Lecture 4. Differential Analysis of Fluid Flow Navier-Stockes equation

Transcription

1 Lecture 4 Differential Analysis of Fluid Flow Navier-Stockes equation

2 Newton second law and conservation of momentum & momentum-of-momentum A jet of fluid deflected by an object puts a force on the object. This force is the result of the change of momentum of the fluid and can happen even though the speed (magnitude of velocity) remains constant.

3 Newton second law and conservation of momentum & momentum-of-momentum In an inertial coordinate system: D Dt VρdV = Fsys sys Rate of change of the momentum of the system On the other hand: D Dt Sum of all external forces acting on the system At a moment when system coincide with control volume: F sys = F contentsof thecontrol volume VρdV = VρdV + VρV n da t sys CV CS t CV VρdV + VρV n da= F CS contents of thecontrol volume

4 Example: Linear momentum Determine anchoring forces required to keep the vane stationary vs angle Q. Neglect gravity and viscosity. A=0.06 m 2 V 1 =10 m/s t t CV CV uρdv + ρuvn da= F CS wρdv + ρwvn da= F CS x y Vρ( V ) A + V cos θρ( V ) A = F ρ( V ) A + V sin θρ( V ) A = F F = V ρ A(1 cos θ ) Ax Az F = V ρa sinθ Az Ax

5 Linear momentum: comments Linear momentum is a vector As normal vector points outwards, momentum flow inside a CV involves negative V n product and moment flow outside of a CV involves a positive V n product. The time rate of change of the linear momentum of the contents of a nondeforming CV is zero for steady flow Forces due to atmospheric pressure on the CV may need to be considered

6 Example: Linear momentum taking into account weight, pressure and change in speed Determine the anchoring force required to hold in place a conical nozzle attached to the end of the laboratorial sink facet. The water flow rate is 0.6 l/s, nozzle mass 0.1kg. The pressure at the section (1) is 464 kpa.

7 Example: Linear momentum taking into account weight, pressure and change in speed t CV wρdv + wρ da= Vn CS = F W p A W + p A A n 1 1 w 2 2 volume of a truncated cone: 1 Vw = π h D + D + DD ( ) pressure distribution:

8 Moment-of-Momentum Equation The net rate of flow of moment-of-momentum through a control surface equals the net torque acting on the contents of the control volume. Water enters the rotating arm of a lawn sprinkler along the axis of rotation with no angular momentum about the axis. Thus, with negligible frictional torque on the rotating arm, the absolute velocity of the water exiting at the end of the arm must be in the radial direction (i.e., with zero angular momentum also). Since the sprinkler arms are angled "backwards", the arms must therefore rotate so that the circumferential velocity of the exit nozzle (radius times angular velocity) equals the oppositely directed circumferential water velocity.

9 D Dt sys The Energy Equation The First Law of thermodynamics eρdv = ( Q + W ) sys Rate of increase of the total stored energy of the system Net rate of energy addition by heat transfer into the system Net rate of energy addition by work transfer into the system Total stored energy per unit mass: 2 V e= u+ + gz 2

10 D Dt sys eρdv = t CV eρdv + CS eρv n da Rate of increase of the total stored energy of the system rate of increase of the total stored energy of the control volume Net rate of energy flow out of the control volume t CV eρdv + eρ da= ( Q + W ) CS net net cv in in Vn

11 Power transfer due to normal and tangential stress Work transfer rate (i.e. power) can be transferred through a rotating shaft (e.g. turbines, propellers etc) W = T w where T shaft torque and w angular velocity shaft shaft or through the work of normal stress on a single particle: integrating: δw = δf V = σnδa V = pv nδa normal stress normal stress W normal = pv nda stress CS δw = δf V = 0 tangential stress: tangential tangential stress stress

12 Power transfer due to normal and tangential stress The first law of themrodynamics can be expressed now as: t eρdv + eρvn da= Q + W pv nda net shaft CV CS in net in CS so, we can obtain the energy equation 2 p V eρdv + u gz ρvn da= Q net + W t ρ 2 CV CS in shaft net in

13 Application of energy equation Let s consider a steady (in the mean, still can be cyclical) flow and take a one stream Product V n is non-zero only where liquid crosses the CS; if we have only one stream entering and leaving control 2 p V volume: ρ V ρvn e d + u gz da= Q net + W t ρ 2 CV CS in p V p V p V ρvn = out ρ 2 ρ 2 ρ 2 CS out in u gz da u gz m u gz m 2 2 p p Vout V in m uout uin g( zout zin ) = Q net + W ρ ρ 2 out in in shaft net in in shaft net in h p = u+ ρ

14 Example: Temperature change at a water fall find the temperature change after a water fall, c water =4.19 kj/kg K p p V V m u u g z z Q 2 2 out in out in ( out in ) = net ρ ρ 2 out in in 150m =0 =0 =0

15 Energy equation vs Bernoulli equation Let s return to our one-stream volume, steady flow (also no shaft power) p p V V m u u g z z Q 2 2 out in out in ( out in ) = net ρ ρ 2 out in in 2 2 pout Vout pin Vin + + gzout = + + gzin ( uout uin qnet ), qnet = ρ 2 ρ 2 in in Q net in m available energy loss Comparing with Bernoulli equation: uout uin qnet = 0 i.e. steady incompressible flow should be also frictionless in

16 Energy transfer Work must be done on the device shown to turn it over because the system gains potential energy as the heavy (dark) liquid is raised above the light (clear) liquid. This potential energy is converted into kinetic energy which is either dissipated due to friction as the fluid flows down the ramp or is converted into power by the turbine and then dissipated by friction. The fluid finally becomes stationary again. The initial work done in turning it over eventually results in a very slight increase in the system temperature

17 Second law of thermodynamics Let s apply stream line energy equation to an infinitesimally thin volume 2 p V m du + d + d + gdz = δq ρ 2 net in For closed system in the absence of additional work: du = TdS pdv 2 dp V + d + gdz = ( Tds δ q net ) ρ 2 in If we take into account Clausius inequality: 1 du = Tds pd( ) ρ ds dq 0 T 2 dp V + d + gdz ρ 2 0

18 An axial-flow ventilating fan is driven by the 0.4kW motor and producing 12m/s speed in a 0.6m diameter channel. Determine the useful effect and efficiency. Air density is 1.23 kg/m 3. Example: Fan Efficiency 2 2 pout V out pin V in wshaft loss= + + gzout + + gzin net in ρ 2 ρ 2 w shaft net in = W shaft net in m

19 Differential analysis of Fluid Flow The aim: to produce differential equation describing the motion of fluid in detail

20 Fluid Element Kinematics Any fluid element motion can be represented as consisting of translation, linear deformation, rotation and angular deformation

21 Velocity and acceleration field Material derivative Dt D z w y v x u t t r V V V V V = = ), a( V ) + = = ( t z w y v x u t Dt D Acceleration Velocity field ˆ ˆ ˆ u v w = + + V i j k

22 Linear motion and deformation Let s consider stretching of a fluid element under velocity gradient in one direction Volumetric dilatation rate: u δxδtδyδz 1 d( δ V ) x u = lim = δv dt δt 0 δxδyδzδt x 1 d( δ V ) u v w = + + = V δ V dt x y z

23 Angular motion and deformation Fluid elements located in a moving fluid move with the fluid and generally undergo a change in shape (angular deformation). A small rectangular fluid element is located in the space between concentric cylinders. The inner wall is fixed. As the outer wall moves, the fluid element undergoes an angular deformation. The rate at which the corner angles change (rate of angular deformation) is related to the shear stress causing the deformation

24 Angular motion and deformation ω ω OA OB δα ( v/ x) δxδt 1 v t x t x = lim = δt 0 δ δ δ u = y Rotation is defined as the average of those velocities: ω z 1 v = 2 x u y

25 Angular motion and deformation 1 v u 1 w v 1 u w ωz = ωx = ωy = 2 x y 2 y z 2 z x iˆ ˆj kˆ 1 1 ω = V= = 2 2 x y z u v w 1 w v ˆ 1 u w ˆ 1 v u = ˆ i + j + k 2 y z 2 z x 2 x y Vorticity is defined as twice the rotation vector ζ = 2ω = V If rotation (and vorticity) is zero flow is called irrotational

26 Angular motion and deformation Rate of shearing strain (or rate of angular deformation) can be defined as sum of fluid element rotations: v u γ = + x y

27 Conservation of mass DM Sys As we found before: = ρdv + ρ da= 0 Dt t CV Vn CV t ρ ρdv = δxδyδz Flow of mass in x-direction: t CV ( ρu) δ x δ y δ z x ρ ( ρu) ( ρv) ( ρw) ρ = 0 + ρv = 0 t x y z t

28 Conservation of mass Incompressible flow u v w V = = 0 x y z Flow in cylindrical coordinates ρ 1 ( rρvr) 1 ( ρvθ ) ( ρvz) = 0 t r r r θ z Incompressible flow in cylindrical coordinates 1 ( rvr) 1 ( vθ ) ( vz) + + = r r r θ z 0

29 Stream function 2D incompressible flow u x v + = 0 y We can define a scalar function such that u ψ = v= y ψ x Stream function Lines along which stream function is const are stream lines: dy v = dx u ψ ψ Indeed: dψ = dx + dy = vdx + udy = x y 0

30 Stream function Flow between streamlines dq = udy vdx ψ ψ dq = dy + dx = dψ y x q = ψ 2 ψ 1

31 Description of forces Body forces distributed through the element, e.g. Gravity δ F = δ mg Normal stress Forces Surface forces result of interaction with the surrounding elements: e.g. Stress Shearing stresses Linear forces: Surface tension

32 Stress acting on a fluidic element normal stress σ n = lim δ A 0 δ F n δ A shearing stresses δ F τ = lim τ = δ A lim δ F δ A δa 0 δa 0

33 Stresses: double subscript notation normal stress: σ xx τ shearing stress: xy τ xz normal to the plane direction of stress sign convention: positive stress is directed in positive axis directions if surface normal is pointing in the positive direction

34 Stress tensor To define stress at a point we need to define stress vector for all 3 perpendicular planes passing through the point σ τ τ τ τ σ τ τ τ σ =

35 Force on a fluid element To find force in each direction we need to sum all forces (normal and shearing) acting in the same direction σ τ xx yx τ zx δ Fx = ( + + ) δxδyδz x y z

36 Problems 6.2 A certain flow field is given by equation: 2 V = (3x + 1) i 6xyj Determine expression for local and convective components of the acceleration in x and y directions Water flows steadily down the inclined pump. Determine: The pressure difference, p 1 -p 2 ; The loss between sections 1 and 2 The axial force exerted on the pipe by water

37 Problems 6.4 The components of the velocity in the flow field are given by: u = x + y + z v= xy+ yz+ z 2 w= 3xz z 2+ 4 Determine the volumetric dilation rate and interpret the results. Determine the expression for the rotation vector. Is the flow irrotational? 6.22 The stream function for an incompressible flow ψ = 2 3x y y 2 3 sketch the streamline passing through the origin; determine of flow across the strait path AB