本文构建了多松弛时间离散玻尔兹曼模型,并使用该模型模拟爆轰现象。相对于我们之前的一个模型[Xu A., Lin C., Zhang G., Li Y., Phys. Rev. E 91 (2015) 043306],本模型在模拟有化学反应或无化学反应流体系统时的计算效率更高。这是...本文构建了多松弛时间离散玻尔兹曼模型,并使用该模型模拟爆轰现象。相对于我们之前的一个模型[Xu A., Lin C., Zhang G., Li Y., Phys. Rev. E 91 (2015) 043306],本模型在模拟有化学反应或无化学反应流体系统时的计算效率更高。这是因为前者使用了24个离散速度,而本模型只使用16个。在模拟部分高马赫物理系统时,本模型表现出更高的数值稳定性。使用该模型,本文分四种情况模拟了爆轰波激发的Richtmyer-Meshkov不稳定性问题。当爆轰波由反应物传向另一种较轻的不反应的物质时,由于突然失去能量补充,温度急剧下降,在物质界附近将会出现一层高密区域。展开更多
A lattice Boltzmann model of KdV-Burgers equation is derived by using the single-relaxation form of the lattice Boltzmann equation. With the present model, we simulate the traveling-wave solutions, the solitary-wave s...A lattice Boltzmann model of KdV-Burgers equation is derived by using the single-relaxation form of the lattice Boltzmann equation. With the present model, we simulate the traveling-wave solutions, the solitary-wave solutions, and the sock-wave solutions of KdV-Burgers equation, and calculate the decay factor and the wavelength of the sock-wave solution, which has exponential decay. The numerical results agree with the analytical solutions quite well.展开更多
文摘本文构建了多松弛时间离散玻尔兹曼模型,并使用该模型模拟爆轰现象。相对于我们之前的一个模型[Xu A., Lin C., Zhang G., Li Y., Phys. Rev. E 91 (2015) 043306],本模型在模拟有化学反应或无化学反应流体系统时的计算效率更高。这是因为前者使用了24个离散速度,而本模型只使用16个。在模拟部分高马赫物理系统时,本模型表现出更高的数值稳定性。使用该模型,本文分四种情况模拟了爆轰波激发的Richtmyer-Meshkov不稳定性问题。当爆轰波由反应物传向另一种较轻的不反应的物质时,由于突然失去能量补充,温度急剧下降,在物质界附近将会出现一层高密区域。
基金partially supported by the National Natural Science Foundation of China (Grant No. 11701576 and 11501053)the Natural Science Foundation of Hunan Province (Grant No. 2019JJ50786)Changsha University of Science and Technology (Grant No.JG2019YB16)。
文摘A lattice Boltzmann model of KdV-Burgers equation is derived by using the single-relaxation form of the lattice Boltzmann equation. With the present model, we simulate the traveling-wave solutions, the solitary-wave solutions, and the sock-wave solutions of KdV-Burgers equation, and calculate the decay factor and the wavelength of the sock-wave solution, which has exponential decay. The numerical results agree with the analytical solutions quite well.
