Chapter 5: Control Volume Approach and Continuity Principle Dr Ali Jawarneh

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1 Chaper 5: Conrol Volume Approach and Coninuiy Principle By Dr Ali Jawarneh Deparmen of Mechanical Engineering Hashemie Universiy 1

2 Ouline Rae of Flow Conrol volume approach. Conservaion of mass he coninuiy equaion. Caviaions Differenial form of he coninuiy equaion 2

3 5.1: Rae of flow - Discharge or Volume flow rae, Q {m 3 /s} For a fluid wih consan velociy: Q V.A The do produc means ha he velociy componen normal o he area For a fluid wih variable velociy: m Q - Mass Flow Rae,, {kg/s} Consan velociy Variable velociy and consan densiy m& V.A m& V.dAQ V.dA - Mean Velociy VV avg Q /A 3

4 Example: Find he volume and mass flow rae of waer. Soluion: π Q VA V 4 D 2 10 π m QQ k/ kg/s 4 m 3 /s Example: Find he volume and mass flow rae of waer. Soluion: Q VAcos m o 4 3 /s m Q kg/s 4

5 Example: The recangular channel is 2 m wide. Wha is he discharge in he channel? Soluion: Q A V.dA y x Vdxdy Q 1 / 3 V.dA y 2dy A 1cos 30 If he case is pipe 0 o m 3 /s Q 1 r0 cos 30 2 V.dA V r 2 π A 0 o r dr 5

6 5.2: Conrol Volume Approach The mehod employed for solving mos problems in fluid mechanics is he conrol volume approach also known as he Reynolds Transpor Theorem. A conrol volume is a seleced volumeric region in space. A conrol surface is he surface enclosing he conrol volume. 6

7 The mass wihin he conrol volume can change wih ime, and a conrol volume can deform wih ime, and move and roae in space. In conras wih he conrol volume, a sysem is defined as a coninuous mass of fluid ha always conains he same fluid paricles. 7

8 A ime : M M Afer ime : sys M m& M m& ou ou in in 8

9 The mass of he sysem a ime : in ou sys M M M M m m M M M M & & m m M M M M in ou sys sys 9

10 Dividing by and aking he limi as 0: dm sys dm m& ou m& in dm sys By definiion: 0 Consequenly: dm m& in m& ou Coninuiy Equaion 10

11 Example: A ank has a hole in he boom wih a cross-secional secional area of m 2. The cross-secional secional area of he ank is 0.1 m 2. The velociy of he liquid flowing ou he boom hole is V2gh 0.5, where h is he heigh of he waer surface in he ank above he oule. A a cerain ime he surface level in he ank is 1 m and rising i a he rae of 0.1 cm/s. The liquid id is incompressible. Find he velociy of he liquid hrough he inle. 11

12 Soluion: dh 0. 1 cm/s dm m in m M V Ah ou m in Vin A in m ou V Aou 2 gh dh A Vin Ain 2gh A A ou ou dm dh A / 100 Vin V in m/s End of he soluion Exercise:Commen if we need o find he ime from h1 o h2: Example: dh/ 2h 12

13 Propery Transpor across he Conrol surface For seady, onedimensional flow in a condui: V. A V.A V.A cs VA VA For consan densiy incompressible flow: V.A V2A2 V1A 1 cs 13

14 Reynolds Transpor Theorem dbsys d b dv b VdA { cs Lagrangian Eulerian B bdm b dv 14

15 5.3: Coninuiy Equaion C Conservaion of fm Mass Depending on he previous analysis, i can be verified ha: dm sys d dv V. da And since he mass of a sysem is consan, hen: cs d V. d A dv cs dm For flow sreams having a uniform velociy across he flow secions: cs V. A m& in d dv m& ou 15

16 cs d V. da dv This equaion saes ha he rae of accumulaion of mass in he CV he ne mass efflux hrough he CS Ne efflux mass flow rae ou of he CV mass flow rae in he CV 16

17 Example: seady, incompressible flow of waer hrough he device. Given: A m 2 A m 2 A m 2 V 1 5 m/s V 3 12 m/s Q m 3 /s 999 kg/m 3 Find V 2 and i s direcion. 17

18 Soluion: 0.0 cs d V.dA dv V.dA 0. 0 V.A 0 cs V. A V. A V. A V. A since he problem is incompressible: cons 3 V 1. A1 V2. A2 V3. A3 V4. A V A V A V A V A V 1 A1 V2 A2 V3 A3 Q V V m / s Because i is ve, he assumpion is no valid, so he V 2 goes inward he. 2 g

19 Example: Find he V max for he given seady, incompressible flow hrough pipe 3. Given: A m 2 V 1 5 m/s A m 2 V 2 3 m/s A m 2 19

20 Soluion: 0.0 V. cs d V.dA dv V.dA A1. cs o find V 3 : le V 3 a b r cs V2. A2 V3.dA3 0 0 V A V A V.dA 0 V 3 V max a V o,v r V 3 V max [1- r/r o ] cs o r π 2 V3.dA3 Vmax 1 2πr dr ro Vmax r cs o o 3 π 2 A d 3 d m ro m π Vmax 0. 0 Vmax m/s b -V max /r o 20

21 Example: Tank of a volume of 0.05 m 3 conain air. A 0.0, 0 air escapes hrough a valve. Air leaves wih speed V300 m/s and densiy of 6 kg/m 3 hrough area of 65 mm 2. Find he rae of change of air densiy in he ank a

22 Soluion: cs V.dA d dv Properies in he ank are uniform, bu ime dependence: 1 V 1 A 1 d f n dv d V V End of he soluion. Exercise: Solve he example ha appears in slide # 11 using his equaion: d d 3 2 d 1 V1 A kg/m /s V 0.05 cs V.dA dv 22

23 Example: Air flows seadily beween wo secions in a long, sraigh porion of 4-in inside diameer pipe. The uniformly disribued emperaure and pressure a each secion are given. If he average air velociy nonuniform velociy disribuion a secion 2 is 1000 f/s, calculae he average air velociy a secion 1. 23

24 Soluion: 24

25 5.4: Caviaion - Caviaion is he phenomenon ha occur when he fluid pressure is reduced d o he local l vapor pressure -Under such condiions vapor bubbles form, grow, and hen collapse, producing shock wave, noise, and dynamics effecs ha lead o decreased equipmen performance and, frequenly, equipmen failure -Caviaion ypically occurs a locaions where he velociy is high -If he pipe area is reduced, he velociy is increased according o he coninuiy equaion and he pressure is reduced as dicaed by he Bernoulli equaion 25

26 - As he flow rae increases, he pressure a he resricion can become sub-amospheric. The pressure can drop no longer han he vapor pressure of he liquid because, a his poin, he liquid will boil and caviaion ensues. Formaion of vapor bubbles in he process of caviaion. i a Caviaion. b Caviaion higher flow rae. 26

27 A a given pressure, he emperaure a which a pure subsance changes phase is called he sauraion emperaure T sa. Likewise, a a given emperaure, he pressure a which a pure subsance changes phase is called he sauraion pressure P sa. A an absolue pressure of 1 sandard amosphere 1 am or kpa, for example, he sauraion emperaure of waer is 100 C. Conversely, a a emperaure of 100 C, he sauraion pressure of waer is 1 am. The vapor pressure P v of a pure subsance is defined as he pressure exered by is vapor in phase equilibrium wih is liquid a a given emperaure. P v is a propery of he pure subsance, and urns ou o be idenical o he sauraion pressure Psa of he liquid P v P sa. 27

28 For phase-change processes beween he liquid and vapor phases of a pure subsance, he sauraion pressure and he vapor pressure are equivalen since he vapor is pure. Noe ha he pressure value would be he samewheher i is measured in he vapor or liquidid phase provided ha i is measured a a locaion close o he liquid vapor inerface o avoid he hydrosaic effecs. Vapor pressure increases wih emperaure. Thus, a subsance a higher emperaures boils a higher pressures. For example, waer boils a 134 C in a pressure cooker operaing a 3 am absolue pressure, bu i boils a 93 C in an ordinary pan a a 2000-m elevaion, where he amospheric pressure is 0.8 am. The sauraion or vapor pressures are given in Appendices for various subsances. A mini able for waer is given in following able for easy reference. 28

29 29

30 5.5: Differenial Form of he Coninuiy Equaion Coninuiy Equaion Applying he coninuiy equaion o a finie volume: cs d d V A V. z y x d y x w z x v z y u Dividing by he volume: 0 w v u 30 0 z w y v x u

31 When he volume approaches zero: pp 0 z w y v x u For seady flow: z y x w v u If he flow is incompressible: 0 z w y v x u If he flow is incompressible: 0 z w y v x u in vecor noaion: z y x 0 V. where is he Del operaor 31 k z j y i x operaor

32 Comparison beween he differenial and Comparison beween he differenial and inegral form of he coninuiy equaion. Differenial Form of he Coninuiy Equaion 0 w v u 0 z w y v x u Inegral Form of he Coninuiy Equaion cs dv d V.dA 32

33 Example: check he following equaion if i saisfies he coninuiy equaion for incompressible flow. Soluion: V u v 0 x y u u 2xy x v 2xy y x 2 y 2 i xy j i saisfies he coninuiy equaion. 33