DIFFERENTIAL ANALYSIS OF FLUID FLOW

Transcription

1 DIFFERENTIAL ANALYSIS OF FLUID FLOW A: Mathematical Fomulation (4.1.1, 4., ) B: Inviscid Flow: Eule Equation/Some Basic, Plane Potential Flows ( ) C: Viscous Flow: Navie-Stokes Equation ( )

2 Diffeential Analysis Intoduction Thee ae situations in which the details of the flow ae impotant, e.g., pessue and shea stess vaiation along the wing. Theefoe, we need to develop elationship that apply at a point o at least in a vey small egion (infinitesimal volume) with a given flow field. This appoach is commonly efeed to as diffeential analysis. The solutions of the equations ae athe difficult. Computational Fluid Dynamic (CFD) can be applied to complex flow poblems.

3 PART A Mathematical Fomulation (Sections 4.1.1, 4., )

4 Fluid Kinematics (4.1.1, 4.) Kinematics involves position, velocity and acceleation, not foces. kinematics of the motion: velocity and acceleation of the fluid, and the desciption and visualization of its motion. The analysis of the specific foce necessay to poduce the motion - the dynamics of the motion.

5 4.1 The Velocity Field A field epesentation epesentations of fluid paametes as functions of spatial coodinate the velocity field V = u ( x, y, z, t) i + v( x, y, z, t) j + w( x, y, z, t)k d dt A uu = V A ( x, y, z t) V = V, ( ) 1 u + v w V = V = + A change in velocity esults in an acceleation which may be due to a change in speed and/o diection.

6 4.1.1 Euleian and Lagangian Flow Desciptions Euleian method: the fluid motion is given by completely pescibing the necessay popeties as functions of space and time. Fom this method, we obtain infomation about the flow in tems of what happens at fixed points in space as the fluid flows past those points. Lagangian method: following individual fluid paticles as they move about and detemining how the fluid popeties associated with these paticles change as a function of time. V4.3 Cylinde-velocity vectos V4.4 Follow the paticles V4.5 Follow the paticles

7 4.1.4 Steamlines, Steaklines and Pathlines A steamline is a line that is eveywhee tangent to the velocity field. A steakline consists of all paticles in a flow that have peviously passed though a common point. A pathline is a line taced out by a given flowing paticle. V4.9 steamlines V4.10 steaklines V4.1 steaklines

8 4.1.4 Steamlines, Steaklines and Pathlines Fo steady flows, steamlines, steaklines and pathlines all coincide. This is not tue fo unsteady flows. Unsteady steamlines ae difficult to geneate expeimentally, but easy to daw in numeical computation. On the contay, steaklines ae moe of a lab tool than an analytical tool. How can you detemine the unsteady pathline of a moving paticle?

9 4. The Acceleation Field The acceleation of a paticle is the time ate change of its velocity. Fo unsteady flows the velocity at a given point in space may vay with time, giving ise to a potion of the fluid acceleation. In addition, a fluid paticle may expeience an acceleation because its velocity changes as it flows fom one point to anothe in space.

10 4..1 The Mateial Deivative Conside a paticle moving along its pathline uu uu u uu V (, ) (), (), () A = VA A t = VA xa t ya t za t, t

11 The Mateial Deivative Thus the acceleation of paticle A, uu uu u uu V ( ) A = VA A, t = VA xa( t), ya( t), za( t), t uu uu uu uu uu dva VA VA dxa VA dya aa () t = = + + dt t x dt y dt uu VA dza + z dt uu uu uu uu VA VA VA VA = + ua + va + wa t x y z

12 Acceleation This is valid fo any paticle u u u u V V V V a= + u + v + w t x y z u u u u ax = + u + v + w t x y z v v v v ay = + u + v + w t x y z w w w w az = + u + v + w t x y z

13 Mateial deivative Acceleation: Associated with time vaiation u u u u u u DV DV V V V V a =, = + u + v + w Dt Dt t x y z Total deivative, mateial deivative o substantial deivative ( ) ( ) ( ) ( ) ( ) D = + u + v + w Dt t x y z ( ) = + ( V )( ) t Associated with space vaiation

14 Mateial deivative The mateial deivative of any vaiable is the ate at which that vaiable changes with time fo a given paticle (as seen by one moving along with the fluid the Lagangian desciptions) If velocity is known, the time ate change of tempeatue can be expessed as, DT T T T T = + u + v + w Dt t x y z T = + ( V ) T t Example: the tempeatue of a passenge expeienced on a tain stating fom Taipei on 9am and aiving at Kaohsiung on 1.

15 Acceleation along a steamline u u u 3 R V V u u V = u( x) i = V0 1 + i, a = + u = + u i 3 x t x t x 3 R 3 4 a = Vo(1 + ) V [ ( 3 )] 3 o R x i x

16 4.. Unsteady Effects Fo steady flow ( )/ t 0, thee is no change in flow paametes at a fixed point in space. Fo unsteady flow ( )/ t 0. spatial (convective) deivative DT T v DT T = + V T (fo an unstied cup of coffee < 0) Dt t Dt t time (local) deivative v v DV V v v = + V V Dt t local acceleation V4.1 Unsteady flow

17 4..3 Convective Effects DT T v = + V T Dt t DT T = 0 + us Dt s Tout T =0+ us Δs in

18 4..3 Convective Effects convective acceleation v v DV V v v = + V V Dt t local acceleation Du Dt u = 0 + u x

19 4..4 Steamline Coodinates In many flow situations it is convenient to use a coodinate system defined in tems of the steamlines of the flow. v v V = Vs v v v DV DV v Ds a = = s + V Dt Dt Dt V V ds V dn v = + + s t s dt n dt v v v s s ds s dn + V + + t s dt n dt V4.13 Steamline coodinates

20 4..4 Steamline Coodinates Steady flow v v V v s a = V s + V V s s V v V v V V = V s + n o as = V, an = s R s R v δs δ s v v v v v δs 1 s δs n Q = = δ s, o =, = lim = v δ δ R s s R s δ s 0 s R

21 6.1 Fluid Element Kinematics Types of motion and defomation fo a fluid element.

22 6.1.1 Velocity and Acceleation Fields Revisited Velocity field epesentation u u u V = V x, y, z, t o V = ui+ vj+ wk ( ) Acceleation of a paticle V V V V a = + u + v + w t x y z u u DV V u u a = = + ( V ) V Dt t ( ) ( ) ( ) ( ) = i + j + k x y z a x a y a z u = t v = t w = t u + u x u + v y u + w z v v v + u + v + w x y z w + u x w + v y w + w z

23 6.1. Linea Motion and Defomation vaiations of the velocity in the diection of u v w velocity,,, cause a linea stetching x y z defomation. Conside the x-component defomation:

24 Linea Motion and Defomation The change in the oiginal volume, δ V = δxδyδz, due to u/ x: u Change in δv = ( δx)( δyδz)( δt) x Rate change of δ V pe unit volume due to u/ x: ( / ) 1 d( δ V ) u x δt u = lim δv dt δ t = 0 δt x If velocity gadient v/ y and w/ z ae also pesent, then 1 d( δ V ) u v w uv = + + = V volumetic dilatation ate V dt x y z δ The volume of a fluid may change as the element moves fom one location to anothe in the flow field. Fo incompessible fluid, the volumetic dilation ate is zeo.

25 6.1.3 Angula Motion and Defomation Conside an element unde otation and angula defomation V6.3 Shea defomation

26 Angula Motion and Defomation the angula velocity of OA is ω OA = δα lim δ t δ t 0 Fo small angles v δ xδ t v tan x δα δα = = δt δ x x so that ω OA ( δ v/ δx) δt δv = lim δt 0 = δt δ x v x (if is positive then will be counteclockwise) ω OA

27 Angula Motion and Defomation the angula velocity of the line OB is ω OB = δβ lim δ t δ t 0 u δyδt y tan δβ δβ = δy so that = u δt y ω OB ( / ) u y δt u = lim = δt 0 δt y u y ( if is positive, will be clockwise) ω OB

28 Angula otation The otation,, of the element about the z axis is defined as the aveage of the angula velocities and, if counteclockwise is consideed to be positive, then, ω z OA ω OB ω = y u x v z 1 ω similaly = z v y w x 1 ω = x w z u y 1 ω, thus k y u x v j x w z u i z v y w w v u z y x k j i V + + = = V culv k j i x y x = = + + = 1 1 ω ω ω ω V4.6 Flow past a wing

29 Definition of voticity Define voticity ξ ξ = ω = V The fluid element will otate about z axis as an undefomed block ( ie: ω = ) only when u y = v x OA ω OB u y v = x Othewise the otation will be associated with an angula defomation. V = 0 If o, then the otation (and the voticity ) ae zeo, and flow fields ae temed iotational.

30 Diffeent types of angula motions Solid body otation φ u = u z = 0 u = Ω ω = Ω ω = 0 ω z z = 1 ω ( u ) Fee votex ω u = k φ = φ 1 u θ u = = 0 φ u z ω = 0 = 0 φ 1 ω ( u ) = 0 = z φ 0 fo

31 Angula Defomation Apat fom otation associated with these deivatives v y and x, these deivatives can cause the element to undego an angula defomation, which esults in a change in shape of the element. The change in the oiginal ight angle is temed the sheaing stain δγ, δγ = δα + δβ whee δγ is consideed to be positive if the oiginal ight angle is deceasing. u

32 Angula Defomation Rate of sheaing stain o ate of angula defomation v u δt δt δγ + x y & γ = lim = lim δt 0δt δt 0 δt v u = + x y The ate of angula defomation is elated to a coesponding sheaing stess which causes the fluid element to change in shape. u v = If y x, the ate of angula defomation is zeo and this condition indicates that the element is simply otating as an undefomed block.

33 6. Consevation of Mass Consevation of mass: DM sys Dt = 0 In contol volume epesentation (continuity equation): t cv ρdv + cs ρv nda = 0 (6.19) To obtain the diffeential fom of the continuity equation, Eq is applied to an infinitesimal contol volume.

34 6..1 Diffeential Fom of Continuity Equation z y x t dv t δ δ δ ρ ρ Net mass flow in the diection z y x x u z y x x u u z y x x u u δ δ δ ρ δ δ δ ρ ρ δ δ δ ρ ρ = + Net mass flow in the y diection z y x y v δ δ δ ρ ρ Net mass flow in the z diection z y x z w δ δ δ ρ ρ Net ate of mass out of flow z y x z w y v x u δ δ δ ρ ρ ρ + +

35 Diffeential Fom of Continuity Equation Thus consevation of mass become In vecto fom Fo steady compessible flow Fo incompessible flow = 0 V = z w y v x u t ρ ρ ρ ρ (continuity equation ) = z w y v x u ρ ρ ρ = z w y v x u = 0 V ρ = 0 + V t ρ ρ

36 6.. Cylindical Pola Coodinates The diffeential fom of continuity equation ρ + t 1 ( ρv ) 1 ( ρv ) + θ θ + ρv z z = 0

37 6..3 The Steam Function Fo -D incompessible plane flow then, u x v + y = 0 Define a steam function then u ψ = v y ψ = x ( xy, ) such that Fo velocity expessed in foms of the steam function, the consevation of mass will be satisfied. ψ ψ ψ ψ ψ + = = 0 x y x y xy yx

38 The Steam Function ψ Lines along constant ae steam lines. Definition of steam line dy = dx Thus change of, fom ( xy, ) to ( x+ dx, y + dy) vdx + udy = 0 dy = dx Thus we can use to plot steamline. v u The actual numeical value of a steam line is not impotant but the change in the value of ψ is elated to the volume flow ate. v u dψ = 0 which is the defining equation fo a steamline. ψ ψ d ψ ψ ψ = dx + dy = vdx udy x y + ψ Along a line of constant of we have

39 The Steam Function Note:Flow neve cosses steamline, since by definition the velocity is tangent to the steamlines. Volume ate of flow (pe unit width pependicula to the x-y plane) ψ > ψ dq = udy vdx ψ ψ = dy + dx = dψ y x = d = ψ1 1 If 1 then q is positive and vice vesa. In cylindical coodinates the incompessible continuity equation becomes, 1 ( v ) 1 v + θ = 0 θ 1 ψ ψ Then, v = vθ θ Ex 6.3 Steam function q = ψ ψ ψ ψ

40 6.3 Consevation of Linea Momentum Linea momentum equation o F = D Dt F Vdm sys contents of the contol volume = V ρdv + t CV CS VρV nda Conside a diffeential system with δm and δv ( ) D Vδm then δ F = Dt Using the system appoach then DV δ F = δm = δma Dt

41 6.3.1 Desciptions of Foce Acting on the Diffeential Element Two types of foces need to be consideed suface foces:which action the sufaces of the diffeential element. body foces:which ae distibuted thoughout the element. Fo simplicity, the only body foce consideed is the weight of the element, u δf b u = δmg o δ Fbx = δmg x δ Fby = δmg y δ Fbz = δmg z

42 Suface foce act on the element as a esult of its inteaction with its suoundings (the components depend on the aea oientation) Whee δ is nomal to the aea A and F and δf ae paallel F n to the aea and othogonal to each othe. δ δ 1

43 The nomal stess σis n defined as, σ Sign of stesses n δf = lim n δa 0 δa and the sheaing stesses ae define as τ = 1 lim δa 0 σ δf 1 δa δf τ = lim δa 0 δa we use fo nomal stesses and fo shea stesses. Positive sign fo the stess as positive coodinate diection on the sufaces fo which the outwad nomal is in the positive coodinate diection. Note:Positive nomal stesses ae tensile stesses, ie, they tend to stetch the mateial. τ

44 Thus δ F δ F δ F sx sy sz σ τ τ x y z τ σ τ = + + x y z xx yx = + + zx xy yy zy τ τ σ x y z xz yx = + + s b zz δxδyδz δxδyδz δxδyδz u δf s = δfsxi+ δfsy j+ δfsz k u uu uu δf = δf + δf

45 6.3. Equation of Motion δ F = δma, δf = δma, δf = δm a x x y y z z δ m = ρδ x δ y δ z Thus ρ ρ ρ σ τ τ u u u u = ρ x y z t x y z τ σ τ v v v v = + + v + x y z t x y z τ τ σ w w w w = x y z t x y z xx yx zx g x u v w xy yy zy g y ρ u w xz yz zz g z ρ u v w (6.50)

46 PART B Inviscid Flow: Eule Equation/Some Basic, Plane Potential Flows (Sections )

47 6.4 Inviscid Flow Eule s Equation of Motion Fo an inviscid flow in which the sheaing stesses ae all zeo, and the nomal stesses ae eplaced by -p, thus the equation of motion becomes p u u u u ρgx = ρ + u + υ + w x t x y z p υ υ υ υ ρgy = ρ + u + υ + w y t x y z p w w w w ρgz = ρ + u + υ + w z t x y z o uv v V uv uv ρg p= ρ + ( V ) V t The main difficulty in solving the equation is the nonlinea tems which appea in the convective acceleation.

48 6.4. The Benoulli Equation Fo steady flow v uv uv ρg p= ρ( V ) V uv g = g z (up being positive) uv uv uv uv uv uv 1 ( V ) V = ( V V) V ( V) thus the equation can be witten as, ρ uv uv uv uv ρg z p= V V V V o p 1 uv uv + ( V ) + g z = V ( V) ρ ( ) ρ ( ) Take the dot poduct of each tem with a diffeential length ds along a steamline p v 1 v v uv uv v ds + ( V ) ds + g z ds = V ( V ) ds ρ

49 Since ds v and V uv ae paallel, theefoe uv v v V V ds = Since ( ) 0 v v v v d s = dxi + dy j + dz k v p p p p d s = dx + dy + dz = dp x y z Thus the equation becomes dp 1 d( V ) gdz 0 ρ + + = whee the change in pv, uv, and z is along the steamline

50 Equation afte integation become dp V gz constant ρ + + = which indicates that the sum of the thee tems on the left side of the equation must emain a constant along a given steamline. Fo inviscid, incompessible flow, the equation become, p V + + gz = const ρ o p V p V γ g γ g z1 = + + z Fo (1) inviscid flow () steady flow (3) incompessible flow (4) flow along a steamline

51 6.4.3 Iotational Flow If the flow is iotational, the analysis of inviscid flow poblem is futhe simplified. The otation of the fluid element is equal 1 V to, and fo iotational flow field, uv V = 0 uv ω uv Since V = ζ, theefoe fo an iotational flow field, the voticity is zeo. The condition of iotationality imposes specific elationships among these velocity gadients. Fo example, 1 υ u ω = = 0 z x y υ u w υ u w =, =, = x y y z z x A geneal flow field would not satisfy these thee equations.

52 Can iotational flow hold in a viscous fluid? Accoding to the -D voticity tanspot equation (cf. Poblem 6.81) Dζ z = ν ζ z Dt Voticity of an fluid element gows along with its motion as long as ν is positive. So, an initially iotatioal flow will eventually tun into otational flow in a viscous fluid. On the othe hand, an initially iotatioal flow emains iotational in an inviscid fluid, if without extenal excitement.

53 6.4.4 The Benoulli Equation fo Iotational Flow In Section 6.4., we have obtained along a steamline that, uv uv V V ds= ( ) 0 uv In an iotational flow, V = 0, so the equation is zeo egadless of the diection of ds v. Consequently, fo iotational flow the Benoulli equation is valid thoughout the flow field. Theefoe, between any flow points in the flow field, dp V + + gz = constant ρ o p V p V γ g γ g z1 = + + z Fo (1) Inviscid flow () Stead flow (3) Incompessible flow (4) Iotational flow

54 6.4.5 The Velocity Potential Fo iotational flow since uv uv V = 0 thus V = φ φ φ φ u =, υ =, w= x y z so that fo an iotational flow the velocity is expessible as the gadient of a scala function φ. The velocity potential is a consequence of the iotationality of the flow field (only valid fo inviscid flow), wheeas the steam function is a consequence of consevation of mass (valid fo inviscid o viscous flow). Velocity potential can be defined fo a geneal theedimensional flow, wheeas the steam function is esticted to two-dimensional flows.

55 Thus fo iotational flow uv uv V = 0 V = φ, futhe with V = 0 fo incomp. flow, φ = 0 φ φ φ In Catesian coodinates, + + = 0 x y z Thus, inviscid, incompessible, iotational flow fields ae govened by Laplace s equation. Cylindical coodinate ( ) v 1 ( ) v ( ) v ( ) = e + e + e θ z φ v 1 φ v φ v φ = e + eθ + ez θ z whee φ = φ(, θ, z) uv v v v Since V = υ e + υ e + υ e θ θ θ z z Thus fo an iotational flow with V z uv = φ 1 φ 1 φ φ + + = θ z 0

56 Example 6.4 ψ = sin θ 1 ψ φ ( ) θ ψ 1 φ = = = = + f θ Thus φ = cosθ + C υ = = 4cosθ = φ = cosθ + f1 θ υθ 4 sinθ φ cosθ The specific value of C is not impotant, theefoe φ = cos θ ( θ) ( θ) V = 4cos + 4sin = 16 p1 V1 p V + = + γ g γ g ( )

57 6.5 Some basic, plane potential flows Since the Laplace equation is a linea diffeential equation, vaious solutions can be added to obtain othe solutions. i.e. φ = φ1+ φ The pactical implication is that if we have basic solutions, we can combine them to obtain moe complicated and inteesting solutions. In this section seveal basic velocity potentials, which descibe some elatively simple flows, will be detemined.

58 Fo simplicity, only two-dimensional flows will be consideed. Defining the velocities in tems of the steam function, consevation of mass is identically satisfied. Now impose the condition of iotationality, Thus 1, o, : velocity potential θ φ φ φ φ θ = = = = v v y v x u, 1 o, steam function : v v x v y u = = = = ψ θ ψ ψ ψ θ x v y u = 0 o = + = y x x x y y ψ ψ ψ ψ

59 Thus fo a two-dimensional iotational flow, the velocity potential and the steam function both satisfy Laplace equation. It is appaent fom these esults that the velocity potential and the steam function ae somehow elated. Along a line of constant ψ, dψ =0 ψ ψ dψ = dx + dy = vdx + udy x y dy v udy = vdx, = dx u Along a line of constant φ, dφ =0 φ φ dφ = dx + dy = udx + vdy x y udx = vdy, dy dx = u v = 0

60 Theefoe, the equations indicate that lines of constant φ (equipotential lines) ae othogonal to lines of constant ψ (steam line) at all points whee they intesect. Q: Why V > V 1? How about p 1 and p?

61 6.5.1 Unifom Flow The simplest plane flow is one fo which the steamlines ae all staight and paallel, and the magnitude of the velocity is constant unifom flow. u = U v= 0 φ φ = U, = 0 x y φ = Ux + C Thus, fo a unifom flow in the positive x diection, φ = Ux The coesponding steam function can be obtained in a simila manne, ψ ψ = U, = 0 ψ = Uy y x

62 The velocity potential and steam function fo a unifom flow at an angle α with the x axis, φ = U ( x cosα + y sinα) ψ = U ( y cosα xsinα)

63 6.5. Souce and Sink- puely adial flow Conside a fluid flowing adially outwad fom a line though the oigin pependicula to the x-y plane. Let m be the volume ate of flow emanating fom the line (pe unit length). Consevation of mass m π ( v) = m o v = π Also, since the flow is puely adial, velocity potential becomes, v θ = 0 φ m 1 φ =, = 0 π θ m φ = ln π

64 Souce and Sink flows If m is positive, the flow is adially outwad, and the flow is consideed to be a souce flow. If m is negative, the flow is towad the oigin, and the flow is consideed to be a sink flow. The flow ate, m, is the stength of the souce o sink. The steam function fo the souce: 1 ψ m ψ =, = 0 ψ = m θ θ π π Note: At =0, the velocity becomes infinite, which is of couse physically impossible and is a singula point.

65 6.5.3 Votex-steamlines ae concentic cicles (v =0) Conside a flow field in which the steamlines ae concentic cicles. i.e. we intechange the velocity potential and steam function fo the souce. Thus, let φ = Kθ and ψ = K ln whee K is a constant. v 1 φ ψ θ = = θ = K (fee votex)

66 Fee and Foced votex Rotation efes to the oientation of a fluid element and not the path followed by the element. Fee votex Foced votex If the fluid wee otating as a igid body, such that v = θ K, this type of votex motion is otational and can not be descibed by a velocity potential. Fee votex: bathtub flow. V6.4 Votex in a beake Foced votex: liquid contained in a tank otating about its axis.

67 Combined votex Combined votex: a foced votex as a cental coe and a fee votex outside the coe. v = ω θ K vθ = > 0 0 whee K and ae constant and 0 coesponds to the adius of cental coe.

68 Ciculation A mathematical concept commonly associated with votex motion is that of ciculation. Γ = V ds (6.89) C The integal is taken aound cuve, C, in the counteclockwise diection. Note: Geen s theoem in the plane dictates ( V) k dxdy = CV ds R Fo an iotational flow V = φ V ds = φ ds = dφ theefoe, Γ = dφ = 0 C Fo an iotational flow, the ciculation will geneally be zeo. Howeve, if thee ae singulaities enclosed within the cuve, the ciculation may not be zeo.

69 Ciculation fo fee votex Fo example, the fee votex with π K Γ= ( d θ) = π K K = 0 v θ = Γ π Note: Howeve Γ along any path which does not include the singula point at the oigin will be zeo. The velocity potential and steam function fo the fee votex ae commonly expessed in tems of ciculation as, K Γ φ = θ π Γ ψ = ln π (6.90) (6.91)

70 Example 6.6 Detemine an expession elating the suface shape to the stength of the votex as specified by ciculation Γ. Γ φ = θ π Fo iotational flow, the Benoulli equation p1 V1 p V + + z1 = + + z p1 = p = γ g γ g V1 V = zs + z1 = 0 g g v z θ s 1 φ Γ = = V1 0 θ π Γ = 8π g 0

71 6.5.4 Doublet Conside potential flow that is fomed by combining a souce and a sink in a special way. Conside a souce-sink pai ψ m πψ tanθ1 tanθ = tan( θ1 θ) = π 1+ tanθ tanθ = ( θ1 θ) tan m 1 sinθ Since tanθ1 = and tanθ cosθ a = sinθ cosθ + a Thus tan πψ m = a sinθ a ψ = m tan π 1 a sinθ a

72 Doublet Fo small values of a m a sinθ ψ = π a = ma sinθ π ( a Doublet is fomed by letting the souce and sink appoach one anothe ( a 0 ) while inceasing the stength m ( m ) so that the poduct ma/π emains constant. ) (6.94) As a 0, /( a ) 1/ Eq educes to: K sinθ ψ = whee K = ma/π is called the stength of the doublet. The coesponding velocity potential is K cosθ φ = (6.95) (6.96)

73 Doublet-steamlines V V θ = = φ 1 ψ K cosθ = = θ 1 φ ψ K sinθ = = θ Steamlines fo a doublet

74 Summay of basic, plane potential flows

75 6.6 Supeposition of Basic, Plane Potential Flows Method of supeposition Any steamline in an inviscid flow field can be consideed as a solid bounday, since the conditions along a solid bounday and a steamline ae the sameno flow though the bounday o the steamline. Theefoe, some basic velocity potential o steam function can be combined to yield a steamline that coesponds to a paticula body shape of inteest. This method is called the method of supeposition.

76 6.6.1 Souce in a Unifom Steam- Half-Body Conside a supeposition of a souce and a unifom flow. The esulting steam function is ψ = ψ unifom flow + ψ souce m = U sinθ + θ π and the coesponding velocity potential is m φ = U cosθ + ln π V6.5 Half-body (6.97)

77 Fo the souce alone v = m π Let the stagnation point occu at x=-b, whee so m b = πu m πb The value of the steam function at the stagnation point can be obtained by evaluating x at =b and θ=π, which yields fom Eq m ψ stagnation = = πbu Thus the equation of the steamline passing though the stagnation point is, b( π θ ) πbu = U sin θ + buθ o = (6.100) sinθ U =

78 The width of the half-body asymptotically appoaches πb. This follows fom Eq , which can be witten as y = b( π θ ) so that as θ 0 o θ π, the half-width appoaches ±bπ. With the steam function (o velocity potential) known, the velocity components at any point can be obtained. 1 ψ m v = = U cosθ + θ π ψ vθ = = U sinθ

79 Thus the squae of the magnitude of the velocity V at any point is, Umcosθ m V = v + vθ = U + + ( ) π π m since b = πu b b V = U 1+ cosθ + (6.101) With the velocity known, the pessue distibution can be detemined fom the Benoulli equation, 1 1 p0 + ρu = p+ ρv (6.10) Note: the velocity tangent to the suface of the body is not zeo; that is, the fluid slips by the bounday.

80 Example 6.7 b b V = U 1+ cosθ + b( π θ) πb on the suface θ= π / = = sinθ 4 Thus V = U 1+ π πb y = p V p V γ g γ g y1 = + + y ρ p p = V V + y y ( ) γ ( ) 1 1 1

81 6.6. Rankine Ovals Conside a souce and a sink of equal stength combined with a unifom flow to fom the flow aound a closed body. The steam function and velocity potential fo this combination ae, sin m ψ = U θ ( θ 1 θ ) π m φ = U cosθ (ln 1 ln ) π

82 As in Section o ψ = U sinθ ψ = Uy m tan π m tan π 1 x 1 + a sinθ ay y a a The steam line ψ=0 foms a closed body. Since the body is closed, all of the flow emanating fom the souce flows into the sink. These bodies have an oval shape and ae temed Rankine ovals. The stagnation points coespond to the points whee the unifom velocity, the souce velocity, and the sink velocity all combine to give a zeo velocity. The location of the stagnation points depend on the value of a, m, and U.

83 The body half length: l souce: 1 1 ma a o l m = + = + 1 πu a πua v m = π Theefoe m m U + = 0 π a π + a ( ) ( ) m a U = 0 π a m 1 m 1 = 0 o a = πu a πu m l = = + a πu 1

84 The body half width, h, can be obtained by detemining the value of y whee the y axis intesects the ψ=0 steamline. Thus, fom Eq with ψ=0, x=0, and y=h, It follows that ψ = Uy m 1 ay m 1 ah tan 0 = Uh tan π x + y a π h a 1 ah Uhπ tan = h a m ah πuh = tan h a m h a πuh h = a tan m h 1 h πuh 1 h πua h = 1 tan = 1 tan a a m a m a

85 Both l/a and h/a ae functions of the dimensionless paamete Ua/m. As l/h becomes lage, flow aound a long slende body is descibed, wheeas fo small value of paamete, flow aound a moe blunt shape is obtained. Downsteam fom the point of maximum body width the suface pessue incease with distance along the suface. In actual viscous flow, an advese pessue gadient will lead to sepaation of the flow fom the suface and esult in a lage low pessue wake on the downsteam side of the body. Howeve, sepaation is not pedicted by potential theoy. Rankine ovals will give a easonable appoximation of the velocity outside the thin, viscous bounday laye and the pessue distibution on the font pat of the body. V6.6 Cicula cylinde V6.8 Cicula cylinde with sepaation V6.9 Potential and viscous flow

86 6.6.3 Flow aound a cicula cylinde When the distance between the souce-sink pai appoaches zeo, the shape of the Rankine oval becomes moe blunt and appoach a cicula shape. A combination of doublet and unifom flow will epesent flow aound a cicula cylinde. K sinθ K steam function: ψ = U sinθ = U K cosθ velocity potential: φ = U cosθ + to detemine K with ψ=0 fo =a, K U = 0 K = Ua a sinθ

87 Thus the steam function and velocity potential fo flow aound a cicula cylinde ae a ψ = U 1 sinθ a φ = U 1 + cosθ The velocity components ae v v θ φ 1 ψ a = = = U 1 sinθ θ 1 φ ψ a = = = U 1 sinθ θ + On the cylinde suface (=a): v = 0 and vθ = U sinθ Potential flow aound a cicula cylinde

88 Theefoe the maximum velocity occus at the top and bottom of the cylinde θ = ±π/ and has a magnitude of twice the upsteam velocity U. The pessue distibution on the cylinde suface is obtained fom the Benoulli equation, 1 1 p0 + ρu = p + ρv 1 ps = p0 + ρu s θ s ( 1 4sin θ ) whee p 0 and U ae pessue and velocity fo point fa fom the cylinde.

89 The figue eveals that only on the upsteam pat of the cylinde is thee appoximate ageement between the potential flow and the expeimental esults.

90 The esulting foce (pe unit length) developed on the cylinde can be detemined by integating the pessue ove the suface. π F = p cosθadθ = 0 x 0 π F = p sinθadθ = 0 Both the dag and lift as pedicted by potential theoy fo a fixed cylinde in a unifom steam ae zeo. since the pessue distibution is symmetical aound the cylinde. In eality, thee is a significant dag developed on a cylinde when it is placed in a moving fluid. (d Alembet paadox) Ex 6.8 Potential flow--cylinde y 0 s s

91 By adding a fee votex to the steam function o velocity potential fo the flow aound a cylinde, then a Γ ψ = U 1 sinθ ln π a φ = U 1 + cosθ + whee Γ is the ciculation Γ θ π Tangential velocity on the suface (=a): v θs ψ = = a = U sinθ + Γ πa (6.119) (6.10) (6.11) This type of flow could be appoximately ceated by placing a otating cylinde in a unifom steam. Because the pesence of viscosity in any eal fluid, the fluid in contact with the otating cylinde would otate with the same velocity as the cylinde, and the esulting flow field would esemble that developed by the combination of a unifom flow past a cylinde and a fee votex.

92 Location of the stagnation point vθ s = 0 = U if Γ = 0 if 1 Γ / 4πUa if Γ / 4πUa Γ sinθ + πa θ = 0 o stag > 1 1 θ stag π θ stag sinθ stag Γ = 4πUa is at some othe location on the suface is located away fom the cylinde

93 The development of this lift on otating bodies is called the Magnus effect. Foce pe unit length developed on the cylinde 1 1 Γ + ρ = + ρ sinθ + π a p0 U ps U 1 Γsinθ Γ ps = p + U + πau 4π a U 0 ρ 1 4sin θ π F = p cosθadθ = 0 x 0 s π π ρuγ Fy pssinθadθ sin θdθ ρu π 0 0 = = = Γ Fo a cylinde with ciculation, lift is developed equal to the poduct of the fluid density, the upsteam velocity, and the ciculation. F = ρuγ y ( ) U( + ) Γ +, counteclockwise the F is downwad y

94 6.7 Othe Aspects of Potential Flow Analysis Exact solutions based in potential theoy will usually povide at best appoximate solutions to eal fluid poblems. Potential theoy will usually povide a easonable appoximation in those cicumstances when we ae dealing with a low viscosity fluid moving at a elatively high velocity, in egions of the flow field in which the flow is acceleating. Outside the bounday laye the velocity distibution and the pessue distibution ae closely appoximated by the potential flow solution. In situation when the flow is deceleating (in the eawad potion of the bluff body expanding egion of a conduit), and advese pessue gadient is educed leading to flow sepaation, a phenomenon that ae not accounted fo by potential theoy. V6.10 Potential flow

95 PART C Viscous Flow: Navie-Stokes Equation (Sections )

96 6.8 Viscous Flow Equation of Motion δ F x = δma x δ F y = δma y δ F z = δma z δ m = ρδ x δ y δ z Thus ρ ρ ρ σ τ τ u u u u x y z t x y z τ σ τ v v v v x y z t x y z τ τ σ w w w w x y z t x y z xx yx zx gx = ρ + u + v + w xy yy zy g y = ρ + u + v + w xz yz zz gz = ρ + u + v + w

97 6.8.1 Stess-Defomation Relationships When a shea stess is applied on a fluid: Fluids continuously defom (stess τ ~ ate of stain) Solids defom o bend (stess τ ~ stain) stain ate ~ velocity gadient d α = dt du dy fom Fox, McDonald and Pitchad, Intoduction to Fluid Mechanics.

98 6.8.1 Stess-Defomation Relationships Fo incompessible Newtonian fluids it is known that the stesses ae linealy elated to the ate of defomation. V1.6 Non-Newtonian behavio τ Fo incompessible, Newtonian fluids, the viscous stesses ae: v x v v x y vz vx μ μ + μ x y x + x z σxx,visc τxy τ xz v v x y vy vy v z τ σ τ μ + μ μ + y x y z y τzx τzy σ zz,visc v v z vx y v z vz μ + μ + μ x z z y z visc, ij yx yy,visc yz

99 6.8.1 Stess-Defomation Relationships But in nomal stesses, thee is additional contibution of pessue p, whee 1 p = ( σ xx + σ yy + σzz ) 3 Consequently, fo nomal stesses σ σ σ xx yy zz u = p + μ x υ = p + μ y w = p + μ z fo sheaing stesses u τxy = τ yx = μ + y υ x υ w τ yz = τzy = μ + z y w u τzx = τxz = μ + x z Can you figue out why the nomal viscous stess σ xx,visc can be expessed as μ u x?

100 Fo viscous fluids in motion the nomal stesses ae not necessaily the same in diffeent diections, thus, the need to define the pessue as the aveage of the thee nomal stesses. Stess-stain elationship in cylindical coodinate 1 z zz p p p z θ θθ υ σ μ υ υ σ μ θ υ σ μ = + = + + = z z z z z z z z θ θ θ θ θ θ υ υ τ τ μ θ υ υ τ τ μ θ υ υ τ τ μ = = + = = + = = + Note: Notation x: plane pependicula to x coodinate y: diection τ xy Stess-Defomation Relationships

101 6.8. The Navie-Stokes Equations The Navie-Stokes equations ae consideed to be the govening diffeential equations of motion fo incompessible Newtonian fluids u u u u p u u u ρ + u + υ + w = + ρgx + μ + + t x y z x x y z υ υ υ υ p υ υ υ ρ + u + υ + w = + ρgy + μ + + t x y z y x y z w w w w p w w w ρ + u + υ + w = + ρgz + μ + + t x y z z x y z

102 The Navie-Stokes Equations In tems of cylindical coodinate υ υ υθ υ υθ υ ρ + υ + + υz t θ z p 1 υ υ 1 υ υθ υ = + ρg + μ + + θ θ z υθ υθ υθ υθ υυ θ υθ ρ + υ υz t θ z 1 p 1 υθ υθ 1 υθ υ υ θ = + ρgθ + μ + θ + θ θ z υz υz υθ υz υz ρ + υ + + υz t θ z p 1 υz 1 υz υ z = + ρgz + μ + + z θ z

103 6.9 Some Simple Solutions fo Viscous, Incompessible Fluids Thee ae no geneal analytical schemes fo solving nonlinea patial diffeential equations, and each poblem must be consideed individually. u u u u p u u u ρ + u + υ + w = + ρgx + μ + + t x y z x x y z υ υ υ υ p υ υ υ ρ + u + υ + w = + ρgy + μ + + t x y z y x y z w w w w p w w w ρ + u + υ + w = + ρgz + μ + + t x y z z x y z Nonlinea tems

104 6.9.1 Steady Lamina Flow Between Fixed Paallel plates υ = 0, w = 0 Thus continuity indicates that u = 0 x fo steady flow, u = u( y) g = 0, g = g and g = o x y z u max u u u u p u u u ρ + u + υ + w = + ρgx + μ + + t x y z x x y z υ υ υ υ p υ υ υ ρ + u + υ + w = + ρgy + μ + + t x y z y x y z w w w w p w w w ρ + u + υ + w = + ρgz + μ + + t x y z z x y z

105 Steady Lamina Flow Between Fixed Paallel plates Thus p u 0 = + μ x y p 0 = ρg p= ρgy+ f1 x y p 0 = z du 1 p = dy μ x du 1 p = y + C 1 dy μ x 1 p = + + μ x u y C1y C ( ) u u u u p u u u ρ + u + υ + w = + ρgx + μ + + t x y z x x y z υ υ υ υ p υ υ υ ρ + u + υ + w = + ρgy + μ + + t x y z y x y z w w w w p w w w ρ + u + υ + w = + ρgz + μ + + t x y z z x y z p ( x is teated as a constant since it is not a function of y)

106 Steady Lamina Flow Between Fixed Paallel plates the constants ae detemined fom the bounday conditions. BCs : u = 0 fo y =± h Thus C = 0 C 1 V6.11 No-slip bounday conditions 1 p = h μ x Thus the velocity distibution becomes, 1 p u = y h μ x ( ) which indicates that the velocity pofile between the two fixed plates is paabolic. V6.13 Lamina flow

107 Steady Lamina Flow Between Fixed Paallel plates The volume ate of flow h h 1 p q= u dy = ( y h ) dy h h μ x 3 h 1 p y q = h y μ x 3 h p y 3 h 3 = h + h μ x h p = 3 μ x The pessue gadient is negative, since the pessue deceases in the diection of the flow.

108 Steady Lamina Flow Between Fixed Paallel plates If Δp epesents the pessue dop between two points a distance apat, then l Δp p = l x Δ Δ q= =, V = = 3 μ x 3μl h 3μl 3 3 h p h p q h p The maximum velocity u max, occus midway y=0 between the two plates, thus h p umax = o umax = μ x 3 V

109 Steady Lamina Flow Between Fixed Paallel plates The pessue field ( ) p = ρgy + f x p f1( x) = x+ p0 x whee is a efeence pessue at x=y=0 p 0 Thus the pessue vaiation thoughout the fluid can be obtained fom p p = ρ gy + x + p0 x ρvh The above analysis is valid fo Re = μ emains below about 1400 Poblem 6.88: 10 tons on 8psi

110 6.9. Couette Flow Theefoe 1 p = + + μ x u y C1y C bounday conditions u=o at y=0, u=u at y=b b 1 p u = U + y by y μ x ( ) o in dimensionless fom u y b p y y = 1 U b μu x b b The actual velocity pofile will depend on the dimensionless paamete b p P = μu x This type of flow is called Couette flow.

111 Couette flow The simplest type of Couette flow is one fo which the pessue gadient is zeo i.e. the fluid motion is caused by the fluid being dagged along by the moving bounday. p = 0 x y Thus u = U b which indicates that the velocity vaies linealy between the two plates. e.g. : Jounal beaing o - i << i The flow in an unloaded jounal beaing might be appoximated by this simple Couette flow.

112 Example 6.9 υ u = w= 0 = 0 υ = υ x y ( ) p p = = 0 x z x = h p = atmospheic pessue dp dp = 0 = 0 dx dz Theefoe d υ 0 = ρg + μ dx d υ γ = dx μ dυ γ = x+ C1 dx μ on the film suface x=h, we assume that the sheaing stess is zeo dυ τxy = μ τxy = 0 at x = h dx γ h C1 = μ

113 nd integation γ γ h υ = + μ μ x x C x= 0 υ = V C = V 0 0 γ γh υ = + μ μ x x V0 γ μ h h q = υ dx= x x V dx 3 γ h q= V0h 3μ γh μ The aveage film velocity q γ h V = = V0 h 3μ γ h Only if V0 >, will thee be a net upwad flow of liquid. 3μ Q: Do you find anything weid in this poblem?

114 6.9.3 Steady, Lamina flow in Cicula Tubes Hagen Poiseuille flow o Poiseuille flow steady, lamina flow though a staight cicula tube of constant coss section Conside the flow though a hoizontal cicula tube of adius R Assume the flow is paallel v = v = z θ z 0 vz = 0 z v = v ( )

115 Steady, Lamina flow in Cicula Tubes Thus 1 p 0 = ρg cosθ θ p 0 = ρg sinθ p 1 0 = + μ z v z g = g sinθ gθ = g cosθ Integation of equations in the and θ diections p = ρg sinθ + f z 1 ( ) = ρ gy + f z 1 ( ) which indicate that the pessue is hydostatically distibuted at any paticula coss section and the z component of the pessue gadient, p / z, is not a function of o θ.

116 Steady, Lamina flow in Cicula Tubes the equation of motion in the z diection 1 vz 1 p = μ z v 1 p = + μ z z C1 1 p = + + z vz C1ln C 4μ Bounday conditions At =0, v z is finite at the cente of the tube, thus C 1 =0 1 p At =R, v z =0, then C = R 4μ z Thus the velocity distibution becomes, 1 p ( v R ) z = 4 μ z That is, at any coss section, the velocity distibution is paabolic.

117 Steady, Lamina flow in Cicula Tubes Volume flow ate dq = v z Q = π (π ) d 0 R v z d = π 0 R 1 p ( 4μ z R ) d 4 πr p = 8μ z Δp p Let =, then l z mean velocity V = v Q R ΔP = πr 8μl maximum velocity max R p R Δp = = 4μ z 4μl Q 4 πr Δp = 8μl Poiseuille s law so vmax = V the velocity distibution in tems of v max v z v max = 1 R

118 6.9.4 Steady, Axial, Lamina Flow in an Annulus 1 p = 4 μ z v z + C1 ln + B.Cs:v z =0 at = o and = i C thus 1 p i 0 vz = 0 + ln ( 0 / i) 4μ z ln ( 0 / i ) volume ate of flow ( o i ) 0 π p Q= v π d = i 4 4 z o i 8μ z ln ( 0 / i ) ( o i ) πδp 4 4 = o i 8μl ln ( 0 / i )

119 The maximum velocity occu at the m, = v = 0 z m o i = ln ( 0 / i ) 1 The maximum velocity does not occu at the mid point of the annulus space, but athe it occus neae the inne cylinde. To detemine Reynolds numbe, it is common pactice to use an effective diamete hydaulic diamete fo on cicula tubes. D h = 4 coss - sctional aea wetted peimete Thus the flow will emain lamina if ρdhv R = e emains below 100. μ

120 6.10 Othe Aspects of Diffeential Analysis V = 0 ρ V t + V V = p + ρg + μ V The solutions of the equations and not eadily available Numeical Methods Finite diffeence method Finite element ( o finite volume ) method Bounday element method V6.15 CFD example