Chapter 6 Differential Analysis of Fluid Flow

Transcription

1 57:00 Mechanics of Fluids and Transpor Processes Chaper 6 Professor Fred Sern Fall Chaper 6 Differenial Analysis of Fluid Flow Fluid Elemen Kinemaics Fluid elemen moion consiss of ranslaion, linear deformaion, roaion, and angular deformaion. Types of moion and deformaion for a fluid elemen. Linear Moion and Deformaion: Translaion of a fluid elemen Linear deformaion of a fluid elemen 1

2 57:00 Mechanics of Fluids and Transpor Processes Chaper 6 Professor Fred Sern Fall 006 Change inδ : u δ = δ x ( δ y δ z ) δ x he rae a which he volume δ is changing per uni volume due o he gradien u/ x is 1 δ ( δ ) ( ) d u x δ u = lim = d δ 0 δ x If velociy gradiens v/ y and w/ z are also presen, hen using a similar analysis i follows ha, in he general case, ( δ ) 1 d u v w = + + = V δ d x y z This rae of change of he volume per uni volume is called he volumeric dilaaion rae. Angular Moion and Deformaion For simpliciy we will consider moion in he x y plane, bu he resuls can be readily exended o he more general case.

3 57:00 Mechanics of Fluids and Transpor Processes Chaper 6 Professor Fred Sern Fall Angular moion and deformaion of a fluid elemen The angular velociy of line OA, ω OA, is δα ωoa = lim δ 0 δ For small angles ( v x) δxδ v anδα δα = δ δ x = x so ha ( v x) δ v ωoa = lim = δ 0 δ x Noe ha if v/ x is posiive, ω OA will be counerclockwise. Similarly, he angular velociy of he line OB is ω OB δβ lim u = = δ 0 δ y In his insance if u/ y is posiive, ω OB will be clockwise. 3

4 57:00 Mechanics of Fluids and Transpor Processes Chaper 6 Professor Fred Sern Fall The roaion, ω z, of he elemen abou he z axis is defined as he average of he angular velociies ω OA and ω OB of he wo muually perpendicular lines OA and OB. Thus, if counerclockwise roaion is considered o be posiive, i follows ha 1 v u ω z = x y Roaion of he field elemen abou he oher wo coordinae axes can be obained in a similar manner: 1 w v ω x = y z 1 u w ω y = z x The hree componens, ω x,ω y, and ω z can be combined o give he roaion vecor, ω, in he form: 1 1 ω = ωxi+ ωyj+ ωzk = curlv = V since i j k 1 1 V = x y z u v w 1 w v 1 u w 1 v u = i+ j+ k y z z x x y 4

5 57:00 Mechanics of Fluids and Transpor Processes Chaper 6 Professor Fred Sern Fall The voriciy, ζ, is defined as a vecor ha is wice he roaion vecor; ha is, ς = ω = V The use of he voriciy o describe he roaional characerisics of he fluid simply eliminaes he (1/) facor associaed wih he roaion vecor. If V = 0, he flow is called irroaional. In addiion o he roaion associaed wih he derivaives u/ y and v/ x, hese derivaives can cause he fluid elemen o undergo an angular deformaion, which resuls in a change in shape of he elemen. The change in he original righ angle formed by he lines OA and OB is ermed he shearing srain, δγ, δγ = δα + δβ The rae of change of δγ is called he rae of shearing srain or he rae of angular deformaion: lim lim Similarly, The rae of angular deformaion is relaed o a corresponding shearing sress which causes he fluid elemen o change in shape. 5

6 57:00 Mechanics of Fluids and Transpor Processes Chaper 6 Professor Fred Sern Fall The Coninuiy Equaion in Differenial Form The governing equaions can be expressed in boh inegral and differenial form. Inegral form is useful for large-scale conrol volume analysis, whereas he differenial form is useful for relaively small-scale poin analysis. Applicaion of RTT o a fixed elemenal conrol volume yields he differenial form of he governing equaions. For example for conservaion of mass ρv A CS = CV ρ dv ne ouflow of mass = rae of decrease across CS of mass wihin CV 6

7 57:00 Mechanics of Fluids and Transpor Processes Chaper 6 Professor Fred Sern Fall Consider a cubical elemen oriened so ha is sides are o he (x,y,z) axes ( ρu) dydz ρ u + x oule mass flux inle mass flux ρudydz Taylor series expansion reaining only firs order erm We assume ha he elemen is infiniesimally small such ha we can assume ha he flow is approximaely one dimensional hrough each face. The mass flux erms occur on all six faces, hree inles, and hree oules. Consider he mass flux on he x faces x = ρu + ( ρu) dydz x ρudydz = ( u) dydz x ρ V flux ouflux influx Similarly for he y and z faces yflux = ( ρv)dydz y z flux = ( ρw)dydz z 7

8 57:00 Mechanics of Fluids and Transpor Processes Chaper 6 Professor Fred Sern Fall The oal ne mass ouflux mus balance he rae of decrease of mass wihin he CV which is ρ dydz Combining he above expressions yields he desired resul ρ + ( ρu) + ( ρv) + ( ρw) dydz = 0 x y z dv ρ + ( ρu) + x ( ρv) + y ( ρw) z = 0 per uni V differenial form of coninuiy equaions ρ + ( ρv) = 0 ρ V + V ρ Dρ + ρ V D = 0 D D = + V Nonlinear 1 s order PDE; ( unless ρ = consan, hen linear) Relaes V o saisfy kinemaic condiion of mass conservaion Simplificaions: 1. Seady flow: ( ρv) = 0. ρ = consan: V = 0 8

9 57:00 Mechanics of Fluids and Transpor Processes Chaper 6 Professor Fred Sern Fall u v w i.e., + + = 0 x y z 3D u x + v y = 0 D The coninuiy equaion in Cylindrical Polar Coordinaes The velociy a some arbirary poin P can be expressed as V = v e + v e + v e r r θ θ z z The coninuiy equaion: ρ 1 ( rρvr) 1 ( ρvθ ) ( ρv z) = 0 r r r θ z For seady, compressible flow 1 ( rρvr) 1 ( ρvθ ) ( ρvz) + + = r r r θ z 0 For incompressible fluids (for seady or unseady flow) 1 ( rvr ) 1 vθ vz + + = 0 r r r θ z 9

10 57:00 Mechanics of Fluids and Transpor Processes Chaper 6 Professor Fred Sern Fall The Sream Funcion Seady, incompressible, plane, wo-dimensional flow represens one of he simples ypes of flow of pracical imporance. By plane, wo-dimensional flow we mean ha here are only wo velociy componens, such as u and v, when he flow is considered o be in he x y plane. For his flow he coninuiy equaion reduces o u v + = 0 x y We sill have wo variables, u and v, o deal wih, bu hey mus be relaed in a special way as indicaed. This equaion suggess ha if we define a funcion ψ(x, y), called he sream funcion, which relaes he velociies as ψ ψ u =, v = y x hen he coninuiy equaion is idenically saisfied: ψ ψ ψ ψ + = = 0 x y y x x y x y Velociy and velociy componens along a sreamline 10

11 57:00 Mechanics of Fluids and Transpor Processes Chaper 6 Professor Fred Sern Fall Anoher paricular advanage of using he sream funcion is relaed o he fac ha lines along which ψ is consan are sreamlines.the change in he value of ψ as we move from one poin (x, y) o a nearby poin (x +, y + dy) along a line of consan ψ is given by he relaionship: ψ ψ dψ = + dy = v + udy = 0 x y and, herefore, along a line of consan ψ dy v = u The flow beween wo sreamlines The acual numerical value associaed wih a paricular sreamline is no of paricular significance, bu he change in he value of ψ is relaed o he volume rae of flow. Le dq represen he volume rae of flow (per uni widh perpendicular o he x y plane) passing beween he wo sreamlines. ψ ψ dq = udy v = + dy = dψ x y Thus, he volume rae of flow, q, beween wo sreamlines such as ψ1 and ψ, can be deermined by inegraing o yield: 11

12 57:00 Mechanics of Fluids and Transpor Processes Chaper 6 Professor Fred Sern Fall q ψ = d = ψ1 ψ ψ ψ 1 In cylindrical coordinaes he coninuiy equaion for incompressible, plane, wo-dimensional flow reduces o 1 ( rvr ) 1 vθ + = 0 r r r θ and he velociy componens, v r and v θ, can be relaed o he sream funcion, ψ(r, θ), hrough he equaions 1 ψ ψ vr =, vθ = r θ r Navier-Sokes Equaions Differenial form of momenum equaion can be derived by applying conrol volume form o elemenal conrol volume The differenial equaion of linear momenum: elemenal fluid volume approach 1

13 57:00 Mechanics of Fluids and Transpor Processes Chaper 6 Professor Fred Sern Fall V (1) = () = = combining and making use of he coninuiy equaion yields where f f f or f 13

14 57:00 Mechanics of Fluids and Transpor Processes Chaper 6 Professor Fred Sern Fall Body forces are due o exernal fields such as graviy or magneics. Here we only consider a graviaional field; ha is, F and i.e., = dfgrav = ρgdydz g = gkˆ for g z f body = ρgkˆ body Surface forces are due o he sresses ha ac on he sides of he conrol surfaces symmeric (σ ij = σ ji ) σ ij = - pδ ij + τ ij nd order ensor δ ij = 1 i = j normal pressure viscous sress δ ij = 0 i j = -p+τ xx τ xy τ xz τ yx -p+τ yy τ yz τ zx τ zy -p+τ zz As shown before for p alone i is no he sresses hemselves ha cause a ne force bu heir gradiens. x σ + y + z df x,surf = ( ) ( ) ( ) dydz xx p + x x τ σ xy + y = ( ) ( ) ( ) dydz xx τ xy σ xz + z τ xz 14

15 57:00 Mechanics of Fluids and Transpor Processes Chaper 6 Professor Fred Sern Fall This can be pu in a more compac form by defining vecor sress on x-face τ x = τ xx î + τ xy ĵ+ τ xz kˆ and noing ha p df x,surf = + τx dydz x p f x,surf = + τx per uni volume x similarly for y and z p f y,surf = + τy τ y = τyxî + τyy ĵ+ τyzkˆ y f z,surf = p + τz τ z = τzxî + τzy ĵ+ τzzkˆ z finally if we define τ = τ î + τ ĵ+ τ kˆ hen ij x y z f surf = p + τij = σij σ ij = pδij + τij 15

16 57:00 Mechanics of Fluids and Transpor Processes Chaper 6 Professor Fred Sern Fall Puing ogeher he above resuls f = f body + f surf = ρ DV D f body = ρgkˆ f surface = p + τij DV V a = = + V V D ρa = ρgkˆ p+ τ ij ineria body force force surface surface force due o force due due o viscous graviy o p shear and normal sresses 16

17 57:00 Mechanics of Fluids and Transpor Processes Chaper 6 Professor Fred Sern Fall For Newonian fluid he shear sress is proporional o he rae of srain, which for incompressible flow can be wrien where, = coefficien of viscosiy = rae of srain ensor = Ex) 1-D flow where, 0 Navier-Sokes Equaion Coninuiy Equaion 17

18 57:00 Mechanics of Fluids and Transpor Processes Chaper 6 Professor Fred Sern Fall Four equaions in four unknowns: V and p Difficul o solve since nd order nonlinear PDE x: y: z: u x + v y + w z = 0 Navier-Sokes equaions can also be wrien in oher coordinae sysems such as cylindrical, spherical, ec. There are abou 80 exac soluions for simple geomeries. For pracical geomeries, he equaions are reduced o algebraic form using finie differences and solved using compuers. 18

19 57:00 Mechanics of Fluids and Transpor Processes Chaper 6 Professor Fred Sern Fall Ex) Exac soluion for laminar incompressible seady flow in a circular pipe Use cylindrical coordinaes wih assumpions 0 : Fully-developed flow 0 : Flow is parallel o he wall Coninuiy equaion: 0 B.C i.e., 0 19

20 57:00 Mechanics of Fluids and Transpor Processes Chaper 6 Professor Fred Sern Fall Momenum equaion: or 0sin 0cos 0 where, sin cos (1) () (3) Equaions (1) and () can be inegraed o give sin pressure is hydrosaic and is no a funcion of or 0

21 57:00 Mechanics of Fluids and Transpor Processes Chaper 6 Professor Fred Sern Fall Equaion (3) can be wrien in he from 1 1 and inegraed (using he fac ha = consan) o give Inegraing again we obain B.C ln a any cross secion he velociy disribuion is parabolic 1

22 57:00 Mechanics of Fluids and Transpor Processes Chaper 6 Professor Fred Sern Fall 006 1) Flow rae : 8 where, If he pressure drops Δ over a lengh l: l Δ 8l ) Mean velociy : 1 Δ Δ 8l 8l 3) Maximum velociy : 0 4 Δ 4l 1

23 57:000 Mechanics of Fluids and Transpor Processes Professor Fred Sern Fall 006 Differenial Analysis of Fluid Flow Chaper 3 6 We now discuss a couple of exac soluions o he Navier- Sokes equaions. Alhough all known exac soluions (abou 80) are for highly simplified geomeries and flow condiions, hey are very valuable as an aid o our under- sanding of he characer of he NS equaions and heir so- luions. Acually he examples o be discussed are for in- ernal flow (Chaper 8) and open channel flow (Chaper 10), bu hey serve o underscore and display viscous flow. Finally, he derivaions o follow uilize differenial analy- sis. See he ex for derivaions using CV analysis. Couee Flow boundary condiions Firs, consider flow due o he relaive moion of wo paral- lel plaes Coninuiy Momenumm u = 0 x 0 = μ d dy u u = u(y) v = o p p = = 0 x y or by CV coninuiy and momenumm equaions: 3

24 57:000 Mechanics of Fluids and Transpor Processes Professor Fred Sern Fall 006 ρu 1 Δy = u 1 = u 1 ρ u Δ y Chaper 4 6 F i.e. F x = uρ V da = ρq( u u 1 ) = 0 dp d τ = pδy p + Δx Δ y τδx + τ + dy Δx = 0 dy d τ = 0 dy d du μ = dy dy d u μ = 0 dy 0 from momenum equaion du μ = C dy C u = y + D μ u(0) = 0 D = 0 U u() = U C = μ U u = y du τ = μ dy μ U = = con nsan 4

25 57:000 Mechanics of Fluids and Transpor Processes Professor Fred Sern Fall 006 Generalizaion for inclined flow wih a consan pressure gradien Chaper 5 6 Coninuiy Momenumm u = 0 x 0 = ( p + γz x d u ) + μ dy u = u(y) v = o p = 0 y d u i.e., μ dy = γ dh h = p/γ +z = consan plaes horizonal dz = 0 plaes verical dz =-1 which can be inegraed wice o yield μ du dy = γ dh y + A 5

26 57:00 Mechanics of Fluids and Transpor Processes Chaper 6 Professor Fred Sern Fall dh y μ u = γ + Ay + B now apply boundary condiions o deermine A and B u(y = 0) = 0 B = 0 u(y = ) = U μ U = γ dh + A A = μu γ dh γ dh y u(y) = μ γ dh μ 1 μu + γ μ U y y + dh = ( ) y This equaion can be pu in non-dimensional form: u γ dh y y y = 1 + U μu define: P = non-dimensional pressure gradien γ dh p = h = + z μu γ Y = y/ u = P Y(1 Y) + Y U parabolic velociy profile γ 1 dp = μu γ z + dz 6

27 57:000 Mechanics of Fluids and Transpor Processes Professor Fred Sern Fall 006 Chaper 7 6 u U = Py Py + y q u = udy = 0 q U 0 = [ ] dy u U = 0 P y P y y + dy P P = + 3 u = U P u = 1 μ γ dh + U 7

28 57:00 Mechanics of Fluids and Transpor Processes Chaper 6 Professor Fred Sern Fall u For laminar flow < 1000 ν Re cri 1000 The maximum velociy occurs a he value of y for which: du d u P P 1 = 0 = 0 = y + dy dy U y = ( P + 1) = u max P P for U = 0, y = / u = u ( y ) UP + 4 U max max = + U 4P noe: if U = 0: u u max = P 6 P 4 = 3 The shape of he velociy profile u(y) depends on P: dh 1. If P > 0, i.e., < 0 he pressure decreases in he direcion of flow (favorable pressure gradien) and he velociy is posiive over he enire widh γ dh = γ d p γ + z = dp γsin θ dp a) < 0 8

29 57:00 Mechanics of Fluids and Transpor Processes Chaper 6 Professor Fred Sern Fall dp b) < γsin θ 1. If P < 0, i.e., dh > 0 he pressure increases in he direcion of flow (adverse pressure gradien) and he velociy over a porion of he widh can become negaive (backflow) near he saionary wall. In his case he dragging acion of he faser layers exered on he fluid paricles near he saionary wall is insufficien o overcome he influence of he adverse pressure gradien. dp dp γsin θ > 0 > γsin θ or γ sin θ < dp dh. If P = 0, i.e., = 0 he velociy profile is linear U u = y dp a) = 0 and θ = 0 Noe: we derived his special case dp b) = γsin θ u For U = 0 he form = PY( 1 Y) + Y is no appropriae U u = UPY(1-Y)+UY γ dh = Y( 1 Y) + UY μ 9

30 57:00 Mechanics of Fluids and Transpor Processes Chaper 6 Professor Fred Sern Fall γ μ 3. Shear sress disribuion dh Now le U = 0: u = Y( 1 Y) Non-dimensional velociy disribuion * u = = ( 1 ) + where u U γ dh P μu y Y * u u P Y Y Y U is he non-dimensional velociy, is he non-dimensional pressure gradien is he non-dimensional coordinae. Shear sress du τ = μ dy In order o see he effec of pressure gradien on shear sress using he non-dimensional velociy disribuion, we define he non-dimensional shear sress: Then * τ τ = 1 ρ U ( ) ( ) * 1 Ud u U μ du τ = μ = 1 ρu d y ρu dy μ = ( PY + P + 1) ρu μ = ( PY + P + 1) ρu = A PY + P+ 1 ( ) * 30

31 57:00 Mechanics of Fluids and Transpor Processes Chaper 6 Professor Fred Sern Fall where μ A > 0 is a posiive consan. ρu So he shear sress always varies linearly wih Y across any secion. Y = 0 : A he lower wall ( ) A he upper wall ( 1) ( 1 ) * τ = + lw A P * τ = ( 1 ) Y = : uw A P For favorable pressure gradien, he lower wall shear sress is always posiive: 1. For small favorable pressure gradien ( 0< P < 1) : * * τ > 0 and τ > 0 lw. For large favorable pressure gradien ( 1) * lw uw τ > 0 and τ < 0 * uw P > : τ τ ( 0< P < 1) ( P > 1) For adverse pressure gradien, he upper wall shear sress is always posiive: 1. For small adverse pressure gradien ( 1< P < 0) : * τ > 0 and τ > 0 * lw uw 31

32 57:00 Mechanics of Fluids and Transpor Processes Chaper 6 Professor Fred Sern Fall For large adverse pressure gradien ( P < 1) : * τ < 0 and τ > 0 * lw uw 3

33 57:00 Mechanics of Fluids and Transpor Processes Chaper 6 Professor Fred Sern Fall τ τ ( 1< P < 0) ( P < 1) For U = 0, i.e., channel flow, he above non-dimensional form of velociy profile is no appropriae. Le s use dimensional form: u = γ dh Y( 1 Y) = γ dh y( y) μ μ Thus he fluid always flows in he direcion of decreasing piezomeric pressure or piezomeric head because γ 0, y 0 μ > > and y > 0. So if dh is negaive, u is posiive; if dh is posiive, u is negaive. Shear sress: du γ dh 1 τ = μ = y dy Since 1 y > 0, he sign of shear sress τ is always opposie o he sign of piezomeric pressure gradien dh, and he magniude of τ is always maximum a boh walls and zero a cenerline of he channel. 33

34 57:000 Mechanics of Fluids and Transpor Processes Professor Fred Sern Fall 006 For favorable pressuree gradien, 0 For adverse pressure gradien, dh < 0 dh >, τ > 0, τ < 0 Chaper 34 6 τ τ dh < 0 dh 0 > Flow down an inclined plane uniform flow velociy and deph do no change in x-direcion Coninuiy du = 0 34

35 57:00 Mechanics of Fluids and Transpor Processes Chaper 6 Professor Fred Sern Fall d u x-momenum 0 = ( p + γz) + μ x dy y-momenum = ( p + γz) y dp = 0 0 hydrosaic pressure variaion d u μ dy = γsin θ du dy γ = μ sin θy + c u γ y = sin θ μ + Cy + D du dy y= d = 0 γ = μ sin θd + c c = + γ sin θd μ u(0) = 0 D = 0 u γ y = sin θ μ + γ sin θdy μ γ μ = sin θ y( d y) 35

36 57:00 Mechanics of Fluids and Transpor Processes Chaper 6 Professor Fred Sern Fall gsin θ ν u(y) = y( d y) q = d 3 γ y udy = sin θ dy 0 μ γ = d sin θ 3 μ d 0 discharge per uni widh V avg q 1 γ gd = = d sin θ = sin θ d 3 μ 3ν in erms of he slope S o = an θ sin θ V gd S = 3 ν o Exp. show Re cri 500, i.e., for Re > 500 he flow will become urbulen p y = γ cosθ Re = Vd cri 500 ν p = γ cosθ y + C ( d) = p = γcosθd C p o + 36

37 57:00 Mechanics of Fluids and Transpor Processes Chaper 6 Professor Fred Sern Fall p = γcosθ d y + p i.e., ( ) o * p(d) > p o * if θ = 0 p = γ(d y) + p o enire weigh of fluid imposed if θ = π/ p = p o no pressure change hrough he fluid 37