Numerical Simulation of the Quasi-2D Turbulence on a Half Soap Bubble Heated at the Bottom
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摘要:
底部加热的肥皂泡是一种全新的二维热对流模型,在实验中已发现肥皂泡上的岛涡运动规律与飓风轨迹规律一致。然而,肥皂泡的曲面特征对其准二维流场的数值模拟以及数据分析造成了较多困难。针对肥皂泡球面几何特征,该文介绍了其直接数值模拟(DNS)方法,及其流场空间波数谱、湍流通量和结构函数的计算分析方法。开展了Ra=3×107,3×109,3×1011的数值计算,并获得了相应的波数谱、通量和湍流结构函数。计算结果表明,肥皂泡上速度的小尺度脉动特征满足Bo59的理论标度律,通过湍动能与拟涡能通量特征,发现在该准二维湍流场中存在湍流能量双级串现象。且随着Rayleigh数的增加,大尺度结构湍能量减小,更小尺度湍流结构能量增加。
Abstract:The soap bubble heated at the bottom was introduced as a novel quasi-2D turbulence system. The curved geometry of the bubble brings challenges for the direct numerical simulations (DNS) of the turbulence on the bubble. In order to overcome the difficulties due to the curved geometry, a numerical method based on the stereographic projection was implemented for the DNS of the soap bubble. The numerical methods to compute the spectrum, the flux and the structure functions of the flows on the bubble were described in detail. Three different Rayleigh numbers Ra=3×107,3×109,3×1011 were used in the simulation based on the present numerical methods. Then, the related spectrum, flux and structure functions were calculated. The results indicate that, both the inversed energy cascade and forward enstrophy cascade can be observed in all the calculation cases. The Bo59 scaling law fits the small-scale fluctuations on the soap bubble. With the increase of the Rayleigh number, the turbulent energy decreases for the large-scale plumes, and the kinetic energy increases for the higher wave number structures.
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Key words:
- soap bubble /
- turbulence /
- structure function /
- direct numerical simulation
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图 5 瞬时流场图:(a) 无量纲瞬时温度场,Ra=3×107;(b) 无量纲瞬时温度场,Ra=3×109;(c) 无量纲瞬时温度场,Ra=3×1011;(d) 无量纲瞬时动能场,Ra=3×107;(e) 无量纲瞬时动能场,Ra=3×109;(f) 无量纲瞬时动能场,Ra=3×1011;(g) 无量纲瞬时拟涡能场,Ra=3×107;(h) 无量纲瞬时拟涡能场,Ra=3×109;(i) 无量纲瞬时拟涡能场,Ra=3×1011
Figure 5. The instantaneous flow field: (a) dimensionless instantaneous temperature field, Ra=3×107; (b) dimensionless instantaneous temperature field, Ra=3×109; (c) dimensionless instantaneous temperature field, Ra=3×1011; (d) dimensionless instantaneous kinetic energy field, Ra=3×107; (e) dimensionless instantaneous kinetic energy field, Ra=3×109; (f) dimensionless instantaneous kinetic energy field, Ra=3×1011; (g) dimensionless instantaneous enstrophy field, Ra=3×107; (h) dimensionless instantaneous enstrophy field, Ra=3×109; (i) dimensionless instantaneous enstrophy field, Ra=3×1011
图 9 纬度与经度方向上2到9阶的温度和速度结构函数:(a) 纬度方向的温度结构函数;(b) 纬度方向的速度结构函数;(c) 经度方向的温度结构函数;(d) 经度方向的速度结构函数
Figure 9. The temperature and velocity structure functions in the latitude and longitude directions, n=2~9: (a) the temperature structure functions in the latitude direction; (b) the velocity structure functions in the latitude direction; (c) the temperature structure functions in the longitude direction; (d) the velocity structure functions in the longitude direction
算法1 共轭梯度算法 1 $ {\boldsymbol{r}} = \bar{\bar{{\boldsymbol{A}}}}{\boldsymbol{T}} - {\boldsymbol{C}} $; 2 $ {\boldsymbol{d}} = {\boldsymbol{r}} $; 3 while $ \| \bar{\bar{{\boldsymbol{A}}}}{\boldsymbol{T}} - {\boldsymbol{C}}\| \geqslant \xi $ do 4 ${\boldsymbol{a}} = \dfrac{{\boldsymbol{r}}^{{\rm{T}}}{\boldsymbol{r}}}{{\boldsymbol{d}}^{{\rm{T}}}\bar{\bar{{\boldsymbol{A}}}}{\boldsymbol{d}}}$; 5 $ {\boldsymbol{T}} = {\boldsymbol{T}} + {\boldsymbol{a}}{\boldsymbol{d}} $; 6 $\,\dot{\beta} = \dfrac{\left|\bar{\bar{{\boldsymbol{A}}}}{\boldsymbol{T}} - {\boldsymbol{C}}\right|^2}{|{\boldsymbol{r}}|^2}$; 7 $ {\boldsymbol{r}} = \bar{\bar{{\boldsymbol{A}}}}{\boldsymbol{T}} - {\boldsymbol{C}} $; 8 $ {\boldsymbol{d}} = {\boldsymbol{r}} + \dot{\beta}{\boldsymbol{d}} $。 算法2 V循环算法 1 while $ \| {\boldsymbol{U}}_p - \bar{\bar{{\boldsymbol{L}}}}_{p} {\boldsymbol{B}}_p\| \geqslant \xi $ do 2 $ {\boldsymbol{U}}_{p} = \widetilde{S}_{n}(\bar{\bar{{\boldsymbol{L}}}}_p, {\boldsymbol{U}}_{p}^{0}, {\boldsymbol{B}}_p) $; /* Correction on coarse grids */ 3 for $ n = q-1, q-2, q-3, \cdots, 2 $ do 4 $ \dot{{\boldsymbol{U}}}_{n}=R^{n+1}_{n}{\boldsymbol{U}}_{n+1} $; 5 $ {\boldsymbol{B}}_i = R^{n+1}_{n}({\boldsymbol{B}}_{n+1}-\bar{\bar{{\boldsymbol{L}}}}_{n+1}{\boldsymbol{U}}_{n+1}) + \bar{\bar{{\boldsymbol{L}}}}_{n} \dot{{\boldsymbol{U}}}_{n} $; 6 $ {\boldsymbol{U}}_{n}=\widetilde{S}_{n}(\bar{\bar{{\boldsymbol{L}}}}_{n}, \dot{{\boldsymbol{U}}}_{n}, {\boldsymbol{B}}_{n}) $; /* Updating the fine grids */ 7 for $ n = 2, 3, 4, \cdots, q $ do 8 $ \ddot{{\boldsymbol{U}}}_{n}={\boldsymbol{U}}_{n} + P^{n}_{n+1}({\boldsymbol{U}}_{n-1} - \dot{{\boldsymbol{U}}}_{n-1}) $; 9 $ {\boldsymbol{U}}_{n}=\widetilde{S}_{n}(\bar{\bar{{\boldsymbol{L}}}}_{n}, \ddot{{\boldsymbol{U}}}_{n}, {\boldsymbol{B}}_{n}) $; 10 $ {\boldsymbol{U}}^{0}_{p} = {\boldsymbol{U}}_{p} $。 // Update initial solution 表 1 算例参数信息
Table 1. Information for the simulated cases
case number $ Ra $ $ Pr $ $\hat{ S}$ $\hat{F}$ resolution 1 $ 3\times10^{7} $ $ 7 $ $0.06$ $0.06$ $ 1\;024\times1\;024 $ 2 $ 3\times10^{9} $ $ 7 $ $0.06$ $0.06$ $ 2\;048\times2\;048 $ 3 $ 3\times10^{11} $ $ 7 $ $0.06$ $0.06$ $ 2\;048\times2\;048 $ -
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