The law of conservation of mass: Mass can be neither created nor destroyed. It can only be transported or stored.

Transcription

1 UDMETL COCEPTS OR LOW LYSIS We covere mehos of analysis of nonflowing fluis in he previous chaper. In his chaper, we evelop he funamenal conceps of flow analysis, incluing he way o escribe flui flow, naural laws ha govern flui flow, ifferen approaches o formulaing mahemaical moels of flui flow, an mehos ha engineers use o solve flow problems. The unamenal Laws Eperience have shown ha all flui moion analysis mus be consisen wih he following funamenal laws of naure. The law of conservaion of mass: Mass can be neiher creae nor esroye. I can only be ranspore or sore. ewon s hree laws of moion: mass remains in a sae of equilibrium, ha is, a res or moving a consan velociy, unless ace on by an unbalance force. The rae of change of linear momenum of mass is equal o he ne force acing on he mass. ny force acion has a force reacion equal in magniue an opposie in irecion. The firs law of hermoynamics (law of conservaion of energy) Energy, like mass, can be neiher creae nor esroye. Energy can be ranspore, change in form, or sore. The secon law of hermoynamics: The enropy of he universe mus increase or, in he ieal case, remain consan in all naural processes. The sae of posulae (law of propery relaions): The various properies of a flui are relae. If a cerain minimum number (usually wo) of flui s properies are specifie, he remainer of he properies can be eermine. 1

2 OTE: These laws apply o all flows. They o no epen on he naure of he flui, he geomery of he bounaries, or anyhing else. s far as we know, hey have always been rue an will coninue o be rue unless hey are suspene by he creaor of he universe. Hence, we can firmly base analysis of all flows on hese laws. Consiuive Relaions In aiion o hese universal laws, several less funamenal laws, such as ewon s law of viscosiy, ourier s law of conucion, are neee o solve flow problems. These laws are rue for some fluis. Mahemaical ormulaion The funamenal laws are he basis of our unersaning of flui moion. However, besies unersaning, an engineer nees o know quaniaively he velociy, an he pressure o calculae he effecs of he flui on surfaces ha i conacs, such as force eere by he flui on a surface, pressure rop in a pipe flow, ec. To obain preicive capabiliy, he funamenal laws mus be epresse mahemaically an hey mus be solve o preic velociy or pressure. To formulae he funamenal laws, we choose boh a poin of view an a mahemaical meho. - Sysem versus Conrol Volume - Differenial versus inegral formulaion 2

3 Sysem versus Conrol Volume We may apply he funamenal laws o eiher a em or a conrol volume. Sysem : a specific flui mass selece for analysis. Conrol Volume: a specific region of space selece for analysis. Sysem an conrol volume may be eiher infiniesimally small or finie. The em poin of view is relae o a Lagrangian escripion of flow. Is avanages is ha all he funamenal laws may be epresse irecly in erms of a specific collecion of mass. Conrol volume poin of view is relae o an Eulerian escripion of flow. Is avanage is ha conrol volumes are easier o use for problem soluion. Thus we aop he em poin of view o formulae he funamenal laws, bu use he conrol volume poin of view o apply hem o problems. orunaely, we can formally connec he wo poins view by purely mahemaical relaionships. 3

4 Differenial versus Inegral ormulaion We mus now consier he level of eail of he resuling flow analysis. We mus choose beween a eaile poin by poin escripion an a global or lumpe escripion. When a poin by poin (local) escripion is esire, funamenal laws are applie o an infiniesimal conrol volume. The resul will be a se of ifferenial equaions wih he flui velociy an pressure as epenen variables an he locaion (, y, z) an ime as inepenen variables. Soluion of hese ifferenial equaions, ogeher wih bounary coniions, will be wo funcions V(, y, z, ), an P(, y, z, ) ha can ell us he velociy an pressure a every poin. When global informaion such as flow rae, force an emperaure change beween inle an oule is esire, he funamenal laws are applie o a finie conrol volume. The resul will be a se of inegral equaions. 4

5 BSIC LWS OR SYSTEM Conservaion of Mass Sysem M em where M mass ( ) m ewon s Secon Law P P : linear momenum P The irs Law of Thermoynamics Q W E in he rae form mass ( ) Vm Q W The oal energy of he em is given by v E em e where e u mass ( ) 2 2 V E gz Q W V The Secon Law of Thermoynamics If an amoun of hea is ransferre o a em a emperaure T, he secon law of hermoynamics saes ha he change in enropy s of he em is given by Q S T on he rae basis Toal enropy of he em is T S S ngular Momenum Principle H T H : angular momenum ( r V ) 1 Q T sm mass ( ) ( ) H mass ( ) s ( r V ) m ( r V ) e 1 s Q T 5

6 RELTIO O SYSTEM DERIVTIVES TO THE COTROL VOLUME ORMULTIO The above equaions involve he ime erivaive of an eensive propery of he em (mass, momenum, energy, enropy). ll he above equaions can be epresse in erms of a general inensive propery. Thus m mass ( ) ( ) Comparing his wih he above equaions, we see ha when =M hen =1 =E hen =e = P hen = V =S hen =s Consier a em an conrol volume whose bounaries coincie a. Objecive: To relae he em o he ime variaions of his propery () associae wih he conrol volume. rom he efiniion of a erivaive, em s s lim (1) +, he em occupies regions II an III, a, he em an he conrol volume coincie, we can wrie 6

7 7 C C s (2) III I C III II s III I C s (3) Subsiuing hese epressions şno he efiniion of he em erivaive C III I C em lim or lim lim lim I III C C em (4) Term 1 in Eq. 4 simplifies o C C lim (5) III III lim lim To evaluae III ) o+, le us look a an enlarge view of a ypical subregion of region III. III III III III l l cos cos

8 oe: The angle will always be less han p/2 over he enire area of he conrol surface bouning region III. In he above epression, l is he isance ravelle by a paricle on he em surface uring he ime inerval along he sreamline ha eise a. III l lim lim cos III l oe: lim V an Hence III lim V cos III (6) The erm 3 in Eq. 4 simplifies o I I lim lim To evaluae I, look a an enlarge view of a ypical subregion ŞEKİL VR l( cos ) lim I lim lim l cos I l cos I V cos I (7) 8

9 Subsiuing Eqs. (5),(6) an (7) ino (4) C V cos III V cos I I III Hence we can wrie, V cos Recognizing ha C V cos V C V I is imporan o recall ha in eriving he above equaion, he limiing process (aking he limi as ) ensure ha he relaion is vali a he insan when he em an conrol volume coincie. C V : he oal rae of change of any arbirary eensive propery of he em. : he ime rae of change of he arbirary eensive propery wihin he conrol volume : he ne rae of flu of he eensive propery hrough he conrol surface. Evaluaing he scalar prouc 9

10 BSIC EQUTIOS OR COTROL VOLUME Basic equaions for a conrol volume are obaine by combining he basic equaions for a em an Reynols ranspor heorem. COSERVTIO O MSS (Coninuiy Equaion) Combining he law of conservaion of mass wih he ranspor heorem yiels one of he mos useful equaions in all flui mechanics: he coninuiy equaion. Recall ha conservaion of mass saes simply ha he mass of a em is consan, M, M SYS The em an conrol volume formulaion of he conservaion of mass are relae by Reynols ranspor heorem. Seing, =M hen=1, we obain M V C C V Coninuiy equaion for a finie conrol volume C V rae of change of mass wihin he conrol volume ne rae of flu hrough he conrol surface OTE: V is he velociy measure relaive o he conrol surface. The sign of he o prouc V epens on he irecion of velociy vecor V, relaive o he area vecor. V is posiive where flow is ou hrough he conrol surface, negaive where flow is in hrough he conrol surface, an zero where flow is angen o surface. 1

11 Special Cases 1. Incompressible low or incompressible flow, = consan V C V V consan consan V is calle he volume flow rae of flow over a secion of he conrol surface. 2. Seay low Hence he coninuiy equaion becomes, V [flow coul be compressible] 3. Uniform low The velociy is consan across he enire area a a secion when ensiy is also consan a a secion, hen a secion n V V V n n n n n n n 11

12 Eample: consan ensiy flui flows in he converging, wo-imensional channel shown in he figure. The wih perpenicular o he paper is quie large compare o he channel heigh. The velociy in he z-irecion is zero. The channel half heigh y an he flui velociy in -irecion are given by u y y 1 l 1 l y Y 2 u 1 where u =1. m/s Show ha he flow fiel saisfies he coninuiy equaion. Soluion: To be complee in class 12

13 Eample: Waer is being ae o a sorage ank a he rae of 2 l/min. he same ime, waer flows hrough a 5 cm insie iameer pipe wih an average velociy of 18 m/s. The sorage ank has an insie iameer of 3 cm. in he rae a which he waer level rises or falls. in flow rae 2 l/min. sorage ank iameer 3 cm ischarge pipe iameer 5 cm ischarge velociy 18 m/s Soluion: To be complee in class 13

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15 MOMETUM EQUTIO OR IERTIL COTROL VOLUME In his secion, we will evelop mahemaical formulaion of ewon s Secon Law for an inerial conrol volume. Inerial conrol volume is he conrol volume ha is no acceleraing relaive o a saionary frame of reference. Recall ha ewon s secon law for a em moving relaive o an inerial coorinae em was P P V linear momenum = oal resulan force Using he relaion beween he em an conrol volume formulaions V C an seing =P an = V, we obain P V C oe: P on VV Since, in eriving he relaion beween he em an conrol volume formulaion, he em an conrol volume coincie a o on on conrol volume Hence, we can wrie, S B V C VV MOMETUM EQUTIO 15

16 16 This equaion saes ha he sum of all forces (surface an boy forces) acing on a nonacceleraing conrol volume is equal o he sum of he rae of he change of momenum insie he conrol volume an he ne rae of efflu of momenum hrough he conrol surface. S p surface force ue o pressure C B g boy force ue o graviy Someimes surface force s may also inclue shear force. The momenum equaion is a vecor equaion. rom his vecor equaion, a scalar componen in each irecion can be wrien, i.e. C B S V u u C B S y V v v y y C B S z V w w z z The momenum equaion is usually use o calculae force ineracions beween a moving flui an soli objecs in conac wih i. C B S V V V MOMETUM EQUTIO

17 Eample: Waer from a saionary nozzle srikes a fla plae as shown. The velociy of he waer leaving he nozzle is 15 m/sec. The nozzle area is.1 m 2. ssuming he waer is irece normal o he plae; eermine he horizonal force on he suppor. Horizonal force K =? Soluion ssumpions: - Seay flow - Incompressible flow - Uniform flow a each secion where flow crosses he CV surface Since he force ineracion beween he flui an he soli objec is he poin of ineres, we have o use momenum equaion. We mus choose a suiable conrol volume. number of possible choices are, Regarless of our choice of conrol volume, he resul shoul be he same. 17

18 I. Use C I Momenum equaion in -irecion S B u uv C Seay flow B =, no boy force in -irecion p p R S S R pressure force pressure force force of he suppor onlef face on righ face on CV ( assume posiive) OTE: Lef an righ faces of he conrol volume are equal. R is he force eere by he suppor (han) on he conrol volume R R uv u V1 1 1 uv 1, V V1, since irecion of area is 18 apar from he velociy. or op an boom surfaces, u= R u 1 V1 1 {properies are uniform over 1 } m kg m R sec m sec R k orce acing on he suppor m {R is he force acing on he conrol volume an i acs opposie o posiive irecion} K R k 18

19 II. Use C II o boy force acs on his conrol volume in -irecion. Momenum equaion in - irecion becomes uv S S p R uv u 1 25 p R K an R p R p 1 k k 2.25 k 1 V 2. k To eermine he ne force on he plae, we nee o ake ino accoun pressure (amospheric) force of he righ face of he plae. ne ne ne K p p k p 19

20 Eample: meal conainer, which has a heigh of.6 m an an insie crosssecional area of.1 m 2, is place on a scale. Waer flows ino he ank a a velociy of 6 m/s hrough an opening a he op wih a cross-secional area of.1 m 2, flows ou he openings on he sie walls wih equal cross-secional areas. Uner seay flow coniions, he heigh of he waer in he ank is.5 m. The pressure is amospheric across all openings, an he conainer weighs 5 when i is empy. If he fricional effecs are negligible hen eermine he reaing on he scale. T =.1 m 2 V 1 = 6 m/s 1 =.1 m 2 2 = 3 h =.5 m K y =? Soluion To be complee in class 2

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22 Eample: shallow circular ish has a sharp-ege orifice a is cener. waer je of spee V srikes he ish concenrically. If he je issuing from he orifice an from he surface of he ish also has spee V, evaluae he eernal force neee o hol he ish in place for V = 5 m/s, D=1 mm an =2 mm. Soluion To be complee in class 22