Chapter 2: Basic Governing Equations

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1 -1 Reynolds Transport Theorem (RTT) - Continuity Equation -3 The Linear Momentum Equation -4 The First Law of Thermodynamics -5 General Equation in Conservative Form -6 General Equation in Non-Conservative Form Chapter Basic Governing Equations

2 -1 Reynolds Transport Theorem (RTT) (1) Conservation of Mass: -1

3 -1 Reynolds Transport Theorem (RTT) () System and fixed control volume. -

4 -1 Reynolds Transport Theorem (RTT) (3) B net B out Bin ρ b( V n ) da (inflow if negative) (11-41) B CV CV General ρb dv : db dt CS sys d dt (11-4) CV ρ b dv CS ρ b ( V n ) da (11-43) -3

5 -1 Reynolds Transport Theorem (RTT) (4) Interpreting the Scalar Product: -4

6 -1 Reynolds Transport Theorem (RTT) (5) Special Case 1: Steady Flow Steady flow : db dt sys CS ρ b ( V n ) da (11-44) Special Case : One-Dimensional Flow One - dim ensiona l flow : db dt db sys dt sys d dt d dt ρb dv CV CV ρb dv out out ρ ebeve Ae for each exit m e b e - in - in m b i ρ ibivi Ai for each exit i (11-46) (11-45) -5

7 -1 Reynolds Transport Theorem (RTT) (6) Extensive and Intensive Properties: -6

8 - Continuity Equation (1) An Application: The Continuity Equation Continuity equation : 0 d dt CV ρ dv CS ρ( V n ) da (11-47) Steady Single flow : out stream, m e m in constant i : (11-48) V1A 1 VA or V1 A1 V A (11-49) -7

9 - Continuity Equation () Conservation of Mass: Incompressible Fluids Steady, Compressible Flow -8

10 - Continuity Equation (3) The amount of mass flowing through a control surface per unit time is called the mass flow rate and is denoted m The dot over a symbol is used to indicate time rate of change. Flow rate across the entire crosssectional area of a pipe or duct is obtained by integration m m V da n c Ac Ac While this expression for is exact, it is not always convenient for m engineering analyses. -9

11 - Continuity Equation (4) Average Velocity and Volume Flow Rate: Integral in ρ and V n m can be replaced with average values of V 1 A avg n c c A c V da For many flows variation of r is very small: Volume flow rate V is given by m V A avg c V V da V A VA A c n c avg c c Note: many textbooks use Q instead of V volume flow rate. Mass and volume flow rates are related by for m V -10

12 - Continuity Equation (5) Conservation of Mass Principle: The conservation of mass principle can be expressed as m in m out dm dt CV -11

13 - Continuity Equation (6) Steady Flow Processes: in m out m For steady flow, the total amount of mass contained in CV is constant. Total amount of mass entering must be equal to total amount of mass leaving in in V A m For incompressible flows, out m V A n n n n out -1

14 - Continuity Equation (7) Divergence Theorem: Divergence theorem allows us to transform a volume integral of the divergence of a vector into an area integral over the surface that defines the volume. -13

15 - Continuity Equation (8) Rewrite conservation of momentum Using divergence theorem, replace area integral with volume integral and collect terms Integral holds for ANY CV, therefore: or -14

16 - Continuity Equation (9) Alternative form -15

17 - Continuity Equation (10) Incompressible flow and = constant -16

18 -3 The Linear Momentum Equation (1) F m a m d V dt d dt (mv ) (11-50) F d ( mv ) dt d dt sys sys VρdV d dt CV ρvdv (11-51) CS ρv ( V n) da (11-5) General : F d dt CV ρ V dv CS ρv ( V n )da (11-53) -17

19 -3 The Linear Momentum Equation () -18

20 -3 The Linear Momentum Equation (3) Stress vector acting on the control surface. -19

21 -3 The Linear Momentum Equation (4) Special Cases Steady flow : One - dimensional F flow : ρv ( V n) da CS (11-54) F d dt CV ρvdv out m e V e - in m i V i (11-55) Steady,one - dimensional flow : F out m e V e - m in i V i (11-56) -0

22 -3 The Linear Momentum Equation (5) Steady,one - dimensional flow (one - inlet,one - exit) : F m (V - V 1 ) (11-57) Along x coordinate : F x m (V,x - V 1, x ) (11-58) -1

23 -3 The Linear Momentum Equation (6) Choosing a Control Volume: CV is arbitrarily chosen by fluid dynamicist, however, selection of CV can either simplify or complicate analysis. Clearly define all boundaries. Analysis is often simplified if CS is normal to flow direction. Clearly identify all fluxes crossing the CS. Clearly identify forces and torques of interest acting on the CV and CS. Fixed, moving, and deforming control volumes. For moving CV, use relative velocity, For deforming CV, use relative velocity all deforming control surfaces, -

24 -3 The Linear Momentum Equation (7) Forces Acting on a CV: Forces acting on CV consist of body forces that act throughout the entire body of the CV (such as gravity, electric, and magnetic forces) and surface forces that act on the control surface (such as pressure and viscous forces, and reaction forces at points of contact). Body forces act on each volumetric portion dv of the CV. Surface forces act on each portion da of the CS. -3

25 -3 The Linear Momentum Equation (8) Body Forces: The most common body force is gravity, which exerts a downward force on every differential element of the CV The different body force Typical convention is that acts in the negative z-direction, Total body force acting on CV -4

26 -3 The Linear Momentum Equation (9) Surface Forces: Surface forces are not as simple to analyze since they include both normal and tangential components Diagonal components xx, yy zz are called normal stresses and are due to pressure and viscous stresses Off-diagonal components xy, xz etc., are called shear stresses and are due solely to viscous stresses Total surface force acting on CS -5

27 -3 The Linear Momentum Equation (10) Body and Surface Forces Surface integrals are cumbersome. Careful selection of CV allows expression of total force in terms of more readily available quantities like weight, pressure, and reaction forces. Goal is to choose CV to expose only the forces to be determined and a minimum number of other forces. -6

28 -3 The Linear Momentum Equation (11) Special Case: Control Volume Moving with Constant Velocity: Momentum Equation for Inertial Control Volume with Rectilinear Acceleration -7

29 -3 The Linear Momentum Equation (1) Divergence theorem Body Force Surface Force -8

30 -3 The Linear Momentum Equation (13) Substituting volume integrals gives, This is Cauchy s Equation -9

31 -3 The Linear Momentum Equation (14) Alternate form of the Cauchy Equation can be derived by introducing Inserting these into Cauchy Equation and rearranging gives or -30

32 -3 The Linear Momentum Equation (15) Unfortunately, this equation is not very useful 10 unknowns Stress tensor, ij : 6 independent components Density Velocity, V : 3 independent components 4 equations (continuity + momentum) 6 more equations required to close problem! -31

33 -3 The Linear Momentum Equation (16) Navier-Stokes Equation First step is to separate ij into pressure and viscous stresses xx xy xz p 0 0 xx xy xz ij yx yy yz 0 p 0 zx zy yx yy yz zz 0 0 p zx zy zz -3

34 -3 The Linear Momentum Equation (17) Reduction in the number of variables is achieved by relating shear stress to strainrate tensor. For Newtonian fluid with constant properties Newtonian fluid includes most common fluids: air, other gases, water, gasoline Newtonian closure is analogous to Hooke s Law for elastic solids -33

35 -3 The Linear Momentum Equation (18) Substituting Newtonian closure into stress tensor gives Using the definition of ij (Chapter 4) -34

36 -3 The Linear Momentum Equation (19) Substituting ij into Cauchy s equation gives the Navier-Stokes equations Incompressible NSE written in vector form This results in a closed system of equations! 4 equations (continuity and momentum equations) 4 unknowns (U, V, W, p) -35

37 -3 The Linear Momentum Equation (0) X-momentum Y-momentum Z-momentum -36

38 -4 The First Law of Thermodynamics (1) Basic Law, and Transport Theorem: -37

39 -4 The First Law of Thermodynamics () General Energy Equation: The energy content of a closed system can be changed by two mechanisms: heat transfer Q and work transfer W. Conservation of energy for a closed system can be expressed in rate form as Q W net, in net, in de Net rate of heat transfer to the system: dt sys Q Q Q net, in in out Net power input to the system: W W W net, in in out -38

40 -4 The First Law of Thermodynamics (3) b=e de Q W ) system edv ev da dt t V e u gz W W W W W shaft normal shear other cv cs -39

41 -4 The First Law of Thermodynamics (4) Where does expression for pressure work come from? When piston moves down ds under the influence of F=PA, the work done on the system is W boundary =PAds. If we divide both sides by dt, we have W PdAV PdA V n pressure For generalized control volumes: ds W W PA PAV dt n pressure boundary piston Note sign conventions: n is outward pointing normal Negative sign ensures that work done is positive when is done on the system. -40

42 -4 The First Law of Thermodynamics (5) (E) Work Involves Shaft Work Work by Shear Stresses at the Control Surface Other Work Recall that P is the flow work, which is the work associated with pushing a fluid into or out of a CV per unit mass. -41

43 -4 The First Law of Thermodynamics (6) As with the mass equation, practical analysis is often facilitated as averages across inlets and exits d P P Qnet, in Wshaft, net, in edv m e m e dt m A C V n da c Since e=u+ke+pe = u+v /+gz CV out in d P V P V Qnet, in Wshaft, net, in edv m u gz m u gz dt out in CV -4

44 -4 The First Law of Thermodynamics (7) Energy Analysis of Steady Flows: V V Qnet, in Wshaft, net, in mh gz mh gz out in For steady flow, time rate of change of the energy content of the CV is zero. This equation states: the net rate of energy transfer to a CV by heat and work transfers during steady flow is equal to the difference between the rates of outgoing and incoming energy flows with mass. -43

45 -4 The First Law of Thermodynamics (8) Energy Analysis of Steady Flows: For single-stream devices, mass flow rate is constant. V V q w h h g z z P V P V w gz gz u u q 1 net, in shaft, net, in shaft, net, in 1 1 net, in 1 P V P V 1 1 gz1 wpump gz wturbine emech, loss 1-44

46 -4 The First Law of Thermodynamics (9) Energy Analysis of Steady Flows: Divide by g to get each term in units of length P V P V 1 1 z1 hpump z hturbine hl 1g g g g Magnitude of each term is now expressed as an equivalent column height of fluid, i.e., Head -45

47 -4 The First Law of Thermodynamics (10) If we neglect piping losses, and have a system without pumps or turbines P V P V 1 1 z1 z 1g g g g This is the Bernoulli equation 3 terms correspond to: Static, dynamic, and hydrostatic head (or pressure). Limitations on the use of the Bernoulli Equation Steady flow: d/dt = 0 Frictionless flow No shaft work: w pump =w turbine =0 Incompressible flow: ρ = constant No heat transfer: q net,in =0 Applied along a streamline (except for irrotational flow) -46

48 -4 The First Law of Thermodynamics (11) Right hand side of (E) t CV V e dv ( h gz)v dv CV V { ( e) ( h t CV V { [ ( h t CV gz)] P t gz) V } dv V ( h gz) V } dv (E.1) Note: e u V gz h V P gz h P V gz -47

49 -4 The First Law of Thermodynamics (1) In rectangular coordinates, the heat conduction vector can be expressed in terms of its components as Qn Qxi Qy j Qzk which can be determined from Fourier s law as T Qx kax x T Qy kay y T Qz kaz z -48

50 -4 The First Law of Thermodynamics (13) Left hand side of (E)--- Q Rate of heat conduction at x - Rate of heat conduction at x+dx + Rate of heat generation inside the element Q Q xdx x Q Q x x D x Q x Q x x E gen, element Q Dx! x 1 x ( Dx)... Q x T T Q xdx Dx ( ka ) Dx ( ka ) Dx x x x x x T ( k ) DxDyDz Neglect higher orders x x -49

51 E dv A x element DyDz egen, gen, element Q -4 The First Law of Thermodynamics (14) A x x Qx D x element dv element Dx DyDzDx DxDyDz e gen dv DxDyDz T, elementdxdydz [ ( k ) e gen, x x element ] dv 3D Heat Conduction Qx Qy Qz Qx Dx QyD y Qz Dz [ ( kt ) e, gen element ] dv E gen, element -50

52 -4 The First Law of Thermodynamics (15) (E): shear W Rate of work=force x velocity dxdydz w u z w u w u x zz zy zx yz yy yx xz xy xx ) ( ) ( ) ( v v y v } 3 { z w y x u z u x w y w z x y u z w y x u v v v v is viscous dissipation and it is important in high speed flow and for very viscous fluids. -51

53 -4 The First Law of Thermodynamics (16) Energy Equation in Differential Form: From (E) E t ( EV ) E h V gz ( kt ) P t S E S E : source term -5

54 -5 General Equation in Conservative Form (1) The General Differential Equation The differential equation obeying the generalized conservation principle can be written by the general differential equation as v s (1) t :dependent variable, such as velocity components (u,v,w), h or T, k, ε concentration, etc. : diffusion coefficients s : source term The four terms of eq.(1) are the unsteady term, the convection term, the diffusion term and the source term. -53

55 -5 General Equation in Conservative Form () Conservation form of the governing equations of fluid flow Continuity : v t v Mometum : t h Energy : t c Species : t 0 vv p v vh kt S vc DC SC T S M -54

56 -6 General Equation in Non-Conservative Form -55

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