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Chapter 4 Continuity Equation and Reynolds Transport Theorem

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Chapter 4 Continuity Equation and Reynolds Transport Theorem Ch. 4 Continuity Equation Chapter 4 Continuity Equation and Reynolds Transport Theorem 4.1 Control Volume 4.2 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 Ch. 4 Continuity Equation 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 4-2 Ch. 4 Continuity Equation → 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 Ch. 4 Continuity Equation 4.2 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 Ch. 4 Continuity Equation Now, define the control volume as marked by the control surface that bounds the region between sections 1 and 2. → To be consistent with the assumption of one-dimensional steady flow, the velocities at sections 1 and 2 are assumed to be uniform. → The control volume comprises volumes I and R. → The control volume is fixed in space, but in dt the system moves downstream. From the conservation of system mass ()()mmIRtROtt mm (1) m For steady flow, the fluid properties at points in space are not functions of time, 0 t → ()()mmR tRtt (2) Substituting (2) into (1) yields ()mmIt ( Ot ) t (3) Inflow Outflow 4-5 Ch. 4 Continuity Equation Express inflow and outflow in terms of the mass of fluid moving across the control surfce in time dt ()mAdsIt 11 1 ()mAdsOt t 22 2 (4) Substituting (4) into (3) yields 11A ds 1 2 A 2 ds 2 Dividing by dt gives mAdsAds 11 1 2 2 2 (4.1) → Continuity equation In steady flow, the mass flow rate, m passing all sections of a stream tube is constant. mAV = constant (kg/sec) dAV()0 (4.2.a) → dAVdAVdVA() () ()0 (5) 4-6 Ch. 4 Continuity Equation Dividing (a) by AV results in ddAdV 0 (4.2.b) AV → 1-D steady compressible fluid flow For incompressible fluid flow; constant density → d 0 t From Eq. (4.2a) dAV()0 dAV()0 (6) Set Q =volume flowrate (m3/s, cms) Then (6) becomes QAVconst. AVAV11 2 2 (4.5) For 2-D flow, flowrate is usually quoted per unit distance normal to the plane of the flow, b → q flowrate per unit distance normal to the plane of flow msm3 4-7 Ch. 4 Continuity Equation QAV qhV bb hV11 hV 2 2 (4.7) [re] For unsteady flow masstt mass t+ inflow outflow ()mmmmR tt ()()() Rt It Ott Divide by dt ()mm () Rt t Rt()(mm ) dt It Ot t Define mm() () m ( vol ) Rt t Rt tdtt Then () vol ()(mm ) t It Ot t 4-8 Ch. 4 Continuity Equation • 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. Q V A QdQvdA AA 1 VvdA 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. A11VAVAA 2 2: 1 2 VV 1 2 4-9 Ch. 4 Continuity Equation [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, vc ? [Sol] dr dA=2rdr r R vc r 2 vvc 1 2 ← equation of parabola R 2 Qr11R V v dA vc 12 r dr A 220 AA R R 32422R 222vvvvR rrrRR ccccrdr → Laminar flow 222220 RRRRR 24 0 2 4 2 [Cf] Turbulent flow → logarithmic velocity distribution 4-10 Ch. 4 Continuity Equation 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 ()()mmIRtROtt mm (a) For steady flow: ()()mmR tRtt Then (a) becomes ()mmIt ( Ot ) t (b) i) Mass in O moving out through control surface ()mdsdA (cos) Ot t C.. S out mass vol area 1 ds dA cos 4-11 Ch. 4 Continuity Equation • Displacement along a streamline; ds vdt (c) Substituting (c) into (b) gives ()m (cos) v dAdt (d) Ot t CSout.. By the way, vcos = normal velocity component normal to C.S. at dA Set n = outward unit normal vector at dA n 1 ← scalar or dot product (e) vvnvn cos Substitute (e) into (d) ()mdtvndAdtvdA Ot t C.. S out C .. S out where dA ndA=directed area element [Cf] tangential component of velocity does not contribute to flow through the C.S. → Circulation ii) Mass flow into I ()mdsdA (cos) 90 cos 0 It CSin.. (cos)vdAdtdtvndA ( ) CSin.. CSin .. 4-12 Ch. 4 Continuity Equation dt v ndA dt v dA CSin.. CSin .. For steady flow, mass in = mass out dt v dA dt v dA CSout.. CSin .. Divide by dt vdA vdA CSin.. CSout .. vdA vdA 0 (f) C.. S out C .. S in Combine c.s. in and c.s. out vdA vndA 0 (4.9) CS.. CS .. where integral around the control surface in the counterclockwise direction CS.. → Continuity equation for 2-D steady flow of compressible fluid [Cf] For unsteady flow (..)mass inside c v = mass flowrate in – mass flowrate out t 4-13 Ch. 4 Continuity Equation (2) Infinitesimal control volume udx udx u u x 2 x 2 Apply (4.9) to control volume ABCD v ndA v ndA v ndA v ndA 0 (f) AB BC CD DA By the way, to first-order accuracy dx AB x 2 dy v dy vndA v dx AB yy22 vdy vn v y 2 dx u dx vndA u dy BC xx22 dy v dy vndA v dx (g) CD yy22 dx u dx vndA u dy DA xx22 4-14 Ch. 4 Continuity Equation Substitute (g) to (f), and expand products, and then retain only terms of lowest order (largest order of magnitude) vdy dy v() dy 2 vdx dx v dx dx yyyy22 4 vdy dy v() dy 2 vdx dx v dx dx yyyy22 4 udx dx u() dx 2 udy dy u dy dy x 22xxx 4 udx dx() dx 2 udy dy u dy dy 0 x 22xxx 4 vu dxdy v dxdy dxdy u dxdy 0 yy xx vu vu0 yy xx ()uv ()0 (4.10) xy → Continuity equation for 2-D steady flow of compressible fluid • Continuity equation of incompressible flow for both steady and unsteady flow ( const.) uv 0 (4.11) xy 4-15 Ch. 4 Continuity Equation [Cf] Continuity equation for unsteady 3-D flow of compressible fluid ()uvw () ( )0 tx y z For steady 3-D flow of incompressible fluid uvw 0 xyz • 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 dvdt VndA vt dr AB 22 dr vr dr VndA vr () r drd BC rr22 dvdt VndA vt dr CD 22 4-16 Ch. 4 Continuity Equation dr v dr VndA v r rd DA r rr22 Substituting these terms yields vd d v() d 2 v drtt dr v dr dr tt22 4 vd d v() d 2 v drtt dr v dr dr tt22 4 vdr vdr vrd vdrd rr rd drd rr rr22 22 dr dr dr vrr dr v vrr rd v drd rd drd rrrrrr222 2 2 vdrrr dr v dr vrr rd rd v rd rd 0 rrrr22 2 vv trd dr v d dr rdrd v rdrd trrr vv11 1 v drdrr() dr22 d v () dr d () dr 3 d 0 rrrrrr22 2 Divide by drd vv v11 v 1 trvvrvdrvdrdr r r 0 trrrrr r22 r rr 2 vv rtrv r v v 0 rrrr t 4-17 Ch.
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