Chapter 4 Continuity Equation and Reynolds Transport Theorem

Transcription

1 Chapter 4 Continuity Equation and Reynolds Transport Theorem 4.1 Control Volume 4. The Continuity Equation for One-Dimensional Steady Flow 4.3 The Continuity Equation for Two-Dimensional Steady Flow 4.4 The Reynolds Transport Theorem Objectives: apply the concept of the control volume to derive equations for the conservation of mass for steady one- and two-dimensional flows derive the Reynolds Transport Theorem for three-dimensional flow show that continuity equation can recovered by simplification of the Reynolds Transport Theorem 4-1

2 4.1 Control Volume Physical system surroundings ~ is defined as a particular collection of matter or a region of space chosen for study ~ is identified as being separated from everything external to the system by closed boundary The boundary of a system: fixed vs. movable boundary Two types of system: closed system (control mass) ~ consists of a fixed mass, no mass can cross its boundary open system (control volume) ~ mass and energy can cross the boundary of a control volume 4-

3 A system-based analysis of fluid flow leads to the Lagrangian equations of motion in which particles of fluid are tracked. A fluid system is mobile and very deformable. A large number of engineering problems involve mass flow in and out of a system. This suggests the need to define a convenient object for analysis. control volume Control volume ~ a volume which is fixed in space and through whose boundary matter, mass, momentum, energy can flow ~ The boundary of control volume is a control surface. ~ The control volume can be any size (finite or infinitesimal), any space. ~ The control volume can be fixed in size and shape. This approach is consistent with the Eulerian view of fluid motion, in which attention is focused on particular points in the space filled by him fluid rather than on the fluid particles. 4-3

4 4. The Continuity Equation for One-Dimensional Steady Flow Principle of conservation of mass The application of principle of conservation of mass to a steady flow in a streamtube results in the continuity equation. Continuity equation ~ describes the continuity of flow from section to section of the streamtube One-dimensional steady flow No flow Fig. 4.1 Consider the element of a finite streamtube - no net velocity normal to a streamline - no fluid can leave or enter the stream tube except at the ends 4-4

5 Now, define the control volume as marked by the control surface that bounds the region between sections 1 and. To be consistent with the assumption of one-dimensional steady flow, the velocities at sections 1 and are assumed to be uniform. The control volume comprises volumes I and R. The control volume is fixed in space, but indt the system moves downstream. From the conservation of system mass ( m m ) ( m m ) I R t R O t t (1) For steady flow, the fluid properties at points in space are not functions of time, 0 ( m ) ( m ) () R t R t t m t Substituting () into (1) yields Inflow ( m ) ( m ) (3) I t O t t Outflow 4-5

6 Express inflow and outflow in terms of the mass of fluid moving across the control surfce in time dt ( m ) I t Ads ( m ) Ads O tt (4) Substituting (4) into (3) yields A ds A ds Dividing by dt gives (4.1) m Ads Ads Continuity equation In steady flow, the mass flow rate, m passing all sections of a stream tube is constant. m AV = constant (kg/sec) d( AV) 0 (4..a) d( AV) da( V) dv( A) 0 (5) 4-6

7 Dividing (a) by AV results in d da dv 0 A V (4..b) 1-D steady compressible fluid flow For incompressible fluid flow; constant density t d 0 From Eq. (4.a) d( AV) 0 d( AV) 0 (6) Set Q =volume flowrate (m 3 /s, cms) Then (6) becomes Q AV const. AV AV (4.5) 1 1 For -D flow, flowrate is usually quoted per unit distance normal to the plane of the flow, b 3 q flowrate per unit distance normal to the plane of flow m s m 4-7

8 Q AV q hv b b hv hv (4.7) 1 1 [re] For unsteady flow mass tt masst+ inflow outflow ( m ) ( m ) ( m ) ( m ) R tt R t I t O tt Divide by dt ( m ) ( m ) dt R t t R t ( m ) ( m ) I t O tt Define m ( mr) tt ( mr) t ( vol) t dt t Then ( vol) t ( m ) ( m ) I t O tt 4-8

9 Non-uniform velocity distribution through flow cross section Eq. (4.5) can be applied. However, velocity in Eq. (4.5) should be the mean velocity. V Q A Q dq vda A A 1 V vda A A The product AV remains constant along a streamline in a fluid of constant density. As the cross-sectional area of stream tube increases, the velocity must decrease. Streamlines widely spaced indicate regions of low velocity, streamlines closely spaced indicate regions of high velocity. AV AV : A A V V

10 [IP 4.3] p. 113 The velocity in a cylindrical pipe of radius R is represented by an axisymmetric parabolic distribution. What is V in terms of maximum velocity, v c? [Sol] dr R r da=rdr v c vv c 1 r R equation of parabola Q 1 1 R r V v da v 1 c r dr A 0 A A R R 3 4 R R c r c r r c R R c 0 r dr v v v v R R R R R Laminar flow [Cf] Turbulent flow logarithmic velocity distribution 4-10

11 4.3 The Continuity Equation for Two-Dimensional Steady Flow (1) Finite control volume Fig. 4.3 Consider a general control volume, and apply conservation of mass ( m m ) ( m m ) (a) I R t R O t t For steady flow: ( m ) ( m ) R t R t t Then (a) becomes ( m ) ( m ) (b) I t O t t i) Mass in O moving out through control surface ( m ) O tt ( dscos ) da C. S. out mass vol area 1 ds dacos 4-11

12 Displacement along a streamline; ds vdt (c) Substituting (c) into (b) gives ( m ) O tt ( vcos ) dadt (d) CSout.. By the way, vcos = normal velocity component normal to C.S. at da Set n = outward unit normal vector at da v vnvcos n n 1 scalar or dot product (e) Substitute (e) into (d) where da ( m O) dt v nda dt v da tt nda C. S. out C. S. out =directed area element [Cf] tangential component of velocity does not contribute to flow through the C.S. Circulation ii) Mass flow into I ( mi) t ( dscos ) da CSin.. 90 cos0 ( cos ) v dadt dt v ( n ) da CSin.. CSin.. 4-1

13 CSin.. CSin.. dt v nda dt v da For steady flow, mass in = mass out dt v da dt v da CSout.. CSin.. Divide by dt vda vda CSin.. CSout.. vda vda0 C. S. out C. S. in (f) Combine c.s. in and c.s. out vda vnda0 CS.. (4.9) CS.. where integral around the control surface in the counterclockwise direction CS.. Continuity equation for -D steady flow of compressible fluid [Cf] For unsteady flow t ( mass inside c. v.) = mass flowrate in mass flowrate out 4-13

14 () Infinitesimal control volume udx u x udx u x Apply (4.9) to control volume ABCD v nda v nda v nda v nda 0 AB BC CD DA (f) By the way, to first-order accuracy dx AB x AB dy v dy vnda v dx y y vdy vnv y BC dx u dx v nda u dy x x CD dy v dy vnda v dx y y (g) DA dx u dx v nda u dy x x 4-14

15 Substitute (g) to (f), and expand products, and then retain only terms of lowest order (largest order of magnitude) vdy dy v( dy) vdx dx v dx dx y y y y 4 vdy dy v( dy) vdx dx v dx dx y y y y 4 udx dx u( dx) udy dy u dy dy x x x x 4 udx dx ( dx) udy dy u dy dy 0 x x x x 4 v u dxdy v dxdy dxdy u dxdy 0 y y x x v u v u 0 y y x x ( u) ( v) 0 x y (4.10) Continuity equation for -D steady flow of compressible fluid Continuity equation of incompressible flow for both steady and unsteady flow ( const.) u x v y 0 (4.11) 4-15

16 [Cf] Continuity equation for unsteady 3-D flow of compressible fluid ( u) ( v) ( w) 0 t x y z For steady 3-D flow of incompressible fluid u v w 0 x y z Continuity equation for polar coordinates D C A B Apply (4.9) to control volume ABCD v nda V nda V nda V nda 0 AB BC CD DA AB BC CD d vt d V nda vt dr dr vr dr V nda vr ( rdr) d r r d vt d V nda vt dr 4-16

17 DA dr vr dr V nda vr rd r r Substituting these terms yields vt d d vt ( d) vtdr dr vt dr dr 4 vt d d vt ( d) vtdr dr vt dr dr 4 vr dr vr dr vrd r vdrd r rd drd r r dr dr dr vr dr vr vr rd vr drd rd drd r r r r r r vr dr dr vr dr vrrd rd vr rd rd 0 r r r r v v d dr v d dr rdrd v rdrd r r t r t r vr 1 1 v 1 vrdrd dr d v dr d dr d r r r r r 3 ( ) r ( ) ( ) 0 Divide by drd vt vr vr 1 1 vr 1 vt vr rvr drvr dr dr 0 r r r r r r vr vt rvr rvr vt 0 r r 4-17

18 Divide by r vr vr vt vr vt 0 r r r r r ( vr) vr ( v t ) 0 r r r (4.1) For incompressible fluid vr vr v t 0 r r r (4.13) 4-18

19 [IP 4.4] p. 117 A mixture of ethanol and gasoline, called "gasohol," is created by pumping the two liquids into the "wye" pipe junction. Find Q eth and V eth 3 mix kg m V 1.08 m s mix No flow in and out Qgas l/ s m /s 3 gas kg m 3 eth kg m n [Sol] A 1 (0.) 0.031m 4 ; A m ; A m 3 V / m/s (4.4) vnda vnda vnda vnda kg/s vnda788.6v V vnda kg/s vnda V cs. (4.9) V 0.43 m/s Q V A l eth 3 3 (0.43)(0.0079) m /s 3.4 /s 4-19

20 4.4 The Reynolds Transport Theorem Reynolds Transport Theorem (RTT) Osborne Reynolds ( ); English engineer a general relationship that converts the laws such as mass conservation and Newton s nd law from the system to the control volume Most principles of fluid mechanics are adopted from solid mechanics, where the physical laws dealing with the time rates of change of extensive properties are expressed for systems. Thesre is a need to relate the changes in a control volume to the changes in a system. Two types of properties Extensive properties ( E ): total system mass, momentum, energy Intensive properties (i ): mass, momentum, energy per unit mass E i system mass, m 1 system momentum, mv v system energy, mv ( v ) (4.14) E i dm i dvol system system 4-0

21 Derivation of RTT Consider time rate of change of a system property E E ( E E ) ( E E ) (a) tdt t R 0 tdt R I t E dt iv da ( 0) t dt c. s. out ( E ) dt iv da I t csin.. ( E ) i dvol R tdt R tdt (b.1) (b.) (b.3) ( E ) i dvol R t R (b.4) t Substitute (b) into (a) and divide by dt idvol idvol Etdt Et 1 dt dt R tdt csout.. csin.. iv da i v da R t 4-1

22 de dt idvol iv da t cv.. cs.. (4.15) 1 de dt = time rate of change of E in the system dvol 3 = time rate change within the control volume unsteady term t i cs.. i cv.. v da = fluxes of E across the control surface Application of RTT to conservation of mass For application of RTT to the conservation of mass, in Eq. (4.15), E dm m, i 1 and 0 because mass is conserved. dt dvol v da v da v da t cv.. cs.. csout.. csin.. (4.16) Unsteady flow: mass within the control volume may change if the density changes For flow of uniform density or steady flow, (4.16) becomes vda vda0 csout.. csin.. ~ same as Eq. (4.9) 4-

23 For one-dimensional flow c. s. out csin.. vda V A vdav A AV AV (4.1) In Ch. 5 & 6, RTT is used to derive the work-energy, impulse-momentum, and moment of momentum principles. 4-3

24 Homework Assignment # 4 Due: 1 week from today Prob. 4.9 Prob. 4.1 Prob Prob. 4.0 Prob