Centro de Estudos de Fenómenos de Transporte, FEUP & Universidade do Minho, Portugal

Transcription

1 DEVELOPING CLOSURES FOR TURBULENT FLOW OF VISCOELASTIC FENE-P FLUIDS F. T. Pinho Centro de Estudos de Fenómenos de Transporte, FEUP & Universidade do Minho, Portugal C. F. Li Dep. Energy, Environmental and Chemical Engineering, Washington University of St. Louis, St Louis, MO, USA B. A. Younis Dep. Civil and Environmental Engineering, University of California, Davis, USA R. Sureshkumar Dep. Energy, Environmental and Chemical Engineering, Washington University of St. Louis, St Louis, MO, USA Acknowledgments:FCT (POCI/EQU/56342/24), Gulbenkian Foundation Conference on Whither rheology? 2 nd -4 th April 27 Lake Vyrnwy Hotel, Mid Wales, UK

2 Summary Relevance: some facts about drag reduction Governing equations for FENE-P in RANS form Reynolds decomposition of FENE-P equations Closure needs and analysis of DNS case (LDR) A simplified closure with a priori testing of DNS data Some concerns regarding limiting cases Some preliminary results Conclusions 2

3 Relevance: drag reduction in turbulent pipe flow 1-1 f Blasius eq. 6 u % PAA Re=429.25% CMC Re= 166.3% CMC Re=43.9/.9% CMC/XG Re= % PAA.2% PAA.25% CMC.2% XG.9%/.9% CMC/XG Virks MDRA asymptote Re Reduction of shear Reynolds stress (DR) - Increase of normal streamwise Reynolds stress - Dampening of normal radial and tangential Reynolds stress Deficit of Reynolds stress y + Escudier et al JNNFM (1999) 3

4 Time-average governing equations: turbulent flow & FENE-P Continuity:!U i!x i = Momentum balance:! "U i "t ^- instantaneous Overbar or capital letter- time-average or small letter- fluctuations "U +!U i k = # "p " +$ 2 U i s # " (!u i u "x k "x i "x k "x k "x k )+ "% ik, p k "x k Rheological constitutive equation: FENE-P! ij = 2" s S ij +! ij, p ˆ! ij, p = " p & # f ( Ĉ kk )Ĉij $ f ( L )% ( ij ) $ f ( Ĉ kk )Ĉij +! "Ĉij "Ĉij +Ûk # "t "x Ĉ "Ûi "Û # jk k "x Ĉik j & ) = f (L) ij % k "x k ( # % $!Ĉij!t +Ûk!Ĉij!x k " Ĉ jk!ûi!û "!x Ĉik j k!x k & ) ( = Ĉij = " ˆ ij, p + p 4

5 DNS case: LDR DNS, DR=18% (LDR) Fully-developed channel flow u 2,y 1,x 2h We! = 25, Re! = 395 " =.9, L 2 = 9 We! = "u 2! Re #! = hu! " 5

6 Reynolds decomposition of conformation tensor ˆB = B + b where b = Time average polymeric stress Function: f ( C kk ) = L2! 3 L 2! C kk! ij, p = " p & # f ( C kk )C ij $ f ( L)% ij ( ) + " p # f ( C + c kk kk )c ij Time average conformation tensor equation f c ij Neglected (see slide) " + #c %! C ij+! ij #u uk $ c i #u - kj + c j ik #x,- k & #x k #x k (. )/ = $ +, f ( C kk )C ij $ f ( L)1 ij + f ( C kk + c kk )c ij. /! "c ij $ "u C ij + uk # c i "u j kj + c ik "x & k % "x k "x k ) ( = # ij, p + p CT ij NLT ij 6

7 Simplifying assumptions: justification from DNS times smaller c 12 /[L 2 -(C kk +c kk )] C 12 /[L 2 -(C kk )] y + f ( C kk )C 12 >> f c 12 = f ( C kk + c kk )c 12 = ( L 2! 3) c 12 ( ) L 2! C kk + c kk 7

8 Polymer stress 15 1 (a) xx component CT NLT M D S= -! p xx xy 1 5 (b) xy component CT NLT M D S = -! p (c) yy component y + CT NLT M D S= -! p y + 2 yy CT(J) aver y + Housiadas et al (25) Li et al (26) JNNFM 8

9 Modeling requirements! $ ij, p %c & = #C ij # ij %u uk + c i %u kj + c j ik " p %x ( k %x k %x k ) + M ij NLT ij : originates in Oldroyd derivative not negligible, must be modeled What about: CT ij 1) Reynolds stresses? 2) Turbulent kinetic energy? : originates in advective term, it is negligible no need for modeling DNS: Housiadas et al (25), Li et al (26) JNNFM 3) Dissipation of turbulent kinetic energy or of Reynolds stresses? 9

10 Transport equation for the Reynolds stresses and k! "u i u j "t "u +!U i u j k = P ij +Q ij +Q V ij + D ij,n + # ij $!% N V ij $!% ij "x k Q V ij =! ( u i " jk, p + u j " ) ik, p!x k! V ij = 1 % " # jk, p & $u i + # ik, p $x k $u j $x k Viscoelastic turbulent transport due to fluctuations polymeric stresses ( ) Viscoelastic work of polymer chains: dissipation of energy plus stored free energy (< ou >)! Dk Dt +!u "U i "k iu k = #!u i # "pu i " 2 k "u +$ s #$ i "u i s + "% ik, p # % ik, p "x k dx i "x i "x i "x i "x k "x k "x k u i "u i "x k ε Ν Q V! V 1

11 Transport equation of ε Ν "u 2! i " $ # Du i s & "x m "x m % Dt )+ 2! s ( "u i "x m " $ & % "x m #uk "U i "x k "u )+ 2! i " $ # "u u i k s & ( "x m "x m % "x k " #"p& "ui 2 " #" & 2 " #"+ "ui ui "ui +2! s % () 2! s % () 2! ik, p % 2 s "x m "x m $ "x i "x m "x m $ "x k "x m "x m $ "x k ) ( & ( = New term As for Newtonian fluids, the whole equation will be approximated 11

12 Budget of Reynolds stress (a) uu Budget Pxx Q N xx Q V xx D N xx!xx " N xx " V xx (c) vv Budget Pyy Q N yy Q V yy D N yy!yy " N yy " V yy y y+ - Need to model viscoelastic stress work - Need to model pressure strain (effect of elasticity)- Advanced mod. - Viscoelastic turbulent transport is not so important 12

13 Viscoelastic work! V " 1 # $ ik, p %u i = & ( p %x k # C f C + c ik mm mm ) ( ) %u i ( ) %u i %x k + c ik f C mm + c mm %x k + -, 35 Assumption: (same reasons) C ik f ( C mm + c mm )!u i < c ik f ( C mm + c mm )!u i!x k!x k 3 C kk 2 c kk C kk > 2 c kk! V " # p $% f C &u ( mm )c i ik = C &x! V k # p $% f ( C mm ) NLT nn y + C! V = 1.76 (from DNS- next slide) Needs model 13

14 Performance of the viscoelastic work model 1 5 (e)! V > Negative sign: different definitions -5-1! ( V + Re " ) ! V (DNS) -1.7f(C kk )NLT kk /2 versus f ( C kk ) NLT ij y + 1 3

15 Modeling NLT ij 1 Key ideas: 1) Write down exact equation- complex 4 lines long 2) Get inspiration from it: physical insight, trial-and-error, luck 3) Make assumptions 4) Do a priori testing of each term 5) Select appropriate combination and dimensional homogeneity 6) Try in code - under investigation f ( C mm ) NLT ij!!c kj!c kj!c kj u i u m + u i u m " Coef # u i u m!x m!x m!x m $ = function S ij,w ij,c ij," N ij, #u iu j, #C ij, #NLT ij &, M ij,u i u j % #x k #x k #x n ) ( f ( C mm ) NLT ij! #u = f µ1 C k u n #C ij "1 + C u 2 E3 $ C 2 kk u i u j + C & ) 14 u i u k W kn C nj + u j u k W kn C ni + u k u i W jn C nk ( #x n #x k % % ( ), + 15

16 Model for NLT ij 2 Blue model Red model C E3 =.4;C! 1 = 3;C "14 =.15 C E3 =.35;C! 1 = ;C "14 = f(ckk)nltxy f(ckk)nltii Pred f(ckk)nltxy Pred f(ckk)nltii f(ckk)nltyy Pred f(ckk)nltyy 25 2 f(ckk)nltxy f(ckk)nltii Pred f(ckk)nltxy Pred f(ckk)nltii f(ckk)nltyy Pred f(ckk)nltyy y y + 1 f µ1 = ( 1! exp (! y ) ) 2 16

17 Viscoelastic turbulent transport Q V! "# ik, pu i = $ p "x k % " & C ik f ( C mm + c mm )u i + c ik f ( C mm + c mm )u i "x k ( ) C kk > f 2 c kk ( Ĉ ) mm = L 2! 3 ( ) L 2! C mm + c mm Weak coupling between c kk and c ij, u i C ik f ( C mm + c mm )u i < c ik f ( C mm + c mm )u i Neglect of this term is irrelevant because non-neglected term is modeled Q V =! p " # #x $% f ( C mm )CU iik & k Needs model General case ( CU ) ijk 17

18 Model for CU ijk - Same modelling approach as with NLT ij 4 f ( C mm )CU ijk! + % $C = f µ2 "C #1 u i u kj $C m + u j u ik ( m " C # - 7 & $x m $x m )! f C mm, ( ) + ± u 2 C ± u 2 C j ik i jk,-.. / / 3 2 QFii2 Pred QFii2 C!1 =1.3;C!7 =.37 " % f µ2 = 1! exp $! y+ # 26.5 & y + 18

19 Final equations: low Re k-ε type model for channel flow Momentum: d % dy! s & du dy + " # $uv ( p,xy # dp ) dx =! xy, p = " p # f ( C kk )C xy du f ( C kk )C xy =!C yy dy +!NLT xy f ( C kk )C yy =!NLT yy +1 du f ( C kk )C xx = 2!C xy dy +!NLT xx +1 f ( C kk )C zz =!NLT zz +1 f ( C kk ) = L 2! 3 ( ) L 2! C xx + C yy + C zz Reynolds stress: du!"uv = "# T with dy! = C f T µ µ k 2!" N 19

20 k and ε transport equations = d + % dy! + "# ( T s dk. d P k 1 "!2 N 1 "D +! p -,& $ k ) dy/ dy, f ( C mm ) 3 CU nny 2 " d k %! N =! N + D N D N = 2! s $ # dy & ( ). / 1! f C mm p 3 2 NLT nn 2 = d,& dy! + "# T.( s - $ % ) + d!% N dy /!% N 1+ " f 1 C %1 k P k " 2 " f C % N 2 2 % 2 k + "E + E 3 p E =! s " # T 1$ f µ % ( ) d 2 U & dy 2 ( ) 2 f 1 = 1 f 2 = 1!.3exp ( 2!R ) T ( " % + f µ = 1! exp $!y+ - ) # 26.5 &, 2 based on Newtonian model of Nagano & Hishida (1984) 2

21 Some results 1: Re τ = 395; We τ = 25; β=.9, L 2 =9 3 u Virk s asymptote X DNS Black: Newtonian! =! s +! p! =! wall! =! wall ;same!q;re " = Newtonian log law FENE-P simulations C!1 = 1.3;C! 7 =.37;C "14 = 1.5 # 1 $4 C! V = 1.76 With! p C E 3 = 1.93 " 1 # y + Without! p C E 3 = 2.86 " 1 #4 21

22 Some results 2: Re τ = 395; We τ = 25; β=.9, L 2 =9 5 k + k + uv + 1 uv y y 1 22

23 Some results 3: Re τ = 395; We τ = 25; β=.9, L 2 =9 25 NLT ii 2 NLT ii 35 NLT xy 3 NLT xy y y + 23

24 Some results 4: Re τ = 395; We τ = 25; β=.9, L 2 =9 NLT xx NLT yy 15 5 NLT xx NLT yy y y + 24

25 Some results 5: Re τ = 395; We τ = 25; β=.9, L 2 =9 C xy C yy 25 C xy C yy y y + 25

26 Some results 6: Re τ = 395; We τ = 25; β=.9, L 2 =9 5 C xx 1 Stresses + C xx 4 Stresses +.8 -uv + +! p,xy +! s,xy SUMSTRESS du f ( C kk )C xx = 2!C xy dy +!NLT +1 xx y y 26

27 Some results 7: Re τ = 395; We τ = 25; β=.9, L 2 =9 8 + CU iiy CU ii y + 27

28 Conclusions - Developed simplified k-ε model: code and closures are working - Viscoelastic stress power well modeled by NLT ij - Viscoelastic turbulent transport (CU ijk ) is not that relevant at 18% - NLT ij is also required for C ij -Closure for NLT ij has deficiencies and needs significant improvement - Excessive dissipation of turbulence - Need to model viscoelastic turbulence production close to wall - Isotropic turbulence does not allow a good model - Need to consider anisotropic turbulence: anisotr. k-ε and RSM - Closure for CU ijk is fair but also needs improvement: small impact 28