dq 1 (5) q 1 where the previously mentioned limit has been taken.

Transcription

1 1 Vecto Calculus And Continuum Consevation Equations In Cuvilinea Othogonal Coodinates Robet Maska: Novembe 25, 2008 In ode to ewite the consevation equations(continuit, momentum, eneg) to some cuvilinea otogonal coodinate sstem(pola, spheical, elliptical etc.) it is a useful stateg to fist eview vecto analsis. A displacement vecto in some thee-dimensional otogonal metic coodinate sstem (q 1, q 2, q 3 ) can be witten b its base vectos b d = dq 1 ε 1 + dq 2 ε 2 + dq 3 ε 3 (1) whee ε coesponds to a base vecto. The base vectos do not necessail have unit magnitude and dimension length such as is the case in catesian coodinates. One can epand this in tems of unit vectos instead b letting ε i = h i e i whee h i is a scaling facto fo the base vecto, i.e. its length. The e i s ae mutuall othogonal and have unit magnitude. This gives fo the diffeential displacement d = dq 1 h 1 e 1 + dq 2 h 2 e 2 + dq 3 h 3 e 3 An eample of this is the displacement vecto in clindical pola coodinates (q 1, q 2, q 3 ) = (,, ), it can be witten as d = dˆ + dˆ + dˆ whee h 1, h 2, h 3 ae easil ecognied on inspection. It tanspies that these scaling factos does not necessail have unit magnitude and in fact could ca diffeent dimensions. In contast, a displacement vecto in catesian coodinates is witten d = dˆ + dŷ + dẑ whee all the h s have dimension and length of unit. The net diffeential displacement still caies the dimension length. Conside now a diffeential displacement in the space (q 1, q 2, q 3 ), its magnitude is d d = (ds) 2 = (h 1 dq 1 ) 2 + (h 2 dq 2 ) 2 + (h 3 dq 3 ) 2 (2) since the unit vectos ae mutuall othogonal. Gadient The component i of the gadient of a scala ψ is e i ψ = ψ s i = 1 h i ψ q i A genealiation gives the the gadient ψ = X i 1 ψ e i (3) h i q i Fo altenative deivations, see fo eample Afken&Webe[1]. Divegence The divegence of a vecto v is defined[1] H v dσ div v = R lim R (4) dτ 0 dτ Fo convenience, a ight-hand sstem has been chosen. A diffeential aea in the chosen space can be witten dσ = ds 2 ds 3 e 1 + ds 3 ds 1 e 2 + ds 1 ds 2 e 3 whee now, should it not be clea, ds i = h i dq i. In catesian coodinates fo instance, this would simpl educe to dσ = (dd)ˆ + (dd)ŷ + (dd)ẑ which is a slanted squae face. The above equation is a genealiation to cuved faces. This lets us evaluate the suface integal, the vecto poduct v dσ is integated temwise. Doing the v 1 dσ 1 tem fist one finds that the integal fo the two faces in the e 1 -diection is v 1 dσ 1 = v 1 ds 2 ds 3 + v 1 ds 2 ds 3 + (v 1ds 2 ds 3 ) dq 1 (5) whee the peviousl mentioned limit has been taken. ds 3 ds 1 e 2 e 3 ds 2 dσ 1 e 1 FIG. 1: Diffeential volume and sufaces. We will now eplain this, see FIG. 1 fo efeence. The fist tem is simpl the suface integal ove the face opposite the label dσ 1 in FIG. 1. The second tem comes fom a Talo epansion of the vecto component v 1 and the diffeential aea. This just means that the v 1 component changes b v 1 (q 1 + dq 1 ) = v 1 (q 1 ) + v 1 Similal the suface aea has changed b dσ 1 (q 1 + dq 1 ) = dσ 1 (q 1 ) + dσ 1 Fo notational easons we now denote v 1 (q 1 ) b v 1 and a simila fomalism fo dσ 1. Then the suface integal ove the second face is found b multiplication of v 1 (q 1 + dq 1 ) and dσ 1 (q 1 + dq 1 ) = v 1 dσ 1 + v 1 dσ 1 dq 1 + dσ 1 v 1 dq 1 + O(dq 2 1 ) = v 1 dσ 1 + (v 1dσ 1 ) which follows fom the poduct ule. Imposing the limit R dτ 0 leaves onl the fist ode tems and EQ. 5 is fulfilled. EQ. 5 now educes to v 1 dσ 1 = (v 1ds 2 ds 3 ) dq 1 = (v 1h 2 h 3 ) We have now integated ove two of the faces. The esult is of couse the same fo the othe fou faces, with a cclic pemutation of 123. In othe wods v 2 dσ 2 = (v 2ds 3 ds 1 ) dq 2 = (v 2h 3 h 1 ) v 3 dσ 3 = (v 3ds 1 ds 2 ) dq 3 = (v 3h 1 h 2 ) Addition of these thee integals gives the closed suface integal, as in EQ. 4. The diffeential volume is simpl ds 1 ds 2 ds 3 = Robet Maska: Novembe 25, 2008 Depatment Of Eneg And Pocess Engineeing Depatment Of Phsics

2 2 h 1 h 2 h 3. B collecting tems and imposing the limit we conclude that 1 (v1 h 2 h 3 ) div v = + (v 2h 3 h 1 ) + (v 3h 1 h 2 ) (6) h 1 h 2 h 3 q 3 As peviousl mentioned, in catesian coodinates the h i s ae equal to one. This allows us to wite the divegence as an inne poduct of and v. It is howeve common to adopt the notation v = div v even in cuvilinea coodinates. This is bad notation since it suggests that the divegence can be witten as an inne poduct which is not the case. Laplacian The Laplacian of a scala ψ(q 1, q 2, q 3 ) can be developed b taking the gadient of the divegence. That is, we choose the components of v as the vecto components of EQ. 3. This gives without dwelling on details div v = div gad ψ = 1 h2 h 3 ψ + h3 h 1 ψ h 1 h 2 h 3 h 1 h 2 + h1 h 2 ψ q 3 h 3 q 3 This is b definition the Laplacian and is sometimes witten 2 ψ. Some people seem to like confusing the Laplacian of a scala with the Laplacian of a vecto. A component of the Laplacian of a vecto is not the same as the Laplacian of the vecto component, as will be shown late. Cul The cul can be found using Stokes theoem S I (cul v) dσ = (7) v ds (8) One will meit fom doing this component-wise on an infinitesimal aea. Thus let the suface be made up of thee sufaces, each with q 1, q 3 o q 3 kept constant. (4) (q 2, q 3 ) e 2 ds 2 (q 3 ) (1) Suface q 1 = const (3) e (2) 3 ds 3 (q 2 + dq 2 ) FIG. 2: Temwise integation of EQ. 8 Conside fist the suface whee q 1 is constant. Following the loop in the figue counteclockwise fom the leftmost cone, the line integal is I v ds =v 2 ds 2 + v 3 ds 3 + (v 3ds 3 ) v 2 ds 2 + (v 2ds 2 ) = (h 3 v 3 ) dq 2 dq 3 v 3 ds 3 q 3 (h 2 v 2 ) dq 2 dq 3 q 3 This also follows b Talo epanding the vecto components. The fist integal along the line (1) is simpl = v 2 ds 2 Fo the second line integal (2), the infinitesimal length and the vecto component has changed accoding to a Talo seies v 3 (q 2 + dq 2 ) = v 3 (q 2 ) + v 3(q 2 ) dq 2 + O(dq 2 2 ) ds 3 (q 2 + dq 2 ) = ds 3 (q 2 ) + ds 3(q 2 ) dq 2 + O(dq 2 2 ) Using the same fomalism as ealie, v 3 (q 2 ) is denoted v 3 etc., combined with the poduct ule gives that the line integal (2) is = v 3 ds 3 + (v 3ds 3 ) dq 2 + O(dq 2 2) The emaining tems in EQ. 9 is found b identical agumentation on the two emaining line integals. Fo an infinitesimal aea, the left hand side of EQ. 8 can be witten (cul v) dσ = e 1 (cul v)ds 2 ds 3 S since the fist suface is oiented in the q 1 diection. Combining these esults with the limit of an infinitesimal aea gives [cul v] 1 = 1 (h 3 v 3 ) (h 2 v 2 ) h 2 h 3 q 3 since the highe-ode tems vanish in the limit. The emaining two components ae found in a simila wa, o b cclic pemutation of 123. Thus [cul v] 2 = 1 h 3 h 1 [cul v] 3 = 1 h 1 h 2 O in inde notation (h 1 v 1 ) (h 3 v 3 ) q 3 (h 2 v 2 ) (h 1 v 1 ) (9) [cul v] i = 1 h j h k ε ijk j (h k v k ) (10) whee ε ijk denotes the Levi-Civita pseudo-tenso. Occasionall, but ael, it is useful to wite this as a deteminant. It can then be witten e 1 h 1 e 2 h 2 e 3 h 3 1 cul v = h 1 h 2 h (11) 3 q 3 h 1 v 1 h 2 v 2 h 3 v 3 Net we will develop the divegence and cul in clindical and spheical pola coodinates. Vecto Calculus And Continuum Consevation Equations In Cuvilinea Othogonal Coodinates

3 3 θ FIG. 3: Clindical pola coodinates. FIG. 4: Spheical pola coodinates. Clindical Pola Coodinates In clindical pola coodinates (q 1, q 2, q 3 ) = (,, ) the elation between the clindical coodinates and the catesian is = cos ; = sin ; = A diffeential displacement can be witten which again gives One should now ecognie d = dˆ + dˆ + dˆ d d = (ds) 2 = (d) 2 + (d) 2 + (d) 2 h 1 = h = 1; h 2 = h = ; h 3 = h = 1 Going to EQ. 6 and EQ. 11, o EQ. 10, the divegence and cul is found b diect insetion to be div v = 1 (v ) v [cul v] = 1 v v [cul v] = v v [cul v] = 1 (v ) 1 v If one absolutel insists this can be witten in mati fom ˆ ˆ ˆ cul v = 1 v v v Spheical Pola Coodinates In spheical clindical coodinates (q 1, q 2, q 3 ) = (, θ, ) the elation to the catesian is = sin θ cos ; = sin θ sin ; = cos θ A diffeential displacement can be witten One should now ecognie d = dˆ + dθˆθ + d sin θ ˆ h 1 = h = 1; h 2 = h θ = ; h 3 = h = sin θ Going to EQ. 6 gives the divegence of a vecto v div v = 1 2 ( 2 v ) + 1 (v θ sin θ) + 1 sin θ θ sin θ Fom EQ. 10 the cul components can be found [cul v] = 1 (v sin θ) sin θ θ [cul v] θ = 1 sin θ [cul v] = 1 (v θ ) v 1 v θ (v) sin θ v θ This can be witten in mati fom following EQ. 11 ˆ ˆθ sin θ ˆ 1 cul v = 2 sin θ θ v v θ sin θv This is ael seen in pactice. The Consevation Equations In this tet we neglect the ewiting of the eneg equation and focus on the continuit equation and the incompessible Navie- Stokes equation. Fom hee on we adopt the notations cul v = v and div v = v. The full compessible equations ae in thei vecto fom given b[3] + div (v) = 0 (12) t and t + v v = 1 p + g + 1 τ ij (13) whee denotes the densit of the fluid and τ ij denotes the full viscous stess tenso including the second coefficient of viscosit. Including compessibilit into the Navie-Stokes equations makes fo a too cumbesome calculation. In the case of constant densit the final tem on the ight hand of EQ. 13 side educes to ν 2 v. We will stat b ewiting the Navie-Stokes equations[2]. Two vecto identities ae needed (v )v = v ( v) v 2 (14a) 2 v = ( v) ( v) (14b) Robet Maska: Novembe 25, 2008 Depatment Of Eneg And Pocess Engineeing Depatment Of Phsics

4 4 This can be found in an elementa book on calculus, and is in fact quite eas to show using the Levi-Civita tenso. This puts the Navie-Stokes equations on the fom v 1 t v ω+ 2 v 2 = 1 p+g+ν [ ( v) ω] (15) whee ω = v. The fluid velocit vecto is denoted b v and the components b v 1, v 2, v 3. The poblem with the vecto Laplacian is that we have 2 (ˆv + ˆv + ẑv ) = 2 (ˆv + ˆv ) + ẑ 2 v (16) Witing out the fist paanthesis on the ight hand side would equie seveal pages of equations. Convesel, the Laplacian in catesian coodinates is 2 (ˆv + ŷv + ẑv ) = ˆ 2 v + ŷ 2 v + ẑ 2 v This ma seem confusing at fist, but emembe that we ae not looking fo the Laplacian of a component v i, but looking fo component i of the Laplacian of v. Confusing these is a common mistake! Clindical Pola Coodinates In clindical pola coodinates, the gadient is given b EQ. 3, ω and ω ae given b EQ. 10 o EQ. 11. In ode not to confuse the densit with the adial coodinate, we denote the latte b. Stating with the tem ( v), these equations give afte some etensive ewiting [( v)] = 2 v v v v v [( v)] = 1 2 v v v v The tems ω ae(afte some ewiting) 1 2 [ ω] = v v [ ω] = v + v v 1 v v Notice that it is not necessa to calculate [( v) ω] because b EQ. 16 one finds that the vecto Laplacian [ 2 v] = 2 v. This is not tue fo the othe two components which is wh the need to be eplicitl calculated. In addition, the tem v ω is [v ω] = v [v ω] = v 1 (v) v v v v [v ω] = v v v v v 1 (v ) v 1 1 v v v v Thee is now onl one moe vecto tem which needs teatment, and that is the gadient of 1 2 v 2. B EQ. 3 this is 1 v 2 v 2 = v + v + v v 1 2 v 2 = v v + v + v v 1 2 v 2 v = v + v + v v We ae now in position to wite down the momentum equations b use of EQ. 15. Some viscous tems can be put into the Laplacian b EQ. 7, these tems ae put fist on the ight hand side of the equations above. Fo the adial component ˆ one finds v t + (v ) v v2 = 1 p + g + ν Similal the aimuthal component is vv + (v ) v t + 2 v 2 = 1 Finall, the vetical component is 2 v v p + g + ν 2 v v 2 v t + (v ) v = 1 p + g + ν 2 v whee the convective time-deivative is v = v + v + v It is eas to be led asta at this point since one might intuivel think that since [(v )v] = (v )v this also applies in cuvilinea coodinates. This, as in the case of the vecto Laplacian, is not tue. One should now have leaned the diffeence between the Laplacian of a vecto component and the component of a vecto Laplacian. To summaie [ 2 v] = 2 v v [ 2 v] = 2 v v v 2 [ 2 v] = 2 v The equation of continuit can be witten fom EQ. 4 t + 1 (v ) + 1 (v ) + (v ) The eneg equation is left out of this tet. Spheical Pola Coodinates The same tems ae needed, v ω, 1 2 v 2, ( v) and ω. We will howeve not go into the eplicit details on the calculation but cite the esult. B use of EQ. 3, 7, 10 and EQ. 14b one finds afte etensive use of the poduct ule and ewiting that the vecto Laplacian(the final tem in EQ. 15) is [ 2 v] = 2 v 2v 2 2 v θ 2 θ 2v θ 2 tan θ 2 2 sin θ [ 2 v] θ = 2 v θ [ 2 v] = 2 v v θ 2 sin 2 θ v θ v 2 sin 2 θ sin θ 2cos θ 2 sin 2 θ v + 2cos θ 2 sin 2 θ v θ Fo the left hand side of the Navie-Stokes equations one finds 1 v ω + 2 v 2 = (v )v v2 θ + v2 1 v ω + 2 v 2 = (v )v θ vv θ v 2 cot θ θ 1 v ω + 2 v 2 = (v )v vv + v θv cot θ Vecto Calculus And Continuum Consevation Equations In Cuvilinea Othogonal Coodinates

5 5 whee the convective tem is v = v + v θ θ + v sin θ To summaie, the incompessible Navie-Stokes equations ae (stating with the adial component) v t +(v )v v2 θ + v2 = 1 p 2 v 2v v θ θ The ˆθ component: 2v θ 2 tan θ 2 2 sin θ + g + ν v θ t +(v )v θ vv θ v cot θ = 1 p θ + g θ + ν ` 2 v θ v θ 2 sin 2 θ + 2 v 2 θ and finall the aimuthal component 2 cos θ 2 sin 2 θ t +(v )v vv + v θv cot θ + ν 2 v = 1 p sin θ + g v 2 sin 2 θ + 2 v 2 sin θ + 2cos θ 2 sin 2 θ v θ The equation of continuit can be witten fom EQ. 4 and is 1 t + ( 2 v ) (v θ sin θ) + 1 (v ) = 0 sin θ θ sin θ Fo constant densit, vanishes. The Navie-Stokes equations ae quite mess but ae sometimes useful. The eneg equation is best tansfomed b tansfoming each element in the viscous stess tenso. * Refeences [1] Geoge B. Afken and Hans J. Webe. Mathematical Methods Fo Phsicists. Academic Pess, 5th edition, [2] W.P. Gaebel. Advanced Fluid Mechanics. Academic Pess, 1st edition, [3] Hebet Oetel. Pandlt s Essentials of Fluid Mechanics, volume 158. Applied Mathematical Sciences, 2nd edition, Robet Maska: Novembe 25, 2008 Depatment Of Eneg And Pocess Engineeing Depatment Of Phsics