Numerical Study of Turbulent Structure. Hitoshi SUGIYAMA, Mitsunobu AKIYAMA and Mieko TANAKA

Transcription

1 Vol.1 ( ) Numerical Study of Turbulent Structure in Curved Open-Channel Flow with Flood Plains Hitoshi SUGIYAMA, Mitsunobu AKIYAMA and Mieko TANAKA ( ) Ph.D. ( ) ( ) A numerical study of turbulent ow developing in a curved open-channel ow with ood plains is performed by using an algebraic stress model. The ow conguration of interest to this study consists of 18 degree bend with oodplains. This turbulent ow is characterized by interactive motions between pressure driven secondary ow and turbulence driven secondary ow in the curved channel. Therefor,it is indispensable to adopt anisotropic turbulence model to predict its ow preciously. Calculated results are compared with the experimental data available. In order to clarify the generation mechanism of secondary ow, production terms concerned with it, i.e. centrifugal, pressure and turbulence driven forces, are evaluated by using calculated results. As a result of this numerical study, the present method can realize the main features well although agreement is cirtainly not perfect in all detail.as for the genaration mechanism of secondary ow, the calculated results show that turbulence driven forces plays an important role near the boundary between main channel and ood plains. Key Words : numerical analysis, curved open-channel, compound open channel, secondary ow, Reynolds stress model, boundary-tted coordinate system 1 1) ) -- 3) 18 Thomson 4) Ippen-Drinker 5) - 6) Kiely 7) 3 1

2 Sellin-Ervine-Willetts 8) Shiono-Muto 9) 18 Leschziner-Rodi 1) k- 11) -- 3) c 1 c c 1 c 1 c c c : k : L : P k : R : R e : = RU b = r : r c : X 31 X 3 X u i u j : r rc = 6 unit:[cm] U i : i U b : U r : U : U z : u r : u : u z : X i : i z : : : : : ij : : 3 U z U r U θ 3.1-3) -1 cm 8cm cm 18 r c =6cm 31cm41cm R 4 68 R R e =:33 1 3

3 X 1 X X 3 ;X 1 =4R +X 1 =4R 3. 1 Rodi 1) Du i u j Dt = ;(u i u k + u j u k k [(u i u j u k iu k + p ( jku i + ik u j )] + p i j ) j k -1-13) ij1 ij ijw f(l=x w ) 1 X w 3.3 y 1 = c3=4 k 3= y () 14) 1 15) η ξ ξ ij1 + ji1 ;C1 " C +8 ; 11 ij + ji ζ - -1 k u i u j 3 k ij P ij ; P k ij j + i ; 8C ; 11 D ij ; 3 P k ij C 1 = C L 1 + C 1 f C = C L [ ij + Xw + C f Xw ji ]w = + L f k P ij = ;u i u k ; u j u k D ij = ;u i u k ; u j u k P k = ;u k u l - C 1 C C 1 C @ (3) 16) = =9 - R e =:

4 X 1 /4R = θ = X /4R θ = θ = θ = θ = θ = θ = X 1 /4R = X 1 /4R = 44. X /4R X /4R U θ /U b X 1 /4R = θ = X /4R θ = θ = θ = θ = θ = θ = X 1 /4R = X 1 /4R = 44. X /4R X /4R U θ /U b Uθ/Ub X 1/4R =. θ = θ = 3 U θ/ub =.33 θ = 1 θ = 15 θ = 18 k Regular Grid QUICK θ = X 1/4R =. 4 θ = 9 X1/4R = k = U b 1 ;5 = k 3= =4R X 1 =4R = ;: =. 11) 4

5 X 1 /4R = θ = X /4R θ = θ = θ = θ = X 1 /4R = X /4R U b X1/4R X 1 /4R = θ = X /4R θ = θ = θ = θ = X 1 /4R = X /4R U b -5 Ur/Ub X 3 /4R =.38 X1/4R =. θ = θ = 3 θ = 6 θ = 9 θ = Uz/Ub X 3 /4R =.33 X1/4R =. θ = θ = 3 θ = 6 θ = 9 θ = , =3 1.1 = X 3 =4R =:33 5% 3) -5 X 1 =4R = ;: 3% = =3 5

6 X 1 /4R = θ = 3 X /4R θ = (r r u c)/4r θ /U b X 1 /4R = θ = 3 X /4R θ = (r r u c)/4r θ /U b X 1 /4R = θ = 3 X /4R θ = (r r u c)/4r r /U b X 1 /4R = θ = 3 X /4R θ = u (r r r /U b c)/4r X 3/4R X 3/4R X 3/4R X 3/4R X 1 /4R = θ = 3 X/4R θ = u z /U b X 1 /4R = θ = 3 X /4R θ = (r r u c)/4r z /U b =6 (r ; r c )=4R =:4 =6 3) 6

7 X 1 /4R = θ = 3 X /4R θ = k/u (r r b c)/4r X 1 /4R = θ = 3 X /4R θ = k/u (r r b c)/4r uθ /Ub X 3/4R =.34 X 1/4R =. θ = θ = 3 θ = 6 θ = 9 θ = 18 X 1/4R = ur /Ub =.34 X 1/4R =. θ = θ = 3 θ = 6 θ = 9 θ = X1/4R = u z /Ub.5. =.34 X 1/4R =. θ =.5. θ = 3.5. θ = 6.5. θ = θ = X 1/4R = =18 X 1 =4R =: X 1 =4R = ;: -6 X 3 =4R =:38:33 4. q -7,8,9 u q qu r u z -7 = 135.8, -3, =3 7

8 uθur/ub X 3/4R =.34 X1/4R =. θ = θ = 3 θ = 6 θ = 9 θ = 18 X 1/4R = uθuz/ub =.34 X1/4R =. θ = θ = 3 θ = 6 θ = 9 θ = 18 X 1/4R = ) -11 X 3 =4R = :34-1 u u r u u z X 1 =4R = ;: u u r (r ; r c )=4R =: u u =@r X u u r X 1 =4R = =@r u u z u u =@X 3 u u r 4.3 (4) (5)(6) (4)(5)(6) U b U b 4R F rcen = U r F rpre F Zpre (4) (5) 8

9 " F rtur = F ztur = θ = θ = θ = ρU b /4R (F r,cen, ) -13 θ = θ = θ = ρU b /4R (F r,pre, F z,pre ) -14 θ = θ = θ = ρU b /4R (F r,tur, F z,tur ) -15 r u r zu ru u z r@ ; u r ; u r z ; u ru r!#! (6) -13,14,15 5 (1) % () (3) (4) (5) (6) (7) 9

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