Apply mass and momentum conservation to a differential control volume. Simple classical solutions of NS equations

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1 Module 5: Navier-Stokes Equations: Appl mass and momentum conservation to a differential control volume Derive the Navier-Stokes equations Simple classical solutions of NS equations Use dimensional analsis to simplif the NS equations Introduce: inviscid flows, irotational flows and potential flows

2 Mass Conservation 0 = dv V n t CV CS V = ( u, v, w) Mass flu in direction da Term 1: CV dv = t dddz t Term : Sum of flues through all 6 faces. For Eample, mass flu through left face: (u) ddz mass flu through right face: ( u) ( u) ( u) Net mass flu in direction is (u) dddz ddz d (u)

3 Mass Conservation, contd. Similarl, net flu in direction: and z direction: dddz z w z ) ( dddz v ) ( da dv t CS CV = n V 0 dddz z w dddz v dddz u dddz t z = ) ( ) ( ) ( 0 z w v u t = ) ( ) ( ) ( 0 Or in vector form ( ) 0 = V t Continuit equation

4 Mass conservation - Continuit Chain rule... t 1 D V Dt ( V) = = t 0 V V = 0 Stead Flow? Incompressible flow?

5 Mass conservation - Continuit

6 Conservation of momentum F = n t CV Consider first left-hand side: ( V) dv V( V ) da CS F Fbod FPressure = F F bod = g dv = CV m CV Viscous Bod force: acts upon entire mass within element (gravit, magnetism, etc) g = zg Surface force: acts onl on the sides of control surface p σ ij = p 0 pressure 0 τ 0 τ p τ z τ τ τ z τ z τ z τ zz Viscous stress Pressure forces: Viscous forces

7 Conservation of momentum contd. on left surface, the pressure force is p ddz And on the right surface p p ddz = p d So the net force in direction p dddz Similarl in direction p dddz and in zz direction p dddz z ddz

8 Conservation of momentum contd. Viscous stresses: in direction: τ τ ddz τ τ τ τ = τ τ ddz = τ z τ z τ z z z z z dd z z z z z dd = z τ = τ τ z z [ τ τ ] ddz = τ τ ddz = dddz [ ] dddz [ ] dddz

9 Conservation of momentum contd. Net forces in direction Similarl in direction and in zz direction τ τ τ z τ dddz τ dddz τ dddz z τ dddz z z τ dddz z z τ dddz z zz dddz dddz dddz So far F Fbod FPressure = F Viscous ( F) ( F) = g = g p τ p τ p τ z z τ zz ( F) = g z dddz z τ τ τ τ z z z τ z z dddz dddz

10 Conservation of momentum F = V dv V V n t CV CS Now lets look at right-hand side: First term: ( V) dv = ( V)dddz t t Second term of right-hand side: Momentum in direction -faces: ( uu) ( uu) -faces: CV [ ] ddz = ( uu) ( uu) ( uu) uv uv uv ddz = [( ) ( ) ] ( ) dddz z-faces: [( uw) ( uw) ] z z ( ) ( ) da z dd = ( uw) dddz z ddz = ( uu) dddz

11 Conservation of momentum contd. Putting it all together, for the -component... And similarl in and zz direction ( ) ( ) ( ) ( ) dddz z uw uv uu t u dddz p g z = τ τ τ ( ) ( ) CS CV da dv t = n V V V F ( ) ( ) ( ) ( ) z g z p z uw uv uu t u τ τ τ = ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) z zz z z z g z z p z ww wv uw t w g z p z vw uv uv t v τ τ τ τ τ τ = =

12 Constitutive relationship for a Newtonian fluid Not finished et... Need to epress shear stress in terms of velocit... How? This relationship is attributed to Navier & Stokes. Navier apparentl made some mistakes and in fact Stokes onl considerd slow flows Possibl, Poisson & St-Venant also have as strong a claim... but... The basic momentum equation is due to Cauch (man ears earlier) but without a closure for the stresses is not much use Cauch s equation is also the basis for solid mechanics/dnamics, but we don t use the material derivative

13 Incompressible flow in Cartesian Coordinates Continuit -momentum -momentum z-momentum See tet for equations in clindrical and spherical coordinates g z u u u p z u w u v u u t u µ = g z v v v p z v w v v v u t v µ = g z z w w w z p z w w w v w u t w µ = = 0 z w v u

14 Special cases Vector notation for incompressible Newtonian fluid t V t Stead Flow [ V] = 0 [ ] V V = P g µ V Incompressible

15 Boundar conditions Boundar conditions (BC s) are critical to eact, approimate, and computational solutions. BC s used in analtical solutions are discussed here No-slip boundar condition Interface boundar condition These are used in CFD as well, plus there are some BC s which arise due to specific issues in CFD modeling. Inflow and outflow boundar conditions Smmetr and periodic boundar conditions

16 No-slip boundar condition For a fluid in contact with a solid wall, the velocit of the fluid must equal that of the wall

17 Interface boundar conditions When two fluids meet at an interface, the velocit and shear stress balance: Normal stresses differ b an amount proportional to surface tension & inversel proportional to curvature. Negligible surface tension or surface nearl flat:

18 Interface boundar condition Degenerate case of the interface BC occurs at the free surface of a liquid. Same conditions hold Since µ air << µ water, As with general interfaces, if surface tension effects are negligible or the surface is nearl flat P water = P air

19 What we covered: Derived Navier-Stokes & mass conservation equations: Control volume approach, applied to differential volume of fluid t V t [ V] = [ V ] V = P g µ V Special cases, e.g. stead flows, incompressible 0 Discussed different boundar conditions and interfacial conditions

20 Clasical Solutions of NS equations There are about 80 known eact solutions to the NSE The can be classified as: Linear solutions, where the convective term is zero: No acceleration = Stokes flows Nonlinear solutions where convective term is not zero Usuall some other terms are zero, e.g. inviscid flows Solutions also classified b tpe or geometr, e.g. 1. Couette flows. Stead duct/pipe flows 3. Unstead duct/pipe flows 4. Flows with moving boundaries 5. Similarit solutions 6. Asmptotic solutions 7. Ekman flows

21 Eact Solutions of the NS equations Procedure for solving NS equations 1. Set up the problem and geometr, identifing all relevant dimensions and parameters. List all appropriate assumptions, approimations, simplifications, and boundar conditions 3. Simplif the differential equations as much as possible 4. Integrate the equations 5. Appl BC s to solve for constants of integration 6. Verif results

22 Eample 1: Full Developed Couette Flow For the given geometr and BC s, calculate the velocit and pressure fields, and estimate the shear force per unit area acting on the bottom plate Step 1: Geometr, dimensions, and properties

23 Eample: Full Developed Couette Flow Step : Assumptions and BC s Assumptions 1. Plates are infinite in and z. Flow is stead, /t = 0 3. Parallel flow, v=0 4. Incompressible, Newtonian, laminar, constant properties 5. No pressure gradient 6. D, w=0, /z = 0 7. Gravit acts in the -z direction, Boundar conditions 1. Bottom plate (=0) : u=0, v=0, w=0. Top plate (=h) : u=v, v=0, w=0

24 Eample: Full Developed Couette Flow Step 3: Simplif Continuit 3 6 Note: these numbers refer to the assumptions on the previous slide X-momentum This means the flow is full developed or not changing in the direction of flow Cont Cont. 6 U = 0

25 Eample: Full Developed Couette Flow Step 3: Simplif, cont. Y-momentum, , Z-momentum,

26 Eample: Full Developed Couette Flow Step 4: Integrate X-momentum integrate integrate Z-momentum integrate

Eample: Full Developed Couette Flow Step 5: Appl BC s @ = 0, uu = 0 CC = 0 @ = h, uu = VV CC 1 = VV h This gives For pressure, no eplicit BC, therefore CC 3 can remain an arbitrar constant (recall

27 Eample: Full Developed Couette Flow Step 5: Appl BC = 0, uu = 0 CC = = h, uu = VV CC 1 = VV h This gives For pressure, no eplicit BC, therefore CC 3 can remain an arbitrar constant (recall onl pp appears in NSE). Let pp = pp 0 at zz = 0 (CC 3 renamed pp 0 ) 1. Hdrostatic pressure. Pressure acts independentl of flow Step 6: Verif solution b back-substituting into differential equations

28 Eample: Full Developed Couette Flow Finall, calculate shear force on bottom plate Shear force per unit area acting on the wall Note that τ w is equal and opposite to the shear stress acting on the fluid τ (Newton s third law).

29 Eample : Full Developed Planar Poiseuille Flow Find velocit profile of full developed flow of a fluid moving between two fied plate due to a constant pressure gradient GG in direction. Assumptions: 1) stead, incompressible flow ) Newtonian 3) infinitel large plates in z direction (i.e. D flow) 4) full developed flow 5)

30 Eample : Full Developed Planar Poiseuille Flow Continuit: = 0 ffuuuuuuuu dddddddddddddddddd = 0 vv = 0 -mom: -mom: uu vv = μμ uu uu GG μμ uu = 0 uu = GG μμ CC 1 CC uu vv = μμ vv vv = 0, = ±h, uu = 0 CC 1 = 0, and CC = GG μμ h uu = GG μμ h 1 h = uu mmmmmm 1 h

31 Eample 3: Poiseuille Flow with a moving plate Consider Poiseuille problem ecept that the top plate is moving with the speed U. Derive the velocit profile. Everthing remains unchanged, ecept the boundar = 0, uu = 0 = bb, uu = UU uu = GG μμ bb bb bb

32 Eample 4: Rotational Couette Flow The area between two ver long clinder is filled with a viscous fluid. The inner clinder rotates at the angular speed of ωω, while the outer clinder is stationar. Find velocit profile of flow in annulus. Assumptions: 1) stead, incompressible flow ) Newtonian 3) Parallel flow (in a polar sense) 4) Use polar coordinate 5) Long clinders, so no aial flow

33 Eample 4: Rotational Couette Flow Continuit: 1 r θθ-momentum: = 1 rr rruu rr 1 rr uu θθ uu uu θθ rr 1 rr uu θθ = 0, no radial flow uu θθ = 0 uu θθ = uu θθ rr uu θθ uu rr rr gg θθ μμ 1 rr rr uu θθ uu θθ uu rruu θθ rr rrrruu θθ uu θθ rr 1 uu θθ rr θθ uu rr rr uu θθ rr = 0 uu θθ θθ 1 uu θθ rr uu θθ rr = 0 Euler equation: let uu θθ = rr kk rr kk kk kk 1 kk 1 = 0 kk = ±1 uu θθ = CC 1 rr CC rr = rr ii, uu θθ = rr ii ωω = rr oo, uu θθ = 0 uu θθ = ωωrr ii rr oo rr rr rr oo rr ii

34 Eample 5: Stokes first problem Consider a stationar flat plate beneath a semi-infinite laer of initiall quiescent incompressible fluid. At tt = 0, suddenl the plate is pulled with the speed U. initiall stationar fluid Assumptions: 1) unstead, incompressible flow ) Newtonian 3) infinitel large in and positive direction 4) Parallel flow 5) No eternal pressure gradient At tt = 0, the plate is pulled with the speed U.

35 Eample 5: Stokes first problem Continuit: = 0 parallel flow vv = 0 -mom: uu vv = μμ uu uu = μμ uu or uu tt = ννuu = 0, uu = UU and tt = 0, uu = 0 everwhere 4νννν Solution is b change of variable. ηη = (How did we know this change of variable works?) uu tt = uu ηη ηη tt = 4ννννtt uu ηη, and uu = 1 4ννν uu ηηηη uu ηηηη = ηηuu ηη uu ηη = CC 1 ep ηη uu = CC 1 0 ηη ep dddd CC ηη = 0, uu = UU CC = UU ηη =, uu = 0 CC 1 uu = UU 1 ππ 0 ηη 0 ee dddd ep dddd UU = 0 CC 1 = UU/ ππ

36 What we covered: Stud classical eamples of Navier-Stokes equation Poiseuille flow Couette flow Miture of Poiseuille and Couette flow Stokes First problem

37 Dimensional analsis version.0 Scale the fundamental equations: To determine dimensionless variables and dependencies In order to simplif the equations in a rational wa Eample 1: Bernoulli s equation {p} = {force/area}={mass length/time 1/length } = {m/(t L)} {1/V } = {mass/length 3 (length/time) } = {m/(t L)} {gz} = {mass/length 3 length/time length} ={m/(t L)} Non-dimensionalize all variables with scaling parameters

38 Non dimensional equations: Back-substitute p,, V, g, z into dimensional equation: Divide b 0 U 0 and set * = 1 (assume incompressible) Since g* = 1/Fr, where we can write this as: Flow depends on dimensionless parameters

39 Eample : Scale the incompressible Navier- Stokes equations appropriatel for flow of water along a vertical square duct of side D, driven b a stead flow rate Q. How man dimensionless groups govern the velocit field? Conservation of mass U = 0 Conservation of momentum U t [ U ] U = P g µ U

40 Eample 3: Slider bearing Use scaling to simplif the Navier- Stokes equations for stead flow in the slider bearing, as illustrated Hence derive an epression for the force (per unit depth) eerted on the bearing P atm L =h() =0 =L U = 0 P atm h() <<L U 0 [ U ] U = P g µ U

42 Eample 4: Boundar laer flow Fluid flows in the -direction past a semi-infinite flat plate positioned along the -ais for >0. The far-field speed is U. Use scaling arguments to simplif the Navier-Stokes equations in the boundar laer separating the plate from the far-field

44 Inviscid flows Consider the scaled sstem of Navier-Stokes equations Re represents balance of inertial to viscous stresses Re represents inviscid limit We stud onl incompressible inviscid flows Re-derive Bernoulli's equation Define vorticit and irrotational flows Potential flows

45 Incompressible inviscid flows: Approimation to flow of e.g. air & water, far from solid surfaces: Euler equations: Vector form: General vector identit: Align z with gravit: Euler s eqns in vector form:

46 Relation to Bernoulli s equation? Dot product with ds, along streamline: Note that: Differential along streamline: Integrate to get: Restrictions: incompressible, inviscid, stead, along streamline If VV = 0, then we can write Bernoulli's equation between an two points (not necessaril on one streamline) So what is special about VV = 0?

47 Irrotational Flow We define ωω = VV and call it vorticit. Vorticit is a measure of local rotation of fluid particles. If ωω = 0, the flow is called irrotational. For incompressible irrotational flows, we can define VV as the gradient of a potential function VV = (This automaticall satisfies ωω = 0. wh?) Together with continuit, Φ = 0 Laplace s equation, which has man analtical solutions. So VV = uu = Φ, and vv = Φ. Also remember, uu = Ψ Ψ, and vv =. It can be easil shown that lines of iso-potential are normal to stream lines.

48 Some basic potential flows (D) 1) Uniform Flow uu = UU, vv = 0 Φ uu = UU, vv = 0 Ψ ) Source/Sink uu rr = mm ππππ, uu θθ = 0 Φ rr = mm uu rr = mm ππππ, uu θθ = 0 1 rr = UU, Φ = 0 Φ = UUUU = UU, Ψ = 0 Ψ = UUUU Φ, 1 = 0 Φ = mm ln rr ππππ rr θθ ππ Ψ = mm, Ψ = 0 Ψ = mm θθ θθ ππππ rr ππ m characterises the strength of the source and ππ is onl there conventionall. 3) Free Vorte uu θθ = Γ ππππ, uu rr = 0 Φ = Γ ππ Γ θθ, aaaaaa Ψ = ln rr ππ Γ characterises the strength of vorte and ππ is onl there conventionall. Since Laplace equation is linear, we can superimpose these solutions to create comple flows

49 Eample 1: Superimposing a source and a sink Consider a source and sink of equal strength mm locating at a distance aa as shown. Find stream function. Simplif our answer when aa is small. Ψ = Ψ ssssssssssss Ψ ssssssss = mm ππ θθ 1 θθ which can be written (test our trigonometr and prove this) Ψ = mm aaaa sin θθ ππ tan 1 rr aa We know tan 1 εε εε when εε 1, so when aa 1 Ψ mm aaaa sin θθ ππ rr aa = mmmm rr sin θθ ππ rr aa

50 Eample : Doublet In the previous eample, let aa 0, while we increase mm, in such a wa that K = mmmm is kept constant. This flow is ππ called a doublet. Find stream and potential functions of a doublet. rr sin θθ Ψ = K rr aa Let aa 0 KK sin θθ Ψ = rr We can similarl derive KK cos θθ Φ = rr

51 Eample 3: Flow around a fied clinder Superimpose a uniform flow and doublet. Design this configuration such that ou have a flow around a fied clinder of radius aa ΨΨ = ΨΨ uniform ΨΨ doublet = UUUU KK sin θθ rr = UUUU sin θθ KK sin θθ rr = rr sin θθ UU KK rr If this is to be flow around a fied clinder, on the surface of clinder, we must have zero radial = aa, uu rr = 0 Ψ = 0 UU KK = 0 KK = UUaa aa So Ψ = UUUU sin θθ 1 aa rr Potential flow around a clinder

52 Eample 3: Flow around a fied clinder contd. How does pressure changes in the previous eample. Using the epression for pressure find the drag and lift forces: Appling Bernoulli s equation: pp ss 1 uu θθss = pp 1 UU pp ss pp = 1 UU Lift = FF = 0 ππ ppss sin θθ aaaaa = 0 (because of smmetr) Drag = FF = 0 ππ ppss cos θθ aadθθ = 0 (d Alembert s parado) 1 4 sin θθ

ΨΨ = ΨΨ clinder ΨΨ vorte = UUUU sin θθ 1 aa rr Find tangential velocit at rr = aa uu θθθθ = Ψ mm = UU sin θθ

53 Eample 4: Flow around a rotating clinder Let s add a free vorte to the stream function for the flow around a clinder. ΨΨ = ΨΨ clinder ΨΨ vorte = UUUU sin θθ 1 aa rr Find tangential velocit at rr = aa uu θθθθ = Ψ mm = UU sin θθ ππππ uu rrrr = 1 Ψ rr = 0 mm ln rr ππ This resembles potential flow around a rotating clinder. At θθ = sin 1 mm, flow is stationar. We call this stagnation point. 4ππππππ mm = 0 mm 4ππππππ < 1 mm 4ππππππ = 1 mm 4ππππππ > 1

54 Eample 4: Flow around a rotating clinder contd Let s find Pressure pp ss pp = 1 UU 1 4 sin θθ Now if we calculate Drag and Lift mm sin θθ ππππππ mm 4ππ aa UU Drag = FF = 0 ππ (ppss pp ) cos θθ aaaaa = 0 and Lift = FF = (pp ss pp ) sin θθ aaaaa = A counter clockwise rotation create a downward lift. The development of this lift is called Magnus effect.

Chapter 9: Differential Analysis

Chapter 9: Differential Analysis 9-1 Introduction 9-2 Conservation of Mass 9-3 The Stream Function 9-4 Conservation of Linear Momentum 9-5 Navier Stokes Equation 9-6 Differential Analysis Problems Recall 9-1 Introduction (1) Chap 5: Control