at a horizontal distance of 3.6 m and a 1 2 vertical distance of 0.6 m from the nozzle as shown, what is the rate of flow in cubic meters per second? What is the total head of the jet? (See equation (6-11b).) q 30° Top view 3 0.60 m 6.9 AliquidflowsfromAtoBinahorizontalpipelineshownata 3 rateof3ft /swithafrictionlossof0.45ftofflowingfluid.Fora 3.6 m pressureheadatBof24in.,whatwillbethepressureheadatA? 6.15 The pump shown in the figure delivers water at 598Fat a rate of 550 gal/min. The inlet pipe has an inside diameter P A of 5.95 in. and it is 10 ft long. The inlet pipe is submerged 6 ft 24 in. into the water and is vertical. Estimate the pressure inside the pipe at the pump inlet. 12 in. 6 in. AB 6.10 InProblem6.26,computetheupwardforceonthedevice fromwaterandair.UsetheresultsofProblem6.26aswellasany otherdatagiveninthatproblemthatyoumayneed.Explainwhy you cannot profitably use Bernoulli’s equation here for a force calculation. 6.11 A Venturi meter with an inlet diameter of 0.6 m is 3 designed to handle 6 m /s of standard air. What is the required throatdiameterifthisflowistogiveareadingof0.10mofalco- hol in a differential manometer connected to the inlet and the throat? The specific gravity of alcohol may be taken as 0.8.78 Chapter 6 Conservation of Energy: Control-Volume Approach 6.16 In the previous problem, determine the flow rate at compressor to the air. Assume that the air speeds within the which the pump inlet pressure is equal to the vapor pressure cushion are very low. ofthewater.Assumethatfrictioncausesaheadlossof4ft.The vapor pressure of water at 598F is 0.247 psi. 6.17 Using the data of Problem 5.27, determine the velocity headofthefluidleavingtheimpeller.Whatpressurerisewould result from such a velocity head? 6.18 Inordertomaneuveralargeshipwhiledocking,pumps are used to issue a jet of water perpendicular to the bow of the ship as shown in the figure. The pump inlet is located farenoughawayfromtheoutletthattheinletandoutletdonot interact. The inlet is also vertical so that the net thrust of the 6.22 The solution to Problem 5.22 is jets on the ship is independent of the inlet velocity and " pressure.Determinethepumphorsepowerrequiredperpound  1=2 2 h 8v 1 1 of thrust. Assume that the inlet and outlet are at the same h ¼ 1þ 1 2 2 gh 1 depth. Which will produce more thrust per horsepower, a low-volume, high-pressure pump or a high-volume, low- ShowthatBernoulli’sequationappliedbetweensections1and2 pressure pump? doesnotgivethisresult.Deriveallexpressionsforthechangein total head across a hydraulic jump. 6.23 Residentialwateruse,exclusiveoffireprotection,runsabout 80gallonsperpersonperday.Ifthewaterisdeliveredtoaresidence at60psig,estimatethemonthlyenergyrequiredtopumpthewater from atmospheric pressure to the delivery pressure. Neglect line lossesandelevationchanges.Assumethepumpsare75%efficient Jet andaredrivenbyelectricmotorswith90%efficiency. 6.24 A1968Volkswagensedanisdrivingovera7300-ft-high mountain pass at a speed of v m/s into a headwind of W m/s. ComputethegagepressureinmPaatapointontheautowhere Side thevelocityrelativetotheautoisv–Wm/s.Thelocalairdensity 3 is 0.984 kg/m . 6.25 Aliquidisheatedinaverticaltubeofconstantdiameter, 15 m long. The flow is upward. At the entrance the average velocityis1m/s,thepressure340,000Pa,andthedensityis1001 3 kg/m .Iftheincreaseininternalenergyis200,000J/kg,findthe heat added to the fluid. 6.26 Water flows steadily up the vertical pipe and is then deflected to flow outward with a uniform radial velocity. If Top friction is neglected, what is the flow rate of water through the pipe if the pressure at A is 10 psig? 12 in. 6.19 Determine the head loss between stations (1) and (2) in Problem 5.7. 6.20 Multnomah Falls in Oregon has sheer drop of 165 m. 1 in. 2 Estimate the change in water temperature caused by this drop. 8 in. 5 ft 6.21 An ‘‘air cushion’’ vehicle is designed to traverse terrain while floating on a cushion of air. Air, supplied by a compressor, escapes through the clearing between the ground and the skirt of the vehicle. If the skirt has a A rectangular shape 3  9 m, the vehicle mass is 8100 kg and the ground clearance is 3 cm, determine the airflow rate 6.27 Waterflowsthroughthepipecontractionshownatarateof 3 needed to maintain the cushion and the power given by the 1ft /s.CalculatethedifferentialmanometerreadingininchesofProblems 79 mercury,assumingnoenergylossintheflow.Besuretogivethe andthepressureheadatBifthepipehasauniformdiameterof1 correct direction of the manometer reading. in.Howlongwillittakeforthewaterleveltodecreaseby3ft? The tank diameter is 10 ft. B 4 in. 6 in. 4 ft ? 6.28 The figure illustrates the operation of an air lift pump. H O 2 Compressedairisforcedintoaperforatedchambertomixwiththe 10 ft water sothatthe specific gravity of the air–water mixtureabove theairinletis0.5.Neglectinganypressuredropacrosssection(1), compute the discharge velocity v of the air–water mixture. Can Bernoulli’s equation be used across section (1)? 6.32 In Problem 6.31, find the rate of discharge if the fric- u = ? 2 tional head loss in the pipe is 3.2 v /g where v is the flow 0.90 m velocity in the pipe. Compressed air 6.33 Assume that the level of water in the tank remains the same and that there is no friction loss in the pipe, entrance, or nozzle. Determine 1.8 m (a) the volumetric discharge rate from the nozzle; (b) the pressure and velocity at points A, B, C, and D. 1 C 6.29 ReworkProblem withtheassumptionthatthemomentum 20 ft of the incoming air at section (1) is zero. Determine the exit 20 ft 2 in. 23 ft 4 in. Diameter velocity,v,andthemagnitudeofthepressuredropatsection(1). Diameter 3 6.30 Air of density 1.21 kg/m is flowing as shown. If B D 3 ft v¼ 15 m/s, determine the readings on manometers (a) and A (b) in the figures below. 6.34 Water in an open cylindrical tank 15 ft in diameter dis- charges into the atmosphere through a nozzle 2 in. in diameter. Air Neglectingfrictionandtheunsteadinessoftheflow,findthetime u requiredforthewaterinthetanktodropfromalevelof28ftabove the nozzle to the 4-ft level. 6.35 A fluid of density r enters a chamber where the fluid 1 H O 2 isheatedsothatthedensitydecreasestor .Thefluidthenescapes 2 throughaverticalchimneythathasaheightL.Neglectingfriction and treating the flow processes as incompressible except for the heating, determine thevelocity, v, in the stack. The fluidvelocity (a) enteringtheheatingchambermaybeneglectedandthechimneyis immersed in fluid of density r . 1 u V 24 m/s r 2 Oil r 1 L 2 S.G. = 0.86 (b) 6.31 Referringtothefigure,assumetheflowtobefrictionless r 1 Heating section inthesiphon.Findtherateofdischargeincubicfeetpersecond,80 Chapter 6 Conservation of Energy: Control-Volume Approach 2 6.36 Repeatthepreviousproblemwithouttheassumptionthat problemifaheadlossof3v /goccursinthepipewherev isthe thevelocityintheheating section isnegligible. The ratioof the flow velocity in the pipe. flow area of the heating section to the chimney flow area is R. 6.38 Thetankinthepreviousproblemfeedstwolines,a4-cm 6.37 Considera4-cmpipethatrunsbetweenatankopentothe pipethatexits10mbelowthewaterlevelinthetankandasecond atmosphereandastationopentotheatmosphere10mbelowthe line,also 4 cmin diameter runsfrom thetank toa station 20m water surfaceinthetank.Assumingfrictionlessflow,whatwill belowthewaterlevelinthetank.Theexitsofbothlinesareopen bethemassflowrate?Ifanozzlewitha1-cmdiameterisplaced totheatmosphere.Assumingfrictionlessflow,whatisthemass at the pipe exit, what will be the mass flow rate? Repeat the flow rate in each line?Chapter7 Shear Stress in Laminar Flow In the analysis of fluid flow thus far, shear stress has been mentioned, but it has not been related to the fluid or flow properties. We shall now investigate this relation for laminar flow. The shear stress acting on a fluid depends upon the type of flow that exists. In the so-called laminar flow, the fluid flows in smooth layers or lamina, and the shear stress is the result of the (nonobservable) microscopic action of the molecules. Turbulent flow is characterized by the large scale, observable fluctuations in fluid and flow properties, and the shear stress is the result of these fluctuations. The criteria for laminar and turbulent flows will be discussed in Chapters 12 and 13. The shear stress in turbulent flow will be discussed in Chapter 13. 7.1 NEWTON’S VISCOSITY RELATION Inasolid,theresistancetodeformationisthemodulusofelasticity.Theshearmodulusofan elastic solid is given by shear stress shear modulus¼ (7-1) shear strain Just as the shear modulus of an elastic solid is a property of the solid relating shear stress and shear strain, there exists a relation similar to (7-1), which relates the shear stressinaparallel,laminarflowtoapropertyofthefluid.ThisrelationisNewton’slawof viscosity shear stress viscosity¼ (7-2) rate of shear strain Thus, the viscosity is the property of a fluid to resist the rate at which deformation takes placewhenthefluidisacteduponbyshearforces.Asapropertyofthefluid,theviscosity depends upon the temperature, composition, and pressure of the fluid, but is independent of the rate of shear strain. TherateofdeformationinasimpleflowisillustratedinFigure7.1.Theflowparallel to the x axis will deform the element if the velocity at the top of the element is dif- ferent than the velocity at the bottom. Therateofshearstrainatapointisdefinedasdd/dt.FromFigure7.1,itmaybeseenthat dj  dj dd tþDt t ¼lim  dt Dx,Dy,Dt0 Dt  fp/2arctan½(vj vj )Dt/Dygp/2 yþDy y ¼lim (7-3) Dx,Dy,Dt0 Dt In the limit,dd/dt¼ dv/dy¼ rate of shear strain: 8182 Chapter 7 Shear Stress in Laminar Flow y (u – u ) ∆ t ⎜ ⎜ y+∆ y y ∆ y d ∆ x Element at Element at time t time t + ∆ t Figure 7.1 Deformation of a u, x fluid element. Combining equations (7-2) and (7-3) and denoting the viscosity by m, we may write Newton’s law of viscosity as dv t¼ m (7-4) dy Thevelocity profile and shear stress variation in a fluid flowing between two parallel plates 1 isillustratedinFigure7.2.Thevelocityprofile inthiscaseisparabolic;astheshearstressis proportional to the derivative of the velocity, the shear stress varies in a linear manner. h h Figure 7.2 Velocity and shear stress profiles for flow between two parallel plates. ut 7.2 NON-NEWTONIAN FLUIDS t Newton’slawofviscositydoesnotpre- dict the shear stress in all fluids. Fluids are classified as Newtonian or non- Newtonian, depending upon the rela- tion between shear stress and the rate ofshearing strain. InNewtonianfluids, therelationislinear,asshowninFigure 7.3. Yield In non-Newtonian fluids, the shear stress stress depends upon the rate of shear Rate of strain strain.Whilefluidsdeformcontinuously under the action of shear stress, plastics will sustain a shear stress before defor- Figure 7.3 Stress rate-of-strain relation for mationoccurs.The‘‘idealplastic’’hasa Newtonian and non-Newtonian fluids. 1 The derivation of velocity profiles is discussed in Chapter 8. Dilatant Newtonian fluid Ideal plastic Real plastic Pseudo plastic